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Contents
Preface xviii
Nomenclature xxvi
CHAPTER ONE
BASICS OF HEAT TRANSFER 1
1-1 Thermodynamics and Heat Transfer 2
Application Areas of Heat Transfer 3
Historical Background 3
1 -2 Engineering Heat Transfer 4
Modeling in Heat Transfer 5
1 -3 Heat and Other Forms of Energy 6
Specific Heats of Gases, Liquids, and Solids 7
Energy Transfer 9
1 -4 The First Law of Thermodynamics 1 1
Energy Balance for Closed Systems (Fixed Mass) 12
Energy Balance for Steady-Flow Systems 12
Surface Energy Balance 13
1-5 Heat Transfer Mechanisms 17
1-6 Conduction 17
Thermal Conductivity 19
Thermal Diffusivity 23
1 -7 Convection 25
1 -8 Radiation 27
1 -9 Simultaneous Heat Transfer Mechanisms 30
1-10 Problem-Solving Technique 35
A Remark on Significant Digits 37
Engineering Software Packages 38
Engineering Equation Solver (EES) 39
Heat Transfer Tools (HTT) 39
Topic of Special Interest:
Thermal Comfort 40
Summary 46
References and Suggested Reading 47
Problems 47
CHAPTER TWO
HEAT CONDUCTION EQUATION 61
2- 1 Introduction 62
Steady versus Transient Heat Transfer
Multidimensional Heat Transfer 64
Heat Generation 66
2-5
2-6
2-7
2-2 One -Dimensional
Heat Conduction Equation
68
Heat Conduction Equation in a Large Plane Wall 68
Heat Conduction Equation in a Long Cylinder 69
Heat Conduction Equation in a Sphere 71
Combined One-Dimensional
Heat Conduction Equation 72
2-3 General Heat Conduction Equation 74
Rectangular Coordinates 74
Cylindrical Coordinates 75
Spherical Coordinates 76
2-4 Boundary and Initial Conditions 77
1 Specified Temperature Boundary Condition 78
2 Specified Heat Flux Boundary Condition 79
3 Convection Boundary Condition 81
4 Radiation Boundary Condition 82
5 Interface Boundary Conditions 83
6 Generalized Boundary Conditions 84
Solution of Steady One-Dimensional
Heat Conduction Problems 86
Heat Generation in a Solid 97
Variable Thermal Conductivity, k(T) 104
Topic of Special Interest:
A Brief Review of Differential Equations 107
Summary 111
References and Suggested Reading 112
Problems 113
CHAPTER THREE
STEADY HEAT CONDUCTION 1 27
3-1 Steady Heat Conduction in Plane Walls 128
The Thermal Resistance Concept 129
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CONTENTS
Thermal Resistance Network 131
Multilayer Plane Walls 133
3-2 Thermal Contact Resistance 138
3-3 Generalized Thermal Resistance Networks 143
3-4 Heat Conduction in Cylinders and Spheres 146
Multilayered Cylinders and Spheres 148
3-5 Critical Radius of Insulation 153
3-6 Heat Transfer from Finned Surfaces 156
Fin Equation 157
Fin Efficiency 160
Fin Effectiveness 163
Proper Length of a Fin 165
3-7 Heat Transfer in Common Configurations 169
Topic of Special Interest:
Heat Transfer Through Walls and Roofs 175
Summary 185
References and Suggested Reading 186
Problems 187
4 Complications 268
5 Human Nature 268
5-2 Finite Difference Formulation of
Differential Equations 269
5-3 One -Dimensional Steady Heat Conduction 272
Boundary Conditions 274
5-4 Two-Dimensional
Steady Heat Conduction 282
Boundary Nodes 283
Irregular Boundaries 287
5-5 Transient Heat Conduction 291
Transient Heat Conduction in a Plane Wall 293
Two-Dimensional Transient Heat Conduction 304
Topic of Special Interest:
Controlling Numerical Error 309
Summary 312
References and Suggested Reading 314
Problems 314
CHAPTER FOUR
TRANSIENT HEAT CONDUCTION 209
4-1 Lumped System Analysis 210
Criteria for Lumped System Analysis 211
Some Remarks on Heat Transfer in Lumped Systems 213
4-2 Transient Heat Conduction in
Large Plane Walls, Long Cylinders,
and Spheres with Spatial Effects 216
4-3 Transient Heat Conduction in
Semi-Infinite Solids 228
4-4 Transient Heat Conduction in
Multidimensional Systems 231
Topic of Special Interest:
Refrigeration and Freezing of Foods 239
Summary 250
References and Suggested Reading 251
Problems 252
CHAPTER FIVE
NUMERICAL METHODS
IN HEAT CONDUCTION 265
5-1 Why Numerical Methods? 266
1 Limitations 267
2 Better Modeling 267
3 Flexibility 268
CHAPTER SIX
FUNDAMENTALS OF CONVECTION
333
6-1 Physical Mechanism on Convection 334
Nusselt Number 336
6-2 Classification of Fluid Flows 337
Viscous versus I nviscid Flow 337
Internal versus External Flow 337
Compressible versus Incompressible Flow 337
Laminar versus Turbulent Flow 338
Natural (or Unforced) versus Forced Flow 338
Steady versus Unsteady (Transient) Flow 338
One-, Two-, and Three-Dimensional Flows 338
6-3 Velocity Boundary Layer 339
Surface Shear Stress 340
6-4 Thermal Boundary Layer 341
Prandtl Number 341
6-5 Laminar and Turbulent Flows 342
Reynolds Number 343
6-6 Heat and Momentum Transfer
in Turbulent Flow 343
6-7 Derivation of Differential
Convection Equations 345
Conservation of Mass Equation 345
Conservation of Momentum Equations 346
Conservation of Energy Equation 348
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6-8 Solutions of Convection Equations
for a Flat Plate 352
The Energy Equation 354
6-9 Nondimensionalized Convection
Equations and Similarity 356
6-1 Functional Forms of Friction and
Convection Coefficients 357
6-1 1 Analogies between Momentum
and Heat Transfer 358
Summary 361
References and Suggested Reading 362
Problems 362
CHAPTER SEVEN
EXTERNAL FORCED CONVECTION
367
CONTENTS
8-4 General Thermal Analysis 426
Constant Surface Heat Flux [q s = constant) 427
Constant Surface Temperature (T s = constant) 428
8-5 Laminar Flow in Tubes 431
Pressure Drop 433
Temperature Profile and the Nusselt Number 434
Constant Surface Heat Flux 435
Constant Surface Temperature 436
Laminar Flow in Noncircular Tubes 436
Developing Laminar Flow in the Entrance Region 436
8-6 Turbulent Flow in Tubes 441
Rough Surfaces 442
Developing Turbulent Flow in the Entrance Region 443
Turbulent Flow in Noncircular Tubes 443
Flow through Tube Annulus 444
Heat Transfer Enhancement 444
Summary 449
References and Suggested Reading 450
Problems 452
7-1 Drag Force and Heat Transfer
in External Flow 368
Friction and Pressure Drag 368
Heat Transfer 370
7-2 Parallel Flow over Flat Plates 371
Friction Coefficient 372
Heat Transfer Coefficient 373
Flat Plate with Unheated Starting Length 375
Uniform Heat Flux 375
7-3 Flow across Cylinders and Spheres 380
Effect of Surface Roughness 382
Heat Transfer Coefficient 384
7-4 Flow across Tube Banks 389
Pressure Drop 392
Topic of Special Interest:
Reducing Heat Transfer through Surfaces 395
Summary 406
References and Suggested Reading 407
Problems 408
CHAPTER EIGHT
INTERNAL FORCED CONVECTION 419
8-1 Introduction 420
8-2 Mean Velocity and Mean Temperature 420
Laminar and Turbulent Flow in Tubes 422
8-3 The Entrance Region 423
Entry Lengths 425
CHAPTER NINE
NATURAL CONVECTION 459
9-1 Physical Mechanism of
Natural Convection 460
9-2 Equation of Motion and
the Grashof Number 463
The Grashof Number 465
9-3 Natural Convection over Surfaces 466
Vertical Plates (7~ s = constant) 467
Vertical Plates {q s = constant) 467
Vertical Cylinders 467
Inclined Plates 467
Horizontal Plates 469
Horizontal Cylinders and Spheres 469
9-4 Natural Convection from
Finned Surfaces and PCBs 473
Natural Convection Cooling of Finned Surfaces
(T s = constant) 473
Natural Convection Cooling of Vertical PCBs
(q s = constant) 474
Mass Flow Rate through the Space between Plates 475
9-5 Natural Convection inside Enclosures 477
Effective Thermal Conductivity 478
Horizontal Rectangular Enclosures 479
Inclined Rectangular Enclosures 479
Vertical Rectangular Enclosures 480
Concentric Cylinders 480
Concentric Spheres 481
Combined Natural Convection and Radiation 481
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9-6 Combined Natural and Forced Convection 486 1 1 -6 Atmospheric and Solar Radiation 586
Topic of Special Interest:
Heat Transfer through Windows 489
Summary 499
References and Suggested Reading 500
Problems 501
Topic of Special Interest:
Solar Heat Gain through Windows 590
Summary 597
References and Suggested Reading 599
Problems 599
CHAPTER TEN
BOILING AND CONDENSATION
515
CHAPTER TWELVE
RADIATION HEAT TRANSFER 605
10-1 Boiling Heat Transfer 516
10-2 Pool Boiling 518
Boiling Regimes and the Boiling Curve 518
Heat Transfer Correlations in Pool Boiling 522
Enhancement of Heat Transfer in Pool Boiling 526
10-3 Flow Boiling 530
1 0-4 Condensation Heat Transfer 532
1 0-5 Film Condensation 532
Flow Regimes 534
Heat Transfer Correlations for Film Condensation 535
1 0-6 Film Condensation Inside
Horizontal Tubes 545
1 0-7 Dropwise Condensation 545
Topic of Special Interest:
Heat Pipes 546
Summary 551
References and Suggested Reading 553
Problems 553
CHAPTER ELEVEN
FUNDAMENTALS OF THERMAL RADIATION 561
11-1 Introduction 562
1 1 -2 Thermal Radiation 563
1 1 -3 Blackbody Radiation 565
1 1 -4 Radiation Intensity 57 1
Solid Angle 572
Intensity of Emitted Radiation 573
Incident Radiation 574
Radiosity 575
Spectral Quantities 575
1 1 -5 Radiative Properties 577
Emissivity 578
Absorptivity, Reflectivity, and Transmissivity 582
Kirchhoffs Law 584
The Greenhouse Effect 585
12-1 The View Factor 606
1 2-2 View Factor Relations 609
1 The Reciprocity Relation 610
2 The Summation Rule 613
3 The Superposition Rule 615
4 The Symmetry Rule 616
View Factors between Infinitely Long Surfaces:
The Crossed-Strings Method 618
12-3 Radiation Heat Transfer: Black Surfaces 620
1 2-4 Radiation Heat Transfer:
Diffuse, Gray Surfaces 623
Radiosity 623
Net Radiation Heat Transfer to or from a Surface 623
Net Radiation Heat Transfer between Any
Two Surfaces 625
Methods of Solving Radiation Problems 626
Radiation Heat Transfer in Two-Surface Enclosures 627
Radiation Heat Transfer in Three-Surface Enclosures 629
1 2-5 Radiation Shields and the Radiation Effect 635
Radiation Effect on Temperature Measurements 637
1 2-6 Radiation Exchange with Emitting and
Absorbing Gases 639
Radiation Properties of a Participating Medium 640
Emissivity and Absorptivity of Gases and Gas Mixtures 642
Topic of Special Interest:
Heat Transfer from the Human Body 649
Summary 653
References and Suggested Reading 655
Problems 655
CHAPTER THIRTEEN
HEAT EXCHANGERS 667
13-1 Types of Heat Exchangers 668
1 3-2 The Overall Heat Transfer Coefficient 67 1
Fouling Factor 674
1 3-3 Analysis of Heat Exchangers 678
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1 3-4 The Log Mean Temperature
Difference Method 680
Counter-Flow Heat Exchangers 682
Multipass and Cross-Flow Heat Exchangers:
Use of a Correction Factor 683
1 3-5 The Effectiveness-NTU Method 690
13-6 Selection of Heat Exchangers 700
Heat Transfer Rate 700
Cost 700
Pumping Power 701
Size and Weight 701
Type 701
Materials 701
Other Considerations 702
Summary 703
References and Suggested Reading 704
Problems 705
CHAPTER FOURTEEN
MASS TRANSFER 717
14-1 Introduction 718
14-2 Analogy between Heat and Mass Transfer 719
Temperature 720
Conduction 720
Heat Generation 720
Convection 721
14-3 Mass Diffusion 721
1 Mass Basis 722
2 Mole Basis 722
Special Case: Ideal Gas Mixtures 723
Fick's Law of Diffusion: Stationary Medium Consisting
of Two Species 723
14-4 Boundary Conditions 727
1 4-5 Steady Mass Diffusion through a Wall 732
14-6 Water Vapor Migration in Buildings 736
14-7 Transient Mass Diffusion 740
14-8 Diffusion in a Moving Medium 743
Special Case: Gas Mixtures at Constant Pressure
and Temperature 747
Diffusion of Vapor through a Stationary Gas:
Stefan Flow 748
Equimolar Counterdiffusion 750
14-9 Mass Convection 754
Analogy between Friction, Heat Transfer, and Mass
Transfer Coefficients 758
Limitation on the Heat-Mass Convection Analogy 760
Mass Convection Relations 760
CONTENTS
14-10 Simultaneous Heat and Mass Transfer 763
Summary 769
References and Suggested Reading 771
Problems 772
CHAPTER FIFTEEN
COOLING OF ELECTRONIC EQUIPMENT
785
15-1 Introduction and History 786
15-2 Manufacturing of Electronic Equipment 787
The Chip Carrier 787
Printed Circuit Boards 789
The Enclosure 791
15-3 Cooling Load of Electronic Equipment 793
1 5-4 Thermal Environment 794
1 5-5 Electronics Cooling in
Different Applications 795
1 5-6 Conduction Cooling 797
Conduction in Chip Carriers 798
Conduction in Printed Circuit Boards 803
Heat Frames 805
The Thermal Conduction Module (TCM) 810
1 5-7 Air Cooling: Natural Convection
and Radiation 812
15-8 Air Cooling: Forced Convection 820
Fan Selection 823
Cooling Personal Computers 826
15-9 Liquid Cooling 833
15-10 Immersion Cooling 836
Summary 841
References and Suggested Reading 842
Problems 842
APPENDIX 1
PROPERTY TABLES AND CHARTS
(SI UNITS) 855
Table A-1 Molar Mass, Gas Constant, and
Critical-Point Properties 856
Table A-2 Boiling- and Freezing-Point
Properties 857
Table A-3 Properties of Solid Metals 858
Table A-4 Properties of Solid Nonmetals 861
Table A-5 Properties of Building Materials 862
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Table A-6 Properties of Insulating Materials 864
Table A-7 Properties of Common Foods 865
Table A-8 Properties of Miscellaneous
Materials 867
Table A-9 Properties of Saturated Water 868
Ta b I e A- 1 Properties of S aturated
Refrigerant- 134a 869
Table A-11 Properties of Saturated Ammonia 870
Ta b I e A- 1 2 Properties of S aturated Propane 87 1
TableA-13 Properties of Liquids 872
Table A-14 Properties of Liquid Metals 873
Table A-15 Properties of Air at 1 atm Pressure 874
Table A-1 6 Properties of Gases at 1 atm
Pressure 875
Ta b I e A- 1 7 Properties of the Atmosphere at
High Altitude 877
Ta b I e A- 1 8 Emi ssivitiesofS urf aces 878
Table A-1 9 Solar Radiative Properties of
Materials 880
Figure A-20 The Moody Chart for the Friction
Factor for Fully Developed Flow
in Circular Tubes 881
APPENDIX 2
PROPERTY TABLES AND CHARTS
(ENGLISH UNITS) 883
Table A-1 E Molar Mass, Gas Constant, and
Critical-Point Properties 884
Table A-2E Boiling- and Freezing-Point
Properties 885
Table A-3E Properties of Solid Metals 886
Table A-4E Properties of Solid Nonmetals 889
Table A-5E Properties of Building Materials 890
Table A-6E Properties of Insulating Materials 892
Table A-7E Properties of Common Foods 893
Table A-8E Properties of Miscellaneous
Materials 895
Table A-9E Properties of Saturated Water 896
Table A-1 0E Properties of Saturated
Refrigerant- 134a 897
Table A-1 1 E Properties of Saturated Ammonia 898
Table A-1 2E Properties of Saturated Propane 899
Table A-1 3E Properties of Liquids 900
Table A-1 4E Properties of Liquid Metals 901
Table A-1 5E Properties of Air at 1 atm Pressure 902
Table A-1 6E Properties of Gases at 1 atm
Pressure 903
Table A-1 7E Properties of the Atmosphere at
High Altitude 905
APPENDIX 3
INTRODUCTION TO EES 907
INDEX 921
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TABLE OF EXAMPLES:
CHAPTER ONE
BASICS OF HEAT TRANSFER 1
Example 1-1 Heating of a Copper Ball 10
Example 1-2 Heating of Water in an
Electric Teapot 14
Example 1-3 Heat Loss from Heating Ducts
in a Basement 15
Example 1 -4 Electric Heating of a House at
High Elevation 16
Example 1 -5 The Cost of Heat Loss through
a Roof 19
Example 1 -6 Measuring the Thermal Conductivity
of a Material 23
Example 1 -7 Conversion between SI and
English Units 24
Example 1 -8 Measuring Convection Heat
Transfer Coefficient 26
Example 1-9 Radiation Effect on
Thermal Comfort 29
Example 1-10 Heat Loss from a Person 31
Example 1-11 Heat Transfer between
Two Isothermal Plates 32
Example 1-12 Heat Transfer in Conventional
and Microwave Ovens 33
Example 1-13 Heating of a Plate by
Solar Energy 34
Example 1-14 Solving a System of Equations
with EES 39
CHAPTER TWO
HEAT CONDUCTION EQUATION 61
Example 2-1
Heat Gain by a Refrigerator 67
Example 2-2 Heat Generation in a
Hair Dryer 67
Example 2-3 Heat Conduction through the
Bottom of a Pan 72
Example 2-4 Heat Conduction in a
Resistance Heater 72
Exa m p I e 2-5 Cooling of a Hot Metal B all
in Air 73
Example 2-6 Heat Conduction in a
Short Cylinder 76
Example 2-7 Heat Flux Boundary Condition 80
Example 2-8 Convection and Insulation
Boundary Conditions 82
Example 2-9 Combined Convection and
Radiation Condition 84
Example 2-10 Combined Convection, Radiation,
and Heat Flux 85
Exa m p I e 2- 1 1 Heat Conduction in a
Plane Wall 86
Example 2-12 A Wall with Various Sets of
Boundary Conditions 88
Example 2-1 3 Heat Conduction in the Base Plate
of an Iron 90
Exa m p I e 2- 1 4 Heat Conduction in a
Solar Heated Wall 92
Example 2-15 Heat Loss through a
Steam Pipe 94
Example 2-16 Heat Conduction through a
Spherical Shell 96
Example 2-17 Centerline Temperature of a
Resistance Heater 100
Example 2-18 Variation of Temperature in a
Resistance Heater 100
Example 2-19 Heat Conduction in a Two-Layer
Medium 102
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CONTENTS
Example 2-20 Variation of Temperature in a Wall
with k(T) 105
Example 2-21 Heat Conduction through a Wall
with k(T) 106
CHAPTER THREE
STEADY HEAT CONDUCTION 1 27
Example 3-1
Heat Loss through a Wall 134
Example 3-2
Heat Loss through a
Single-Pane Window 135
Example 3-3
Heat Loss through
Double-Pane Windows 136
Example 3-4
Equivalent Thickness for
Contact Resistance 140
Example 3-5
Contact Resistance of
Transistors 141
Example 3-6
Heat Loss through a
Composite Wall 144
Example 3-7
Heat Transfer to a
Spherical Container 149
:
)
Example 3-8
Heat Loss through an Insulated
Steam Pipe 151
Example 3-9
Heat Loss from an Insulated
Electric Wire 154
Example 3-10
Maximum Power Dissipation of
a Transistor 166
Example 3-1 1
Selecting a Heat Sink for a
Transistor 167
Example 3-12
Effect of Fins on Heat Transfer from
Steam Pipes 168
Example 3-13
Heat Loss from Buried
Steam Pipes 170
Example 3-14
Heat Transfer between Hot and
Cold Water Pipes 173
Example 3-15
Cost of Heat Loss through Walls
in Winter 174
Example 3-16
The i?-Value of a Wood
Frame Wall 179
Example 3-17
The R -Value of a Wall with
Rigid Foam 180
Example 3-18
The #-Value of a Masonry Wall 181
Example 3-19
The R- Value of a Pitched Roof 1 82
CHAPTER FOUR
TRANSIENT HEAT CONDUCTION 209
Example 4-1 Temperature Measurement by
Thermocouples 214
Example 4-2 Predicting the Time of Death 215
Example 4-3 Boiling Eggs 224
Example 4-4 Heating of Large Brass Plates
in an Oven 225
Example 4-5 Cooling of a Long Stainless Steel
Cylindrical Shaft 226
Example 4-6 Minimum Burial Depth of Water
Pipes to Avoid Freezing 230
Exa m p I e 4-7 Cooling of a S hort Brass
Cylinder 234
Example 4-8 Heat Transfer from a Short
Cylinder 235
Example 4-9 Cooling of a Long Cylinder
by Water 236
Example 4-10 Refrigerating Steaks while
Avoiding Frostbite 238
Example 4-1 1 Chilling of Beef Carcasses in a
Meat Plant 248
CHAPTER FIVE
NUMERICAL METHODS IN
HEAT CONDUCTION 265
Example 5-1 Steady Heat Conduction in a Large
Uranium Plate 277
Example 5-2 Heat Transfer from
Triangular Fins 279
Example 5-3 Steady Two-Dimensional Heat
Conduction in L-Bars 284
Example 5-4 Heat Loss through Chimneys 287
Example 5-5 Transient Heat Conduction in a Large
Uranium Plate 296
Example 5-6 Solar Energy Storage in
Trombe Walls 300
Example 5-7 Transient Two-Dimensional Heat
Conduction in L-Bars 305
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CHAPTER SIX
FUNDAMENTALS OF CONVECTION
333
Example 8-6
CONTENTS
Heat Loss from the Ducts of a
Heating System 448
Example 6-1 Temperature Rise of Oil in a
Journal Bearing 350
Example 6-2 Finding Convection Coefficient from
Drag Measurement 360
CHAPTER SEVEN
EXTERNAL FORCED CONVECTION
367
Exa m p I e 7 - 1 Flow of Hot Oil over a
Flat Plate 376
Exa m p I e 7 -2 Cooling of a Hot B lock by Forced Air
at High Elevation 377
Example 7-3 Cooling of Plastic Sheets by
Forced Air 378
Example 7-4 Drag Force Acting on a Pipe
in a River 383
Example 7-5 Heat Loss from a Steam Pipe
in Windy Air 386
Example 7-6 Cooling of a Steel Ball by
Forced Air 387
Example 7-7 Preheating Air by Geothermal Water
in a Tube Bank 393
Example 7-8 Effect of Insulation on
Surface Temperature 402
Example 7-9 Optimum Thickness of
Insulation 403
CHAPTER EIGHT
INTERNAL FORCED CONVECTION 41 9
Example 8-1 Heating of Water in a Tube
by Steam 430
Example 8-2 Pressure Drop in a Pipe 438
Example 8-3 Flow of Oil in a Pipeline through
a Lake 439
Example 8-4 Pressure Drop in a Water Pipe 445
Example 8-5 Heating of Water by Resistance
Heaters in a Tube 446
CHAPTER NINE
NATURAL CONVECTION 459
Example 9-1 Heat Loss from Hot
Water Pipes 470
Example 9-2 Cooling of a Plate in
Different Orientations 47 1
Example 9-3 Optimum Fin Spacing of a
Heat Sink 476
Example 9-4 Heat Loss through a Double-Pane
Window 482
Example 9-5 Heat Transfer through a
Spherical Enclosure 483
Example 9-6 Heating Water in a Tube by
Solar Energy 484
Example 9-7 [/-Factor for Center-of-Glass Section
of Windows 496
Example 9-8 Heat Loss through Aluminum Framed
Windows 497
Example 9-9 [/-Factor of a Double-Door
Window 498
CHAPTER TEN
BOILING AND CONDENSATION 515
Example 1 0-1 Nucleate Boiling Water
in a Pan 526
Example 10-2 Peak Heat Flux in
Nucleate Boiling 528
Example 1 0-3 Film Boiling of Water on a
Heating Element 529
Example 10-4 Condensation of Steam on a
Vertical Plate 541
Example 10-5 Condensation of Steam on a
Tilted Plate 542
Example 10-6 Condensation of Steam on
Horizontal Tubes 543
Example 10-7 Condensation of Steam on
Horizontal Tube Banks 544
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Example 1 0-8 Replacing a Heat Pipe by a
Copper Rod 550
CHAPTER ELEVEN
FUNDAMENTALS OF THERMAL RADIATION 561
Example 11-1 Radiation Emission from a
Black Ball 568
Example 1 1 -2 Emission of Radiation from
aLightbulb 571
Example 1 1 -3 Radiation Incident on a
Small Surface 576
Example 1 1 -4 Emissivity of a Surface
and Emissive Power 581
Example 1 1 -5 Selective Absorber and
Reflective Surfaces 589
Example 1 1 -6 Installing Reflective Films
on Windows 596
CHAPTER TWELVE
RADIATION HEAT TRANSFER 605
Example 12-1
Example 12-2
Example 12-3
Example 12-4
Example 12-5
Example 12-6
Example 12-7
Example 12-8
Example 12-9
Example 12-10
Example 12-11
View Factors Associated with
Two Concentric Spheres 614
Fraction of Radiation Leaving
through an Opening 615
View Factors Associated with
a Tetragon 617
View Factors Associated with a
Triangular Duct 617
The Crossed-Strings Method for
View Factors 619
Radiation Heat Transfer in a
Black Furnace 621
Radiation Heat Transfer between
Parallel Plates 627
Radiation Heat Transfer in a
Cylindrical Furnace 630
Radiation Heat Transfer in a
Triangular Furnace 63 1
Heat Transfer through a Tubular
Solar Collector 632
Radiation Shields 638
Example 12-12 Radiation Effect on Temperature
Measurements 639
Example 12-13 Effective Emissivity of
Combustion Gases 646
Example 1 2-14 Radiation Heat Transfer in a
Cylindrical Furnace 647
Example 12-15 Effect of Clothing on Thermal
Comfort 652
CHAPTER THIRTEEN
HEAT EXCHANGERS 667
Example 13-1
Example 13-2
Example 13-3
Example 13-4
Example 13-5
Example 13-6
Example 13-7
Example 13-8
Example 13-9
Example 13-10
Overall Heat Transfer Coefficient of
a Heat Exchanger 675
Effect of Fouling on the Overall Heat
Transfer Coefficient 677
The Condensation of Steam in
a Condenser 685
Heating Water in a Counter-Flow
Heat Exchanger 686
Heating of Glycerin in a Multipass
Heat Exchanger 687
Cooling of an
Automotive Radiator 688
Upper Limit for Heat Transfer
in a Heat Exchanger 69 1
Using the Effectiveness-
NTU Method 697
Cooling Hot Oil by Water in a
Multipass Heat Exchanger 698
Installing a Heat Exchanger to Save
Energy and Money 702
CHAPTER FOURTEEN
MASS TRANSFER 717
Example 14-1 Determining Mass Fractions from
Mole Fractions 727
Example 14-2 Mole Fraction of Water Vapor at
the Surface of a Lake 728
Example 14-3 Mole Fraction of Dissolved Air
in Water 730
Example 1 4-4 Diffusion of Hydrogen Gas into
a Nickel Plate 732
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Example 14-5
Example 14-6
Example 14-7
Example 14-8
Example 14-9
Example 14-10
Example 14-11
Example 14-12
Example 14-13
Diffusion of Hydrogen through a
Spherical Container 735
Condensation and Freezing of
Moisture in the Walls 738
Hardening of Steel by the Diffusion
of Carbon 742
Venting of Helium in the Atmosphere
by Diffusion 751
Measuring Diffusion Coefficient by
the Stefan Tube 752
Mass Convection inside a
Circular Pipe 761
Analogy between Heat and
Mass Transfer 762
Evaporative Cooling of a
Canned Drink 765
Heat Loss from Uncovered Hot
Water Baths 766
CHAPTER FIFTEEN
COOLING OF ELECTRONIC EQUIPMENT 785
Example 1 5-1 Predicting the Junction Temperature
of a Transistor 788
Example 15-2 Determining the Junction-to-Case
Thermal Resistance 789
Example 15-3 Analysis of Heat Conduction in
a Chip 799
Example 1 5-4 Predicting the Junction Temperature
of a Device 802
Example 15-5
Example 15-6
Example 15-7
Example 15-8
Example 15-9
Example 15-10
Example 15-1 1
Example 15-12
Example 15-13
Example 15-14
Example 15-15
Example 15-16
Example 15-17
Example 15-18
Example 15-19
CONTENTS
Heat Conduction along a PCB with
Copper Cladding 804
Thermal Resistance of an Epoxy
Glass Board 805
Planting Cylindrical Copper Fillings
in an Epoxy Board 806
Conduction Cooling of PCB s by a
Heat Frame 807
Cooling of Chips by the Thermal
Conduction Module 812
Cooling of a Sealed
Electronic Box 816
Cooling of a Component by
Natural Convection 817
Cooling of a PCB in a Box by
Natural Convection 818
Forced- Air Cooling of a
Hollow-Core PCB 826
Forced- Air Cooling of a Transistor
Mounted on a PCB 828
Choosing a Fan to Cool
a Computer 830
Cooling of a Computer
by a Fan 831
Cooling of Power Transistors on
a Cold Plate by Water 835
Immersion Cooling of
a Logic Chip 840
Cooling of a Chip by Boiling 840
cen5 8 93 3_fm.qxd 9/11/2002 10:56 AM Page xviii
Preface
objectives
Heat transfer is a basic science that deals with the rate of transfer of ther-
mal energy. This introductory text is intended for use in a first course in
heat transfer for undergraduate engineering students, and as a reference
book for practicing engineers. The objectives of this text are
• To cover the basic principles of heat transfer.
• To present a wealth of real-world engineering applications to give stu-
dents a feel for engineering practice.
• To develop an intuitive understanding of the subject matter by empha-
sizing the physics and physical arguments.
Students are assumed to have completed their basic physics and calculus se-
quence. The completion of first courses in thermodynamics, fluid mechanics,
and differential equations prior to taking heat transfer is desirable. The rele-
vant concepts from these topics are introduced and reviewed as needed.
In engineering practice, an understanding of the mechanisms of heat trans-
fer is becoming increasingly important since heat transfer plays a crucial role
in the design of vehicles, power plants, refrigerators, electronic devices, build-
ings, and bridges, among other things. Even a chef needs to have an intuitive
understanding of the heat transfer mechanism in order to cook the food "right"
by adjusting the rate of heat transfer. We may not be aware of it, but we al-
ready use the principles of heat transfer when seeking thermal comfort. We in-
sulate our bodies by putting on heavy coats in winter, and we minimize heat
gain by radiation by staying in shady places in summer. We speed up the cool-
ing of hot food by blowing on it and keep warm in cold weather by cuddling
up and thus minimizing the exposed surface area. That is, we already use heat
transfer whether we realize it or not.
GENERAL APPROACH
This text is the outcome of an attempt to have a textbook for a practically ori-
ented heat transfer course for engineering students. The text covers the stan-
dard topics of heat transfer with an emphasis on physics and real-world
applications, while de-emphasizing intimidating heavy mathematical aspects.
This approach is more in line with students' intuition and makes learning the
subject matter much easier.
The philosophy that contributed to the warm reception of the first edition of
this book has remained unchanged. The goal throughout this project has been
to offer an engineering textbook that
cen58 933_fm.qxd 9/11/2002 10:56 AM Page xix
• Talks directly to the minds of tomorrow's engineers in a simple yet pre-
cise manner.
• Encourages creative thinking and development of a deeper understand-
ing of the subject matter.
• Is read by students with interest and enthusiasm rather than being used
as just an aid to solve problems.
Special effort has been made to appeal to readers' natural curiosity and to help
students explore the various facets of the exciting subject area of heat transfer.
The enthusiastic response we received from the users of the first edition all
over the world indicates that our objectives have largely been achieved.
Yesterday's engineers spent a major portion of their time substituting values
into the formulas and obtaining numerical results. However, now formula ma-
nipulations and number crunching are being left to computers. Tomorrow's
engineer will have to have a clear understanding and a firm grasp of the basic
principles so that he or she can understand even the most complex problems,
formulate them, and interpret the results. A conscious effort is made to em-
phasize these basic principles while also providing students with a look at
how modern tools are used in engineering practice.
NEW IN THIS EDITION
All the popular features of the previous edition are retained while new ones
are added. The main body of the text remains largely unchanged except that
the coverage of forced convection is expanded to three chapters and the cov-
erage of radiation to two chapters. Of the three applications chapters, only the
Cooling of Electronic Equipment is retained, and the other two are deleted to
keep the book at a reasonable size. The most significant changes in this edi-
tion are highlighted next.
EXPANDED COVERAGE OF CONVECTION
Forced convection is now covered in three chapters instead of one. In Chapter
6, the basic concepts of convection and the theoretical aspects are introduced.
Chapter 7 deals with the practical analysis of external convection while Chap-
ter 8 deals with the practical aspects of internal convection. See the Content
Changes and Reorganization section for more details.
ADDITIONAL CHAPTER ON RADIATION
Radiation is now covered in two chapters instead of one. The basic concepts
associated with thermal radiation, including radiation intensity and spectral
quantities, are covered in Chapter 11. View factors and radiation exchange be-
tween surfaces through participating and nonparticipating media are covered
in Chapter 12. See the Content Changes and Reorganization section for more
details.
TOPICS OF SPECIAL INTEREST
Most chapters now contain a new end-of-chapter optional section called
"Topic of Special Interest" where interesting applications of heat transfer are
discussed. Some existing sections such as A Brief Review of Differential
Equations in Chapter 2, Thermal Insulation in Chapter 7, and Controlling Nu-
merical Error in Chapter 5 are moved to these sections as topics of special
cen5 8 93 3_fm.qxd 9/11/2002 10:56 AM Page xx
interest. Some sections from the two deleted chapters such as the Refrigera-
tion and Freezing of Foods, Solar Heat Gain through Windows, and Heat
Transfer through the Walls and Roofs are moved to the relevant chapters as
special topics. Most topics selected for these sections provide real-world
applications of heat transfer, but they can be ignored if desired without a loss
in continuity.
COMPREHENSIVE PROBLEMS WITH PARAMETRIC STUDIES
A distinctive feature of this edition is the incorporation of about 130 compre-
hensive problems that require conducting extensive parametric studies, using
the enclosed EES (or other suitable) software. Students are asked to study the
effects of certain variables in the problems on some quantities of interest, to
plot the results, and to draw conclusions from the results obtained. These
problems are designated by computer-EES and EES -CD icons for easy recog-
nition, and can be ignored if desired. Solutions of these problems are given in
the Instructor's Solutions Manual.
CONTENT CHANGES AND REORGANIZATION
With the exception of the changes already mentioned, the main body of the
text remains largely unchanged. This edition involves over 500 new or revised
problems. The noteworthy changes in various chapters are summarized here
for those who are familiar with the previous edition.
• In Chapter 1, surface energy balance is added to Section 1-4. In a new
section Problem-Solving Technique, the problem-solving technique is
introduced, the engineering software packages are discussed, and
overviews of EES (Engineering Equation Solver) and HTT (Heat Trans-
fer Tools) are given. The optional Topic of Special Interest in this chap-
ter is Thermal Comfort.
• In Chapter 2, the section A Brief Review of Differential Equations is
moved to the end of chapter as the Topic of Special Interest.
• In Chapter 3, the section on Thermal Insulation is moved to Chapter 7,
External Forced Convection, as a special topic. The optional Topic of
Special Interest in this chapter is Heat Transfer through Walls and
Roofs.
• Chapter 4 remains mostly unchanged. The Topic of Special Interest in
this chapter is Refrigeration and Freezing of Foods.
• In Chapter 5, the section Solutions Methods for Systems of Algebraic
Equations and the FORTRAN programs in the margin are deleted, and
the section Controlling Numerical Error is designated as the Topic of
Special Interest.
• Chapter 6, Forced Convection, is now replaced by three chapters: Chap-
ter 6 Fundamentals of Convection, where the basic concepts of convec-
tion are introduced and the fundamental convection equations and
relations (such as the differential momentum and energy equations and
the Reynolds analogy) are developed; Chapter 7 External Forced Con-
vection, where drag and heat transfer for flow over surfaces, including
flow over tube banks, are discussed; and Chapter 8 Internal Forced
Convection, where pressure drop and heat transfer for flow in tubes are
cen58 933_fm.qxd 9/11/2002 10:56 AM Page xxi
presented. Reducing Heat Transfer through Surfaces is added to Chap-
ter 7 as the Topic of Special Interest.
• Chapter 7 (now Chapter 9) Natural Convection is completely rewritten.
The Grashof number is derived from a momentum balance on a differ-
ential volume element, some Nusselt number relations (especially those
for rectangular enclosures) are updated, and the section Natural Con-
vection from Finned Surfaces is expanded to include heat transfer from
PCBs. The optional Topic of Special Interest in this chapter is Heat
Transfer through Windows.
• Chapter 8 (now Chapter 10) Boiling and Condensation remained largely
unchanged. The Topic of Special Interest in this chapter is Heat Pipes.
• Chapter 9 is split in two chapters: Chapter 11 Fundamentals of Thermal
Radiation, where the basic concepts associated with thermal radiation,
including radiation intensity and spectral quantities, are introduced, and
Chapter 12 Radiation Heat Transfer, where the view factors and radia-
tion exchange between surfaces through participating and nonparticipat-
ing media are discussed. The Topic of Special Interest are Solar Heat
Gain through Windows in Chapter 11, and Heat Transfer from the Hu-
man Body in Chapter 12.
• There are no significant changes in the remaining three chapters of Heat
Exchangers, Mass Transfer, and Cooling of Electronic Equipment.
• In the appendices, the values of the physical constants are updated; new
tables for the properties of saturated ammonia, refrigerant- 134a, and
propane are added; and the tables on the properties of air, gases, and liq-
uids (including liquid metals) are replaced by those obtained using EES .
Therefore, property values in tables for air, other ideal gases, ammonia,
refrigerant- 134a, propane, and liquids are identical to those obtained
from EES.
LEARNING TOOLS
EMPHASIS ON PHYSICS
A distinctive feature of this book is its emphasis on the physical aspects of
subject matter rather than mathematical representations and manipulations.
The author believes that the emphasis in undergraduate education should re-
main on developing a sense of underlying physical mechanism and a mastery
of solving practical problems an engineer is likely to face in the real world.
Developing an intuitive understanding should also make the course a more
motivating and worthwhile experience for the students.
EFFECTIVE USE OF ASSOCIATION
An observant mind should have no difficulty understanding engineering sci-
ences. After all, the principles of engineering sciences are based on our every-
day experiences and experimental observations. A more physical, intuitive
approach is used throughout this text. Frequently parallels are drawn between
the subject matter and students' everyday experiences so that they can relate
the subject matter to what they already know. The process of cooking, for ex-
ample, serves as an excellent vehicle to demonstrate the basic principles of
heat transfer.
cen58 93 3_fm.qxd 9/11/2002 10:56 AM Page xxii
SELF-INSTRUCTING
The material in the text is introduced at a level that an average student can
follow comfortably. It speaks to students, not over students. In fact, it is self-
instructive. Noting that the principles of sciences are based on experimental
observations, the derivations in this text are based on physical arguments, and
thus they are easy to follow and understand.
EXTENSIVE USE OF ARTWORK
Figures are important learning tools that help the students "get the picture."
The text makes effective use of graphics. It contains more figures and illus-
trations than any other book in this category. Figures attract attention and
stimulate curiosity and interest. Some of the figures in this text are intended to
serve as a means of emphasizing some key concepts that would otherwise go
unnoticed; some serve as paragraph summaries.
CHAPTER OPENERS AND SUMMARIES
Each chapter begins with an overview of the material to be covered and its re-
lation to other chapters. A summary is included at the end of each chapter for
a quick review of basic concepts and important relations.
NUMEROUS W0RKED-0UT EXAMPLES
Each chapter contains several worked-out examples that clarify the material
and illustrate the use of the basic principles. An intuitive and systematic ap-
proach is used in the solution of the example problems, with particular atten-
tion to the proper use of units.
A WEALTH OF REAL-WORLD END-OF-CHAPTER PROBLEMS
The end-of-chapter problems are grouped under specific topics in the order
they are covered to make problem selection easier for both instructors and stu-
dents. The problems within each group start with concept questions, indicated
by "C," to check the students' level of understanding of basic concepts. The
problems under Review Problems are more comprehensive in nature and are
not directly tied to any specific section of a chapter. The problems under the
Design and Essay Problems title are intended to encourage students to make
engineering judgments, to conduct independent exploration of topics of inter-
est, and to communicate their findings in a professional manner. Several eco-
nomics- and safety-related problems are incorporated throughout to enhance
cost and safety awareness among engineering students. Answers to selected
problems are listed immediately following the problem for convenience to
students.
A SYSTEMATIC SOLUTION PROCEDURE
A well-structured approach is used in problem solving while maintaining an
informal conversational style. The problem is first stated and the objectives
are identified, and the assumptions made are stated together with their justifi-
cations. The properties needed to solve the problem are listed separately. Nu-
merical values are used together with their units to emphasize that numbers
without units are meaningless, and unit manipulations are as important as
manipulating the numerical values with a calculator. The significance of the
findings is discussed following the solutions. This approach is also used
consistently in the solutions presented in the Instructor's Solutions Manual.
cen5 8 93 3_fm.qxd 9/11/2002 10:56 AM Page xxiii
A CHOICE OF SI ALONE OR SI/ENGLISH UNITS
In recognition of the fact that English units are still widely used in some in-
dustries, both SI and English units are used in this text, with an emphasis on
SI. The material in this text can be covered using combined Si/English units
or SI units alone, depending on the preference of the instructor. The property
tables and charts in the appendices are presented in both units, except the ones
that involve dimensionless quantities. Problems, tables, and charts in English
units are designated by "E" after the number for easy recognition, and they
can be ignored easily by the SI users.
CONVERSION FACTORS
Frequently used conversion factors and the physical constants are listed on the
inner cover pages of the text for easy reference.
SUPPLEMENTS
These supplements are available to the adopters of the book.
COSMOS SOLUTIONS MANUAL
Available to instructors only.
The detailed solutions for all text problems will be delivered in our
new electronic Complete Online Solution Manual Organization System
(COSMOS). COSMOS is a database management tool geared towards as-
sembling homework assignments, tests and quizzes. No longer do instructors
need to wade through thick solutions manuals and huge Word files. COSMOS
helps you to quickly find solutions and also keeps a record of problems as-
signed to avoid duplication in subsequent semesters. Instructors can contact
their McGraw-Hill sales representative at http://www.mhhe.com/catalogs/rep/
to obtain a copy of the COSMOS solutions manual.
EES SOFTWARE
Developed by Sanford Klein and William Beckman from the University of
Wisconsin-Madison, this software program allows students to solve prob-
lems, especially design problems, and to ask "what if questions. EES (pro-
nounced "ease") is an acronym for Engineering Equation Solver. EES is very
easy to master since equations can be entered in any form and in any order.
The combination of equation-solving capability and engineering property data
makes EES an extremely powerful tool for students.
EES can do optimization, parametric analysis, and linear and nonlinear re-
gression and provides publication-quality plotting capability. Equations can be
entered in any form and in any order. EES automatically rearranges the equa-
tions to solve them in the most efficient manner. EES is particularly useful for
heat transfer problems since most of the property data needed for solving such
problems are provided in the program. For example, the steam tables are im-
plemented such that any thermodynamic property can be obtained from a
built-in function call in terms of any two properties. Similar capability is pro-
vided for many organic refrigerants, ammonia, methane, carbon dioxide, and
many other fluids. Air tables are built-in, as are psychrometric functions and
JANAF table data for many common gases. Transport properties are also pro-
vided for all substances. EES also allows the user to enter property data or
functional relationships with look-up tables, with internal functions written
cen5 8 93 3_fm.qxd 9/11/2002 10:56 AM Page xxiv
with EES, or with externally compiled functions written in Pascal, C, C + + ,
or FORTRAN.
The Student Resources CD that accompanies the text contains the Limited
Academic Version of the EES program and the scripted EES solutions of about
30 homework problems (indicated by the "EES-CD" logo in the text). Each
EES solution provides detailed comments and on-line help, and can easily be
modified to solve similar problems. These solutions should help students
master the important concepts without the calculational burden that has been
previously required.
HEAT TRANSFER TOOLS (HTT)
One software package specifically designed to help bridge the gap between
the textbook fundamentals and commercial software packages is Heat Trans-
fer Tools, which can be ordered "bundled" with this text (Robert J. Ribando,
ISBN 0-07-246328-7). While it does not have the power and functionality of
the professional, commercial packages, HTT uses research-grade numerical
algorithms behind the scenes and modern graphical user interfaces. Each
module is custom designed and applicable to a single, fundamental topic in
heat transfer.
BOOK-SPECIFIC WEBSITE
The book website can be found at www.mhhe.com/cengel/. Visit this site for
book and supplement information, author information, and resources for fur-
ther study or reference. At this site you will also find PowerPoints of selected
text figures.
ACKNOWLEDGMENTS
I would like to acknowledge with appreciation the numerous and valuable
comments, suggestions, criticisms, and praise of these academic evaluators:
Sanjeev Chandra
University of Toronto, Canada
Fan-Bill Cheung
The Pennsylvania State University
Nicole DeJong
San Jose State University
David M. Doner
West Virginia University Institute of
Technology
Mark J. Holowach
The Pennsylvania State University
Mehmet Kanoglu
Gaziantep University, Turkey
Francis A. Kulacki
University of Minnesota
Sai C. Lau
Texas A&M University
Joseph Majdalani
Marquette University
Jed E. Marquart
Ohio Northern University
Robert J. Ribando
University of Virginia
Jay M. Ochterbeck
Clemson University
James R. Thomas
Virginia Polytechnic Institute and
State University
John D. Wellin
Rochester Institute of Technology
cen58 933_fm.qxd 9/11/2002 10:56 AM Page xxv
Their suggestions have greatly helped to improve the quality of this text. I also
would like to thank my students who provided plenty of feedback from their
perspectives. Finally, I would like to express my appreciation to my wife
Zehra and my children for their continued patience, understanding, and sup-
port throughout the preparation of this text.
Yunus A. Cengel
cen58933_ch01.qxd 9/10/2002 8:29 AM Page 1
BASICS OF HEAT TRANSFER
CHAPTER
The science of thermodynamics deals with the amount of heat transfer as
a system undergoes a process from one equilibrium state to another, and
makes no reference to how long the process will take. But in engineer-
ing, we are often interested in the rate of heat transfer, which is the topic of
the science of heat transfer.
We start this chapter with a review of the fundamental concepts of thermo-
dynamics that form the framework for heat transfer. We first present the
relation of heat to other forms of energy and review the first law of thermo-
dynamics. We then present the three basic mechanisms of heat transfer, which
are conduction, convection, and radiation, and discuss thermal conductivity.
Conduction is the transfer of energy from the more energetic particles of a
substance to the adjacent, less energetic ones as a result of interactions be-
tween the particles. Convection is the mode of heat transfer between a solid
surface and the adjacent liquid or gas that is in motion, and it involves the
combined effects of conduction and fluid motion. Radiation is the energy
emitted by matter in the form of electromagnetic waves (or photons) as a re-
sult of the changes in the electronic configurations of the atoms or molecules.
We close this chapter with a discussion of simultaneous heat transfer.
CONTENTS
1-1 Thermodynamics and
Heat Transfer 2
1-2 Engineering Heat Transfer 4
1-3 Heat and Other Forms
of Energy 6
1-4 The First Law of
Thermodynamics 11
1-5 Heat Transfer
Mechanisms 17
1-6 Conduction 17
1-7 Convection 25
1-8 Radiation 27
1-9 Simultaneous Heat Transfer
Mechanism 30
1-10 Problem-Solving Technique 35
Topic of Special Interest:
Thermal Comfort 40
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HEAT TRANSFER
1-1 - THERMODYNAMICS AND HEAT TRANSFER
Thermos
bottle
TU
Hot
coffee
±
^- Insulation
FIGURE 1-1
We are normally interested in how long
it takes for the hot coffee in a thermos to
cool to a certain temperature, which
cannot be determined from a
thermodynamic analysis alone.
Cool
environment
20°C
1 Heat
FIGURE 1-2
Heat flows in the direction of
decreasing temperature.
We all know from experience that a cold canned drink left in a room warms up
and a warm canned drink left in a refrigerator cools down. This is accom-
plished by the transfer of energy from the warm medium to the cold one. The
energy transfer is always from the higher temperature medium to the lower
temperature one, and the energy transfer stops when the two mediums reach
the same temperature.
You will recall from thermodynamics that energy exists in various forms. In
this text we are primarily interested in heat, which is the form of energy that
can be transferred from one system to another as a result of temperature dif-
ference. The science that deals with the determination of the rates of such en-
ergy transfers is heat transfer.
You may be wondering why we need to undertake a detailed study on heat
transfer. After all, we can determine the amount of heat transfer for any sys-
tem undergoing any process using a thermodynamic analysis alone. The rea-
son is that thermodynamics is concerned with the amount of heat transfer as a
system undergoes a process from one equilibrium state to another, and it gives
no indication about how long the process will take. A thermodynamic analysis
simply tells us how much heat must be transferred to realize a specified
change of state to satisfy the conservation of energy principle.
In practice we are more concerned about the rate of heat transfer (heat trans-
fer per unit time) than we are with the amount of it. For example, we can de-
termine the amount of heat transferred from a thermos bottle as the hot coffee
inside cools from 90°C to 80°C by a thermodynamic analysis alone. But a typ-
ical user or designer of a thermos is primarily interested in how long it will be
before the hot coffee inside cools to 80°C, and a thermodynamic analysis can-
not answer this question. Determining the rates of heat transfer to or from a
system and thus the times of cooling or heating, as well as the variation of the
temperature, is the subject of heat transfer (Fig. 1-1).
Thermodynamics deals with equilibrium states and changes from one equi-
librium state to another. Heat transfer, on the other hand, deals with systems
that lack thermal equilibrium, and thus it is a nonequilibrium phenomenon.
Therefore, the study of heat transfer cannot be based on the principles of
thermodynamics alone. However, the laws of thermodynamics lay the frame-
work for the science of heat transfer. The first law requires that the rate of
energy transfer into a system be equal to the rate of increase of the energy of
that system. The second law requires that heat be transferred in the direction
of decreasing temperature (Fig. 1-2). This is like a car parked on an inclined
road that must go downhill in the direction of decreasing elevation when its
brakes are released. It is also analogous to the electric current flowing in the
direction of decreasing voltage or the fluid flowing in the direction of de-
creasing total pressure.
The basic requirement for heat transfer is the presence of a temperature dif-
ference. There can be no net heat transfer between two mediums that are at the
same temperature. The temperature difference is the driving force for heat
transfer, just as the voltage difference is the driving force for electric current
flow and pressure difference is the driving force for fluid flow. The rate of heat
transfer in a certain direction depends on the magnitude of the temperature
gradient (the temperature difference per unit length or the rate of change of
cen58933_ch01.qxd 9/10/2002 8:29 AM Page 3
CHAPTER 1
temperature) in that direction. The larger the temperature gradient, the higher
the rate of heat transfer.
Application Areas of Heat Transfer
Heat transfer is commonly encountered in engineering systems and other as-
pects of life, and one does not need to go very far to see some application ar-
eas of heat transfer. In fact, one does not need to go anywhere. The human
body is constantly rejecting heat to its surroundings, and human comfort is
closely tied to the rate of this heat rejection. We try to control this heat trans-
fer rate by adjusting our clothing to the environmental conditions.
Many ordinary household appliances are designed, in whole or in part, by
using the principles of heat transfer. Some examples include the electric or gas
range, the heating and air-conditioning system, the refrigerator and freezer, the
water heater, the iron, and even the computer, the TV, and the VCR. Of course,
energy-efficient homes are designed on the basis of minimizing heat loss in
winter and heat gain in summer. Heat transfer plays a major role in the design
of many other devices, such as car radiators, solar collectors, various compo-
nents of power plants, and even spacecraft. The optimal insulation thickness
in the walls and roofs of the houses, on hot water or steam pipes, or on water
heaters is again determined on the basis of a heat transfer analysis with eco-
nomic consideration (Fig. 1-3).
Historical Background
Heat has always been perceived to be something that produces in us a sensa-
tion of warmth, and one would think that the nature of heat is one of the first
things understood by mankind. But it was only in the middle of the nineteenth
The human body
Water in
■ ll
08
30D[
nnnDn
Air-conditioning
systems
Water out
oj i n I
Circuit boards
Car radiators
Power plants
Refrigeration systems
FIGURE 1-3
Some application areas of heat transfer.
cen58933_ch01.qxd 9/10/2002 8:29 AM Page 4
HEAT TRANSFER
FIGURE 1-4
In the early nineteenth century, heat was
thought to be an invisible fluid called the
caloric that flowed from warmer bodies
to the cooler ones.
century that we had a true physical understanding of the nature of heat, thanks
to the development at that time of the kinetic theory, which treats molecules
as tiny balls that are in motion and thus possess kinetic energy. Heat is then
defined as the energy associated with the random motion of atoms and mole-
cules. Although it was suggested in the eighteenth and early nineteenth cen-
turies that heat is the manifestation of motion at the molecular level (called the
live force), the prevailing view of heat until the middle of the nineteenth cen-
tury was based on the caloric theory proposed by the French chemist Antoine
Lavoisier (1743-1794) in 1789. The caloric theory asserts that heat is a fluid-
like substance called the caloric that is a massless, colorless, odorless, and
tasteless substance that can be poured from one body into another (Fig. 1-4).
When caloric was added to a body, its temperature increased; and when
caloric was removed from a body, its temperature decreased. When a body
could not contain any more caloric, much the same way as when a glass of
water could not dissolve any more salt or sugar, the body was said to be satu-
rated with caloric. This interpretation gave rise to the terms saturated liquid
and saturated vapor that are still in use today.
The caloric theory came under attack soon after its introduction. It main-
tained that heat is a substance that could not be created or destroyed. Yet it
was known that heat can be generated indefinitely by rubbing one's hands to-
gether or rubbing two pieces of wood together. In 1798, the American Ben-
jamin Thompson (Count Rumford) (1753-1814) showed in his papers that
heat can be generated continuously through friction. The validity of the caloric
theory was also challenged by several others. But it was the careful experi-
ments of the Englishman James P. Joule (1818-1889) published in 1843 that
finally convinced the skeptics that heat was not a substance after all, and thus
put the caloric theory to rest. Although the caloric theory was totally aban-
doned in the middle of the nineteenth century, it contributed greatly to the de-
velopment of thermodynamics and heat transfer.
1-2 - ENGINEERING HEAT TRANSFER
Heat transfer equipment such as heat exchangers, boilers, condensers, radia-
tors, heaters, furnaces, refrigerators, and solar collectors are designed pri-
marily on the basis of heat transfer analysis. The heat transfer problems
encountered in practice can be considered in two groups: (1) rating and
(2) sizing problems. The rating problems deal with the determination of the
heat transfer rate for an existing system at a specified temperature difference.
The sizing problems deal with the determination of the size of a system in
order to transfer heat at a specified rate for a specified temperature difference.
A heat transfer process or equipment can be studied either experimentally
(testing and taking measurements) or analytically (by analysis or calcula-
tions). The experimental approach has the advantage that we deal with the
actual physical system, and the desired quantity is determined by measure-
ment, within the limits of experimental error. However, this approach is ex-
pensive, time-consuming, and often impractical. Besides, the system we are
analyzing may not even exist. For example, the size of a heating system of
a building must usually be determined before the building is actually built
on the basis of the dimensions and specifications given. The analytical ap-
proach (including numerical approach) has the advantage that it is fast and
cen58933_ch01.qxd 9/10/2002 8:29 AM Page 5
CHAPTER 1
inexpensive, but the results obtained are subject to the accuracy of the
assumptions and idealizations made in the analysis. In heat transfer studies,
often a good compromise is reached by reducing the choices to just a few by
analysis, and then verifying the findings experimentally.
Modeling in Heat Transfer
The descriptions of most scientific problems involve expressions that relate
the changes in some key variables to each other. Usually the smaller the
increment chosen in the changing variables, the more general and accurate
the description. In the limiting case of infinitesimal or differential changes in
variables, we obtain differential equations that provide precise mathematical
formulations for the physical principles and laws by representing the rates of
changes as derivatives. Therefore, differential equations are used to investi-
gate a wide variety of problems in sciences and engineering, including heat
transfer. However, most heat transfer problems encountered in practice can be
solved without resorting to differential equations and the complications asso-
ciated with them.
The study of physical phenomena involves two important steps. In the first
step, all the variables that affect the phenomena are identified, reasonable as-
sumptions and approximations are made, and the interdependence of these
variables is studied. The relevant physical laws and principles are invoked,
and the problem is formulated mathematically. The equation itself is very in-
structive as it shows the degree of dependence of some variables on others,
and the relative importance of various terms. In the second step, the problem
is solved using an appropriate approach, and the results are interpreted.
Many processes that seem to occur in nature randomly and without any or-
der are, in fact, being governed by some visible or not-so-visible physical
laws. Whether we notice them or not, these laws are there, governing consis-
tently and predictably what seem to be ordinary events. Most of these laws are
well defined and well understood by scientists. This makes it possible to pre-
dict the course of an event before it actually occurs, or to study various aspects
of an event mathematically without actually running expensive and time-
consuming experiments. This is where the power of analysis lies. Very accu-
rate results to meaningful practical problems can be obtained with relatively
little effort by using a suitable and realistic mathematical model. The prepara-
tion of such models requires an adequate knowledge of the natural phenomena
involved and the relevant laws, as well as a sound judgment. An unrealistic
model will obviously give inaccurate and thus unacceptable results.
An analyst working on an engineering problem often finds himself or her-
self in a position to make a choice between a very accurate but complex
model, and a simple but not-so-accurate model. The right choice depends on
the situation at hand. The right choice is usually the simplest model that yields
adequate results. For example, the process of baking potatoes or roasting a
round chunk of beef in an oven can be studied analytically in a simple way by
modeling the potato or the roast as a spherical solid ball that has the properties
of water (Fig. 1-5). The model is quite simple, but the results obtained are suf-
ficiently accurate for most practical purposes. As another example, when we
analyze the heat losses from a building in order to select the right size for a
heater, we determine the heat losses under anticipated worst conditions and
select a furnace that will provide sufficient heat to make up for those losses.
Oven
( Potato ) ■* Actual
175°C
Water ■* Ideal
FIGURE 1-5
Modeling is a powerful engineering
tool that provides great insight and
simplicity at the expense of
some accuracy.
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6
HEAT TRANSFER
Often we tend to choose a larger furnace in anticipation of some future ex-
pansion, or just to provide a factor of safety. A very simple analysis will be ad-
equate in this case.
When selecting heat transfer equipment, it is important to consider the ac-
tual operating conditions. For example, when purchasing a heat exchanger
that will handle hard water, we must consider that some calcium deposits will
form on the heat transfer surfaces over time, causing fouling and thus a grad-
ual decline in performance. The heat exchanger must be selected on the basis
of operation under these adverse conditions instead of under new conditions.
Preparing very accurate but complex models is usually not so difficult. But
such models are not much use to an analyst if they are very difficult and time-
consuming to solve. At the minimum, the model should reflect the essential
features of the physical problem it represents. There are many significant real-
world problems that can be analyzed with a simple model. But it should al-
ways be kept in mind that the results obtained from an analysis are as accurate
as the assumptions made in simplifying the problem. Therefore, the solution
obtained should not be applied to situations for which the original assump-
tions do not hold.
A solution that is not quite consistent with the observed nature of the prob-
lem indicates that the mathematical model used is too crude. In that case, a
more realistic model should be prepared by eliminating one or more of the
questionable assumptions. This will result in a more complex problem that, of
course, is more difficult to solve. Thus any solution to a problem should be in-
terpreted within the context of its formulation.
1-3 - HEAT AND OTHER FORMS OF ENERGY
Energy can exist in numerous forms such as thermal, mechanical, kinetic, po-
tential, electrical, magnetic, chemical, and nuclear, and their sum constitutes
the total energy E (or e on a unit mass basis) of a system. The forms of energy
related to the molecular structure of a system and the degree of the molecular
activity are referred to as the microscopic energy. The sum of all microscopic
forms of energy is called the internal energy of a system, and is denoted by
U (or u on a unit mass basis).
The international unit of energy is joule (J) or kilojoule (1 kJ = 1000 J).
In the English system, the unit of energy is the British thermal unit (Btu),
which is defined as the energy needed to raise the temperature of 1 lbm of
water at 60°F by 1°F. The magnitudes of kJ and Btu are almost identical
(1 Btu = 1.055056 kJ). Another well-known unit of energy is the calorie
(1 cal = 4.1868 J), which is defined as the energy needed to raise the temper-
ature of 1 gram of water at 14.5°C by 1°C.
Internal energy may be viewed as the sum of the kinetic and potential ener-
gies of the molecules. The portion of the internal energy of a system asso-
ciated with the kinetic energy of the molecules is called sensible energy or
sensible heat. The average velocity and the degree of activity of the mole-
cules are proportional to the temperature. Thus, at higher temperatures the
molecules will possess higher kinetic energy, and as a result, the system will
have a higher internal energy.
The internal energy is also associated with the intermolecular forces be-
tween the molecules of a system. These are the forces that bind the molecules
cen58933_ch01.qxd 9/10/2002 8:29 AM Page 7
CHAPTER 1
to each other, and, as one would expect, they are strongest in solids and weak-
est in gases. If sufficient energy is added to the molecules of a solid or liquid,
they will overcome these molecular forces and simply break away, turning the
system to a gas. This is a phase change process and because of this added en-
ergy, a system in the gas phase is at a higher internal energy level than it is in
the solid or the liquid phase. The internal energy associated with the phase of
a system is called latent energy or latent heat.
The changes mentioned above can occur without a change in the chemical
composition of a system. Most heat transfer problems fall into this category,
and one does not need to pay any attention to the forces binding the atoms in
a molecule together. The internal energy associated with the atomic bonds in
a molecule is called chemical (or bond) energy, whereas the internal energy
associated with the bonds within the nucleus of the atom itself is called nu-
clear energy. The chemical and nuclear energies are absorbed or released dur-
ing chemical or nuclear reactions, respectively.
In the analysis of systems that involve fluid flow, we frequently encounter
the combination of properties u and Pv. For the sake of simplicity and conve-
nience, this combination is defined as enthalpy /;. That is, h = u + Pv where
the term Pv represents the//ow energy of the fluid (also called the flow work),
which is the energy needed to push a fluid and to maintain flow. In the energy
analysis of flowing fluids, it is convenient to treat the flow energy as part of
the energy of the fluid and to represent the microscopic energy of a fluid
stream by enthalpy h (Fig. 1-6).
• Energy = h
Stationary
fluid
Energy ;
FIGURE 1-6
The internal energy u represents the mi-
croscopic energy of a nonflowing fluid,
whereas enthalpy h represents the micro-
scopic energy of a flowing fluid.
Specific Heats of Gases, Liquids, and Solids
You may recall that an ideal gas is defined as a gas that obeys the relation
Pv = RT
pRT
(1-1)
where P is the absolute pressure, v is the specific volume, T is the absolute
temperature, p is the density, and R is the gas constant. It has been experi-
mentally observed that the ideal gas relation given above closely approxi-
mates the P-v-T behavior of real gases at low densities. At low pressures and
high temperatures, the density of a gas decreases and the gas behaves like an
ideal gas. In the range of practical interest, many familiar gases such as air,
nitrogen, oxygen, hydrogen, helium, argon, neon, and krypton and even heav-
ier gases such as carbon dioxide can be treated as ideal gases with negligible
error (often less than one percent). Dense gases such as water vapor in
steam power plants and refrigerant vapor in refrigerators, however, should not
always be treated as ideal gases since they usually exist at a state near
saturation.
You may also recall that specific heat is defined as the energy required to
raise the temperature of a unit mass of a substance by one degree (Fig. 1-7).
In general, this energy depends on how the process is executed. In thermo-
dynamics, we are interested in two kinds of specific heats: specific heat at
constant volume C, and specific heat at constant pressure C p . The specific
heat at constant volume C, can be viewed as the energy required to raise the
temperature of a unit mass of a substance by one degree as the volume is held
constant. The energy required to do the same as the pressure is held constant
is the specific heat at constant pressure C p . The specific heat at constant
m =
1kg
AT =
1°C
Specific heat
1
= 5 kJ/kg-
°C
5kJ
FIGURE 1-7
Specific heat is the energy required to
raise the temperature of a unit mass
of a substance by one degree in a
specified way.
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THERMODYNAMICS
Air
Air
in = 1 kg
m = 1 kg
300^301 K
f
1000 -> 1001 K
f
I
0.718 kJ
I
0.855 kJ
FIGURE 1-8
The specific heat of a
substance changes
with temperatur
e.
pressure C p is greater than C r because at constant pressure the system is al-
lowed to expand and the energy for this expansion work must also be supplied
to the system. For ideal gases, these two specific heats are related to each
other by C p = C, + R.
A common unit for specific heats is kJ/kg • °C or kJ/kg • K. Notice that these
two units are identical since AT(°C) = AT(K), and 1°C change in temperature
is equivalent to a change of 1 K. Also,
1 kJ/kg • °C = 1 J/g • °C = 1 kJ/kg • K = 1 J/g • K
The specific heats of a substance, in general, depend on two independent
properties such as temperature and pressure. For an ideal gas, however, they
depend on temperature only (Fig. 1-8). At low pressures all real gases ap-
proach ideal gas behavior, and therefore their specific heats depend on tem-
perature only.
The differential changes in the internal energy u and enthalpy h of an ideal
gas can be expressed in terms of the specific heats as
du = C,,dT
and
dh
C p dT
(1-2)
The finite changes in the internal energy and enthalpy of an ideal gas during a
process can be expressed approximately by using specific heat values at the
average temperature as
Am
C AT
and
Ah
C AT
(J/g)
(1-3)
or
At/ = mC,. Ar
and
AH
mC pmc AT
(J)
(1-4)
FIGURE 1-9
The C„ and C p values of incompressible
substances are identical and are
denoted by C.
where m is the mass of the system.
A substance whose specific volume (or density) does not change with tem-
perature or pressure is called an incompressible substance. The specific vol-
umes of solids and liquids essentially remain constant during a process, and
thus they can be approximated as incompressible substances without sacrific-
ing much in accuracy.
The constant- volume and constant-pressure specific heats are identical for
incompressible substances (Fig. 1-9). Therefore, for solids and liquids the
subscripts on C r and C p can be dropped and both specific heats can be rep-
resented by a single symbol, C. That is, C p = C, = C. This result could also
be deduced from the physical definitions of constant- volume and constant-
pressure specific heats. Specific heats of several common gases, liquids, and
solids are given in the Appendix.
The specific heats of incompressible substances depend on temperature
only. Therefore, the change in the internal energy of solids and liquids can be
expressed as
AU
mC mc AT
(J)
(1-5)
cen58933_ch01.qxd 9/10/2002 8:29 AM Page 9
CHAPTER 1
where C ave is the average specific heat evaluated at the average temperature.
Note that the internal energy change of the systems that remain in a single
phase (liquid, solid, or gas) during the process can be determined very easily
using average specific heats.
Energy Transfer
Energy can be transferred to or from a given mass by two mechanisms: heat
Q and work W. An energy interaction is heat transfer if its driving force is a
temperature difference. Otherwise, it is work. Arising piston, a rotating shaft,
and an electrical wire crossing the system boundaries are all associated with
work interactions. Work done per unit time is called power, and is denoted
by W. The unit of power is W or hp (1 hp = 746 W). Car engines and hy-
draulic, steam, and gas turbines produce work; compressors, pumps, and
mixers consume work. Notice that the energy of a system decreases as it does
work, and increases as work is done on it.
In daily life, we frequently refer to the sensible and latent forms of internal
energy as heat, and we talk about the heat content of bodies (Fig. 1-10). In
thermodynamics, however, those forms of energy are usually referred to as
thermal energy to prevent any confusion with heat transfer.
The term heat and the associated phrases such as heat flow, heat addition,
heat rejection, heat absorption, heat gain, heat loss, heat storage, heat gener-
ation, electrical heating, latent heat, body heat, and heat source are in com-
mon use today, and the attempt to replace heat in these phrases by thermal
energy had only limited success. These phrases are deeply rooted in our vo-
cabulary and they are used by both the ordinary people and scientists without
causing any misunderstanding. For example, the phrase body heat is under-
stood to mean the thermal energy content of a body. Likewise, heat flow is
understood to mean the transfer of thermal energy, not the flow of a fluid-like
substance called heat, although the latter incorrect interpretation, based on the
caloric theory, is the origin of this phrase. Also, the transfer of heat into a sys-
tem is frequently referred to as heat addition and the transfer of heat out of a
system as heat rejection.
Keeping in line with current practice, we will refer to the thermal energy as
heat and the transfer of thermal energy as heat transfer. The amount of heat
transferred during the process is denoted by Q. The amount of heat transferred
per unit time is called heat transfer rate, and is denoted by Q. The overdot
stands for the time derivative, or "per unit time." The heat transfer rate Q has
the unit J/s, which is equivalent to W.
When the rate of heat transfer Q is available, then the total amount of heat
transfer Q during a time interval Af can be determined from
Vapor
80°C
1 ^^- Heat
transfer
Liquid
25°C
80°C
k_
)
FIGURE 1-10
The sensible and latent forms of internal
energy can be transferred as a result of
a temperature difference, and they are
referred to as heat or thermal energy.
Q
Qdt
(J)
(1-6)
provided that the variation of Q with time is known. For the special case of
Q = constant, the equation above reduces to
Q = QM
(J)
(1-7)
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10
HEAT TRANSFER
:24W
: const.
3 m
Q _ 24 W .
^ 6 m 2
4 W/m 2
FIGURE 1-11
Heat flux is heat transfer per unit
time and per unit area, and is equal
to q = QIA when Q is uniform over
the area A.
A= kD 2
FIGURE 1-12
Schematic for Example 1—1.
The rate of heat transfer per unit area normal to the direction of heat transfer
is called heat flux, and the average heat flux is expressed as (Fig. 1-11)
Q
A
(W/m 2 )
(1-8)
where A is the heat transfer area. The unit of heat flux in English units is
Btu/h • ft 2 . Note that heat flux may vary with time as well as position on a
surface.
" EXAMPLE 1-1 Heating of a Copper Ball
A 10-cm diameter copper ball is to be heated from 100°C to an average tem-
I perature of 150 C C in 30 minutes (Fig. 1-12). Taking the average density and
E specific heat of copper in this temperature range to be p = 8950 kg/m 3 and
C p = 0.395 kJ/kg • C C, respectively, determine (a) the total amount of heat
transfer to the copper ball, (b) the average rate of heat transfer to the ball, and
(c) the average heat flux.
SOLUTION The copper ball is to be heated from 100°C to 150°C. The total
heat transfer, the average rate of heat transfer, and the average heat flux are to
be determined.
Assumptions Constant properties can be used for copper at the average
temperature.
Properties The average density and specific heat of copper are given to be
p = 8950 kg/m 3 and C p = 0.395 kJ/kg ■ °C.
Analysis (a) The amount of heat transferred to the copper ball is simply the
change in its internal energy, and is determined from
Energy transfer to the system = Energy increase of the system
Q = AU = mC me (T 2 - TO
where
m = pV = ^pD 3 = ^(8950 kg/m 3 )(0.1 m) 3 = 4.69 kg
6 o
Substituting,
Q = (4.69 kg)(0.395 kJ/kg • °C)(150 - 100)°C = 92.6 kj
Therefore, 92.6 kJ of heat needs to be transferred to the copper ball to heat it
from 100°Cto 150°C.
(b) The rate of heat transfer normally changes during a process with time. How-
ever, we can determine the average rate of heat transfer by dividing the total
amount of heat transfer by the time interval. Therefore,
e a
Q = 92.6 kJ
A? 1800 s
0.0514 kJ/s = 51.4 W
cen58933_ch01.qxd 9/10/2002 8:29 AM Page 11
(c) Heat flux is defined as the heat transfer per unit time per unit area, or the
rate of heat transfer per unit area. Therefore, the average heat flux in this
case is
s^avi
x£ave
^D 2
51.4 W
tt(0.1 m) 2
1636 W/m 2
Discussion Note that heat flux may vary with location on a surface. The value
calculated above is the average heat flux over the entire surface of the ball.
1^ ■ THE FIRST LAW OF THERMODYNAMICS
The first law of thermodynamics, also known as the conservation of energy
principle, states that energy can neither be created nor destroyed; it can only
change forms. Therefore, every bit of energy must be accounted for during a
process. The conservation of energy principle (or the energy balance) for any
system undergoing any process may be expressed as follows: The net change
(increase or decrease) in the total energy of the system during a process is
equal to the difference between the total energy entering and the total energy
leaving the system during that process. That is,
(Total energy
entering the
system
Total energy
leaving the
system
Change in the \
total energy of
the system /
(1-9)
Noting that energy can be transferred to or from a system by heat, work, and
mass flow, and that the total energy of a simple compressible system consists
of internal, kinetic, and potential energies, the energy balance for any system
undergoing any process can be expressed as
Net energy transfer
by heat, work, and mass
A£„
Change in internal, kinetic,
potential, etc., energies
(J)
(1-10)
n
CHAPTER 1
or, in the rate form, as
Rate of net energy transfer
by heat, work, and mass
"^■system'"*
Rate of change in internal
kinetic, potential, etc., energies
(W)
(1-11)
Energy is a property, and the value of a property does not change unless the
state of the system changes. Therefore, the energy change of a system is zero
(A,E system = 0) if the state of the system does not change during the process,
that is, the process is steady. The energy balance in this case reduces to
(Fig. 1-13)
Steady, rate form:
(1-12)
Rate of net energy transfer in
by heat, work, and mass
Rate of net energy transfer out
by heat, work, and mass
In the absence of significant electric, magnetic, motion, gravity, and surface
tension effects (i.e., for stationary simple compressible systems), the change
FIGURE 1-13
In steady operation, the rate of energy
transfer to a system is equal to the rate
of energy transfer from the system.
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12
HEAT TRANSFER
mC(T.-T„)
FIGURE 1-14
In the absence of any work interactions,
the change in the energy content of a
closed system is equal to the net
heat transfer.
in the total energy of a system during a process is simply the change in its in-
ternal energy. That is, AE,
system
AC/,
system*
In heat transfer analysis, we are usually interested only in the forms of en-
ergy that can be transferred as a result of a temperature difference, that is, heat
or thermal energy. In such cases it is convenient to write a heat balance and
to treat the conversion of nuclear, chemical, and electrical energies into ther-
mal energy as heat generation. The energy balance in that case can be ex-
pressed as
Gin " Gou,
Net heat
transfer
AE
Heat
generation
thermal, system
Change in thermal
energy of the system
(J)
(1-13)
Energy Balance for Closed Systems (Fixed Mass)
A closed system consists of a fixed mass. The total energy E for most systems
encountered in practice consists of the internal energy U. This is especially the
case for stationary systems since they don't involve any changes in their ve-
locity or elevation during a process. The energy balance relation in that case
reduces to
Stationary closed system:
AU = mCAT
(J)
(1-14)
where we expressed the internal energy change in terms of mass m, the spe-
cific heat at constant volume C r , and the temperature change AT of the sys-
tem. When the system involves heat transfer only and no work interactions
across its boundary, the energy balance relation further reduces to (Fig. 1-14)
Stationary closed system, no work:
Q = mC r AT
(J)
(1-15)
where Q is the net amount of heat transfer to or from the system. This is the
form of the energy balance relation we will use most often when dealing with
a fixed mass.
Energy Balance for Steady-Flow Systems
A large number of engineering devices such as water heaters and car radiators
involve mass flow in and out of a system, and are modeled as control volumes.
Most control volumes are analyzed under steady operating conditions. The
term steady means no change with time at a specified location. The opposite
of steady is unsteady or transient. Also, the term uniform implies no change
with position throughout a surface or region at a specified time. These mean-
ings are consistent with their everyday usage (steady girlfriend, uniform
distribution, etc.). The total energy content of a control volume during a
steady-flow process remains constant (E CY = constant). That is, the change
in the total energy of the control volume during such a process is zero
(A£ cv = 0). Thus the amount of energy entering a control volume in all forms
(heat, work, mass transfer) for a steady-flow process must be equal to the
amount of energy leaving it.
The amount of mass flowing through a cross section of a flow device per
unit time is called the mass flow rate, and is denoted by rh. A fluid may flow
in and out of a control volume through pipes or ducts. The mass flow rate of a
fluid flowing in a pipe or duct is proportional to the cross-sectional area A c of
cen58933_ch01.qxd 9/10/2002 8:29 AM Page 13
the pipe or duct, the density p, and the velocity T of the fluid. The mass flow
rate through a differential area dA c can be expressed as 8m = pT„ dA c where
Y„ is the velocity component normal to dA c . The mass flow rate through the
entire cross-sectional area is obtained by integration over A c .
The flow of a fluid through a pipe or duct can often be approximated to be
one-dimensional. That is, the properties can be assumed to vary in one direc-
tion only (the direction of flow). As a result, all properties are assumed to be
uniform at any cross section normal to the flow direction, and the properties
are assumed to have bulk average values over the entire cross section. Under
the one-dimensional flow approximation, the mass flow rate of a fluid flow-
ing in a pipe or duct can be expressed as (Fig. 1-15)
P TA r
(kg/s)
(1-16)
where p is the fluid density, T is the average fluid velocity in the flow direc-
tion, and A c is the cross-sectional area of the pipe or duct.
The volume of a fluid flowing through a pipe or duct per unit time is called
the volume flow rate V, and is expressed as
V = TA r
m
P
(m 3 /s)
(1-17)
13
CHAPTER 1
A = kD 2 /4 —
c
for a circular pipe
m =pVA
FIGURE 1-15
The mass flow rate of a fluid at a cross
section is equal to the product of the
fluid density, average fluid velocity,
and the cross-sectional area.
Note that the mass flow rate of a fluid through a pipe or duct remains constant
during steady flow. This is not the case for the volume flow rate, however, un-
less the density of the fluid remains constant.
For a steady-flow system with one inlet and one exit, the rate of mass flow
into the control volume must be equal to the rate of mass flow out of it. That
is, m in = m out = m. When the changes in kinetic and potential energies are
negligible, which is usually the case, and there is no work interaction, the en-
ergy balance for such a steady-flow system reduces to (Fig. 1-16)
Q = ihAh
rhC p AT
(kJ/s)
(1-18)
where Q is the rate of net heat transfer into or out of the control volume. This
is the form of the energy balance relation that we will use most often for
steady-flow systems.
s- Control volume
r
~l
m
4 T' 2
L
___#_
transfer /A 2 V
FIGURE 1-16
Under steady conditions, the net rate of
energy transfer to a fluid in a control
volume is equal to the rate of increase in
the energy of the fluid stream flowing
through the control volume.
Surface Energy Balance
As mentioned in the chapter opener, heat is transferred by the mechanisms of
conduction, convection, and radiation, and heat often changes vehicles as it is
transferred from one medium to another. For example, the heat conducted to
the outer surface of the wall of a house in winter is convected away by the
cold outdoor air while being radiated to the cold surroundings. In such cases,
it may be necessary to keep track of the energy interactions at the surface, and
this is done by applying the conservation of energy principle to the surface.
A surface contains no volume or mass, and thus no energy. Thereore, a sur-
face can be viewed as a fictitious system whose energy content remains con-
stant during a process (just like a steady-state or steady-flow system). Then
the energy balance for a surface can be expressed as
Surface energy balance:
(1-19)
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14
HEAT TRANSFER
WALL
Control
i/ surface
radiation
conduction
1^****' Qi
e,
N*
i convection
FIGURE 1-17
Energy interactions at the outer wall
surface of a house.
This relation is valid for both steady and transient conditions, and the surface
energy balance does not involve heat generation since a surface does not have
a volume. The energy balance for the outer surface of the wall in Fig. 1-17,
for example, can be expressed as
61 = 62 + 63
(1-20)
where Q , is conduction through the wall to the surface, Q 2 is convection from
the surface to the outdoor air, and <2 3 is net radiation from the surface to the
surroundings.
When the directions of interactions are not known, all energy interactions
can be assumed to be towards the surface, and the surface energy balance can
be expressed as 2 E in = 0. Note that the interactions in opposite direction will
end up having negative values, and balance this equation.
Electric
heating
element
FIGURE 1-18
Schematic for Example 1-2.
EXAMPLE 1-2 Heating of Water in an Electric Teapot
1.2 kg of liquid water initially at 15°C is to be heated to 95°C in a teapot
equipped with a 1200-W electric heating element inside (Fig. 1-18). The
teapot is 0.5 kg and has an average specific heat of 0.7 kJ/kg • °C. Taking the
specific heat of water to be 4.18 kJ/kg • °C and disregarding any heat loss from
the teapot, determine how long it will take for the water to be heated.
SOLUTION Liquid water is to be heated in an electric teapot. The heating time
is to be determined.
Assumptions 1 Heat loss from the teapot is negligible. 2 Constant properties
can be used for both the teapot and the water.
Properties The average specific heats are given to be 0.7 kJ/kg • °C for the
teapot and 4.18 kJ/kg • °C for water.
Analysis We take the teapot and the water in it as the system, which is
a closed system (fixed mass). The energy balance in this case can be ex-
pressed as
AF
ulJ system
system ^ ^ water
AU
teapot
Then the amount of energy needed to raise the temperature of water and the
teapot from 15°C to 95°C is
E m = (mCAT),
(mCAT
teapot
= (1.2 kg)(4.18 kJ/kg • °C)(95 - 15)°C + (0.5 kg)(0.7 kJ/kg • °C)
(95 - 15)°C
= 429.3 kJ
The 1200-W electric heating unit will supply energy at a rate of 1.2 kW or
1.2 kJ per second. Therefore, the time needed for this heater to supply
429.3 kJ of heat is determined from
At
Total energy transferred E in 429 3 ^ j
Rate of energy transfer E 1 .2 kJ/s
CJ ^ transfer
358 s = 6.0 min
cen58933_ch01.qxd 9/10/2002 8:29 AM Page 15
Discussion In reality, it will take more than 6 minutes to accomplish this heat-
ing process since some heat loss is inevitable during heating.
15
CHAPTER 1
EXAMPLE 1-3 Heat Loss from Heating Ducts in a Basement
A 5-m-long section of an air heating system of a house passes through an un-
heated space in the basement (Fig. 1-19). The cross section of the rectangular
duct of the heating system is 20 cm X 25 cm. Hot air enters the duct at
100 kPa and 60°C at an average velocity of 5 m/s. The temperature of the air
in the duct drops to 54°C as a result of heat loss to the cool space in the base-
ment. Determine the rate of heat loss from the air in the duct to the basement
under steady conditions. Also, determine the cost of this heat loss per hour if
the house is heated by a natural gas furnace that has an efficiency of 80 per-
cent, and the cost of the natural gas in that area is $0.60/therm (1 therm =
100,000 Btu = 105,500 kJ).
54°C
- loss
FIGURE 1-19
Schematic for Example 1-3.
SOLUTION The temperature of the air in the heating duct of a house drops as
a result of heat loss to the cool space in the basement. The rate of heat loss
from the hot air and its cost are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Air can be treated as an
ideal gas with constant properties at room temperature.
Properties The constant pressure specific heat of air at the average tempera-
ture of (54 + 60)/2 = 57°C is 1.007 kJ/kg • °C (Table A-15).
Analysis We take the basement section of the heating system as our system,
which is a steady-flow system. The rate of heat loss from the air in the duct can
be determined from
Q
mC p AT
where rh is the mass flow rate and A T is the temperature drop. The density of
air at the inlet conditions is
P_
RT
100 kPa
(0.287 kPa ■ m 3 /kg • K)(60 + 273)K
1.046 kg/m 3
The cross-sectional area of the duct is
A c = (0.20 m)(0.25 m) = 0.05 m 2
Then the mass flow rate of air through the duct and the rate of heat loss
become
m = pTA c = (1.046 kg/m 3 )(5 m/s)(0.05 m 2 ) = 0.2615 kg/s
and
Q loss ~~ " Z< --p(i m T out )
= (0.2615 kg/s)(1.007 kJ/kg ■ °C)(60 - 54)°C
= 1.580 kj/s
cen58933_ch01.qxd 9/10/2002 8:29 AM Page 16
16
HEAT TRANSFER
or 5688 kJ/h. The cost of this heat loss to the home owner is
(Rate of heat loss)(Unit cost of energy input)
Cost of heat loss
Furnace efficiency
(5688 kJ/h)($0.60/therm)/ i therm
0.80
105,500 kJ
= $0.040/h
Discussion The heat loss from the heating ducts in the basement is costing the
home owner 4 cents per hour. Assuming the heater operates 2000 hours during
a heating season, the annual cost of this heat loss adds up to $80. Most of this
money can be saved by insulating the heating ducts in the unheated areas.
9 ft
FIGURE 1-20
Schematic for Example 1-4.
-
EXAMPLE 1-4 Electric Heating of a House at High Elevation
Consider a house that has a floor space of 2000 ft 2 and an average height of 9
ft at 5000 ft elevation where the standard atmospheric pressure is 12.2 psia
(Fig. 1-20). Initially the house is at a uniform temperature of 50°F. Now the
electric heater is turned on, and the heater runs until the air temperature in the
house rises to an average value of 70°F. Determine the amount of energy trans-
ferred to the air assuming (a) the house is air-tight and thus no air escapes dur-
ing the heating process and (£>) some air escapes through the cracks as the
heated air in the house expands at constant pressure. Also determine the cost
of this heat for each case if the cost of electricity in that area is $0.075/kWh.
SOLUTION The air in the house is heated from 50°F to 70°F by an electric
heater. The amount and cost of the energy transferred to the air are to be de-
termined for constant-volume and constant-pressure cases.
Assumptions 1 Air can be treated as an ideal gas with constant properties at
room temperature. 2 Heat loss from the house during heating is negligible.
3 The volume occupied by the furniture and other things is negligible.
Properties The specific heats of air at the average temperature of (50 + 70)/2
60°F are C„ = 0.240 Btu/lbm ■ °F and C„
C„- R
0.171 Btu/lbm • °F
(Tables A-1E and A-15E).
Analysis The volume and the mass of the air in the house are
V = (Floor area)(Height) = (2000 ft 2 )(9 ft)
PV (12.2psia)(18,000ft 3 )
18,000 ft 3
m
RT (0.3704 psia • ftMbm ■ R)(50 + 460)R
1162 lbm
(a) The amount of energy transferred to air at constant volume is simply the
change in its internal energy, and is determined from
F — F = AF
in out system
^in, constant volume ^^air — ^C,, 111
= (1162 lbm)(0.171 Btu/lbm • °F)(70 - 50)°F
= 3974 Btu
At a unit cost of $0.075/kWh, the total cost of this energy is
cen58933_ch01.qxd 9/10/2002 8:29 AM Page 17
Cost of energy = (Amount of energy)(Unit cost of energy)
1 kWh
(3974 Btu)($0.075/kWh)
$0,087
3412 Btu
(b) The amount of energy transferred to air at constant pressure is the change
in its enthalpy, and is determined from
in, constant pressure air iil^pLxl
= (1162 lbm)(0.240 Btu/lbm • °F)(70 - 50)°F
= 5578 Btu
At a unit cost of $0.075/kWh, the total cost of this energy is
Cost of energy = (Amount of energy)(Unit cost of energy)
1 kWh
(5578 Btu)($0.075/kWh)
$0,123
3412 Btu
Discussion It will cost about 12 cents to raise the temperature of the air in
this house from 50°F to 70°F. The second answer is more realistic since every
house has cracks, especially around the doors and windows, and the pressure in
the house remains essentially constant during a heating process. Therefore, the
second approach is used in practice. This conservative approach somewhat
overpredicts the amount of energy used, however, since some of the air will es-
cape through the cracks before it is heated to 70°F.
17
CHAPTER 1
1-5 - HEAT TRANSFER MECHANISMS
In Section 1-1 we defined heat as the form of energy that can be transferred
from one system to another as a result of temperature difference. A thermo-
dynamic analysis is concerned with the amount of heat transfer as a system
undergoes a process from one equilibrium state to another. The science that
deals with the determination of the rates of such energy transfers is the heat
transfer. The transfer of energy as heat is always from the higher-temperature
medium to the lower-temperature one, and heat transfer stops when the two
mediums reach the same temperature.
Heat can be transferred in three different modes: conduction, convection,
and radiation. All modes of heat transfer require the existence of a tempera-
ture difference, and all modes are from the high-temperature medium to a
lower-temperature one. Below we give a brief description of each mode. A de-
tailed study of these modes is given in later chapters of this text.
1-6 - CONDUCTION
Conduction is the transfer of energy from the more energetic particles of a
substance to the adjacent less energetic ones as a result of interactions be-
tween the particles. Conduction can take place in solids, liquids, or gases. In
gases and liquids, conduction is due to the collisions and diffusion of the
cen58933_ch01.qxd 9/10/2002 8:29 AM Page 16
18
HEAT TRANSFER
FIGURE 1-21
Heat conduction through a large plane
wall of thickness Ax and area A.
molecules during their random motion. In solids, it is due to the combination
of vibrations of the molecules in a lattice and the energy transport by free
electrons. A cold canned drink in a warm room, for example, eventually
warms up to the room temperature as a result of heat transfer from the room
to the drink through the aluminum can by conduction.
The rate of heat conduction through a medium depends on the geometry of
the medium, its thickness, and the material of the medium, as well as the tem-
perature difference across the medium. We know that wrapping a hot water
tank with glass wool (an insulating material) reduces the rate of heat loss from
the tank. The thicker the insulation, the smaller the heat loss. We also know
that a hot water tank will lose heat at a higher rate when the temperature of the
room housing the tank is lowered. Further, the larger the tank, the larger the
surface area and thus the rate of heat loss.
Consider steady heat conduction through a large plane wall of thickness
Ax = L and area A, as shown in Fig. 1—21. The temperature difference across
the wall is AT = T 2 — T { . Experiments have shown that the rate of heat trans-
fer Q through the wall is doubled when the temperature difference AT across
the wall or the area A normal to the direction of heat transfer is doubled, but is
halved when the wall thickness L is doubled. Thus we conclude that the rate
of heat conduction through a plane layer is proportional to the temperature
difference across the layer and the heat transfer area, but is inversely propor-
tional to the thickness of the layer. That is,
Rate of heat conduction «
(Area)(Temperature difference)
Thickness
30°C
lm
20°C
4 = 4010W/m 2
{a) Copper (Jfc = 401 W/m-°C)
30°C
lm
20°C
g=1480W/m 2
(b) Silicon (k = 148 W/m-°C)
FIGURE 1-22
The rate of heat conduction through a
solid is directly proportional to
its thermal conductivity.
or,
xl, cond
kA
7\
Ax
-kA
AT
Ax
(W)
(1-21)
where the constant of proportionality k is the thermal conductivity of the
material, which is a measure of the ability of a material to conduct heat
(Fig. 1-22). In the limiting case of Ax — > 0, the equation above reduces to the
differential form
Oc
-kA
(IT
dx
(W)
(1-22)
which is called Fourier's law of heat conduction after J. Fourier, who ex-
pressed it first in his heat transfer text in 1822. Here dT/dx is the temperature
gradient, which is the slope of the temperature curve on a T-x diagram (the
rate of change of T with x), at location x. The relation above indicates that the
rate of heat conduction in a direction is proportional to the temperature gradi-
ent in that direction. Heat is conducted in the direction of decreasing tem-
perature, and the temperature gradient becomes negative when temperature
decreases with increasing x. The negative sign in Eq. 1-22 ensures that heat
transfer in the positive x direction is a positive quantity.
The heat transfer area A is always normal to the direction of heat transfer.
For heat loss through a 5-m-long, 3-m-high, and 25-cm-thick wall, for exam-
ple, the heat transfer area is A = 15 m 2 . Note that the thickness of the wall has
no effect on A (Fig. 1-23).
cen58933_ch01.qxd 9/10/2002 8:29 AM Page 19
19
CHAPTER 1
EXAMPLE 1-5 The Cost of Heat Loss through a Roof
The roof of an electrically heated home is 6 m long, 8 m wide, and 0.25 m
thick, and is made of a flat layer of concrete whose thermal conductivity is
k = 0.8 W/m • °C (Fig. 1-24). The temperatures of the inner and the outer sur-
faces of the roof one night are measured to be 15°C and 4°C, respectively, for a
period of 10 hours. Determine (a) the rate of heat loss through the roof that
night and (b) the cost of that heat loss to the home owner if the cost of elec-
tricity is $0.08/kWh.
SOLUTION The inner and outer surfaces of the flat concrete roof of an electri-
cally heated home are maintained at specified temperatures during a night. The
heat loss through the roof and its cost that night are to be determined.
Assumptions 1 Steady operating conditions exist during the entire night since
the surface temperatures of the roof remain constant at the specified values.
2 Constant properties can be used for the roof.
Properties The thermal conductivity of the roof is given to be k = 0.8
W/m ■ °C.
Analysis (a) Noting that heat transfer through the roof is by conduction and
the area of the roof is/4=6mX8m = 48 m 2 , the steady rate of heat trans-
fer through the roof is determined to be
Q =kA
(0.8 W/m • °C)(48 m 2 )
(15 - 4)°C
0.25 m
1690 W = 1.69 kW
(b) The amount of heat lost through the roof during a 10-hour period and its
cost are determined from
Q = Q At = (1.69 kW)(10 h) = 16.9 kWh
Cost = (Amount of energy)(Unit cost of energy)
= (16.9 kWh)($0.08/kWh) = $1.35
Discussion The cost to the home owner of the heat loss through the roof that
night was $1.35. The total heating bill of the house will be much larger since
the heat losses through the walls are not considered in these calculations.
FIGURE 1-23
In heat conduction analysis, A represents
the area normal to the direction
of heat transfer.
Concrete roof
8 m
0.25 m
FIGURE 1-24
Schematic for Example 1-5.
Thermal Conductivity
We have seen that different materials store heat differently, and we have de-
fined the property specific heat C p as a measure of a material's ability to store
thermal energy. For example, C p = 4.18 kJ/kg • °C for water and C p = 0.45
kJ/kg • °C for iron at room temperature, which indicates that water can store
almost 10 times the energy that iron can per unit mass. Likewise, the thermal
conductivity A; is a measure of a material's ability to conduct heat. For exam-
ple, k = 0.608 W/m • °C for water and k = 80.2 W/m • °C for iron at room
temperature, which indicates that iron conducts heat more than 100 times
faster than water can. Thus we say that water is a poor heat conductor relative
to iron, although water is an excellent medium to store thermal energy.
Equation 1-22 for the rate of conduction heat transfer under steady condi-
tions can also be viewed as the defining equation for thermal conductivity.
Thus the thermal conductivity of a material can be defined as the rate of
cen58933_ch01.qxd 9/10/2002 8:29 AM Page 20
20
HEAT TRANSFER
TABLE
1-1
The thermal conductivities of some
materials at room temperature
Materia
k, W/m
°c*
Diamond
2300
Silver
429
Copper
401
Gold
317
Aluminum
237
Iron
80.2
Mercury (I)
8.54
Glass
0.78
Brick
0.72
Water (I)
0.613
Human skin
0.37
Wood (oak)
0.17
Helium (g)
0.152
Soft rubber
0.13
Glass fiber
0.043
Air (g)
0.026
Urethane, rigid foam
0.026
♦Multiply by 0.5778 to convert to Btu/h • ft • °F.
Sample
material
■TV)
Q
A(T r
FIGURE 1-25
A simple experimental setup to
determine the thermal conductivity
of a material.
heat transfer through a unit thickness of the material per unit area per unit
temperature difference. The thermal conductivity of a material is a measure of
the ability of the material to conduct heat. A high value for thermal conduc-
tivity indicates that the material is a good heat conductor, and a low value
indicates that the material is a poor heat conductor or insulator. The thermal
conductivities of some common materials at room temperature are given in
Table 1—1. The thermal conductivity of pure copper at room temperature is
k = 401 W/m • °C, which indicates that a 1-m-thick copper wall will conduct
heat at a rate of 401 W per m 2 area per °C temperature difference across the
wall. Note that materials such as copper and silver that are good electric con-
ductors are also good heat conductors, and have high values of thermal con-
ductivity. Materials such as rubber, wood, and styrofoam are poor conductors
of heat and have low conductivity values.
A layer of material of known thickness and area can be heated from one side
by an electric resistance heater of known output. If the outer surfaces of the
heater are well insulated, all the heat generated by the resistance heater will be
transferred through the material whose conductivity is to be determined. Then
measuring the two surface temperatures of the material when steady heat
transfer is reached and substituting them into Eq. 1-22 together with other
known quantities give the thermal conductivity (Fig. 1-25).
The thermal conductivities of materials vary over a wide range, as shown in
Fig. 1-26. The thermal conductivities of gases such as air vary by a factor of
10 4 from those of pure metals such as copper. Note that pure crystals and met-
als have the highest thermal conductivities, and gases and insulating materials
the lowest.
Temperature is a measure of the kinetic energies of the particles such as the
molecules or atoms of a substance. In a liquid or gas, the kinetic energy of the
molecules is due to their random translational motion as well as their
vibrational and rotational motions. When two molecules possessing differ-
ent kinetic energies collide, part of the kinetic energy of the more energetic
(higher-temperature) molecule is transferred to the less energetic (lower-
temperature) molecule, much the same as when two elastic balls of the same
mass at different velocities collide, part of the kinetic energy of the faster
ball is transferred to the slower one. The higher the temperature, the faster the
molecules move and the higher the number of such collisions, and the better
the heat transfer.
The kinetic theory of gases predicts and the experiments confirm that the
thermal conductivity of gases is proportional to the square root of the abso-
lute temperature T, and inversely proportional to the square root of the molar
mass M. Therefore, the thermal conductivity of a gas increases with increas-
ing temperature and decreasing molar mass. So it is not surprising that the
thermal conductivity of helium (M = 4) is much higher than those of air
(M = 29) and argon (M = 40).
The thermal conductivities of gases at 1 arm pressure are listed in Table
A- 16. However, they can also be used at pressures other than 1 atm, since the
thermal conductivity of gases is independent of pressure in a wide range of
pressures encountered in practice.
The mechanism of heat conduction in a liquid is complicated by the fact that
the molecules are more closely spaced, and they exert a stronger intermolecu-
lar force field. The thermal conductivities of liquids usually lie between those
cen58933_ch01.qxd 9/10/2002 8:29 AM Page 21
NONMETALLIC
CRYSTALS
1000
k,
W/m°C
100
10
0.1
0.0 I
Diamond
Graphite
Silicon
carbide
Beryllium
oxide
Quartz
METAL
ALLOYS
PURE
METALS
Silver
Copper
Iron
Manganese
Aluminum
alloys
Bronze
Steel
Nichrome
NONMETALLIC
SOLIDS
Oxides
Rock
Food
Rubber
LIQUIDS
Mercury
Water
Oils
NSULATORS
GASES
Fibers
Wood
Foams
Hydrogen
Helium
Aii-
Carbon
dioxide
21
CHAPTER 1
FIGURE 1-26
The range of thermal conductivity of
various materials at room temperature.
of solids and gases. The thermal conductivity of a substance is normally high-
est in the solid phase and lowest in the gas phase. Unlike gases, the thermal
conductivities of most liquids decrease with increasing temperature, with wa-
ter being a notable exception. Like gases, the conductivity of liquids decreases
with increasing molar mass. Liquid metals such as mercury and sodium have
high thermal conductivities and are very suitable for use in applications where
a high heat transfer rate to a liquid is desired, as in nuclear power plants.
In solids, heat conduction is due to two effects: the lattice vibrational waves
induced by the vibrational motions of the molecules positioned at relatively
fixed positions in a periodic manner called a lattice, and the energy trans-
ported via the free flow of electrons in the solid (Fig. 1-27). The ther-
mal conductivity of a solid is obtained by adding the lattice and electronic
components. The relatively high thermal conductivities of pure metals are pri-
marily due to the electronic component. The lattice component of thermal
conductivity strongly depends on the way the molecules are arranged. For ex-
ample, diamond, which is a highly ordered crystalline solid, has the highest
known thermal conductivity at room temperature.
Unlike metals, which are good electrical and heat conductors, crystalline
solids such as diamond and semiconductors such as silicon are good heat con-
ductors but poor electrical conductors. As a result, such materials find wide-
spread use in the electronics industry. Despite their higher price, diamond heat
sinks are used in the cooling of sensitive electronic components because of the
FIGURE 1-27
The mechanisms of heat conduction in
different phases of a substance.
cen58933_ch01.qxd 9/10/2002 8:29 AM Page 22
22
HEAT TRANSFER
TABLE 1-2
The thermal conductivity of an
alloy is usually much lower than
the thermal conductivity of either
metal of which it is composed
Pure metal or k, W/m • °C,
alloy at 300 K
Copper
401
Nickel
91
Constantan
(55% Cu, 45% Ni)
23
Copper
401
Aluminum
237
Commercial bronze
(90% Cu, 10% Al)
52
TABLE
1-3
Thermal conductivities
vary with temperature
of materials
T, K
Copper
Aluminum
100
482
200
413
300
401
400
393
600
379
800
366
302
237
237
240
231
218
FIGURE 1-28
The variation of the thermal
conductivity of various solids,
liquids, and gases with temperature
(from White, Ref. 10).
excellent thermal conductivity of diamond. Silicon oils and gaskets are com-
monly used in the packaging of electronic components because they provide
both good thermal contact and good electrical insulation.
Pure metals have high thermal conductivities, and one would think that
metal alloys should also have high conductivities. One would expect an alloy
made of two metals of thermal conductivities k { and k 2 to have a conductivity
k between k [ and k 2 . But this turns out not to be the case. The thermal conduc-
tivity of an alloy of two metals is usually much lower than that of either metal,
as shown in Table 1-2. Even small amounts in a pure metal of "foreign" mol-
ecules that are good conductors themselves seriously disrupt the flow of heat
in that metal. For example, the thermal conductivity of steel containing just
1 percent of chrome is 62 W/m • °C, while the thermal conductivities of iron
and chromium are 83 and 95 W/m • °C, respectively.
The thermal conductivities of materials vary with temperature (Table 1-3).
The variation of thermal conductivity over certain temperature ranges is neg-
ligible for some materials, but significant for others, as shown in Fig. 1-28.
The thermal conductivities of certain solids exhibit dramatic increases at tem-
peratures near absolute zero, when these solids become superconductors. For
example, the conductivity of copper reaches a maximum value of about
20,000 W/m • °C at 20 K, which is about 50 times the conductivity at room
temperature. The thermal conductivities and other thermal properties of vari-
ous materials are given in Tables A-3 to A- 16.
10,000
k,
W/m-°C
1000
100
10
0.1
0.01
^^v^Diamonds
s / v \ s- ^" v Type Ha
"\/ ' Type lib
• Type I
- Solids
. Liquids
Gases
Silver r>„„„„.
" Aluminum ,
Tungsten .
Gold ■ J ■
*v^ ■ Platinum
' Iron
■ Pyroceram glass
Aluminum oxide
Clear fused quartz
^-- -. Water
Helium
— _ Carbon tetrachloride
""• -. Steam
Air
Argon
200
400
600
800
1000
1200 1400
cen58933_ch01.qxd 9/10/2002 8:29 AM Page 23
The temperature dependence of thermal conductivity causes considerable
complexity in conduction analysis. Therefore, it is common practice to evalu-
ate the thermal conductivity k at the average temperature and treat it as a con-
stant in calculations.
In heat transfer analysis, a material is normally assumed to be isotropic; that
is, to have uniform properties in all directions. This assumption is realistic for
most materials, except those that exhibit different structural characteristics in
different directions, such as laminated composite materials and wood. The
thermal conductivity of wood across the grain, for example, is different than
that parallel to the grain.
Thermal Diffusivity
The product pC p , which is frequently encountered in heat transfer analysis, is
called the heat capacity of a material. Both the specific heat C p and the heat
capacity pC p represent the heat storage capability of a material. But C p ex-
presses it per unit mass whereas pC p expresses it per unit volume, as can be
noticed from their units J/kg • °C and J/m 3 • °C, respectively.
Another material property that appears in the transient heat conduction
analysis is the thermal diffusivity, which represents how fast heat diffuses
through a material and is defined as
Heat conducted
Heat stored
k
pC P
(m 2 /s)
(1-23)
Note that the thermal conductivity k represents how well a material con-
ducts heat, and the heat capacity pC p represents how much energy a material
stores per unit volume. Therefore, the thermal diffusivity of a material can be
viewed as the ratio of the heat conducted through the material to the heat
stored per unit volume. A material that has a high thermal conductivity or a
low heat capacity will obviously have a large thermal diffusivity. The larger
the thermal diffusivity, the faster the propagation of heat into the medium.
A small value of thermal diffusivity means that heat is mostly absorbed by the
material and a small amount of heat will be conducted further.
The thermal diffusivities of some common materials at 20°C are given in
Table 1-4. Note that the thermal diffusivity ranges from a = 0.14 X IO 6 m 2 /s
for water to 174 X 10~ 6 m 2 /s for silver, which is a difference of more than a
thousand times. Also note that the thermal diffusivities of beef and water are
the same. This is not surprising, since meat as well as fresh vegetables and
fruits are mostly water, and thus they possess the thermal properties of water.
EXAMPLE 1-6 Measuring the Thermal Conductivity of a Material
A common way of measuring the thermal conductivity of a material is to sand-
wich an electric thermofoil heater between two identical samples of the ma-
terial, as shown in Fig. 1-29. The thickness of the resistance heater, including
its cover, which is made of thin silicon rubber, is usually less than 0.5 mm.
A circulating fluid such as tap water keeps the exposed ends of the samples
at constant temperature. The lateral surfaces of the samples are well insulated
to ensure that heat transfer through the samples is one-dimensional. Two
I thermocouples are embedded into each sample some distance L apart, and a
23
CHAPTER 1
TABLE 1-4
The thermal diffusivities of some
materials at room temperature
Material
a,
m 2 /s*
Silver
149
X
10- 5
Gold
127
X
10- 5
Copper
113
X
io- 5
Aluminum
97.5
X
io- 6
Iron
22.8 X
io- 5
Mercury (I)
4.7
X
io- 5
Marble
1.2
X
io- 5
Ice
1.2
X
io- 6
Concrete
0.75
X
io- 5
Brick
0.52
X
io- 6
Heavy soil (dry)
0.52
X
io- 5
Glass
0.34
X
io- 5
Glass wool
0.23
X
IO" 5
Water (I)
0.14
X
IO" 5
Beef
0.14
X
IO" 5
Wood (oak)
0.13
X
IO" 5
*Mu Iti ply by 10.76 to convert to ft 2 /s.
, Cooling
r fluid
Insulation
Sample
<
Thermocouple
/
X
l }Ar,
Resistance
heater
a
-z?=
Sample
a
L W,
1
, Cooling
r fluid
FIGURE 1-29
Apparatus to measure the thermal
conductivity of a material using two
identical samples and a thin resistance
heater (Example 1-6).
cen58933_ch01.qxd 9/10/2002 8:29 AM Page 24
24
HEAT TRANSFER
differential thermometer reads the temperature drop AT across this distance
along each sample. When steady operating conditions are reached, the total
rate of heat transfer through both samples becomes equal to the electric power
drawn by the heater, which is determined by multiplying the electric current by
the voltage.
In a certain experiment, cylindrical samples of diameter 5 cm and length
10 cm are used. The two thermocouples in each sample are placed 3 cm apart.
After initial transients, the electric heater is observed to draw 0.4 A at 110 V,
and both differential thermometers read a temperature difference of 15°C. De-
termine the thermal conductivity of the sample.
SOLUTION The thermal conductivity of a material is to be determined by en-
suring one-dimensional heat conduction, and by measuring temperatures when
steady operating conditions are reached.
Assumptions 1 Steady operating conditions exist since the temperature
readings do not change with time. 2 Heat losses through the lateral surfaces
of the apparatus are negligible since those surfaces are well insulated, and
thus the entire heat generated by the heater is conducted through the samples.
3 The apparatus possesses thermal symmetry.
Analysis The electrical power consumed by the resistance heater and con-
verted to heat is
W.
VI = (110V)(0.4A) = 44W
The rate of heat flow through each sample is
Q = \ W e = \ X (44 W) = 22 W
since only half of the heat generated will flow through each sample because of
symmetry. Reading the same temperature difference across the same distance
in each sample also confirms that the apparatus possesses thermal symmetry.
The heat transfer area is the area normal to the direction of heat flow, which is
the cross-sectional area of the cylinder in this case:
A = \ ttD 2 = \ tt(0.05 m) 2 = 0.00196 m 2
Noting that the temperature drops by 15 C C within 3 cm in the direction of heat
flow, the thermal conductivity of the sample is determined to be
Q =kA
AT
QL
(22 W)(0.03 m)
A AT (0.00196 m 2 )(15°C)
22.4 W/m • °C
Discussion Perhaps you are wondering if we really need to use two samples in
the apparatus, since the measurements on the second sample do not give any
additional information. It seems like we can replace the second sample by in-
sulation. Indeed, we do not need the second sample; however, it enables us to
verify the temperature measurements on the first sample and provides thermal
symmetry, which reduces experimental error.
EXAMPLE 1-7 Conversion between SI and English Units
An engineer who is working on the heat transfer analysis of a brick building in
English units needs the thermal conductivity of brick. But the only value he can
cen58933_ch01.qxd 9/10/2002 8:29 AM Page 25
find from his handbooks is 0.72 W/m • °C, which is in SI units. To make mat-
i ters worse, the engineer does not have a direct conversion factor between the
two unit systems for thermal conductivity. Can you help him out?
25
CHAPTER 1
SOLUTION The situation this engineer is facing is not unique, and most engi-
neers often find themselves in a similar position. A person must be very careful
during unit conversion not to fall into some common pitfalls and to avoid some
costly mistakes. Although unit conversion is a simple process, it requires utmost
care and careful reasoning.
The conversion factors for W and m are straightforward and are given in con-
version tables to be
1 W
lm
3.41214 Btu/h
3.2808 ft
But the conversion of C C into C F is not so simple, and it can be a source of er-
ror if one is not careful. Perhaps the first thought that comes to mind is to re-
place °C by (°F - 32V1.8 since 7"(°C) = [T(°F) - 32]/1.8. But this will be
wrong since the °C in the unit W/m • °C represents per °C change in tempera-
ture. Noting that 1°C change in temperature corresponds to 1.8°F, the proper
conversion factor to be used is
1°C = 1.8°F
Substituting, we get
1 w/m • ° c = S^ = °- 5778 Btu/h ■ ft ■ ° F
which is the desired conversion factor. Therefore, the thermal conductivity of
the brick in English units is
= 0.72 X (0.5778 Btu/h • ft • °F)
= 0.42 Btu/h • ft • °F
Discussion Note that the thermal conductivity value of a material in English
units is about half that in SI units (Fig. 1-30). Also note that we rounded the
result to two significant digits (the same number in the original value) since ex-
pressing the result in more significant digits (such as 0.4160 instead of 0.42)
would falsely imply a more accurate value than the original one.
k = 0.72 W/m-°C
= 0.42 Btu/h-ft-°F
FIGURE 1-30
The thermal conductivity value in
English units is obtained by multiplying
the value in SI units by 0.5778.
1-7 ■ CONVECTION
Convection is the mode of energy transfer between a solid surface and the
adjacent liquid or gas that is in motion, and it involves the combined effects of
conduction and fluid motion. The faster the fluid motion, the greater the
convection heat transfer. In the absence of any bulk fluid motion, heat trans-
fer between a solid surface and the adjacent fluid is by pure conduction. The
presence of bulk motion of the fluid enhances the heat transfer between the
solid surface and the fluid, but it also complicates the determination of heat
transfer rates.
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26
HEAT TRANSFER
Velocity
variation
of air
Temperature
variation
of air
Hot Block
FIGURE 1-31
Heat transfer from a hot
surface to air by convection.
Natural
convection
Air
\ Xhot egg) v. /
FIGURE 1-32
The cooling of a boiled egg
by forced and natural convection.
TABLE 1-5
Typical values of convection heat
transfer coefficient
Type of
convection
h, W/m 2
Free convection of
gases
Free convection of
liquids
Forced convection
of gases
Forced convection
of liquids
Boiling and
condensation
2-25
10-1000
25-250
50-20,000
2500-100,000
•Multiply by 0.176 to convert to Btu/h • ft 2 ■ °F.
Consider the cooling of a hot block by blowing cool air over its top surface
(Fig. 1-31). Energy is first transferred to the air layer adjacent to the block by
conduction. This energy is then carried away from the surface by convection,
that is, by the combined effects of conduction within the air that is due to ran-
dom motion of air molecules and the bulk or macroscopic motion of the air
that removes the heated air near the surface and replaces it by the cooler air.
Convection is called forced convection if the fluid is forced to flow over
the surface by external means such as a fan, pump, or the wind. In contrast,
convection is called natural (or free) convection if the fluid motion is caused
by buoyancy forces that are induced by density differences due to the varia-
tion of temperature in the fluid (Fig. 1-32). For example, in the absence of a
fan, heat transfer from the surface of the hot block in Fig. 1-31 will be by nat-
ural convection since any motion in the air in this case will be due to the rise
of the warmer (and thus lighter) air near the surface and the fall of the cooler
(and thus heavier) air to fill its place. Heat transfer between the block and the
surrounding air will be by conduction if the temperature difference between
the air and the block is not large enough to overcome the resistance of air to
movement and thus to initiate natural convection currents.
Heat transfer processes that involve change of phase of a fluid are also con-
sidered to be convection because of the fluid motion induced during the
process, such as the rise of the vapor bubbles during boiling or the fall of the
liquid droplets during condensation.
Despite the complexity of convection, the rate of convection heat transfer is
observed to be proportional to the temperature difference, and is conveniently
expressed by Newton's law of cooling as
6 c
hA s (T s - r„)
(W)
(1-24)
where h is the convection heat transfer coefficient in W/m 2 • °C or Btu/h • ft 2 ■ °F,
A s is the surface area through which convection heat transfer takes place, T s is
the surface temperature, and T m is the temperature of the fluid sufficiently far
from the surface. Note that at the surface, the fluid temperature equals the sur-
face temperature of the solid.
The convection heat transfer coefficient h is not a property of the fluid. It is
an experimentally determined parameter whose value depends on all the vari-
ables influencing convection such as the surface geometry, the nature of fluid
motion, the properties of the fluid, and the bulk fluid velocity. Typical values
of h are given in Table 1-5.
Some people do not consider convection to be a fundamental mechanism of
heat transfer since it is essentially heat conduction in the presence of fluid mo-
tion. But we still need to give this combined phenomenon a name, unless we
are willing to keep referring to it as "conduction with fluid motion." Thus, it
is practical to recognize convection as a separate heat transfer mechanism de-
spite the valid arguments to the contrary.
EXAMPLE 1-8 Measuring Convection Heat Transfer Coefficient
A 2-m-long, 0.3-cm-diameter electrical wire extends across a room at 15°C, as
shown in Fig. 1-33. Heat is generated in the wire as a result of resistance heat-
■ ing, and the surface temperature of the wire is measured to be 152°C in steady
cen58933_ch01.qxd 9/10/2002 8:29 AM Page 27
operation. Also, the voltage drop and electric current through the wire are mea-
sured to be 60 V and 1.5 A, respectively. Disregarding any heat transfer by
radiation, determine the convection heat transfer coefficient for heat transfer
between the outer surface of the wire and the air in the room.
SOLUTION The convection heat transfer coefficient for heat transfer from an
electrically heated wire to air is to be determined by measuring temperatures
when steady operating conditions are reached and the electric power consumed.
Assumptions 1 Steady operating conditions exist since the temperature read-
ings do not change with time. 2 Radiation heat transfer is negligible.
Analysis When steady operating conditions are reached, the rate of heat loss
from the wire will equal the rate of heat generation in the wire as a result of
resistance heating. That is,
Q = generated = W = (60 V)(1.5 A) = 90 W
The surface area of the wire is
A s = ttDL = tt(0.003 m)(2 m) = 0.01885 m 2
Newton's law of cooling for convection heat transfer is expressed as
Gco„v = hA s (T s - zy
Disregarding any heat transfer by radiation and thus assuming all the heat loss
from the wire to occur by convection, the convection heat transfer coefficient is
determined to be
h
Gc
90 W
A£T S - r„) (0.01885 m 2 )(152 - 15)°C
34.9 W/m 2
Discussion Note that the simple setup described above can be used to deter-
mine the average heat transfer coefficients from a variety of surfaces in air.
Also, heat transfer by radiation can be eliminated by keeping the surrounding
surfaces at the temperature of the wire.
27
CHAPTER 1
T = 15°C
1.5 A
152°C
■60 V-
FIGURE 1-33
Schematic for Example 1-8.
1-8 - RADIATION
Radiation is the energy emitted by matter in the form of electromagnetic
waves (or photons) as a result of the changes in the electronic configurations
of the atoms or molecules. Unlike conduction and convection, the transfer of
energy by radiation does not require the presence of an intervening medium.
In fact, energy transfer by radiation is fastest (at the speed of light) and it
suffers no attenuation in a vacuum. This is how the energy of the sun reaches
the earth.
In heat transfer studies we are interested in thermal radiation, which is the
form of radiation emitted by bodies because of their temperature. It differs
from other forms of electromagnetic radiation such as x-rays, gamma rays,
microwaves, radio waves, and television waves that are not related to temper-
ature. All bodies at a temperature above absolute zero emit thermal radiation.
Radiation is a volumetric phenomenon, and all solids, liquids, and gases
emit, absorb, or transmit radiation to varying degrees. However, radiation is
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28
HEAT TRANSFER
FIGURE 1-34
Blackbody radiation represents the
maximum amount of radiation that
can be emitted from a surface
at a specified temperature.
TABLE 1-6
Emissivities of some materials
at 300 K
Materia]
Emissivity
Aluminum foil
0.07
Anodized aluminum
0.82
Polished copper
0.03
Polished gold
0.03
Polished silver
0.02
Polished stainless steel
0.17
Black paint
0.98
White paint
0.90
White paper
0.92-0.97
Asphalt pavement
0.85-0.93
Red brick
0.93-0.96
Human skin
0.95
Wood
0.82-0.92
Soil
0.93-0.96
Water
0.96
Vegetation
0.92-0.96
-V*
*^ref ^ "incid
Q ,. = a 6 - A ,
*--abs ^incident
FIGURE 1-35
The absorption of radiation incident on
an opaque surface of absorptivity a.
usually considered to be a surface phenomenon for solids that are opaque to
thermal radiation such as metals, wood, and rocks since the radiation emitted
by the interior regions of such material can never reach the surface, and the
radiation incident on such bodies is usually absorbed within a few microns
from the surface.
The maximum rate of radiation that can be emitted from a surface at an ab-
solute temperature T s (in K or R) is given by the Stefan-Boltzmann law as
£?e
uAJt
(W)
(1-25)
where a = 5.67 X 1(T 8 W/m 2 • K 4 or 0.1714 X 1(T 8 Btu/h • ft 2 • R 4 is the
Stefan-Boltzmann constant. The idealized surface that emits radiation at this
maximum rate is called a blackbody, and the radiation emitted by a black-
body is called blackbody radiation (Fig. 1-34). The radiation emitted by all
real surfaces is less than the radiation emitted by a blackbody at the same tem-
perature, and is expressed as
C?e
saATj
(W)
(1-26)
where e is the emissivity of the surface. The property emissivity, whose value
is in the range ^ e < 1, is a measure of how closely a surface approximates
a blackbody for which e = 1 . The emissivities of some surfaces are given in
Table 1-6.
Another important radiation property of a surface is its absorptivity a,
which is the fraction of the radiation energy incident on a surface that is ab-
sorbed by the surface. Like emissivity, its value is in the range ^ a < 1.
A blackbody absorbs the entire radiation incident on it. That is, a blackbody is
a perfect absorber (a = 1) as it is a perfect emitter.
In general, both e and a of a surface depend on the temperature and the
wavelength of the radiation. Kirchhoff 's law of radiation states that the emis-
sivity and the absorptivity of a surface at a given temperature and wavelength
are equal. In many practical applications, the surface temperature and the
temperature of the source of incident radiation are of the same order of mag-
nitude, and the average absorptivity of a surface is taken to be equal to its av-
erage emissivity. The rate at which a surface absorbs radiation is determined
from (Fig. 1-35)
Q,
a 2incid
(W)
(1-27)
where 2 incident is the rate at which radiation is incident on the surface and a is
the absorptivity of the surface. For opaque (nontransparent) surfaces, the
portion of incident radiation not absorbed by the surface is reflected back.
The difference between the rates of radiation emitted by the surface and the
radiation absorbed is the net radiation heat transfer. If the rate of radiation ab-
sorption is greater than the rate of radiation emission, the surface is said to be
gaining energy by radiation. Otherwise, the surface is said to be losing energy
by radiation. In general, the determination of the net rate of heat transfer by ra-
diation between two surfaces is a complicated matter since it depends on the
properties of the surfaces, their orientation relative to each other, and the in-
teraction of the medium between the surfaces with radiation.
cen58933_ch01.qxd 9/10/2002 8:29 AM Page 29
When a surface of emissivity e and surface area A s at an absolute tempera-
ture T s is completely enclosed by a much larger (or black) surface at absolute
temperature T sun separated by a gas (such as air) that does not intervene with
radiation, the net rate of radiation heat transfer between these two surfaces is
given by (Fig. 1-36)
Q,
BaA,(T}-T*^
(W)
(1-28)
In this special case, the emissivity and the surface area of the surrounding sur-
face do not have any effect on the net radiation heat transfer.
Radiation heat transfer to or from a surface surrounded by a gas such as air
occurs parallel to conduction (or convection, if there is bulk gas motion) be-
tween the surface and the gas. Thus the total heat transfer is determined by
adding the contributions of both heat transfer mechanisms. For simplicity and
convenience, this is often done by defining a combined heat transfer co-
efficient ^combined that includes the effects of both convection and radiation.
Then the total heat transfer rate to or from a surface by convection and radia-
tion is expressed as
fit
K
iA s (1 s I a,)
(W)
(1-29)
Note that the combined heat transfer coefficient is essentially a convection
heat transfer coefficient modified to include the effects of radiation.
Radiation is usually significant relative to conduction or natural convection,
but negligible relative to forced convection. Thus radiation in forced convec-
tion applications is usually disregarded, especially when the surfaces involved
have low emissivities and low to moderate temperatures.
29
CHAPTER 1
Q mi = £GA S (T A S -Ti m )
FIGURE 1-36
Radiation heat transfer between a
surface and the surfaces surrounding it.
EXAMPLE 1-9 Radiation Effect on Thermal Comfort
It is a common experience to feel "chilly" in winter and "warm" in summer in
our homes even when the thermostat setting is kept the same. This is due to the
so called "radiation effect" resulting from radiation heat exchange between our
bodies and the surrounding surfaces of the walls and the ceiling.
Consider a person standing in a room maintained at 22°C at all times. The
inner surfaces of the walls, floors, and the ceiling of the house are observed to
be at an average temperature of 10 C C in winter and 25°C in summer. Determine
the rate of radiation heat transfer between this person and the surrounding sur-
faces if the exposed surface area and the average outer surface temperature of
the person are 1.4 m 2 and 30°C, respectively (Fig. 1-37).
SOLUTION The rates of radiation heat transfer between a person and the sur-
rounding surfaces at specified temperatures are to be determined in summer
and winter.
Assumptions 1 Steady operating conditions exist. 2 Heat transfer by convection
is not considered. 3 The person is completely surrounded by the interior sur-
faces of the room. 4 The surrounding surfaces are at a uniform temperature.
Properties The emissivity of a person is e = 0.95 (Table 1-6).
Analysis The net rates of radiation heat transfer from the body to the sur-
rounding walls, ceiling, and floor in winter and summer are
FIGURE 1-37
Schematic for Example 1-9.
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30
HEAT TRANSFER
rad, winter oun j \ l s -* surr, winter/
= (0.95)(5.67 X l(T 8 W/m 2 • K 4 )(1.4nr)
X [(30 + 273) 4 - (10 + 273) 4 ] K 4
= 152 W
and
= (0.95)(5.67 X 10~ 8 W/m 2 • K 4 )(1.4 m 2 )
X [(30 + 273) 4 - (25 + 273) 4 ] K 4
= 40.9 W
Discussion Note that we must use absolute temperatures in radiation calcula-
tions. Also note that the rate of heat loss from the person by radiation is almost
four times as large in winter than it is in summer, which explains the "chill" we
feel in winter even if the thermostat setting is kept the same.
OPAQUE
SOLID
Conduction
1 mode
Conduction or
convection
2 modes
VACUUM
Radiation
1 mode
FIGURE 1-38
Although there are three mechanisms of
heat transfer, a medium may involve
only two of them simultaneously.
1-9 - SIMULTANEOUS HEAT TRANSFER
MECHANISMS
We mentioned that there are three mechanisms of heat transfer, but not all
three can exist simultaneously in a medium. For example, heat transfer is
only by conduction in opaque solids, but by conduction and radiation in
semitransparent solids. Thus, a solid may involve conduction and radiation
but not convection. However, a solid may involve heat transfer by convection
and/or radiation on its surfaces exposed to a fluid or other surfaces. For
example, the outer surfaces of a cold piece of rock will warm up in a warmer
environment as a result of heat gain by convection (from the air) and radiation
(from the sun or the warmer surrounding surfaces). But the inner parts of the
rock will warm up as this heat is transferred to the inner region of the rock by
conduction.
Heat transfer is by conduction and possibly by radiation in a still fluid (no
bulk fluid motion) and by convection and radiation in a flowing fluid. In the
absence of radiation, heat transfer through a fluid is either by conduction or
convection, depending on the presence of any bulk fluid motion. Convection
can be viewed as combined conduction and fluid motion, and conduction in a
fluid can be viewed as a special case of convection in the absence of any fluid
motion (Fig. 1-38).
Thus, when we deal with heat transfer through & fluid, we have either con-
duction or convection, but not both. Also, gases are practically transparent to
radiation, except that some gases are known to absorb radiation strongly at
certain wavelengths. Ozone, for example, strongly absorbs ultraviolet radia-
tion. But in most cases, a gas between two solid surfaces does not interfere
with radiation and acts effectively as a vacuum. Liquids, on the other hand,
are usually strong absorbers of radiation.
Finally, heat transfer through a vacuum is by radiation only since conduc-
tion or convection requires the presence of a material medium.
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31
CHAPTER 1
EXAMPLE 1-10 Heat Loss from a Person
Consider a person standing in a breezy room at 20°C. Determine the total rate
of heat transfer from this person if the exposed surface area and the average
outer surface temperature of the person are 1.6 m 2 and 29°C, respectively, and
the convection heat transfer coefficient is 6 W/m 2 • °C (Fig. 1-39).
SOLUTION The total rate of heat transfer from a person by both convection
and radiation to the surrounding air and surfaces at specified temperatures is to
be determined.
Assumptions 1 Steady operating conditions exist. 2 The person is completely
surrounded by the interior surfaces of the room. 3 The surrounding surfaces are
at the same temperature as the air in the room. 4 Heat conduction to the floor
through the feet is negligible.
Properties The emissivity of a person is e = 0.95 (Table 1-6).
Analysis The heat transfer between the person and the air in the room will be
by convection (instead of conduction) since it is conceivable that the air in the
vicinity of the skin or clothing will warm up and rise as a result of heat transfer
from the body, initiating natural convection currents. It appears that the exper-
imentally determined value for the rate of convection heat transfer in this case
is 6 W per unit surface area (m 2 ) per unit temperature difference (in K or °C)
between the person and the air away from the person. Thus, the rate of convec-
tion heat transfer from the person to the air in the room is
Gconv = hA s (T s - TJ
= (6 W/m 2 • °C)(1.6 m 2 )(29 - 20)°C
= 86.4 W
The person will also lose heat by radiation to the surrounding wall surfaces.
We take the temperature of the surfaces of the walls, ceiling, and floor to be
equal to the air temperature in this case for simplicity, but we recognize that
this does not need to be the case. These surfaces may be at a higher or lower
temperature than the average temperature of the room air, depending on the
outdoor conditions and the structure of the walls. Considering that air does not
intervene with radiation and the person is completely enclosed by the sur-
rounding surfaces, the net rate of radiation heat transfer from the person to the
surrounding walls, ceiling, and floor is
Srad = evA s (Tf- r s 4 urr )
= (0.95)(5.67 X 10-
X [(29 + 273) 4 -
= 81.7W
B W/m 2 -K 4 )(1.6m 2 )
(20 + 273) 4 ] K 4
Note that we must use absolute temperatures in radiation calculations. Also
note that we used the emissivity value for the skin and clothing at room tem-
perature since the emissivity is not expected to change significantly at a slightly
higher temperature.
Then the rate of total heat transfer from the body is determined by adding
these two quantities:
fit
x£ conv x£r
(86.4 + 81.7) W = 168.1 W
Q A
FIGURE 1-39
Heat transfer from the person
described in Example 1-10.
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32
HEAT TRANSFER
Discussion The heat transfer would be much higher if the person were not
dressed since the exposed surface temperature would be higher. Thus, an im-
portant function of the clothes is to serve as a barrier against heat transfer.
In these calculations, heat transfer through the feet to the floor by conduc-
tion, which is usually very small, is neglected. Heat transfer from the skin by
perspiration, which is the dominant mode of heat transfer in hot environments,
is not considered here.
T l = 300 K
-L=\ cm
~e = V
FIGURE 1-40
Schematic for Example 1—11.
r 2 = 200K 'EXAMPLE 1-11 Heat Transfer between Two Isothermal Plates
Consider steady heat transfer between two large parallel plates at constant
I temperatures of 7"! = 300 K and 7~ z = 200 K that are L = 1 cm apart, as shown
I in Fig. 1-40. Assuming the surfaces to be black (emissivity e = 1), determine
I the rate of heat transfer between the plates per unit surface area assuming the
gap between the plates is (a) filled with atmospheric air, (£>) evacuated, (c) filled
with urethane insulation, and (d) filled with superinsulation that has an appar-
ent thermal conductivity of 0.00002 W/m • °C.
SOLUTION The total rate of heat transfer between two large parallel plates at
specified temperatures is to be determined for four different cases.
Assumptions 1 Steady operating conditions exist. 2 There are no natural con-
vection currents in the air between the plates. 3 The surfaces are black and
thus e = 1.
Properties The thermal conductivity at the average temperature of 250 K is
k = 0.0219 W/m • °C for air (Table A-l 1), 0.026 W/m ■ °C for urethane insula-
tion (Table A-6), and 0.00002 W/m ■ °C for the superinsulation.
Analysis (a) The rates of conduction and radiation heat transfer between the
plates through the air layer are
6c
M
T 2
(0.0219 W/m -°C)(lm 2 )
(300 - 200)°C
0.01m
219W
and
!2 rad = svMTf - 7\ 4 )
= (1)(5.67 X 10- 8 W/m 2 • K 4 )(l m 2 )[(300 K) 4
(200 K) 4 ] = 368 W
Therefore,
Qu
fico„d + e ra d = 219 + 368 = 587W
The heat transfer rate in reality will be higher because of the natural convection
currents that are likely to occur in the air space between the plates.
(b) When the air space between the plates is evacuated, there will be no con-
duction or convection, and the only heat transfer between the plates will be by
radiation. Therefore,
G t
G,
368 W
(c) An opaque solid material placed between two plates blocks direct radiation
heat transfer between the plates. Also, the thermal conductivity of an insulating
material accounts for the radiation heat transfer that may be occurring through
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33
CHAPTER 1
300 K
200 K 300 K
200 K 300 K
2 = 587W
1 cm
Q = 368 W
1 cm
200 K 300 K
I
!j
Q = 260 W
1 cm
200 K
2 = 0.2W
4
1 cm
(a) Air space (b) Vacuum (c) Insulation (d) Superinsulation
FIGURE 1-41
Different ways of reducing heat transfer between two isothermal plates, and their effectiveness.
the voids in the insulating material. The rate of heat transfer through the ure-
thane insulation is
Qu
Gcond = kA
. (300 - 200)°C
(0.026 W/m • °C)(1 m 2 )- — — : — = 260 W
0.01 m
Note that heat transfer through the urethane material is less than the heat
transfer through the air determined in (a), although the thermal conductivity of
the insulation is higher than that of air. This is because the insulation blocks
the radiation whereas air transmits it.
(d) The layers of the superinsulation prevent any direct radiation heat transfer
between the plates. However, radiation heat transfer between the sheets of su-
perinsulation does occur, and the apparent thermal conductivity of the super-
insulation accounts for this effect. Therefore,
e„
kA
T 2
(0.00002 W/m •°C)(lm 2 )
(300 - 200)°C
0.01m
0.2 W
which is yi_ of the heat transfer through the vacuum. The results of this ex-
ample are summarized in Fig. 1-41 to put them into perspective.
Discussion This example demonstrates the effectiveness of superinsulations,
which are discussed in the next chapter, and explains why they are the insula-
tion of choice in critical applications despite their high cost.
EXAMPLE 1-12 Heat Transfer in Conventional
and Microwave Ovens
The fast and efficient cooking of microwave ovens made them one of the es-
sential appliances in modern kitchens (Fig. 1-42). Discuss the heat transfer
mechanisms associated with the cooking of a chicken in microwave and con-
ventional ovens, and explain why cooking in a microwave oven is more efficient.
SOLUTION Food is cooked in a microwave oven by absorbing the electromag-
netic radiation energy generated by the microwave tube, called the magnetron.
®
}
FIGURE 1-42
A chicken being cooked in a
microwave oven (Example 1-12).
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HEAT TRANSFER
The radiation emitted by the magnetron is not thermal radiation, since its emis-
sion is not due to the temperature of the magnetron; rather, it is due to the
conversion of electrical energy into electromagnetic radiation at a specified
wavelength. The wavelength of the microwave radiation is such that it is re-
flected by metal surfaces; transmitted by the cookware made of glass, ceramic,
or plastic; and absorbed and converted to internal energy by food (especially the
water, sugar, and fat) molecules.
In a microwave oven, the radiation that strikes the chicken is absorbed by
the skin of the chicken and the outer parts. As a result, the temperature of the
chicken at and near the skin rises. Heat is then conducted toward the inner
parts of the chicken from its outer parts. Of course, some of the heat absorbed
by the outer surface of the chicken is lost to the air in the oven by convection.
In a conventional oven, the air in the oven is first heated to the desired tem-
perature by the electric or gas heating element. This preheating may take sev-
eral minutes. The heat is then transferred from the air to the skin of the chicken
by natural convection in most ovens or by forced convection in the newer con-
vection ovens that utilize a fan. The air motion in convection ovens increases
the convection heat transfer coefficient and thus decreases the cooking time.
Heat is then conducted toward the inner parts of the chicken from its outer
parts as in microwave ovens.
Microwave ovens replace the slow convection heat transfer process in con-
ventional ovens by the instantaneous radiation heat transfer. As a result, micro-
wave ovens transfer energy to the food at full capacity the moment they are
turned on, and thus they cook faster while consuming less energy.
a = 0.6
25°C
FIGURE 1-43
Schematic for Example 1-13.
I
2 EXAMPLE 1-13 Heating of a Plate by Solar Energy
A thin metal plate is insulated on the back and exposed to solar radiation at the
I front surface (Fig. 1-43). The exposed surface of the plate has an absorptivity
■ of 0.6 for solar radiation. If solar radiation is incident on the plate at a rate of
■ 700 W/m 2 and the surrounding air temperature is 25 C C, determine the surface
temperature of the plate when the heat loss by convection and radiation equals
the solar energy absorbed by the plate. Assume the combined convection and
radiation heat transfer coefficient to be 50 W/m 2 ■ C C.
SOLUTION The back side of the thin metal plate is insulated and the front
side is exposed to solar radiation. The surface temperature of the plate is to be
determined when it stabilizes.
Assumptions 1 Steady operating conditions exist. 2 Heat transfer through the
insulated side of the plate is negligible. 3 The heat transfer coefficient remains
constant.
Properties The solar absorptivity of the plate is given to be a = 0.6.
Analysis The absorptivity of the plate is 0.6, and thus 60 percent of the solar
radiation incident on the plate will be absorbed continuously. As a result, the
temperature of the plate will rise, and the temperature difference between the
plate and the surroundings will increase. This increasing temperature difference
will cause the rate of heat loss from the plate to the surroundings to increase.
At some point, the rate of heat loss from the plate will equal the rate of solar
cen58933_ch01.qxd 9/10/2002 8:30 AM Page 35
energy absorbed, and the temperature of the plate will no longer change. The
temperature of the plate when steady operation is established is deter-
mined from
J gained
OF CLA S q incident, solar ^combined ™s V-* s * °°J
Solving for 7" s and substituting, the plate surface temperature is determined
to be
T„ + a-
^i incide
K,
25°C +
0.6 X (700 W/m 2 )
50 W/m 2 ■ °C
33.4°C
Discussion Note that the heat losses will prevent the plate temperature from
rising above 33.4°C. Also, the combined heat transfer coefficient accounts for
the effects of both convection and radiation, and thus it is very convenient
to use in heat transfer calculations when its value is known with reasonable
accuracy.
35
CHAPTER 1
SOLUTION
1-10 - PROBLEM-SOLVING TECHNIQUE
The first step in learning any science is to grasp the fundamentals, and to gain
a sound knowledge of it. The next step is to master the fundamentals by
putting this knowledge to test. This is done by solving significant real-world
problems. Solving such problems, especially complicated ones, requires a sys-
tematic approach. By using a step-by-step approach, an engineer can reduce
the solution of a complicated problem into the solution of a series of simple
problems (Fig. 1-44). When solving a problem, we recommend that you use
the following steps zealously as applicable. This will help you avoid some of
the common pitfalls associated with problem solving.
Step 1: Problem Statement
In your own words, briefly state the problem, the key information given, and
the quantities to be found. This is to make sure that you understand the prob-
lem and the objectives before you attempt to solve the problem.
Step 2: Schematic
Draw a realistic sketch of the physical system involved, and list the relevant
information on the figure. The sketch does not have to be something elaborate,
but it should resemble the actual system and show the key features. Indicate
any energy and mass interactions with the surroundings. Listing the given in-
formation on the sketch helps one to see the entire problem at once. Also,
check for properties that remain constant during a process (such as tempera-
ture during an isothermal process), and indicate them on the sketch.
Step 3: Assumptions
State any appropriate assumptions made to simplify the problem to make it
possible to obtain a solution. Justify the questionable assumptions. Assume
reasonable values for missing quantities that are necessary. For example, in
the absence of specific data for atmospheric pressure, it can be taken to be
<&
4?
<P
%
£
PROBLEM
FIGURE 1-44
A step-by-step approach can greatly
simplify problem solving.
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36
HEAT TRANSFER
FIGURE 1-45
The assumptions made while solving
an engineering problem must be
reasonable and justifiable.
FIGURE 1-46
The results obtained from
an engineering analysis must
be checked for reasonableness.
1 atm. However, it should be noted in the analysis that the atmospheric pres-
sure decreases with increasing elevation. For example, it drops to 0.83 atm in
Denver (elevation 1610 m) (Fig. 1-45).
Step 4: Physical Laws
Apply all the relevant basic physical laws and principles (such as the conser-
vation of energy), and reduce them to their simplest form by utilizing the as-
sumptions made. However, the region to which a physical law is applied must
be clearly identified first. For example, the heating or cooling of a canned
drink is usually analyzed by applying the conservation of energy principle to
the entire can.
Step 5: Properties
Determine the unknown properties at known states necessary to solve the
problem from property relations or tables. List the properties separately, and
indicate their source, if applicable.
Step 6: Calculations
Substitute the known quantities into the simplified relations and perform the
calculations to determine the unknowns. Pay particular attention to the units
and unit cancellations, and remember that a dimensional quantity without a
unit is meaningless. Also, don't give a false implication of high accuracy by
copying all the digits from the screen of the calculator — round the results to
an appropriate number of significant digits.
Step 7: Reasoning, Verification, and Discussion
Check to make sure that the results obtained are reasonable and intuitive, and
verify the validity of the questionable assumptions. Repeat the calculations
that resulted in unreasonable values. For example, insulating a water heater
that uses $80 worth of natural gas a year cannot result in savings of $200 a
year (Fig. 1-46).
Also, point out the significance of the results, and discuss their implications.
State the conclusions that can be drawn from the results, and any recommen-
dations that can be made from them. Emphasize the limitations under which
the results are applicable, and caution against any possible misunderstandings
and using the results in situations where the underlying assumptions do not
apply. For example, if you determined that wrapping a water heater with a
$20 insulation jacket will reduce the energy cost by $30 a year, indicate that
the insulation will pay for itself from the energy it saves in less than a year.
However, also indicate that the analysis does not consider labor costs, and that
this will be the case if you install the insulation yourself.
Keep in mind that you present the solutions to your instructors, and any en-
gineering analysis presented to others is a form of communication. Therefore
neatness, organization, completeness, and visual appearance are of utmost im-
portance for maximum effectiveness. Besides, neatness also serves as a great
checking tool since it is very easy to spot errors and inconsistencies in a neat
work. Carelessness and skipping steps to save time often ends up costing more
time and unnecessary anxiety.
cen58933_ch01.qxd 9/10/2002 8:30 AM Page 37
The approach just described is used in the solved example problems with-
out explicitly stating each step, as well as in the Solutions Manual of this text.
For some problems, some of the steps may not be applicable or necessary.
However, we cannot overemphasize the importance of a logical and orderly
approach to problem solving. Most difficulties encountered while solving a
problem are not due to a lack of knowledge; rather, they are due to a lack of
coordination. You are strongly encouraged to follow these steps in problem
solving until you develop your own approach that works best for you.
37
CHAPTER 1
A Remark on Significant Digits
In engineering calculations, the information given is not known to more than
a certain number of significant digits, usually three digits. Consequently, the
results obtained cannot possibly be accurate to more significant digits. Re-
porting results in more significant digits implies greater accuracy than exists,
and it should be avoided.
For example, consider a 3.75-L container filled with gasoline whose density
is 0.845 kg/L, and try to determine its mass. Probably the first thought that
comes to your mind is to multiply the volume and density to obtain 3.16875
kg for the mass, which falsely implies that the mass determined is accurate to
six significant digits. In reality, however, the mass cannot be more accurate
than three significant digits since both the volume and the density are accurate
to three significant digits only. Therefore, the result should be rounded to three
significant digits, and the mass should be reported to be 3.17 kg instead of
what appears in the screen of the calculator. The result 3.16875 kg would be
correct only if the volume and density were given to be 3.75000 L and
0.845000 kg/L, respectively. The value 3.75 L implies that we are fairly con-
fident that the volume is accurate within ±0.01 L, and it cannot be 3.74 or
3.76 L. However, the volume can be 3.746, 3.750, 3.753, etc., since they all
round to 3.75 L (Fig. 1-47). It is more appropriate to retain all the digits dur-
ing intermediate calculations, and to do the rounding in the final step since
this is what a computer will normally do.
When solving problems, we will assume the given information to be accu-
rate to at least three significant digits. Therefore, if the length of a pipe is
given to be 40 m, we will assume it to be 40.0 m in order to justify using three
significant digits in the final results. You should also keep in mind that all ex-
perimentally determined values are subject to measurement errors, and such
errors will reflect in the results obtained. For example, if the density of a sub-
stance has an uncertainty of 2 percent, then the mass determined using this
density value will also have an uncertainty of 2 percent.
You should also be aware that we sometimes knowingly introduce small er-
rors in order to avoid the trouble of searching for more accurate data. For ex-
ample, when dealing with liquid water, we just use the value of 1000 kg/m 3
for density, which is the density value of pure water at 0°C. Using this value
at 75°C will result in an error of 2.5 percent since the density at this tempera-
ture is 975 kg/m 3 . The minerals and impurities in the water will introduce ad-
ditional error. This being the case, you should have no reservation in rounding
the final results to a reasonable number of significant digits. Besides, having
a few percent uncertainty in the results of engineering analysis is usually the
norm, not the exception.
FIGURE 1-47
A result with more significant digits than
that of given data falsely implies
more accuracy.
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38
HEAT TRANSFER
FIGURE 1-48
An excellent word-processing
program does not make a person a good
writer; it simply makes a good writer
a better and more efficient writer.
Engineering Software Packages
Perhaps you are wondering why we are about to undertake a painstaking study
of the fundamentals of heat transfer. After all, almost all such problems we are
likely to encounter in practice can be solved using one of several sophisticated
software packages readily available in the market today. These software pack-
ages not only give the desired numerical results, but also supply the outputs in
colorful graphical form for impressive presentations. It is unthinkable to prac-
tice engineering today without using some of these packages. This tremen-
dous computing power available to us at the touch of a button is both a
blessing and a curse. It certainly enables engineers to solve problems easily
and quickly, but it also opens the door for abuses and misinformation. In the
hands of poorly educated people, these software packages are as dangerous as
sophisticated powerful weapons in the hands of poorly trained soldiers.
Thinking that a person who can use the engineering software packages
without proper training on fundamentals can practice engineering is like
thinking that a person who can use a wrench can work as a car mechanic. If it
were true that the engineering students do not need all these fundamental
courses they are taking because practically everything can be done by com-
puters quickly and easily, then it would also be true that the employers would
no longer need high-salaried engineers since any person who knows how to
use a word-processing program can also learn how to use those software pack-
ages. However, the statistics show that the need for engineers is on the rise,
not on the decline, despite the availability of these powerful packages.
We should always remember that all the computing power and the engi-
neering software packages available today are just tools, and tools have mean-
ing only in the hands of masters. Having the best word-processing program
does not make a person a good writer, but it certainly makes the job of a good
writer much easier and makes the writer more productive (Fig. 1-48). Hand
calculators did not eliminate the need to teach our children how to add or sub-
tract, and the sophisticated medical software packages did not take the place
of medical school training. Neither will engineering software packages re-
place the traditional engineering education. They will simply cause a shift in
emphasis in the courses from mathematics to physics. That is, more time will
be spent in the classroom discussing the physical aspects of the problems in
greater detail, and less time on the mechanics of solution procedures.
All these marvelous and powerful tools available today put an extra burden
on today's engineers. They must still have a thorough understanding of the
fundamentals, develop a "feel" of the physical phenomena, be able to put the
data into proper perspective, and make sound engineering judgments, just like
their predecessors. However, they must do it much better, and much faster, us-
ing more realistic models because of the powerful tools available today. The
engineers in the past had to rely on hand calculations, slide rules, and later
hand calculators and computers. Today they rely on software packages. The
easy access to such power and the possibility of a simple misunderstanding or
misinterpretation causing great damage make it more important today than
ever to have a solid training in the fundamentals of engineering. In this text we
make an extra effort to put the emphasis on developing an intuitive and phys-
ical understanding of natural phenomena instead of on the mathematical de-
tails of solution procedures.
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39
CHAPTER 1
Engineering Equation Solver (EES)
EES is a program that solves systems of linear or nonlinear algebraic or dif-
ferential equations numerically. It has a large library of built-in thermody-
namic property functions as well as mathematical functions, and allows the
user to supply additional property data. Unlike some software packages, EES
does not solve thermodynamic problems; it only solves the equations supplied
by the user. Therefore, the user must understand the problem and formulate it
by applying any relevant physical laws and relations. EES saves the user con-
siderable time and effort by simply solving the resulting mathematical equa-
tions. This makes it possible to attempt significant engineering problems not
suitable for hand calculations, and to conduct parametric studies quickly and
conveniently. EES is a very powerful yet intuitive program that is very easy to
use, as shown in the examples below. The use and capabilities of EES are ex-
plained in Appendix 3.
Heat Transfer Tools (HTT)
One software package specifically designed to help bridge the gap between
the textbook fundamentals and these powerful software packages is Heat
Transfer Tools, which may be ordered "bundled" with this text. The software
included in that package was developed for instructional use only and thus is
applicable only to fundamental problems in heat transfer. While it does not
have the power and functionality of the professional, commercial packages,
HTT uses research-grade numerical algorithms behind the scenes and modern
graphical user interfaces. Each module is custom designed and applicable to a
single, fundamental topic in heat transfer to ensure that almost all time at the
computer is spent learning heat transfer. Nomenclature and all inputs and
outputs are consistent with those used in this and most other textbooks in
the field. In addition, with the capability of testing parameters so readily
available, one can quickly gain a physical feel for the effects of all the non-
dimensional numbers that arise in heat transfer.
EXAMPLE 1-14 Solving a System of Equations with EES
The difference of two numbers is 4, and the sum of the squares of these
two numbers is equal to the sum of the numbers plus 20. Determine these two
numbers.
SOLUTION Relations are given for the difference and the sum of the squares
of two numbers. They are to be determined.
Analysis We start the EES program by double-clicking on its icon, open a new
file, and type the following on the blank screen that appears:
x-y=4
x A 2+y A 2=x+y+20
which is an exact mathematical expression of the problem statement with
x and y denoting the unknown numbers. The solution to this system of two
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HEAT TRANSFER
nonlinear equations with two unknowns is obtained by a single click on the
"calculator" symbol on the taskbar. It gives
x=5 and y=l
Discussion Note that all we did is formulate the problem as we would on pa-
per; EES took care of all the mathematical details of solution. Also note that
equations can be linear or nonlinear, and they can be entered in any order with
unknowns on either side. Friendly equation solvers such as EES allow the user
to concentrate on the physics of the problem without worrying about the mathe-
matical complexities associated with the solution of the resulting system of
equations.
Throughout the text, problems that are unsuitable for hand calculations and
are intended to be solved using EES are indicated by a computer icon.
TOPIC OF SPECIAL INTEREST
FIGURE 1-49
Most animals come into this world with
built-in insulation, but human beings
come with a delicate skin.
Thermal Comfort
Unlike animals such as a fox or a bear that are born with built-in furs, hu-
man beings come into this world with little protection against the harsh en-
vironmental conditions (Fig. 1^9). Therefore, we can claim that the search
for thermal comfort dates back to the beginning of human history. It is be-
lieved that early human beings lived in caves that provided shelter as well
as protection from extreme thermal conditions. Probably the first form of
heating system used was open fire, followed by fire in dwellings through
the use of a chimney to vent out the combustion gases. The concept of cen-
tral heating dates back to the times of the Romans, who heated homes by
utilizing double-floor construction techniques and passing the fire's fumes
through the opening between the two floor layers. The Romans were also
the first to use transparent windows made of mica or glass to keep the wind
and rain out while letting the light in. Wood and coal were the primary en-
ergy sources for heating, and oil and candles were used for lighting. The ru-
ins of south-facing houses indicate that the value of solar heating was
recognized early in the history.
The term air-conditioning is usually used in a restricted sense to imply
cooling, but in its broad sense it means to condition the air to the desired
level by heating, cooling, humidifying, dehumidifying, cleaning, and de-
odorizing. The purpose of the air-conditioning system of a building is to
provide complete thermal comfort for its occupants. Therefore, we need to
understand the thermal aspects of the human body in order to design an ef-
fective air-conditioning system.
The building blocks of living organisms are cells, which resemble minia-
ture factories performing various functions necessary for the survival of
organisms. The human body contains about 100 trillion cells with an aver-
age diameter of 0.01 mm. In a typical cell, thousands of chemical reactions
"This section can be skipped without a loss in continuity.
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CHAPTER 1
occur every second during which some molecules are broken down and en-
ergy is released and some new molecules are formed. The high level of
chemical activity in the cells that maintain the human body temperature at
a temperature of 37.0°C (98.6°F) while performing the necessary bodily
functions is called the metabolism. In simple terms, metabolism refers to
the burning of foods such as carbohydrates, fat, and protein. The metabo-
lizable energy content of foods is usually expressed by nutritionists in
terms of the capitalized Calorie. One Calorie is equivalent to 1 Cal = 1
kcal = 4.1868 kJ.
The rate of metabolism at the resting state is called the basal metabolic
rate, which is the rate of metabolism required to keep a body performing
the necessary bodily functions such as breathing and blood circulation at
zero external activity level. The metabolic rate can also be interpreted as
the energy consumption rate for a body. For an average man (30 years old,
70 kg, 1.73 m high, 1.8 m 2 surface area), the basal metabolic rate is 84 W.
That is, the body is converting chemical energy of the food (or of the body
fat if the person had not eaten) into heat at a rate of 84 J/s, which is then
dissipated to the surroundings. The metabolic rate increases with the level
of activity, and it may exceed 10 times the basal metabolic rate when some-
one is doing strenuous exercise. That is, two people doing heavy exercising
in a room may be supplying more energy to the room than a 1-kW resis-
tance heater (Fig. 1-50). An average man generates heat at a rate of 108 W
while reading, writing, typing, or listening to a lecture in a classroom in a
seated position. The maximum metabolic rate of an average man is 1250 W
at age 20 and 730 at age 70. The corresponding rates for women are about
30 percent lower. Maximum metabolic rates of trained athletes can exceed
2000 W.
Metabolic rates during various activities are given in Table 1-7 per unit
body surface area. The surface area of a nude body was given by D.
DuBois in 1916 as
FIGURE1-50
Two fast-dancing people supply
more heat to a room than a
1-kW resistance heater.
0.202m - 425 h
D.425 1,0.725
(m 2 )
(1-30)
where m is the mass of the body in kg and h is the height in m. Clothing in-
creases the exposed surface area of a person by up to about 50 percent. The
metabolic rates given in the table are sufficiently accurate for most pur-
poses, but there is considerable uncertainty at high activity levels. More ac-
curate values can be determined by measuring the rate of respiratory
oxygen consumption, which ranges from about 0.25 L/min for an average
resting man to more than 2 L/min during extremely heavy work. The entire
energy released during metabolism can be assumed to be released as heat
(in sensible or latent forms) since the external mechanical work done by the
muscles is very small. Besides, the work done during most activities such
as walking or riding an exercise bicycle is eventually converted to heat
through friction.
The comfort of the human body depends primarily on three environmen-
tal factors: the temperature, relative humidity, and air motion. The temper-
ature of the environment is the single most important index of comfort.
Extensive research is done on human subjects to determine the "thermal
comfort zone" and to identify the conditions under which the body feels
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HEAT TRANSFER
TABLE 1-7
Metabolic rates during various
activities (from ASHRAE
Handbook of Fundamentals,
Ret. 1, Chap. 8, Table 4).
Metabolic
rate*
Activity
W/m 2
Resting:
Sleeping
40
Reclining
45
Seated, quiet
60
Standing, relaxed
70
Walking (on the level):
2 mph (0.89 m/s)
115
3 mph (1.34 m/s)
150
4 mph (1.79 m/s)
220
Office Activities:
Reading, seated
55
Writing
60
Typing
65
Filing, seated
70
Filing, standing
80
Walking about
100
Lifting/packing
120
Driving/Flying:
Car
60-115
Aircraft, routine
70
Heavy vehicle
185
Miscellaneous Occupational
Activities:
Cooking
95-115
Cleaning house
115-140
Machine work:
Light
115-140
Heavy
235
Handling 50-kg bags
235
Pick and shovel work
235-280
Miscellaneous Leisure
Activities:
Dancing, social
140-255
Calisthenics/exercise
175-235
Tennis, singles
210-270
Basketball
290-440
Wrestling, competitive
410-505
*M ultiply by 1.8 m 2 to obtain metabolic rates for
an average man. Multiply by 0.3171 to convert
to Btu/h • ft 2 .
comfortable in an environment. It has been observed that most normally
clothed people resting or doing light work feel comfortable in the operative
temperature (roughly, the average temperature of air and surrounding sur-
faces) range of 23°C to 27°C or 73°C to 80°F (Fig. 1-51). For unclothed
people, this range is 29°C to 31°C. Relative humidity also has a con-
siderable effect on comfort since it is a measure of air's ability to absorb
moisture and thus it affects the amount of heat a body can dissipate by
evaporation. High relative humidity slows down heat rejection by evapora-
tion, especially at high temperatures, and low relative humidity speeds it
up. The desirable level of relative humidity is the broad range of 30 to
70 percent, with 50 percent being the most desirable level. Most people at
these conditions feel neither hot nor cold, and the body does not need to
activate any of the defense mechanisms to maintain the normal body tem-
perature (Fig. 1-52).
Another factor that has a major effect on thermal comfort is excessive air
motion or draft, which causes undesired local cooling of the human body.
Draft is identified by many as a most annoying factor in work places, auto-
mobiles, and airplanes. Experiencing discomfort by draft is most common
among people wearing indoor clothing and doing light sedentary work, and
least common among people with high activity levels. The air velocity
should be kept below 9 m/min (30 ft/min) in winter and 15 m/min
(50 ft/min) in summer to minimize discomfort by draft, especially when the
air is cool. A low level of air motion is desirable as it removes the warm,
moist air that builds around the body and replaces it with fresh air. There-
fore, air motion should be strong enough to remove heat and moisture from
the vicinity of the body, but gentle enough to be unnoticed. High speed air
motion causes discomfort outdoors as well. For example, an environment
at 10°C (50°F) with 48 km/h winds feels as cold as an environment at
— 7°C (20°F) with 3 km/h winds because of the chilling effect of the air
motion (the wind-chill factor).
A comfort system should provide uniform conditions throughout the
living space to avoid discomfort caused by nonuniformities such as drafts,
asymmetric thermal radiation, hot or cold floors, and vertical temperature
stratification. Asymmetric thermal radiation is caused by the cold sur-
faces of large windows, uninsulated walls, or cold products and the warm
surfaces of gas or electric radiant heating panels on the walls or ceiling,
solar-heated masonry walls or ceilings, and warm machinery. Asymmetric
radiation causes discomfort by exposing different sides of the body to sur-
faces at different temperatures and thus to different heat loss or gain by
radiation. A person whose left side is exposed to a cold window, for exam-
ple, will feel like heat is being drained from that side of his or her body
(Fig. 1-53). For thermal comfort, the radiant temperature asymmetry
should not exceed 5°C in the vertical direction and 10°C in the horizontal
direction. The unpleasant effect of radiation asymmetry can be minimized
by properly sizing and installing heating panels, using double-pane win-
dows, and providing generous insulation at the walls and the roof.
Direct contact with cold or hot floor surfaces also causes localized dis-
comfort in the feet. The temperature of the floor depends on the way it is
constructed (being directly on the ground or on top of a heated room, being
made of wood or concrete, the use of insulation, etc.) as well as the floor
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CHAPTER 1
covering used such as pads, carpets, rugs, and linoleum. A floor tempera-
ture of 23 to 25°C is found to be comfortable to most people. The floor
asymmetry loses its significance for people with footwear. An effective and
economical way of raising the floor temperature is to use radiant heating
panels instead of turning the thermostat up. Another nonuniform condition
that causes discomfort is temperature stratification in a room that ex-
poses the head and the feet to different temperatures. For thermal comfort,
the temperature difference between the head and foot levels should not ex-
ceed 3°C. This effect can be minimized by using destratification fans.
It should be noted that no thermal environment will please everyone. No
matter what we do, some people will express some discomfort. The thermal
comfort zone is based on a 90 percent acceptance rate. That is, an environ-
ment is deemed comfortable if only 10 percent of the people are dissatis-
fied with it. Metabolism decreases somewhat with age, but it has no effect
on the comfort zone. Research indicates that there is no appreciable differ-
ence between the environments preferred by old and young people. Exper-
iments also show that men and women prefer almost the same environment.
The metabolism rate of women is somewhat lower, but this is compensated
by their slightly lower skin temperature and evaporative loss. Also, there is
no significant variation in the comfort zone from one part of the world to
another and from winter to summer. Therefore, the same thermal comfort
conditions can be used throughout the world in any season. Also, people
cannot acclimatize themselves to prefer different comfort conditions.
In a cold environment, the rate of heat loss from the body may exceed
the rate of metabolic heat generation. Average specific heat of the human
body is 3.49 kJ/kg • °C, and thus each 1°C drop in body temperature corre-
sponds to a deficit of 244 kJ in body heat content for an average 70-kg
man. A drop of 0.5°C in mean body temperature causes noticeable but ac-
ceptable discomfort. A drop of 2.6°C causes extreme discomfort. A sleep-
ing person will wake up when his or her mean body temperature drops by
1.3°C (which normally shows up as a 0.5°C drop in the deep body and 3°C
in the skin area). The drop of deep body temperature below 35°C may dam-
age the body temperature regulation mechanism, while a drop below 28°C
may be fatal. Sedentary people reported to feel comfortable at a mean skin
temperature of 33.3°C, uncomfortably cold at 31°C, shivering cold at
30°C, and extremely cold at 29°C. People doing heavy work reported to
feel comfortable at much lower temperatures, which shows that the activity
level affects human performance and comfort. The extremities of the body
such as hands and feet are most easily affected by cold weather, and their
temperature is a better indication of comfort and performance. A hand-skin
temperature of 20°C is perceived to be uncomfortably cold, 15°C to be
extremely cold, and 5°C to be painfully cold. Useful work can be per-
formed by hands without difficulty as long as the skin temperature of fin-
gers remains above 16°C (ASHRAE Handbook of Fundamentals, Ref. 1,
Chapter 8).
The first line of defense of the body against excessive heat loss in a cold
environment is to reduce the skin temperature and thus the rate of heat loss
from the skin by constricting the veins and decreasing the blood flow to the
skin. This measure decreases the temperature of the tissues subjacent to
the skin, but maintains the inner body temperature. The next preventive
2.0
& 1.5
20
25
30
1.0
3 0.5
u
Sedentary
50% RH
v '■.. T < 30 fpm
Heavy
>. \ (0.15 m/s)
clothing
^ X.
Winter
- s \
^ X
N X
v x.
^ X
clothing
^ X
^ X
S X.
Summer
■..clothing
s x_
s X
s
\
64
72
76
Operative temperature
Upper acceptability limit
Optimum
Lower acceptability limit
FIGURE 1-51
The effect of clothing on
the environment temperature
that feels comfortable (1 clo =
0.155 m 2 • °C/W = 0.880 ft 2 ■ °F ■ h/Btu)
(from ASHRAE Standard 55-1981).
FIGURE 1-52
A thermally comfortable environment.
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44
HEAT TRANSFER
FIGURE 1-53
Cold surfaces cause excessive heat loss
from the body by radiation, and thus
discomfort on that side of the body.
Shivering
FIGURE 1-54
The rate of metabolic heat generation
may go up by six times the resting
level during total body shivering
in cold weather.
measure is increasing the rate of metabolic heat generation in the body by
shivering, unless the person does it voluntarily by increasing his or her
level of activity or puts on additional clothing. Shivering begins slowly in
small muscle groups and may double the rate of metabolic heat production
of the body at its initial stages. In the extreme case of total body shivering,
the rate of heat production may reach six times the resting levels (Fig.
1-54). If this measure also proves inadequate, the deep body temperature
starts falling. Body parts furthest away from the core such as the hands and
feet are at greatest danger for tissue damage.
In hot environments, the rate of heat loss from the body may drop be-
low the metabolic heat generation rate. This time the body activates the op-
posite mechanisms. First the body increases the blood flow and thus heat
transport to the skin, causing the temperature of the skin and the subjacent
tissues to rise and approach the deep body temperature. Under extreme heat
conditions, the heart rate may reach 180 beats per minute in order to main-
tain adequate blood supply to the brain and the skin. At higher heart rates,
the volumetric efficiency of the heart drops because of the very short time
between the beats to fill the heart with blood, and the blood supply to the
skin and more importantly to the brain drops. This causes the person to
faint as a result of heat exhaustion. Dehydration makes the problem worse.
A similar thing happens when a person working very hard for a long time
stops suddenly. The blood that has flooded the skin has difficulty returning
to the heart in this case since the relaxed muscles no longer force the blood
back to the heart, and thus there is less blood available for pumping to the
brain.
The next line of defense is releasing water from sweat glands and resort-
ing to evaporative cooling, unless the person removes some clothing and
reduces the activity level (Fig. 1-55). The body can maintain its core tem-
perature at 37°C in this evaporative cooling mode indefinitely, even in en-
vironments at higher temperatures (as high as 200°C during military
endurance tests), if the person drinks plenty of liquids to replenish his or
her water reserves and the ambient air is sufficiently dry to allow the sweat
to evaporate instead of rolling down the skin. If this measure proves inad-
equate, the body will have to start absorbing the metabolic heat and the
deep body temperature will rise. A person can tolerate a temperature rise of
1.4°C without major discomfort but may collapse when the temperature
rise reaches 2.8°C. People feel sluggish and their efficiency drops consid-
erably when the core body temperature rises above 39°C. A core tempera-
ture above 41°C may damage hypothalamic proteins, resulting in cessation
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45
CHAPTER 1
of sweating, increased heat production by shivering, and a heat stroke with
irreversible and life-threatening damage. Death can occur above 43°C.
A surface temperature of 46°C causes pain on the skin. Therefore, direct
contact with a metal block at this temperature or above is painful. How-
ever, a person can stay in a room at 100°C for up to 30 min without any
damage or pain on the skin because of the convective resistance at the skin
surface and evaporative cooling. We can even put our hands into an oven at
200°C for a short time without getting burned.
Another factor that affects thermal comfort, health, and productivity is
ventilation. Fresh outdoor air can be provided to a building naturally by
doing nothing, ox forcefully by a mechanical ventilation system. In the first
case, which is the norm in residential buildings, the necessary ventilation is
provided by infiltration through cracks and leaks in the living space and by
the opening of the windows and doors. The additional ventilation needed in
the bathrooms and kitchens is provided by air vents with dampers or ex-
haust fans. With this kind of uncontrolled ventilation, however, the fresh
air supply will be either too high, wasting energy, or too low, causing poor
indoor air quality. But the current practice is not likely to change for resi-
dential buildings since there is not a public outcry for energy waste or air
quality, and thus it is difficult to justify the cost and complexity of me-
chanical ventilation systems.
Mechanical ventilation systems are part of any heating and air condi-
tioning system in commercial buildings, providing the necessary amount of
fresh outdoor air and distributing it uniformly throughout the building. This
is not surprising since many rooms in large commercial buildings have no
windows and thus rely on mechanical ventilation. Even the rooms with
windows are in the same situation since the windows are tightly sealed and
cannot be opened in most buildings. It is not a good idea to oversize the
ventilation system just to be on the "safe side" since exhausting the heated
or cooled indoor air wastes energy. On the other hand, reducing the venti-
lation rates below the required minimum to conserve energy should also be
avoided so that the indoor air quality can be maintained at the required lev-
els. The minimum fresh air ventilation requirements are listed in Table 1-8.
The values are based on controlling the C0 2 and other contaminants with
an adequate margin of safety, which requires each person be supplied with
at least 7.5 L/s (15 ftVmin) of fresh air.
Another function of the mechanical ventilation system is to clean the air
by filtering it as it enters the building. Various types of filters are available
for this purpose, depending on the cleanliness requirements and the allow-
able pressure drop.
Evaporation
FIGURE 1-55
In hot environments, a body can
dissipate a large amount of metabolic
heat by sweating since the sweat absorbs
the body heat and evaporates.
TABLE 1-8
Minimum fresh air requirements
in buildings (from ASHRAE
Standard 62-1989)
Requirement
(per person)
Application
L/s
ft 3 /min
Classrooms,
libraries,
supermarkets
8
15
Dining rooms,
conference
rooms, offices
10
20
Hospital
rooms
13
25
Hotel rooms 15
(per room)
30
(per room)
Smoking
lounges
30
60
Retail stores
1.0-1.5
(per m 2 )
0.2-0.3
(per ft 2 )
Residential 0.35 air change per
buildings hour, but not less than
7.5 L/s (or 15ft 3 /min)
per person
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46
HEAT TRANSFER
SUMMARY
In this chapter, the basics of heat transfer are introduced and
discussed. The science of thermodynamics deals with the
amount of heat transfer as a system undergoes a process from
one equilibrium state to another, whereas the science of heat
transfer deals with the rate of heat transfer, which is the main
quantity of interest in the design and evaluation of heat transfer
equipment. The sum of all forms of energy of a system is called
total energy, and it includes the internal, kinetic, and potential
energies. The internal energy represents the molecular energy
of a system, and it consists of sensible, latent, chemical, and
nuclear forms. The sensible and latent forms of internal energy
can be transferred from one medium to another as a result of a
temperature difference, and are referred to as heat or thermal
energy. Thus, heat transfer is the exchange of the sensible and
latent forms of internal energy between two mediums as a re-
sult of a temperature difference. The amount of heat transferred
per unit time is called heat transfer rate and is denoted by Q.
The rate of heat transfer per unit area is called heat flux, q.
A system of fixed mass is called a closed system and a sys-
tem that involves mass transfer across its boundaries is called
an open system or control volume. The first law of thermody-
namics or the energy balance for any system undergoing any
process can be expressed as
When a stationary closed system involves heat transfer only
and no work interactions across its boundary, the energy bal-
ance relation reduces to
Q = mC r AT
where Q is the amount of net heat transfer to or from the sys-
tem. When heat is transferred at a constant rate of Q, the
amount of heat transfer during a time interval At can be deter-
mined from Q = Q At.
Under steady conditions and in the absence of any work in-
teractions, the conservation of energy relation for a control vol-
ume with one inlet and one exit with negligible changes in
kinetic and potential energies can be expressed as
e c
kA
Q
hi C p AT
where m = p°VA c is the mass flow rate and Q is the rate of net
heat transfer into or out of the control volume.
Heat can be transferred in three different modes: conduction,
convection, and radiation. Conduction is the transfer of energy
from the more energetic particles of a substance to the adjacent
less energetic ones as a result of interactions between the parti-
cles, and is expressed by Fourier's law of heat conduction as
cIT
dx
where k is the thermal conductivity of the material, A is the
area normal to the direction of heat transfer, and dT/dx is the
temperature gradient. The magnitude of the rate of heat con-
duction across a plane layer of thickness L is given by
6 c
kA
AT
where AT is the temperature difference across the layer.
Convection is the mode of heat transfer between a solid sur-
face and the adjacent liquid or gas that is in motion, and in-
volves the combined effects of conduction and fluid motion.
The rate of convection heat transfer is expressed by Newton 's
law of cooling as
e
convection
hA, (T. - TJ
where h is the convection heat transfer coefficient in W/m 2 • °C
or Btu/h • ft 2 ■ °F, A s is the surface area through which con-
vection heat transfer takes place, T s is the surface temperature,
and T^ is the temperature of the fluid sufficiently far from the
surface.
Radiation is the energy emitted by matter in the form of
electromagnetic waves (or photons) as a result of the changes
in the electronic configurations of the atoms or molecules. The
maximum rate of radiation that can be emitted from a surface
at an absolute temperature T s is given by the Stefan-Boltzmann
law as 2 crait . raax = uAJ* where <r = 5.67 X 10" 8 W/m 2 • K 4
or 0.1714 X 10" 8 Btu/h • ft 2 • R 4 is the Stefan-Boltzmann
constant.
When a surface of emissivity 8 and surface area A s at an ab-
solute temperature T s is completely enclosed by a much larger
(or black) surface at absolute temperature T SUII separated by a
gas (such as air) that does not intervene with radiation, the net
rate of radiation heat transfer between these two surfaces is
given by
e rad = e aA s (r s 4 -r 8 4 urr )
In this case, the emissivity and the surface area of the sur-
rounding surface do not have any effect on the net radiation
heat transfer.
The rate at which a surface absorbs radiation is determined
from g absorbed = afiinciden, where g lncidcnt is the rate at which ra-
diation is incident on the surface and a is the absorptivity of
the surface.
cen58933_ch01.qxd 9/10/2002 8:30 AM Page 47
REFERENCES AND SUGGESTED READING
47
CHAPTER 1
1. American Society of Heating, Refrigeration, and Air-
Conditioning Engineers, Handbook of Fundamentals.
Atlanta: ASHRAE, 1993.
2. Y. A. Cengel and R. H. Turner. Fundamentals of Thermal-
Fluid Sciences. New York: McGraw-Hill, 2001 .
3. Y. A. Cengel and M. A. Boles. Thermodynamics — An
Engineering Approach. 4th ed. New York: McGraw-Hill,
2002.
4. J. P. Holman. Heat Transfer. 9th ed. New York: McGraw-
Hill, 2002.
5. F. P. Incropera and D. P. DeWitt. Introduction to Heat
Transfer. 4th ed. New York: John Wiley & Sons, 2002.
6. F. Kreith and M. S. Bohn. Principles of Heat Transfer. 6th
ed. Pacific Grove, CA: Brooks/Cole, 2001.
7. A. F. Mills. Basic Heat and Mass Transfer. 2nd ed.
Upper Saddle River, NJ: Prentice-Hall, 1999.
8. M. N. Ozisik. Heat Transfer — A Basic Approach. New
York: McGraw-Hill, 1985.
9. Robert J. Ribando. Heat Transfer Tools. New York:
McGraw-Hill, 2002.
10. F M. White. Heat and Mass Transfer. Reading, MA:
Addison-Wesley, 1988.
PROBLEMS
Thermodynamics and Heat Transfer
1-1 C How does the science of heat transfer differ from the
science of thermodynamics?
1-2C What is the driving force for (a) heat transfer, (b) elec-
tric current flow, and (c) fluid flow?
1-3C What is the caloric theory? When and why was it
abandoned?
1— 4C How do rating problems in heat transfer differ from the
sizing problems?
1-5C What is the difference between the analytical and ex-
perimental approach to heat transfer? Discuss the advantages
and disadvantages of each approach.
1-6C What is the importance of modeling in engineering?
How are the mathematical models for engineering processes
prepared?
1-7C When modeling an engineering process, how is the
right choice made between a simple but crude and a complex
but accurate model? Is the complex model necessarily a better
choice since it is more accurate?
Heat and Other Forms of Energy
1-8C What is heat flux? How is it related to the heat trans-
fer rate?
*Problems designated by a "C" are concept questions, and
students are encouraged to answer them all. Problems designated
by an "E" are in English units, and the SI users can ignore them.
Problems with a CD-EES icon ® are solved using EES, and
complete solutions together with parametric studies are included
on the enclosed CD. Problems with a computer-EES icon H are
comprehensive in nature, and are intended to be solved with a
computer, preferably using the EES software that accompanies
this text.
1-9C What are the mechanisms of energy transfer to a closed
system? How is heat transfer distinguished from the other
forms of energy transfer?
1-10C How are heat, internal energy, and thermal energy
related to each other?
1-11C An ideal gas is heated from 50°C to 80°C (a) at con-
stant volume and (b) at constant pressure. For which case do
you think the energy required will be greater? Why?
1-12 A cylindrical resistor element on a circuit board dis-
sipates 0.6 W of power. The resistor is 1.5 cm long, and has a
diameter of 0.4 cm. Assuming heat to be transferred uniformly
from all surfaces, determine (a) the amount of heat this resistor
dissipates during a 24-hour period, (b) the heat flux, and (c) the
fraction of heat dissipated from the top and bottom surfaces.
1-13E A logic chip used in a computer dissipates 3 W of
power in an environment at 120°F, and has a heat transfer sur-
face area of 0.08 in 2 . Assuming the heat transfer from the sur-
face to be uniform, determine (a) the amount of heat this chip
dissipates during an eight-hour work day, in kWh, and (b) the
heat flux on the surface of the chip, in W/in 2 .
1-14 Consider a 150-W incandescent lamp. The filament
of the lamp is 5 cm long and has a diameter of 0.5 mm. The
diameter of the glass bulb of the lamp is 8 cm. Determine the
heat flux, in W/m 2 , (a) on the surface of the filament and (b) on
the surface of the glass bulb, and (c) calculate how much it will
cost per year to keep that lamp on for eight hours a day every
day if the unit cost of electricity is $0.08/kWh.
Answers: (a) 1.91 x 10 6 W/m 2 , (b) 7500 W/m 2 , (c) $35.04/yr
1-15 A 1200-W iron is left on the ironing board with its base
exposed to the air. About 90 percent of the heat generated
in the iron is dissipated through its base whose surface area is
150 cm 2 , and the remaining 10 percent through other surfaces.
Assuming the heat transfer from the surface to be uniform,
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48
HEAT TRANSFER
8 cm
FIGURE P1 -14
determine (a) the amount of heat the iron dissipates during a
2-hour period, in kWh, (b) the heat flux on the surface of the
iron base, in W/m 2 , and (c) the total cost of the electrical en-
ergy consumed during this 2-hour period. Take the unit cost of
electricity to be $0.07/kWh.
1-16 A 15-cm X 20-cm circuit board houses on its surface
120 closely spaced logic chips, each dissipating 0.12 W. If the
heat transfer from the back surface of the board is negligible,
determine (a) the amount of heat this circuit board dissipates
during a 10-hour period, in kWh, and (b) the heat flux on the
surface of the circuit board, in W/m 2 .
15 cm
Chips
FIGURE P1 -16
1-17 A 15-cm-diameter aluminum ball is to be heated from
80°C to an average temperature of 200°C. Taking the average
density and specific heat of aluminum in this temperature
range to be p = 2700 kg/m 3 and C p = 0.90 kJ/kg • °C, respec-
tively, determine the amount of energy that needs to be trans-
ferred to the aluminum ball. Answer: 515 kJ
1-18 The average specific heat of the human body is 3.6
kJ/kg • °C. If the body temperature of a 70-kg man rises from
37°C to 39°C during strenuous exercise, determine the increase
in the thermal energy content of the body as a result of this rise
in body temperature.
1-19 Infiltration of cold air into a warm house during winter
through the cracks around doors, windows, and other openings
is a major source of energy loss since the cold air that enters
needs to be heated to the room temperature. The infiltration is
often expressed in terms of ACH (air changes per hour). An
ACH of 2 indicates that the entire air in the house is replaced
twice every hour by the cold air outside.
Consider an electrically heated house that has a floor space
of 200 m 2 and an average height of 3 m at 1000 m elevation,
where the standard atmospheric pressure is 89.6 kPa. The
house is maintained at a temperature of 22°C, and the infiltra-
tion losses are estimated to amount to 0.7 ACH. Assuming the
pressure and the temperature in the house remain constant, de-
termine the amount of energy loss from the house due to infil-
tration for a day during which the average outdoor temperature
is 5°C. Also, determine the cost of this energy loss for that day
if the unit cost of electricity in that area is $0.082/kWh.
Answers: 53.8 kWh/day, $4.41/day
1-20 Consider a house with a floor space of 200 m 2 and an
average height of 3 m at sea level, where the standard atmos-
pheric pressure is 101 .3 kPa. Initially the house is at a uniform
temperature of 10°C. Now the electric heater is turned on, and
the heater runs until the air temperature in the house rises to an
average value of 22°C. Determine how much heat is absorbed
by the air assuming some air escapes through the cracks as the
heated air in the house expands at constant pressure. Also, de-
termine the cost of this heat if the unit cost of electricity in that
area is $0.075/kWh.
1-21E Consider a 60-gallon water heater that is initially
filled with water at 45°F. Determine how much energy needs to
be transferred to the water to raise its temperature to 140°F.
Take the density and specific heat of water to be 62 lbm/ft 3 and
1 .0 Btu/lbm ■ °F, respectively.
The First Law of Thermodynamics
1-22C On a hot summer day, a student turns his fan on when
he leaves his room in the morning. When he returns in the
evening, will his room be warmer or cooler than the neighbor-
ing rooms? Why? Assume all the doors and windows are kept
closed.
1-23C Consider two identical rooms, one with a refrigerator
in it and the other without one. If all the doors and windows are
closed, will the room that contains the refrigerator be cooler or
warmer than the other room? Why?
1-24C Define mass and volume flow rates. How are they re-
lated to each other?
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CHAPTER 1
1-25 Two 800-kg cars moving at a velocity of 90 km/h have
a head-on collision on a road. Both cars come to a complete
rest after the crash. Assuming all the kinetic energy of cars is
converted to thermal energy, determine the average tempera-
ture rise of the remains of the cars immediately after the crash.
Take the average specific heat of the cars to be 0.45 kJ/kg • °C.
1-26 A classroom that normally contains 40 people is to be
air-conditioned using window air-conditioning units of 5-kW
cooling capacity. A person at rest may be assumed to dissipate
heat at a rate of 360 kJ/h. There are 10 lightbulbs in the room,
each with a rating of 100 W. The rate of heat transfer to the
classroom through the walls and the windows is estimated to
be 15,000 kJ/h. If the room air is to be maintained at a constant
temperature of 21°C, determine the number of window air-
conditioning units required. Answer: two units
1-27E A rigid tank contains 20 lbm of air at 50 psia and
80°F. The air is now heated until its pressure is doubled. Deter-
mine (a) the volume of the tank and (b) the amount of heat
transfer. Answers: (a) 80 ft 3 , (b) 2035 Btu
1-28 A 1-m 3 rigid tank contains hydrogen at 250 kPa and
420 K. The gas is now cooled until its temperature drops to 300
K. Determine (a) the final pressure in the tank and (b) the
amount of heat transfer from the tank.
1-29 A 4-m X 5-m X 6-m room is to be heated by a base-
board resistance heater. It is desired that the resistance heater
be able to raise the air temperature in the room from 7°C to
25°C within 15 minutes. Assuming no heat losses from the
room and an atmospheric pressure of 100 kPa, determine the
required power rating of the resistance heater. Assume constant
specific heats at room temperature. Answer: 3.01 kW
1-30 A 4-m X 5-m X 7-m room is heated by the radiator of
a steam heating system. The steam radiator transfers heat at a
rate of 10,000 kJ/h and a 100-W fan is used to distribute the
warm air in the room. The heat losses from the room are esti-
mated to be at a rate of about 5000 kJ/h. If the initial tempera-
ture of the room air is 10°C, determine how long it will take for
the air temperature to rise to 20°C. Assume constant specific
heats at room temperature.
5000 kJ/h
Steam
Ro
Jin
4 n
i X 5 m x7 m
»- «
4
10,000 kJ/h
*■ $
•< 4
Room
4mx6mx6m
FIGURE P1-31
1-31 A student living in a 4-m X 6-m X 6-m dormitory
room turns his 150-W fan on before she leaves her room on a
summer day hoping that the room will be cooler when she
comes back in the evening. Assuming all the doors and win-
dows are tightly closed and disregarding any heat transfer
through the walls and the windows, determine the temperature
in the room when she comes back 10 hours later. Use specific
heat values at room temperature and assume the room to be at
100 kPa and 15°C in the morning when she leaves.
Answer: 58.1°C
1-32E A 10-ft 3 tank contains oxygen initially at 14.7 psia
and 80°F. A paddle wheel within the tank is rotated until the
pressure inside rises to 20 psia. During the process 20 Btu of
heat is lost to the surroundings. Neglecting the energy stored in
the paddle wheel, determine the work done by the paddle
wheel.
1-33 A room is heated by a baseboard resistance heater.
When the heat losses from the room on a winter day amount to
7000 kJ/h, it is observed that the air temperature in the room
remains constant even though the heater operates continuously.
Determine the power rating of the heater, in kW.
1-34 A 50-kg mass of copper at 70°C is dropped into an in-
sulated tank containing 80 kg of water at 25°C. Determine the
final equilibrium temperature in the tank.
1-35 A 20-kg mass of iron at 100°C is brought into contact
with 20 kg of aluminum at 200°C in an insulated enclosure.
Determine the final equilibrium temperature of the combined
system. Answer: 168°C
1-36 An unknown mass of iron at 90 C C is dropped into an
insulated tank that contains 80 L of water at 20°C. At the same
Water
n
w
FIGURE P 1-30
FIGURE P1-36
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HEAT TRANSFER
time, a paddle wheel driven by a 200-W motor is activated to
stir the water. Thermal equilibrium is established after 25 min-
utes with a final temperature of 27°C. Determine the mass of
the iron. Neglect the energy stored in the paddle wheel, and
take the density of water to be 1000 kg/m 3 . Answer: 72.1 kg
1-37E A 90-lbm mass of copper at 160°F and a 50-lbm mass
of iron at 200°F are dropped into a tank containing 1 80 lbm of
water at 70°F. If 600 Btu of heat is lost to the surroundings dur-
ing the process, determine the final equilibrium temperature.
1-38 A 5-m X 6-m X 8-m room is to be heated by an elec-
trical resistance heater placed in a short duct in the room. Ini-
tially, the room is at 15°C, and the local atmospheric pressure
is 98 kPa. The room is losing heat steadily to the outside at a
rate of 200 kJ/min. A 200-W fan circulates the air steadily
through the duct and the electric heater at an average mass flow
rate of 50 kg/min. The duct can be assumed to be adiabatic, and
there is no air leaking in or out of the room. If it takes 15 min-
utes for the room air to reach an average temperature of 25°C,
find (a) the power rating of the electric heater and (b) the tem-
perature rise that the air experiences each time it passes
through the heater.
1-39 A house has an electric heating system that consists of
a 300-W fan and an electric resistance heating element placed
in a duct. Air flows steadily through the duct at a rate of 0.6
kg/s and experiences a temperature rise of 5°C. The rate of heat
loss from the air in the duct is estimated to be 250 W. De-
termine the power rating of the electric resistance heating
element.
1-40 A hair dryer is basically a duct in which a few layers of
electric resistors are placed. A small fan pulls the air in and
forces it to flow over the resistors where it is heated. Air enters
a 1200-W hair dryer at 100 kPa and 22°C, and leaves at 47°C.
The cross-sectional area of the hair dryer at the exit is 60 cm 2 .
Neglecting the power consumed by the fan and the heat losses
through the walls of the hair dryer, determine (a) the volume
flow rate of air at the inlet and (b) the velocity of the air at the
exit. Answers: (a) 0.0404 m 3 /s, (b) 7.30 m/s
T- = 47°C
: 60
cm ~ rAVS
= 100 kPa
:22°C
W =1200W
e
FIGURE P1-40
1-41 The ducts of an air heating system pass through an un-
heated area. As a result of heat losses, the temperature of the air
in the duct drops by 3°C. If the mass flow rate of air is 120
kg/min, determine the rate of heat loss from the air to the cold
environment.
1-42E Air enters the duct of an air-conditioning system at 1 5
psia and 50°F at a volume flow rate of 450 ft'/min. The diam-
eter of the duct is 10 inches and heat is transferred to the air in
the duct from the surroundings at a rate of 2 Btu/s. Determine
(a) the velocity of the air at the duct inlet and (b) the tempera-
ture of the air at the exit. Answers: (a) 825 ft/min, (£>) 64°F
1-43 Water is heated in an insulated, constant diameter tube
by a 7-kW electric resistance heater. If the water enters the
heater steadily at 15°C and leaves at 70°C, determine the mass
flow rate of water.
Water M
LvC \?
i-AWvWVWvWV^n
\70°C
Resistance
heater, 7 kW
FIGURE P1 -43
Heat Transfer Mechanisms
1-44C Define thermal conductivity and explain its signifi-
cance in heat transfer.
1-45C What are the mechanisms of heat transfer? How are
they distinguished from each other?
1-46C What is the physical mechanism of heat conduction in
a solid, a liquid, and a gas?
1-47C Consider heat transfer through a windowless wall of
a house in a winter day. Discuss the parameters that affect the
rate of heat conduction through the wall.
1-48C Write down the expressions for the physical laws that
govern each mode of heat transfer, and identify the variables
involved in each relation.
1-49C How does heat conduction differ from convection?
1-50C Does any of the energy of the sun reach the earth by
conduction or convection?
1-51 C How does forced convection differ from natural
convection?
1-52C Define emissivity and absorptivity. What is Kirch-
hoff's law of radiation?
1-53C What is a blackbody? How do real bodies differ from
blackbodies?
1-54C Judging from its unit W/m • °C, can we define ther-
mal conductivity of a material as the rate of heat transfer
through the material per unit thickness per unit temperature
difference? Explain.
1-55C Consider heat loss through the two walls of a house
on a winter night. The walls are identical, except that one of
them has a tightly fit glass window. Through which wall will
the house lose more heat? Explain.
1-56C Which is a better heat conductor, diamond or silver?
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51
CHAPTER 1
1-57C Consider two walls of a house that are identical ex-
cept that one is made of 10-cm-thick wood, while the other is
made of 25-cm-thick brick. Through which wall will the house
lose more heat in winter?
1-58C How do the thermal conductivity of gases and liquids
vary with temperature?
1-59C Why is the thermal conductivity of superinsulation
orders of magnitude lower than the thermal conductivity of
ordinary insulation?
1-60C Why do we characterize the heat conduction ability
of insulators in terms of their apparent thermal conductivity
instead of the ordinary thermal conductivity?
1-61 C Consider an alloy of two metals whose thermal con-
ductivities are k t and k 2 . Will the thermal conductivity of the
alloy be less than k t , greater than k 2 , or between k x and k{l
1-62 The inner and outer surfaces of a 5-m X 6-m brick wall
of thickness 30 cm and thermal conductivity 0.69 W/m ■ °C are
maintained at temperatures of 20°C and 5°C, respectively.
Determine the rate of heat transfer through the wall, in W.
Answer: 1035 W
20°C
- —
Brick
wall
30 cm
5°C
FIGURE P1-62
1-63 The inner and outer surfaces of a 0.5-cm-thick 2-m X
2-m window glass in winter are 10°C and 3°C, respectively. If
the thermal conductivity of the glass is 0.78 W/m • °C, deter-
mine the amount of heat loss, in kJ, through the glass over a
period of 5 hours. What would your answer be if the glass were
1 cm thick? Answers: 78,624 kJ, 39,312 kJ
1-64 [JJ^l Reconsider Problem 1-63. Using EES (or other)
b^2 software, plot the amount of heat loss through the
glass as a function of the window glass thickness in the range
of 0. 1 cm to 1 .0 cm. Discuss the results.
1-65 An aluminum pan whose thermal conductivity is
237 W/m ■ °C has a flat bottom with diameter 20 cm and thick-
ness 0.4 cm. Heat is transferred steadily to boiling water in the
pan through its bottom at a rate of 800 W. If the inner surface
of the bottom of the pan is at 105°C, determine the temperature
of the outer surface of the bottom of the pan.
ki (,
X<
5°C
0.4 cm
I I 1 I I I I III I t
800 W
FIGURE P1 -65
1-66E The north wall of an electrically heated home is 20 ft
long, 10 ft high, and 1 ft thick, and is made of brick whose
thermal conductivity is k = 0.42 Btu/h • ft • °F. On a certain
winter night, the temperatures of the inner and the outer sur-
faces of the wall are measured to be at about 62°F and 25°F,
respectively, for a period of 8 hours. Determine (a) the rate of
heat loss through the wall that night and (b) the cost of that heat
loss to the home owner if the cost of electricity is $0.07/kWh.
1-67 In a certain experiment, cylindrical samples of diameter
4 cm and length 7 cm are used (see Fig. 1-29). The two
thermocouples in each sample are placed 3 cm apart. After ini-
tial transients, the electric heater is observed to draw 0.6 A at
110 V, and both differential thermometers read a temperature
difference of 10°C. Determine the thermal conductivity of the
sample. Answer: 78.8 W/m • °C
1-68 One way of measuring the thermal conductivity of a
material is to sandwich an electric thermofoil heater between
two identical rectangular samples of the material and to heavily
insulate the four outer edges, as shown in the figure. Thermo-
couples attached to the inner and outer surfaces of the samples
record the temperatures.
During an experiment, two 0.5-cm-thick samples 10 cm X
10 cm in size are used. When steady operation is reached, the
heater is observed to draw 35 W of electric power, and the tem-
perature of each sample is observed to drop from 82°C at the
inner surface to 74°C at the outer surface. Determine the ther-
mal conductivity of the material at the average temperature.
Samples
"- Insulation
- Insulation
Source
FIGURE P1 -68
0.5 cm
1-69 Repeat Problem 1-68 for an electric power consump-
tion of 28 W.
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HEAT TRANSFER
1-70 A heat flux meter attached to the inner surface of a
3-cm-thick refrigerator door indicates a heat flux of 25 W/m 2
through the door. Also, the temperatures of the inner and the
outer surfaces of the door are measured to be 7°C and 15°C,
respectively. Determine the average thermal conductivity of
the refrigerator door. Answer: 0.0938 W/m • °C
1-71 Consider a person standing in a room maintained at
20°C at all times. The inner surfaces of the walls, floors, and
ceiling of the house are observed to be at an average tempera-
ture of 12°C in winter and 23°C in summer. Determine the
rates of radiation heat transfer between this person and the sur-
rounding surfaces in both summer and winter if the exposed
surface area, emissivity, and the average outer surface temper-
ature of the person are 1.6 m 2 , 0.95, and 32°C, respectively.
1-72 [7(^1 Reconsider Problem 1-71. Using EES (or other)
t^S software, plot the rate of radiation heat transfer in
winter as a function of the temperature of the inner surface of
the room in the range of 8°C to 18°C. Discuss the results.
1-73 For heat transfer purposes, a standing man can be mod-
eled as a 30-cm-diameter, 1 70-cm-long vertical cylinder with
both the top and bottom surfaces insulated and with the side
surface at an average temperature of 34°C. For a convection
heat transfer coefficient of 15 W/m 2 ■ °C, determine the rate of
heat loss from this man by convection in an environment at
20°C. Answer: 336 W
1-74 Hot air at 80°C is blown over a 2-m X 4-m flat surface
at 30 C C. If the average convection heat transfer coefficient is
55 W/m 2 • °C, determine the rate of heat transfer from the air to
the plate, in kW. Answer: 22 kW
1-75 rSi'M Reconsider Problem 1-74. Using EES (or other)
b^2 software, plot the rate of heat transfer as a func-
tion of the heat transfer coefficient in the range of 20 W/m 2 ■ °C
to 100 W/m 2 • °C. Discuss the results.
1-76 The heat generated in the circuitry on the surface of a
silicon chip (k = 130 W/m • °C) is conducted to the ceramic
substrate to which it is attached. The chip is 6 mm X 6 mm in
size and 0.5 mm thick and dissipates 3 W of power. Disregard-
ing any heat transfer through the 0.5-mm-high side surfaces,
determine the temperature difference between the front and
back surfaces of the chip in steady operation.
Silicon
chip
0.5 mm
1-77 A 50-cm-long, 800-W electric resistance heating ele-
ment with diameter 0.5 cm and surface temperature 120°C is
immersed in 60 kg of water initially at 20°C. Determine how
long it will take for this heater to raise the water temperature to
80°C. Also, determine the convection heat transfer coefficients
at the beginning and at the end of the heating process.
1-78 A 5 -cm-external-diameter, 10-m-long hot water pipe at
80°C is losing heat to the surrounding air at 5°C by natural
convection with a heat transfer coefficient of 25 W/m 2 ■ °C.
Determine the rate of heat loss from the pipe by natural con-
vection, in W. Answer: 2945 W
1-79 A hollow spherical iron container with outer diameter
20 cm and thickness 0.4 cm is filled with iced water at 0°C. If
the outer surface temperature is 5°C, determine the approxi-
mate rate of heat loss from the sphere, in kW, and the rate at
which ice melts in the container. The heat from fusion of water
is 333.7 kJ/kg.
5 C C
0.4 cm
Ceramic
substrate
FIGURE P1-76
FIGURE P1-79
1-80 VcgM Reconsider Problem 1-79. Using EES (or other)
t£^ software, plot the rate at which ice melts as a
function of the container thickness in the range of 0.2 cm to
2.0 cm. Discuss the results.
1-81E The inner and outer glasses of a 6-ft X 6-ft double-
pane window are at 60°F and 42°F, respectively. If the 0.25-in.
space between the two glasses is filled with still air, determine
the rate of heat transfer through the window.
Answer: 439 Btu/h
1-82 Two surfaces of a 2-cm-thick plate are maintained at
0°C and 80°C, respectively. If it is determined that heat is
transferred through the plate at a rate of 500 W/m 2 , determine
its thermal conductivity.
1-83 Four power transistors, each dissipating 15 W, are
mounted on a thin vertical aluminum plate 22 cm X 22 cm in
size. The heat generated by the transistors is to be dissipated by
both surfaces of the plate to the surrounding air at 25 °C, which
is blown over the plate by a fan. The entire plate can be as-
sumed to be nearly isothermal, and the exposed surface area of
the transistor can be taken to be equal to its base area. If the
average convection heat transfer coefficient is 25 W/m 2 ■ °C,
determine the temperature of the aluminum plate. Disregard
any radiation effects.
cen58933_ch01.qxd 9/10/2002 8:30 AM Page 53
1-84 An ice chest whose outer dimensions are 30 cm X
40 cm X 40 cm is made of 3-cm-thick Styrofoam (k = 0.033
W/m ■ °C). Initially, the chest is filled with 40 kg of ice at 0°C,
and the inner surface temperature of the ice chest can be taken
to be 0°C at all times. The heat of fusion of ice at 0°C is 333.7
kJ/kg, and the surrounding ambient air is at 30°C. Disregarding
any heat transfer from the 40-cm X 40-cm base of the ice
chest, determine how long it will take for the ice in the chest to
melt completely if the outer surfaces of the ice chest are at 8°C.
Answer: 32.7 days
:30°C
.
,
ts « °
o Ice chest Q O
pop O
Q o
-3 cm
Styrofoam
FIGURE P1-84
1-85 A transistor with a height of 0.4 cm and a diameter of
0.6 cm is mounted on a circuit board. The transistor is cooled
by air flowing over it with an average heat transfer coefficient
of 30 W/m 2 • °C. If the air temperature is 55°C and the tran-
sistor case temperature is not to exceed 70°C, determine the
amount of power this transistor can dissipate safely. Disregard
any heat transfer from the transistor base.
Air
55°C
Power
transistor
T <70°C
0.6 cm
0.4 cm -
FIGURE P1-85
1-86 [Z?vfl Reconsider Problem 1-85. Using EES (or other)
b^2 software, plot the amount of power the transistor
can dissipate safely as a function of the maximum case tem-
perature in the range of 60°C to 90°C. Discuss the results.
53
CHAPTER 1
1-87E A 200-ft-long section of a steam pipe whose outer di-
ameter is 4 inches passes through an open space at 50°F. The
average temperature of the outer surface of the pipe is mea-
sured to be 280°F, and the average heat transfer coefficient on
that surface is determined to be 6 Btu/h ■ ft 2 • °F. Determine
(a) the rate of heat loss from the steam pipe and (b) the annual
cost of this energy loss if steam is generated in a natural gas
furnace having an efficiency of 86 percent, and the price of nat-
ural gas is $0.58/therm (1 therm = 100,000 Btu).
Answers: (a) 289,000 Btu/h, (b) $17,074/yr
1-88 The boiling temperature of nitrogen at atmospheric
pressure at sea level (1 atm) is — 196°C. Therefore, nitrogen is
commonly used in low temperature scientific studies since the
temperature of liquid nitrogen in a tank open to the atmosphere
will remain constant at — 196°C until the liquid nitrogen in
the tank is depleted. Any heat transfer to the tank will result
in the evaporation of some liquid nitrogen, which has a heat of
vaporization of 198 kJ/kg and a density of 810 kg/m 3 at 1 atm.
Consider a 4-m-diameter spherical tank initially filled
with liquid nitrogen at 1 atm and — 196°C. The tank is ex-
posed to 20°C ambient air with a heat transfer coefficient of
25 W/m 2 • °C. The temperature of the thin-shelled spherical
tank is observed to be almost the same as the temperature of
the nitrogen inside. Disregarding any radiation heat exchange,
determine the rate of evaporation of the liquid nitrogen in the
tank as a result of the heat transfer from the ambient air.
FIGURE P1 -88
1-89 Repeat Problem 1-88 for liquid oxygen, which has
a boiling temperature of — 183°C, a heat of vaporization of
213 kJ/kg, and a density of 1 140 kg/m 3 at 1 atm pressure.
1-90 rSi'M Reconsider Problem 1-88. Using EES (or other)
1^2 software, plot the rate of evaporation of liquid
nitrogen as a function of the ambient air temperature in the
range of 0°C to 35°C. Discuss the results.
1-91 Consider a person whose exposed surface area is
1.7 m 2 , emissivity is 0.7, and surface temperature is 32°C.
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HEAT TRANSFER
Determine the rate of heat loss from that person by radiation in
a large room having walls at a temperature of (a) 300 K and
(b) 280 K. Answers: (a) 37.4 W, (b) 169.2 W
1-92 A 0.3-cm-thick, 12-cm-high, and 18-cm-long circuit
board houses 80 closely spaced logic chips on one side, each
dissipating 0.06 W. The board is impregnated with copper fill-
ings and has an effective thermal conductivity of 16 W/m • °C.
All the heat generated in the chips is conducted across the cir-
cuit board and is dissipated from the back side of the board to
the ambient air. Determine the temperature difference between
the two sides of the circuit board. Answer-. 0.042°C
1-93 Consider a sealed 20-cm-high electronic box whose
base dimensions are 40 cm X 40 cm placed in a vacuum cham-
ber. The emissivity of the outer surface of the box is 0.95. If the
electronic components in the box dissipate a total of 100 W of
power and the outer surface temperature of the box is not to ex-
ceed 55°C, determine the temperature at which the surrounding
surfaces must be kept if this box is to be cooled by radiation
alone. Assume the heat transfer from the bottom surface of the
box to the stand to be negligible.
FIGURE P1-93
1-94 Using the conversion factors between W and Btu/h, m
and ft, and K and R, express the Stefan-Boltzmann constant
o- = 5.67 X 10- 8 W/m 2 • K 4 in the English unit Btu/h • ft 2 ■ R 4 .
1-95 An engineer who is working on the heat transfer analy-
sis of a house in English units needs the convection heat trans-
fer coefficient on the outer surface of the house. But the only
value he can find from his handbooks is 20 W/m 2 • °C, which
is in SI units. The engineer does not have a direct conversion
factor between the two unit systems for the convection heat
transfer coefficient. Using the conversion factors between
W and Btu/h, m and ft, and °C and °F, express the given con-
vection heat transfer coefficient in Btu/h • ft 2 ■ °F.
Answer: 3.52 Btu/h • ft 2 • °F
Simultaneous Heat Transfer Mechanisms
1-96C Can all three modes of heat transfer occur simultane-
ously (in parallel) in a medium?
1-97C Can a medium involve (a) conduction and con-
vection, (b) conduction and radiation, or (c) convection and ra-
diation simultaneously? Give examples for the "yes" answers.
1-98C The deep human body temperature of a healthy
person remains constant at 37°C while the temperature and
the humidity of the environment change with time. Discuss the
heat transfer mechanisms between the human body and the en-
vironment both in summer and winter, and explain how a per-
son can keep cooler in summer and wanner in winter.
1-99C We often turn the fan on in summer to help us cool.
Explain how a fan makes us feel cooler in the summer. Also
explain why some people use ceiling fans also in winter.
1-100 Consider a person standing in a room at 23°C. Deter-
mine the total rate of heat transfer from this person if the ex-
posed surface area and the skin temperature of the person are
1 .7 m 2 and 32°C, respectively, and the convection heat transfer
coefficient is 5 W/m 2 ■ °C. Take the emissivity of the skin and
the clothes to be 0.9, and assume the temperature of the inner
surfaces of the room to be the same as the air temperature.
Answer: 161 W
1-101 Consider steady heat transfer between two large
parallel plates at constant temperatures of T t = 290 K and
T 2 = 1 50 K that are L = 2 cm apart. Assuming the surfaces to
be black (emissivity e = 1 ), determine the rate of heat transfer
between the plates per unit surface area assuming the gap
between the plates is (a) filled with atmospheric air, (b) evacu-
ated, (c) filled with fiberglass insulation, and (rf) filled with
superinsulation having an apparent thermal conductivity of
0.00015 W/m - C.
1-102 A 1 .4-m-long, 0.2-cm-diameter electrical wire extends
across a room that is maintained at 20°C. Heat is generated in
the wire as a result of resistance heating, and the surface tem-
perature of the wire is measured to be 240°C in steady op-
eration. Also, the voltage drop and electric current through
the wire are measured to be 110 V and 3 A, respectively. Dis-
regarding any heat transfer by radiation, determine the con-
vection heat transfer coefficient for heat transfer between the
outer surface of the wire and the air in the room.
Answer: 170.5 W/m 2 • °C
Room
20°C
x- 240°C
^ Electric resistance heater
FIGURE P1 -102
1-103
Reconsider Problem 1-102. Using EES (or
other) software, plot the convection heat trans-
fer coefficient as a function of the wire surface temperature in
the range of 100°C to 300°C. Discuss the results.
1-104E A 2-in-diameter spherical ball whose surface is
maintained at a temperature of 170°F is suspended in the mid-
dle of a room at 70°F. If the convection heat transfer coefficient
is 12 Btu/h ■ ft 2 ■ °F and the emissivity of the surface is 0.8, de-
termine the total rate of heat transfer from the ball.
cen58933_ch01.qxd 9/10/2002 8:30 AM Page 55
1-105 fl&\ A 1000-W iron is left on the iron board with its
Yss£y base exposed to the air at 20°C. The convection
heat transfer coefficient between the base surface and the sur-
rounding air is 35 W/m 2 • °C. If the base has an emissivity of
0.6 and a surface area of 0.02 m 2 , determine the temperature of
the base of the iron. Answer: 674°C
20°C
FIGURE P1-105
1-106 The outer surface of a spacecraft in space has an emis-
sivity of 0.8 and a solar absorptivity of 0.3. If solar radiation is
incident on the spacecraft at a rate of 950 W/m 2 , determine the
surface temperature of the spacecraft when the radiation emit-
ted equals the solar energy absorbed.
1-107 A 3-m-internal-diameter spherical tank made of 1 -cm-
thick stainless steel is used to store iced water at 0°C. The tank
is located outdoors at 25°C. Assuming the entire steel tank to
be at 0°C and thus the thermal resistance of the tank to be neg-
ligible, determine (a) the rate of heat transfer to the iced water
in the tank and (b) the amount of ice at 0°C that melts during a
24-hour period. The heat of fusion of water at atmospheric pres-
sure is h jf = 333.7 kJ/kg. The emissivity of the outer surface of
the tank is 0.6, and the convection heat transfer coefficient on
the outer surface can be taken to be 30 W/m 2 ■ °C. Assume the
average surrounding surface temperature for radiation ex-
change to be 15°C. Answer: 5898 kg
1-108 fJb\ The roof of a house consists of a 15-cm-thick
W concrete slab (k = 2 W/m • °C) that is 15 m
wide and 20 m long. The emissivity of the outer surface of the
roof is 0.9, and the convection heat transfer coefficient on that
surface is estimated to be 15 W/m 2 • °C. The inner surface of
the roof is maintained at 15°C. On a clear winter night, the am-
bient air is reported to be at 10°C while the night sky tempera-
ture for radiation heat transfer is 255 K. Considering both
radiation and convection heat transfer, determine the outer sur-
face temperature and the rate of heat transfer through the roof.
If the house is heated by a furnace burning natural gas with
an efficiency of 85 percent, and the unit cost of natural gas is
$0.60/fherm (1 therm = 105,500 kJ of energy content), de-
termine the money lost through the roof that night during a
14-hour period.
1-109E Consider a flat plate solar collector placed horizon-
tally on the flat roof of a house. The collector is 5 ft wide and
15 ft long, and the average temperature of the exposed surface
55
CHAPTER 1
of the collector is 100°F. The emissivity of the exposed sur-
face of the collector is 0.9. Determine the rate of heat loss from
the collector by convection and radiation during a calm day
when the ambient air temperature is 70°F and the effective
sky temperature for radiation exchange is 50°F. Take the con-
vection heat transfer coefficient on the exposed surface to be
2.5 Btu/h ■ ft 2 • °F.
FIGURE P1-109E
Problem Solving Technique and EES
1-110C What is the value of the engineering software pack-
ages in (a) engineering education and (b) engineering practice?
| Determine a positive real root of the following
equation using EES:
2x 3 - lOx 05 - 3x = -3
1-112
[J3 Solve the following system of two equations
with two unknowns using EES:
x* - y
3xy + y
7.75
3.5
1-113
Solve the following system of three equations
with three unknowns using EES:
2x — y + z
3x 2 + 2y
xy + 2z
5
z + 2
1-114
[tt3 Solve the following system of three equations
with three unknowns using EES:
1
Special Topic: Thermal Comfort
1-115C What is metabolism? What is the range of metabolic
rate for an average man? Why are we interested in metabolic
x 2 y -
- z
3y ' 5 +
xz
x + y -
- z
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HEAT TRANSFER
rate of the occupants of a building when we deal with heating
and air conditioning?
1-116C Why is the metabolic rate of women, in general,
lower than that of men? What is the effect of clothing on the
environmental temperature that feels comfortable?
1-117C What is asymmetric thermal radiation? How does it
cause thermal discomfort in the occupants of a room?
1-118C How do (a) draft and (b) cold floor surfaces cause
discomfort for a room's occupants?
1-119C What is stratification? Is it likely to occur at places
with low or high ceilings? How does it cause thermal discom-
fort for a room's occupants? How can stratification be pre-
vented?
1-120C Why is it necessary to ventilate buildings? What is
the effect of ventilation on energy consumption for heating in
winter and for cooling in summer? Is it a good idea to keep the
bathroom fans on all the time? Explain.
Review Problems
1-121 2.5 kg of liquid water initially at 18°C is to be heated
to 96°C in a teapot equipped with a 1200-W electric heating
element inside. The teapot is 0.8 kg and has an average specific
heat of 0.6 kJ/kg ■ °C. Taking the specific heat of water to be
4.18 kJ/kg • °C and disregarding any heat loss from the teapot,
determine how long it will take for the water to be heated.
1-122 A 4-m-long section of an air heating system of a house
passes through an unheated space in the attic. The inner diam-
eter of the circular duct of the heating system is 20 cm. Hot air
enters the duct at 100 kPa and 65 °C at an average velocity of
3 m/s. The temperature of the air in the duct drops to 60°C as a
result of heat loss to the cool space in the attic. Determine the
rate of heat loss from the air in the duct to the attic under steady
conditions. Also, determine the cost of this heat loss per hour if
the house is heated by a natural gas furnace having an effi-
ciency of 82 percent, and the cost of the natural gas in that area
is $0.58/therm (1 therm = 105,500 kJ).
Answers: 0.488 kJ/s, $0.012/h
4 m
65 °C
3 m/s
Hot air
FIGURE P1-122
1-123
Reconsider Problem 1-122. Using EES (or
other) software, plot the cost of the heat loss per
hour as a function of the average air velocity in the range of
1 m/s to 10 m/s. Discuss the results.
1-124 Water flows through a shower head steadily at a rate
of 10 L/min. An electric resistance heater placed in the water
pipe heats the water from 16°C to 43°C. Taking the density of
Resistance
heater
FIGURE P1-1 24
water to be 1 kg/L, determine the electric power input to the
heater, in kW.
In an effort to conserve energy, it is proposed to pass the
drained warm water at a temperature of 39°C through a heat
exchanger to preheat the incoming cold water. If the heat ex-
changer has an effectiveness of 0.50 (that is, it recovers only
half of the energy that can possibly be transferred from the
drained water to incoming cold water), determine the electric
power input required in this case. If the price of the electric en-
ergy is 8.5 0/kWh, determine how much money is saved during
a 10-minute shower as a result of installing this heat exchanger.
Answers: 18.8 kW, 10.8 kW, $0.0113
1-125 It is proposed to have a water heater that consists of an
insulated pipe of 5 cm diameter and an electrical resistor in-
side. Cold water at 15°C enters the heating section steadily at a
rate of 18 L/min. If water is to be heated to 50°C, determine
(a) the power rating of the resistance heater and (b) the average
velocity of the water in the pipe.
1-126 A passive solar house that is losing heat to the out-
doors at an average rate of 50,000 kJ/h is maintained at 22°C at
all times during a winter night for 10 hours. The house is to be
heated by 50 glass containers each containing 20 L of water
heated to 80°C during the day by absorbing solar energy.
A thermostat-controlled 15-kW back-up electric resistance
heater turns on whenever necessary to keep the house at 22°C.
(a) How long did the electric heating system run that night?
(b) How long would the electric heater have run that night if
the house incorporated no solar heating?
Answers: (a) 4.77 h, (b) 9.26 h
1-127 It is well known that wind makes the cold air feel
much colder as a result of the windchill effect that is due to the
increase in the convection heat transfer coefficient with in-
creasing air velocity. The windchill effect is usually expressed
in terms of the windchill factor, which is the difference be-
tween the actual air temperature and the equivalent calm-air
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50,000 kJ/h
22°C
FIGURE P1 126
temperature. For example, a windchill factor of 20°C for an
actual air temperature of 5°C means that the windy air at 5°C
feels as cold as the still air at — 15°C. In other words, a person
will lose as much heat to air at 5°C with a windchill factor of
20°C as he or she would in calm air at — 15°C.
For heat transfer purposes, a standing man can be modeled
as a 30-cm -diameter, 170-cm-long vertical cylinder with both
the top and bottom surfaces insulated and with the side surface
at an average temperature of 34 °C. For a convection heat trans-
fer coefficient of 15 W/m 2 • °C, determine the rate of heat loss
from this man by convection in still air at 20°C. What would
your answer be if the convection heat transfer coefficient is in-
creased to 50 W/m 2 • °C as a result of winds? What is the wind-
chill factor in this case? Answers: 336 W, 1120 W, 32.7°C
1-128 A thin metal plate is insulated on the back and ex-
posed to solar radiation on the front surface. The exposed sur-
face of the plate has an absorptivity of 0.7 for solar radiation. If
solar radiation is incident on the plate at a rate of 700 W/m 2
57
CHAPTER 1
and the surrounding air temperature is 10°C, determine the sur-
face temperature of the plate when the heat loss by convection
equals the solar energy absorbed by the plate. Take the convec-
tion heat transfer coefficient to be 30 W/m 2 • °C, and disregard
any heat loss by radiation.
1-129 A 4-m X 5-m X 6-m room is to be heated by one ton
(1000 kg) of liquid water contained in a tank placed in the
room. The room is losing heat to the outside at an average rate
of 10,000 kJ/h. The room is initially at 20°C and 100 kPa, and
is maintained at an average temperature of 20°C at all times. If
the hot water is to meet the heating requirements of this room
for a 24-hour period, determine the minimum temperature of
the water when it is first brought into the room. Assume con-
stant specific heats for both air and water at room temperature.
Answer: 77.4°C
1-130 Consider a 3-m X 3-m X 3-m cubical furnace whose
top and side surfaces closely approximate black surfaces at a
temperature of 1200 K. The base surface has an emissivity of
e = 0.7, and is maintained at 800 K. Determine the net rate
of radiation heat transfer to the base surface from the top and
side surfaces. Answer: 594,400 W
1-131 Consider a refrigerator whose dimensions are 1 .8 m X
1.2 m X 0.8 m and whose walls are 3 cm thick. The refrigera-
tor consumes 600 W of power when operating and has a COP
of 2.5. It is observed that the motor of the refrigerator remains
on for 5 minutes and then is off for 15 minutes periodically. If
the average temperatures at the inner and outer surfaces of the
refrigerator are 6°C and 17°C, respectively, determine the av-
erage thermal conductivity of the refrigerator walls. Also, de-
termine the annual cost of operating this refrigerator if the unit
cost of electricity is $0.08/kWh.
FIGURE P1-1 28
FIGURE P1-131
1-132 A 0.2-L glass of water at 20°C is to be cooled with
ice to 5°C. Determine how much ice needs to be added to
the water, in grams, if the ice is at 0°C. Also, determine how
much water would be needed if the cooling is to be done with
cold water at 0°C. The melting temperature and the heat of
fusion of ice at atmospheric pressure are 0°C and 333.7 kJ/kg,
respectively, and the density of water is 1 kg/L.
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58
HEAT TRANSFER
FIGURE P1-132
1-133 Tu'M Reconsider Problem 1-132. Using EES (or
I^S other) software, plot the amount of ice that
needs to be added to the water as a function of the ice temper-
ature in the range of — 24°C to 0°C. Discuss the results.
1-134E In order to cool 1 short ton (2000 lbm) of water at
70°F in a tank, a person pours 160 lbm of ice at 25°F into the
water. Determine the final equilibrium temperature in the tank.
The melting temperature and the heat of fusion of ice at atmo-
spheric pressure are 32°F and 143.5 Btu/lbm, respectively.
Answer: 56.3T
1-135 Engine valves (C p = 440 J/kg • °C and p = 7840
kg/m 3 ) are to be heated from 40°C to 800°C in 5 minutes in the
heat treatment section of a valve manufacturing facility. The
valves have a cylindrical stem with a diameter of 8 mm and a
length of 10 cm. The valve head and the stem may be assumed
to be of equal surface area, with a total mass of 0.0788 kg. For
a single valve, determine (a) the amount of heat transfer,
(b) the average rate of heat transfer, and (c) the average heat
flux, (d) the number of valves that can be heat treated per day if
the heating section can hold 25 valves, and it is used 10 hours
per day.
1-136 The hot water needs of a household are met by an
electric 60-L hot water tank equipped with a 1 .6-kW heating
element. The tank is initially filled with hot water at 80°C, and
the cold water temperature is 20°C. Someone takes a shower
by mixing constant flow rates of hot and cold waters. After a
showering period of 8 minutes, the average water temperature
in the tank is measured to be 60°C. The heater is kept on during
the shower and hot water is replaced by cold water. If the cold
water is mixed with the hot water stream at a rate of 0.06 kg/s,
determine the flow rate of hot water and the average tempera-
ture of mixed water used during the shower.
1-137 Consider a flat plate solar collector placed at the roof
of a house. The temperatures at the inner and outer surfaces of
glass cover are measured to be 28°C and 25°C, respectively.
The glass cover has a surface area of 2.2. m 2 and a thickness of
0.6 cm and a thermal conductivity of 0.7 W/m • C. Heat is lost
from the outer surface of the cover by convection and radiation
with a convection heat transfer coefficient of 10 W/m 2 • °C and
an ambient temperature of 15°C. Determine the fraction of heat
lost from the glass cover by radiation.
1-138 The rate of heat loss through a unit surface area of
a window per unit temperature difference between the in-
doors and the outdoors is called the {/-factor. The value of
the [/-factor ranges from about 1.25 W/m 2 ■ °C (or 0.22
Btu/h ■ ft 2 • °F) for low-e coated, argon-filled, quadruple -pane
windows to 6.25 W/m 2 ■ °C (or 1.1 Btu/h ■ ft 2 • °F) for a single-
pane window with aluminum frames. Determine the range for
the rate of heat loss through a 1.2-m X 1.8-m window of a
house that is maintained at 20°C when the outdoor air temper-
ature is -8°C.
Indoors
20°Cfc
M.
Outdoors
C C
FIGURE P1-138
1-139
Reconsider Problem 1-138. Using EES (or
other) software, plot the rate of heat loss
through the window as a function of the [/-factor. Discuss
the results.
Design and Essay Problems
1-140 Write an essay on how microwave ovens work, and
explain how they cook much faster than conventional ovens.
Discuss whether conventional electric or microwave ovens
consume more electricity for the same task.
1-141 Using information from the utility bills for the coldest
month last year, estimate the average rate of heat loss from
your house for that month. In your analysis, consider the con-
tribution of the internal heat sources such as people, lights, and
appliances. Identify the primary sources of heat loss from your
house and propose ways of improving the energy efficiency of
your house.
1-142 Design a 1200-W electric hair dryer such that the air
temperature and velocity in the dryer will not exceed 50°C and
3/ms, respectively.
1-143 Design an electric hot water heater for a family of
four in your area. The maximum water temperature in the tank
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and the power consumption are not to exceed 60°C and 4 kW,
respectively. There are two showers in the house, and the
flow rate of water through each of the shower heads is about
10 L/min. Each family member takes a 5 -minute shower every
morning. Explain why a hot water tank is necessary, and deter-
mine the proper size of the tank for this family.
1-144 Conduct this experiment to determine the heat transfer
coefficient between an incandescent lightbulb and the sur-
rounding air using a 60-W lightbulb. You will need an indoor-
outdoor thermometer, which can be purchased for about $ 1 in
59
CHAPTER 1
a hardware store, and a metal glue. You will also need a piece
of string and a ruler to calculate the surface area of the light-
bulb. First, measure the air temperature in the room, and then
glue the tip of the thermocouple wire of the thermometer to the
glass of the lightbulb. Turn the light on and wait until the tem-
perature reading stabilizes. The temperature reading will give
the surface temperature of the lightbulb. Assuming 10 percent
of the rated power of the bulb is converted to light, calculate
the heat transfer coefficient from Newton's law of cooling.
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HEAT CONDUCTION
EOUATION
CHAPTER
Heat transfer has direction as well as magnitude. The rate of heat con-
duction in a specified direction is proportional to the temperature gra-
dient, which is the change in temperature per unit length in that
direction. Heat conduction in a medium, in general, is three-dimensional and
time dependent. That is, T = T(x, y, z, t) and the temperature in a medium
varies with position as well as time. Heat conduction in a medium is said to be
steady when the temperature does not vary with time, and unsteady or tran-
sient when it does. Heat conduction in a medium is said to be one-dimensional
when conduction is significant in one dimension only and negligible in the
other two dimensions, two-dimensional when conduction in the third dimen-
sion is negligible, and three-dimensional when conduction in all dimensions
is significant.
We start this chapter with a description of steady, unsteady, and multi-
dimensional heat conduction. Then we derive the differential equation that
governs heat conduction in a large plane wall, a long cylinder, and a sphere,
and generalize the results to three-dimensional cases in rectangular, cylin-
drical, and spherical coordinates. Following a discussion of the boundary con-
ditions, we present the formulation of heat conduction problems and their
solutions. Finally, we consider heat conduction problems with variable ther-
mal conductivity.
This chapter deals with the theoretical and mathematical aspects of heat
conduction, and it can be covered selectively, if desired, without causing a
significant loss in continuity. The more practical aspects of heat conduction
are covered in the following two chapters.
CONTENTS
2-1 Introduction 62
1-1 One-Dimensional Heat
Conduction Equation 68
2-3 General Heat
Conduction Equation 74
2-4 Boundary and
Initial Conditions 77
2-5 Solution of Steady
One-Dimensional Heat
Conduction Problems 86
2-6 Heat Generation in a Solid 97
1-1 Variable Thermal
Conductivity k(T) 104
Topic of Special Interest:
A Brief Review of
Differential Equations 107
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62
HEAT TRANSFER
Magnitude of
temperature
at a point A
(no direction)
80 W/m 2
Magnitude and
direction of heat
flux at the same
point
FIGURE 2-1
Heat transfer has direction as well
as magnitude, and thus it is
a vector quantity.
500 W
^►•2 =
Hot
edium
Cold
medium
L
■ e=-
Cold
tedium
Hot
medium
L
500 W
FIGURE 2-2
Indicating direction for heat transfer
(positive in the positive direction;
negative in the negative direction).
2-1 - INTRODUCTION
In Chapter 1 heat conduction was defined as the transfer of thermal energy
from the more energetic particles of a medium to the adjacent less energetic
ones. It was stated that conduction can take place in liquids and gases as well
as solids provided that there is no bulk motion involved.
Although heat transfer and temperature are closely related, they are of a dif-
ferent nature. Unlike temperature, heat transfer has direction as well as mag-
nitude, and thus it is a vector quantity (Fig. 2-1). Therefore, we must specify
both direction and magnitude in order to describe heat transfer completely at
a point. For example, saying that the temperature on the inner surface of a
wall is 18°C describes the temperature at that location fully. But saying that
the heat flux on that surface is 50 W/m 2 immediately prompts the question "in
what direction?" We can answer this question by saying that heat conduction
is toward the inside (indicating heat gain) or toward the outside (indicating
heat loss).
To avoid such questions, we can work with a coordinate system and indicate
direction with plus or minus signs. The generally accepted convention is that
heat transfer in the positive direction of a coordinate axis is positive and in the
opposite direction it is negative. Therefore, a positive quantity indicates heat
transfer in the positive direction and a negative quantity indicates heat trans-
fer in the negative direction (Fig. 2-2).
The driving force for any form of heat transfer is the temperature difference,
and the larger the temperature difference, the larger the rate of heat transfer.
Some heat transfer problems in engineering require the determination of the
temperature distribution (the variation of temperature) throughout the
medium in order to calculate some quantities of interest such as the local heat
transfer rate, thermal expansion, and thermal stress at some critical locations
at specified times. The specification of the temperature at a point in a medium
first requires the specification of the location of that point. This can be done
by choosing a suitable coordinate system such as the rectangular, cylindrical,
or spherical coordinates, depending on the geometry involved, and a conve-
nient reference point (the origin).
The location of a point is specified as (x, y, z) in rectangular coordinates, as
(r, 4>, z) in cylindrical coordinates, and as (r, <j>, 6) in spherical coordinates,
where the distances x, y, z, and r and the angles cj) and 8 are as shown in Fig-
ure 2-3. Then the temperature at a point (x, y, z) at time t in rectangular coor-
dinates is expressed as T(x, y, z, t). The best coordinate system for a given
geometry is the one that describes the surfaces of the geometry best.
For example, a parallelepiped is best described in rectangular coordinates
since each surface can be described by a constant value of the x-, y-, or
z-coordinates. A cylinder is best suited for cylindrical coordinates since its lat-
eral surface can be described by a constant value of the radius. Similarly, the
entire outer surface of a spherical body can best be described by a constant
value of the radius in spherical coordinates. For an arbitrarily shaped body, we
normally use rectangular coordinates since it is easier to deal with distances
than with angles.
The notation just described is also used to identify the variables involved
in a heat transfer problem. For example, the notation T(x, y, z, implies that
the temperature varies with the space variables x, y, and z as well as time. The
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63
CHAPTER 2
(a) Rectangular coordinates (b) Cylindrical coordinates
(c) Spherical coordinates
FIGURE 2-3
The various distances
and angles involved when
describing the location of a point
in different coordinate systems.
notation T{x), on the other hand, indicates that the temperature varies in the
x-direction only and there is no variation with the other two space coordinates
or time.
Steady versus Transient Heat Transfer
Heat transfer problems are often classified as being steady (also called steady-
state) or transient (also called unsteady). The term steady implies no change
with time at any point within the medium, while transient implies variation
with time or time dependence. Therefore, the temperature or heat flux remains
unchanged with time during steady heat transfer through a medium at any lo-
cation, although both quantities may vary from one location to another
(Fig. 2-4). For example, heat transfer through the walls of a house will be
steady when the conditions inside the house and the outdoors remain constant
for several hours. But even in this case, the temperatures on the inner and
outer surfaces of the wall will be different unless the temperatures inside and
outside the house are the same. The cooling of an apple in a refrigerator, on
the other hand, is a transient heat transfer process since the temperature at any
fixed point within the apple will change with time during cooling. During
transient heat transfer, the temperature normally varies with time as well as
position. In the special case of variation with time but not with position, the
temperature of the medium changes uniformly with time. Such heat transfer
systems are called lumped systems. A small metal object such as a thermo-
couple junction or a thin copper wire, for example, can be analyzed as a
lumped system during a heating or cooling process.
Most heat transfer problems encountered in practice are transient in nature,
but they are usually analyzed under some presumed steady conditions since
steady processes are easier to analyze, and they provide the answers to our
questions. For example, heat transfer through the walls and ceiling of a typi-
cal house is never steady since the outdoor conditions such as the temperature,
the speed and direction of the wind, the location of the sun, and so on, change
constantly. The conditions in a typical house are not so steady either. There-
fore, it is almost impossible to perform a heat transfer analysis of a house ac-
curately. But then, do we really need an in-depth heat transfer analysis? If the
Time = 2 PM
Time = 5 PM
15°C
7°C 12°C
5'C
■e,*e,
(a) Transient
15°C
7°C 15°C
7°C
■e, = e,
(b) Steady-state
FIGURE 2-4
Steady and transient heat
conduction in a plane wall.
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64
HEAT TRANSFER
••65°C
FIGURE 2-5
Two-dimensional heat transfer
in a long rectangular bar.
Primary
direction of
heat transfer
FIGURE 2-6
Heat transfer through the window
of a house can be taken to be
one-dimensional.
purpose of a heat transfer analysis of a house is to determine the proper size of
a heater, which is usually the case, we need to know the maximum rate of heat
loss from the house, which is determined by considering the heat loss from the
house under worst conditions for an extended period of time, that is, during
steady operation under worst conditions. Therefore, we can get the answer to
our question by doing a heat transfer analysis under steady conditions. If the
heater is large enough to keep the house warm under the presumed worst con-
ditions, it is large enough for all conditions. The approach described above is
a common practice in engineering.
Multidimensional Heat Transfer
Heat transfer problems are also classified as being one-dimensional, two-
dimensional, or three-dimensional, depending on the relative magnitudes of
heat transfer rates in different directions and the level of accuracy desired. In
the most general case, heat transfer through a medium is three-dimensional.
That is, the temperature varies along all three primary directions within the
medium during the heat transfer process. The temperature distribution
throughout the medium at a specified time as well as the heat transfer rate at
any location in this general case can be described by a set of three coordinates
such as the x, y, and z in the rectangular (or Cartesian) coordinate system;
the r, <|>, and z in the cylindrical coordinate system; and the r, <j), and in the
spherical (or polar) coordinate system. The temperature distribution in this
case is expressed as T(x, y, z, t), T(r, 4>, z, t), and T(r, 4>, 9, t) in the respective
coordinate systems.
The temperature in a medium, in some cases, varies mainly in two primary
directions, and the variation of temperature in the third direction (and thus
heat transfer in that direction) is negligible. A heat transfer problem in that
case is said to be two-dimensional. For example, the steady temperature dis-
tribution in a long bar of rectangular cross section can be expressed as T(x, y)
if the temperature variation in the z-direction (along the bar) is negligible and
there is no change with time (Fig. 2-5).
A heat transfer problem is said to be one-dimensional if the temperature in
the medium varies in one direction only and thus heat is transferred in one di-
rection, and the variation of temperature and thus heat transfer in other direc-
tions are negligible or zero. For example, heat transfer through the glass of a
window can be considered to be one-dimensional since heat transfer through
the glass will occur predominantly in one direction (the direction normal to
the surface of the glass) and heat transfer in other directions (from one side
edge to the other and from the top edge to the bottom) is negligible (Fig. 2-6).
Likewise, heat transfer through a hot water pipe can be considered to be one-
dimensional since heat transfer through the pipe occurs predominantly in the
radial direction from the hot water to the ambient, and heat transfer along the
pipe and along the circumference of a cross section (z- and cjvdirections) is
typically negligible. Heat transfer to an egg dropped into boiling water is also
nearly one-dimensional because of symmetry. Heat will be transferred to the
egg in this case in the radial direction, that is, along straight lines passing
through the midpoint of the egg.
We also mentioned in Chapter 1 that the rate of heat conduction through a
medium in a specified direction (say, in the A-direction) is proportional to the
temperature difference across the medium and the area normal to the direction
cen58933_ch02.qxd 9/10/2002 8:46 AM Page 65
of heat transfer, but is inversely proportional to the distance in that direction.
This was expressed in the differential form by Fourier's law of heat conduc-
tion for one-dimensional heat conduction as
2 c
-kA
dT
dx
(W)
(2-1)
where k is the thermal conductivity of the material, which is a measure of the
ability of a material to conduct heat, and dT/dx is the temperature gradient,
which is the slope of the temperature curve on a T-x diagram (Fig. 2-7). The
thermal conductivity of a material, in general, varies with temperature. But
sufficiently accurate results can be obtained by using a constant value for
thermal conductivity at the average temperature.
Heat is conducted in the direction of decreasing temperature, and thus
the temperature gradient is negative when heat is conducted in the positive
x-direction. The negative sign in Eq. 2-1 ensures that heat transfer in the posi-
tive x-direction is a positive quantity.
To obtain a general relation for Fourier's law of heat conduction, consider a
medium in which the temperature distribution is three-dimensional. Figure
2-8 shows an isothermal surface in that medium. The heat flux vector at a
point P on this surface must be perpendicular to the surface, and it must point
in the direction of decreasing temperature. If n is the normal of the isothermal
surface at point P, the rate of heat conduction at that point can be expressed by
Fourier's law as
Qn
-kA
BT
dn
(W)
(2-2)
65
CHAPTER 2
FIGURE 2-7
The temperature gradient dT/dx is
simply the slope of the temperature
curve on a T-x diagram.
In rectangular coordinates, the heat conduction vector can be expressed in
terms of its components as
Qn = QJ +Qyj +Q-J
(2-3)
where i, j, and k are the unit vectors, and Q x , Q y , and Q z are the magnitudes
of the heat transfer rates in the x-, y-, and z-directions, which again can be de-
termined from Fourier's law as
Q x = -kA x
dT
dx'
Q y = ~kA y
dT
By'
and Q.
-kA
BT
z Bz
(2-4)
Here A x , A y and A z are heat conduction areas normal to the x-, y-, and
Z-directions, respectively (Fig. 2-8).
Most engineering materials are isotropic in nature, and thus they have the
same properties in all directions. For such materials we do not need to be con-
cerned about the variation of properties with direction. But in anisotropic ma-
terials such as the fibrous or composite materials, the properties may change
with direction. For example, some of the properties of wood along the grain
are different than those in the direction normal to the grain. In such cases the
thermal conductivity may need to be expressed as a tensor quantity to account
for the variation with direction. The treatment of such advanced topics is be-
yond the scope of this text, and we will assume the thermal conductivity of a
material to be independent of direction.
FIGURE 2-8
The heat transfer vector is
always normal to an isothermal
surface and can be resolved into its
components like any other vector.
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66
HEAT TRANSFER
FIGURE 2-9
Heat is generated in the heating coils
of an electric range as a result of the
conversion of electrical energy to heat.
-^tl Sun
Solar
radiation
X
'
Wh
IV
- Solar energy
■T
absorbed by
■V
water
Water
1+
1
L «(*) =
's, absorbed^ '
FIGURE 2-10
The absorption of solar radiation by
water can be treated as heat
generation.
Heat Generation
A medium through which heat is conducted may involve the conversion of
electrical, nuclear, or chemical energy into heat (or thermal) energy. In heat
conduction analysis, such conversion processes are characterized as heat
generation.
For example, the temperature of a resistance wire rises rapidly when elec-
tric current passes through it as a result of the electrical energy being con-
verted to heat at a rate of I 2 R, where / is the current and R is the electrical
resistance of the wire (Fig. 2-9). The safe and effective removal of this heat
away from the sites of heat generation (the electronic circuits) is the subject
of electronics cooling, which is one of the modern application areas of heat
transfer.
Likewise, a large amount of heat is generated in the fuel elements of nuclear
reactors as a result of nuclear fission that serves as the heat source for the nu-
clear power plants. The natural disintegration of radioactive elements in nu-
clear waste or other radioactive material also results in the generation of heat
throughout the body. The heat generated in the sun as a result of the fusion of
hydrogen into helium makes the sun a large nuclear reactor that supplies heat
to the earth.
Another source of heat generation in a medium is exothermic chemical re-
actions that may occur throughout the medium. The chemical reaction in this
case serves as a heat source for the medium. In the case of endothermic reac-
tions, however, heat is absorbed instead of being released during reaction, and
thus the chemical reaction serves as a heat sink. The heat generation term be-
comes a negative quantity in this case.
Often it is also convenient to model the absorption of radiation such as so-
lar energy or gamma rays as heat generation when these rays penetrate deep
into the body while being absorbed gradually. For example, the absorption of
solar energy in large bodies of water can be treated as heat generation
throughout the water at a rate equal to the rate of absorption, which varies
with depth (Fig. 2-10). But the absorption of solar energy by an opaque body
occurs within a few microns of the surface, and the solar energy that pene-
trates into the medium in this case can be treated as specified heat flux on the
surface.
Note that heat generation is a volumetric phenomenon. That is, it occurs
throughout the body of a medium. Therefore, the rate of heat generation in a
medium is usually specified per unit volume and is denoted by g, whose unit
is W/m 3 or Btu/h • ft 3 .
The rate of heat generation in a medium may vary with time as well as po-
sition within the medium. When the variation of heat generation with position
is known, the total rate of heat generation in a medium of volume V can be de-
termined from
l#v
(W)
(2-5)
In the special case of uniform heat generation, as in the case of electric resis-
tance heating throughout a homogeneous material, the relation in Eq. 2-5
reduces to G = gV, where g is the constant rate of heat generation per unit
volume.
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67
CHAPTER 2
EXAMPLE 2-1 Heat Gain by a Refrigerator
In order to size the compressor of a new refrigerator, it is desired to determine
the rate of heat transfer from the kitchen air into the refrigerated space through
the walls, door, and the top and bottom section of the refrigerator (Fig. 2-11).
In your analysis, would you treat this as a transient or steady-state heat transfer
problem? Also, would you consider the heat transfer to be one-dimensional or
multidimensional? Explain.
SOLUTION The heat transfer process from the kitchen air to the refrigerated
space is transient in nature since the thermal conditions in the kitchen and the
refrigerator, in general, change with time. However, we would analyze this prob-
lem as a steady heat transfer problem under the worst anticipated conditions
such as the lowest thermostat setting for the refrigerated space, and the antic-
ipated highest temperature in the kitchen (the so-called design conditions). If
the compressor is large enough to keep the refrigerated space at the desired
temperature setting under the presumed worst conditions, then it is large
enough to do so under all conditions by cycling on and off.
Heat transfer into the refrigerated space is three-dimensional in nature since
heat will be entering through all six sides of the refrigerator. However, heat
transfer through any wall or floor takes place in the direction normal to the sur-
face, and thus it can be analyzed as being one-dimensional. Therefore, this
problem can be simplified greatly by considering the heat transfer to be one-
dimensional at each of the four sides as well as the top and bottom sections,
and then by adding the calculated values of heat transfer at each surface.
Heat transfer
Hi
nz\
B
FIGURE 2-1 1
Schematic for Example 2-1 .
EXAMPLE 2-2 Heat Generation in a Hair Dryer
The resistance wire of a 1200-W hair dryer is 80 cm long and has a diameter of
D = 0.3 cm (Fig. 2-12). Determine the rate of heat generation in the wire per
unit volume, in W/cm 3 , and the heat flux on the outer surface of the wire as a
result of this heat generation.
SOLUTION The power consumed by the resistance wire of a hair dryer is given.
The heat generation and the heat flux are to be determined.
Assumptions Heat is generated uniformly in the resistance wire.
Analysis A 1200-W hair dryer will convert electrical energy into heat in the
wire at a rate of 1200 W. Therefore, the rate of heat generation in a resistance
wire is equal to the power consumption of a resistance heater. Then the rate of
heat generation in the wire per unit volume is determined by dividing the total
rate of heat generation by the volume of the wire,
1200 W
V„
(ttD 2 /4)L [tt(0.3 cm) 2 /4](80 cm)
212 W/cm 3
Similarly, heat flux on the outer surface of the wire as a result of this heat gen-
eration is determined by dividing the total rate of heat generation by the surface
area of the wire,
1200 W
A wire ttDL
tt(0.3 cm)(80 cm)
15.9 W/cm 2
Hair dryer
1200 W
FIGURE 2-12
Schematic for Example 2-2.
cen58933_ch02.qxd 9/10/2002 8:46 AM Page 68
68
HEAT TRANSFER
Discussion Note that heat generation is expressed per unit volume in W/cm 3 or
Btu/h • ft 3 , whereas heat flux is expressed per unit surface area in W/cm 2 or
Btu/h • ft 2 .
G - Volume
/ element
V A' + Ai' £
FIGURE 2-13
One-dimensional heat conduction
through a volume element
in a large plane wall.
2-2 - ONE-DIMENSIONAL
HEAT CONDUCTION EQUATION
Consider heat conduction through a large plane wall such as the wall of a
house, the glass of a single pane window, the metal plate at the bottom of
a pressing iron, a cast iron steam pipe, a cylindrical nuclear fuel element,
an electrical resistance wire, the wall of a spherical container, or a spherical
metal ball that is being quenched or tempered. Heat conduction in these
and many other geometries can be approximated as being one-dimensional
since heat conduction through these geometries will be dominant in one
direction and negligible in other directions. Below we will develop the one-
dimensional heat conduction equation in rectangular, cylindrical, and spheri-
cal coordinates.
Heat Conduction Equation in a Large Plane Wall
Consider a thin element of thickness Ax in a large plane wall, as shown in Fig-
ure 2-13. Assume the density of the wall is p, the specific heat is C, and the
area of the wall normal to the direction of heat transfer is A. An energy bal-
ance on this thin element during a small time interval At can be expressed as
/Rate of heat\
conduction
I atjc /
or
/Rate of heat\
conduction
\ at x + Ax J
\lx x!x + Ax
/ Rate of heat \
generation
inside the
element
/ Rate of change \
of the energy
content of the
element
A£„
At
(2-6)
But the change in the energy content of the element and the rate of heat gen-
eration within the element can be expressed as
A£eien,e„t = E, + Af - E, = mC(T, + A , - T,) = pCAAxiT, + Al - T,)
^element — § 'element — §Al±X
(2-7)
(2-8)
Substituting into Equation 2-6, we get
Q x -Q I + tu + gAAx = pCAAx-
Dividing by AAx gives
1 2.V + A.V Qx , . _ T, + \,
+ g = pC
At
Ax
At
(2-9)
(2-10)
cen58933_ch02.qxd 9/10/2002 8:46 AM Page 69
Taking the limit as Ax — > and At — > yields
1 d Ua dT \ j- ■ r dT
(2-11)
69
CHAPTER 2
since, from the definition of the derivative and Fourier's law of heat conduc-
tion,
e, + A, - e, dQ
lim
Ax->
Ax
dx dx \ dx
(2-12)
Noting that the area A is constant for a plane wall, the one-dimensional tran-
sient heat conduction equation in a plane wall becomes
Variable conductivity:
d , dT\ , . n dT
(2-13)
The thermal conductivity k of a material, in general, depends on the tempera-
ture T (and therefore x), and thus it cannot be taken out of the derivative.
However, the thermal conductivity in most practical applications can be as-
sumed to remain constant at some average value. The equation above in that
case reduces to
Constant conductivity:
PT
dx 2
]_dT
a dt
(2-14)
where the property a = k/pC is the thermal diffusivity of the material and
represents how fast heat propagates through a material. It reduces to the fol-
lowing forms under specified conditions (Fig. 2-14):
(1) Steady-state:
(d/dt = 0)
(2) Transient, no heat generation:
(8 = 0)
(3) Steady-state, no heat generation:
{d/dt = and g = 0)
d 2 T
S
dx 2
k
d 2 T
1 dT
dx 2
a dt
d 2 T
=
dx 2
(2-15)
(2-16)
(2-17)
Note that we replaced the partial derivatives by ordinary derivatives in the
one-dimensional steady heat conduction case since the partial and ordinary
derivatives of a function are identical when the function depends on a single
variable only [T = T(x) in this case].
Heat Conduction Equation in a Long Cylinder
Now consider a thin cylindrical shell element of thickness Ar in a long cylin-
der, as shown in Figure 2-15. Assume the density of the cylinder is p, the spe-
cific heat is C, and the length is L. The area of the cylinder normal to the
direction of heat transfer at any location is A = 2irrL where r is the value of
the radius at that location. Note that the heat transfer area A depends on r in
this case, and thus it varies with location. An energy balance on this thin
cylindrical shell element during a small time interval A; can be expressed as
General, one dimensional:
No Steady-
generation state
Steady, one-dimensional:
d 2 T
dx 2
=
FIGURE 2-14
The simplification of the one-
dimensional heat conduction equation
in a plane wall for the case of constant
conductivity for steady conduction
with no heat generation.
■ Volume element
FIGURE 2-15
One-dimensional heat conduction
through a volume element
in a long cylinder.
cen58933_ch02.qxd 9/10/2002 8:46 AM Page 7C
70
HEAT TRANSFER
' Rate of heat \
conduction
\ at r
( Rate of heat \
conduction +
1 at r + Ar j
Rate of heat \
generation
inside the
element
/ Rate of change \
of the energy
content of the
element
or
Qr ~ Qr + Ar + G c | e
A£„
Ar
(2-18)
The change in the energy content of the element and the rate of heat genera-
tion within the element can be expressed as
A£„
gV A
E, = mC(T, + A , - T,) = pCAAr(T, + A , - T,) (2-19)
= gAAr (2-20)
Substituting into Eq. 2-18, we get
Q r -Q r + Ar + gAAr = pCAAr
At
(2-21)
where A = 2ittL. You may be tempted to express the area at the middle of the
element using the average radius as A = 2ir(r + Ar/2)L. But there is nothing
we can gain from this complication since later in the analysis we will take the
limit as Ar — > and thus the term Ar/2 will drop out. Now dividing the equa-
tion above by AAr gives
1 8 r+ A, ~ Qr
A Ar
P C-
Ar
(2-22)
Taking the limit as Ar — > and Ar — > yields
Adr( kA Tr +8 = PC V
(2-23)
since, from the definition of the derivative and Fourier's law of heat
conduction,
lim
Ar->0
g,. + Ar " Qr dQ
d
Ar
dr dr
-kA
dT
dr
(2-24)
Noting that the heat transfer area in this case is A = 2irrL, the one-
dimensional transient heat conduction equation in a cylinder becomes
Variable conductivity:
1 d / , dT\,
rTr V k Tr ]
P C
dT
dt
(2-25)
For the case of constant thermal conductivity, the equation above reduces to
1 dT
Constant conductivity:
ld_ dT\ .__
>' drV dr k a dt
(2-26)
cen58933_ch02.qxd 9/10/2002 8:46 AM Page 71
71
CHAPTER 2
where again the property a = k/pC is the thermal diffusivity of the material.
Equation 2-26 reduces to the following forms under specified conditions
(Fig. 2-16):
(1) Steady-state: l_d/ dT\ g
(d/dt = 0) '" dr \ r dr) k
(2) Transient, no heat generation: Id/ dT\ _ 1 dT
(g = 0) r dr\ r dr ) ~~ « dt
(3) Steady-state, no heat generation: d I dT
(d/dt = and g = 0) d~r\ r ~dr
(2-27)
(2-28)
(2-29)
Note that we again replaced the partial derivatives by ordinary derivatives in
the one-dimensional steady heat conduction case since the partial and ordinary
derivatives of a function are identical when the function depends on a single
variable only [T = T(r) in this case].
Heat Conduction Equation in a Sphere
Now consider a sphere with density p, specific heat C, and outer radius R. The
area of the sphere normal to the direction of heat transfer at any location is
A = 4ttt 2 , where r is the value of the radius at that location. Note that the heat
transfer area A depends on r in this case also, and thus it varies with location.
By considering a thin spherical shell element of thickness Ar and repeating
the approach described above for the cylinder by using A = 4 tit 2 instead of
A = 2irrL, the one-dimensional transient heat conduction equation for a
sphere is determined to be (Fig. 2-17)
Variable conductivity:
1 d
, r l k —
dr \ dr
P C
dT
dt
(2-30)
(a) The form that is ready to integrate
my-"
(b) The equivalent alternative form
d 2 T dT „
FIGURE 2-16
Two equivalent forms of the
differential equation for the one-
dimensional steady heat conduction in
a cylinder with no heat generation.
FIGURE 2-17
One-dimensional heat conduction
through a volume element in a sphere.
which, in the case of constant thermal conductivity, reduces to
1 dT
Constant conductivity:
L± r 2§T + _.
r 2 dr\ dr k a dt
(2-31)
where again the property a = k/pC is the thermal diffusivity of the material.
It reduces to the following forms under specified conditions:
(1) Steady-state:
(d/dt = 0)
(2) Transient,
no heat generation:
(g = 0)
(3) Steady-state,
no heat generation:
(d/dt = and g = 0)
11 [ r idT
r 2 dr\ dr
\_d_l 2 dT
r 2 dr\ r dr
d_ 2 dT
dr \ dr
~k =
\_dT
a dt
dr 2 dr
(2-32)
(2-33)
(2-34)
where again we replaced the partial derivatives by ordinary derivatives in the
one-dimensional steady heat conduction case.
cen58933_ch02.qxd 9/10/2002 8:46 AM Page 72
72
HEAT TRANSFER
Combined One-Dimensional
Heat Conduction Equation
An examination of the one-dimensional transient heat conduction equations
for the plane wall, cylinder, and sphere reveals that all three equations can be
expressed in a compact form as
d_
; dr
r" k
dT
dr
P C
dT
dt
(2-35)
where n = for a plane wall, n = 1 for a cylinder, and n = 2 for a sphere. In
the case of a plane wall, it is customary to replace the variable r by x. This
equation can be simplified for steady-state or no heat generation cases as
described before.
800 W
FIGURE 2-18
Schematic for Example 2-3.
EXAMPLE 2-3 Heat Conduction through the Bottom of a Pan
Consider a steel pan placed on top of an electric range to cook spaghetti (Fig.
2-18). The bottom section of the pan is L = 0.4 cm thick and has a diameter
of D = 18 cm. The electric heating unit on the range top consumes 800 W of
power during cooking, and 80 percent of the heat generated in the heating ele-
ment is transferred uniformly to the pan. Assuming constant thermal conduc-
tivity, obtain the differential equation that describes the variation of the
temperature in the bottom section of the pan during steady operation.
SOLUTION The bottom section of the pan has a large surface area relative to
its thickness and can be approximated as a large plane wall. Heat flux is ap-
plied to the bottom surface of the pan uniformly, and the conditions on the
inner surface are also uniform. Therefore, we expect the heat transfer through
the bottom section of the pan to be from the bottom surface toward the top,
and heat transfer in this case can reasonably be approximated as being one-
dimensional. Taking the direction normal to the bottom surface of the pan to be
the x-axis, we will have T = T(x) during steady operation since the temperature
in this case will depend on x only.
The thermal conductivity is given to be constant, and there is no heat gener-
ation in the medium (within the bottom section of the pan). Therefore, the dif-
ferential equation governing the variation of temperature in the bottom section
of the pan in this case is simply Eq. 2-17,
d 2 T _
dx 2 '
which is the steady one-dimensional heat conduction equation in rectangular
coordinates under the conditions of constant thermal conductivity and no heat
generation. Note that the conditions at the surface of the medium have no ef-
fect on the differential equation.
EXAMPLE 2-4
Heat Conduction in a Resistance Heater
A 2-kW resistance heater wire with thermal conductivity k = 15 W/m ■ °C, di-
ameter D = 0.4 cm, and length L = 50 cm is used to boil water by immersing
cen58933_ch02.qxd 9/10/2002 8:46 AM Page 73
it in water (Fig. 2-19). Assuming the variation of the thermal conductivity of the
wire with temperature to be negligible, obtain the differential equation that de-
scribes the variation of the temperature in the wire during steady operation.
SOLUTION The resistance wire can be considered to be a very long cylinder
since its length is more than 100 times its diameter. Also, heat is generated
uniformly in the wire and the conditions on the outer surface of the wire are uni-
form. Therefore, it is reasonable to expect the temperature in the wire to vary in
the radial r direction only and thus the heat transfer to be one-dimensional.
Then we will have T = T(r) during steady operation since the temperature in
this case will depend on ronly.
The rate of heat generation in the wire per unit volume can be determined
from
G
G
2000 W
v„
(ttD 2 /4)L [tt(0.004 m) 2 /4](0.5 cm)
0.318 X 10"W/m 3
Noting that the thermal conductivity is given to be constant, the differential
equation that governs the variation of temperature in the wire is simply
Eq. 2-27,
1 d I dT
r dr\ dr
which is the steady one-dimensional heat conduction equation in cylindrical co-
ordinates for the case of constant thermal conductivity. Note again that the con-
ditions at the surface of the wire have no effect on the differential equation.
73
CHAPTER 2
FIGURE 2-19
Schematic for Example 2-4.
EXAMPLE 2-5 Cooling of a Hot Metal Ball in Air
A spherical metal ball of radius R is heated in an oven to a temperature of
600°F throughout and is then taken out of the oven and allowed to cool in am-
bient air at 7" x = 75°F by convection and radiation (Fig. 2-20). The thermal
conductivity of the ball material is known to vary linearly with temperature. As-
suming the ball is cooled uniformly from the entire outer surface, obtain the dif-
ferential equation that describes the variation of the temperature in the ball
during cooling.
SOLUTION The ball is initially at a uniform temperature and is cooled uni-
formly from the entire outer surface. Also, the temperature at any point in the
ball will change with time during cooling. Therefore, this is a one-dimensional
transient heat conduction problem since the temperature within the ball will
change with the radial distance rand the time t. That is, T = T(r, t).
The thermal conductivity is given to be variable, and there is no heat genera-
tion in the ball. Therefore, the differential equation that governs the variation of
temperature in the ball in this case is obtained from Eq. 2-30 by setting the
heat generation term equal to zero. We obtain
r 2 dr
r 2 k
§T
dr
PC
dT
dt
75°F
FIGURE 2-20
Schematic for Example 2-5.
cen58933_ch02.qxd 9/10/2002 8:46 AM Page 74
74
HEAT TRANSFER
which is the one-dimensional transient heat conduction equation in spherical
coordinates under the conditions of variable thermal conductivity and no heat
generation. Note again that the conditions at the outer surface of the ball have
no effect on the differential equation.
FIGURE 2-21
Three-dimensional heat conduction
through a rectangular volume element.
2-3 - GENERAL HEAT CONDUCTION EQUATION
In the last section we considered one-dimensional heat conduction and
assumed heat conduction in other directions to be negligible. Most heat trans-
fer problems encountered in practice can be approximated as being one-
dimensional, and we will mostly deal with such problems in this text.
However, this is not always the case, and sometimes we need to consider heat
transfer in other directions as well. In such cases heat conduction is said to be
multidimensional, and in this section we will develop the governing differen-
tial equation in such systems in rectangular, cylindrical, and spherical coordi-
nate systems.
Rectangular Coordinates
Consider a small rectangular element of length Ax, width Ay, and height Az,
as shown in Figure 2-21. Assume the density of the body is p and the specific
heat is C. An energy balance on this element during a small time interval At
can be expressed as
\ / Rate of heat \
generation
inside the
/ Rate of heat \
conduction at
\ x, y, and z J
Rate of heat
conduction
at x + Ax,
y + Ay, and z + Az I
element
Rate of change \
of the energy
content of
the element
or
Qt+Qy + Qz-Qx + Ax-Qy
Q-.
A£„
At
(2-36)
Noting that the volume of the element is V e i ement = AxAyAz, the change in the
energy content of the element and the rate of heat generation within the ele-
ment can be expressed as
A£„
E-i + At
o 'element
E, = mC(T, + A ,
= gAxAyAz
T,) = pCAxAyAz{T, + At -T t )
Substituting into Eq. 2-36, we get
Qx+Qy + Qz-Qx
Qy + .
Q-.
g AxAyAz = pCAxAyAz
T'f + Af
Ar
Dividing by AxAyAz gives
G.Y+A.V _ Qx 1 Gy + Ay
1
AyAz
e,
Ax
AxAz
Av
1 Q l + Az-
AxAy Az
Qz
PC
T t +At
At
(2-37)
cen58933_ch02.qxd 9/10/2002 8:46 AM Page 75
Noting that the heat transfer areas of the element for heat conduction in the
x, y, and z directions are A x = AyAz, A y = AxAz, and A, = AxAy, respectively,
and taking the limit as Ax, A_y, Az and At — > yields
dx\ dx I dy \ dy
+ l*fH
pC
dT
dt
(2-38)
75
CHAPTER 2
since, from the definition of the derivative and Fourier's law of heat
conduction,
l
Q
+ Ax
Q x
Ax-»
AyAz
Ax
1
Q.
+ A.v ~
e,
Ay-»0
AxAz
Ay
1
Q
+ Az —
Qz
Az^oAxAy Az
1 dQ x
AyAz dx
1 9Qy
AxAz dy
1 dQ z
AxAy dz AxAy dz
1
d
AyAz
dx
1
d
AxAz dy
1
a
-kAyAz
-kAxAz
-kAxAy
i)T
dx
8T
dy
dT
dz
dx \ dx
^- k-
dy\ dy
dz r dz
Equation 2-38 is the general heat conduction equation in rectangular coordi-
nates. In the case of constant thermal conductivity, it reduces to
d 2 T d 2 T d 2 T 8 = 1 dT
dx 2 dy 2 dz 2 k « dt
(2-39)
where the property a = k/pC is again the thermal dijfusivity of the material.
Equation 2-39 is known as the Fourier-Biot equation, and it reduces to these
forms under specified conditions:
(1) Steady-state:
(called the Poisson equation)
(2) Transient, no heat generation:
(called the diffusion equation)
(3) Steady-state, no heat generation:
(called the Laplace equation)
Note that in the special case of one-dimensional heat transfer in the
x-direction, the derivatives with respect to y and z drop out and the equations
above reduce to the ones developed in the previous section for a plane wall
(Fig. 2-22).
d 2 T | d 2 T | d 2 T | g _
dx 2 dy 2 dz 2 k
(2-40)
d 2 T d 2 T d 2 T 1 dT
dx 2 3y 2 dz 2 ~ a dt
(2-41)
d_^ + djT + d^T =0
dx 2 9y 2 dz 2
(2-42)
Cylindrical Coordinates
The general heat conduction equation in cylindrical coordinates can be ob-
tained from an energy balance on a volume element in cylindrical coordinates,
shown in Figure 2-23, by following the steps just outlined. It can also be ob-
tained directly from Eq. 2-38 by coordinate transformation using the follow-
ing relations between the coordinates of a point in rectangular and cylindrical
coordinate systems:
x = r cos <f>, y = r sin 4>, and z = Z
FIGURE 2-22
The three-dimensional heat
conduction equations reduce to the
one-dimensional ones when the
temperature varies in one
dimension only.
FIGURE 2-23
A differential volume element in
cylindrical coordinates.
cen58933_ch02.qxd 9/10/2002 8:46 AM Page 76
76
HEAT TRANSFER
FIGURE 2-24
A differential volume element in
spherical coordinates.
After lengthy manipulations, we obtain
r dr
kr
dT
dr
,11 , dT
r 2 acp \ 5cj>
d_
dz
dT
dz
+ g
n dT
P c s7
(2-43)
Spherical Coordinates
The general heat conduction equations in spherical coordinates can be ob-
tained from an energy balance on a volume element in spherical coordinates,
shown in Figure 2-24, by following the steps outlined above. It can also be
obtained directly from Eq. 2-38 by coordinate transformation using the fol-
lowing relations between the coordinates of a point in rectangular and spheri-
cal coordinate systems:
x = r cos 4> sin G, y = r sin 4> sin G, and
cos G
Again after lengthy manipulations, we obtain
r 2 dr \ dr
1
d
dT
r 2 sin 2 G d<t> V" ^
6 96
k sin 6
dT
ae
P c
dT
dt
(2-44)
Obtaining analytical solutions to these differential equations requires a
knowledge of the solution techniques of partial differential equations, which
is beyond the scope of this introductory text. Here we limit our consideration
to one-dimensional steady-state cases or lumped systems, since they result in
ordinary differential equations.
r
Metal
Heat
loss
600°F
billet
>
T„ = 65°F
FIGURE 2-25
Schematic for Example 2-6.
EXAMPLE 2-6 Heat Conduction in a Short Cylinder
A short cylindrical metal billet of radius R and height h is heated in an oven to
a temperature of 600 C F throughout and is then taken out of the oven and al-
lowed to cool in ambient air at 7" x = 65°F by convection and radiation. Assum-
ing the billet is cooled uniformly from all outer surfaces and the variation of the
thermal conductivity of the material with temperature is negligible, obtain the
differential equation that describes the variation of the temperature in the bil-
let during this cooling process.
SOLUTION The billet shown in Figure 2-25 is initially at a uniform tempera-
ture and is cooled uniformly from the top and bottom surfaces in the z-direction
as well as the lateral surface in the radial r-direction. Also, the temperature at
any point in the ball will change with time during cooling. Therefore, this is a
two-dimensional transient heat conduction problem since the temperature
within the billet will change with the radial and axial distances rand zand with
time t. That is, T= T(r, z, f).
The thermal conductivity is given to be constant, and there is no heat gener-
ation in the billet. Therefore, the differential equation that governs the variation
of temperature in the billet in this case is obtained from Eq. 2-43 by setting
the heat generation term and the derivatives with respect to 4> equal to zero. We
obtain
IA
r dr
kr
+ f
dT
dz
P c
dT
dt
cen58933_ch02.qxd 9/10/2002 8:46 AM Page 77
In the
case
of constc
nt therma
cond
JCt
vity,
it reduces
to
1A
r dr
+
dz 2
_ 1 dT
a dt
which
is the desired
equation.
77
CHAPTER 2
2-A - BOUNDARY AND INITIAL CONDITIONS
The heat conduction equations above were developed using an energy balance
on a differential element inside the medium, and they remain the same re-
gardless of the thermal conditions on the surfaces of the medium. That is, the
differential equations do not incorporate any information related to the condi-
tions on the surfaces such as the surface temperature or a specified heat flux.
Yet we know that the heat flux and the temperature distribution in a medium
depend on the conditions at the surfaces, and the description of a heat transfer
problem in a medium is not complete without a full description of the thermal
conditions at the bounding surfaces of the medium. The mathematical expres-
sions of the thermal conditions at the boundaries are called the boundary
conditions.
From a mathematical point of view, solving a differential equation is essen-
tially a process of removing derivatives, or an integration process, and thus
the solution of a differential equation typically involves arbitrary constants
(Fig. 2-26). It follows that to obtain a unique solution to a problem, we need
to specify more than just the governing differential equation. We need to spec-
ify some conditions (such as the value of the function or its derivatives at
some value of the independent variable) so that forcing the solution to satisfy
these conditions at specified points will result in unique values for the arbi-
trary constants and thus a unique solution. But since the differential equation
has no place for the additional information or conditions, we need to supply
them separately in the form of boundary or initial conditions.
Consider the variation of temperature along the wall of a brick house in
winter. The temperature at any point in the wall depends on, among other
things, the conditions at the two surfaces of the wall such as the air tempera-
ture of the house, the velocity and direction of the winds, and the solar energy
incident on the outer surface. That is, the temperature distribution in a medium
depends on the conditions at the boundaries of the medium as well as the heat
transfer mechanism inside the medium. To describe a heat transfer problem
completely, two boundary conditions must be given for each direction of
the coordinate system along which heat transfer is significant (Fig. 2-27).
Therefore, we need to specify two boundary conditions for one-dimensional
problems, four boundary conditions for two-dimensional problems, and six
boundary conditions for three-dimensional problems. In the case of the wall
of a house, for example, we need to specify the conditions at two locations
(the inner and the outer surfaces) of the wall since heat transfer in this case is
one-dimensional. But in the case of a parallelepiped, we need to specify six
boundary conditions (one at each face) when heat transfer in all three dimen-
sions is significant.
The differential equation:
d 2 T
dx 2
C,JE
T
c,
General solution:
T(x)
Arbitrary constants
Some specific solutions:
T(x) = 2x + 5
T(x) = -x + 12
T(x) = -3
T(x) = 6.2*
FIGURE 2-26
The general solution of a typical
differential equation involves
arbitrary constants, and thus an
infinite number of solutions.
50°C
Some solutions of
d 2 T
dx 2 '
:0
15°C
The only solution
that satisfies
the conditions
7/(0) = 50°C
and T(L) = 15°C.
FIGURE 2-27
To describe a heat transfer problem
completely, two boundary conditions
must be given for each direction along
which heat transfer is significant.
cen58933_ch02.qxd 9/10/2002 8:46 AM Page 7E
78
HEAT TRANSFER
The physical argument presented above is consistent with the mathematical
nature of the problem since the heat conduction equation is second order (i.e.,
involves second derivatives with respect to the space variables) in all direc-
tions along which heat conduction is significant, and the general solution of a
second-order linear differential equation involves two arbitrary constants for
each direction. That is, the number of boundary conditions that needs to be
specified in a direction is equal to the order of the differential equation in that
direction.
Reconsider the brick wall already discussed. The temperature at any point
on the wall at a specified time also depends on the condition of the wall at the
beginning of the heat conduction process. Such a condition, which is usually
specified at time t = 0, is called the initial condition, which is a mathemati-
cal expression for the temperature distribution of the medium initially. Note
that we need only one initial condition for a heat conduction problem regard-
less of the dimension since the conduction equation is first order in time (it in-
volves the first derivative of temperature with respect to time).
In rectangular coordinates, the initial condition can be specified in the gen-
eral form as
T(x, y, z, 0) = f(x, y, z)
(2-45)
150°C
T(x, t)
m
70°C
7X0, = 150°C
T(L, f) = 70°C
FIGURE 2-28
Specified temperature boundary
conditions on both surfaces
of a plane wall.
where the function /(x, y, z) represents the temperature distribution throughout
the medium at time t = 0. When the medium is initially at a uniform tem-
perature of T h the initial condition of Eq. 2-45 can be expressed as
T(x, y, z, 0) = Tj. Note that under steady conditions, the heat conduction equa-
tion does not involve any time derivatives, and thus we do not need to specify
an initial condition.
The heat conduction equation is first order in time, and thus the initial con-
dition cannot involve any derivatives (it is limited to a specified temperature).
However, the heat conduction equation is second order in space coordinates,
and thus a boundary condition may involve first derivatives at the boundaries
as well as specified values of temperature. Boundary conditions most com-
monly encountered in practice are the specified temperature, specified heat
flux, convection, and radiation boundary conditions.
1 Specified Temperature Boundary Condition
The temperature of an exposed surface can usually be measured directly and
easily. Therefore, one of the easiest ways to specify the thermal conditions on
a surface is to specify the temperature. For one-dimensional heat transfer
through a plane wall of thickness L, for example, the specified temperature
boundary conditions can be expressed as (Fig. 2-28)
7/(0, t) = T x
T(L, t) = T 2
(2-46)
where T { and T 2 are the specified temperatures at surfaces at x = and x = L,
respectively. The specified temperatures can be constant, which is the case for
steady heat conduction, or may vary with time.
cen58933_ch02.qxd 9/10/2002 8:46 AM Page 79
2 Specified Heat Flux Boundary Condition
When there is sufficient information about energy interactions at a surface, it
may be possible to determine the rate of heat transfer and thus the heat flux q
(heat transfer rate per unit surface area, W/m 2 ) on that surface, and this infor-
mation can be used as one of the boundary conditions. The heat flux in the
positive x-direction anywhere in the medium, including the boundaries, can be
expressed by Fourier's law of heat conduction as
dT = Heat flux in the \
dx ^positive ^-direction/
(W/m 2 )
(2-47)
Then the boundary condition at a boundary is obtained by setting the specified
heat flux equal to —k(dT/dx) at that boundary. The sign of the specified heat
flux is determined by inspection: positive if the heat flux is in the positive di-
rection of the coordinate axis, and negative if it is in the opposite direction.
Note that it is extremely important to have the correct sign for the specified
heat flux since the wrong sign will invert the direction of heat transfer and
cause the heat gain to be interpreted as heat loss (Fig. 2-29).
For a plate of thickness L subjected to heat flux of 50 W/m 2 into the medium
from both sides, for example, the specified heat flux boundary conditions can
be expressed as
37(0, t)
-k—i = 50
dx
and
dT(L, t)
'' dx
-50
(2-48)
79
CHAPTER 2
Heat
flux
%■■
Conduction
. 37X0,
dx
Conduction
.dT(L,t)
Heat
flux
9 L
FIGURE 2-29
Specified heat flux
boundary conditions on both
surfaces of a plane wall.
Note that the heat flux at the surface at x = L is in the negative x-direction,
and thus it is —50 W/m 2 .
Special Case: Insulated Boundary
Some surfaces are commonly insulated in practice in order to minimize heat
loss (or heat gain) through them. Insulation reduces heat transfer but does not
totally eliminate it unless its thickness is infinity. However, heat transfer
through a properly insulated surface can be taken to be zero since adequate
insulation reduces heat transfer through a surface to negligible levels. There-
fore, a well-insulated surface can be modeled as a surface with a specified
heat flux of zero. Then the boundary condition on a perfectly insulated surface
(at x = 0, for example) can be expressed as (Fig. 2-30)
3T(0, t)
dx
dT(0,
dx
(2-49)
That is, on an insulated surface, the first derivative of temperature with re-
spect to the space variable (the temperature gradient) in the direction normal
to the insulated surface is zero. This also means that the temperature function
must be perpendicular to an insulated surface since the slope of temperature at
the surface must be zero.
Another Special Case: Thermal Symmetry
Some heat transfer problems possess thermal symmetry as a result of the
symmetry in imposed thermal conditions. For example, the two surfaces of a
large hot plate of thickness L suspended vertically in air will be subjected to
Insulation
T(x, t)
97/(0, t)
60°C
dx
T(L, t) = 60°C
FIGURE 2-30
A plane wall with insulation
and specified temperature
boundary conditions.
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HEAT TRANSFER
L^— Center plane
Zero I
slope
Temperature
distribution
(symmetric
about center
plane)
:()
I —
I 2
dT(LI2, f)
dx
FIGURE 2-31
Thermal symmetry boundary
condition at the center plane
of a plane wall.
the same thermal conditions, and thus the temperature distribution in one half
of the plate will be the same as that in the other half. That is, the heat transfer
problem in this plate will possess thermal symmetry about the center plane at
x = L/2. Also, the direction of heat flow at any point in the plate will be
toward the surface closer to the point, and there will be no heat flow across the
center plane. Therefore, the center plane can be viewed as an insulated sur-
face, and the thermal condition at this plane of symmetry can be expressed as
(Fig. 2-31)
dT(L/2, t)
dx
(2-50)
which resembles the insulation or zero heat flux boundary condition. This
result can also be deduced from a plot of temperature distribution with a
maximum, and thus zero slope, at the center plane.
In the case of cylindrical (or spherical) bodies having thermal symmetry
about the center line (or midpoint), the thermal symmetry boundary condition
requires that the first derivative of temperature with respect to r (the radial
variable) be zero at the centerline (or the midpoint).
L
Water
110°C
I!
%
FIGURE 2-32
Schematic for Example 2-7.
EXAMPLE 2-7 Heat Flux Boundary Condition
Consider an aluminum pan used to cook beef stew on top of an electric range.
The bottom section of the pan is L = 0.3 cm thick and has a diameter of D =
20 cm. The electric heating unit on the range top consumes 800 W of power
during cooking, and 90 percent of the heat generated in the heating element is
transferred to the pan. During steady operation, the temperature of the inner
surface of the pan is measured to be 110 C C. Express the boundary conditions
for the bottom section of the pan during this cooking process.
SOLUTION The heat transfer through the bottom section of the pan is from the
bottom surface toward the top and can reasonably be approximated as being
one-dimensional. We take the direction normal to the bottom surfaces of the
pan as the x axis with the origin at the outer surface, as shown in Figure 2-32.
Then the inner and outer surfaces of the bottom section of the pan can be rep-
resented by x = and x = L, respectively. During steady operation, the tem-
perature will depend on x only and thus T = T(x).
The boundary condition on the outer surface of the bottom of the pan at
x = can be approximated as being specified heat flux since it is stated that
90 percent of the 800 W (i.e., 720 W) is transferred to the pan at that surface.
Therefore,
dT(0)
dx
q a
where
<?o
Heat transfer rate
0.720 kW
Bottom surface area tt(0.1 m) 2
22.9 kW/m 2
cen58933_ch02.qxd 9/10/2002 8:46 AM Page 81
The temperature at the inner surface of the bottom of the pan is specified to be
110°C. Then the boundary condition on this surface can be expressed as
T(L) = 110°C
where L = 0.003 m. Note that the determination of the boundary conditions
may require some reasoning and approximations.
81
CHAPTER 2
3 Convection Boundary Condition
Convection is probably the most common boundary condition encountered
in practice since most heat transfer surfaces are exposed to an environment at
a specified temperature. The convection boundary condition is based on a sur-
face energy balance expressed as
I Heat conduction \
at the surface in a
\ selected direction/
' Heat convection \
at the surface in
Uhe same direction/
For one-dimensional heat transfer in the x-direction in a plate of thickness L,
the convection boundary conditions on both surfaces can be expressed as
and
dT(0, t)
dx
dT(L, t)
dx
hJT^- 7X0, r)]
h 2 [T(L, t) - r„ 2 ]
(2-51a)
(2-51 b)
where /?, and h 2 are the convection heat transfer coefficients and T rj3l and T^ 2
are the temperatures of the surrounding mediums on the two sides of the plate,
as shown in Figure 2-33.
In writing Eqs. 2-5 1 for convection boundary conditions, we have selected
the direction of heat transfer to be the positive x-direction at both surfaces. But
those expressions are equally applicable when heat transfer is in the opposite
direction at one or both surfaces since reversing the direction of heat transfer
at a surface simply reverses the signs of both conduction and convection terms
at that surface. This is equivalent to multiplying an equation by — 1 , which has
no effect on the equality (Fig. 2-34). Being able to select either direction as
the direction of heat transfer is certainly a relief since often we do not know
the surface temperature and thus the direction of heat transfer at a surface in
advance. This argument is also valid for other boundary conditions such as the
radiation and combined boundary conditions discussed shortly.
Note that a surface has zero thickness and thus no mass, and it cannot store
any energy. Therefore, the entire net heat entering the surface from one side
must leave the surface from the other side. The convection boundary condi-
tion simply states that heat continues to flow from a body to the surrounding
medium at the same rate, and it just changes vehicles at the surface from con-
duction to convection (or vice versa in the other direction). This is analogous
to people traveling on buses on land and transferring to the ships at the shore.
Convection
Conduction
. dm t)
dx
Conduction
Convection
-k d y ( L ' f) = h 2 lT(L, - 7^.,]
L x
FIGURE 2-33
Convection boundary conditions on
the two surfaces of a plane wall.
Convection
Conduction
hjr^-no,
o,=-»«
*,, r-,
Convection
Conduction
h l [m,f)-T
»i J " * dx
L
X
FIGURE 2-34
The assumed direction of heat transfer
at a boundary has no effect on the
boundary condition expression.
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HEAT TRANSFER
If the passengers are not allowed to wander around at the shore, then the rate
at which the people are unloaded at the shore from the buses must equal the
rate at which they board the ships. We may call this the conservation of "peo-
ple" principle.
Also note that the surface temperatures T(0, t) and T(L, t) are not known
(if they were known, we would simply use them as the specified temperature
boundary condition and not bother with convection). But a surface tempera-
ture can be determined once the solution T(x, t) is obtained by substituting the
value of x at that surface into the solution.
Insulation
FIGURE 2-35
Schematic for Example 2-
EXAMPLE 2-8 Convection and Insulation Boundary Conditions
Steam flows through a pipe shown in Figure 2-35 at an average temperature of
Ty, = 200°C. The inner and outer radii of the pipe are r x = 8 cm and r 2 =
8.5 cm, respectively, and the outer surface of the pipe is heavily insulated. If
the convection heat transfer coefficient on the inner surface of the pipe is
h = 65 W/m 2 • °C, express the boundary conditions on the inner and outer sur-
faces of the pipe during transient periods.
SOLUTION During initial transient periods, heat transfer through the pipe ma-
terial will predominantly be in the radial direction, and thus can be approxi-
mated as being one-dimensional. Then the temperature within the pipe material
will change with the radial distance r and the time t. That is, T = T(r, t).
It is stated that heat transfer between the steam and the pipe at the inner
surface is by convection. Then taking the direction of heat transfer to be the
positive ^direction, the boundary condition on that surface can be expressed as
dT(r u t)
dr
h[T a - T(r,)]
The pipe is said to be well insulated on the outside, and thus heat loss through
the outer surface of the pipe can be assumed to be negligible. Then the bound-
ary condition at the outer surface can be expressed as
dT(r 2 , t)
dr
That is, the temperature gradient must be zero on the outer surface of the pipe
at all times.
4 Radiation Boundary Condition
In some cases, such as those encountered in space and cryogenic applications,
a heat transfer surface is surrounded by an evacuated space and thus there is
no convection heat transfer between a surface and the surrounding medium. In
such cases, radiation becomes the only mechanism of heat transfer between
the surface under consideration and the surroundings. Using an energy bal-
ance, the radiation boundary condition on a surface can be expressed as
(Heat conduction |
at the surface in a
selected direction
[ Radiation exchange \
at the surface in
\ the same direction /
cen58933_ch02.qxd 9/10/2002 8:46 AM Page 83
For one-dimensional heat transfer in the x-direction in a plate of thickness L,
the radiation boundary conditions on both surfaces can be expressed as
(Fig. 2-36)
and
dT(0, t)
dx
dT(L, t)
dx
e i CT U SUIT, 1
e 2 (T[T(L, t) 4
T(0, f) 4
(2-52a)
(2-52b)
5.67 X
tr, 2 are the
where e { and e 2 are the emissivities of the boundary surfaces, a
10~ 8 W/m 2 • K 4 is the Stefan-Boltzmann constant, and T sum , and T,
average temperatures of the surfaces surrounding the two sides of the plate,
respectively. Note that the temperatures in radiation calculations must be ex-
pressed in K or R (not in °C or °F).
The radiation boundary condition involves the fourth power of temperature,
and thus it is a nonlinear condition. As a result, the application of this bound-
ary condition results in powers of the unknown coefficients, which makes it
difficult to determine them. Therefore, it is tempting to ignore radiation ex-
change at a surface during a heat transfer analysis in order to avoid the com-
plications associated with nonlinearity. This is especially the case when heat
transfer at the surface is dominated by convection, and the role of radiation is
minor.
83
CHAPTER 2
Radiation Conduction
37X0, Q
dx
Conduction
surr, 2
Radiation
37X^0 =4 _ r 4 ,
dx 2 K ' surr - 2
L x
FIGURE 2-36
Radiation boundary conditions on
both surfaces of a plane wall.
5 Interface Boundary Conditions
Some bodies are made up of layers of different materials, and the solution of
a heat transfer problem in such a medium requires the solution of the heat
transfer problem in each layer. This, in turn, requires the specification of the
boundary conditions at each interface.
The boundary conditions at an interface are based on the requirements that
(1) two bodies in contact must have the same temperature at the area of con-
tact and (2) an interface (which is a surface) cannot store any energy, and thus
the heat flux on the two sides of an interface must be the same. The boundary
conditions at the interface of two bodies A and B in perfect contact at x = x
can be expressed as (Fig. 2-37)
T A (x , t) = T B (x , t)
and
dT A (x , t)
dx
dT B (x Q , t)
' dx
(2-53)
(2-54)
Material
A
Interface
Material
B
T A (x ,t) = T B (x ,t)
TJx, t)
dT A (x Q , 1) dT B (x Q ,t)
K. - - "d
A dx
B dx
L x
FIGURE 2-37
Boundary conditions at the interface
of two bodies in perfect contact.
where k A and k B are the thermal conductivities of the layers A and B, respec-
tively. The case of imperfect contact results in thermal contact resistance,
which is considered in the next chapter.
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HEAT TRANSFER
6 Generalized Boundary Conditions
So far we have considered surfaces subjected to single mode heat transfer,
such as the specified heat flux, convection, or radiation for simplicity. In gen-
eral, however, a surface may involve convection, radiation, and specified heat
flux simultaneously. The boundary condition in such cases is again obtained
from a surface energy balance, expressed as
IHeat transfer \
to the surface
in all modes /
I Heat transfer \
from the surface
\ in all modes /
(2-55)
This is illustrated in Examples 2-9 and 2-10.
T = 78°F
FIGURE 2-38
Schematic for Example 2-9.
EXAMPLE 2-9
Combined Convection and Radiation Condition
A spherical metal ball of radius r is heated in an oven to a temperature of
600°F throughout and is then taken out of the oven and allowed to cool in am-
bient air at 7"„ = 78°F, as shown in Figure 2-38. The thermal conductivity of
the ball material is k = 8.3 Btu/h • ft • °F, and the average convection heat
transfer coefficient on the outer surface of the ball is evaluated to be h = 4.5
Btu/h ■ ft 2 • °F. The emissivity of the outer surface of the ball is e = 0.6, and the
average temperature of the surrounding surfaces is T smr = 525 R. Assuming the
ball is cooled uniformly from the entire outer surface, express the initial and
boundary conditions for the cooling process of the ball.
SOLUTION The ball is initially at a uniform temperature and is cooled uni-
formly from the entire outer surface. Therefore, this is a one-dimensional tran-
sient heat transfer problem since the temperature within the ball will change
with the radial distance rand the time f. That is, T = T(r, t). Taking the mo-
ment the ball is removed from the oven to be t = 0, the initial condition can be
expressed as
T(r, 0)
600°F
The problem possesses symmetry about the midpoint (r = 0) since the
isotherms in this case will be concentric spheres, and thus no heat will be
crossing the midpoint of the ball. Then the boundary condition at the midpoint
can be expressed as
dT(0, t)
dr
The heat conducted to the outer surface of the ball is lost to the environment by
convection and radiation. Then taking the direction of heat transfer to be the
positive r direction, the boundary condition on the outer surface can be ex-
pressed as
, d1\r , t)
dr
h[T(r ) - rj + eo-[r(r ) 4 - TU
cen58933_ch02.qxd 9/10/2002 8:46 AM Page 85
All the quantities in the above relations are known except the temperatures
and their derivatives at r = and r . Also, the radiation part of the boundary
condition is often ignored for simplicity by modifying the convection heat trans-
fer coefficient to account for the contribution of radiation. The convection coef-
ficient h in that case becomes the combined heat transfer coefficient.
85
CHAPTER 2
EXAMPLE 2-10 Combined Convection, Radiation, and Heat Flux
Consider the south wall of a house that is L = 0.2 m thick. The outer surface of
the wall is exposed to solar radiation and has an absorptivity of a = 0.5 for so-
lar energy. The interior of the house is maintained at 7" xl = 20 C C, while the am-
bient air temperature outside remains at T a2 = 5°C. The sky, the ground, and
the surfaces of the surrounding structures at this location can be modeled as a
surface at an effective temperature of 7" sky = 255 K for radiation exchange on
the outer surface. The radiation exchange between the inner surface of the wall
and the surfaces of the walls, floor, and ceiling it faces is negligible. The con-
vection heat transfer coefficients on the inner and the outer surfaces of the wall
are h 1 = 6 W/m 2 • °C and h 2 = 25 W/m 2 • °C, respectively. The thermal con-
ductivity of the wall material is k = 0.7 W/m • °C, and the emissivity of the
outer surface is e z = 0.9. Assuming the heat transfer through the wall to be
steady and one-dimensional, express the boundary conditions on the inner and
the outer surfaces of the wall.
SOLUTION We take the direction normal to the wall surfaces as the x-axis with
the origin at the inner surface of the wall, as shown in Figure 2-39. The heat
transfer through the wall is given to be steady and one-dimensional, and thus
the temperature depends on x only and not on time. That is, T = T(x).
The boundary condition on the inner surface of the wall at x = is a typical
convection condition since it does not involve any radiation or specified heat
flux. Taking the direction of heat transfer to be the positive x-direction, the
boundary condition on the inner surface can be expressed as
dT(0)
dx
AiLT., - 7X0)]
The boundary condition on the outer surface at x = is quite general as it in-
volves conduction, convection, radiation, and specified heat flux. Again taking
the direction of heat transfer to be the positive x-direction, the boundary condi-
tion on the outer surface can be expressed as
dT{L)
dx
h 2 [T(L) - KJ + e 2 o-[r(L) 4 - TLA - a# solar
where q solar is the incident solar heat flux. Assuming the opposite direction for
heat transfer would give the same result multiplied by -1, which is equivalent
to the relation here. All the quantities in these relations are known except the
temperatures and their derivatives at the two boundaries.
FIGURE 2-39
Schematic for Example 2-10.
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HEAT TRANSFER
Note that a heat transfer problem may involve different kinds of boundary
conditions on different surfaces. For example, a plate may be subject to heat
flux on one surface while losing or gaining heat by convection from the other
surface. Also, the two boundary conditions in a direction may be specified at
the same boundary, while no condition is imposed on the other boundary. For
example, specifying the temperature and heat flux at x = of a plate of thick-
ness L will result in a unique solution for the one-dimensional steady temper-
ature distribution in the plate, including the value of temperature at the surface
x = L. Although not necessary, there is nothing wrong with specifying more
than two boundary conditions in a specified direction, provided that there is
no contradiction. The extra conditions in this case can be used to verify the
results.
S-
Heat transfer problem \
I
Mathematical formulation
(Differential equation and
boundary conditions)
I
General solution of differential equation
I
Application of boundary conditions
I
Solution of the problem
FIGURE 2-40
Basic steps involved in the solution of
heat transfer problems.
2-5 - SOLUTION OF STEADY ONE-DIMENSIONAL
HEAT CONDUCTION PROBLEMS
So far we have derived the differential equations for heat conduction in
various coordinate systems and discussed the possible boundary conditions.
A heat conduction problem can be formulated by specifying the applicable
differential equation and a set of proper boundary conditions.
In this section we will solve a wide range of heat conduction problems in
rectangular, cylindrical, and spherical geometries. We will limit our attention
to problems that result in ordinary differential equations such as the steady
one-dimensional heat conduction problems. We will also assume constant
thermal conductivity, but will consider variable conductivity later in this chap-
ter. If you feel rusty on differential equations or haven't taken differential
equations yet, no need to panic. Simple integration is all you need to solve the
steady one-dimensional heat conduction problems.
The solution procedure for solving heat conduction problems can be sum-
marized as (1) formulate the problem by obtaining the applicable differential
equation in its simplest form and specifying the boundary conditions, (2) ob-
tain the general solution of the differential equation, and (3) apply the bound-
ary conditions and determine the arbitrary constants in the general solution
(Fig. 2-40). This is demonstrated below with examples.
Plane
wall
^
T
2
L x
FIGURE 2-41
Schematic for Example 2-11.
EXAMPLE 2-11
Heat Conduction in a Plane Wall
Consider a large plane wall of thickness L = 0.2 m, thermal conductivity k =
1.2 W/m ■ °C, and surface area A = 15 m 2 . The two sides of the wall are main-
tained at constant temperatures of 7", = 120 C C and T 2 = 50°C, respectively, as
shown in Figure 2-41. Determine (a) the variation of temperature within the
wall and the value of temperature at x = 0.1 m and (b) the rate of heat con-
duction through the wall under steady conditions.
SOLUTION A plane wall with specified surface temperatures is given. The vari-
ation of temperature and the rate of heat transfer are to be determined.
Assumptions 1 Heat conduction is steady. 2 Heat conduction is one-
dimensional since the wall is large relative to its thickness and the thermal
cen58933_ch02.qxd 9/10/2002 8:46 AM Page 87
conditions on both sides are uniform. 3 Thermal conductivity is constant.
4 There is no heat generation.
Properties The thermal conductivity is given to be k = 1.2 W/m • °C.
Analysis (a) Taking the direction normal to the surface of the wall to be the
^-direction, the differential equation for this problem can be expressed as
d-T _
dx 2 '
with boundary conditions
7\0)
T(L) = T 2 = 50°C
The differential equation is linear and second order, and a quick inspection of
it reveals that it has a single term involving derivatives and no terms involving
the unknown function 7"as a factor. Thus, it can be solved by direct integration.
Noting that an integration reduces the order of a derivative by one, the general
solution of the differential equation above can be obtained by two simple suc-
cessive integrations, each of which introduces an integration constant.
Integrating the differential equation once with respect to x yields
dT
dx
C,
where C 1 is an arbitrary constant. Notice that the order of the derivative went
down by one as a result of integration. As a check, if we take the derivative of
this equation, we will obtain the original differential equation. This equation is
not the solution yet since it involves a derivative.
Integrating one more time, we obtain
T(x)
Co
which is the general solution of the differential equation (Fig. 2-42). The gen-
eral solution in this case resembles the general formula of a straight line whose
slope is C x and whose value at x = is C 2 . This is not surprising since the sec-
ond derivative represents the change in the slope of a function, and a zero sec-
ond derivative indicates that the slope of the function remains constant.
Therefore, any straight line is a solution of this differential equation.
The general solution contains two unknown constants Cj and C 2 , and thus we
need two equations to determine them uniquely and obtain the specific solu-
tion. These equations are obtained by forcing the general solution to satisfy the
specified boundary conditions. The application of each condition yields one
equation, and thus we need to specify two conditions to determine the con-
stants C 1 and C 2 .
When applying a boundary condition to an equation, all occurrences of the
dependent and independent variables and any derivatives are replaced by the
specified values. Thus the only unknowns in the resulting equations are the ar-
bitrary constants.
The first boundary condition can be interpreted as in the general solution, re-
place all the x's by zero and T(x) by T x . That is (Fig. 2-43),
7X0) = C, X + c 2
c,
87
CHAPTER 2
Differential equation:
dx 1
=
Integrate:
dT
dx
= c,
Integrate again:
T(x) =
General
C t x + C,
Arbitrary
solution
constants
FIGURE 2-42
Obtaining the general solution of a
simple second order differential
equation by integration.
Boundary condition:
T(0) = r,
General solution:
T(x) = C,x + C 2
Applying the boundary condition:
C,
T(x)
1
C,x
1
Substituting:
T, = C, X + C, ->
c,
It cannot involve x or T(x) after the
boundary condition is applied.
FIGURE 2-43
When applying a boundary condition
to the general solution at a specified
point, all occurrences of the dependent
and independent variables should be
replaced by their specified values
at that point.
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88
HEAT TRANSFER
The second boundary condition can be interpreted as in the general solution, re-
place all the x's by L and T{x) by T 2 . That is,
T(L) = C X L + C 2
C,L + T,
C,
r,
Substituting the C x and C 2 expressions into the general solution, we obtain
T — T
T(x)= 2 L - x + T, (2-56)
which is the desired solution since it satisfies not only the differential equation
but also the two specified boundary conditions. That is, differentiating Eq.
2-56 with respect to x twice will give d 2 Tldx 2 , which is the given differential
equation, and substituting x = and x = L into Eq. 2-56 gives 7(0) = T x and
T(L) = T z , respectively, which are the specified conditions at the boundaries.
Substituting the given information, the value of the temperature at x = 0.1 m
is determined to be
(50 - 120)°C
T(0.l m) = — — (0.1 m) + 120°C = 85°C
(b) The rate of heat conduction anywhere in the wall is determined from
Fourier's law to be
Gv
-kA
dT
dx
-kAC,
-kA-
r,
kA-
T 2
(2-57)
The numerical value of the rate of heat conduction through the wall is deter-
mined by substituting the given values to be
T,-T 2 „ (120 - 50)°C
Q = kA ' = (1.2W/m-°C)(15m 2 ) — — = 6300 W
Discussion Note that under steady conditions, the rate of heat conduction
through a plane wall is constant.
EXAMPLE 2-12 A Wall with Various Sets of Boundary Conditions
Consider steady one-dimensional heat conduction in a large plane wall of thick-
ness L and constant thermal conductivity k with no heat generation. Obtain ex-
pressions for the variation of temperature within the wall for the following pairs
of boundary conditions (Fig. 2-44):
(a) -k
(b) -k
(c) -k
dT(0)
dx
dT(0)
dx
dT(0)
dx
q = 40 W/cm 2
and
7T[0) = T
= 15°C
4 = 40 W/cm 2
and
k dT(L)
dx
= q L = -25 W/cm
q = 40 W/cm 2
and
dT(L)
k ^
= q a = 40 W/cm 2
cen58933_ch02.qxd 9/10/2002 8:46 AM Page 89
15°C
40 W/cm 7
(a)
Plane
wall
T(x)
I W/cm 7
Plane
.. —
wall
- —
■"
T(x)
- — 2f
0<
- —
L
X
(l>)
40 W/cm 7
25 W/cm 2
(c)
89
CHAPTER 2
Plane
wall
T(x)
40 W/cnr
FIGURE 2-44
Schematic for Example 2-12.
SOLUTION This is a steady one-dimensional heat conduction problem with
constant thermal conductivity and no heat generation in the medium, and the
heat conduction equation in this case can be expressed as (Eq. 2-17)
cV-T _
dx 2 '
whose general solution was determined in the previous example by direct inte-
gration to be
T(x)
C,
where C 1 and C 2 are two arbitrary integration constants. The specific solutions
corresponding to each specified pair of boundary conditions are determined as
follows.
(a) In this case, both boundary conditions are specified at the same boundary
at x = 0, and no boundary condition is specified at the other boundary at x = L.
Noting that
dT
dx
C,
the application of the boundary conditions gives
rf7T0)
dx
<?o
and
ZY0)
-kCi = q
c,xo + c,
c,
<7o
' k
Co
Substituting, the specific solution in this case is determined to be
T(x) =~t+T
Therefore, the two boundary conditions can be specified at the same boundary,
and it is not necessary to specify them at different locations. In fact, the fun-
damental theorem of linear ordinary differential equations guarantees that a
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90
HEAT TRANSFER
Differential equation:
T"(x) =
General solution:
T{x) = C,x + C
i
(a) Unique solution:
-kT'(0) = q } T . ._ _
r(0) = r J K '
la
k
x + T
(b) No solution:
-kT'(0) = q o \
-kT(L) = q L \ 1(Xy
= None
(c) Multiple solutions:
-kT (L) = q \
la
k
x + C,
f
Arbitrary
FIGURE 2-45
A boundary-value problem may have a
unique solution, infinitely many
solutions, or no solutions at all.
Resistance heater
1200 W
Base plate
FIGURE 2-46
Schematic for Example 2-13.
unique solution exists when both conditions are specified at the same location.
But no such guarantee exists when the two conditions are specified at different
boundaries, as you will see below.
(b) In this case different heat fluxes are specified at the two boundaries. The
application of the boundary conditions gives
and
dT(0)
dx
dT{L)
dx
Qo
1l
-kC x = q
-kCi = 4l
c,
c,
k
Ik
k
Since q + q L and the constant Cj cannot be equal to two different things at
the same time, there is no solution in this case. This is not surprising since this
case corresponds to supplying heat to the plane wall from both sides and ex-
pecting the temperature of the wall to remain steady (not to change with time).
This is impossible.
(c) In this case, the same values for heat flux are specified at the two bound-
aries. The application of the boundary conditions gives
and
dT(0)
dx
dT(L)
dx
<7o
9o
-kC x = q
kC x — q
C,
c,
<7o
' k
<7o
' k
Thus, both conditions result in the same value for the constant C 1 , but no value
for C z . Substituting, the specific solution in this case is determined to be
T(x)
4o
C,
which is not a unique solution since C 2 is arbitrary. This solution represents a
family of straight lines whose slope is -q /k. Physically, this problem corre-
sponds to requiring the rate of heat supplied to the wall at x = be equal to the
rate of heat removal from the other side of the wall at x = L. But this is a con-
sequence of the heat conduction through the wall being steady, and thus the
second boundary condition does not provide any new information. So it is not
surprising that the solution of this problem is not unique. The three cases dis-
cussed above are summarized in Figure 2-45.
EXAMPLE 2-13
Heat Conduction in the Base Plate of an Iron
Consider the base plate of a 1200-W household iron that has a thickness of
L = 0.5 cm, base area of A = 300 cm 2 , and thermal conductivity of k =
15 W/m ■ C C. The inner surface of the base plate is subjected to uniform heat
flux generated by the resistance heaters inside, and the outer surface loses
heat to the surroundings at 7~ x = 20°C by convection, as shown in Figure 2-46.
cen58933_ch02.qxd 9/10/2002 8:46 AM Page 91
Taking the convection heat transfer coefficient to be h = 80 W/m 2 • °C and
disregarding heat loss by radiation, obtain an expression for the variation of
temperature in the base plate, and evaluate the temperatures at the inner and
the outer surfaces.
SOLUTION The base plate of an iron is considered. The variation of tempera-
ture in the plate and the surface temperatures are to be determined.
Assumptions 1 Heat transfer is steady since there is no change with time.
2 Heat transfer is one-dimensional since the surface area of the base plate is
large relative to its thickness, and the thermal conditions on both sides are uni-
form. 3 Thermal conductivity is constant. 4 There is no heat generation in the
medium. 5 Heat transfer by radiation is negligible. 6 The upper part of the iron
is well insulated so that the entire heat generated in the resistance wires is
transferred to the base plate through its inner surface.
Properties The thermal conductivity is given to be k = 15 W/m • °C.
Analysis The inner surface of the base plate is subjected to uniform heat flux
at a rate of
1o
Qo
1200 W
0.03 m 2
40,000 W/m 2
The outer side of the plate is subjected to the convection condition. Taking the
direction normal to the surface of the wall as the x-direction with its origin on
the inner surface, the differential equation for this problem can be expressed as
(Fig. 2-47)
d 2 T
dx 2 '
with the boundary conditions
dlXP)
dx
q = 40,000 W/m 2
dT(L)
_,_ = /![r(L) _ rj
The general solution of the differential equation is again obtained by two suc-
cessive integrations to be
dT
dx
C,
and
T(x) = C x x + C 2 (a)
where C 1 and C 2 are arbitrary constants. Applying the first boundary condition,
dx
qo
-kC x = q Q
C,
Noting that dT/dx = C 1 and T(L) = C 1 L + C 2 , the application of the second
boundary condition gives
91
CHAPTER 2
Heat flux
%--
Base plate
Conduction
rf7\0)
dx
Conduction
h
T-,
Convection
-k^&-=h[T(L)-T a ,
FIGURE 2-47
The boundary conditions on the base
plate of the iron discussed
in Example 2-13.
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92
HEAT TRANSFER
dT(L)
kC x = h[(C t L + C 2 ) - rj
Substituting C 1 = - q /k and solving for C 2 , we obtain
Qo <7o
h k
Now substituting Cj and C 2 into the general solution (a) gives
'L- jc , V
T(x) = T m + 4,
h
(W
which is the solution for the variation of the temperature in the plate. The tem-
peratures at the inner and outer surfaces of the plate are determined by substi-
tuting x = and x = L, respectively, into the relation (£>):
7X0) = T^ + q
20°C + (40,000 W/m 2 )
1
0.005 m
15W/m-°C 80W/m 2 -°C
533°C
and
T(L)
«.0 +
20°C +
40,000 W/m 2
80 W/m 2 • °C
520°C
Discussion Note that the temperature of the inner surface of the base plate
will be 13°C higher than the temperature of the outer surface when steady op-
erating conditions are reached. Also note that this heat transfer analysis enables
us to calculate the temperatures of surfaces that we cannot even reach. This ex-
ample demonstrates how the heat flux and convection boundary conditions are
applied to heat transfer problems.
FIGURE 2-48
Schematic for Example 2-14.
EXAMPLE 2-14
Heat Conduction in a Solar Heated Wall
Consider a large plane wall of thickness L = 0.06 m and thermal conductivity
k = 1.2 W/m • °C in space. The wall is covered with white porcelain tiles that
have an emissivity of e = 0.85 and a solar absorptivity of a = 0.26, as shown
in Figure 2-48. The inner surface of the wall is maintained at T x = 300 K at all
times, while the outer surface is exposed to solar radiation that is incident at a
rate of q s
800 W/m 2 . The outer surface is also losing heat by radiation to
deep space at K. Determine the temperature of the outer surface of the wall
and the rate of heat transfer through the wall when steady operating conditions
are reached. What would your response be if no solar radiation was incident on
the surface?
SOLUTION A plane wall in space is subjected to specified temperature on one
side and solar radiation on the other side. The outer surface temperature and
the rate of heat transfer are to be determined.
cen58933_ch02.qxd 9/10/2002 8:46 AM Page 93
Assumptions 1 Heat transfer is steady since there is no change with time.
2 Heat transfer is one-dimensional since the wall is large relative to its
thickness, and the thermal conditions on both sides are uniform. 3 Thermal
conductivity is constant. 4 There is no heat generation.
Properties The thermal conductivity is given to be k = 1.2 W/m • °C.
Analysis Taking the direction normal to the surface of the wall as the
^-direction with its origin on the inner surface, the differential equation for this
problem can be expressed as
d 2 T _
dx 2 '
with boundary conditions
7X0)
dT(L)
-k—j- L = su[T(Lf
300 K
spaceJ ^s(
where 7" space = 0. The general solution of the differential equation is again ob-
tained by two successive integrations to be
T(x)
C,
(a)
where C x and C 2 are arbitrary constants. Applying the first boundary condition
yields
7X0)
C,
C,
Noting that dT/dx = C x and T(L) = C^ + C 2 = CjL + T lt the application of
the second boundary conditions gives
dT(L)
dx
euT(L) 4 - aq st
kC [ = s(j{C x L + T { ) 4 — <xq si
Although C 1 is the only unknown in this equation, we cannot get an explicit ex-
pression for it because the equation is nonlinear, and thus we cannot get a
closed-form expression for the temperature distribution. This should explain
why we do our best to avoid nonlinearities in the analysis, such as those asso-
ciated with radiation.
Let us back up a little and denote the outer surface temperature by T{L) = T L
instead of T[L) = C X L + T v The application of the second boundary condition
in this case gives
dT(L)
~k—H L = saTiLf
dx
a( 7solar
-kC,
atfs,
Solving for C x gives
C,
a^soiar ~ eoT L 4
Now substituting Cj and C 2 into the general solution (a), we obtain
K^solar ~ «t7 l 4
T(x)
■x + T,
(«
(c)
93
CHAPTER 2
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94
HEAT TRANSFER
(1) Rearrange the equation to be solved:
T L = 310.4 - 0.240975(— £-)
\100/
The equation is in the proper form since the
left side consists of T L only.
(2) Guess the value ofT D say 300 K, and
substitute into the right side of the equation.
It gives
T L = 290.2 K
(3) Now substitute this value ofT L into the
right side of the equation and get
T L = 293.1 K
(4) Repeat step (3) until convergence to
desired accuracy is achieved. The
subsequent iterations give
T L = 292.6 K
T L = 292.7 K
T L = 292.7 K
Therefore, the solution is T L = 292.7 K. The
result is independent of the initial guess.
FIGURE 2-49
A simple method of solving a
nonlinear equation is to arrange the
equation such that the unknown is
alone on the left side while everything
else is on the right side, and to iterate
after an initial guess until
convergence.
FIGURE 2-50
Schematic for Example 2-15.
which is the solution for the variation of the temperature in the wall in terms of
the unknown outer surface temperature T L . At x = £ it becomes
°"?solar
suTt
L+T,
(d)
which is an implicit relation for the outer surface temperature T L . Substituting
the given values, we get
0.26 X (800 W/m 2 ) - 0.85 X (5.67 X 10- 8 W/m 2 ■ K 4 ) Tt
1.2 W/m • K
(0.06 m) + 300 K
which simplifies to
0.240975
100/
This equation can be solved by one of the several nonlinear equation solvers
available (or by the old fashioned trial-and-error method) to give (Fig. 2-49)
T L = 292.7 K
Knowing the outer surface temperature and knowing that it must remain con-
stant under steady conditions, the temperature distribution in the wall can be
determined by substituting the T L value above into Eq. (c):
m
0.26 X (800 W/m 2 ) - 0.85 X (5.67 X 10~ 8 W/m 2 • K 4 )(292.7 K) 4
which simplifies to
1.2 W/m ■ K
T(x) = (-121.5 K/m)x + 300 K
300 K
Note that the outer surface temperature turned out to be lower than the inner
surface temperature. Therefore, the heat transfer through the wall will be toward
the outside despite the absorption of solar radiation by the outer surface. Know-
ing both the inner and outer surface temperatures of the wall, the steady rate of
heat conduction through the wall can be determined from
(1.2 W/m -K)
(300 - 292.7) K
0.06 m
146 W/m 2
Discussion In the case of no incident solar radiation, the outer surface tem-
perature, determined from Eq. (d) by setting <j S0 | ar = 0, will be T L = 284.3 K. It
is interesting to note that the solar energy incident on the surface causes the
surface temperature to increase by about 8 K only when the inner surface tem-
perature of the wall is maintained at 300 K.
EXAMPLE 2-15 Heat Loss through a Steam Pipe
Consider a steam pipe of length L = 20 m, inner radius r x = 6 cm, outer radius
r 2 = 8 cm, and thermal conductivity k = 20 W/m • °C, as shown in Figure
2-50. The inner and outer surfaces of the pipe are maintained at average tem-
peratures of 7*1 = 150°C and T 2 = 60°C, respectively. Obtain a general relation
cen58933_ch02.qxd 9/10/2002 8:46 AM Page 95
95
CHAPTER 2
for the temperature distribution inside the pipe under steady conditions, and
determine the rate of heat loss from the steam through the pipe.
SOLUTION A steam pipe is subjected to specified temperatures on its
surfaces. The variation of temperature and the rate of heat transfer are to be
determined.
Assumptions 1 Heat transfer is steady since there is no change with time.
2 Heat transfer is one-dimensional since there is thermal symmetry about the
centerline and no variation in the axial direction, and thus T = T(r). 3 Thermal
conductivity is constant. 4 There is no heat generation.
Properties The thermal conductivity is given to be k = 20 W/m • °C.
Analysis The mathematical formulation of this problem can be expressed as
d_ dT
dr \ dr
with boundary conditions
T(r 2 )
150°C
60°C
Integrating the differential equation once with respect to rgives
dT „
where Cj is an arbitrary constant. We now divide both sides of this equation by
a - to bring it to a readily integrable form,
dT
dr
C,
Again integrating with respect to /"gives (Fig. 2-51)
T(r) = C, In r + C 2
(a)
We now apply both boundary conditions by replacing all occurrences of rand
T(r) in Eq. (a) with the specified values at the boundaries. We get
T(A) = r,
T(r 2 ) = T 2
C l In r t + C 2 = Tj
C, In r, + C = T 2
which are two equations in two unknowns, C : and C 2 . Solving them simultane-
ously gives
C,
ln(r 2 /7-i)
and
C,
t 2 - r,
ln(r 2 /r{)
Substituting them into Eq. (a) and rearranging, the variation of temperature
within the pipe is determined to be
\ln(r 2 'rij "
T { ) + T {
(2-58)
The rate of heat loss from the steam is simply the total rate of heat conduction
through the pipe, and is determined from Fourier's law to be
Differential equation:
Integrate:
m=°
dT
dr
Divide by r (r + 0):
dT = C 1
dr ''
Integrate again:
T(r) = C, In r + C 2
which is the general solution.
FIGURE 2-51
Basic steps involved in the solution
of the steady one-dimensional
heat conduction equation in
cylindrical coordinates.
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96
HEAT TRANSFER
x£ cylinder
-kA
dT
dr
-k(2irrL) ■
-2<nkLC, = 2-nkL , , , '
ln(r 2 /r,)
(2-59)
The numerical value of the rate of heat conduction through the pipe is deter-
mined by substituting the given values
(150 - 60)°C
Q = 2.(20 W/m ■ °C)(20 m) ln(Q Q8/Q Q6) = 786 kW
DISCUSSION Note that the total rate of heat transfer through a pipe is con-
stant, but the heat flux is not since it decreases in the direction of heat trans-
fer with increasing radius since q = Q/(2nrL).
FIGURE 2-52
Schematic for Example 2-16.
EXAMPLE 2-16 Heat Conduction through a Spherical Shell
Consider a spherical container of inner radius r x = 8 cm, outer radius r z =
10 cm, and thermal conductivity k = 45 W/m • °C, as shown in Figure 2-52.
The inner and outer surfaces of the container are maintained at constant tem-
peratures of 7i = 200°C and T 2 = 80°C, respectively, as a result of some chem-
ical reactions occurring inside. Obtain a general relation for the temperature
distribution inside the shell under steady conditions, and determine the rate of
heat loss from the container.
SOLUTION A spherical container is subjected to specified temperatures on its
surfaces. The variation of temperature and the rate of heat transfer are to be
determined.
Assumptions 1 Heat transfer is steady since there is no change with time.
2 Heat transfer is one-dimensional since there is thermal symmetry about the
midpoint, and thus T = T[r). 3 Thermal conductivity is constant. 4 There is no
heat generation.
Properties The thermal conductivity is given to be k = 45 W/m • °C.
Analysis The mathematical formulation of this problem can be expressed as
d_l 2 dT
dr \ dr
with boundary conditions
T{r0 = T { = 200°C
T(r 2 ) = T 2 = 80°C
Integrating the differential equation once with respect to /-yields
,dT
dr
C,
where C x is an arbitrary constant. We now divide both sides of this equation by
r 2 to bring it to a readily integrable form,
dT = C±
dr r 2
cen58933_ch02.qxd 9/10/2002 8:46 AM Page 97
Again integrating with respect to r gives
T(r) = - -y- + C 2
(a)
We now apply both boundary conditions by replacing all occurrences of rand
T(r) in the relation above by the specified values at the boundaries. We get
T(r 2 ) = T 2
C,
r 2
Co
which are two equations in two unknowns, C, and C 2 . Solving them simultane-
ously gives
C,
r,
■ (Tj — T 2 ) and C 2
r 2 T 2 - r,T,
Substituting into Eq. (a), the variation of temperature within the spherical shell
is determined to be
T(r)
r(r 2 - /-,)
(T t
r 2 T 2 - r.r,
(2-60)
The rate of heat loss from the container is simply the total rate of heat conduc-
tion through the container wall and is determined from Fourier's law
Sc spher
sphere
-kA
dT
dr
Q
-fe(4irr 2 ) — = -\iskC x = A^kr x r 2 ■
r-
(2-61)
The numerical value of the rate of heat conduction through the wall is deter-
mined by substituting the given values to be
(200 - 80)°C
Q = 4tt(45 W/m • °C)(0.08 m)(0.10 m) ( q 10 _ g } m = 27,140 W
Discussion Note that the total rate of heat transfer through a spherical shell is
constant, but the heat flux, q = QIAur 2 , is not since it decreases in the direc-
tion of heat transfer with increasing radius as shown in Figure 2-53.
2-6 HEAT GENERATION IN A SOLID
Many practical heat transfer applications involve the conversion of some form
of energy into thermal energy in the medium. Such mediums are said to in-
volve internal heat generation, which manifests itself as a rise in temperature
throughout the medium. Some examples of heat generation are resistance
heating in wires, exothermic chemical reactions in a solid, and nuclear reac-
tions in nuclear fuel rods where electrical, chemical, and nuclear energies are
converted to heat, respectively (Fig. 2-54). The absorption of radiation
throughout the volume of a semitransparent medium such as water can also be
considered as heat generation within the medium, as explained earlier.
97
CHAPTER 2
«2-
Q 2
27.14kW
27.14kW
: 337.5 kW/m-
: 216.0 kW/m-
\ 2 4rc(0.10mr
FIGURE 2-53
During steady one-dimensional
heat conduction in a spherical (or
cylindrical) container, the total rate
of heat transfer remains constant,
but the heat flux decreases with
increasing radius.
Electric
resistance
wires
FIGURE 2-54
Heat generation in solids is
commonly encountered in practice.
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98
HEAT TRANSFER
Heat generation is usually expressed per unit volume of the medium, and is
denoted by g, whose unit is W/m 3 . For example, heat generation in an electri-
cal wire of outer radius r and length L can be expressed as
V„
I 2 R e
vr}L
(W/m 3 )
(2-62)
h,T,
Q = E
FIGURE 2-55
At steady conditions, the entire heat
generated in a solid must leave the
solid through its outer surface.
where / is the electric current and R e is the electrical resistance of the wire.
The temperature of a medium rises during heat generation as a result of the
absorption of the generated heat by the medium during transient start-up
period. As the temperature of the medium increases, so does the heat transfer
from the medium to its surroundings. This continues until steady operating
conditions are reached and the rate of heat generation equals the rate of heat
transfer to the surroundings. Once steady operation has been established, the
temperature of the medium at any point no longer changes.
The maximum temperature T mm in a solid that involves uniform heat gener-
ation will occur at a location farthest away from the outer surface when the
outer surface of the solid is maintained at a constant temperature T s . For ex-
ample, the maximum temperature occurs at the midplane in a plane wall, at
the centerline in a long cylinder, and at the midpoint in a sphere. The temper-
ature distribution within the solid in these cases will be symmetrical about the
center of symmetry.
The quantities of major interest in a medium with heat generation are the
surface temperature T s and the maximum temperature T max that occurs in the
medium in steady operation. Below we develop expressions for these two
quantities for common geometries for the case of uniform heat generation
(g = constant) within the medium.
Consider a solid medium of surface area A s , volume V, and constant thermal
conductivity k, where heat is generated at a constant rate of g per unit volume.
Heat is transferred from the solid to the surrounding medium at T„, with a
constant heat transfer coefficient of h. All the surfaces of the solid are main-
tained at a common temperature T s . Under steady conditions, the energy bal-
ance for this solid can be expressed as (Fig. 2-55)
( Rate of ^
heat transfer
I from the solid /
Rate of 1
energy generation
\ within the solid j
(2-63)
or
Q =gV
(W)
(2-64)
Disregarding radiation (or incorporating it in the heat transfer coefficient h),
the heat transfer rate can also be expressed from Newton's law of cooling as
Q = hA s (T, - r„)
(W)
(2-65)
Combining Eqs. 2-64 and 2-65 and solving for the surface temperature
T s gives
T=T^ +
gV
hA,
(2-66)
cen58933_ch02.qxd 9/10/2002 8:46 AM Page 99
For a large plane wall of thickness 2L (A s = 2j4 wall and V = 2LA wall ), a long
solid cylinder of radius r (A s = 2irr L and V = Tir^L), and a solid sphere of
radius r (A s = 4ttt„ and V = |irr|), Eq. 2-66 reduces to
h
2h
T — T A-
J s, sphere * <» ni.
T = T 4-
-* 5, plane wall x °° ' /
T = T +
-* 5, cylinder L °o ' ^^
(2-67)
(2-68)
(2-69)
Note that the rise in surface temperature T s is due to heat generation in the
solid.
Reconsider heat transfer from a long solid cylinder with heat generation.
We mentioned above that, under steady conditions, the entire heat generated
within the medium is conducted through the outer surface of the cylinder.
Now consider an imaginary inner cylinder of radius r within the cylinder
(Fig. 2-56). Again the heat generated within this inner cylinder must be equal
to the heat conducted through the outer surface of this inner cylinder. That is,
from Fourier's law of heat conduction,
-kA
dT
' dr '
gV r
(2-70)
99
CHAPTER 2
FIGURE 2-56
Heat conducted through a cylindrical
shell of radius r is equal to the heat
generated within a shell.
where A,. = 2irrL and V r = irr 2 L at any location r. Substituting these expres-
sions into Eq. 2-70 and separating the variables, we get
-k{2TtrL) =£■ = e(Trr 2 L)
ar
dT
2k
rdr
Integrating from r = where T(0) = T to r = r where T(r ) = T s yields
AT = T
max, cylinder o
8[l
Ak
(2-71)
where T is the centerline temperature of the cylinder, which is the maximum
temperature, and AT max is the difference between the centerline and the sur-
face temperatures of the cylinder, which is the maximum temperature rise in
the cylinder above the surface temperature. Once Ar max is available, the cen-
terline temperature can easily be determined from (Fig. 2-57)
T + AT
(2-72)
The approach outlined above can also be used to determine the maximum tem-
perature rise in a plane wall of thickness 2L and a solid sphere of radius r ,
with these results:
AT,
max, plane wall r\ t
max, sphere sik.
AT,
(2-73)
(2-74)
r* Symmetry
line
FIGURE 2-57
The maximum temperature in
a symmetrical solid with uniform
heat generation occurs at its center.
cen58933_ch02.qxd 9/10/2002 8:46 AM Page 1C
100
HEAT TRANSFER
Again the maximum temperature at the center can be determined from
Eq. 2-72 by adding the maximum temperature rise to the surface temperature
of the solid.
FIGURE 2-58
Schematic for Example 2-17.
EXAMPLE 2-17 Centerline Temperature of a Resistance Heater
A 2-kW resistance heater wire whose thermal conductivity is k = 15 W/m • °C
has a diameter of D = 4 mm and a length of L = 0.5 m, and is used to boil
water (Fig. 2-58). If the outer surface temperature of the resistance wire is T s =
105°C, determine the temperature at the center of the wire.
SOLUTION The surface temperature of a resistance heater submerged in water
is to be determined.
Assumptions 1 Heat transfer is steady since there is no change with time.
2 Heat transfer is one-dimensional since there is thermal symmetry about the
centerline and no change in the axial direction. 3 Thermal conductivity is con-
stant. 4 Heat generation in the heater is uniform.
Properties The thermal conductivity is given to be k = 15 W/m • °C.
Analysis The 2-kW resistance heater converts electric energy into heat at a rate
of 2 kW. The heat generation per unit volume of the wire is
G«
2000 W
irrJL ir(0.002 m) 2 (0.5 m)
0.318 X 10 9 W/m 3
Then the center temperature of the wire is determined from Eq. 2-71 to be
T = T
4k
(0.318 X 10 9 W/m 3 )(0.002m) 2
105°C + - ttttt^tt^^ = 126°C
4 X (15 W/m • °C)
Discussion Note that the temperature difference between the center and the
surface of the wire is 21°C.
FIGURE 2-59
Schematic for Example 2-18.
We have developed these relations using the intuitive energy balance ap-
proach. However, we could have obtained the same relations by setting up the
appropriate differential equations and solving them, as illustrated in Examples
2-18 and 2-19.
EXAMPLE 2-18 Variation of Temperature in a Resistance Heater
A long homogeneous resistance wire of radius r = 0.2 in. and thermal con-
ductivity k = 7.8 Btu/h • ft • °F is being used to boil water at atmospheric pres-
sure by the passage of electric current, as shown in Figure 2-59. Heat is
generated in the wire uniformly as a result of resistance heating at a rate of g =
2400 Btu/h ■ in 3 . If the outer surface temperature of the wire is measured to be
T s = 226 C F, obtain a relation for the temperature distribution, and determine
the temperature at the centerline of the wire when steady operating conditions
are reached.
cen58933_ch02.qxd 9/10/2002 8:46 AM Page 101
SOLUTION This heat transfer problem is similar to the problem in Example
2-17, except that we need to obtain a relation for the variation of temperature
within the wire with r. Differential equations are well suited for this purpose.
Assumptions 1 Heat transfer is steady since there is no change with time.
2 Heat transfer is one-dimensional since there is no thermal symmetry about
the centerline and no change in the axial direction. 3 Thermal conductivity is
constant. 4 Heat generation in the wire is uniform.
Properties The thermal conductivity is given to be k = 7.8 Btu/h • ft ■ C F.
Analysis The differential equation which governs the variation of temperature
in the wire is simply Eq. 2-27,
ld_( dT\ l
r drV dr k
This is a second-order linear ordinary differential equation, and thus its general
solution will contain two arbitrary constants. The determination of these con-
stants requires the specification of two boundary conditions, which can be
taken to be
T(r ) = T S = 226°F
and
dT(0)
dr
The first boundary condition simply states that the temperature of the outer
surface of the wire is 226°F. The second boundary condition is the symmetry
condition at the centerline, and states that the maximum temperature in the
wire will occur at the centerline, and thus the slope of the temperature at r =
must be zero (Fig. 2-60). This completes the mathematical formulation of the
problem.
Although not immediately obvious, the differential equation is in a form that
can be solved by direct integration. Multiplying both sides of the equation by r
and rearranging, we obtain
d_l dT\_l
dry dr k r
Integrating with respect to r gives
dT
r Tr =
k 2
C,
(a)
since the heat generation is constant, and the integral of a derivative of a func-
tion is the function itself. That is, integration removes a derivative. It is conve-
nient at this point to apply the second boundary condition, since it is related to
the first derivative of the temperature, by replacing all occurrences of rand
dT/dr in Eq. (a) by zero. It yields
X
dT(0)
dr
2k
X0 + C,
C, =
101
CHAPTER 2
T J(r)
dT(0)
dr
FIGURE 2-60
The thermal symmetry condition at the
centerline of a wire in which heat
is generated uniformly.
cen58933_ch02.qxd 9/10/2002 8:47 AM Page 102
102
HEAT TRANSFER
Thus C x cancels from the solution. We now divide Eq. (a) by rto bring it to a
readily integrable form,
dT =
dr
Again integrating with respect to r gives
g_
4k
T{r)
2k
r 2 + C 7
m
We now apply the first boundary condition by replacing all occurrences of r by
r and all occurrences of T by 7" s . We get
Jk r ° + C >
c,
\k rs
Substituting this C 2 relation into Eq. (b) and rearranging give
T{r)
T - + -hM-'*>
(c)
which is the desired solution for the temperature distribution in the wire as a
function of r. The temperature at the centerline (r = 0) is obtained by replacing
r in Eq. (c) by zero and substituting the known quantities,
„ m rr , 8 2 „„, ot , . 2400 Btu/h - in 3 /l2inA . , .,„„
HO) = T s + - ^ = 226 F + 4x(78Btu/h . ft .o F) (T7TJ (a2 m ' } = 263 F
Discussion The temperature of the centerline will be 37 C F above the tempera-
ture of the outer surface of the wire. Note that the expression above for the cen-
terline temperature is identical to Eq. 2-71, which was obtained using an
energy balance on a control volume.
Interface
45°C
Ceramic layer
FIGURE 2-61
Schematic for Example 2-19.
EXAMPLE 2-19 Heat Conduction in a Two-Layer Medium
Consider a long resistance wire of radius r x = 0.2 cm and thermal conductivity
/f wire = 15 W/m • °C in which heat is generated uniformly as a result of re-
sistance heating at a constant rate of g = 50 W/cm 3 (Fig. 2-61). The wire
is embedded in a 0.5-cm-thick layer of ceramic whose thermal conductivity is
Ceramic =1-2 W/m • °C. If the outer surface temperature of the ceramic layer
is measured to be 7" s = 45°C, determine the temperatures at the center of the
resistance wire and the interface of the wire and the ceramic layer under steady
conditions.
SOLUTION The surface and interface temperatures of a resistance wire cov-
ered with a ceramic layer are to be determined.
Assumptions 1 Heat transfer is steady since there is no change with time.
2 Heat transfer is one-dimensional since this two-layer heat transfer problem
possesses symmetry about the centerline and involves no change in the axial di-
rection, and thus T = T(r). 3 Thermal conductivities are constant. 4 Heat gen-
eration in the wire is uniform.
Properties It is given that /c wire =15 W/m • °C and k a
= 1.2 W/m • ° C.
cen58933_ch02.qxd 9/10/2002 8:47 AM Page 103
Analysis Letting 7, denote the unknown interface temperature, the heat trans-
fer problem in the wire can be formulated as
]_Cl_ I rf?w,re
r drV dr
with
-* wire( r l) ~~ Tj
rfr wirc (0)
dr
This problem was solved in Example 2-18, and its solution was determined
to be
* wirev) ~~ -'/
4fc»
W " r 2 )
(a)
Noting that the ceramic layer does not involve any heat generation and its
outer surface temperature is specified, the heat conduction problem in that
layer can be expressed as
d I dT a
dr
with
dr
to = T,
■ to = T s
45°C
This problem was solved in Example 2-15, and its solution was determined
to be
ln(r/r t )
(W
We have already utilized the first interface condition by setting the wire and ce-
ramic layer temperatures equal to T, at the interface r = r x . The interface tem-
perature 7", is determined from the second interface condition that the heat flux
in the wire and the ceramic layer at r = r 1 must be the same:
gmre to
dr
dT r ,
,ic to
dr
9
T >(l
ln(r 2 to
Solving for 7, and substituting the given values, the interface temperature is de-
termined to be
gr{
2k,
ceramic
r 2
In t + T s
M
(50 X 10 6 W/m 3 )(0.002m) 2 0.007 m
2(1.2 W/m • °C)
In
0.002 m
+ 45° C = 149.4°C
Knowing the interface temperature, the temperature at the centerline (r = 0) is
obtained by substituting the known quantities into Eq. (a),
:(0)
grr
149.4°C
(50 X 10 6 W/m 3 )(0.002 m) 2
4 X (15 W/m • °C)
152.7°C
103
CHAPTER 2
cen58933_ch02.qxd 9/10/2002 8:47 AM Page 104
104
HEAT TRANSFER
Thus the temperature of the centerline will be slightly above the interface
temperature.
Discussion This example demonstrates how steady one-dimensional heat con-
duction problems in composite media can be solved. We could also solve this
problem by determining the heat flux at the interface by dividing the total heat
generated in the wire by the surface area of the wire, and then using this value
as the specifed heat flux boundary condition for both the wire and the ceramic
layer. This way the two problems are decoupled and can be solved separately.
500
400
300
200
£
100
50
-g 20
£
10
,J^ Copper _
'T^Gold
T
in
?s
ten
I
-s
tainles
AISI
s St
30
ee
4
,-
Fuse
d quart
z
1
100 300 500 1000 2000 4000
Temperature (K)
FIGURE 2-62
Variation of the thermal conductivity
of some solids with temperature.
2-7 ■ VARIABLE THERMAL CONDUCTIVITY, k(T)
You will recall from Chapter 1 that the thermal conductivity of a material, in
general, varies with temperature (Fig. 2-62). However, this variation is mild
for many materials in the range of practical interest and can be disregarded. In
such cases, we can use an average value for the thermal conductivity and treat
it as a constant, as we have been doing so far. This is also common practice for
other temperature-dependent properties such as the density and specific heat.
When the variation of thermal conductivity with temperature in a specified
temperature interval is large, however, it may be necessary to account for this
variation to minimize the error. Accounting for the variation of the thermal
conductivity with temperature, in general, complicates the analysis. But in the
case of simple one-dimensional cases, we can obtain heat transfer relations in
a straightforward manner.
When the variation of thermal conductivity with temperature k(T) is known,
the average value of the thermal conductivity in the temperature range be-
tween T, and T 2 can be determined from
k(T)dT
T 2
(2-75)
This relation is based on the requirement that the rate of heat transfer through
a medium with constant average thermal conductivity & ave equals the rate of
heat transfer through the same medium with variable conductivity k(T). Note
that in the case of constant thermal conductivity k(T) = k, Eq. 2-75 reduces to
£ ave = k, as expected.
Then the rate of steady heat transfer through a plane wall, cylindrical layer,
or spherical layer for the case of variable thermal conductivity can be deter-
mined by replacing the constant thermal conductivity k in Eqs. 2-57, 2-59,
and 2-61 by the k awe expression (or value) from Eq. 2-75:
2= k A
plane wall ave
Ti - T 2 A rT,
k{T)dT
H cylinder •^' 1 ™ave ^
L
T, - T,
2-rrL rr,
Q
sphere
4TrL vp r,r.
ln(r 2 /r,) ln(r 2 /r,) J r ,
T, - To 4m-,/-., rr,
k(T)dT
ave' V 2 f.
k{T)dT
(2-76)
(2-77)
(2-78)
cen58933_ch02.qxd 9/10/2002 8:47 AM Page 105
The variation in thermal conductivity of a material with temperature in the
temperature range of interest can often be approximated as a linear function
and expressed as
k{T) = *o(l + $T)
(2-79)
where (3 is called the temperature coefficient of thermal conductivity. The
average value of thermal conductivity in the temperature range T x to T 2 in this
case can be determined from
*o(l + $T)dT
k 1 + ■
To + T,
k(lme)
(2-80)
Note that the average thermal conductivity in this case is equal to the thermal
conductivity value at the average temperature.
We have mentioned earlier that in a plane wall the temperature varies
linearly during steady one-dimensional heat conduction when the thermal
conductivity is constant. But this is no longer the case when the thermal con-
ductivity changes with temperature, even linearly, as shown in Figure 2-63.
105
CHAPTER 2
T
Plane wall
k{T) = k Q (l+PT)
V
p>o
0-
j3<0
■ T 2
L x
FIGURE 2-63
The variation of temperature
in a plane wall during steady
one-dimensional heat conduction
for the cases of constant and variable
thermal conductivity.
EXAMPLE 2-20 Variation of Temperature in a Wall with k[T)
Consider a plane wall of thickness L whose thermal conductivity varies linearly
in a specified temperature range as k(T) = k (l + p7") where k and p are con-
stants. The wall surface at x = is maintained at a constant temperature of 7"!
while the surface at x = £ is maintained at T z , as shown in Figure 2-64.
Assuming steady one-dimensional heat transfer, obtain a relation for (a) the
heat transfer rate through the wall and (b) the temperature distribution T{x) in
the wall.
SOLUTION A plate with variable conductivity is subjected to specified tem-
peratures on both sides. The variation of temperature and the rate of heat trans-
fer are to be determined.
Assumptions 1 Heat transfer is given to be steady and one-dimensional.
2 Thermal conductivity varies linearly. 3 There is no heat generation.
Properties The thermal conductivity is given to be k(T) = k (\ + p7").
Analysis (a) The rate of heat transfer through the wall can be determined from
Q
If A —
^ave ^ J
where A is the heat conduction area of the wall and
£(r ave ) = k 1 + (3
is the average thermal conductivity (Eq. 2-80).
{b) To determine the temperature distribution in the wall, we begin with
Fourier's law of heat conduction, expressed as
Q = -k(T)A
dT
dx
k(T) = kJi+pr>
,nyi
J-* Y
Plane
wall
()•
T
L x
FIGURE 2-64
Schematic for Example 2-20.
cen58933_ch02.qxd 9/10/2002 8:47 AM Page 106
106
HEAT TRANSFER
where the rate of conduction heat transfer Q and the area A are constant.
Separating variables and integrating from x = where 7~(0) = 7"! to any x where
T(x) = T, we get
" Qdx = -A ( k{T)dT
Substituting k(T) = k Q (\ + fJ7") and performing the integrations we obtain
Qx = -Ak [(T - 7\) + P(J- - T?)/2]
Substituting the Q expression from part (a) and rearranging give
t 2 + |r-
2^ave £
P^o L
(r, - r 2 ) - r, 2
P
o
which is a quadratic equation in the unknown temperature T. Using the qua-
dratic formula, the temperature distribution T(x) in the wall is determined to be
T(x)
1
P
1
2k
P 2 P*o L
(Xi
T 2 ) + n + -r l
The proper sign of the square root term (+ or -) is determined from the re-
quirement that the temperature at any point within the medium must remain
between 7"! and T z . This result explains why the temperature distribution in a
plane wall is no longer a straight line when the thermal conductivity varies with
temperature.
.k(T) = k n (l+pT)
Bronze
plate
FIGURE 2-65
Schematic for Example 2-21.
EXAMPLE 2-21 Heat Conduction through a Wall with k(T)
Consider a 2-m-high and 0.7-m-wide bronze plate whose thickness is 0.1 m.
One side of the plate is maintained at a constant temperature of 600 K while
the other side is maintained at 400 K, as shown in Figure 2-65. The thermal
conductivity of the bronze plate can be assumed to vary linearly in that temper-
ature range as k(T) = k (\ + (57) where k = 38 W/m • K and (5 = 9.21 X
10~ 4 K _1 . Disregarding the edge effects and assuming steady one-dimensional
heat transfer, determine the rate of heat conduction through the plate.
SOLUTION A plate with variable conductivity is subjected to specified tem-
peratures on both sides. The rate of heat transfer is to be determined.
Assumptions 1 Heat transfer is given to be steady and one-dimensional.
2 Thermal conductivity varies linearly. 3 There is no heat generation.
Properties The thermal conductivity is given to be k(T) = k (l + p7").
Analysis The average thermal conductivity of the medium in this case is sim-
ply the value at the average temperature and is determined from
/ T 2 + r,
^ave — k(T ave) — *o( 1 + P 9
(38 W/m -K)
55.5 W/m ■ K
1 + (9.21 X lO^Kr 1 )
(600 + 400) K
cen58933_ch02.qxd 9/10/2002 8:47 AM Page 107
Then the rate of heat conduction through the plate can be determined from Eq.
2-76 to be
Q ~ k-aveA j
(600 - 400)K
= (55.5 W/m • K)(2 m X 0.7 m) — : — = 155,400 W
0.1 m
Discussion We would have obtained the same result by substituting the given
k{T) relation into the second part of Eq. 2-76 and performing the indicated
integration.
107
CHAPTER 2
TOPIC OF SPECIAL INTEREST
A Brief Review of Differential Equations*
As we mentioned in Chapter 1, the description of most scientific problems
involves relations that involve changes in some key variables with respect
to each other. Usually the smaller the increment chosen in the changing
variables, the more general and accurate the description. In the limiting
case of infinitesimal or differential changes in variables, we obtain differ-
ential equations, which provide precise mathematical formulations for the
physical principles and laws by representing the rates of change as deriva-
tives. Therefore, differential equations are used to investigate a wide vari-
ety of problems in science and engineering, including heat transfer.
Differential equations arise when relevant physical laws and principles
are applied to a problem by considering infinitesimal changes in the vari-
ables of interest. Therefore, obtaining the governing differential equation
for a specific problem requires an adequate knowledge of the nature of the
problem, the variables involved, appropriate simplifying assumptions, and
the applicable physical laws and principles involved, as well as a careful
analysis (Fig. 2-66).
An equation, in general, may involve one or more variables. As the name
implies, a variable is a quantity that may assume various values during a
study. A quantity whose value is fixed during a study is called a constant.
Constants are usually denoted by the earlier letters of the alphabet such as
a, b, c, and d, whereas variables are usually denoted by the later ones such
as t, x, y, and z- A variable whose value can be changed arbitrarily is called
an independent variable (or argument). A variable whose value depends
on the value of other variables and thus cannot be varied independently is
called a dependent variable (or a function).
A dependent variable y that depends on a variable x is usually denoted as
y(x) for clarity. However, this notation becomes very inconvenient and
cumbersome when y is repeated several times in an expression. In such
cases it is desirable to denote y(x) simply as y when it is clear that y is a
function of x. This shortcut in notation improves the appearance and the
Physical problem
Identify
important
variables
Make
reasonable
Apply
relevant
assumptions and
approximations
jhysical laws
' '
A differential equation
Apply
applicable
solution
technique
Boundary
and initial
conditions
Solution of the problem
FIGURE 2-66
Mathematical modeling
of physical problems.
*This section can be skipped if desired without a loss in continuity.
cen58933_ch02.qxd 9/10/2002 8:47 AM Page 1C
108
HEAT TRANSFER
X x + Ax
FIGURE 2-67
The derivative of a function at a point
represents the slope of the tangent
line of the function at that point.
FIGURE 2-68
Graphical representation of
partial derivative dzldx.
readability of the equations. The value of y at a fixed number a is denoted
by y(a).
The derivative of a function y(x) at a point is equivalent to the slope of
the tangent line to the graph of the function at that point and is defined as
(Fig. 2-67)
y'W
dy(x)
dx
hm —
Ax->0 AX
lim
y(x + Ax) - y(x)
Ax
(2-81)
Here Ax represents a (small) change in the independent variable x and is
called an increment of x. The corresponding change in the function y is
called an increment of y and is denoted by Ay. Therefore, the derivative of
a function can be viewed as the ratio of the increment Ay of the function to
the increment Ax of the independent variable for very small Ax. Note that
Ay and thus y'(x) will be zero if the function y does not change with x.
Most problems encountered in practice involve quantities that change
with time t, and their first derivatives with respect to time represent the rate
of change of those quantities with time. For example, if N(t) denotes the
population of a bacteria colony at time /, then the first derivative N' =
dNIdt represents the rate of change of the population, which is the amount
the population increases or decreases per unit time.
The derivative of the first derivative of a function y is called the second
derivative of y, and is denoted by y" or d 2 y/dx 2 . In general, the derivative of
the (n — l)st derivative of y is called the nth derivative of y and is denoted
by y (,,) or d n y/dx". Here, n is a positive integer and is called the order of the
derivative. The order n should not be confused with the degree of a deriva-
tive. For example, y'" is the third-order derivative of y, but (y') 3 is the third
degree of the first derivative of y. Note that the first derivative of a function
represents the slope or the rate of change of the function with the indepen-
dent variable, and the second derivative represents the rate of change of the
slope of the function with the independent variable.
When a function y depends on two or more independent variables such
as x and t, it is sometimes of interest to examine the dependence of the
function on one of the variables only. This is done by taking the derivative
of the function with respect to that variable while holding the other vari-
ables constant. Such derivatives are called partial derivatives. The first
partial derivatives of the function y(x, t) with respect to x and t are defined
as (Fig. 2-68)
dy y(x + Ax, t) - y(x, t)
— = hm -.
OX Ax -^ o Ax
y(x, t + At) - y(x, t)
dy
T^ = llm
at \t^o
At
(2-82)
(2-83)
Note that when finding dy/dx we treat (asa constant and differentiate y
with respect to x. Likewise, when finding dy/dt we treat x as a constant and
differentiate y with respect to /.
Integration can be viewed as the inverse process of differentiation. Inte-
gration is commonly used in solving differential equations since solving a
differential equation is essentially a process of removing the derivatives
cen58933_ch02.qxd 9/10/2002 8:47 AM Page 109
from the equation. Differentiation is the process of finding y'(x) when a
function y(x) is given, whereas integration is the process of finding the
function y(x) when its derivative y'(x) is given. The integral of this deriva-
tive is expressed as
J y'(x)dx = J dy = y(x) + C
(2-84)
since y'(x)dx = dy and the integral of the differential of a function is the
function itself (plus a constant, of course). In Eq. 2-84, x is the integration
variable and C is an arbitrary constant called the integration constant.
The derivative of y(x) + C is y'(x) no matter what the value of the con-
stant C is. Therefore, two functions that differ by a constant have the same
derivative, and we always add a constant C during integration to recover
this constant that is lost during differentiation. The integral in Eq. 2-84 is
called an indefinite integral since the value of the arbitrary constant C is
indefinite. The described procedure can be extended to higher-order deriv-
atives (Fig. 2-69). For example,
109
CHAPTER 2
I
dy-
= y + C
.(>'
dx =
-y + C
J"
dx =
= y' + c
\'
dx =
= y" + C
I"
dx =
--y<"-» + C
y"{x)dx = y'(x) + C
(2-85)
FIGURE 2-69
Some indefinite integrals
that involve derivatives.
This can be proved by defining a new variable u(x) = y'(x), differentiating
it to obtain u'(x) = y"{x), and then applying Eq. 2-84. Therefore, the order
of a derivative decreases by one each time it is integrated.
Classification of Differential Equations
A differential equation that involves only ordinary derivatives is called an
ordinary differential equation, and a differential equation that involves
partial derivatives is called a partial differential equation. Then it follows
that problems that involve a single independent variable result in ordinary
differential equations, and problems that involve two or more independent
variables result in partial differential equations. A differential equation may
involve several derivatives of various orders of an unknown function. The
order of the highest derivative in a differential equation is the order of the
equation. For example, the order of /" + (y") 4 = 7x 5 is 3 since it contains
no fourth or higher order derivatives.
You will remember from algebra that the equation 3x — 5 = is much
easier to solve than the equation x 4 + 3x — 5 = because the first equation
is linear whereas the second one is nonlinear. This is also true for differen-
tial equations. Therefore, before we start solving a differential equation, we
usually check for linearity. A differential equation is said to be linear if the
dependent variable and all of its derivatives are of the first degree and their
coefficients depend on the independent variable only. In other words, a dif-
ferential equation is linear if it can be written in a form that does not in-
volve (1) any powers of the dependent variable or its derivatives such as y 3
or (y') 2 , (2) any products of the dependent variable or its derivatives such
as yy' or y'y'", and (3) any other nonlinear functions of the dependent vari-
able such as sin y or e>. If any of these conditions apply, it is nonlinear
(Fig. 2-70).
(a) A nonlinear equation:
30") 2 -4vv' + e 2
Power Product
-6x l
Other
nonlinear
functions
(b) A linear equation:
3x y" - 4xy' + e y = 6x
FIGURE 2-70
A differential equation that is
(a) nonlinear and (b) linear. When
checking for linearity, we examine the
dependent variable only.
cen58933_ch02.qxd 9/10/2002 8:47 AM Page 11C
110
HEAT TRANSFER
(a)
With constant coefficients:
y" + 6y' — 2y = xe
-2x
v
V
Constant
(b)
With
variable coefficients:
y
■> i 2
xe 2 *
AT X — 1
V
V
Variable
FIGURE 2-71
A differential equation with
(a) constant coefficients and
(b) variable coefficients.
(a) An algebraic equation:
y 2 - 7v - 10 =
Solution: y = 2 and y = 5
(b) A differential equation:
y' - v.v = o
Solution: y = e 7 '
FIGURE 2-72
Unlike those of algebraic equations,
the solutions of differential equations
are typically functions instead of
discrete values.
A linear differential equation, however, may contain (1) powers or non-
linear functions of the independent variable, such as x 2 and cos x and
(2) products of the dependent variable (or its derivatives) and functions of
the independent variable, such as x 3 y' , x 2 y, and e^^y". A linear differential
equation of order n can be expressed in the most general form as
yM +f l (x)yf» -'>+■•■ +f„_ l (x)y' +f„(x)y = R(x)
(2-86)
A differential equation that cannot be put into this form is nonlinear. A
linear differential equation in y is said to be homogeneous as well if
R(x) = 0. Otherwise, it is nonhomogeneous. That is, each term in a linear
homogeneous equation contains the dependent variable or one of its deriv-
atives after the equation is cleared of any common factors. The term R(x) is
called the nonhomogeneous term.
Differential equations are also classified by the nature of the coefficients
of the dependent variable and its derivatives. A differential equation is said
to have constant coefficients if the coefficients of all the terms that involve
the dependent variable or its derivatives are constants. If, after clearing any
common factors, any of the terms with the dependent variable or its deriv-
atives involve the independent variable as a coefficient, that equation is
said to have variable coefficients (Fig. 2-71). Differential equations with
constant coefficients are usually much easier to solve than those with vari-
able coefficients.
Solutions of Differential Equations
Solving a differential equation can be as easy as performing one or more
integrations; but such simple differential equations are usually the excep-
tion rather than the rule. There is no single general solution method appli-
cable to all differential equations. There are different solution techniques,
each being applicable to different classes of differential equations. Some-
times solving a differential equation requires the use of two or more tech-
niques as well as ingenuity and mastery of solution methods. Some
differential equations can be solved only by using some very clever tricks.
Some cannot be solved analytically at all.
In algebra, we usually seek discrete values that satisfy an algebraic equa-
tion such as x 1 — Ix —10 = 0. When dealing with differential equations,
however, we seek functions that satisfy the equation in a specified interval.
For example, the algebraic equation x 2 — Ix — 10 = is satisfied by two
numbers only: 2 and 5. But the differential equation y' — ly = is satis-
fied by the function e lx for any value of x (Fig. 2-72).
Consider the algebraic equation x 3 — 6x 2 + llx — 6 = 0. Obviously,
x = 1 satisfies this equation, and thus it is a solution. However, it is not the
only solution of this equation. We can easily show by direct substitution
that x = 2 and x = 3 also satisfy this equation, and thus they are solutions
as well. But there are no other solutions to this equation. Therefore, we
say that the set 1, 2, and 3 forms the complete solution to this algebraic
equation.
The same line of reasoning also applies to differential equations. Typi-
cally, differential equations have multiple solutions that contain at least one
arbitrary constant. Any function that satisfies the differential equation on an
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CHAPTER 2
interval is called a solution of that differential equation in that interval.
A solution that involves one or more arbitrary constants represents a fam-
ily of functions that satisfy the differential equation and is called a general
solution of that equation. Not surprisingly, a differential equation may
have more than one general solution. A general solution is usually referred
to as the general solution or the complete solution if every solution of the
equation can be obtained from it as a special case. A solution that can be
obtained from a general solution by assigning particular values to the arbi-
trary constants is called a specific solution.
You will recall from algebra that a number is a solution of an algebraic
equation if it satisfies the equation. For example, 2 is a solution of the equa-
tion x 3 — 8 = because the substitution of 2 for x yields identically zero.
Likewise, a function is a solution of a differential equation if that function
satisfies the differential equation. In other words, a solution function yields
identity when substituted into the differential equation. For example, it
can be shown by direct substitution that the function 3e~ 2v is a solution of
y" - 4y = (Fig. 2-73).
Function:/ = 3e~ 2>
Differential equation: y" —
4y =
Derivatives off:
r --
= -6e-
2.v
/" =
= 12e- 2
Substituting into y"
-4y =
0:
f"-
4/i
He- 2 * -
4X 3e
-2x ?
=
Therefore, the function 3e~
^is £
solution of
the differential equation y"
-4y
= 0.
FIGURE 2-73
Verifying that a given function is a
solution of a differential equation.
SUMMARY
In this chapter we have studied the heat conduction equation
and its solutions. Heat conduction in a medium is said to be
steady when the temperature does not vary with time and un-
steady or transient when it does. Heat conduction in a medium
is said to be one-dimensional when conduction is significant
in one dimension only and negligible in the other two di-
mensions. It is said to be two-dimensional when conduction in
the third dimension is negligible and three-dimensional when
conduction in all dimensions is significant. In heat transfer
analysis, the conversion of electrical, chemical, or nuclear
energy into heat (or thermal) energy is characterized as heat
generation.
The heat conduction equation can be derived by performing
an energy balance on a differential volume element. The one-
dimensional heat conduction equation in rectangular, cylindri-
cal, and spherical coordinate systems for the case of constant
thermal conductivities are expressed as
d 2 T , 8 1 dT
i a
d.\ 2
dT
r dr \ dr
1 d I , dT
dr
dr
k « dt
8_ = }_dT
k a dt
8_ = }_dT
k a dt
where the property a = k/pC is the thermal diffusivity of the
material.
The solution of a heat conduction problem depends on the
conditions at the surfaces, and the mathematical expressions
for the thermal conditions at the boundaries are called the
boundary conditions. The solution of transient heat conduction
problems also depends on the condition of the medium at the
beginning of the heat conduction process. Such a condition,
which is usually specified at time t = 0, is called the initial
condition, which is a mathematical expression for the temper-
ature distribution of the medium initially. Complete mathemat-
ical description of a heat conduction problem requires the
specification of two boundary conditions for each dimension
along which heat conduction is significant, and an initial con-
dition when the problem is transient. The most common
boundary conditions are the specified temperature, specified
heat flux, convection, and radiation boundary conditions. A
boundary surface, in general, may involve specified heat flux,
convection, and radiation at the same time.
For steady one-dimensional heat transfer through a plate of
thickness L, the various types of boundary conditions at the
surfaces at x = and x = L can be expressed as
Specified temperature:
T(0) = T {
and
T(L)
where T { and T 2 are the specified temperatures at surfaces at
x = and x = L.
Specified heat flux:
,_dT(0)
k dx
4o
and
dT(L)
dx
1l
where q and q L are the specified heat fluxes at surfaces at
x = and x = L.
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HEAT TRANSFER
Insulation or thermal symmetry:
dT(0) dT(L)
dx
and
Convection
dT(0)
dx
h,[T-M - T(0)] and
dx
-k-^ = h 2 [T(L)-T oc2 ]
where h x and h 2 are the convection heat transfer coefficients
and T^, and T„ 2 are the temperatures of the surrounding medi-
ums on the two sides of the plate.
Radiation:
-k^ 1 = sMTL, l -T(0)' i ] and
dT(L)
-k^^ = ^MT(L) 4 -Ti urr J
where h is the convection heat transfer coefficient. The maxi-
mum temperature rise between the surface and the midsection
of a medium is given by
AT,
= *kL
max, plane wall n y
AT,
max, cylinder a k
max, sphere /- r.
When the variation of thermal conductivity with temperature
k(T) is known, the average value of the thermal conductivity in
the temperature range between T { and T 2 can be determined
from
f : k(T)dT
T 2 -T t
where s t and e 2 are the emissivities of the boundary surfaces,
o- = 5.67 X 1CT 8 W/m 2 • K 4 is the Stefan-Boltzmann constant,
and r surr , and r slllT 2 are the average temperatures of the sur-
faces surrounding the two sides of the plate. In radiation calcu-
lations, the temperatures must be in K or R.
Interface of two bodies A and B in perfect contact at x = x :
T A (x ) = T B (x ) and
AT A (*o)
dx
dT B (jc )
' dx
Then the rate of steady heat transfer through a plane wall,
cylindrical layer, or spherical layer can be expressed as
r, - T 2 _ A f T <
G plane wall ~~ K ave A 7 — T I k(T)dT
Q cylinder ^'"'^ave-^
Q
4irk„„ r r,r
T, ~ T 2 2ttL f 7 '
\r\{r 2 lr{) ln(r 2 /r,) ) T ,
r, - T 2 4<ir/-,r 2 f r,
sphere ^'"' v ave'l'2 f„ — }• y 2 ~ T
Fi \ T k(T)dT
where k A and k B are the thermal conductivities of the layers
A and B.
Heat generation is usually expressed per unit volume of the
medium and is denoted by g, whose unit is W/m 3 . Under steady
conditions, the surface temperature T s of a plane wall of thick-
ness 2L, a cylinder of outer radius r , and a sphere of radius r
in which heat is generated at a constant rate of g per unit vol-
ume in a surrounding medium at T„ can be expressed as
Ik
ii
gr a
2h
gr
s, sphere <» o/.
7* = T +
5, plane wall <» j^
T = T -f-
5, cylinder ro ^ /^
The variation of thermal conductivity of a material with
temperature can often be approximated as a linear function and
expressed as
k{T) = k (l + p7)
where (3 is called the temperature coefficient of thermal
conductivity.
REFERENCES AND SUGGESTED READING
1. W. E. Boyce and R. C. Diprima. Elementary Differential
Equations and Boundary Value Problems. 4th ed.
New York: John Wiley & Sons, 1986.
2. J. R Holman. Heat Transfer. 9th ed. New York:
McGraw-Hill, 2002.
cen58933_ch02.qxd 9/10/2002 8:47 AM Page 113
3. F. P. Incropera and D. P. DeWitt. Introduction to Heat
Transfer. 4th ed. New York: John Wiley & Sons, 2002.
4. S. S. Kutateladze. Fundamentals of Heat Transfer.
New York: Academic Press, 1963.
113
CHAPTER 1
5. M. N. Ozisik. Heat Transfer — A Basic Approach.
New York: McGraw-Hill, 1985.
6. F. M. White. Heat and Mass Transfer. Reading, MA:
Addison-Wesley, 1988.
PROBLEMS
Introduction
2-1C Is heat transfer a scalar or vector quantity? Explain.
Answer the same question for temperature.
2-2C How does transient heat transfer differ from steady
heat transfer? How does one-dimensional heat transfer differ
from two-dimensional heat transfer?
2-3C Consider a cold canned drink left on a dinner table.
Would you model the heat transfer to the drink as one-, two-, or
three-dimensional? Would the heat transfer be steady or tran-
sient? Also, which coordinate system would you use to analyze
this heat transfer problem, and where would you place the ori-
gin? Explain.
2-4C Consider a round potato being baked in an oven.
Would you model the heat transfer to the potato as one-, two-,
or three-dimensional? Would the heat transfer be steady or
transient? Also, which coordinate system would you use to
solve this problem, and where would you place the origin?
Explain.
FIGURE P2-4
*Problems designated by a "C" are concept questions, and
students are encouraged to answer them all. Problems designated
by an "E" are in English units, and the SI users can ignore them.
Problems with an EES-CD icon ® are solved using EES, and
complete solutions together with parametric studies are included
on the enclosed CD. Problems with a computer-EES icon H are
comprehensive in nature, and are intended to be solved with a
computer, preferably using the EES software that accompanies
this text.
2-5C Consider an egg being cooked in boiling water in a
pan. Would you model the heat transfer to the egg as one-,
two-, or three-dimensional? Would the heat transfer be steady
or transient? Also, which coordinate system would you use to
solve this problem, and where would you place the origin?
Explain.
2-6C Consider a hot dog being cooked in boiling water in a
pan. Would you model the heat transfer to the hot dog as one-,
two-, or three-dimensional? Would the heat transfer be steady
or transient? Also, which coordinate system would you use to
solve this problem, and where would you place the origin?
Explain.
FIGURE P2-6
2-7C Consider the cooking process of a roast beef in an
oven. Would you consider this to be a steady or transient heat
transfer problem? Also, would you consider this to be one-,
two-, or three-dimensional? Explain.
2-8C Consider heat loss from a 200-L cylindrical hot water
tank in a house to the surrounding medium. Would you con-
sider this to be a steady or transient heat transfer problem?
Also, would you consider this heat transfer problem to be one-,
two-, or three-dimensional? Explain.
2-9C Does a heat flux vector at a point P on an isothermal
surface of a medium have to be perpendicular to the surface at
that point? Explain.
2-10C From a heat transfer point of view, what is the differ-
ence between isotropic and unisotropic materials?
2-11C What is heat generation in a solid? Give examples.
2-12C Heat generation is also referred to as energy genera-
tion or thermal energy generation. What do you think of these
phrases?
2-13C In order to determine the size of the heating element
of a new oven, it is desired to determine the rate of heat trans-
fer through the walls, door, and the top and bottom section of
the oven. In your analysis, would you consider this to be a
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HEAT TRANSFER
steady or transient heat transfer problem? Also, would you con-
sider the heat transfer to be one -dimensional or multidimen-
sional? Explain.
2-14E The resistance wire of a 1000-W iron is 15 in. long
and has a diameter of D = 0.08 in. Determine the rate of heat
generation in the wire per unit volume, in Btu/h • ft 3 , and the
heat flux on the outer surface of the wire, in Btu/h ■ ft 2 , as a re-
sult of this heat generation.
FIGURE P2-14E
ity and heat generation in its simplest form, and indicate what
each variable represents.
2-20 Write down the one -dimensional transient heat conduc-
tion equation for a long cylinder with constant thermal con-
ductivity and heat generation, and indicate what each variable
represents.
2-21 Starting with an energy balance on a rectangular vol-
ume element, derive the one-dimensional transient heat con-
duction equation for a plane wall with constant thermal
conductivity and no heat generation.
2-22 Starting with an energy balance on a cylindrical shell
volume element, derive the steady one-dimensional heat con-
duction equation for a long cylinder with constant thermal con-
ductivity in which heat is generated at a rate of g.
2-15E [T^vfl Reconsider Problem 2-14E. Using EES (or
h^2 other) software, evaluate and plot the surface
heat flux as a function of wire diameter as the diameter varies
from 0.02 to 0.20 in. Discuss the results.
2-16 In a nuclear reactor, heat is generated uniformly in the
5-cm-diameter cylindrical uranium rods at a rate of 7 X 10 7
W/m 3 . If the length of the rods is 1 m, determine the rate of
heat generation in each rod. Answer: 137 A kW
2-17 In a solar pond, the absorption of solar energy can be
modeled as heat generation and can be approximated by g =
g e~ bx , where g Q is the rate of heat absorption at the top surface
per unit volume and b is a constant. Obtain a relation for the to-
tal rate of heat generation in a water layer of surface area A and
thickness L at the top of the pond.
Radiation
beam being
absorbed
FIGURE P2-1 7
2-18 Consider a large 3-cm-thick stainless steel plate in
which heat is generated uniformly at a rate of 5 X 10 6 W/m 3 .
Assuming the plate is losing heat from both sides, determine
the heat flux on the surface of the plate during steady opera-
tion. Answer: 75,000 W/m 2
Heat Conduction Equation
2-19 Write down the one-dimensional transient heat conduc-
tion equation for a plane wall with constant thermal conductiv-
FIGURE P2-22
2-23 Starting with an energy balance on a spherical shell
volume element, derive the one-dimensional transient heat
conduction equation for a sphere with constant thermal con-
ductivity and no heat generation.
FIGURE P2-23
2-24 Consider a medium in which the heat conduction equa-
tion is given in its simplest form as
dx 1
]_dT
a dt
cen58933_ch02.qxd 9/10/2002 8:47 AM Page 115
(«) Is heat transfer steady or transient?
(b) Is heat transfer one-, two-, or three-dimensional?
(c) Is there heat generation in the medium?
(d) Is the thermal conductivity of the medium constant or
variable?
2-25 Consider a medium in which the heat conduction equa-
tion is given in its simplest form as
\d_
r dr
rk
dT
dr
115
CHAPTER 1
FIGURE P2-29
(a) Is heat transfer steady or transient?
(b) Is heat transfer one-, two-, or three-dimensional?
(c) Is there heat generation in the medium?
(d) Is the thermal conductivity of the medium constant or
variable?
2-26 Consider a medium in which the heat conduction equa-
tion is given in its simplest form as
r 2 dr
_ 2 ar
dr
\_dT
a dt
(a)
(b)
(c)
(d)
Is heat transfer steady or transient?
Is heat transfer one-, two-, or three-dimensional?
Is there heat generation in the medium?
Is the thermal conductivity of the medium constant or
variable?
2-27 Consider a medium in which the heat conduction equa-
tion is given in its simplest form as
d 2 T dT
' dr 2 dr
(fl)
(b)
(c)
(d)
Is heat transfer steady or transient?
Is heat transfer one-, two-, or three-dimensional?
Is there heat generation in the medium?
Is the thermal conductivity of the medium constant or
variable?
2-28 Starting with an energy balance on a volume element,
derive the two-dimensional transient heat conduction equation
in rectangular coordinates for T(x, y, f) for the case of constant
thermal conductivity and no heat generation.
2-29 Starting with an energy balance on a ring-shaped vol-
ume element, derive the two-dimensional steady heat conduc-
tion equation in cylindrical coordinates for T(r, z) for the case
of constant thermal conductivity and no heat generation.
2-30 Starting with an energy balance on a disk volume ele-
ment, derive the one-dimensional transient heat conduction
equation for T(z, t) in a cylinder of diameter D with an insu-
lated side surface for the case of constant thermal conductivity
with heat generation.
2-31 Consider a medium in which the heat conduction equa-
tion is given in its simplest form as
Disk
Insulation
FIGURE P2-30
d 2 T d 2 T = 1 dT
dx 2 dy 2 ~ a dt
(a) Is heat transfer steady or transient?
(b) Is heat transfer one-, two-, or three-dimensional?
(c) Is there heat generation in the medium?
(d) Is the thermal conductivity of the medium constant or
variable?
2-32 Consider a medium in which the heat conduction equa-
tion is given in its simplest form as
r dr
kr
dT
dr
d_
dz
dT
dz
(a) Is heat transfer steady or transient?
(b) Is heat transfer one-, two-, or three-dimensional?
(c) Is there heat generation in the medium?
(d) Is the thermal conductivity of the medium constant or
variable?
2-33 Consider a medium in which the heat conduction equa-
tion is given in its simplest form as
j_d_
r 2 dr
dT
dt
1
d 2 T
sin 2 6 d§ 2
\_dT
a dt
(a) Is heat transfer steady or transient?
(b) Is heat transfer one-, two-, or three-dimensional?
(c) Is there heat generation in the medium?
(d) Is the thermal conductivity of the medium constant or
variable?
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HEAT TRANSFER
Boundary and Initial Conditions;
Formulation of Heat Conduction Problems
2-34C What is a boundary condition? How many boundary
conditions do we need to specify for a two-dimensional heat
transfer problem?
2-35C What is an initial condition? How many initial condi-
tions do we need to specify for a two-dimensional heat transfer
problem?
2-36C What is a thermal symmetry boundary condition?
How is it expressed mathematically?
2-37C How is the boundary condition on an insulated sur-
face expressed mathematically?
2-38C It is claimed that the temperature profile in a medium
must be perpendicular to an insulated surface. Is this a valid
claim? Explain.
2-39C Why do we try to avoid the radiation boundary con-
ditions in heat transfer analysis?
2-40 Consider a spherical container of inner radius r x , outer
radius r 2 , and thermal conductivity k. Express the boundary
condition on the inner surface of the container for steady one-
dimensional conduction for the following cases: (a) specified
temperature of 50°C, {b) specified heat flux of 30 W/m 2 toward
the center, (c) convection to a medium at 7V_ with a heat trans-
fer coefficient of h.
Spherical container
FIGURE P2-40
2-41 Heat is generated in a long wire of radius r at a con-
stant rate of g per unit volume. The wire is covered with a
plastic insulation layer. Express the heat flux boundary condi-
tion at the interface in terms of the heat generated.
2-42 Consider a long pipe of inner radius r lt outer radius r 2 ,
and thermal conductivity k. The outer surface of the pipe is
subjected to convection to a medium at T a with a heat transfer
coefficient of h, but the direction of heat transfer is not known.
Express the convection boundary condition on the outer sur-
face of the pipe.
2-43 Consider a spherical shell of inner radius r u outer ra-
dius r 2 , thermal conductivity k, and emissivity e. The outer sur-
face of the shell is subjected to radiation to surrounding
Express the radiation boundary condition on the outer surface
of the shell.
2-44 A container consists of two spherical layers, A and B,
that are in perfect contact. If the radius of the interface is r ,
express the boundary conditions at the interface.
2-45 Consider a steel pan used to boil water on top of an
electric range. The bottom section of the pan is L = 0.5 cm
thick and has a diameter of D = 20 cm. The electric heating
unit on the range top consumes 1000 W of power during cook-
ing, and 85 percent of the heat generated in the heating element
is transferred uniformly to the pan. Heat transfer from the top
surface of the bottom section to the water is by convection with
a heat transfer coefficient of h. Assuming constant thermal
conductivity and one-dimensional heat transfer, express the
mathematical formulation (the differential equation and the
boundary conditions) of this heat conduction problem during
steady operation. Do not solve.
Steel pan
FIGURE P2-45
2-46E A 2-kW resistance heater wire whose thermal con-
ductivity is k = 10.4 Btu/h ■ ft • °F has a radius of r = 0.06 in.
and a length of L = 15 in., and is used for space heating. As-
suming constant thermal conductivity and one-dimensional
heat transfer, express the mathematical formulation (the differ-
ential equation and the boundary conditions) of this heat con-
duction problem during steady operation. Do not solve.
2-47 Consider an aluminum pan used to cook stew on top of
an electric range. The bottom section of the pan is L = 0.25 cm
thick and has a diameter of D = 18 cm. The electric heating
unit on the range top consumes 900 W of power during cook-
ing, and 90 percent of the heat generated in the heating element
Aluminum pan
FIGURE P2-47
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CHAPTER 1
is transferred to the pan. During steady operation, the temper-
ature of the inner surface of the pan is measured to be 108°C.
Assuming temperature-dependent thermal conductivity and
one-dimensional heat transfer, express the mathematical for-
mulation (the differential equation and the boundary condi-
tions) of this heat conduction problem during steady operation.
Do not solve.
2-48 Water flows through a pipe at an average temperature
of T„ = 50°C. The inner and outer radii of the pipe are r { =
6 cm and r 2 = 6.5 cm, respectively. The outer surface of
the pipe is wrapped with a thin electric heater that consumes
300 W per m length of the pipe. The exposed surface of the
heater is heavily insulated so that the entire heat generated in
the heater is transferred to the pipe. Heat is transferred from the
inner surface of the pipe to the water by convection with a heat
transfer coefficient of h = 55 W/m 2 ■ °C. Assuming constant
thermal conductivity and one-dimensional heat transfer, ex-
press the mathematical formulation (the differential equation
and the boundary conditions) of the heat conduction in the pipe
during steady operation. Do not solve.
Insulation
Electric heater
FIGURE P2-48
2-49 A spherical metal ball of radius r is heated in an oven
to a temperature of T, throughout and is then taken out of the
oven and dropped into a large body of water at T„ where it is
cooled by convection with an average convection heat transfer
coefficient of h. Assuming constant thermal conductivity and
transient one-dimensional heat transfer, express the mathemat-
ical formulation (the differential equation and the boundary
and initial conditions) of this heat conduction problem. Do not
solve.
2-50 A spherical metal ball of radius r is heated in an oven
to a temperature of T, throughout and is then taken out of the
oven and allowed to cool in ambient air at T„ by convection
and radiation. The emissivity of the outer surface of the cylin-
der is e, and the temperature of the surrounding surfaces is
r surr . The average convection heat transfer coefficient is esti-
mated to be h. Assuming variable thermal conductivity and
transient one-dimensional heat transfer, express the mathemat-
ical formulation (the differential equation and the boundary
Radiation
FIGURE P2-50
and initial conditions) of this heat conduction problem. Do not
solve.
2-51 Consider the north wall of a house of thickness L. The
outer surface of the wall exchanges heat by both convection
and radiation. The interior of the house is maintained at T al ,
while the ambient air temperature outside remains at T^ 2 - The
sky, the ground, and the surfaces of the surrounding structures
at this location can be modeled as a surface at an effective tem-
perature of T sky for radiation exchange on the outer surface.
The radiation exchange between the inner surface of the wall
and the surfaces of the walls, floor, and ceiling it faces is neg-
ligible. The convection heat transfer coefficients on the inner
and outer surfaces of the wall are h { and h 2 , respectively. The
thermal conductivity of the wall material is k and the emissiv-
ity of the outer surface is e 2 . Assuming the heat transfer
through the wall to be steady and one-dimensional, express the
mathematical formulation (the differential equation and the
boundary and initial conditions) of this heat conduction prob-
lem. Do not solve.
Wall
h 2
T
FIGURE P2-51
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HEAT TRANSFER
Solution of Steady One-Dimensional
Heat Conduction Problems
2-52C Consider one-dimensional heat conduction through a
large plane wall with no heat generation that is perfectly insu-
lated on one side and is subjected to convection and radiation
on the other side. It is claimed that under steady conditions, the
temperature in a plane wall must be uniform (the same every-
where). Do you agree with this claim? Why?
2-53C It is stated that the temperature in a plane wall with
constant thermal conductivity and no heat generation varies
linearly during steady one-dimensional heat conduction. Will
this still be the case when the wall loses heat by radiation from
its surfaces?
2-54C Consider a solid cylindrical rod whose ends are main-
tained at constant but different temperatures while the side sur-
face is perfectly insulated. There is no heat generation. It is
claimed that the temperature along the axis of the rod varies
linearly during steady heat conduction. Do you agree with this
claim? Why?
2-55C Consider a solid cylindrical rod whose side surface is
maintained at a constant temperature while the end surfaces are
perfectly insulated. The thermal conductivity of the rod mater-
ial is constant and there is no heat generation. It is claimed that
the temperature in the radial direction within the rod will not
vary during steady heat conduction. Do you agree with this
claim? Why?
2-56 Consider a large plane wall of thickness L = 0.4 m,
thermal conductivity k = 2.3 W/m • °C, and surface area A =
20 m 2 . The left side of the wall is maintained at a constant tem-
perature of T { = 80°C while the right side loses heat by con-
vection to the surrounding air at T„ = 15°C with a heat transfer
coefficient of h = 24 W/m 2 • °C. Assuming constant thermal
conductivity and no heat generation in the wall, (a) express the
differential equation and the boundary conditions for steady
one-dimensional heat conduction through the wall, (b) obtain a
relation for the variation of temperature in the wall by solving
the differential equation, and (c) evaluate the rate of heat trans-
fer through the wall. Answer-, (c) 6030 W
2-57 Consider a solid cylindrical rod of length 0.15 m and
diameter 0.05 m. The top and bottom surfaces of the rod are
maintained at constant temperatures of 20°C and 95°C, re-
spectively, while the side surface is perfectly insulated. Deter-
mine the rate of heat transfer through the rod if it is made of
(a) copper, k = 380 W/m • °C, (b) steel, k = 18 W/m • °C, and
(c) granite, k = 1.2 W/m ■ °C.
2-58 rSpM Reconsider Problem 2-57. Using EES (or other)
b^2 software, plot the rate of heat transfer as a func-
tion of the thermal conductivity of the rod in the range of
1 W/m • °C to 400 W/m • °C. Discuss the results.
2-59 Consider the base plate of a 800-W household iron with
a thickness of L = 0.6 cm, base area of A = 160 cm 2 , and ther-
Base
plate
-85°C
FIGURE P2-59
mal conductivity of k = 20 W/m ■ °C. The inner surface of the
base plate is subjected to uniform heat flux generated by the re-
sistance heaters inside. When steady operating conditions are
reached, the outer surface temperature of the plate is measured
to be 85°C. Disregarding any heat loss through the upper part
of the iron, (a) express the differential equation and the bound-
ary conditions for steady one-dimensional heat conduction
through the plate, (b) obtain a relation for the variation of tem-
perature in the base plate by solving the differential equation,
and (c) evaluate the inner surface temperature.
Answer: (c) 100°C
2-60 Repeat Problem 2-59 for a 1200-W iron.
2-61 ta'M Reconsider Problem 2-59. Using the relation ob-
1^2 tained for the variation of temperature in the base
plate, plot the temperature as a function of the distance x in the
range of x = to x = L, and discuss the results. Use the EES
(or other) software.
2-62E Consider a steam pipe of length L = 1 5 ft, inner ra-
dius r x = 2 in., outer radius r 2 = 2.4 in., and thermal conduc-
tivity k = 7.2 Btu/h • ft • °F. Steam is flowing through the pipe
at an average temperature of 250°F, and the average convection
heat transfer coefficient on the inner surface is given to be h =
1 .25 Btu/h • ft 2 • °F . If the average temperature on the outer
FIGURE P2-62E
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CHAPTER 1
surfaces of the pipe is T 2 = 160°F, (a) express the differential
equation and the boundary conditions for steady one-
dimensional heat conduction through the pipe, (b) obtain a re-
lation for the variation of temperature in the pipe by solving the
differential equation, and (c) evaluate the rate of heat loss from
the steam through the pipe. Answer: (c) 16,800 Btu/h
2-63 A spherical container of inner radius r { = 2 m, outer ra-
dius r 2 = 2.1 m, and thermal conductivity k = 30 W/m • °C is
filled with iced water at 0°C. The container is gaining heat by
convection from the surrounding air at T-^ = 25 °C with a heat
transfer coefficient of h = 18 W/m 2 ■ °C. Assuming the inner
surface temperature of the container to be 0°C, (a) express the
differential equation and the boundary conditions for steady
one -dimensional heat conduction through the container, (b) ob-
tain a relation for the variation of temperature in the container
by solving the differential equation, and (c) evaluate the rate of
heat gain to the iced water.
2-64 Consider a large plane wall of thickness L = 0.3 m,
thermal conductivity k = 2.5 W/m • °C, and surface area A =
12 m 2 . The left side of the wall at x = is subjected to a net
heat flux of q = 700 W/m 2 while the temperature at that sur-
face is measured to be T x = 80°C. Assuming constant thermal
conductivity and no heat generation in the wall, (a) express the
differential equation and the boundary conditions for steady
one-dimensional heat conduction through the wall, (b) obtain a
relation for the variation of temperature in the wall by solving
the differential equation, and (c) evaluate the temperature of
the right surface of the wall at x = L. Answer-, (c) -4°C
V
FIGURE P2-64
2-65 Repeat Problem 2-64 for a heat flux of 950 W/m 2 and
a surface temperature of 85 °C at the left surface at x = 0.
2-66E A large steel plate having a thickness of L = 4 in.,
thermal conductivity of k = 7.2 Btu/h • ft • °F, and an emissiv-
ity of e = 0.6 is lying on the ground. The exposed surface of
the plate at x = L is known to exchange heat by convection
with the ambient air at !T„ = 90°F with an average heat transfer
coefficient of h = 12 Btu/h ■ ft 2 • °F as well as by radiation with
the open sky with an equivalent sky temperature of r sky =
5 1 R. Also, the temperature of the upper surface of the plate is
measured to be 75°F. Assuming steady one-dimensional heat
transfer, (a) express the differential equation and the boundary
conditions for heat conduction through the plate, (b) obtain a
relation for the variation of temperature in the plate by solving
Radiation
75°F,
jU
h, T x
Convection
Plate
Ground
FIGURE P2-66E
the differential equation, and (c) determine the value of the
lower surface temperature of the plate at x = 0.
2-67E Repeat Problem 2-66E by disregarding radiation heat
transfer.
2-68 When a long section of a compressed air line passes
through the outdoors, it is observed that the moisture in the
compressed air freezes in cold weather, disrupting and even
completely blocking the air flow in the pipe. To avoid this
problem, the outer surface of the pipe is wrapped with electric
strip heaters and then insulated.
Consider a compressed air pipe of length L = 6m, inner ra-
dius /•[ = 3.7 cm, outer radius r 2 = 4.0 cm, and thermal con-
ductivity k = 14 W/m • °C equipped with a 300-W strip heater.
Air is flowing through the pipe at an average temperature of
— 10°C, and the average convection heat transfer coefficient on
the inner surface is h = 30 W/m 2 • °C. Assuming 15 percent of
the heat generated in the strip heater is lost through the insula-
tion, (a) express the differential equation and the boundary
conditions for steady one-dimensional heat conduction through
the pipe, (b) obtain a relation for the variation of temperature in
the pipe material by solving the differential equation, and
(c) evaluate the inner and outer surface temperatures of the
pipe. Answers: (c) -3.91°C, -3.87°C
Electric heater
Compressed air ■
-10°C
I
Insulation
FIGURE P2-68
2-69
Reconsider Problem 2-68. Using the relation ob-
tained for the variation of temperature in the pipe
material, plot the temperature as a function of the radius r in
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120
HEAT TRANSFER
the range of r = r x to r = r 2 , and discuss the results. Use the
EES (or other) software.
2-70 In a food processing facility, a spherical container of
inner radius r x = 40 cm, outer radius r 2 = 41 cm, and thermal
conductivity k = 1 .5 W/m • °C is used to store hot water and to
keep it at 100°C at all times. To accomplish this, the outer sur-
face of the container is wrapped with a 500-W electric strip
heater and then insulated. The temperature of the inner surface
of the container is observed to be nearly 100°C at all times. As-
suming 10 percent of the heat generated in the heater is lost
through the insulation, (a) express the differential equation and
the boundary conditions for steady one-dimensional heat con-
duction through the container, (b) obtain a relation for the vari-
ation of temperature in the container material by solving the
differential equation, and (c) evaluate the outer surface tem-
perature of the container. Also determine how much water at
100°C this tank can supply steadily if the cold water enters
at 20°C.
Insulation
Spherical
container
FIGURE P2-70
2-71
Reconsider Problem 2-70. Using the relation ob-
tained for the variation of temperature in the con-
tainer material, plot the temperature as a function of the radius
r in the range of r = r { to r = r 2 , and discuss the results. Use
the EES (or other) software.
Heat Generation in a Solid
2-72C Does heat generation in a solid violate the first law of
thermodynamics, which states that energy cannot be created or
destroyed? Explain.
2-73C What is heat generation? Give some examples.
2-74C An iron is left unattended and its base temperature
rises as a result of resistance heating inside. When will the rate
of heat generation inside the iron be equal to the rate of heat
loss from the iron?
2-75C Consider the uniform heating of a plate in an envi-
ronment at a constant temperature. Is it possible for part of the
heat generated in the left half of the plate to leave the plate
through the right surface? Explain.
2-76C Consider uniform heat generation in a cylinder and a
sphere of equal radius made of the same material in the same
environment. Which geometry will have a higher temperature
at its center? Why?
2-77 A 2-kW resistance heater wire with thermal conductiv-
ity of k = 20 W/m ■ °C, a diameter of D = 5 mm, and a length
of L = 0.7 m is used to boil water. If the outer surface temper-
ature of the resistance wire is T s = 110°C, determine the tem-
perature at the center of the wire.
110°C
^D^
I
FIGURE P2-77
2-78 Consider a long solid cylinder of radius r = 4 cm and
thermal conductivity k = 25 W/m ■ °C. Heat is generated in the
cylinder uniformly at a rate of g = 35 W/cm 3 . The side surface
of the cylinder is maintained at a constant temperature of T s =
80°C. The variation of temperature in the cylinder is given by
T(r)
grp
k
1
+ T,
Based on this relation, determine (a) if the heat conduction is
steady or transient, (b) if it is one-, two-, or three-dimensional,
and (c) the value of heat flux on the side surface of the cylinder
at r = r .
2-79 TtPM Reconsider Problem 2-78. Using the relation
fc^S obtained for the variation of temperature in the
cylinder, plot the temperature as a function of the radius r in
the range of r = to r = r , and discuss the results. Use the
EES (or other) software.
2-80E A long homogeneous resistance wire of radius r =
0.25 in. and thermal conductivity k = 8.6 Btu/h ■ ft • °F is being
used to boil water at atmospheric pressure by the passage of
Water
h
_ _,
►
I"
^ Resistance
heater
FIGURE P2-80E
cen58933_ch02.qxd 9/10/2002 8:47 AM Page 121
electric current. Heat is generated in the wire uniformly as a
result of resistance heating at a rate of g = 1800 Btu/h • in 3 .
The heat generated is transferred to water at 212°F by con-
vection with an average heat transfer coefficient of h = 820
Btu/h • ft 2 ■ °F. Assuming steady one-dimensional heat transfer,
(a) express the differential equation and the boundary condi-
tions for heat conduction through the wire, (b) obtain a relation
for the variation of temperature in the wire by solving the dif-
ferential equation, and (c) determine the temperature at the
centerline of the wire. Answer: (c) 290. 8°F
2-81E [?(,">! Reconsider Problem 2-80E. Using the relation
I^S obtained for the variation of temperature in the
wire, plot the temperature at the centerline of the wire as a
function of the heat generation g in the range of 400 Btu/h • in 3
to 2400 Btu/h • in 3 , and discuss the results. Use the EES (or
other) software.
2-82 In a nuclear reactor, 1 -cm -diameter cylindrical uranium
rods cooled by water from outside serve as the fuel. Heat is
generated uniformly in the rods (k = 29.5 W/m • °C) at a rate
of 7 X 10 7 W/m 3 . If the outer surface temperature of rods is
175°C, determine the temperature at their center.
s- 175°C
Uranium rod
FIGURE P2-82
2-83 Consider a large 3-cm-thick stainless steel plate (k =
15.1 W/m • °C) in which heat is generated uniformly at a rate
of 5 X 10 5 W/m 3 . Both sides of the plate are exposed to an en-
vironment at 30°C with a heat transfer coefficient of 60 W/m 2
■ °C. Explain where in the plate the highest and the lowest tem-
peratures will occur, and determine their values.
2-84 Consider a large 5-cm-thick brass plate (k = 111
W/m • °C) in which heat is generated uniformly at a rate of
2 X 10 5 W/m 3 . One side of the plate is insulated while the other
side is exposed to an environment at 25°C with a heat transfer
Brass
plate
h
T
0"
121
CHAPTER 1
coefficient of 44 W/m 2 ■ °C. Explain where in the plate the
highest and the lowest temperatures will occur, and determine
their values.
2-85 Tu'M Reconsider Problem 2-84. Using EES (or other)
k^^ software, investigate the effect of the heat trans-
fer coefficient on the highest and lowest temperatures in the
plate. Let the heat transfer coefficient vary from 20 W/m 2 ■ °C
to 100 W/m 2 ■ °C. Plot the highest and lowest temperatures as
a function of the heat transfer coefficient, and discuss the
results.
2-86 A 6-m-long 2-kW electrical resistance wire is made of
0.2-cm-diameter stainless steel (k = 15.1 W/m • °C). The re-
sistance wire operates in an environment at 30°C with a heat
transfer coefficient of 140 W/m 2 • °C at the outer surface. De-
termine the surface temperature of the wire (a) by using the ap-
plicable relation and (b) by setting up the proper differential
equation and solving it. Answers: (a) 409°C, (b) 409°C
2-87E Heat is generated uniformly at a rate of 3 kW per ft
length in a 0.08-in. -diameter electric resistance wire made of
nickel steel {k = 5.8 Btu/h • ft • °F). Determine the temperature
difference between the centerline and the surface of the wire.
2-88E Repeat Problem 2-87E for a manganese wire (k =
4.5 Btu/h ■ ft ■ °F).
2-89 Consider a homogeneous spherical piece of radioactive
material of radius r = 0.04 m that is generating heat at a con-
stant rate of g = 4 X 10 7 W/m 3 . The heat generated is dissi-
pated to the environment steadily. The outer surface of the
sphere is maintained at a uniform temperature of 80°C and
the thermal conductivity of the sphere is k = 15 W/m ■ °C. As-
suming steady one -dimensional heat transfer, (a) express the
differential equation and the boundary conditions for heat con-
duction through the sphere, (b) obtain a relation for the varia-
tion of temperature in the sphere by solving the differential
equation, and (c) determine the temperature at the center of the
sphere.
FIGURE P2-89
2-90
FIGURE P2-84
rSi'M Reconsider Problem 2-89. Using the relation ob-
1^2 tained for the variation of temperature in the
sphere, plot the temperature as a function of the radius r in the
range of r = to r = r . Also, plot the center temperature of
the sphere as a function of the thermal conductivity in the
range of 10 W/m ■ °C to 400 W/m • °C. Discuss the results. Use
the EES (or other) software.
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HEAT TRANSFER
2-91 A long homogeneous resistance wire of radius r =
5 mm is being used to heat the air in a room by the passage of
electric current. Heat is generated in the wire uniformly at a
rate of g = 5 X 10 7 W/m 3 as a result of resistance heating. If
the temperature of the outer surface of the wire remains at
1 80°C, determine the temperature at r = 2 mm after steady op-
eration conditions are reached. Take the thermal conductivity
of the wire to be k = 8 W/m ■ °C. Answer: 212. 8°C
180°C
FIGURE P2-91
2-92 Consider a large plane wall of thickness L = 0.05 m.
The wall surface at x = is insulated, while the surface at x =
L is maintained at a temperature of 30°C. The thermal conduc-
tivity of the wall is k = 30 W/m • °C, and heat is generated in
the wall at a rate of g = g <?~°' 5,/L W/m 3 where j = 8X 10 6
W/m 3 . Assuming steady one-dimensional heat transfer, (a) ex-
press the differential equation and the boundary conditions for
heat conduction through the wall, (b) obtain a relation for the
variation of temperature in the wall by solving the differential
equation, and (c) determine the temperature of the insulated
surface of the wall. Answer: (c) 314°C
2-93 [ft^S Reconsider Problem 2-92. Using the relation
H^«2 given for the heat generation in the wall, plot the
heat generation as a function of the distance x in the range of
x = to x = L, and discuss the results. Use the EES (or other)
software.
always equivalent to the conductivity value at the average tem-
perature?
2-99 Consider a plane wall of thickness L whose thermal
conductivity varies in a specified temperature range as k(T) =
k (l + (3T 2 ) where k and (3 are two specified constants. The
wall surface at x = is maintained at a constant temperature of
7*!, while the surface at x = L is maintained at T 2 . Assuming
steady one-dimensional heat transfer, obtain a relation for the
heat transfer rate through the wall.
2-100 Consider a cylindrical shell of length L, inner radius
r u and outer radius r 2 whose thermal conductivity varies
linearly in a specified temperature range as k(T) = k (l + (37)
where k and (3 are two specified constants. The inner surface
of the shell is maintained at a constant temperature of 7\, while
the outer surface is maintained at T 2 . Assuming steady one-
dimensional heat transfer, obtain a relation for (a) the heat
transfer rate through the wall and (b) the temperature distribu-
tion T(r) in the shell.
Cylindrical
FIGURE P2-1 00
Variable Thermal Conductivity, k[T)
2-94C Consider steady one-dimensional heat conduction in
a plane wall, long cylinder, and sphere with constant thermal
conductivity and no heat generation. Will the temperature in
any of these mediums vary linearly? Explain.
2-95C Is the thermal conductivity of a medium, in general,
constant or does it vary with temperature?
2-96C Consider steady one-dimensional heat conduction in
a plane wall in which the thermal conductivity varies linearly.
The error involved in heat transfer calculations by assuming
constant thermal conductivity at the average temperature is
(a) none, {b) small, or (c) significant.
2-97C The temperature of a plane wall during steady one-
dimensional heat conduction varies linearly when the thermal
conductivity is constant. Is this still the case when the ther-
mal conductivity varies linearly with temperature?
2-98C When the thermal conductivity of a medium varies
linearly with temperature, is the average thermal conductivity
2-101 Consider a spherical shell of inner radius r x and outer
radius r 2 whose thermal conductivity varies linearly in a speci-
fied temperature range as k(T) = k (l + (37) where k and (3
are two specified constants. The inner surface of the shell is
maintained at a constant temperature of 7\ while the outer sur-
face is maintained at T 2 . Assuming steady one-dimensional
heat transfer, obtain a relation for (a) the heat transfer rate
through the shell and (b) the temperature distribution T(r) in
the shell.
2-102 Consider a 1.5-m-high and 0.6-m-wide plate whose
thickness is 0.15 m. One side of the plate is maintained at a
constant temperature of 500 K while the other side is main-
tained at 350 K. The thermal conductivity of the plate can be
assumed to vary linearly in that temperature range as k(T) =
k {\ + 07) where k = 25 W/m • K and = 8.7 X 10~ 4 K" 1 .
Disregarding the edge effects and assuming steady one-
dimensional heat transfer, determine the rate of heat conduc-
tion through the plate. Answer: 30,800 W
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CHAPTER 1
2-103 PT^| Reconsider Problem 2-102. Using EES (or
1^13 other) software, plot the rate of heat conduction
through the plate as a function of the temperature of the hot
side of the plate in the range of 400 K to 700 K. Discuss the
results.
Special Topic: Review of Differential Equations
2-104C Why do we often utilize simplifying assumptions
when we derive differential equations?
2-105C What is a variable? How do you distinguish a de-
pendent variable from an independent one in a problem?
2-106C Can a differential equation involve more than one
independent variable? Can it involve more than one dependent
variable? Give examples.
2-107C What is the geometrical interpretation of a deriva-
tive? What is the difference between partial derivatives and or-
dinary derivatives?
2-108C What is the difference between the degree and the
order of a derivative?
2-109C Consider a function /(x, y) and its partial derivative
df/dx. Under what conditions will this partial derivative be
equal to the ordinary derivative df/dx?
2-110C Consider a function f(x) and its derivative df/dx.
Does this derivative have to be a function of x?
2-111C How is integration related to derivation?
2-112C What is the difference between an algebraic equa-
tion and a differential equation?
2-113C What is the difference between an ordinary differen-
tial equation and a partial differential equation?
2-114C How is the order of a differential equation deter-
mined?
2-115C How do you distinguish a linear differential equation
from a nonlinear one?
2-116C How do you recognize a linear homogeneous differ-
ential equation? Give an example and explain why it is linear
and homogeneous.
2-117C How do differential equations with constant coeffi-
cients differ from those with variable coefficients? Give an ex-
ample for each type.
2-118C What kind of differential equations can be solved by
direct integration?
2-119C Consider a third order linear and homogeneous dif-
ferential equation. How many arbitrary constants will its gen-
eral solution involve?
Review Problems
2-120 Consider a small hot metal object of mass m and spe-
cific heat C that is initially at a temperature of T t . Now the ob-
ject is allowed to cool in an environment at r„ by convection
FIGURE P2-1 20
with a heat transfer coefficient of h. The temperature of the
metal object is observed to vary uniformly with time during
cooling. Writing an energy balance on the entire metal object,
derive the differential equation that describes the variation of
temperature of the ball with time, Tit). Assume constant ther-
mal conductivity and no heat generation in the object. Do not
solve.
2-121 Consider a long rectangular bar of length a in the
x-direction and width b in the y-direction that is initially at a
uniform temperature of T t . The surfaces of the bar at x = and
y = are insulated, while heat is lost from the other two sur-
faces by convection to the surrounding medium at temperature
r„ with a heat transfer coefficient of h. Assuming constant
thermal conductivity and transient two-dimensional heat trans-
fer with no heat generation, express the mathematical formula-
tion (the differential equation and the boundary and initial
conditions) of this heat conduction problem. Do not solve.
FIGURE P2-1 21
2-122 Consider a short cylinder of radius r and height H in
which heat is generated at a constant rate of g . Heat is lost
from the cylindrical surface at r = r by convection to the sur-
rounding medium at temperature T„ with a heat transfer coeffi-
cient of h. The bottom surface of the cylinder at z = is
insulated, while the top surface at z = H is subjected to uni-
form heat flux q h . Assuming constant thermal conductivity and
steady two-dimensional heat transfer, express the mathematical
formulation (the differential equation and the boundary condi-
tions) of this heat conduction problem. Do not solve.
2-123E Consider a large plane wall of thickness L = 0.5 ft
and thermal conductivity k = 1.2 Btu/h ■ ft • °F. The wall
is covered with a material that has an emissivity of e = 0.80
and a solar absorptivity of a = 0.45. The inner surface of the
wall is maintained at T x = 520 R at all times, while the outer
surface is exposed to solar radiation that is incident at a rate of
<7soiar = 300 Btu/h • ft 2 . The outer surface is also losing heat by
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HEAT TRANSFER
Plate
520 R
. 'solai
FIGURE P2-123E
radiation to deep space at K. Determine the temperature of
the outer surface of the wall and the rate of heat transfer
through the wall when steady operating conditions are reached.
Answers: 530.9 R, 26.2 Btu/h • ft 2
2-124E Repeat Problem 2-123E for the case of no solar
radiation incident on the surface.
2-125 Consider a steam pipe of length L, inner radius r u
outer radius r 2 , and constant thermal conductivity k. Steam
flows inside the pipe at an average temperature of T, with a
convection heat transfer coefficient of h t . The outer surface of
the pipe is exposed to convection to the surrounding air at a
temperature of T with a heat transfer coefficient of h B . Assum-
ing steady one-dimensional heat conduction through the pipe,
(a) express the differential equation and the boundary condi-
tions for heat conduction through the pipe material, (b) obtain
a relation for the variation of temperature in the pipe material
by solving the differential equation, and (c) obtain a relation
for the temperature of the outer surface of the pipe.
FIGURE P2-1 25
2-126 The boiling temperature of nitrogen at atmospheric
pressure at sea level (1 atm pressure) is — 196°C. Therefore, ni-
trogen is commonly used in low temperature scientific studies
since the temperature of liquid nitrogen in a tank open to the at-
mosphere will remain constant at — 196°C until the liquid ni-
trogen in the tank is depleted. Any heat transfer to the tank will
result in the evaporation of some liquid nitrogen, which has a
heat of vaporization of 198 kJ/kg and a density of 810 kg/m 3 at
1 atm.
Consider a thick-walled spherical tank of inner radius r x =
2 m, outer radius r 2 = 2.1 m , and constant thermal conductiv-
ity k = 18 W/m • °C. The tank is initially filled with liquid
nitrogen at 1 atm and — 196°C, and is exposed to ambient air
at T^ = 20°C with a heat transfer coefficient of h = 25
W/m 2 • °C. The inner surface temperature of the spherical tank
is observed to be almost the same as the temperature of the ni-
trogen inside. Assuming steady one -dimensional heat transfer,
(a) express the differential equation and the boundary condi-
tions for heat conduction through the tank, (b) obtain a relation
for the variation of temperature in the tank material by solving
the differential equation, and (c) determine the rate of evapora-
tion of the liquid nitrogen in the tank as a result of the heat
transfer from the ambient air. Answer: (c) 1.32 kg/s
2-127 Repeat Problem 2-126 for liquid oxygen, which has
a boiling temperature of — 183°C, a heat of vaporization of
213 kJ/kg, and a density of 1 140 kg/m 3 at 1 atm.
2-128 Consider a large plane wall of thickness L = 0.4 m
and thermal conductivity k = 8.4 W/m • °C. There is no access
to the inner side of the wall at x = and thus the thermal con-
ditions on that surface are not known. However, the outer sur-
face of the wall at x = L, whose emissivity is e = 0.7, is known
to exchange heat by convection with ambient air at T m = 25°C
with an average heat transfer coefficient of h = 14 W/m 2 • °C
as well as by radiation with the surrounding surfaces at an av-
erage temperature of T smT = 290 K. Further, the temperature of
the outer surface is measured to be T 2 = 45 C C. Assuming
steady one-dimensional heat transfer, (a) express the differen-
tial equation and the boundary conditions for heat conduction
through the plate, (b) obtain a relation for the temperature of
the outer surface of the plate by solving the differential equa-
tion, and (c) evaluate the inner surface temperature of the wall
at x = 0. Answer: (c) 64.3°C
Plane
wall
45°C
h
T
FIGURE P2-1 28
cen58933_ch02.qxd 9/10/2002 8:47 AM Page 125
2-129 A 1000-W iron is left on the iron board with its base
exposed to ambient air at 20°C. The base plate of the iron has
a thickness of L = 0.5 cm, base area of A = 150 cm 2 , and ther-
mal conductivity of k = 18 W/m • °C. The inner surface of the
base plate is subjected to uniform heat flux generated by the re-
sistance heaters inside. The outer surface of the base plate
whose emissivity is e = 0.7, loses heat by convection to ambi-
ent air at 7U = 22° C with an average heat transfer coefficient
of h = 30 W/m 2 • °C as well as by radiation to the surrounding
surfaces at an average temperature of T mn = 290 K. Dis-
regarding any heat loss through the upper part of the iron,
(a) express the differential equation and the boundary con-
ditions for steady one-dimensional heat conduction through
the plate, (b) obtain a relation for the temperature of the outer
surface of the plate by solving the differential equation, and
(c) evaluate the outer surface temperature.
Iron
base
plate
/;
T
FIGURE P2-1 29
2-130 Repeat Problem 2-1 29 for a 1 500-W iron.
2-131E The roof of a house consists of a 0.8-ft-thick con-
crete slab (k = 1.1 Btu/h • ft • °F) that is 25 ft wide and 35 ft
long. The emissivity of the outer surface of the roof is 0.8, and
the convection heat transfer coefficient on that surface is esti-
mated to be 3.2 Btu/h • ft 2 • °F. On a clear winter night, the am-
bient air is reported to be at 50°F, while the night sky
temperature for radiation heat transfer is 310 R. If the inner
125
CHAPTER 1
surface temperature of the roof is T t = 62°F, determine the
outer surface temperature of the roof and the rate of heat loss
through the roof when steady operating conditions are reached.
2-132 Consider a long resistance wire of radius r x = 0.3 cm
and thermal conductivity A; wire =18 W/m • °C in which heat is
generated uniformly at a constant rate of g = 1.5 W/cm 3 as a
result of resistance heating. The wire is embedded in a 0.4-cm-
thick layer of plastic whose thermal conductivity is A: plastic =1.8
W/m ■ °C. The outer surface of the plastic cover loses heat by
convection to the ambient air at T a = 25°C with an average
combined heat transfer coefficient of h = 14 W/m 2 ■ °C. As-
suming one -dimensional heat transfer, determine the tempera-
tures at the center of the resistance wire and the wire -plastic
layer interface under steady conditions.
/lnswers;97.1°C, 97.3°C
- Plastic cover
FIGURE P2-1 32
2-133 Consider a cylindrical shell of length L, inner radius
r u and outer radius r 2 whose thermal conductivity varies in
a specified temperature range as k(T) = fc (l + pr 2 ) where
k and p are two specified constants. The inner surface of the
shell is maintained at a constant temperature of 7", while
the outer surface is maintained at T 2 . Assuming steady one-
dimensional heat transfer, obtain a relation for the heat transfer
rate through the shell.
2-134 In a nuclear reactor, heat is generated in 1 -cm-
diameter cylindrical uranium fuel rods at a rate of 4 X
10 7 W/m 3 . Determine the temperature difference between the
center and the surface of the fuel rod. Answer: 9.0°C
• v t
h
Concrete
FIGURE P2-1 31 E
D
Fuel rod
FIGURE P2-1 34
2-135 Consider a 20-cm-thick large concrete plane wall
(k = 0.77 W/m ■ °C) subjected to convection on both sides with
r„, = 27°C and h x = 5 W/m 2 • °C on the inside, and T x2 = 8°C
and h 2 = 12 W/m 2 • °C on the outside. Assuming constant
thermal conductivity with no heat generation and negligible
cen58933_ch02.qxd 9/10/2002 8:47 AM Page 126
126
HEAT TRANSFER
radiation, (a) express the differential equations and the bound-
ary conditions for steady one-dimensional heat conduction
through the wall, {b) obtain a relation for the variation of tem-
perature in the wall by solving the differential equation, and
(c) evaluate the temperatures at the inner and outer surfaces of
the wall.
2-136 Consider a water pipe of length L = 12 m, inner ra-
dius r, = 15 cm, outer radius r 2 = 20 cm, and thermal conduc-
tivity k = 20 W/m ■ °C. Heat is generated in the pipe material
uniformly by a 25-kW electric resistance heater. The inner and
outer surfaces of the pipe are at T t = 60°C and T 2 = 80°C, re-
spectively. Obtain a general relation for temperature distribu-
tion inside the pipe under steady conditions and determine the
temperature at the center plane of the pipe.
2-137 Heat is generated uniformly at a rate of 2.6 X 10 6
W/m 3 in a spherical ball (k = 45 W/m • °C) of diameter 30 cm.
The ball is exposed to iced-water at 0°C with a heat transfer co-
efficient of 1200 W/m 2 • °C. Determine the temperatures at the
center and the surface of the ball.
Computer, Design, and Essay Problems
2-138 Write an essay on heat generation in nuclear fuel rods.
Obtain information on the ranges of heat generation, the varia-
tion of heat generation with position in the rods, and the ab-
sorption of emitted radiation by the cooling medium.
2-139 f^tb Write an interactive computer program to calcu-
xifv7 late the heat transfer rate and the value of tem-
perature anywhere in the medium for steady one-dimensional
heat conduction in a long cylindrical shell for any combination
of specified temperature, specified heat flux, and convection
boundary conditions. Run the program for five different sets of
specified boundary conditions.
2-140 Write an interactive computer program to calculate the
heat transfer rate and the value of temperature anywhere in
the medium for steady one-dimensional heat conduction in
a spherical shell for any combination of specified tempera-
ture, specified heat flux, and convection boundary conditions.
Run the program for five different sets of specified boundary
conditions.
2-141 Write an interactive computer program to calculate the
heat transfer rate and the value of temperature anywhere in the
medium for steady one-dimensional heat conduction in a plane
wall whose thermal conductivity varies linearly as k(T) =
k (l + (37) where the constants k and p are specified by the
user for specified temperature boundary conditions.
cen58933_ch03.qxd 9/10/2002 8:58 AM Page 127
STEADY HEAT CONDUCTION
CHAPTER
In heat transfer analysis, we are often interested in the rate of heat transfer
through a medium under steady conditions and surface temperatures. Such
problems can be solved easily without involving any differential equations
by the introduction of thermal resistance concepts in an analogous manner to
electrical circuit problems. In this case, the thermal resistance corresponds
to electrical resistance, temperature difference corresponds to voltage, and the
heat transfer rate corresponds to electric current.
We start this chapter with one-dimensional steady heat conduction in
a plane wall, a cylinder, and a sphere, and develop relations for thermal resis-
tances in these geometries. We also develop thermal resistance relations for
convection and radiation conditions at the boundaries. We apply this concept
to heat conduction problems in multilayer plane walls, cylinders, and spheres
and generalize it to systems that involve heat transfer in two or three dimen-
sions. We also discuss the thermal contact resistance and the overall heat
transfer coefficient and develop relations for the critical radius of insulation
for a cylinder and a sphere. Finally, we discuss steady heat transfer from
finned surfaces and some complex geometries commonly encountered in
practice through the use of conduction shape factors.
CONTENTS
3-1 Steady Heat Conduction
in Plane Walls 128
3-2 Thermal Contact
Resistance 138
3-3 Generalized Thermal
Resistance Networks 143
3-4 Heat Conduction in
Cylinders and Spheres 146
3-5 Critical Radius
of Insulation 153
3-6 Heat Transfer from
Finned Surfaces 156
3-7 Heat Transfer in
Common Configurations 169
Topic of Special Interest:
Heat Transfer Through
Walls and Roofs 175
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128
HEAT TRANSFER
20°C
20°C
20°C
20°C
20°C
20°C
20°C,
11°C
20=
11°C
f ♦ <
ire
♦ <
• 3°C
• 3°C
ire
♦ <
11°C
L ♦ *
\j(x)
11°C\
11°C
' ♦ » <
• 3°C
• 3°C
3°C
At
♦ 3°C
3°C
3°C
3'C
3'C
3'C
3'C
+ Q
3'C
3'C
FIGURE 3-1
Heat flow through a wall is one-
dimensional when the temperature of
the wall varies in one direction only.
3-1 - STEADY HEAT CONDUCTION IN PLANE WALLS
Consider steady heat conduction through the walls of a house during a winter
day. We know that heat is continuously lost to the outdoors through the wall.
We intuitively feel that heat transfer through the wall is in the normal direc-
tion to the wall surface, and no significant heat transfer takes place in the wall
in other directions (Fig. 3-1).
Recall that heat transfer in a certain direction is driven by the temperature
gradient in that direction. There will be no heat transfer in a direction in which
there is no change in temperature. Temperature measurements at several loca-
tions on the inner or outer wall surface will confirm that a wall surface is
nearly isothermal. That is, the temperatures at the top and bottom of a wall
surface as well as at the right or left ends are almost the same. Therefore, there
will be no heat transfer through the wall from the top to the bottom, or from
left to right, but there will be considerable temperature difference between the
inner and the outer surfaces of the wall, and thus significant heat transfer in
the direction from the inner surface to the outer one.
The small thickness of the wall causes the temperature gradient in that
direction to be large. Further, if the air temperatures in and outside the house
remain constant, then heat transfer through the wall of a house can be modeled
as steady and one-dimensional. The temperature of the wall in this case
will depend on one direction only (say the x-direction) and can be expressed
as T(x).
Noting that heat transfer is the only energy interaction involved in this case
and there is no heat generation, the energy balance for the wall can be ex-
pressed as
( Rate of \
heat transfer
linto the wall/
1 Rate of \
heat transfer
\ out of the wall/
/Rate of change]
of the energy
\ of the wall J
or
i£in t^out
dE^
dt
(3-1)
But dE mU /dt = for steady operation, since there is no change in the temper-
ature of the wall with time at any point. Therefore, the rate of heat transfer into
the wall must be equal to the rate of heat transfer out of it. In other words, the
rate of heat transfer through the wall must be constant, Q cond wall = constant.
Consider a plane wall of thickness L and average thermal conductivity k.
The two surfaces of the wall are maintained at constant temperatures of
T { and T 2 . For one-dimensional steady heat conduction through the wall,
we have T(x). Then Fourier's law of heat conduction for the wall can be
expressed as
fie
-kA
dT
dx
(W)
(3-2)
where the rate of conduction heat transfer Q cond wall and the wall area A are
constant. Thus we have dTldx = constant, which means that the temperature
cen58933_ch03.qxd 9/10/2002 8:58 AM Page 129
through the wall varies linearly with x. That is, the temperature distribution in
the wall under steady conditions is a straight line (Fig. 3-2).
Separating the variables in the above equation and integrating from x = 0,
where T(0) = T h tox = L, where T(L) = T 2 , we get
[I
Gc
dx
kAdT
Performing the integrations and rearranging gives
xl- cond
kA-
(W)
(3-3)
which is identical to Eq. 3—1. Again, the rate of heat conduction through
a plane wall is proportional to the average thermal conductivity, the wall
area, and the temperature difference, but is inversely proportional to the
wall thickness. Also, once the rate of heat conduction is available, the tem-
perature T(x) at any location x can be determined by replacing T 2 in Eq. 3-3
by T, and L by x.
The Thermal Resistance Concept
Equation 3-3 for heat conduction through a plane wall can be rearranged as
Q
r,
cond, wall
(W)
(3-4)
129
CHAPTER 3
L x
FIGURE 3-2
Under steady conditions,
the temperature distribution in
a plane wall is a straight line.
where
L_
kA
(°C/W)
(3-5)
is the thermal resistance of the wall against heat conduction or simply the
conduction resistance of the wall. Note that the thermal resistance of a
medium depends on the geometry and the thermal properties of the medium.
The equation above for heat flow is analogous to the relation for electric
current flow I, expressed as
R,
(3-6)
where R e = Llu e A is the electric resistance and \ l — V 2 is the voltage differ-
ence across the resistance (<j e is the electrical conductivity). Thus, the rate of
heat transfer through a layer corresponds to the electric current, the thermal
resistance corresponds to electrical resistance, and the temperature difference
corresponds to voltage difference across the layer (Fig. 3-3).
Consider convection heat transfer from a solid surface of area A s and tem-
perature T s to a fluid whose temperature sufficiently far from the surface is !T m ,
with a convection heat transfer coefficient h. Newton's law of cooling for con-
vection heat transfer rate Q com = hA s (T s — T^) can be rearranged as
2c
r«
(W)
(3-7)
• T, - T.
Q = —
T i — -WWW — ■ T 2
R
V -V
. M 12
R
(a) Heat flow
v i« VWW\A
-v,
(b) Electric current flow
FIGURE 3-3
Analogy between thermal
and electrical resistance concepts.
cen58933_ch03.qxd 9/10/2002 8:59 AM Page 13C
130
HEAT TRANSFER
Solid
r.
AAAAAA^
->T X
R.
1
com M ^
FIGURE 3-4
Schematic for convection
resistance at a surface.
where
R,,
hA,
(°C/W)
(3-8)
is the thermal resistance of the surface against heat convection, or simply the
convection resistance of the surface (Fig. 3-4). Note that when the convec-
tion heat transfer coefficient is very large (h — > °°), the convection resistance
becomes zero and T s ~ T m . That is, the surface offers no resistance to convec-
tion, and thus it does not slow down the heat transfer process. This situation is
approached in practice at surfaces where boiling and condensation occur. Also
note that the surface does not have to be a plane surface. Equation 3-8 for
convection resistance is valid for surfaces of any shape, provided that the as-
sumption of h = constant and uniform is reasonable.
When the wall is surrounded by a gas, the radiation effects, which we have
ignored so far, can be significant and may need to be considered. The rate of
radiation heat transfer between a surface of emissivity e and area A s at tem-
perature T s and the surrounding surfaces at some average temperature T smr can
be expressed as
Q md = suAAT? - r«J = h mi A s (T s - T m ) = T ' Tsm
"rad
(W)
(3-9)
where
1
"rad"™j
(K/W)
(3-10)
is the thermal resistance of a surface against radiation, or the radiation re-
sistance, and
Q,
-"■A-* s -*■ surr/
eCT (r s 2 + r s 2 urr )(7; + r surr )
(W/m 2 • K)
(3-11)
V
T<-
r^A/VvW — * T °°
R
Solid
rad
Q=Q +Q ,
FIGURE 3-5
Schematic for convection and
radiation resistances at a surface.
is the radiation heat transfer coefficient. Note that both T s and T^ must be
in K in the evaluation of h md . The definition of the radiation heat transfer co-
efficient enables us to express radiation conveniently in an analogous manner
to convection in terms of a temperature difference. But h lad depends strongly
on temperature while /i conv usually does not.
A surface exposed to the surrounding air involves convection and radiation
simultaneously, and the total heat transfer at the surface is determined by
adding (or subtracting, if in the opposite direction) the radiation and convec-
tion components. The convection and radiation resistances are parallel to each
other, as shown in Fig. 3-5, and may cause some complication in the thermal
resistance network. When T smr ~ T m , the radiation effect can properly be ac-
counted for by replacing h in the convection resistance relation by
K,
(W/m 2 • K)
(3-12)
where /z combined is the combined heat transfer coefficient. This way all the
complications associated with radiation are avoided.
cen58933_ch03.qxd 9/10/2002 8:59 AM Page 131
R , + R „ + R -,
conv, 1 wall conv, 2
T".,-
conv, 1
AWvW
wall
AAA/VW^
131
CHAPTER 3
A/VWW^
-*T,
Thermal
^2 network
T, -T,
R , +R ,+fi ,
e, 1 e, 2 (*, 3
Tj-
=1 AAAAAA^
«,2
AAAAAAA
AAAAMA
-.T,
Electrical
analogy
FIGURE 3-6
The thermal resistance network for heat transfer through a plane wall subjected to convection on both sides,
and the electrical analogy.
Thermal Resistance Network
Now consider steady one-dimensional heat flow through a plane wall of thick-
ness L, area A, and thermal conductivity k that is exposed to convection on
both sides to fluids at temperatures T m \ and T m2 with heat transfer coefficients
hi and h 2 , respectively, as shown in Fig. 3-6. Assuming T m2 < T m{ , the varia-
tion of temperature will be as shown in the figure. Note that the temperature
varies linearly in the wall, and asymptotically approaches T^ and T„, 2 in the
fluids as we move away from the wall.
Under steady conditions we have
/ Rate of \ / Rate of \ / Rate of \
heat convection = heat conduction = heat convection
\ into the wall / \ through the wall / \ from the wall /
or
Q =h t A(T xl -T l ) = kA-
T, -T 7
h 2 A(T 2 - T x2 )
which can be rearranged as
Q
l/h,A LlkA
\lh 2 A
r„i - r, _ r, -t 2 _t 2 - t^ 2
D
**rnnv 1
"wall ^xonv, 2
Adding the numerators and denominators yields (Fig. 3-7)
Q
R„
(W)
(3-13)
(3-14)
(3-15)
FIGURE 3-7
A useful mathematical identity.
cen58933_ch03.qxd 9/10/2002 8:59 AM Page 132
132
HEAT TRANSFER
FIGURE 3-8
The temperature drop across a layer is
proportional to its thermal resistance.
►e= iow
conv, 1 I
»i* — VWWV^
2°C/W
wall
15°C/W
AT=QR
Y conv, 2
-^^WVVW — ♦ r »2
3°C/W
where
"total
7T7 + 7T + 7^7 (° C/W )
/?,A kA h-iA
(3-16)
Note that the heat transfer area A is constant for a plane wall, and the rate of
heat transfer through a wall separating two mediums is equal to the tempera-
ture difference divided by the total thermal resistance between the mediums.
Also note that the thermal resistances are in series, and the equivalent thermal
resistance is determined by simply adding the individual resistances, just like
the electrical resistances connected in series. Thus, the electrical analogy still
applies. We summarize this as the rate of steady heat transfer between two
surfaces is equal to the temperature difference divided by the total thermal re-
sistance between those two surfaces.
Another observation that can be made from Eq. 3-15 is that the ratio of the
temperature drop to the thermal resistance across any layer is constant, and
thus the temperature drop across any layer is proportional to the thermal
resistance of the layer. The larger the resistance, the larger the temperature
drop. In fact, the equation Q = AT/R can be rearranged as
AT = QR
(°C)
(3-17)
which indicates that the temperature drop across any layer is equal to the rate
of heat transfer times the thermal resistance across that layer (Fig. 3-8). You
may recall that this is also true for voltage drop across an electrical resistance
when the electric current is constant.
It is sometimes convenient to express heat transfer through a medium in an
analogous manner to Newton's law of cooling as
Q = UA AT
(W)
(3-18)
where U is the overall heat transfer coefficient. A comparison of Eqs. 3-15
and 3-18 reveals that
cen58933_ch03.qxd 9/10/2002 8:59 AM Page 133
r »i« WWW WWW
R, = L l
1 k t A
-A/WWW
R 1= L 2
-AA/WW » r «2
conv ' 2 h 2 A
133
CHAPTER 3
FIGURE 3-9
The thermal resistance network for
heat transfer through a two-layer
plane wall subjected to
convection on both sides.
UA
R„
(3-19)
Therefore, for a unit area, the overall heat transfer coefficient is equal to the
inverse of the total thermal resistance.
Note that we do not need to know the surface temperatures of the wall in or-
der to evaluate the rate of steady heat transfer through it. All we need to know
is the convection heat transfer coefficients and the fluid temperatures on both
sides of the wall. The surface temperature of the wall can be determined as
described above using the thermal resistance concept, but by taking the
surface at which the temperature is to be determined as one of the terminal
surfaces. For example, once Q is evaluated, the surface temperature T x can be
determined from
Q
l//i, A
(3-20)
Multilayer Plane Walls
In practice we often encounter plane walls that consist of several layers of dif-
ferent materials. The thermal resistance concept can still be used to determine
the rate of steady heat transfer through such composite walls. As you may
have already guessed, this is done by simply noting that the conduction resis-
tance of each wall is LlkA connected in series, and using the electrical analogy.
That is, by dividing the temperature difference between two surfaces at known
temperatures by the total thermal resistance between them.
Consider a plane wall that consists of two layers (such as a brick wall with
a layer of insulation). The rate of steady heat transfer through this two-layer
composite wall can be expressed as (Fig. 3-9)
Q
(3-21)
cen58933_ch03.qxd 9/10/2002 8:59 AM Page 134
134
HEAT TRANSFER
To find 7,: G=^r L
K conv,l
To find T 2 : Q=I^zJl
To find Ty Q
«...
FIGURE 3-10
The evaluation of the surface and
interface temperatures when T„ { and
7^2 are given and Q is calculated.
where i? tota i is the total thermal resistance, expressed as
R„-
R
conv,
l
+ R*
"wall, 2 ' R*
conv. 2
(3-22)
The subscripts 1 and 2 in the R waU relations above indicate the first and the
second layers, respectively. We could also obtain this result by following the
approach used above for the single-layer case by noting that the rate of steady
heat transfer Q through a multilayer medium is constant, and thus it must be
the same through each layer. Note from the thermal resistance network that
the resistances are in series, and thus the total thermal resistance is simply the
arithmetic sum of the individual thermal resistances in the path of heat flow.
This result for the two-layer case is analogous to the single-layer case, ex-
cept that an additional resistance is added for the additional layer. This result
can be extended to plane walls that consist of three or more layers by adding
an additional resistance for each additional layer.
Once Q is known, an unknown surface temperature Tj at any surface or in-
terface j can be determined from
Q
v total, i—j
total, ( — j
(3-23)
is the total thermal
where T t is a known temperature at location i and R
resistance between locations i and j. For example, when the fluid temperatures
T^ and T m2 for the two-layer case shown in Fig. 3-9 are available and Q is
calculated from Eq. 3-21, the interface temperature T 2 between the two walls
can be determined from (Fig. 3-10)
Q
T-,
r,
-"conv, 1 "T "wall, 1
(3-24)
16°C
Wall
•2°C
3 m
L = 0.3m
FIGURE 3-1 1
Schematic for Example 3-1.
The temperature drop across a layer is easily determined from Eq. 3-17 by
multiplying Q by the thermal resistance of that layer.
The thermal resistance concept is widely used in practice because it is intu-
itively easy to understand and it has proven to be a powerful tool in the solu-
tion of a wide range of heat transfer problems. But its use is limited to systems
through which the rate of heat transfer Q remains constant; that is, to systems
involving steady heat transfer with no heat generation (such as resistance
heating or chemical reactions) within the medium.
EXAMPLE 3-1 Heat Loss through a Wall
Consider a 3-m-high, 5-m-wide, and 0.3-m-thick wall whose thermal con-
ductivity is k = 0.9 W/m • °C (Fig. 3-11). On a certain day, the temperatures of
the inner and the outer surfaces of the wall are measured to be 16°C and 2 C C,
respectively. Determine the rate of heat loss through the wall on that day.
cen58933_ch03.qxd 9/10/2002 8:59 AM Page 135
SOLUTION The two surfaces of a wall are maintained at specified tempera-
tures. The rate of heat loss through the wall is to be determined.
Assumptions 1 Heat transfer through the wall is steady since the surface
temperatures remain constant at the specified values. 2 Heat transfer is one-
dimensional since any significant temperature gradients will exist in the direc-
tion from the indoors to the outdoors. 3 Thermal conductivity is constant.
Properties The thermal conductivity is given to be k = 0.9 W/m ■ C C.
Analysis Noting that the heat transfer through the wall is by conduction and
the area of the wall is/! = 3mX5m=15 m 2 , the steady rate of heat transfer
through the wall can be determined from Eq. 3-3 to be
Q=kA-
T 2 , (16
— = (0.9 W/m • °C)(15 m 2 )
2)°C
0.3 m
630 W
We could also determine the steady rate of heat transfer through the wall by
making use of the thermal resistance concept from
Q
Ar„
where
ft„
0.3 m
lwa " kA (0.9 W/m • °C)(15m 2 )
Substituting, we get
Q
0.02222°C/W
(16 - 2)°C
0.02222°C/W
630 W
Discussion This is the same result obtained earlier. Note that heat conduction
through a plane wall with specified surface temperatures can be determined
directly and easily without utilizing the thermal resistance concept. However,
the thermal resistance concept serves as a valuable tool in more complex heat
transfer problems, as you will see in the following examples.
EXAMPLE 3-2 Heat Loss through a Single-Pane Window
Consider a 0.8-m-high and 1.5-m-wide glass window with a thickness of 8 mm
and a thermal conductivity of k = 0.78 W/m • °C. Determine the steady rate of
heat transfer through this glass window and the temperature of its inner surface
for a day during which the room is maintained at 20 C C while the temperature of
the outdoors is — 10°C. Take the heat transfer coefficients on the inner and
outer surfaces of the window to be h x = 10 W/m 2 • °C and h 2 = 40 W/m 2 • °C,
which includes the effects of radiation.
SOLUTION Heat loss through a window glass is considered. The rate of
heat transfer through the window and the inner surface temperature are to be
determined.
135
CHAPTER 3
cen58933_ch03.qxd 9/10/2002 8:59 AM Page 136
136
HEAT TRANSFER
20°C-
h, = 10 W/m 2 -°C
glass
♦-WWW-h — WWW-^^WWW— ♦
Glass
-10°C
h 2 = 40 W/m 2 -°C
L = 8 mm
FIGURE 3-1 2
Schematic for Example 3-2.
Assumptions 1 Heat transfer through the window is steady since the surface
temperatures remain constant at the specified values. 2 Heat transfer through
the wall is one-dimensional since any significant temperature gradients will ex-
ist in the direction from the indoors to the outdoors. 3 Thermal conductivity is
constant.
Properties The thermal conductivity is given to be k = 0.78 W/m • °C.
Analysis This problem involves conduction through the glass window and con-
vection at its surfaces, and can best be handled by making use of the thermal
resistance concept and drawing the thermal resistance network, as shown in
Fig. 3-12. Noting that the area of the window is A = 0.8 m X 1.5 m = 1.2 m 2 ,
the individual resistances are evaluated from their definitions to be
1
1
R
h { A (10W/m 2 • °C)(1.2m 2 )
0.008 m
0.08333°C/W
slass kA (0.78 W/m • °C)( 1.2 m 2 )
1 1
Rn — R n
• 2 h 2 A (40 W/m 2 • °C)(1.2m 2 )
0.00855°C/W
0.02083°C/W
Noting that all three resistances are in series, the total resistance is
glass
0.1127°C/W
R„
0.08333 + 0.00855 + 0.02083
Then the steady rate of heat transfer through the window becomes
T^-T-^ [20-(-10)]°C
Q = -^ " = L TTT77^7^7- = 266 W
Knowing the rate of heat transfer, the inner surface temperature of the window
glass can be determined from
Q
T m , - T,
*conv, 1
-> Tj - r„] G^conv, 1
= 20°C - (266 W)(0.08333°C/W)
= -2.2°C
Discussion Note that the inner surface temperature of the window glass will be
-2.2°C even though the temperature of the air in the room is maintained at
20°C. Such low surface temperatures are highly undesirable since they cause
the formation of fog or even frost on the inner surfaces of the glass when the
humidity in the room is high.
EXAMPLE 3-3 Heat Loss through Double-Pane Windows
Consider a 0.8-m-high and 1.5-m-wide double-pane window consisting of two
4-mm-thick layers of glass (k = 0.78 W/m • °C) separated by a 10-mm-wide
stagnant air space (k = 0.026 W/m • °C). Determine the steady rate of heat
cen58933_ch03.qxd 9/10/2002 8:59 AM Page 137
transfer through this double-pane window and the temperature of its inner sur-
face for a day during which the room is maintained at 20°C while the tempera-
ture of the outdoors is - 10°C. Take the convection heat transfer coefficients on
the inner and outer surfaces of the window to be /?, = 10 W/m 2 • °C and h z =
40 W/m 2 • °C, which includes the effects of radiation.
SOLUTION A double-pane window is considered. The rate of heat transfer
through the window and the inner surface temperature are to be determined.
Analysis This example problem is identical to the previous one except that
the single 8-mm-thick window glass is replaced by two 4-mm-thick glasses that
enclose a 10-mm-wide stagnant air space. Therefore, the thermal resistance
network of this problem will involve two additional conduction resistances cor-
responding to the two additional layers, as shown in Fig. 3-13. Noting that the
area of the window is again A = 0.8 m X 1.5 m = 1.2 m 2 , the individual re-
sistances are evaluated from their definitions to be
1
1
R ' ' Rconv • 1 h y A (10 W/m 2 • °C)(1.2m 2 )
R t — R 3 — R^
0.004 m
M (0.78 W/m ■ °C)(1.2m 2 )
0.01 m
0.08333°C/W
0.00427°C/W
Rn — Rr.
k 2 A (0.026 W/m ■ °C)(1 .2 m 2 )
1 1
'■ h 2 A (40 W/m 2 • °C)(1.2m 2 )
0.3205°C/W
0.02083°C/W
Noting that all three resistances are in series, the total resistance is
R.
R,„- + R.
Rr,
glass, 1 air glass, 2 conv, 2
= 0.08333 + 0.00427 + 0.3205 + 0.00427 + 0.02083
= 0.4332°C/W
Then the steady rate of heat transfer through the window becomes
Q
R„
r„ 2 [2o-(-io)]°c
0.4332°C/W
69.2 W
which is about one-fourth of the result obtained in the previous example. This
explains the popularity of the double- and even triple-pane windows in cold
climates. The drastic reduction in the heat transfer rate in this case is due to
the large thermal resistance of the air layer between the glasses.
The inner surface temperature of the window in this case will be
QRc
20°C - (69.2 W)(0.08333°C/W) = 14.2°C
which is considerably higher than the -2.2°C obtained in the previous ex-
ample. Therefore, a double-pane window will rarely get fogged. A double-pane
window will also reduce the heat gain in summer, and thus reduce the air-
conditioning costs.
137
CHAPTER 3
Glass Glass
20°C
• '\AAAAA/-
10°C
AAAAAV •
3 o ^-2
FIGURE 3-13
Schematic for Example 3-3.
cen58933_ch03.qxd 9/10/2002 8:59 AM Page 13E
138
HEAT TRANSFER
FIGURE 3-14
Temperature distribution and heat flow
lines along two solid plates pressed
against each other for the case of
perfect and imperfect contact.
(a) Ideal (perfect) thermal contact
(b) Actual (imperfect) thermal contact
Applied load
Loading shaft
Alignment collar
Top plate
Steel ball
Pencil heaters
Heaters block
Upper test specimen —
Lower test specimen —
Lower heat flux meter
-SC
Thermocouples
Interface
y=
-t
Cold
fluid
ZP*
Cold plate
Load cell -
Steel ball -
Bottom plate — H x I
Bell jar *
base plate
FIGURE 3-15
A typical experimental setup for
the determination of thermal contact
resistance (from Song et al., Ref. 11).
3-2 ■ THERMAL CONTACT RESISTANCE
In the analysis of heat conduction through multilayer solids, we assumed
"perfect contact" at the interface of two layers, and thus no temperature drop
at the interface. This would be the case when the surfaces are perfectly smooth
and they produce a perfect contact at each point. In reality, however, even flat
surfaces that appear smooth to the eye turn out to be rather rough when ex-
amined under a microscope, as shown in Fig. 3-14, with numerous peaks and
valleys. That is, a surface is microscopically rough no matter how smooth it
appears to be.
When two such surfaces are pressed against each other, the peaks will form
good material contact but the valleys will form voids filled with air. As a re-
sult, an interface will contain numerous air gaps of varying sizes that act as
insulation because of the low thermal conductivity of air. Thus, an interface
offers some resistance to heat transfer, and this resistance per unit interface
area is called the thermal contact resistance, R c . The value of R c is deter-
mined experimentally using a setup like the one shown in Fig. 3-15, and as
expected, there is considerable scatter of data because of the difficulty in char-
acterizing the surfaces.
Consider heat transfer through two metal rods of cross-sectional area A that
are pressed against each other. Heat transfer through the interface of these two
rods is the sum of the heat transfers through the solid contact spots and the
gaps in the noncontact areas and can be expressed as
q =e t
+ e E
(3-25)
It can also be expressed in an analogous manner to Newton's law of cooling as
Q =ft f AAr interfacc (3-26)
cen58933_ch03.qxd 9/10/2002 8:59 AM Page 139
where A is the apparent interface area (which is the same as the cross-sectional
area of the rods) and AT interface is the effective temperature difference at the
interface. The quantity h c , which corresponds to the convection heat transfer
coefficient, is called the thermal contact conductance and is expressed as
QIA
K = ^= (W/m 2 • °C) (3-27)
^-* interface
It is related to thermal contact resistance by
h c QIA
That is, thermal contact resistance is the inverse of thermal contact conduc-
tance. Usually, thermal contact conductance is reported in the literature, but
the concept of thermal contact resistance serves as a better vehicle for ex-
plaining the effect of interface on heat transfer. Note that R c represents ther-
mal contact resistance per unit area. The thermal resistance for the entire
interface is obtained by dividing R c by the apparent interface area A.
The thermal contact resistance can be determined from Eq. 3-28 by
measuring the temperature drop at the interface and dividing it by the heat
flux under steady conditions. The value of thermal contact resistance depends
on the surface roughness and the material properties as well as the tem-
perature and pressure at the interface and the type of fluid trapped at the
interface. The situation becomes more complex when plates are fastened by
bolts, screws, or rivets since the interface pressure in this case is nonuniform.
The thermal contact resistance in that case also depends on the plate thick-
ness, the bolt radius, and the size of the contact zone. Thermal contact
resistance is observed to decrease with decreasing surface roughness
and increasing interface pressure, as expected. Most experimentally deter-
mined values of the thermal contact resistance fall between 0.000005 and
0.0005 m 2 • °C/W (the corresponding range of thermal contact conductance
is 2000 to 200,000 W/m 2 • °C).
When we analyze heat transfer in a medium consisting of two or more lay-
ers, the first thing we need to know is whether the thermal contact resistance
is significant or not. We can answer this question by comparing the magni-
tudes of the thermal resistances of the layers with typical values of thermal
contact resistance. For example, the thermal resistance of a 1-cm-thick layer
of an insulating material per unit surface area is
r, £j U.U1 111 „ __ 2 Of~< f\\J
K c , insulation ~ J ~ 0.04 W/m ■ °C ~ ^ '
whereas for a 1-cm-thick layer of copper, it is
^•»«- = t = 386W/m m °C = °- 000026 m ' ■ ° C/W
Comparing the values above with typical values of thermal contact resistance,
we conclude that thermal contact resistance is significant and can even domi-
nate the heat transfer for good heat conductors such as metals, but can be
139
CHAPTER 3
cen58933_ch03.qxd 9/10/2002 8:59 AM Page 14C
140
HEAT TRANSFER
TABLE 3-1
Thermal contact conductance
for aluminum plates with different
fluids at the interface for a surface
roughness of 10 |xm and interface
pressure of 1 atm (from Fried,
Ref. 5)
Contact
Fluid at the
Conductance, h c ,
Interface
W/m 2 • °C
Air
3640
Helium
9520
Hydrogen
13,900
Silicone oil
19,000
Glycerin
37,700
LU-
LU"
10--
Contact pressure (psi)
10 2 10 3
Coated with
tin/nickel alloy
Bronze
Coated with
nickel alloy
Coated
aluminum
alloy
Jji-^T
10 4
10-'
io-
10 J
10 J 10'
Contact pressure (kN/irr)
■ Uncoated
■ Coated
FIGURE 3-16
Effect of metallic coatings on
thermal contact conductance
(from Peterson, Ref. 10).
disregarded for poor heat conductors such as insulations. This is not surpris-
ing since insulating materials consist mostly of air space just like the inter-
face itself.
The thermal contact resistance can be minimized by applying a thermally
conducting liquid called a thermal grease such as silicon oil on the surfaces
before they are pressed against each other. This is commonly done when at-
taching electronic components such as power transistors to heat sinks. The
thermal contact resistance can also be reduced by replacing the air at the in-
terface by a better conducting gas such as helium or hydrogen, as shown in
Table 3-1.
Another way to minimize the contact resistance is to insert a soft metallic
foil such as tin, silver, copper, nickel, or aluminum between the two surfaces.
Experimental studies show that the thermal contact resistance can be reduced
by a factor of up to 7 by a metallic foil at the interface. For maximum effec-
tiveness, the foils must be very thin. The effect of metallic coatings on thermal
contact conductance is shown in Fig. 3-16 for various metal surfaces.
There is considerable uncertainty in the contact conductance data reported
in the literature, and care should be exercised when using them. In Table 3-2
some experimental results are given for the contact conductance between sim-
ilar and dissimilar metal surfaces for use in preliminary design calculations.
Note that the thermal contact conductance is highest (and thus the contact re-
sistance is lowest) for soft metals with smooth surfaces at high pressure.
EXAMPLE 3-4 Equivalent Thickness for Contact Resistance
The thermal contact conductance at the interface of two 1-cm-thick aluminum
plates is measured to be 11,000 W/m 2 ■ °C. Determine the thickness of the alu-
minum plate whose thermal resistance is equal to the thermal resistance of the
interface between the plates (Fig. 3-17).
SOLUTION The thickness of the aluminum plate whose thermal resistance
is equal to the thermal contact resistance is to be determined.
Properties The thermal conductivity of aluminum at room temperature is
k = 237 W/m • °C (Table A-3).
Analysis Noting that thermal contact resistance is the inverse of thermal con-
tact conductance, the thermal contact resistance is
R,
1
1
h c 1 1,000 W/nr
0.909 X 10- 4 m 2 -°C/W
For a unit surface area, the thermal resistance of a flat plate is defined as
R
L
k
where L is the thickness of the plate and k is the thermal conductivity. Setting
R = R cl the equivalent thickness is determined from the relation above to be
kR r
(237 W/m • °C)(0.909 X 10~ 4 nr • °C/W) = 0.0215 m = 2.15 cm
cen58 93 3_ch03.qxd 9/10/2002 8:59 AM Page 141
141
CHAPTER 3
TABLE 3-2
Thermal contact conductance of some metal surfaces in air (from various sources)
Surface Rough- Tempera-
Material Condition ness, |xm ture, °C
Pressure,
MPa
W/m 2
°C
Identical Metal Pairs
416 Stainless steel
Ground
2.54
90-200
0.3-2.5
3800
304 Stainless steel
Ground
1.14
20
4-7
1900
Aluminum
Ground
2.54
150
1.2-2.5
11,400
Copper
Ground
1.27
20
1.2-20
143,000
Copper
Milled
3.81
20
1-5
55,500
Copper (vacuum)
Milled
0.25
30
0.7-7
11,400
Dissimilar Metal Pairs
Stainless steel-
10
2900
Aluminum
20-30
20
20
3600
Stainless steel-
Aluminum
1.0-2.0
20
10
20
16,400
20,800
Steel Ct-30-
Aluminum
Ground
1.4-2.0
20
10
15-35
50,000
59,000
Steel Ct-30-
Aluminum
Milled
4.5-7.2
20
10
30
4800
8300
Aluminum-Copper
Ground
1.3-1.4
20
5
15
42,000
56,000
Aluminum-Copper
Milled
4.4-4.5
20
10
20-35
12,000
22,000
*Divide the given values by 5.678 to convert to Btu/h • ft 2 • °F.
Discussion Note that the interface between the two plates offers as much re-
sistance to heat transfer as a 2.3-cm-thick aluminum plate. It is interesting
that the thermal contact resistance in this case is greater than the sum of the
thermal resistances of both plates.
EXAMPLE 3-5
Contact Resistance of Transistors
Four identical power transistors with aluminum casing are attached on one side
of a 1-cm-thick 20-cm X 20-cm square copper plate (k = 386 W/m • °C) by
screws that exert an average pressure of 6 MPa (Fig. 3-18). The base area of
each transistor is 8 cm 2 , and each transistor is placed at the center of a 10-cm
X 10-cm quarter section of the plate. The interface roughness is estimated to
be about 1.5 |xm. All transistors are covered by a thick Plexiglas layer, which is
a poor conductor of heat, and thus all the heat generated at the junction of the
transistor must be dissipated to the ambient at 20°C through the back surface
of the copper plate. The combined convection/radiation heat transfer coefficient
at the back surface can be taken to be 25 W/m 2 • °C. If the case temperature of
Plate
1
1 cm
Plate
2
1 cm
Interface
Plate
1
1 cm
Equivalent ] Plate
aluminum i 2
layer
2.15 cm i 1 cm
FIGURE 3-17
Schematic for Example 3-4.
cen58933_ch03.qxd 9/10/2002 8:59 AM Page 142
142
HEAT TRANSFER
20°C
Copper
plate
FIGURE 3-
Schematic
70°C
18
for Example
Plexiglas cover
3-5.
the transistor is not to exceed 70°C, determine the maximum power each
transistor can dissipate safely, and the temperature jump at the case-plate
interface.
SOLUTION Four identical power transistors are attached on a copper plate. For
a maximum case temperature of 70°C, the maximum power dissipation and the
temperature jump at the interface are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Heat transfer can be ap-
proximated as being one-dimensional, although it is recognized that heat con-
duction in some parts of the plate will be two-dimensional since the plate area
is much larger than the base area of the transistor. But the large thermal con-
ductivity of copper will minimize this effect. 3 All the heat generated at the
junction is dissipated through the back surface of the plate since the transistors
are covered by a thick Plexiglas layer. 4 Thermal conductivities are constant.
Properties The thermal conductivity of copper is given to be k = 386
W/m ■ °C. The contact conductance is obtained from Table 3-2 to be h c =
42,000 W/m 2 • °C, which corresponds to copper-aluminum interface for the
case of 1.3-1.4 |xm roughness and 5 MPa pressure, which is sufficiently close
to what we have.
Analysis The contact area between the case and the plate is given to be 8 cm 2 ,
and the plate area for each transistor is 100 cm 2 . The thermal resistance net-
work of this problem consists of three resistances in series (interface, plate, and
convection), which are determined to be
R
1
I
aterface ^ (42,000 W/m 2 • °C)(8 X 10- 4 m 2 )
L 0.01 m
0.030°C/W
plate kA (386 W/m ■ °C)(0.01 m 2 )
1 1
h B A (25 W/m 2 • °C)(0.01 m 2 )
0.0026°C/W
4.0°C/W
The total thermal resistance is then
R„
D _(_ p _i_ p
"•interface ' Opiate ' "~a\
0.030 + 0.0026 + 4.0 = 4.0326°C/W
Note that the thermal resistance of a copper plate is very small and can be
ignored altogether. Then the rate of heat transfer is determined to be
Q
AT (70 - 20)°C
R,.
4.0326°C/W
12.4 W
Therefore, the power transistor should not be operated at power levels greater
than 12.4 W if the case temperature is not to exceed 70°C.
The temperature jump at the interface is determined from
AT
QRm
(12.4 W)(0.030°C/W) = 0.37°C
which is not very large. Therefore, even if we eliminate the thermal contact re-
sistance at the interface completely, we will lower the operating temperature of
the transistor in this case by less than 0.4°C.
cen58933_ch03.qxd 9/10/2002 8:59 AM Page 143
3-3 - GENERALIZED THERMAL RESISTANCE
NETWORKS
The thermal resistance concept or the electrical analogy can also be used to
solve steady heat transfer problems that involve parallel layers or combined
series-parallel arrangements. Although such problems are often two- or even
three-dimensional, approximate solutions can be obtained by assuming one-
dimensional heat transfer and using the thermal resistance network.
Consider the composite wall shown in Fig. 3-19, which consists of two par-
allel layers. The thermal resistance network, which consists of two parallel re-
sistances, can be represented as shown in the figure. Noting that the total heat
transfer is the sum of the heat transfers through each layer, we have
Q=Q l + Qi
Utilizing electrical analogy, we get
R 7
(7, - T 2 )
-1 + 1
R, R.
where
1
^total
Q
R, R,
T 2
"> K
RiR 2
r7Tr~,
(3-29)
(3-30)
(3-31)
since the resistances are in parallel.
Now consider the combined series-parallel arrangement shown in Fig.
3-20. The total rate of heat transfer through this composite system can again
be expressed as
Q
(3-32)
143
CHAPTER 3
Insulation
-V
,4,
CD *,
© k 2
« L ►
— www
R 2
Q = Q\+Q 2
FIGURE 3-1 9
Thermal resistance
network for two parallel layers.
Insulation
<D *i
fc.
L, = L n
h, r„
where
and
^12 + ^3 + ^conv
R 2
L 2
R t R 2
/?! + R 2
L 3
k 3 A 3 '
R 3 + R D
1
(3-33)
(3-34)
Once the individual thermal resistances are evaluated, the total resistance and
the total rate of heat transfer can easily be determined from the relations
above.
The result obtained will be somewhat approximate, since the surfaces of the
third layer will probably not be isothermal, and heat transfer between the first
two layers is likely to occur.
Two assumptions commonly used in solving complex multidimensional
heat transfer problems by treating them as one-dimensional (say, in the
-MAMAA-
Qi-
■AWI/VV^
Q
w —
*3 *c
-vww vwvw — ♦
conv
FIGURE 3-20
Thermal resistance network for
combined series-parallel arrangement.
cen58933_ch03.qxd 9/10/2002 8:59 AM Page 144
144
HEAT TRANSFER
Foam
Plaster ■
Brick
1.5 cm
22 cm
1.5 cm
FIGURE 3-21
Schematic for Example 3-6.
x-direction) using the thermal resistance network are (1) any plane wall nor-
mal to the x-axis is isothermal (i.e., to assume the temperature to vary in the
x-direction only) and (2) any plane parallel to the x-axis is adiabatic (i.e., to
assume heat transfer to occur in the x-direction only). These two assumptions
result in different resistance networks, and thus different (but usually close)
values for the total thermal resistance and thus heat transfer. The actual result
lies between these two values. In geometries in which heat transfer occurs pre-
dominantly in one direction, either approach gives satisfactory results.
EXAMPLE 3-6 Heat Loss through a Composite Wall
A 3-m-high and 5-m-wide wall consists of long 16-cm X 22-cm cross section
horizontal bricks (k = 0.72 W/m • °C) separated by 3-cm-thick plaster layers
(k = 0.22 W/m • °C). There are also 2-cm-thick plaster layers on each side of
the brick and a 3-cm-thick rigid foam (k = 0.026 W/m ■ °C) on the inner side
of the wall, as shown in Fig. 3-21. The indoor and the outdoor temperatures are
20°C and -10°C, and the convection heat transfer coefficients on the inner
and the outer sides are h 1 = 10 W/m 2 • °C and h 2 = 25 W/m 2 • °C, respectively.
Assuming one-dimensional heat transfer and disregarding radiation, determine
the rate of heat transfer through the wall.
SOLUTION The composition of a composite wall is given. The rate of heat
transfer through the wall is to be determined.
Assumptions 1 Heat transfer is steady since there is no indication of change
with time. 2 Heat transfer can be approximated as being one-dimensional since
it is predominantly in the x-direction. 3 Thermal conductivities are constant.
4 Heat transfer by radiation is negligible.
Properties The thermal conductivities are given to be k = 0.72 W/m • °C
for bricks, k = 0.22 W/m • °C for plaster layers, and k = 0.026 W/m • °C for the
rigid foam.
Analysis There is a pattern in the construction of this wall that repeats itself
every 25-cm distance in the vertical direction. There is no variation in the hori-
zontal direction. Therefore, we consider a 1-m-deep and 0.25-m-high portion of
the wall, since it is representative of the entire wall.
Assuming any cross section of the wall normal to the x-direction to be
isothermal, the thermal resistance network for the representative section of
the wall becomes as shown in Fig. 3-21. The individual resistances are eval-
uated as:
1
1
Ri ' Rconv ' 1 h y A (10 W/m 2 • °C)(0.25 X 1 m 2 )
= L__ 0.03 m
1 - foam - M - (Q Q26 w/m . o C)( q 25 xlm 2)
L 0.02 m
: 0.4°C/W
4.6°C/W
Rl Rb ****«"» kA (0.22 W/m • °C)(0.25 X 1 m 2 )
= 0.36°C/W
L 0.16m
K } - K 5 - K
5 plaster, center ^ (0.22 W/m ' °C)(0.015 X 1 III 2 )
48.48°C/W
cen58933_ch03.qxd 9/10/2002 8:59 AM Page 145
^4 — ^h,
0.16m
kA (0.72 W/m ■ °C)(0.22 X 1 m 2
1
1
h 2 A (25 W/m 2 • °C)(0.25 X 1 m 2 )
1.01°C/W
0.16°C/W
The three resistances R 3 , ff 4 , and R 5 in the middle are parallel, and their equiv-
alent resistance is determined from
R mi<l R 3 R A R 5 48.48 1.01 48.48
1.03 W/°C
which gives
tfmid = 0.97°C/W
Now all the resistances are in series, and the total resistance is
^total = Ri + Rl + R 2 + R m U + ^6 + Ro
= 0.4 + 4.6 + 0.36 + 0.97 + 0.36 + 0.16
= 6.85°C/W
Then the steady rate of heat transfer through the wall becomes
Q
[20 - (-10)]°C
6.85°C/W
4.38 W (per 0.25 m 2 surface area)
or 4.38/0.25 = 17.5 W per m 2 area. The total area of the wall is A = 3 m X 5
m = 15 m 2 . Then the rate of heat transfer through the entire wall becomes
fit,
(17.5 W/m 2 )(15 m 2 ) = 263 W
Of course, this result is approximate, since we assumed the temperature within
the wall to vary in one direction only and ignored any temperature change (and
thus heat transfer) in the other two directions.
Discussion In the above solution, we assumed the temperature at any cross
section of the wall normal to the x-direction to be isothermal. We could also
solve this problem by going to the other extreme and assuming the surfaces par-
allel to the x-direction to be adiabatic. The thermal resistance network in this
case will be as shown in Fig. 3-22. By following the approach outlined above,
the total thermal resistance in this case is determined to be ff tota | = 6.97°C/W,
which is very close to the value 6.85°C/W obtained before. Thus either ap-
proach would give roughly the same result in this case. This example demon-
strates that either approach can be used in practice to obtain satisfactory
results.
145
CHAPTER 3
T
V
1,
Adiabatic
lines
R j
■>!♦— VW-
-<wv — vw — vw — vw vw— ♦
FIGURE 3-22
Alternative thermal resistance
network for Example 3-6 for the
case of surfaces parallel to the
primary direction of heat
transfer being adiabatic.
cen58933_ch03.qxd 9/10/2002 8:59 AM Page 146
146
HEAT TRANSFER
FIGURE 3-23
Heat is lost from a hot water pipe to
the air outside in the radial direction,
and thus heat transfer from a long
pipe is one-dimensional.
FIGURE 3-24
A long cylindrical pipe (or spherical
shell) with specified inner and outer
surface temperatures T l and T 2 .
3^ - HEAT CONDUCTION IN
CYLINDERS AND SPHERES
Consider steady heat conduction through a hot water pipe. Heat is continu-
ously lost to the outdoors through the wall of the pipe, and we intuitively feel
that heat transfer through the pipe is in the normal direction to the pipe surface
and no significant heat transfer takes place in the pipe in other directions
(Fig. 3-23). The wall of the pipe, whose thickness is rather small, separates
two fluids at different temperatures, and thus the temperature gradient in the
radial direction will be relatively large. Further, if the fluid temperatures in-
side and outside the pipe remain constant, then heat transfer through the pipe
is steady. Thus heat transfer through the pipe can be modeled as steady and
one-dimensional. The temperature of the pipe in this case will depend on one
direction only (the radial r-direction) and can be expressed as T = T(r). The
temperature is independent of the azimuthal angle or the axial distance. This
situation is approximated in practice in long cylindrical pipes and spherical
containers.
In steady operation, there is no change in the temperature of the pipe with
time at any point. Therefore, the rate of heat transfer into the pipe must be
equal to the rate of heat transfer out of it. In other words, heat transfer through
the pipe must be constant, Q cond cyl = constant.
Consider a long cylindrical layer (such as a circular pipe) of inner radius r x ,
outer radius r 2 , length L, and average thermal conductivity k (Fig. 3-24). The
two surfaces of the cylindrical layer are maintained at constant temperatures
Ti and T 2 . There is no heat generation in the layer and the thermal conductiv-
ity is constant. For one-dimensional heat conduction through the cylindrical
layer, we have T(r). Then Fourier's law of heat conduction for heat transfer
through the cylindrical layer can be expressed as
e
cond, cyl
-kA
dT
dr
(W)
(3-35)
where A = 2ittL is the heat transfer area at location r. Note that A depends on
r, and thus it varies in the direction of heat transfer. Separating the variables
in the above equation and integrating from r = r u where T(r t ) = 7\, to r = r 2 ,
where T{r 2 ) = T 2 , gives
r 2 Q cond.
L
ey I
dr :
kdT
(3-36)
Substituting A = 2tjtL and performing the integrations give
T — T
Q coai ^ = 2itLk^j^ (W)
(3-37)
since Q
cond, cyl
constant. This equation can be rearranged as
xl, con
r.
d, cyl
R.
cy I
(W)
(3-38)
cen58933_ch03.qxd 9/10/2002 8:59 AM Page 147
where
147
CHAPTER 3
v cyl
ln(r 2 lr x ) ln(Outer radius/Inner radius)
2irLk 2tt X (Length) X (Thermal conductivity)
(3-39)
is the thermal resistance of the cylindrical layer against heat conduction, or
simply the conduction resistance of the cylinder layer.
We can repeat the analysis above for a spherical layer by taking A = 4irr 2
and performing the integrations in Eq. 3-36. The result can be expressed as
Q
cond, sph
(3-40)
v sp h
where
Outer radius — Inner radius
sph A-ur^k
4Tr(Outer radius)(Inner radius)(Thermal conductivity)
(3-41)
is the thermal resistance of the spherical layer against heat conduction, or sim-
ply the conduction resistance of the spherical layer.
Now consider steady one-dimensional heat flow through a cylindrical or
spherical layer that is exposed to convection on both sides to fluids at temper-
atures r„[ and T m2 with heat transfer coefficients h { and h 2 , respectively, as
shown in Fig. 3-25. The thermal resistance network in this case consists of
one conduction and two convection resistances in series, just like the one for
the plane wall, and the rate of heat transfer under steady conditions can be ex-
pressed as
Q
(3-42)
: ^conv,l +S cyI +R conv,:
FIGURE 3-25
The thermal resistance network
for a cylindrical (or spherical)
shell subjected to convection from
both the inner and the outer sides.
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148
HEAT TRANSFER
where
Motel ^conv, 1
l
v cyl Jv conv, 2
( ln(r 2 /r,) |
1
(2-nT|L)/i| 2ttL& (2-nr 2 L)/7 2
(3-43)
for a cylindrical layer, and
^,,
#
conv, 1
l
v sph
1
(4irr 1 2 )/ !l
4m\r 2 k (4TTr?)h 2
(3-44)
/or a spherical layer. Note that A in the convection resistance relation R com =
1/liA is the surface area at which convection occurs. It is equal to A = 2vrL
for a cylindrical surface and A = 4irr 2 for a spherical surface of radius r. Also
note that the thermal resistances are in series, and thus the total thermal resis-
tance is determined by simply adding the individual resistances, just like the
electrical resistances connected in series.
Multilayered Cylinders and Spheres
Steady heat transfer through multilayered cylindrical or spherical shells can be
handled just like multilayered plane walls discussed earlier by simply add-
ing an additional resistance in series for each additional layer. For example,
the steady heat transfer rate through the three-layered composite cylinder
of length L shown in Fig. 3-26 with convection on both sides can be ex-
pressed as
Q
(3-45)
R
cyl,3
WWW <- L A/WWV ♦ T^
R
conv, 2
FIGURE 3-26
The thermal resistance network for heat transfer through a three-layered composite cylinder
subjected to convection on both sides.
cen58933_ch03.qxd 9/10/2002 8:59 AM Page 149
where i? tota i is the total thermal resistance, expressed as
R„
R
conv,
l
+ R,
cyl, 1
^cyl.2 + R
cyl, 3
R
conv, 2
ln(r 2 /r,) ln(r 3 /r 2 ) ln(r 4 /r 3 ) i
h,A,
2TxLk,
2 r nLk 1
2 r nLk i h 2 A 4
(3-46)
where A x = litr^L and A 4 = 2irr 4 L. Equation 3-46 can also be used for a
three-layered spherical shell by replacing the thermal resistances of cylindri-
cal layers by the corresponding spherical ones. Again, note from the thermal
resistance network that the resistances are in series, and thus the total thermal
resistance is simply the arithmetic sum of the individual thermal resistances in
the path of heat flow.
Once Q is known, we can determine any intermediate temperature T- } by ap-
plying the relation Q = (T t — T)/R total j_j across any layer or layers such that
Tj is a known temperature at location i and 7? total , ■ _ ■ is the total thermal resis-
tance between locations i andy (Fig. 3-27). For example, once Q has been
calculated, the interface temperature T 2 between the first and second cylindri-
cal layers can be determined from
Q
T x
"conv, 1 "r °cyl, 1
1
\n(r 2 lr x )
h x (2ixr x L) 2nLk x
(3-47)
149
CHAPTER 3
"-, T l T 2 T 3 T^
«-JVWvVv^-»- J VWWv^-^ J WVWV^^ J VWvW^»
R 2
Tv,
i-r,
conv, 1
r„
i-r 2
R conv,l +R l
r,
-T,
*i
+ R,
h
-T 3
«2
h
-T„ 2
Rt + R„,
FIGURE 3-27
The ratio AT/R across any layer is
equal to Q , which remains constant in
one-dimensional steady conduction.
We could also calculate T 2 from
t 2 - r«2
r 2
R 7 + R,
R,,
ln(r 3 /r 2 ) to(r 4 /r 3 )
1
2irLk ?
2irLk 3 h (2irr 4 L)
(3-48)
Although both relations will give the same result, we prefer the first one since
it involves fewer terms and thus less work.
The thermal resistance concept can also be used for other geometries, pro-
vided that the proper conduction resistances and the proper surface areas in
convection resistances are used.
EXAMPLE 3-7 Heat Transfer to a Spherical Container
A 3-m internal diameter spherical tank made of 2-cm-thick stainless steel
(k = 15 W/m • °C) is used to store iced water at 7"^ = 0°C. The tank is located
in a room whose temperature is T m2 = 22°C. The walls of the room are also at
22°C. The outer surface of the tank is black and heat transfer between the outer
surface of the tank and the surroundings is by natural convection and radiation.
The convection heat transfer coefficients at the inner and the outer surfaces of
the tank are h 1 = 80 W/m 2 • °C and h 2 = 10 W/m 2 • °C, respectively. Determine
(a) the rate of heat transfer to the iced water in the tank and (b) the amount of
ice at C C that melts during a 24-h period.
SOLUTION A spherical container filled with iced water is subjected to convec-
tion and radiation heat transfer at its outer surface. The rate of heat transfer
and the amount of ice that melts per day are to be determined.
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150
HEAT TRANSFER
'Vad
I — WWW 1
* — WWW — ♦ — www
Rj R,
i — WWW — i
FIGURE 3-28
Schematic for Example 3-7.
Assumptions 1 Heat transfer is steady since the specified thermal conditions at
the boundaries do not change with time. 2 Heat transfer is one-dimensional
since there is thermal symmetry about the midpoint. 3 Thermal conductivity is
constant.
Properties The thermal conductivity of steel is given to be k = 15 W/m • °C.
The heat of fusion of water at atmospheric pressure is h :f = 333.7 kJ/kg. The
outer surface of the tank is black and thus its emissivity is e = 1.
Analysis (a) The thermal resistance network for this problem is given in
Fig. 3-28. Noting that the inner diameter of the tank is D x = 3 m and the outer
diameter is D, = 3.04 m, the inner and the outer surface areas of the tank are
A, = irD? = tt(3 m) 2 = 28.3 m 2
A 2 = ttD 2 2 = ir(3.04 m) 2 = 29.0 m 2
Also, the radiation heat transfer coefficient is given by
h mi = eu(Ti+ T* 2 )(T 2 + T a2 )
But we do not know the outer surface temperature T 2 of the tank, and thus we
cannot calculate /? rad . Therefore, we need to assume a T 2 value now and check
the accuracy of this assumption later. We will repeat the calculations if neces-
sary using a revised value for T 2 .
We note that T 2 must be between 0°C and 22 C C, but it must be closer
to 0°C, since the heat transfer coefficient inside the tank is much larger. Taking
7", = 5 C C = 278 K, the radiation heat transfer coefficient is determined to be
h mi = (1)(5.67 X 10- 8 W/m 2 • K 4 )[(295 K) 2
= 5.34 W/m 2 ■ K = 5.34 W/m 2 • °C
Then the individual thermal resistances become
1 1
(278 K) 2 ][(295 + 278) K]
Ri ~ ^conv, 1
R, — R s ,
hi A, (80 W/m 2 • °C)(28.3 m 2 )
0.000442°C/W
r 2 - r {
(1.52 - 1.50) m
1 - sphere ~ ^^^ ~ 4^ (15 W / m . °Q(1.52 m)(1.50 Hi)
0.000047°C/W
1 1
R ° Rc ° m ' 2 h 2 A 2 (10 W/m 2 • °C)(29.0 m 2 )
0.00345°C/W
1
1
^^2 (5.34 W/m 2 • °C)(29.0m 2 )
0.00646°C/W
The two parallel resistances R and ff rad can be replaced by an equivalent resis-
tance ff equiv determined from
111 1
#equiv Ro -Rrad 0.00345 0.00646
444.7 W/°C
which gives
flequiv = 0.00225°C/W
cen58933_ch03.qxd 9/10/2002 8:59 AM Page 151
Now all the resistances are in series, and the total resistance is determined
to be
fltotai = Ri + Ri + fleqiuv = 0.000442 + 0.000047 + 0.00225 = 0.00274°C/W
Then the steady rate of heat transfer to the iced water becomes
Q
7^ - r„
(22 - 0)°C
0.00274°C/W
8029 W (or Q = 8.027 kJ/s)
To check the validity of our original assumption, we now determine the outer
surface temperature from
Q
T«a - T 2
*2 — T^ 2 QR cquiv
= 22°C - (8029 W)(0.00225°C/W) = 4°C
which is sufficiently close to the 5°C assumed in the determination of the radi-
ation heat transfer coefficient. Therefore, there is no need to repeat the calcu-
lations using 4°C for T z .
(b) The total amount of heat transfer during a 24-h period is
Q = QAt
.029 kJ/s)(24 X 3600 s) = 673,700 kJ
Noting that it takes 333.7 kJ of energy to melt 1 kg of ice at 0°C, the amount
of ice that will melt during a 24-h period is
Q _ 673,700 kJ
~h iS ~ 333.7 kJ/kg
2079 kg
Therefore, about 2 metric tons of ice will melt in the tank every day.
Discussion An easier way to deal with combined convection and radiation at a
surface when the surrounding medium and surfaces are at the same tempera-
ture is to add the radiation and convection heat transfer coefficients and to treat
the result as the convection heat transfer coefficient. That is, to take h = 10 +
5.34 = 15.34 W/m 2 • °C in this case. This way, we can ignore radiation since
its contribution is accounted for in the convection heat transfer coefficient. The
convection resistance of the outer surface in this case would be
1
1
Combined ^2 (15.34 W/m 2 • °C)(29.0 m 2 )
0.00225°C/W
which is identical to the value obtained for the equivalent resistance for the par-
allel convection and the radiation resistances.
151
CHAPTER 3
EXAMPLE 3-8 Heat Loss through an Insulated Steam Pipe
Steam at 7" xl = 320°C flows in a cast iron pipe (k = 80 W/m • °C) whose inner
and outer diameters are Dj = 5 cm and D 2 = 5.5 cm, respectively. The pipe is
covered with 3-cm-thick glass wool insulation with k = 0.05 W/m • °C. Heat is
lost to the surroundings at 7" x2 = 5°C by natural convection and radiation, with
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152
HEAT TRANSFER
Insulation
T l T 2 T 3
FIGURE 3-29
Schematic for Example 3-8.
a combined heat transfer coefficient of h 2 = 18 W/m 2 • C C. Taking the heat
transfer coefficient inside the pipe to be h 1 = 60 W/m 2 • °C, determine the rate
of heat loss from the steam per unit length of the pipe. Also determine the tem-
perature drops across the pipe shell and the insulation.
SOLUTION A steam pipe covered with glass wool insulation is subjected to
convection on its surfaces. The rate of heat transfer per unit length and the
temperature drops across the pipe and the insulation are to be determined.
Assumptions 1 Heat transfer is steady since there is no indication of any
change with time. 2 Heat transfer is one-dimensional since there is thermal
symmetry about the centerline and no variation in the axial direction. 3 Thermal
conductivities are constant. 4 The thermal contact resistance at the interface is
negligible.
Properties The thermal conductivities are given to be k = 80 W/m • °C for cast
iron and k = 0.05 W/m • °C for glass wool insulation.
Analysis The thermal resistance network for this problem involves four resis-
tances in series and is given in Fig. 3-29. Taking L = 1 m, the areas of the
surfaces exposed to convection are determined to be
A:
A 3
2vr x L = 2tt(0.025 m)(l m) = 0.157 m 2
2irr 3 L = 2tt(0.0575 m)(l m) = 0.361 m 2
Then the individual thermal resistances become
1
1
R t = R
"2 — "insulati'
h,A (60 W/m 2 • °C)(0.157m 2 )
ln(r 2 //-,)_ ln(2.75/2.5)
pipe ~ IvkyL ~ 2ir(80 W/m ■ °C)(1 m) =
m(r 3 /r 2 ) ln(5.75/2.75)
2iTk 2 L 2tt(0.05 W/m ■ °C)(1 m)
1 1
0.106°C/W
0.0002°C/W
2.35°C/W
R " /?conv ' 2 h 2 A 3 (18 W/m 2 • °C)(0.361m 2 )
0.154°C/W
Noting that all resistances are in series, the total resistance is determined to be
fltotai = Ri + Ri + R 2 + Ro = 0.106 + 0.0002 + 2.35 + 0.154 = 2.61°C/W
Then the steady rate of heat loss from the steam becomes
Q
R„
(320 - 5)°C
2.61 °C/W
121 W (per m pipe length)
The heat loss for a given pipe length can be determined by multiplying the
above quantity by the pipe length L.
The temperature drops across the pipe and the insulation are determined
from Eq. 3-17 to be
A7/ pipe = gflpipe = (121 W)(0.0002°C/W) = 0.02°C
Ar insulation = efl insulalion = (121 W)(2.35°C/W) = 284°C
That is, the temperatures between the inner and the outer surfaces of the pipe
differ by 0.02°C, whereas the temperatures between the inner and the outer
surfaces of the insulation differ by 284°C.
cen58933_ch03.qxd 9/10/2002 8:59 AM Page 153
Discussion Note that the thermal resistance of the pipe is too small relative to
the other resistances and can be neglected without causing any significant
error. Also note that the temperature drop across the pipe is practically zero,
and thus the pipe can be assumed to be isothermal. The resistance to heat flow
in insulated pipes is primarily due to insulation.
153
CHAPTER 3
3-5 - CRITICAL RADIUS OF INSULATION
We know that adding more insulation to a wall or to the attic always decreases
heat transfer. The thicker the insulation, the lower the heat transfer rate. This
is expected, since the heat transfer area A is constant, and adding insulation
always increases the thermal resistance of the wall without increasing the
convection resistance.
Adding insulation to a cylindrical pipe or a spherical shell, however, is a dif-
ferent matter. The additional insulation increases the conduction resistance of
the insulation layer but decreases the convection resistance of the surface be-
cause of the increase in the outer surface area for convection. The heat trans-
fer from the pipe may increase or decrease, depending on which effect
dominates.
Consider a cylindrical pipe of outer radius r, whose outer surface tempera-
ture T x is maintained constant (Fig. 3-30). The pipe is now insulated with a
material whose thermal conductivity is k and outer radius is r 2 . Heat is lost
from the pipe to the surrounding medium at temperature T^, with a convection
heat transfer coefficient h. The rate of heat transfer from the insulated pipe to
the surrounding air can be expressed as (Fig. 3-31)
Q
S;„ + R n
1
lnf/j/ri)
2-nLk h(2-nr 2 L)
(3-49)
The variation of Q with the outer radius of the insulation r 2 is plotted in
Fig. 3-31. The value of r 2 at which Q reaches a maximum is determined from
the requirement that dQldr 2 = (zero slope). Performing the differentiation
and solving for r 2 yields the critical radius of insulation for a cylindrical
body to be
' cr, cylinder
/;
(m)
(3-50)
Note that the critical radius of insulation depends on the thermal conductivity
of the insulation k and the external convection heat transfer coefficient h.
The rate of heat transfer from the cylinder increases with the addition of insu-
lation for r 2 < r a , reaches a maximum when r 2 = r cr , and starts to decrease for
r 2 > r a . Thus, insulating the pipe may actually increase the rate of heat trans-
fer from the pipe instead of decreasing it when r 2 < r cr
The important question to answer at this point is whether we need to be con-
cerned about the critical radius of insulation when insulating hot water pipes
or even hot water tanks. Should we always check and make sure that the outer
Insulation
FIGURE 3-30
An insulated cylindrical pipe
exposed to convection from the outer
surface and the thermal resistance
network associated with it.
FIGURE 3-31
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154
HEAT TRANSFER
radius of insulation exceeds the critical radius before we install any insula-
tion? Probably not, as explained here.
The value of the critical radius r a . will be the largest when k is large and h is
small. Noting that the lowest value of h encountered in practice is about
5 W/m 2 • °C for the case of natural convection of gases, and that the thermal
conductivity of common insulating materials is about 0.05 W/m 2 • °C, the
largest value of the critical radius we are likely to encounter is
"W, insulation __ 0.05 W/m ■ °C
~h~ 5 W/m 2 • °C
0.01 m = 1 cm
This value would be even smaller when the radiation effects are considered.
The critical radius would be much less in forced convection, often less than
1 mm, because of much larger h values associated with forced convection.
Therefore, we can insulate hot water or steam pipes freely without worrying
about the possibility of increasing the heat transfer by insulating the pipes.
The radius of electric wires may be smaller than the critical radius. There-
fore, the plastic electrical insulation may actually enhance the heat transfer
from electric wires and thus keep their steady operating temperatures at lower
and thus safer levels.
The discussions above can be repeated for a sphere, and it can be shown in
a similar manner that the critical radius of insulation for a spherical shell is
= 2k
*cr, sphere r.
(3-51)
where k is the thermal conductivity of the insulation and h is the convection
heat transfer coefficient on the outer surface.
EXAMPLE 3-9
Heat Loss from an Insulated Electric Wire
A 3-mm-diameter and 5-m-long electric wire is tightly wrapped with a 2-mm-
thick plastic cover whose thermal conductivity is k = 0.15 W/m • C C. Electrical
measurements indicate that a current of 10 A passes through the wire and there
is a voltage drop of 8 V along the wire. If the insulated wire is exposed to a
medium at 7" x = 30°C with a heat transfer coefficient of h = 12 W/m 2 • °C, de-
termine the temperature at the interface of the wire and the plastic cover in
steady operation. Also determine whether doubling the thickness of the plastic
cover will increase or decrease this interface temperature.
SOLUTION An electric wire is tightly wrapped with a plastic cover. The inter-
face temperature and the effect of doubling the thickness of the plastic cover
on the interface temperature are to be determined.
Assumptions 1 Heat transfer is steady since there is no indication of any
change with time. 2 Heat transfer is one-dimensional since there is thermal
symmetry about the centerline and no variation in the axial direction. 3 Thermal
conductivities are constant. 4 The thermal contact resistance at the interface is
negligible. 5 Heat transfer coefficient incorporates the radiation effects, if any.
Properties The thermal conductivity of plastic is given to be k = 0.15
W/m • °C.
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155
CHAPTER 3
Analysis Heat is generated in the wire and its temperature rises as a result of
resistance heating. We assume heat is generated uniformly throughout the wire
and is transferred to the surrounding medium in the radial direction. In steady
operation, the rate of heat transfer becomes equal to the heat generated within
the wire, which is determined to be
Q = W e = VI = (8 V)(10 A) = 80 W
The thermal resistance network for this problem involves a conduction resis-
tance for the plastic cover and a convection resistance for the outer surface in
series, as shown in Fig. 3-32. The values of these two resistances are deter-
mined to be
A 2 = (2irr 2 )L = 2tt(0.0035 m)(5 m) = 0.110 m 2
1 1
R,
hA 2 (12W/m 2 • °C)(0.110m 2 )
ln(r 2 /r,) ln(3.5/1.5)
plastic
2-nkL 2tt(0.15 W/m ■ °C)(5 m)
0.76°C/W
0.18°C/W
and therefore
#,o, a I = plastic + *co„v = 0.76 + 0.18 = 0.94°C/W
Then the interface temperature can be determined from
Q
T, - 71
= 30°C + (80 W)(0.94°C/W) = 105°C
Note that we did not involve the electrical wire directly in the thermal resistance
network, since the wire involves heat generation.
To answer the second part of the question, we need to know the critical radius
of insulation of the plastic cover. It is determined from Eq. 3-50 to be
k = 0.15 W/m ■ °C
h 12 W/m 2 ■ °C
0.0125 m = 12.5 mm
which is larger than the radius of the plastic cover. Therefore, increasing the
thickness of the plastic cover will enhance heat transfer until the outer radius
of the cover reaches 12.5 mm. As a result, the rate of heat transfer Q will in-
crease when the interface temperature T x is held constant, or T x will decrease
when Q is held constant, which is the case here.
Discussion It can be shown by repeating the calculations above for a 4-mm-
thick plastic cover that the interface temperature drops to 90.6°C when the
thickness of the plastic cover is doubled. It can also be shown in a similar man-
ner that the interface reaches a minimum temperature of 83°C when the outer
radius of the plastic cover equals the critical radius.
T 2
^vwvw — • — www » T„
D D
plastic conv
FIGURE 3-32
Schematic for Example 3-9.
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156
HEAT TRANSFER
FIGURE 3-33
The thin plate fins of a car radiator
greatly increase the rate of
heat transfer to the air (photo by
Yunus £engel and James Kleiser).
3-6 - HEAT TRANSFER FROM FINNED SURFACES
The rate of heat transfer from a surface at a temperature T s to the surrounding
medium at T m is given by Newton's law of cooling as
6 c
hA s (T s -T x )
FIGURE 3-34
Some innovative fin designs.
where A s is the heat transfer surface area and h is the convection heat transfer
coefficient. When the temperatures T s and T„ are fixed by design considera-
tions, as is often the case, there are two ways to increase the rate of heat trans-
fer: to increase the convection heat transfer coefficient h or to increase the
surface area A s . Increasing h may require the installation of a pump or fan, or
replacing the existing one with a larger one, but this approach may or may not
be practical. Besides, it may not be adequate. The alternative is to increase the
surface area by attaching to the surface extended surfaces called fins made of
highly conductive materials such as aluminum. Finned surfaces are manu-
factured by extruding, welding, or wrapping a thin metal sheet on a surface.
Fins enhance heat transfer from a surface by exposing a larger surface area to
convection and radiation.
Finned surfaces are commonly used in practice to enhance heat transfer, and
they often increase the rate of heat transfer from a surface severalfold. The car
radiator shown in Fig. 3-33 is an example of a finned surface. The closely
packed thin metal sheets attached to the hot water tubes increase the surface
area for convection and thus the rate of convection heat transfer from the tubes
to the air many times. There are a variety of innovative fin designs available
in the market, and they seem to be limited only by imagination (Fig. 3-34).
In the analysis of fins, we consider steady operation with no heat generation
in the fin, and we assume the thermal conductivity k of the material to remain
constant. We also assume the convection heat transfer coefficient h to be con-
stant and uniform over the entire surface of the fin for convenience in the
analysis. We recognize that the convection heat transfer coefficient h, in gen-
eral, varies along the fin as well as its circumference, and its value at a point
is a strong function of the fluid motion at that point. The value of h is usually
much lower at the fin base than it is at the fin tip because the fluid is sur-
rounded by solid surfaces near the base, which seriously disrupt its motion to
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157
CHAPTER 3
the point of "suffocating" it, while the fluid near the fin tip has little contact
with a solid surface and thus encounters little resistance to flow. Therefore,
adding too many fins on a surface may actually decrease the overall heat
transfer when the decrease in h offsets any gain resulting from the increase in
the surface area.
Fin Equation
Consider a volume element of a fin at location x having a length of Ax, cross-
sectional area of A c , and a perimeter of/?, as shown in Fig. 3-35. Under steady
conditions, the energy balance on this volume element can be expressed as
/ Rate of heat \ / Rate of heat \ / Rate of heat \
conduction into = conduction from the + convection from
I the element at xj I element at x + Ax / I the element /
or
where
tjcond.x t- cond, X + Ax ' Qc
Q conv = h(pAx)(T-T.J
Substituting and dividing by Ax, we obtain
xi cond. X + A.Y x£ cond, X
Volume
element
FIGURE 3-35
Volume element of a fin at location x
having a length of Ax, cross-sectional
area of A c , and perimeter of p.
Ax
hp(T - 7/J =
(3-52)
Taking the limit as Ax — > gives
"Q cond
dx
hp(T - T„) =
(3-53)
From Fourier's law of heat conduction we have
dT
«- cond
-kA,.
dx
(3-54)
where A c is the cross-sectional area of the fin at location x. Substitution of this
relation into Eq. 3-53 gives the differential equation governing heat transfer
in fins,
d_
dx
kA
dT
dx
hp(T - T a ) =
(3-55)
In general, the cross-sectional area A c and the perimeter/; of a fin vary with x,
which makes this differential equation difficult to solve. In the special case of
constant cross section and constant thermal conductivity, the differential
equation 3-55 reduces to
dx 1
a 2 e = o
(3-56)
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158
HEAT TRANSFER
where
hp
~kA~.
(3-57)
and = 7 — r„ is the temperature excess. At the fin base we have
0/, — T b — T<n-
Equation 3-56 is a linear, homogeneous, second-order differential equation
with constant coefficients. A fundamental theory of differential equations
states that such an equation has two linearly independent solution functions,
and its general solution is the linear combination of those two solution func-
tions. A careful examination of the differential equation reveals that subtract-
ing a constant multiple of the solution function from its second derivative
yields zero. Thus we conclude that the function and its second derivative
must be constant multiples of each other. The only functions whose deriva-
tives are constant multiples of the functions themselves are the exponential
functions (or a linear combination of exponential functions such as sine and
cosine hyperbolic functions). Therefore, the solution functions of the dif-
ferential equation above are the exponential functions e~" x or e ax or constant
multiples of them. This can be verified by direct substitution. For example,
the second derivative of e~ ax is a 2 e~ ax , and its substitution into Eq. 3-56 yields
zero. Therefore, the general solution of the differential equation Eq. 3-56 is
Q(x)
CV
(3-58)
~- Specified
temperature
(a) Specified temperature
(b) Negligible heat loss
(c) Convection
(d) Convection and radiation
FIGURE 3-36
Boundary conditions at the
fin base and the fin tip.
where C { and C 2 are arbitrary constants whose values are to be determined
from the boundary conditions at the base and at the tip of the fin. Note that we
need only two conditions to determine C\ and C 2 uniquely.
The temperature of the plate to which the fins are attached is normally
known in advance. Therefore, at the fin base we have a specified temperature
boundary condition, expressed as
Boundary condition at fin base:
6(0)
(3-59)
At the fin tip we have several possibilities, including specified temperature,
negligible heat loss (idealized as an insulated tip), convection, and combined
convection and radiation (Fig. 3-36). Next, we consider each case separately.
1 Infinitely Long Fin (r fintip = 7" J
For a sufficiently long fin of uniform cross section (A c = constant), the tem-
perature of the fin at the fin tip will approach the environment temperature 7^
and thus will approach zero. That is,
Boundary condition at fin tip: 0(L) = T(L) — T m
This condition will be satisfied by the function e~ ax , but not by the other
prospective solution function e ax since it tends to infinity as x gets larger.
Therefore, the general solution in this case will consist of a constant multiple
of e ax . The value of the constant multiple is determined from the require-
ment that at the fin base where x = the value of will be Q b . Noting that
cen58933_ch03.qxd 9/10/2002 8:59 AM Page 159
e -ax — e o — ^ tne p r0 p er value of the constant is Q b , and the solution function
we are looking for is 0(x) = Q b e~ ax . This function satisfies the differential
equation as well as the requirements that the solution reduce to Q b at the fin
base and approach zero at the fin tip for large x. Noting that = T — T m and
a = \JhplkA c , the variation of temperature along the fin in this case can be
expressed as
Very long fin:
T(x)
(3-60)
Note that the temperature along the fin in this case decreases exponentially
from T b to T m as shown in Fig. 3-37. The steady rate of heat transfer from the
entire fin can be determined from Fourier's law of heat conduction
Very long fin:
do
ng fin
-kA.
dT
c dx
y/hpkA c (T b - r.)
(3-61)
where p is the perimeter, A c is the cross-sectional area of the fin, and x is the
distance from the fin base. Alternatively, the rate of heat transfer from the fin
could also be determined by considering heat transfer from a differential
volume element of the fin and integrating it over the entire surface of the fin.
That is,
Q
,„ = f h[T(x
) - rj dA fm
f A8(.
JA,„
x) dA fln
(3-62)
The two approaches described are equivalent and give the same result since,
under steady conditions, the heat transfer from the exposed surfaces of the fin
is equal to the heat transfer to the fin at the base (Fig. 3-38).
2 Negligible Heat Loss from the Fin Tip
(Insulated fin tip, Q fin tip = 0)
Fins are not likely to be so long that their temperature approaches the sur-
rounding temperature at the tip. A more realistic situation is for heat transfer
from the fin tip to be negligible since the heat transfer from the fin is propor-
tional to its surface area, and the surface area of the fin tip is usually a negli-
gible fraction of the total fin area. Then the fin tip can be assumed to be
insulated, and the condition at the fin tip can be expressed as
Boundary condition at fin tip:
dx
(3-63)
159
CHAPTER 3
(p = kD, A c = kD /4 for a cylindrical fin)
FIGURE 3-37
A long circular fin of uniform cross
section and the variation of
temperature along it.
i e fin
I / / > / 4 1
^base *"iin
FIGURE 3-38
Under steady conditions, heat transfer
from the exposed surfaces of the
fin is equal to heat conduction
to the fin at the base.
The condition at the fin base remains the same as expressed in Eq. 3-59. The
application of these two conditions on the general solution (Eq. 3-58) yields,
after some manipulations, this relation for the temperature distribution:
Adiabatic fin tip:
T(x)
cosh a(L — x)
cosh ah
(3-64)
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160
HEAT TRANSFER
The rate of heat transfer from the fin can be determined again from Fourier's
law of heat conduction:
Adiabatic fin tip:
Q
insulated tip
-M,
dT
dx
VhpkA c (T b - T r _) tanh ah
(3-65)
Note that the heat transfer relations for the very long fin and the fin with
negligible heat loss at the tip differ by the factor tanh ah, which approaches 1
as L becomes very large.
e fin
\- L H
Convection
(a) Actual fin with
convection at the tip
^
Insulated
(b) Equivalent fin with insulated tip
FIGURE 3-39
Corrected fin length L c is defined such
that heat transfer from a fin of length
L c with insulated tip is equal to heat
transfer from the actual fin of length L
with convection at the fin tip.
3 Convection (or Combined Convection and Radiation)
from Fin Tip
The fin tips, in practice, are exposed to the surroundings, and thus the proper
boundary condition for the fin tip is convection that also includes the effects
of radiation. The fin equation can still be solved in this case using the convec-
tion at the fin tip as the second boundary condition, but the analysis becomes
more involved, and it results in rather lengthy expressions for the temperature
distribution and the heat transfer. Yet, in general, the fin tip area is a small
fraction of the total fin surface area, and thus the complexities involved can
hardly justify the improvement in accuracy.
A practical way of accounting for the heat loss from the fin tip is to replace
the fin length L in the relation for the insulated tip case by a corrected length
defined as (Fig. 3-39)
Corrected fin length:
L +
(3-66)
where A c is the cross-sectional area and/7 is the perimeter of the fin at the tip.
Multiplying the relation above by the perimeter gives A corrected = A fm n atera i) +
A tip , which indicates that the fin area determined using the corrected length is
equivalent to the sum of the lateral fin area plus the fin tip area.
The corrected length approximation gives very good results when the vari-
ation of temperature near the fin tip is small (which is the case when ah > 1)
and the heat transfer coefficient at the fin tip is about the same as that at the
lateral surface of the fin. Therefore, fins subjected to convection at their tips
can be treated as fins with insulated tips by replacing the actual fin length by
the corrected length in Eqs. 3-64 and 3-65.
Using the proper relations for A c and p, the corrected lengths for rectangu-
lar and cylindrical fins are easily determined to be
T = f
c, rectangular fin
and
D
*^c, cylindrical fin I
where t is the thickness of the rectangular fins and D is the diameter of the
cylindrical fins.
Fin Efficiency
Consider the surface of a plane wall at temperature T b exposed to a medium at
temperature T^. Heat is lost from the surface to the surrounding medium by
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161
CHAPTER 3
convection with a heat transfer coefficient of h. Disregarding radiation or
accounting for its contribution in the convection coefficient /;, heat transfer
from a surface area A s is expressed as Q = hA s (T s — T^).
Now let us consider a fin of constant cross-sectional area A c = A b and length
L that is attached to the surface with a perfect contact (Fig. 3-40). This time
heat will flow from the surface to the fin by conduction and from the fin to the
surrounding medium by convection with the same heat transfer coefficient h.
The temperature of the fin will be T b at the fin base and gradually decrease to-
ward the fin tip. Convection from the fin surface causes the temperature at any
cross section to drop somewhat from the midsection toward the outer surfaces.
However, the cross-sectional area of the fins is usually very small, and thus
the temperature at any cross section can be considered to be uniform. Also, the
fin tip can be assumed for convenience and simplicity to be insulated by using
the corrected length for the fin instead of the actual length.
In the limiting case of zero thermal resistance or infinite thermal conduc-
tivity (& — > oo), the temperature of the fin will be uniform at the base value of
T b . The heat transfer from the fin will be maximum in this case and can be
expressed as
(a) Surface without fins
Q fin. r
hA fm (T b - r.)
(3-67)
In reality, however, the temperature of the fin will drop along the fin, and
thus the heat transfer from the fin will be less because of the decreasing tem-
perature difference T(x) — T«, toward the fin tip, as shown in Fig. 3-41. To ac-
count for the effect of this decrease in temperature on heat transfer, we define
a fin efficiency as
%in
e,
2fin, r
Actual heat transfer rate from the fin
Ideal heat transfer rate from the fin
if the entire fin were at base temperature
(3-68)
or
Q fin — "Hfin Q fin, r
%.n hA fm (T b - 7/„)
(3-69)
where A Fm is the total surface area of the fin. This relation enables us to deter-
mine the heat transfer from a fin when its efficiency is known. For the cases
of constant cross section of very long fins and fins with insulated tips, the fin
efficiency can be expressed as
Mlong fin
6 fin
*£■ fin, m£
Vtip~kA c (T b - r.) _ i jkAc
M fln (T b - rj Lihp
aL
(3-70)
and
'linsulated t
Q An \ / hpkA c (T b - r„) tanh aL t anh aL
Q fin, r
M fm (T b - T x )
aL
(3-71)
since A fm = pL for fins with constant cross section. Equation 3-71 can also be
used for fins subjected to convection provided that the fin length L is replaced
by the corrected length L c .
(b) Surface with a fin
: 2 X w X L
s 2 X w X L
FIGURE 3-40
Fins enhance heat transfer from
a surface by enhancing surface area.
(a) Ideal
80°C
(b) Actual
56°C
FIGURE 3-41
Ideal and actual
temperature distribution in a fin.
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HEAT TRANSFER
FIGURE 3-42
Efficiency of circular, rectangular, and
triangular fins on a plain surface of
width w (from Gardner, Ref. 6).
Fin efficiency relations are developed for fins of various profiles and are
plotted in Fig. 3-42 for fins on a plain surface and in Fig. 3-43 for circular
fins of constant thickness. The fin surface area associated with each profile is
also given on each figure. For most fins of constant thickness encountered in
practice, the fin thickness t is too small relative to the fin length L, and thus
the fin tip area is negligible.
100
100
FIGURE 3-43
Efficiency of circular fins of length L
and constant thickness t (from
Gardner, Ref. 6).
60
40
20
tmmmtL.
r 2 + ±t
^ h
\2
y\
:4
— h
^2
^Jt
h\ A fm = 2K(rl-,j) + 27ir 2 t
2_
0.5
1.0 1.5
2.0
2.5
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163
CHAPTER 3
Note that fins with triangular and parabolic profiles contain less material
and are more efficient than the ones with rectangular profiles, and thus are
more suitable for applications requiring minimum weight such as space appli-
cations.
An important consideration in the design of finned surfaces is the selection
of the proper fin length L. Normally the longer the fin, the larger the heat
transfer area and thus the higher the rate of heat transfer from the fin. But also
the larger the fin, the bigger the mass, the higher the price, and the larger the
fluid friction. Therefore, increasing the length of the fin beyond a certain
value cannot be justified unless the added benefits outweigh the added cost.
Also, the fin efficiency decreases with increasing fin length because of the de-
crease in fin temperature with length. Fin lengths that cause the fin efficiency
to drop below 60 percent usually cannot be justified economically and should
be avoided. The efficiency of most fins used in practice is above 90 percent.
Fin Effectiveness
Fins are used to enhance heat transfer, and the use of fins on a surface cannot
be recommended unless the enhancement in heat transfer justifies the added
cost and complexity associated with the fins. In fact, there is no assurance that
adding fins on a surface will enhance heat transfer. The performance of the
fins is judged on the basis of the enhancement in heat transfer relative to the
no-fin case. The performance of fins expressed in terms of the/w effectiveness
e fin is defined as (Fig. 3-44)
gf,n
x£ no fin
e,
hA b (T b - ly
Heat transfer rate from
the fin of base area A b
Heat transfer rate from
the surface of area A h
(3-72)
Here, A b is the cross-sectional area of the fin at the base and Q no fin represents
the rate of heat transfer from this area if no fins are attached to the surface.
An effectiveness of e fin = 1 indicates that the addition of fins to the surface
does not affect heat transfer at all. That is, heat conducted to the fin through
the base area A b is equal to the heat transferred from the same area A b to the
surrounding medium. An effectiveness of e fln < 1 indicates that the fin actu-
ally acts as insulation, slowing down the heat transfer from the surface. This
situation can occur when fins made of low thermal conductivity materials are
used. An effectiveness of e fln > 1 indicates that fins are enhancing heat trans-
fer from the surface, as they should. However, the use of fins cannot be justi-
fied unless e fin is sufficiently larger than 1. Finned surfaces are designed on
the basis of maximizing effectiveness for a specified cost or minimizing cost
for a desired effectiveness.
Note that both the fin efficiency and fin effectiveness are related to the per-
formance of the fin, but they are different quantities. However, they are
related to each other by
-no fin
FIGURE 3-44
The effectiveness of a fin.
Qi
G fi „
%,„ hA fm (T b - r„) A fl
6,
hA„ (T b - r„) hA b (T b - T m )
A,
■ %,n
(3-73)
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HEAT TRANSFER
Therefore, the fin effectiveness can be determined easily when the fin effi-
ciency is known, or vice versa.
The rate of heat transfer from a sufficiently long fin of uniform cross section
under steady conditions is given by Eq. 3-61. Substituting this relation into
Eq. 3-72, the effectiveness of such a long fin is determined to be
-"long fin
6 fin _ Vh^kA c (T b -T x
fi„„ fin hA b {T b -T a )
kp
\n~A c
(3-74)
^^
- 3 x (r x w)
A fln =2XLXw + tXw (one fin)
~2x Lxw
FIGURE 3-45
Various surface areas associated with a
rectangular surface with three fins.
since A c = A b in this case. We can draw several important conclusions from
the fin effectiveness relation above for consideration in the design and selec-
tion of the fins:
• The thermal conductivity k of the fin material should be as high as
possible. Thus it is no coincidence that fins are made from metals, with
copper, aluminum, and iron being the most common ones. Perhaps the
most widely used fins are made of aluminum because of its low cost and
weight and its resistance to corrosion.
• The ratio of the perimeter to the cross-sectional area of the fin p/A c
should be as high as possible. This criterion is satisfied by thin plate fins
and slender pin fins.
• The use of fins is most effective in applications involving a low
convection heat transfer coefficient. Thus, the use of fins is more easily
justified when the medium is a gas instead of a liquid and the heat
transfer is by natural convection instead of by forced convection.
Therefore, it is no coincidence that in liquid-to-gas heat exchangers such
as the car radiator, fins are placed on the gas side.
When determining the rate of heat transfer from a finned surface, we must
consider the unfinned portion of the surface as well as the fins. Therefore, the
rate of heat transfer for a surface containing n fins can be expressed as
Q total, fin — 2unfin + Q fin
= ^u„r,n (T b ~ 7V) + r\ {m hA Sm (T b
= KKniin + THfin^finXTft ~ TJ)
r„)
(3-75)
We can also define an overall effectiveness for a finned surface as the ratio
of the total heat transfer from the finned surface to the heat transfer from the
same surface if there were no fins,
Q total, fin h(A mfw + % in A fm )(7; - F„)
J fin. overall
Q
total, no fin
^nofin^i- J"-)
(3-76)
where A no fin is the area of the surface when there are no fins, A fln is the total
surface area of all the fins on the surface, and A unfin is the area of the unfinned
portion of the surface (Fig. 3-45). Note that the overall fin effectiveness
depends on the fin density (number of fins per unit length) as well as the
effectiveness of the individual fins. The overall effectiveness is a better mea-
sure of the performance of a finned surface than the effectiveness of the indi-
vidual fins.
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CHAPTER 3
Proper Length of a Fin
An important step in the design of a fin is the determination of the appropriate
length of the fin once the fin material and the fin cross section are specified.
You may be tempted to think that the longer the fin, the larger the surface area
and thus the higher the rate of heat transfer. Therefore, for maximum heat
transfer, the fin should be infinitely long. However, the temperature drops
along the fin exponentially and reaches the environment temperature at some
length. The part of the fin beyond this length does not contribute to heat trans-
fer since it is at the temperature of the environment, as shown in Fig. 3-46.
Therefore, designing such an "extra long" fin is out of the question since it
results in material waste, excessive weight, and increased size and thus in-
creased cost with no benefit in return (in fact, such a long fin will hurt perfor-
mance since it will suppress fluid motion and thus reduce the convection heat
transfer coefficient). Fins that are so long that the temperature approaches the
environment temperature cannot be recommended either since the little in-
crease in heat transfer at the tip region cannot justify the large increase in the
weight and cost.
To get a sense of the proper length of a fin, we compare heat transfer from
a fin of finite length to heat transfer from an infinitely long fin under the same
conditions. The ratio of these two heat transfers is
High
heat
transfer
r^t-+-t
% % 1
FIGURE 3-46
Because of the gradual temperature
drop along the fin, the region
near the fin tip makes little or
no contribution to heat transfer.
Heat transfer
ratio:
g fln VhJkA.(T b -T^tanhaL
G,
VhpkA-(T b -TJ
tanh ah
(3-77)
Using a hand calculator, the values of tanh aL are evaluated for some values
of aL and the results are given in Table 3-3. We observe from the table that
heat transfer from a fin increases with aL almost linearly at first, but the curve
reaches a plateau later and reaches a value for the infinitely long fin at about
aL = 5. Therefore, a fin whose length is L = ^a can be considered to be an
infinitely long fin. We also observe that reducing the fin length by half in that
case (from aL = 5 to aL = 2.5) causes a drop of just 1 percent in heat trans-
fer. We certainly would not hesitate sacrificing 1 percent in heat transfer per-
formance in return for 50 percent reduction in the size and possibly the cost of
the fin. In practice, a fin length that corresponds to about aL = 1 will transfer
76.2 percent of the heat that can be transferred by an infinitely long fin, and
thus it should offer a good compromise between heat transfer performance
and the fin size.
A common approximation used in the analysis of fins is to assume the fin
temperature varies in one direction only (along the fin length) and the tem-
perature variation along other directions is negligible. Perhaps you are won-
dering if this one-dimensional approximation is a reasonable one. This is
certainly the case for fins made of thin metal sheets such as the fins on a car
radiator, but we wouldn't be so sure for fins made of thick materials. Studies
have shown that the error involved in one-dimensional fin analysis is negligi-
ble (less than about 1 percent) when
TABLE 3-3
The variation of heat transfer from
a fin relative to that from an
infinitely long fin
aL
- tanh aL
"Hong fin
0.1
0.100
0.2
0.197
0.5
0.462
1.0
0.762
1.5
0.905
2.0
0.964
2.5
0.987
3.0
0.995
4.0
0.999
5.0
1.000
hh
<0.2
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HEAT TRANSFER
where 8 is the characteristic thickness of the fin, which is taken to be the plate
thickness t for rectangular fins and the diameter D for cylindrical ones.
Specially designed finned surfaces called heat sinks, which are commonly
used in the cooling of electronic equipment, involve one-of-a-kind complex
geometries, as shown in Table 3-4. The heat transfer performance of heat
sinks is usually expressed in terms of their thermal resistances R in °C/W,
which is defined as
G,
R
hA {m T| fin {T b - r„)
(3-78)
A small value of thermal resistance indicates a small temperature drop across
the heat sink, and thus a high fin efficiency.
FIGURE 3-47
Schematic for Example 3-10.
EXAMPLE 3-1 Maximum Power Dissipation of a Transistor
Power transistors that are commonly used in electronic devices consume large
amounts of electric power. The failure rate of electronic components increases
almost exponentially with operating temperature. As a rule of thumb, the failure
rate of electronic components is halved for each 10°C reduction in the junction
operating temperature. Therefore, the operating temperature of electronic com-
ponents is kept below a safe level to minimize the risk of failure.
The sensitive electronic circuitry of a power transistor at the junction is pro-
tected by its case, which is a rigid metal enclosure. Heat transfer characteris-
tics of a power transistor are usually specified by the manufacturer in terms of
the case-to-ambient thermal resistance, which accounts for both the natural
convection and radiation heat transfers.
The case-to-ambient thermal resistance of a power transistor that has a max-
imum power rating of 10 W is given to be 20°C/W. If the case temperature of
the transistor is not to exceed 85°C, determine the power at which this transis-
tor can be operated safely in an environment at 25°C.
SOLUTION The maximum power rating of a transistor whose case temperature
is not to exceed 85°C is to be determined.
Assumptions 1 Steady operating conditions exist. 2 The transistor case is iso-
thermal at 85°C.
Properties The case-to-ambient thermal resistance is given to be 20°C/W.
Analysis The power transistor and the thermal resistance network associated
with it are shown in Fig. 3-47. We notice from the thermal resistance network
that there is a single resistance of 20°C/W between the case at T c = 85°C and
the ambient at 7" x = 25°C, and thus the rate of heat transfer is
6
AT]
R /
/ c
R,
(85 - 25)°C
20°C/W
3W
Therefore, this power transistor should not be operated at power levels above
3 W if its case temperature is not to exceed 85°C.
Discussion This transistor can be used at higher power levels by attaching it to
a heat sink (which lowers the thermal resistance by increasing the heat transfer
surface area, as discussed in the next example) or by using a fan (which lowers
the thermal resistance by increasing the convection heat transfer coefficient).
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CHAPTER 3
TABLE 3-
Combined natural convection and radiation thermal resistance of various
heat sinks used in the cooling of electronic devices between the heat sink and
the surroundings. All fins are made of aluminum 6063T-5, are black anodized,
and are 76 mm (3 in.) long (courtesy of Vemaline Products, Inc.).
HS 5030
R= 0.9°C/W (vertical)
R = 1.2-C/W (horizontal)
Dimensions: 76 mm X 105 mm X 44 mm
Surface area: 677 cm 2
HS606S
R= 5°C/W
Dimensions: 76 mm X 38 mm X 24 mm
Surface area: 387 cm 2
HS 6071
R = 1.4°C/W (vertical)
R= 1.8-C/W (horizontal)
Dimensions: 76 mm X 92 mm X 26 mm
Surface area: 968 cm 2
HS6105
R= 1.8°C/W (vertical)
R= 2. 1°C/W (horizontal)
Dimensions: 76 mm X 127 mm X 91 mm
Surface area: 677 cm 2
HS6115
R= 1.1°C/W (vertical)
R= 1.3°C/W (horizontal)
Dimensions: 76 mm X 102 mm X 25 mm
Surface area: 929 cm 2
MS 7030
R= 2.9°C/W (vertical)
R= 3. 1°C/W (horizontal)
Dimensions: 76 mm X 97 mm X 19 mm
Surface area: 290 cm 2
EXAMPLE 3-11 Selecting a Heat Sink for a Transistor
A 60-W power transistor is to be cooled by attaching it to one of the commer-
cially available heat sinks shown in Table 3-4. Select a heat sink that will allow
the case temperature of the transistor not to exceed 90°C in the ambient air
at 30°C.
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HEAT TRANSFER
SOLUTION A commercially available heat sink from Table 3-4 is to be se-
lected to keep the case temperature of a transistor below 90°C.
Assumptions 1 Steady operating conditions exist. 2 The transistor case is iso-
thermal at 90°C. 3 The contact resistance between the transistor and the heat
sink is negligible.
Analysis The rate of heat transfer from a 60-W transistor at full power is
Q = 60 W. The thermal resistance between the transistor attached to the heat
sink and the ambient air for the specified temperature difference is determined
to be
Q
AT
R
AT
Q
(90 - 30)°C
60 W
1.0°C/W
Therefore, the thermal resistance of the heat sink should be below 1.0°C/W.
An examination of Table 3-4 reveals that the HS 5030, whose thermal resis-
tance is 0.9°C/W in the vertical position, is the only heat sink that will meet
this requirement.
FIGURE 3-48
Schematic for Example 3-12.
EXAMPLE 3-12 Effect of Fins on Heat Transfer from Steam Pipes
Steam in a heating system flows through tubes whose outer diameter is
D 1 = 3 cm and whose walls are maintained at a temperature of 120°C. Circu-
lar aluminum fins (k = 180 W/m ■ °C) of outer diameter D 2 = 6 cm and con-
stant thickness f = 2 mm are attached to the tube, as shown in Fig. 3-48. The
space between the fins is 3 mm, and thus there are 200 fins per meter length
of the tube. Heat is transferred to the surrounding air at 7" x = 25°C, with a com-
bined heat transfer coefficient of h = 60 W/m 2 • °C. Determine the increase in
heat transfer from the tube per meter of its length as a result of adding fins.
SOLUTION Circular aluminum fins are to be attached to the tubes of a heating
system. The increase in heat transfer from the tubes per unit length as a result
of adding fins is to be determined.
Assumptions 1 Steady operating conditions exist. 2 The heat transfer coeffi-
cient is uniform over the entire fin surfaces. 3 Thermal conductivity is constant.
4 Heat transfer by radiation is negligible.
Properties The thermal conductivity of the fins is given to be k = 180
W/m • °C.
Analysis In the case of no fins, heat transfer from the tube per meter of its
length is determined from Newton's law of cooling to be
A
2 no fin ~~ hA nofin (T b
- -nDiL = tt(0.03 m)(l m) = 0.0942 m 2
(60 W/m 2 ■ °C)(0.0942 m 2 )(120 - 25)°C
537 W
The efficiency of the circular fins attached to a circular tube is plotted in Fig.
3-43. Noting that L = \{D 2 - D x ) = 1(0.06 - 0.03) = 0.015 m in this case,
we have
cen58933_ch03.qxd 9/10/2002 8:59 AM Page 169
r 2 + if (0.03 + i X 0.002) m
2.07
(L + U)
0.015 m
(0.015 +i X 0.002) m X
60W/m 2 -°C
V (180 W/m • °C)(0.002m)
'%.„
0.207
Aa, = 2ir(r 2 2 - /f) + 2irr 2 f
= 2tt[(0.03 m) 2 - (0.015 m) 2 ] + 2tt(0.03 m)(0.002 m)
= 0.00462 m 2
6 fin = THfinGfin, max = T >fin' J Afin (T b — Tj)
= 0.95(60 W/m 2 • °C)(0.00462 m 2 )(120 - 25)°C
= 25.0 W
Heat transfer from the unfinned portion of the tube is
A unfln = ttDiS = ir(0.03 m)(0.003 m) = 0.000283 m 2
Gunfin = hA nn f m {T b — 7„)
= (60 W/m 2 ■ °C)(0.000283 m 2 )(120 - 25)°C
= 1.60 W
Noting that there are 200 fins and thus 200 interfin spacings per meter length
of the tube, the total heat transfer from the finned tube becomes
Q
total, fin
«(2f,„ + Gu„fi„) = 200(25.0 + 1.6) W = 5320 W
Therefore, the increase in heat transfer from the tube per meter of its length as
a result of the addition of fins is
Q ia^e = Q total, fin " 6 „o fin = 5320 - 537 = 4783 W (per m tube length)
Discussion The overall effectiveness of the finned tube is
G total, fin 5320 W
Q
total, no fin
537 W
9.9
That is, the rate of heat transfer from the steam tube increases by a factor of
almost 10 as a result of adding fins. This explains the widespread use of finned
surfaces.
169
CHAPTER 3
0.95
3-7 ■ HEAT TRANSFER IN
COMMON CONFIGURATIONS
So far, we have considered heat transfer in simple geometries such as large
plane walls, long cylinders, and spheres. This is because heat transfer in such
geometries can be approximated as one-dimensional, and simple analytical
solutions can be obtained easily. But many problems encountered in practice
are two- or three-dimensional and involve rather complicated geometries for
which no simple solutions are available.
cen58933_ch03.qxd 9/10/2002 8:59 AM Page 17C
170
HEAT TRANSFER
An important class of heat transfer problems for which simple solutions are
obtained encompasses those involving two surfaces maintained at constant
temperatures T [ and T 2 . The steady rate of heat transfer between these two sur-
faces is expressed as
Q = Sk(T t - T 2 )
(3-79)
where S is the conduction shape factor, which has the dimension of length,
and k is the thermal conductivity of the medium between the surfaces. The
conduction shape factor depends on the geometry of the system only.
Conduction shape factors have been determined for a number of configura-
tions encountered in practice and are given in Table 3-5 for some common
cases. More comprehensive tables are available in the literature. Once the
value of the shape factor is known for a specific geometry, the total steady
heat transfer rate can be determined from the equation above using the speci-
fied two constant temperatures of the two surfaces and the thermal conductiv-
ity of the medium between them. Note that conduction shape factors are
applicable only when heat transfer between the two surfaces is by conduction.
Therefore, they cannot be used when the medium between the surfaces is a
liquid or gas, which involves natural or forced convection currents.
A comparison of Equations 3-4 and 3-79 reveals that the conduction shape
factor S is related to the thermal resistance R by R = 1/kS or S = 1/kR. Thus,
these two quantities are the inverse of each other when the thermal conduc-
tivity of the medium is unity. The use of the conduction shape factors is illus-
trated with examples 3-13 and 3-14.
s-T 2 = 10°C
Z = 0.5m
I_
- ^ | D = 10 cm )
[' ■' ■ ■■' -L = 30 m - — -4
FIGURE 3-49
Schematic for Example 3-13.
EXAMPLE 3-13 Heat Loss from Buried Steam Pipes
A 30-m-long, 10-cm-diameter hot water pipe of a district heating system is
buried in the soil 50 cm below the ground surface, as shown in Figure 3-49.
The outer surface temperature of the pipe is 80°C. Taking the surface tempera-
ture of the earth to be 10°C and the thermal conductivity of the soil at that lo-
cation to be 0.9 W/m • °C, determine the rate of heat loss from the pipe.
SOLUTION The hot water pipe of a district heating system is buried in the soil.
The rate of heat loss from the pipe is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Heat transfer is two-
dimensional (no change in the axial direction). 3 Thermal conductivity of the
soil is constant.
Properties The thermal conductivity of the soil is given to be k = 0.9 W/m • °C.
Analysis The shape factor for this configuration is given in Table 3-5 to be
S
2ttL
ln(4z/D)
since z > 1.5D, where z is the distance of the pipe from the ground surface,
and D is the diameter of the pipe. Substituting,
2ir X (30 m)
ln(4 X 0.5/0.1)
62.9 m
cen58933_ch03.qxd 9/10/2002 8:59 AM Page 171
171
CHAPTER 3
TABLE 3-5
Conduction shape factors S for several configurations for use in Q = kS{Ti - T 2 ) to determine the steady rate of heat
transfer through a medium of thermal conductivity k between the surfaces at temperatures 7"! and T 2
( 1 ) Isothermal cylinder of length L
buried in a semi-infinite medium
(L»D and z>l.5D)
rl
2%L
" In (4z/D)
G
m
> ) ■■
(2) Vertical isothermal cylinder of length L
buried in a semi-infinite medium
(L»D)
2%L
"ln(4L/D)
IX
(3) Two parallel isothermal cylinders
placed in an infinite medium
(L»D l ,D 1 ,z)
2kL
cosh
Az 2 ~D]
~D\
(4) A row of equally spaced parallel isothermal
cylinders buried in a semi-infinite medium
(L»D, z andw>1.5D)
j£±
2kL
ln|^ sinh 2^Z
(per cylinder)
-j-^-W ; >|» '- w ■ ' »]■■ ; 'Wr»|-
(5) Circular isothermal cylinder of length L
in the midplane of an infinite wall
(Z > 0.5£>)
2kL
ln(8ztoD)
(6) Circular isothermal cylinder of length L
at the center of a square solid bar of the
same length
2%L
In (1.08 wID)
(7) Eccentric circular isothermal cylinder
of length L in a cylinder of the same
length (L > £>,) j
2kL
cosh
D]+D\-4z 2
(8) Large plane wall
-T,
(continued)
cen58933_ch03.qxd 9/10/2002 8:59 AM Page 172
172
HEAT TRANSFER
TABLE 3-5 (CONCLUDED)
(9) A long cylindrical layer
2kL
In (D/D.)
(10) A square flow passage
(a) For alb > 1.4,
„ 2%L
T 2\/\ 1
J
0.93 In (0.948 alb)
(b) For alb < 1.41,
/A
c 2%L
L
0.785 In (alb)
\^b^
-> a
(1 1) A spherical layer
2%D X D 1
(12) Disk buried parallel to
the surface in a semi-infinite
medium (z » D)
D^-D
1 -u x
^h_
S = 4D
(5 = 2£)whenz = 0)
&
(13) The edge of two adjoining
walls of equal thickness
5 = 0.54 w
(14) Corner of three walls
of equal thickness
5 = 0.15L
(15) Isothermal sphere buried in a
semi-infinite medium
^
2kD
1 - 0.25D/Z
(16) Isothermal sphere buried
in a semi-infinite medium at T 2
whose surface is insulated
2kD
Insulated
1 + 0.25D/Z
cen58933_ch03.qxd 9/10/2002 8:59 AM Page 173
Then the steady rate of heat transfer from the pipe becomes
Q = Sk(T, - T 2 ) = (62.9 m)(0.9 W/m • °C)(80 - 10)°C = 3963 W
Discussion Note that this heat is conducted from the pipe surface to the sur-
face of the earth through the soil and then transferred to the atmosphere by
convection and radiation.
173
CHAPTER 3
EXAMPLE 3-14 Heat Transfer between Hot and Cold Water Pipes
A 5-m-long section of hot and cold water pipes run parallel to each other in a
thick concrete layer, as shown in Figure 3-50. The diameters of both pipes are
5 cm, and the distance between the centerline of the pipes is 30 cm. The sur-
face temperatures of the hot and cold pipes are 70°C and 15°C, respectively.
Taking the thermal conductivity of the concrete to be k = 0.75 W/m • °C, de-
termine the rate of heat transfer between the pipes.
SOLUTION Hot and cold water pipes run parallel to each other in a thick con-
crete layer. The rate of heat transfer between the pipes is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Heat transfer is two-
dimensional (no change in the axial direction). 3 Thermal conductivity of the
concrete is constant.
Properties The thermal conductivity of concrete is given to be k = 0.75
W/m • °C.
Analysis The shape factor for this configuration is given in Table 3-5 to be
2ttL
4z 2 -D]-D}
cosh
where z is the distance between the centerlines of the pipes and L is their
length. Substituting,
2tt X (5 m)
4 X 0.3 2 - 0.05 2 - 0.05 2
cosh
2 X 0.05 X 0.05
6.34 m
Then the steady rate of heat transfer between the pipes becomes
Q = Sk(T, - T 2 ) = (6.34 m)(0.75 W/m ■ °C)(70 - 15°)C = 262 W
Discussion We can reduce this heat loss by placing the hot and cold water
pipes further away from each other.
15°C
Z = 30 cm
FIGURE 3-50
Schematic for Example 3-14.
It is well known that insulation reduces heat transfer and saves energy and
money. Decisions on the right amount of insulation are based on a heat trans-
fer analysis, followed by an economic analysis to determine the "monetary
value" of energy loss. This is illustrated with Example 3-15.
cen58933_ch03.qxd 9/10/2002 8:59 AM Page 174
174
HEAT TRANSFER
75°F-
Wall, R= 13
■ 45°F
7/cc. ♦-^vwvw^ i — ww- < -^vwwv^-^ 7^ 9
FIGURE 3-51
Schematic for Example 3-15.
EXAMPLE 3-15 Cost of Heat Loss through Walls in Winter
Consider an electrically heated house whose walls are 9 ft high and have an
R-\ia\ue of insulation of 13 (i.e., a thickness-to-thermal conductivity ratio of
L/k = 13 h • ft 2 • °F/Btu). Two of the walls of the house are 40 ft long and the
others are 30 ft long. The house is maintained at 75°F at all times, while
the temperature of the outdoors varies. Determine the amount of heat lost
through the walls of the house on a certain day during which the average tem-
perature of the outdoors is 45°F. Also, determine the cost of this heat loss to the
homeowner if the unit cost of electricity is $0.075/kWh. For combined convec-
tion and radiation heat transfer coefficients, use the ASHRAE (American Soci-
ety of Heating, Refrigeration, and Air Conditioning Engineers) recommended
values of h, = 1.46 Btu/h • ft 2 • °F for the inner surface of the walls and h =
4.0 Btu/h ■ ft 2 • °F for the outer surface of the walls under 15 mph wind condi-
tions in winter.
SOLUTION An electrically heated house with R-13 insulation is considered.
The amount of heat lost through the walls and its cost are to be determined.
Assumptions 1 The indoor and outdoor air temperatures have remained at the
given values for the entire day so that heat transfer through the walls is steady.
2 Heat transfer through the walls is one-dimensional since any significant
temperature gradients in this case will exist in the direction from the indoors
to the outdoors. 3 The radiation effects are accounted for in the heat transfer
coefficients.
Analysis This problem involves conduction through the wall and convection at
its surfaces and can best be handled by making use of the thermal resistance
concept and drawing the thermal resistance network, as shown in Fig. 3-51.
The heat transfer area of the walls is
A = Circumference X Height = (2 X 30 ft + 2 X 40 ft)(9 ft) = 1260 ft 2
Then the individual resistances are evaluated from their definitions to be
0.00054 h ■ °F/Btu
R=R = J_ = 1
conv '' h,A (1.46 Btu/h ■ ft 2 ■ °F)( 1260 ft 2 )
R»
L /{-value 13 h ■ ft 2 ■ °F/Btu
kA A
1260 ft 2
1
0.01032 h-°F/Btu
0.00020 h • °F/Btu
1
h c A (4.0 Btu/h • ft 2 ■ °F)(1260 ft 2 )
Noting that all three resistances are in series, the total resistance is
fltotai = R i + #waii + R = 0.00054 + 0.01032 + 0.00020 = 0.01106 h • °F/Btu
Then the steady rate of heat transfer through the walls of the house becomes
Q
*Mntnl
(75 - 45)°F
0.01 106 h • °F/Btu
2712 Btu/h
Finally, the total amount of heat lost through the walls during a 24-h period and
its cost to the home owner are
Q = Q At = (2712 Btu/h)(24-h/day) = 65,099 Btu/day = 19.1 kWh/day
cen58933_ch03.qxd 9/10/2002 8:59 AM Page 175
since 1 kWh = 3412 Btu, and
Heating cost = (Energy lost)(Cost of energy) = (19.1 kWh/day)($0.075/kWh)
= $1.43/day
Discussion The heat losses through the walls of the house that day will cost
the home owner $1.43 worth of electricity.
175
CHAPTER 3
TOPIC OF SPECIAL INTEREST
Heat Transfer Through Walls and Roofs
Under steady conditions, the rate of heat transfer through any section of a
building wall or roof can be determined from
Q = UA(Ti - T )
MTj ~ T )
R
(3-80)
where T t and T are the indoor and outdoor air temperatures, A is the heat
transfer area, U is the overall heat transfer coefficient (the [/-factor), and
R = l/U is the overall unit thermal resistance (the R-value). Walls and
roofs of buildings consist of various layers of materials, and the structure
and operating conditions of the walls and the roofs may differ significantly
from one building to another. Therefore, it is not practical to list the
^-values (or [/-factors) of different kinds of walls or roofs under different
conditions. Instead, the overall 7?-value is determined from the thermal
resistances of the individual components using the thermal resistance net-
work. The overall thermal resistance of a structure can be determined most
accurately in a lab by actually assembling the unit and testing it as a whole,
but this approach is usually very time consuming and expensive. The ana-
lytical approach described here is fast and straightforward, and the results
are usually in good agreement with the experimental values.
The unit thermal resistance of a plane layer of thickness L and thermal
conductivity k can be determined from R = Llk. The thermal conductivity
and other properties of common building materials are given in the appen-
dix. The unit thermal resistances of various components used in building
structures are listed in Table 3-6 for convenience.
Heat transfer through a wall or roof section is also affected by the con-
vection and radiation heat transfer coefficients at the exposed surfaces. The
effects of convection and radiation on the inner and outer surfaces of walls
and roofs are usually combined into the combined convection and radiation
heat transfer coefficients (also called surface conductances) h t and h a ,
respectively, whose values are given in Table 3-7 for ordinary surfaces
(e = 0.9) and reflective surfaces (e = 0.2 or 0.05). Note that surfaces hav-
ing a low emittance also have a low surface conductance due to the reduc-
tion in radiation heat transfer. The values in the table are based on a surface
TABLE 3-7
Combined convection and radiation
heat transfer coefficients at window,
wall, or roof surfaces (from ASHRAE
Handbook of Fundamentals, Ref. 1,
Chap. 22, Table 1).
Direc-
tion of
Posi- Heat
h, W/m 2 • c
Surface
Emittance
C*
e
tion Flow
0.90
0.20
0.05
Still air (both indoors and
outdoors)
Horiz. UpT
9.26
5.17
4.32
Horiz. Down I
6.13
2.10
1.25
45° slope UpT
9.09
5.00
4.15
45° slope Down I
7.50
3.41
2.56
Vertical Horiz. — >
8.29
4.20
3.35
Moving air (any position, a
ny direction)
Winter condition
(winds at 15 mpr
or 24 km/h)
34.0
—
—
Summer condition
(winds at 7.5 mph
or 12 km/h)
22.7
—
—
*This section can be skipped without a loss of continuity.
♦Multiply by 0.176 to convert to Btu/h • ft 2 ■ °F.
Surface resistance can be obtained from Ft = llh.
cen58933_ch03.qxd 9/10/2002 8:59 AM Page 176
176
HEAT TRANSFER
TABLE 3-6
Unit thermal resistance (the ff-value) of common components used in buildings
R-
value
R- value
Component
-n 2 • °C/W ft 2
• h • °F/Btu
Component
-n 2 • °C/W ft 2 • h • °F/Btu
Outside surface (winter)
0.030
0.17
Wood stud, nominal 2 in. X
Outside surface (summer)
0.044
0.25
6 in. (5.5 in. or 140 mm wide)
0.98
5.56
Inside surface, still air
0.12
0.68
Clay tile, 100 mm (4 in.)
0.18
1.01
Plane air space, vertical,
orc
inary
surfaces (e eff =
0.82):
Acoustic tile
0.32
1.79
13 mm (1 in.)
0.16
0.90
Asphalt shingle roofing
0.077
0.44
20 mm (| in.)
0.17
0.94
Building paper
0.011
0.06
40 mm (1.5 in.)
0.16
0.90
Concrete block, 100 mm (4 in.):
90 mm (3.5 in.)
0.16
0.91
Lightweight
0.27
1.51
Insulation, 25 mm (1 in
)
Heavyweight
0.13
0.71
Glass fiber
0.70
4.00
Plaster or gypsum board,
Mineral fiber batt
0.66
3.73
13 mm (i in.)
0.079
0.45
Urethane rigid foam
0.98
5.56
Wood fiberboard, 13 mm (i in.)
0.23
1.31
Stucco, 25 mm (1 in.)
0.037
0.21
Plywood, 13 mm (| in.)
0.11
0.62
Face brick, 100 mm (4 in.)
0.075
0.43
Concrete, 200 mm (8 in.):
Common brick, 100 mm
(4
n.)
0.12
0.79
Lightweight
1.17
6.67
Steel siding
0.00
0.00
Heavyweight
0.12
0.67
Slag, 13 mm (1 in.)
0.067
0.38
Cement mortar, 13 mm (1/2 in.)
0.018
0.10
Wood, 25 mm (1 in.)
0.22
1.25
Wood bevel lapped siding,
Wood stud, nominal 2 in
. X
13 mm X 200 mm
4 in. (3.5 in. or 90 mm w
de)
0.63
3.58
(1/2 in. X 8 in.)
0.14
0.81
temperature of 21°C (72°F) and a surface-air temperature difference of
5.5°C (10°F). Also, the equivalent surface temperature of the environment
is assumed to be equal to the ambient air temperature. Despite the conve-
nience it offers, this assumption is not quite accurate because of the addi-
tional radiation heat loss from the surface to the clear sky. The effect of sky
radiation can be accounted for approximately by taking the outside tem-
perature to be the average of the outdoor air and sky temperatures.
The inner surface heat transfer coefficient h t remains fairly constant
throughout the year, but the value of h varies considerably because of its
dependence on the orientation and wind speed, which can vary from less
than 1 km/h in calm weather to over 40 km/h during storms. The com-
monly used values of ft, and h for peak load calculations are
h, = 8.29 W/m 2 • °C = 1 .46 Btu/h • ft 2 ■ °F
r 34.0 W/m 2 ■ °C = 6.0 Btu/h ■ ft 2 ■ °F
//„
22.7 W/m 2 ■ °C = 4.0 Btu/h ■ ft 2
(winter and summer)
(winter)
(summer)
which correspond to design wind conditions of 24 km/h (15 mph) for win-
ter and 12 km/h (7.5 mph) for summer. The corresponding surface thermal
resistances (^-values) are determined from Rj = l/hj and R = l/h a . The
surface conductance values under still air conditions can be used for inte-
rior surfaces as well as exterior surfaces in calm weather.
cen58933_ch03.qxd 9/10/2002 8:59 AM Page 177
177
CHAPTER 3
Building components often involve trapped air spaces between various
layers. Thermal resistances of such air spaces depend on the thickness of
the layer, the temperature difference across the layer, the mean air temper-
ature, the emissivity of each surface, the orientation of the air layer, and the
direction of heat transfer. The emissivities of surfaces commonly encoun-
tered in buildings are given in Table 3-8. The effective emissivity of a
plane-parallel air space is given by
Effective e l e 2
1
(3-81)
where s { and e 2 are the emissivities of the surfaces of the air space. Table
3-8 also lists the effective emissivities of air spaces for the cases where
(1) the emissivity of one surface of the air space is e while the emissivity
of the other surface is 0.9 (a building material) and (2) the emissivity of
both surfaces is e. Note that the effective emissivity of an air space between
building materials is 0.82/0.03 = 27 times that of an air space between sur-
faces covered with aluminum foil. For specified surface temperatures, ra-
diation heat transfer through an air space is proportional to effective
emissivity, and thus the rate of radiation heat transfer in the ordinary sur-
face case is 27 times that of the reflective surface case.
Table 3-9 lists the thermal resistances of 20-mm-, 40-mm-, and 90-mm-
(0.75-in., 1.5-in., and 3.5-in.) thick air spaces under various conditions. The
thermal resistance values in the table are applicable to air spaces of uniform
thickness bounded by plane, smooth, parallel surfaces with no air leakage.
Thermal resistances for other temperatures, emissivities, and air spaces can
be obtained by interpolation and moderate extrapolation. Note that the
presence of a low-emissivity surface reduces radiation heat transfer across
an air space and thus significantly increases the thermal resistance. The
thermal effectiveness of a low-emissivity surface will decline, however, if
the condition of the surface changes as a result of some effects such as con-
densation, surface oxidation, and dust accumulation.
The 7?-value of a wall or roof structure that involves layers of uniform
thickness is determined easily by simply adding up the unit thermal re-
sistances of the layers that are in series. But when a structure involves
components such as wood studs and metal connectors, then the ther-
mal resistance network involves parallel connections and possible two-
dimensional effects. The overall R-yalue in this case can be determined by
assuming (1) parallel heat flow paths through areas of different construc-
tion or (2) isothermal planes normal to the direction of heat transfer. The
first approach usually overpredicts the overall thermal resistance, whereas
the second approach usually underpredicts it. The parallel heat flow path
approach is more suitable for wood frame walls and roofs, whereas the
isothermal planes approach is more suitable for masonry or metal frame
walls.
The thermal contact resistance between different components of building
structures ranges between 0.01 and 0.1 m 2 ■ °C/W, which is negligible in
most cases. However, it may be significant for metal building components
such as steel framing members.
TABLE 3-8
Emissivities s of various surfaces
and the effective emissivity of air
spaces (from ASHRAE Handbook
of Fundamentals, Ref. 1, Chap. 22,
Table 3).
Effective
EmissK
ity of
?
Air Space
1 = s
e 1 = e
Surface
E £
2 = 0.9
£ 2 = E
Aluminum foil,
bright
0.05*
0.05
0.03
Aluminum
sheet
0.12
0.12
0.06
Aluminum-
coated
paper,
polished
0.20
0.20
0.11
Steel, galvan
ized,
bright
0.25
0.24
0.15
Aluminum
paint
0.50
0.47
0.35
Building materials:
Wood, paper,
masonry,
nonmetallic
paints
0.90
0.82
0.82
Ordinary glass 0.84
0.77
0.72
*Surface emissivity of aluminum foil
increases to 0.30 with barely visible
condensation, and to 0.70 with clearly
visible condensation.
cen58933_ch03.qxd 9/10/2002 8:59 AM Page 17E
TABLE 3-9
Unit thermal resistances (ff-values) of well-sealed plane air spaces (from ASHRAE Handbook of Fundamentals, Ref. 1,
Chap. 22, Table 2)
(a) SI units (in m 2 • °C/W)
Position Direction
of Air of Heat
Space Flow
Mean
Temp.,
°C
Temp.
Diff.,
°C
20-mm Air Space
40-mm Air Space
90-mm Air Spa
Effective
Emissivity, e e
ce
Effective
Emissivity, e ef
Effective
Emissivity, e ef
t
0.03
0.05
0.5
0.82
0.03
0.05
0.5
0.82
0.03
0.05
0.5
0.82
32.2
5.6
0.41
0.39
0.18
0.13
0.45
0.42
0.19
0.14
0.50
0.47
0.20
0.14
10.0
16.7
0.30
0.29
0.17
0.14
0.33
0.32
0.18
0.14
0.27
0.35
0.19
0.15
Horizontal Up T
10.0
5.6
0.40
0.39
0.20
0.15
0.44
0.42
0.21
0.16
0.49
0.47
0.23
0.16
-17.8
11.1
0.32
0.32
0.20
0.16
0.35
0.34
0.22
0.17
0.40
0.38
0.23
0.18
32.2
5.6
0.52
0.49
0.20
0.14
0.51
0.48
0.20
0.14
0.56
0.52
0.21
0.14
10.0
16.7
0.35
0.34
0.19
0.14
0.38
0.36
0.20
0.15
0.40
0.38
0.20
0.15
45° slope UpT
10.0
5.6
0.51
0.48
0.23
0.17
0.51
0.48
0.23
0.17
0.55
0.52
0.24
0.17
-17.8
11.1
0.37
0.36
0.23
0.18
0.40
0.39
0.24
0.18
0.43
0.41
0.24
0.19
32.2
5.6
0.62
0.57
0.21
0.15
0.70
0.64
0.22
0.15
0.65
0.60
0.22
0.15
10.0
16.7
0.51
0.49
0.23
0.17
0.45
0.43
0.22
0.16
0.47
0.45
0.22
0.16
Vertical Horizontal
-> 10.0
5.6
0.65
0.61
0.25
0.18
0.67
0.62
0.26
0.18
0.64
0.60
0.25
0.18
-17.8
11.1
0.55
0.53
0.28
0.21
0.49
0.47
0.26
0.20
0.51
0.49
0.27
0.20
32.2
5.6
0.62
0.58
0.21
0.15
0.89
0.80
0.24
0.16
0.85
0.76
0.24
0.16
10.0
16.7
0.60
0.57
0.24
0.17
0.63
0.59
0.25
0.18
0.62
0.58
0.25
0.18
45° slope Down i
10.0
5.6
0.67
0.63
0.26
0.18
0.90
0.82
0.28
0.19
0.83
0.77
0.28
0.19
-17.8
11.1
0.66
0.63
0.30
0.22
0.68
0.64
0.31
0.22
0.67
0.64
0.31
0.22
32.2
5.6
0.62
0.58
0.21
0.15
1.07
0.94
0.25
0.17
1.77
1.44
0.28
0.18
10.0
16.7
0.66
0.62
0.25
0.18
1.10
0.99
0.30
0.20
1.69
1.44
0.33
0.21
Horizontal Down i
10.0
5.6
0.68
0.63
0.26
0.18
1.16
1.04
0.30
0.20
1.96
1.63
0.34
0.22
V
J
-17.8
11.1
0.74
0.70
0.32
0.23
1.24
1.13
0.39
0.26
1.92
1.68
0.43
0.29
(b) English units (in h
■ ft 2 ■ °F/Btu)
Position Direction
of Air of Heat
Space Flow
Mean
Temp.,
°F
Temp.
Diff.,
°F
0.75-in.
Air Space
1.5-in. A
ir Space
3.5-in. Air Space
Effective
Emissivity, e ef
Effective
Emissivity, e ef
Effective
Emissivity, e eff
0.03
0.05
0.5
0.82
0.03
0.05
0.5
0.82
0.03
0.05
0.5
0.82
90
10
2.34
2.22
1.04
0.75
2.55
2.41
1.08
0.77
2.84
2.66
1.13
0.80
50
30
1.71
1.66
0.99
0.77
1.87
1.81
1.04
0.80
2.09
2.01
1.10
0.84
Horizontal Up T
50
10
2.30
2.21
1.16
0.87
2.50
2.40
1.21
0.89
2.80
2.66
1.28
0.93
20
1.83
1.79
1.16
0.93
2.01
1.95
1.23
0.97
2.25
2.18
1.32
1.03
90
10
2.96
2.78
1.15
0.81
2.92
2.73
1.14
0.80
3.18
2.96
1.18
0.82
50
30
1.99
1.92
1.08
0.82
2.14
2.06
1.12
0.84
2.26
2.17
1.15
0.86
45° slope UpT
50
10
2.90
2.75
1.29
0.94
2.88
2.74
1.29
0.94
3.12
2.95
1.34
0.96
20
2.13
2.07
1.28
1.00
2.30
2.23
1.34
1.04
2.42
2.35
1.38
1.06
90
10
3.50
3.24
1.22
0.84
3.99
3.66
1.27
0.87
3.69
3.40
1.24
0.85
50
30
2.91
2.77
1.30
0.94
2.58
2.46
1.23
0.90
2.67
2.55
1.25
0.91
Vertical Horizontal
-* 50
10
3.70
3.46
1.43
1.01
3.79
3.55
1.45
1.02
3.63
3.40
1.42
1.01
20
3.14
3.02
1.58
1.18
2.76
2.66
1.48
1.12
2.88
2.78
1.51
1.14
90
10
3.53
3.27
1.22
0.84
5.07
4.55
1.36
0.91
4.81
4.33
1.34
0.90
50
30
3.43
3.23
1.39
0.99
3.58
3.36
1.42
1.00
3.51
3.30
1.40
1.00
45° slope Down i
50
10
3.81
3.57
1.45
1.02
5.10
4.66
1.60
1.09
4.74
4.36
1.57
1.08
20
3.75
3.57
1.72
1.26
3.85
3.66
1.74
1.27
3.81
3.63
1.74
1.27
90
10
3.55
3.29
1.22
0.85
6.09
5.35
1.43
0.94
10.07
8.19
1.57
1.00
50
30
3.77
3.52
1.44
1.02
6.27
5.63
1.70
1.14
9.60
8.17
1.88
1.22
Horizontal Down i
50
10
3.84
3.59
1.45
1.02
6.61
5.90
1.73
1.15
11.15
9.27
1.93
1.24
20
4.18
3.96
1.81
1.30
7.03
6.43
2.19
1.49
10.90
9.52
2.47
1.62
178
cen58933_ch03.qxd 9/10/2002 8:59 AM Page 179
EXAMPLE 3-16
The /?- Value of a Wood Frame Wall
Determine the overall unit thermal resistance (the ff-value) and the overall heat
transfer coefficient (the L/-factor) of a wood frame wall that is built around
38-mm X 90-mm (2x4 nominal) wood studs with a center-to-center distance
of 400 mm. The 90-mm-wide cavity between the studs is filled with glass fiber
insulation. The inside is finished with 13-mm gypsum wallboard and the out-
side with 13-mm wood fiberboard and 13-mm X 200-mm wood bevel lapped
siding. The insulated cavity constitutes 75 percent of the heat transmission
area while the studs, plates, and sills constitute 21 percent. The headers con-
stitute 4 percent of the area, and they can be treated as studs.
Also, determine the rate of heat loss through the walls of a house whose
perimeter is 50 m and wall height is 2.5 m in Las Vegas, Nevada, whose winter
design temperature is -2°C. Take the indoor design temperature to be 22°C
and assume 20 percent of the wall area is occupied by glazing.
179
CHAPTER 3
SOLUTION The ff-value and the ^/-factor of a wood frame wall as well as the
rate of heat loss through such a wall in Las Vegas are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Heat transfer through the
wall is one-dimensional. 3 Thermal properties of the wall and the heat transfer
coefficients are constant.
Properties The ff-values of different materials are given in Table 3-6.
Analysis The schematic of the wall as well as the different elements used in its
construction are shown here. Heat transfer through the insulation and through
the studs will meet different resistances, and thus we need to analyze the ther-
mal resistance for each path separately. Once the unit thermal resistances and
the L/-factors for the insulation and stud sections are available, the overall av-
erage thermal resistance for the entire wall can be determined from
''nv era 1 1 1 / L/ rivers 1 1
where
overall V ./area/insi
(U X/ arca ) stud
and the value of the area fraction f area is 0.75 for the insulation section and
0.25 for the stud section since the headers that constitute a small part of the
wall are to be treated as studs. Using the available ff-values from Table 3-6 and
calculating others, the total ff-values for each section can be determined in a
systematic manner in the table in this sample.
We conclude that the overall unit thermal resistance of the wall is 2.23
m 2 • °C/W, and this value accounts for the effects of the studs and headers. It
corresponds to an ff-value of 2.23 X 5.68 = 12.7 (or nearly ff-13) in English
units. Note that if there were no wood studs and headers in the wall, the over-
all thermal resistance would be 3.05 m 2 ■ °C/W, which is 37 percent greater
than 2.23 m 2 • °C/W. Therefore, the wood studs and headers in this case serve
as thermal bridges in wood frame walls, and their effect must be considered in
the thermal analysis of buildings.
cen58933_ch03.qxd 9/10/2002 8:59 AM Page IE
180
HEAT TRANSFER
Schematic
ft- value,
m 2 • °C/W
Construction
Between At
Studs Studs
1.
Outside surface,
24 km/h wind
0.030
0.030
2.
Wood bevel lapped
siding
0.14
0.14
3.
Wood fiberboard
sheeting, 13 mm
0.23
0.23
4a.
Glass fiber
insulation, 90 mm
2.45
—
4b.
Wood stud, 38 mm X
6
90 mm
—
0.63
5.
Gypsum wall board,
13 mm
0.079
0.079
6.
Inside surface, still air
0.12
0.12
Total unit thermal resistance of each section, R (in m 2 • °C/W) 3.05 1.23
The U-factor of each section, U = 1/ff, in W/m 2 • °C 0.328 0.813
Area fraction of each section, 4rea 0.75 0.25
Overall U-factor: U= 2f area , U, = 0.75 X 0.328 + 0.25 X 0.813
= 0.449 W/m 2 • °C
Overall unit thermal resistance: R = l/U= 2.23 m 2 • °C/W
The perimeter of the building is 50 m and the height of the walls is 2.5 m.
Noting that glazing constitutes 20 percent of the walls, the total wall area is
A wa „ = 0.80(Perimeter)(Height) = 0.80(50 m)(2.5 m) = 100 m 2
Then the rate of heat loss through the walls under design conditions becomes
Swan = (HA) wall {T i - T )
= (0.449 W/m 2 • °C)(100 m 2 )[22 - (-2)°C]
= 1078 W
Discussion Note that a 1-kW resistance heater in this house will make up al-
most all the heat lost through the walls, except through the doors and windows,
when the outdoor air temperature drops to -2°C.
EXAMPLE 3-17 The ff-Value of a Wall with Rigid Foam
The 13-mm-thick wood fiberboard sheathing of the wood stud wall discussed in
the previous example is replaced by a 25-mm-thick rigid foam insulation. De-
termine the percent increase in the R -value of the wall as a result.
cen58933_ch03.qxd 9/10/2002 8:59 AM Page 181
SOLUTION The overall ff-value of the existing wall was determined in Example
3-16 to be 2.23 m 2 • °C/W. Noting that the ff-values of the fiberboard and the
foam insulation are 0.23 m z • °C/W and 0.98 m 2 • °C/W, respectively, and
the added and removed thermal resistances are in series, the overall ff-value of
the wall after modification becomes
Rr,
R„
old removed added
2.23 - 0.23 + 0.98
2.98 m 2 • °C/W
This represents an increase of (2.98 - 2.23)/2.23 = 0.34 or 34 percent in
the ff-value of the wall. This example demonstrated how to evaluate the new
ff-value of a structure when some structural members are added or removed.
EXAMPLE 3-18 The ff-Value of a Masonry Wall
Determine the overall unit thermal resistance (the ff-value) and the overall heat
transfer coefficient (the ^/-factor) of a masonry cavity wall that is built around
6-in. -thick concrete blocks made of lightweight aggregate with 3 cores filled
with perlite (ff = 4.2 h • ft 2 • °F/Btu). The outside is finished with 4-in. face
brick with i-in. cement mortar between the bricks and concrete blocks. The in-
side finish consists of | in. gypsum wallboard separated from the concrete block
by f-in.-thick (1-in. X 3-in. nominal) vertical furring (ff = 4.2 h • ft 2 • °F/Btu)
whose center-to-center distance is 16 in. Both sides of the |-in. -thick air space
between the concrete block and the gypsum board are coated with reflective
aluminum foil (e = 0.05) so that the effective emissivity of the air space is
0.03. For a mean temperature of 50°F and a temperature difference of 30°F,
the ff-value of the air space is 2.91 h • ft 2 ■ °F/Btu. The reflective air space con-
stitutes 80 percent of the heat transmission area, while the vertical furring con-
stitutes 20 percent.
SOLUTION The ff-value and the U-factor of a masonry cavity wall are to be
determined.
Assumptions 1 Steady operating conditions exist. 2 Heat transfer through the
wall is one-dimensional. 3 Thermal properties of the wall and the heat transfer
coefficients are constant.
Properties The ff-values of different materials are given in Table 3-6.
Analysis The schematic of the wall as well as the different elements used in its
construction are shown below. Following the approach described here and using
the available ff-values from Table 3-6, the overall ff-value of the wall is deter-
mined in this table.
181
CHAPTER 3
cen58933_ch03.qxd 9/10/2002 8:59 AM Page 182
182
HEAT TRANSFER
Schematic
l
fl-va
ue,
h • ft 2 •
°F/Btu
Between
At
Construction
Furring
Furring
1.
Outside surface,
15 mph wind
0.17
0.17
2.
Face brick, 4 in.
0.43
0.43
3.
Cement mortar,
0.5 in.
0.10
0.10
4.
Concrete block, 6 in
. 4.20
4.20
5a.
Reflective air space,
fin-
2.91
—
^ 5b.
Nominal 1x3
<\
vertical furring
—
0.94
Gypsum wall board,
0.5 in.
0.45
0.45
7.
Inside surface,
still air
0.68
0.68
Total unit thermal resistance of each section, R 8.94 6.97
The U-factor of each section, U = 1/ff, in Btu/h ■ ft 2 • °F 0.112 0.143
Area fraction of each section, 4rea 0.80 0.20
Overall £/-f actor: U = 2f area ,U, = 0.80 X 0.112 + 0.20 X 0.143
= 0.1 18 Btu/h- ft 2 -°F
Overall unit thermal resistance: R= IIU= 8.46 h • ft 2 • °F/Btu
Therefore, the overall unit thermal resistance of the wall is 8.46 h • ft 2 • C F/Btu
and the overall l/-factor is 0.118 Btu/h • ft 2 • °F. These values account for the
effects of the vertical furring.
EXAMPLE 3-19
The ff-Value of a Pitched Roof
Determine the overall unit thermal resistance (the ff -value) and the overall heat
transfer coefficient (the ^/-factor) of a 45° pitched roof built around nominal
2-in. X 4-in. wood studs with a center-to-center distance of 16 in. The 3.5-in.-
wide air space between the studs does not have any reflective surface and thus
its effective emissivity is 0.84. For a mean temperature of 90°F and a temper-
ature difference of 30°F, the ff-value of the air space is 0.86 h • ft 2 • °F/Btu.
The lower part of the roof is finished with |-in. gypsum wallboard and the upper
part with |-in. plywood, building paper, and asphalt shingle roofing. The air
space constitutes 75 percent of the heat transmission area, while the studs and
headers constitute 25 percent.
SOLUTION The ff-value and the ^/-factor of a 45° pitched roof are to be
determined.
Assumptions 1 Steady operating conditions exist. 2 Heat transfer through the
roof is one-dimensional. 3 Thermal properties of the roof and the heat transfer
coefficients are constant.
Properties The ff-values of different materials are given in Table 3-6.
cen58933_ch03.qxd 9/10/2002 9:00 AM Page 183
183
CHAPTER 3
Analysis The schematic
of the pitched roof as well as the
different elements
used in its construction
are
shown below. Following the a
oproach described
above and using the avai
able ff-values from Table 3-6, the
overall R-
/alue of
the roof can be determined in
the table here.
Schematic
fl-va
h • ft 2 •
Between
ue,
°F/Btu
At
. .
Construction
Studs
Studs
1.
Outside surface,
/ " V \ X \ X \^
15 mph wind
0.17
0.17
/ "^Cv^O^N^/
2.
Asphalt shingle
/ /Q/^f^^w/
roofing
0.44
0.44
/ /r^j^^^A 4 /
3.
Building paper
0.10
0.10
TrCyA
4.
Plywood deck, | in.
0.78
0.78
I I \ \ \ \ C )
5a
Nonreflective air
1 2 3 4 5a 5b 6 7
space, 3.5 in.
0.86
—
5b
Wood stud, 2 in. by 4 in.
—
3.58
6.
Gypsum wallboard, 0.5 in.
0.45
0.45
7.
Inside surface,
45° slope, still air
0.63
0.63
Total unit thermal resista
ice of each section, R
3.43
6.15
The Ofactor of each sect
ion,
U= 1/fl, in Btu/h -ft 2 - °F
0.292
0.163
Area fraction of each seel
ion,
'area
0.75
0.25
Overall U-i actor: U= 2f area
iUr-
= 0.75 X 0.292 + 0.25 X 0.163
= 0.260 Btu/h • ft 2 • °F
Overall unit thermal resistance: R = 1/U =
3.85 h • fl :
• °F/Btu
Therefore, the overall
unit
thermal resistance of this
pitched
roof is
3.85 h • ft 2 • °F/Btu and the overall tZ-factor is 0.260 Btu/h
• ft 2 • °F. Note that
the wood studs offer much larger thermal resistance to heat flow than
the air
space between the studs.
The construction of wood frame flat ceilings typically involve 2-in. X
6-in. joists on 400-mm (16-in.) or 600-mm (24-in.) centers. The fraction of
framing is usually taken to be 0.10 for joists on 400-mm centers and 0.07
for joists on 600-mm centers.
Most buildings have a combination of a ceiling and a roof with an attic
space in between, and the determination of the 7?-value of the roof-attic-
ceiling combination depends on whether the attic is vented or not. For ad-
equately ventilated attics, the attic air temperature is practically the same as
the outdoor air temperature, and thus heat transfer through the roof is gov-
erned by the /lvalue of the ceiling only. However, heat is also transferred
between the roof and the ceiling by radiation, and it needs to be considered
(Fig. 3-52). The major function of the roof in this case is to serve as a ra-
diation shield by blocking off solar radiation. Effectively ventilating the at-
tic in summer should not lead one to believe that heat gain to the building
through the attic is greatly reduced. This is because most of the heat trans-
fer through the attic is by radiation.
Aii-
exhaust
.^ppRadiant
s^' barrier
K/\/\/w\/w\/\/\/w\/\/\/y
K
n
Ail-
intake
Air
intake
FIGURE 3-52
Ventilation paths for a naturally
ventilated attic and the appropriate
size of the flow areas around the
radiant barrier for proper air
circulation (from DOE/CE-0335P,
U.S. Dept. of Energy).
cen58933_ch03.qxd 9/10/2002 9:00 AM Page 184
184
HEAT TRANSFER
- Roof decking
Air space
Roof decking
- Roof decking
Joist -
Insulation
Joist — Insulation -
(b) At the bottom of rafters
(a) Under the roof deck
FIGURE 3-53
Three possible locations for an attic radiant barrier (from DOE/CE-0335P, U.S. Dept. of Energy).
Joist — Insulation -
(c) On top of attic floor insulation
Shingles
Deck
^Rafter/^ %\
y^ Attic < "
A
r ceiling
attic \
Ceiling joist >
' ^ceiling
?i
FIGURE 3-54
Thermal resistance network for a
pitched roof-attic-ceiling combination
for the case of an unvented attic.
Radiation heat transfer between the ceiling and the roof can be mini-
mized by covering at least one side of the attic (the roof or the ceiling side)
by a reflective material, called radiant barrier, such as aluminum foil or
aluminum-coated paper. Tests on houses with R- 19 attic floor insulation
have shown that radiant barriers can reduce summer ceiling heat gains by
16 to 42 percent compared to an attic with the same insulation level and no
radiant barrier. Considering that the ceiling heat gain represents about 15 to
25 percent of the total cooling load of a house, radiant barriers will reduce
the air conditioning costs by 2 to 10 percent. Radiant barriers also reduce
the heat loss in winter through the ceiling, but tests have shown that the
percentage reduction in heat losses is less. As a result, the percentage
reduction in heating costs will be less than the reduction in the air-
conditioning costs. Also, the values given are for new and undusted radiant
barrier installations, and percentages will be lower for aged or dusty radi-
ant barriers.
Some possible locations for attic radiant barriers are given in Figure
3-53. In whole house tests on houses with R-\9 attic floor insulation, radi-
ant barriers have reduced the ceiling heat gain by an average of 35 percent
when the radiant barrier is installed on the attic floor, and by 24 percent
when it is attached to the bottom of roof rafters. Test cell tests also demon-
strated that the best location for radiant barriers is the attic floor, provided
that the attic is not used as a storage area and is kept clean.
For unvented attics, any heat transfer must occur through (1) the ceiling,
(2) the attic space, and (3) the roof (Fig. 3-54). Therefore, the overall
K-value of the roof-ceiling combination with an unvented attic depends on
the combined effects of the 7?-value of the ceiling and the K-value of the
roof as well as the thermal resistance of the attic space. The attic space can
be treated as an air layer in the analysis. But a more practical way of ac-
counting for its effect is to consider surface resistances on the roof and ceil-
ing surfaces facing each other. In this case, the ^-values of the ceiling and
the roof are first determined separately (by using convection resistances for
the still-air case for the attic surfaces). Then it can be shown that the over-
all i?-value of the ceiling-roof combination per unit area of the ceiling can
be expressed as
cen58933_ch03.qxd 9/10/2002 9:00 AM Page 185
/? = /? H- /?
"■ "ceiling "roof
^ceiling
^roof
(3-82)
185
CHAPTER 3
where i4 cei | in g and A mof are the ceiling and roof areas, respectively. The area
ratio is equal to 1 for flat roofs and is less than 1 for pitched roofs. For a 45°
pitched roof, the area ratio is ^ ce iiin g /^roof = l/\/2 = 0.707. Note that the
pitched roof has a greater area for heat transfer than the flat ceiling, and the
area ratio accounts for the reduction in the unit lvalue of the roof when
expressed per unit area of the ceiling. Also, the direction of heat flow is up
in winter (heat loss through the roof) and down in summer (heat gain
through the roof).
The 7?-value of a structure determined by analysis assumes that the ma-
terials used and the quality of workmanship meet the standards. Poor work-
manship and substandard materials used during construction may result in
^-values that deviate from predicted values. Therefore, some engineers use
a safety factor in their designs based on experience in critical applications.
SUMMARY
One-dimensional heat transfer through a simple or composite
body exposed to convection from both sides to mediums at
temperatures T^ t and T rjJ1 can be expressed as
Q
(W)
mediums. For a plane wall exposed to convection on both
sides, the total resistance is expressed as
R„
R,,
-"wiiii *~ Rr,
1
1
This relation can be extended to plane walls that consist of two
or more layers by adding an additional resistance for each ad-
ditional layer. The elementary thermal resistance relations can
be expressed as follows:
where h c is the thermal contact conductance, R c is the thermal
contact resistance, and the radiation heat transfer coefficient is
defined as
/>,,
s(j(T; + Tl Tr )(T s + T sulr )
Once the rate of heat transfer is available, the temperature drop
across any layer can be determined from
AT= QR
The thermal resistance concept can also be used to solve steady
heat transfer problems involving parallel layers or combined
series-parallel arrangements.
Adding insulation to a cylindrical pipe or a spherical shell
will increase the rate of heat transfer if the outer radius of
the insulation is less than the critical radius of insulation,
defined as
Interface resistance:
Radiation resistance:
v cyl
Conduction resistance (plane wall): R wa[[
Conduction resistance (cylinder): R
Conduction resistance (sphere):
Convection resistance:
kA
Info//-))
2nLk
R
sph
R
R
■"rad
4ir r x r 2 k
J_
: onv hA
j_ = Rc
nterface ^ £
1
"rad "■
cr, cylinder
h
2k,
* cr, sphere
h
The effectiveness of an insulation is often given in terms of
its R-value, the thermal resistance of the material per unit sur-
face area, expressed as
lvalue
L
(flat insulation)
where L is the thickness and k is the thermal conductivity of the
material.
cen58933_ch03.qxd 9/10/2002 9:00 AM Page 186
186
HEAT TRANSFER
Finned surfaces are commonly used in practice to enhance
heat transfer. Fins enhance heat transfer from a surface by ex-
posing a larger surface area to convection. The temperature
distribution along the fin for very long fins and for fins with
negligible heat transfer at the fin are given by
Very long fin:
Adiabatic fin tip:
T(x) - T a
T(x)
- T a
p—x\ hplkA
cosh a(L — x)
cosh ah
where a = \/hplkA c , p is the perimeter, and A c is the cross
sectional area of the fin. The rates of heat transfer for both
cases are given to be
Very
long ei ongfin = ~kA c
fin:
Adiabatic
Jm ^c insulated tip — ~~ ^c
tip:
VhpkAc(T b -TJ
\/kpkA c (T b - r„) tanh ah
Fins exposed to convection at their tips can be treated as fins
with insulated tips by using the corrected length L c = L + AJp
instead of the actual fin length.
The temperature of a fin drops along the fin, and thus the
heat transfer from the fin will be less because of the decreasing
temperature difference toward the fin tip. To account for the ef-
fect of this decrease in temperature on heat transfer, we define
fin efficiency as
%i„
Q<
Q
fin, max
Actual heat transfer rate from the fin
Ideal heat transfer rate from the fin if
the entire fin were at base temperature
The performance of the fins is judged on the basis of the en-
hancement in heat transfer relative to the no-fin case and is ex-
pressed in terms of the fin effectiveness s fln , defined as
e f ,»
g f ,„
e» of ,„ hA t {T t -TJ
Heat transfer rate from
the fin of base area A b
Heat transfer rate from
the surface of area A h
Here, A h is the cross-sectional area of the fin at the base and
Q no fin represents the rate of heat transfer from this area if no
fins are attached to the surface. The overall effectiveness for a
finned surface is defined as the ratio of the total heat transfer
from the finned surface to the heat transfer from the same sur-
face if there were no fins,
Q total, fin h(A unfm + TlflnAaJfTj, - TJ)
Q
total, no fin
"A no fin (T b T„)
Fin efficiency and fin effectiveness are related to each other by
Afl„
A,
%,„
Certain multidimensional heat transfer problems involve two
surfaces maintained at constant temperatures T { and T 2 . The
steady rate of heat transfer between these two surfaces is ex-
pressed as
Q = Sk(T { - T 2 )
where S is the conduction shape factor that has the dimen-
sion of length and k is the thermal conductivity of the medium
between the surfaces.
When the fin efficiency is available, the rate of heat transfer
from a fin can be determined from
Gfi„ = infi„2f
■nfin^ fin (T b - T„)
REFERENCES AND SUGGESTED READING
1. American Society of Heating, Refrigeration, and Air
Conditioning Engineers. Handbook of Fundamentals.
Atlanta: ASHRAE, 1993.
2. R. V. Andrews. "Solving Conductive Heat Transfer
Problems with Electrical-Analogue Shape Factors."
Chemical Engineering Progress 5 (1955), p. 67.
3. R. Barron. Cryogenic Systems. New York: McGraw-Hill,
1967.
4. L. S. Fletcher. "Recent Developments in Contact
Conductance Heat Transfer." Journal of Heat Transfer
110, no. 4B (1988), pp. 1059-79.
5. E. Fried. "Thermal Conduction Contribution to Heat
Transfer at Contacts." Thermal Conductivity, vol. 2, ed.
R. R Tye. London: Academic Press, 1969.
6. K. A. Gardner. "Efficiency of Extended Surfaces." Trans.
ASME 67 (1945), pp. 621-31.
7. F. P. Incropera and D. P. DeWitt. Introduction to Heat
Transfer. 4th ed. New York: John Wiley & Sons, 2002.
8. D. Q. Kern and A. D. Kraus. Extended Surface Heat
Transfer. New York: McGraw-Hill, 1972.
cen58933_ch03.qxd 9/10/2002 9:00 AM Page 187
9. M. N. Ozisik. Heat Transfer — A Basic Approach. New
York: McGraw-Hill, 1985.
10. G. P. Peterson. "Thermal Contact Resistance in Waste
Heat Recovery Systems." Proceedings of the 18th
ASME/ETCE Hydrocarbon Processing Symposium.
Dallas, TX, 1987, pp. 45-51.
11. S. Song, M. M. Yovanovich, and F. O. Goodman.
"Thermal Gap Conductance of Conforming Surfaces in
Contact." Journal of Heat Transfer 115 (1993), p. 533.
PROBLEMS
187
CHAPTER 3
12. J. E. Sunderland and K. R. Johnson. "Shape Factors for
Heat Conduction through Bodies with Isothermal or
Convective Boundary Conditions," Trans. ASME 10
(1964), pp. 237-41.
13. N. V. Suryanarayana. Engineering Heat Transfer. St. Paul,
MN: West Publishing, 1995.
Steady Heat Conduction in Plane Walls
3-1 C Consider one-dimensional heat conduction through
a cylindrical rod of diameter D and length L. What is the
heat transfer area of the rod if (a) the lateral surfaces of the rod
are insulated and (b) the top and bottom surfaces of the rod are
insulated?
3-2C Consider heat conduction through a plane wall. Does
the energy content of the wall change during steady heat con-
duction? How about during transient conduction? Explain.
3-3C Consider heat conduction through a wall of thickness L
and area A. Under what conditions will the temperature distri-
butions in the wall be a straight line?
3-4C What does the thermal resistance of a medium
represent?
3-5C How is the combined heat transfer coefficient defined?
What convenience does it offer in heat transfer calculations?
3-6C Can we define the convection resistance per unit
surface area as the inverse of the convection heat transfer
coefficient?
3-7C Why are the convection and the radiation resistances at
a surface in parallel instead of being in series?
3-8C Consider a surface of area A at which the convection
and radiation heat transfer coefficients are h com and /z rad , re-
spectively. Explain how you would determine (a) the single
equivalent heat transfer coefficient, and (b) the equivalent ther-
mal resistance. Assume the medium and the surrounding sur-
faces are at the same temperature.
3-9C How does the thermal resistance network associated
with a single-layer plane wall differ from the one associated
with a five-layer composite wall?
*Problems designated by a "C" are concept questions, and
students are encouraged to answer them all. Problems designated
by an "E" are in English units, and the SI users can ignore them.
Problems with an EES-CD icon ® are solved using EES, and
complete solutions together with parametric studies are included
on the enclosed CD. Problems with a computer-EES icon H are
comprehensive in nature, and are intended to be solved with a
computer, preferably using the EES software that accompanies
this text.
3-10C Consider steady one-dimensional heat transfer
through a multilayer medium. If the rate of heat transfer Q is
known, explain how you would determine the temperature
drop across each layer.
3-11C Consider steady one-dimensional heat transfer
through a plane wall exposed to convection from both sides to
environments at known temperatures T rA and T^ 2 with known
heat transfer coefficients h { and h 2 . Once the rate of heat trans-
fer Q has been evaluated, explain how you would determine
the temperature of each surface.
3-12C Someone comments that a microwave oven can be
viewed as a conventional oven with zero convection resistance
at the surface of the food. Is this an accurate statement?
3-13C Consider a window glass consisting of two 4-mm-
thick glass sheets pressed tightly against each other. Compare
the heat transfer rate through this window with that of one con-
sisting of a single 8-mm-thick glass sheet under identical con-
ditions.
3-14C Consider steady heat transfer through the wall of a
room in winter. The convection heat transfer coefficient at the
outer surface of the wall is three times that of the inner surface
as a result of the winds. On which surface of the wall do you
think the temperature will be closer to the surrounding air tem-
perature? Explain.
3-1 5C The bottom of a pan is made of a 4-mm-thick alu-
minum layer. In order to increase the rate of heat transfer
through the bottom of the pan, someone proposes a design for
the bottom that consists of a 3-mm-thick copper layer sand-
wiched between two 2-mm-thick aluminum layers. Will the
new design conduct heat better? Explain. Assume perfect con-
tact between the layers.
3-16C Consider two cold canned drinks, one wrapped in a
blanket and the other placed on a table in the same room.
Which drink will warm up faster?
3-17 Consider a 4-m-high, 6-m-wide, and 0.3-m-thick brick
wall whose thermal conductivity is k = 0.8 W/m ■ °C . On a
certain day, the temperatures of the inner and the outer surfaces
of the wall are measured to be 14°C and 6°C, respectively. De-
termine the rate of heat loss through the wall on that day.
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HEAT TRANSFER
Aluminum
FIGURE P3-15C
Copper
3-18 Consider a 1.2-m-high and 2-m-wide glass window
whose thickness is 6 mm and thermal conductivity is k = 0.78
W/m ■ °C. Determine the steady rate of heat transfer through
this glass window and the temperature of its inner surface for a
day during which the room is maintained at 24°C while the
temperature of the outdoors is — 5°C. Take the convection heat
transfer coefficients on the inner and outer surfaces of the win-
dow to be hi = 10 W/m 2 • °C and h 2 = 25 W/m 2 • °C, and dis-
regard any heat transfer by radiation.
3-19 Consider a 1 .2-m-high and 2-m-wide double-pane win-
dow consisting of two 3-mm-thick layers of glass (k = 0.78
W/m • °C) separated by a 12-mm-wide stagnant air space (k =
0.026 W/m • °C). Determine the steady rate of heat transfer
through this double-pane window and the temperature of its
inner surface for a day during which the room is maintained
at 24°C while the temperature of the outdoors is — 5°C. Take
the convection heat transfer coefficients on the inner and outer
Glass
surfaces of the window to be h { = 10 W/m 2 • °C and h 2 =
25 W/m 2 • °C, and disregard any heat transfer by radiation.
Answers: 114 W, 19.2°C
3-20 Repeat Problem 3-19, assuming the space between the
two glass layers is evacuated.
3-21 Ta'M Reconsider Problem 3-19. Using EES (or other)
1^2 software, plot the rate of heat transfer through the
window as a function of the width of air space in the range of
2 mm to 20 mm, assuming pure conduction through the air.
Discuss the results.
3-22E Consider an electrically heated brick house (k = 0.40
Btu/h • ft • °F) whose walls are 9 ft high and 1 ft thick. Two of
the walls of the house are 40 ft long and the others are 30 ft
long. The house is maintained at 70°F at all times while the
temperature of the outdoors varies. On a certain day, the tem-
perature of the inner surface of the walls is measured to be at
55°F while the average temperature of the outer surface is ob-
served to remain at 45°F during the day for 10 h and at 35°F at
night for 14 h. Determine the amount of heat lost from the
house that day. Also determine the cost of that heat loss to the
homeowner for an electricity price of $0.09/kWh.
FIGURE P3-1 9
FIGURE P3-22E
3-23 A cylindrical resistor element on a circuit board dissi-
pates 0. 1 5 W of power in an environment at 40°C. The resistor
is 1 .2 cm long, and has a diameter of 0.3 cm. Assuming heat to
be transferred uniformly from all surfaces, determine (a) the
amount of heat this resistor dissipates during a 24-h period,
(b) the heat flux on the surface of the resistor, in W/m 2 , and
(c) the surface temperature of the resistor for a combined con-
vection and radiation heat transfer coefficient of 9 W/m 2 • °C.
3-24 Consider a power transistor that dissipates 0.2 W of
power in an environment at 30°C. The transistor is 0.4 cm long
and has a diameter of 0.5 cm. Assuming heat to be transferred
uniformly from all surfaces, determine (a) the amount of heat
this resistor dissipates during a 24-h period, in kWh; (b) the
heat flux on the surface of the transistor, in W/m 2 ; and (c) the
surface temperature of the resistor for a combined convection
and radiation heat transfer coefficient of 18 W/m 2 ■ °C.
3-25 A 12-cm X 18-cm circuit board houses on its surface
100 closely spaced logic chips, each dissipating 0.07 W in an
environment at 40°C. The heat transfer from the back surface
of the board is negligible. If the heat transfer coefficient on the
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CHAPTER 3
30°C
Power
transistor
0.2 W
0.5 cm
— 0.4 cm
IGURE P3-24
surface of the board is 10 W/m 2 • °C, determine (a) the heat
flux on the surface of the circuit board, in W/m 2 ; (b) the surface
temperature of the chips; and (c) the thermal resistance be-
tween the surface of the circuit board and the cooling medium,
in °C/W.
3-26 Consider a person standing in a room at 20°C with an
exposed surface area of 1.7 m 2 . The deep body temperature of
the human body is 37°C, and the thermal conductivity of the
human tissue near the skin is about 0.3 W/m • °C. The body is
losing heat at a rate of 150 W by natural convection and radia-
tion to the surroundings. Taking the body temperature 0.5 cm
beneath the skin to be 37°C, determine the skin temperature of
the person. Answer: 35.5° C
3-27 Water is boiling in a 25-cm-diameter aluminum pan {k =
237 W/m ■ °C) at 95°C. Heat is transferred steadily to the boil-
ing water in the pan through its 0.5-cm-thick flat bottom at a
rate of 800 W. If the inner surface temperature of the bottom of
the pan is 108°C, determine (a) the boiling heat transfer coeffi-
cient on the inner surface of the pan, and (b) the outer surface
temperature of the bottom of the pan.
3-28E A wall is constructed of two layers of 0.5-in-thick
sheetrock (k = 0.10 Btu/h • ft • °F), which is a plasterboard
made of two layers of heavy paper separated by a layer of
gypsum, placed 5 in. apart. The space between the sheetrocks
Fiberglass
insulation
Sheetrock
0.5 in.
5 in.
0.5 in.
is filled with fiberglass insulation (k = 0.020 Btu/h • ft • °F).
Determine (a) the thermal resistance of the wall, and (b) its
R- value of insulation in English units.
3-29 The roof of a house consists of a 3-cm-thick concrete
slab (k = 2 W/m ■ °C) that is 15 m wide and 20 m long. The
convection heat transfer coefficients on the inner and outer sur-
faces of the roof are 5 and 12 W/m 2 • °C, respectively. On a
clear winter night, the ambient air is reported to be at 10°C,
while the night sky temperature is 100 K. The house and the in-
terior surfaces of the wall are maintained at a constant temper-
ature of 20 C C. The emissivity of both surfaces of the concrete
roof is 0.9. Considering both radiation and convection heat
transfers, determine the rate of heat transfer through the roof,
and the inner surface temperature of the roof.
If the house is heated by a furnace burning natural gas with
an efficiency of 80 percent, and the price of natural gas is
$0.60/therm (1 therm = 105,500 kJ of energy content), deter-
mine the money lost through the roof that night during a 14-h
period.
r sky = iooK
T. = 10°C
15 cm
FIGURE P3-28E
FIGURE P3-29
3-30 A 2-m X 1 .5-m section of wall of an industrial furnace
burning natural gas is not insulated, and the temperature at the
outer surface of this section is measured to be 80°C. The tem-
perature of the furnace room is 30°C, and the combined con-
vection and radiation heat transfer coefficient at the surface of
the outer furnace is 10 W/m 2 • °C. It is proposed to insulate this
section of the furnace wall with glass wool insulation (k =
0.038 W/m • °C) in order to reduce the heat loss by 90 percent.
Assuming the outer surface temperature of the metal section
still remains at about 80°C, determine the thickness of the in-
sulation that needs to be used.
The furnace operates continuously and has an efficiency of
78 percent. The price of the natural gas is $0.55/therm (1 therm
= 105,500 kJ of energy content). If the installation of the insu-
lation will cost $250 for materials and labor, determine how
long it will take for the insulation to pay for itself from the en-
ergy it saves.
3-31 Repeat Problem. 3-30 for expanded perlite insulation
assuming conductivity is k = 0.052 W/m • °C.
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HEAT TRANSFER
3-32 [?(.*)! Reconsider Problem 3-30. Using EES (or other)
|^£^ software, investigate the effect of thermal con-
ductivity on the required insulation thickness. Plot the thick-
ness of insulation as a function of the thermal conductivity of
the insulation in the range of 0.02 W/m ■ °C to 0.08 W/m • °C,
and discuss the results.
3-33E Consider a house whose walls are 12 ft high and 40 ft
long. Two of the walls of the house have no windows, while
each of the other two walls has four windows made of 0.25-in.-
thick glass (k = 0.45 Btu/h ■ ft ■ °F), 3 ft X 5 ft in size. The
walls are certified to have an /{-value of 19 (i.e., an Llk value of
19 h • ft 2 • °F/Btu). Disregarding any direct radiation gain or
loss through the windows and taking the heat transfer coef-
ficients at the inner and outer surfaces of the house to be 2 and
4 Btu/h • ft 2 • °F, respectively, determine the ratio of the heat
transfer through the walls with and without windows.
Attic
space
12 ft
Sheet metal
40 ft
FIGURE P3-33E
3-34 Consider a house that has a 10-m X 20-m base and a
4-m-high wall. All four walls of the house have an /{-value of
2.31 m 2 • °C/W. The two 10-m X 4-m walls have no windows.
The third wall has five windows made of 0.5-cm-thick glass
(k = 0.78 W/m • °C), 1.2 m X 1.8 m in size. The fourth wall
has the same size and number of windows, but they are double-
paned with a 1.5-cm-thick stagnant air space (k = 0.026
W/m • °C) enclosed between two 0.5-cm-thick glass layers.
The thermostat in the house is set at 22°C and the average tem-
perature outside at that location is 8°C during the seven-month-
long heating season. Disregarding any direct radiation gain or
loss through the windows and taking the heat transfer coeffi-
cients at the inner and outer surfaces of the house to be 7 and
15 W/m 2 ■ °C, respectively, determine the average rate of heat
transfer through each wall.
If the house is electrically heated and the price of electricity
is $0.08/kWh, determine the amount of money this household
will save per heating season by converting the single-pane win-
dows to double -pane windows.
3-35 The wall of a refrigerator is constructed of fiberglass in-
sulation (k = 0.035 W/m • °C) sandwiched between two layers
of 1-mm-thick sheet metal (k = 15.1 W/m • °C). The refriger-
ated space is maintained at 3°C, and the average heat transfer
coefficients at the inner and outer surfaces of the wall are
Kitchen
air
25°C
10°C
1 mm]
Insulation
Refrigerated
space
3°C
1 mm
FIGURE P3-35
4 W/m 2 • °C and 9 W/m 2 • °C, respectively. The kitchen tem-
perature averages 25°C. It is observed that condensation occurs
on the outer surfaces of the refrigerator when the temperature
of the outer surface drops to 20°C. Determine the minimum
thickness of fiberglass insulation that needs to be used in the
wall in order to avoid condensation on the outer surfaces.
3-36 rSi'M Reconsider Problem 3-35. Using EES (or other)
1^2 software, investigate the effects of the thermal
conductivities of the insulation material and the sheet metal on
the thickness of the insulation. Let the thermal conductivity
vary from 0.02 W/m • °C to 0.08 W/m • °C for insulation and
10 W/m • °C to 400 W/m • °C for sheet metal. Plot the thick-
ness of the insulation as the functions of the thermal con-
ductivities of the insulation and the sheet metal, and discuss
the results.
3-37 Heat is to be conducted along a circuit board that has a
copper layer on one side. The circuit board is 15 cm long and
15 cm wide, and the thicknesses of the copper and epoxy lay-
ers are 0.1 mm and 1.2 mm, respectively. Disregarding heat
transfer from side surfaces, determine the percentages of heat
conduction along the copper (k = 386 W/m • °C) and epoxy
(k = 0.26 W/m ■ °C) layers. Also determine the effective ther-
mal conductivity of the board.
Answers: 0.8 percent, 99.2 percent, and 29.9 W/m • °C
3-38E A 0.03-in-thick copper plate (k = 223 Btu/h • ft ■ °F)
is sandwiched between two 0.1 -in. -thick epoxy boards (k =
0.15 Btu/h • ft • °F) that are 7 in. X 9 in. in size. Determine the
effective thermal conductivity of the board along its 9-in.-long
side. What fraction of the heat conducted along that side is con-
ducted through copper?
Thermal Contact Resistance
3-39C What is thermal contact resistance? How is it related
to thermal contact conductance?
3-40C Will the thermal contact resistance be greater for
smooth or rough plain surfaces?
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CHAPTER 3
Epoxy
boards
Copper
plate
Plexiglas
9 in.
0.03 in.
IGURI 38E
3-41C A wall consists of two layers of insulation pressed
against each other. Do we need to be concerned about the ther-
mal contact resistance at the interface in a heat transfer analy-
sis or can we just ignore it?
3-42C A plate consists of two thin metal layers pressed
against each other. Do we need to be concerned about the
thermal contact resistance at the interface in a heat transfer
analysis or can we just ignore it?
3-43C Consider two surfaces pressed against each other.
Now the air at the interface is evacuated. Will the thermal con-
tact resistance at the interface increase or decrease as a result?
3-44C Explain how the thermal contact resistance can be
minimized.
3—45 The thermal contact conductance at the interface of two
1-cm-thick copper plates is measured to be 18,000 W/m 2 • °C.
Determine the thickness of the copper plate whose thermal
resistance is equal to the thermal resistance of the interface
between the plates.
3-46 Six identical power transistors with aluminum casing
are attached on one side of a 1 .2-cm-thick 20-cm X 30-cm
copper plate (k = 386 W/m • °C) by screws that exert an aver-
age pressure of 10 MPa. The base area of each transistor is
9 cm 2 , and each transistor is placed at the center of a 10-cm X
10-cm section of the plate. The interface roughness is esti-
mated to be about 1 .4 jjim. All transistors are covered by a thick
Plexiglas layer, which is a poor conductor of heat, and thus all
the heat generated at the junction of the transistor must be dis-
sipated to the ambient at 15°C through the back surface of the
copper plate. The combined convection/radiation heat transfer
coefficient at the back surface can be taken to be 30 W/m 2 ■ °C.
If the case temperature of the transistor is not to exceed 85°C,
determine the maximum power each transistor can dissipate
safely, and the temperature jump at the case-plate interface.
Copper
Transistor plate
15°C
1.2 cm
FIGURE P3^6
3-47 Two 5-cm-diameter, 15-cm-long aluminum bars (k =
176 W/m ■ °C) with ground surfaces are pressed against each
other with a pressure of 20 atm. The bars are enclosed in an in-
sulation sleeve and, thus, heat transfer from the lateral surfaces
is negligible. If the top and bottom surfaces of the two-bar sys-
tem are maintained at temperatures of 150°C and 20°C, re-
spectively, determine (a) the rate of heat transfer along the
cylinders under steady conditions and (b) the temperature drop
at the interface. Answers: (a) 142.4 W, (b) 6.4°C
3-48 A 1 -mm-thick copper plate (k = 386 W/m ■ °C) is sand-
wiched between two 5-mm-thick epoxy boards (k = 0.26
W/m • °C) that are 15 cm X 20 cm in size. If the thermal con-
tact conductance on both sides of the copper plate is estimated
to be 6000 W/m • °C, determine the error involved in the total
thermal resistance of the plate if the thermal contact conduc-
tances are ignored.
Copper
Epoxy
FIGURE P3-48
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HEAT TRANSFER
Generalized Thermal Resistance Networks
3-49C When plotting the thermal resistance network associ-
ated with a heat transfer problem, explain when two resistances
are in series and when they are in parallel.
3-50C The thermal resistance networks can also be used
approximately for multidimensional problems. For what kind
of multidimensional problems will the thermal resistance
approach give adequate results?
3-51 C What are the two approaches used in the develop-
ment of the thermal resistance network for two-dimensional
problems?
3-52 A 4-m-high and 6-m-wide wall consists of a long
18-cm X 30-cm cross section of horizontal bricks (k = 0.72
W/m ■ °C) separated by 3-cm-thick plaster layers (k = 0.22
W/m • °C). There are also 2-cm-thick plaster layers on each
side of the wall, and a 2-cm-thick rigid foam (k =
0.026 W/m • °C) on the inner side of the wall. The indoor and
the outdoor temperatures are 22°C and — 4°C, and the convec-
tion heat transfer coefficients on the inner and the outer sides
are /z, = 10 W/m 2 • °C and h 2 = 20 W/m 2 • °C, respectively.
Assuming one-dimensional heat transfer and disregarding radi-
ation, determine the rate of heat transfer through the wall.
Foam
Brick
2 2
FIGURE P3-52
18 cm
1.5 cm
30 cm
1.5 cm
3-53
Reconsider Problem 3-52. Using EES (or other)
software, plot the rate of heat transfer through
the wall as a function of the thickness of the rigid foam in the
range of 1 cm to 10 cm. Discuss the results.
3-54 A 10-cm-thick wall is to be constructed with 2.5-m-
long wood studs (k = 0.11 W/m • °C) that have a cross section
of 10 cm X 10 cm. At some point the builder ran out of those
studs and started using pairs of 2.5-m-long wood studs that
have a cross section of 5 cm X 10 cm nailed to each other
instead. The manganese steel nails (k = 50 W/m • °C) are
10 cm long and have a diameter of 0.4 cm. A total of 50 nails
are used to connect the two studs, which are mounted to the
wall such that the nails cross the wall. The temperature differ-
ence between the inner and outer surfaces of the wall is 8°C.
Assuming the thermal contact resistance between the two
layers to be negligible, determine the rate of heat transfer
(a) through a solid stud and (b) through a stud pair of equal
length and width nailed to each other, (c) Also determine the
effective conductivity of the nailed stud pair.
3-55 A 12-m-long and 5-m-high wall is constructed of two
layers of 1 -cm-thick sheetrock (k = 0.17 W/m • °C) spaced
12 cm by wood studs (k = 0.11 W/m • °C) whose cross section
is 12 cm X 5 cm. The studs are placed vertically 60 cm apart,
and the space between them is filled with fiberglass insulation
(k = 0.034 W/m • °C). The house is maintained at 20°C and the
ambient temperature outside is — 5°C. Taking the heat transfer
coefficients at the inner and outer surfaces of the house to be
8.3 and 34 W/m 2 • °C, respectively, determine (a) the thermal
resistance of the wall considering a representative section of it
and (b) the rate of heat transfer through the wall.
3-56E A 10-in. -thick, 30-ft-long, and 10-ft-high wall is
to be constructed using 9-in.-long solid bricks (k = 0.40
Btu/h ■ ft • °F) of cross section 7 in. X 7 in., or identical size
bricks with nine square air holes (k = 0.015 Btu/h • ft ■ °F) that
are 9 in. long and have a cross section of 1.5 in. X 1.5 in. There
is a 0.5-in. -thick plaster layer (k = 0.10 Btu/h ■ ft ■ °F) between
two adjacent bricks on all four sides and on both sides of the
wall. The house is maintained at 80°F and the ambient tem-
perature outside is 30°F. Taking the heat transfer coefficients
at the inner and outer surfaces of the wall to be 1.5 and
4 Btu/h • ft 2 • °F, respectively, determine the rate of heat
transfer through the wall constructed of (a) solid bricks and
(b) bricks with air holes.
Air channels
1.5 in. X 1.5 in. x9 in.
Brick
0.5 in.
FIGURE P3-56E
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3-57 Consider a 5-m-high, 8-m-long, and 0.22-m-thick wall
whose representative cross section is as given in the figure. The
thermal conductivities of various materials used, in W/m • °C,
are k.
2, k B = 8, k c = 20, k D = 15, and k E = 35. The
left and right surfaces of the wall are maintained at uniform
temperatures of 300°C and 100°C, respectively. Assuming heat
transfer through the wall to be one-dimensional, determine
(a) the rate of heat transfer through the wall; (b) the tem-
perature at the point where the sections B, D, and E meet; and
(c) the temperature drop across the section F. Disregard any
contact resistances at the interfaces.
100°C
300°C
1 cm
FIGURE P3-57
3-58 Repeat Problem 3-57 assuming that the thermal contact
resistance at the interfaces D-F and E-F is 0.00012 m 2 • °C/W.
3-59 Clothing made of several thin layers of fabric with
trapped air in between, often called ski clothing, is commonly
used in cold climates because it is light, fashionable, and a very
effective thermal insulator. So it is no surprise that such cloth-
ing has largely replaced thick and heavy old-fashioned coats.
Consider a jacket made of five layers of 0.1-mm-thick syn-
thetic fabric {k = 0.13 W/m • °C) with 1.5-mm-thick air space
(k = 0.026 W/m • °C) between the layers. Assuming the inner
surface temperature of the jacket to be 28°C and the surface
area to be 1.1 m 2 , determine the rate of heat loss through the
jacket when the temperature of the outdoors is — 5°C and the
heat transfer coefficient at the outer surface is 25 W/m 2 • °C.
Multilayered
ski jacket
193
CHAPTER 3
What would your response be if the jacket is made of a sin-
gle layer of 0.5-mm-thick synthetic fabric? What should be the
thickness of a wool fabric (k = 0.035 W/m • °C) if the person
is to achieve the same level of thermal comfort wearing a thick
wool coat instead of a five-layer ski jacket?
3-60 Repeat Problem 3-59 assuming the layers of the jacket
are made of cotton fabric (k = 0.06 W/m • °C).
3-61 A 5-m-wide, 4-m-high, and 40-m-long kiln used to cure
concrete pipes is made of 20-cm-thick concrete walls and ceil-
ing (k = 0.9 W/m • °C). The kiln is maintained at 40°C by in-
jecting hot steam into it. The two ends of the kiln, 4 m X 5 m
in size, are made of a 3-mm-thick sheet metal covered with
2-cm-thick Styrofoam (k = 0.033 W/m • °C). The convection
heat transfer coefficients on the inner and the outer surfaces of
the kiln are 3000 W/m 2 • °C and 25 W/m 2 • °C, respectively.
Disregarding any heat loss through the floor, determine the rate
of heat loss from the kiln when the ambient air is at — 4°C.
-4°C
FIGURE P3-59
4 m
5 in
FIGURE P3-61
3-62 TtTM Reconsider Problem 3-61. Using EES (or other)
1^2 software, investigate the effects of the thickness
of the wall and the convection heat transfer coefficient on the
outer surface of the rate of heat loss from the kiln. Let the
thickness vary from 10 cm to 30 cm and the convection heat
transfer coefficient from 5 W/m 2 • °C to 50 W/m 2 ■ °C. Plot the
rate of heat transfer as functions of wall thickness and the con-
vection heat transfer coefficient, and discuss the results.
3-63E Consider a 6-in. X 8-in. epoxy glass laminate (k =
0.10 Btu/h ■ ft • °F) whose thickness is 0.05 in. In order to re-
duce the thermal resistance across its thickness, cylindrical
copper fillings (k = 223 Btu/h ■ ft • °F) of 0.02 in. diameter are
to be planted throughout the board, with a center-to-center
distance of 0.06 in. Determine the new value of the thermal
resistance of the epoxy board for heat conduction across its
thickness as a result of this modification.
Answer: 0.00064 h ■ °F/Btu
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HEAT TRANSFER
0.02 in.
0.06 in.-
Copper filling Epoxy board
FIGURE P3-63E
Heat Conduction in Cylinders and Spheres
3-64C What is an infinitely long cylinder? When is it proper
to treat an actual cylinder as being infinitely long, and when
is it not?
3-65C Consider a short cylinder whose top and bottom sur-
faces are insulated. The cylinder is initially at a uniform tem-
perature Tj and is subjected to convection from its side surface
to a medium at temperature T^, with a heat transfer coefficient
of /?. Is the heat transfer in this short cylinder one- or two-
dimensional? Explain.
3-66C Can the thermal resistance concept be used for a solid
cylinder or sphere in steady operation? Explain.
3-67 A 5-m-internal-diameter spherical tank made of
1.5-cm -thick stainless steel (k = 15 W/m • °C) is used to store
iced water at 0°C. The tank is located in a room whose temper-
ature is 30°C. The walls of the room are also at 30°C. The outer
surface of the tank is black (emissivity e = 1), and heat trans-
fer between the outer surface of the tank and the surroundings
is by natural convection and radiation. The convection heat
O Iced water VJ
■o?.£W Q
O
1.5 cm
vQ-OIQ'PIO
transfer coefficients at the inner and the outer surfaces of the
tank are 80 W/m 2 ■ °C and 10 W/m 2 ■ °C, respectively. Deter-
mine (a) the rate of heat transfer to the iced water in the tank
and (b) the amount of ice at 0°C that melts during a 24-h
period. The heat of fusion of water at atmospheric pressure is
h if = 333.7 kJ/kg.
3-68 Steam at 320°C flows in a stainless steel pipe (k =
1 5 W/m • °C) whose inner and outer diameters are 5 cm and
5.5 cm, respectively. The pipe is covered with 3-cm-thick glass
wool insulation (k = 0.038 W/m • °C). Heat is lost to the sur-
roundings at 5°C by natural convection and radiation, with
a combined natural convection and radiation heat transfer co-
efficient of 15 W/m 2 • °C. Taking the heat transfer coefficient
inside the pipe to be 80 W/m 2 • °C, determine the rate of heat
loss from the steam per unit length of the pipe. Also determine
the temperature drops across the pipe shell and the insulation.
3-69 fitt'M Reconsider Problem 3-68. Using EES (or other)
1^2 software, investigate the effect of the thickness of
the insulation on the rate of heat loss from the steam and the
temperature drop across the insulation layer. Let the insulation
thickness vary from 1 cm to 10 cm. Plot the rate of heat loss
and the temperature drop as a function of insulation thickness,
and discuss the results.
3-70 (Jb\ A 50-m-long section of a steam pipe whose outer
^<UP diameter is 10 cm passes through an open space
at 15°C. The average temperature of the outer surface of the
pipe is measured to be 150°C. If the combined heat transfer co-
efficient on the outer surface of the pipe is 20 W/m 2 • °C, de-
termine (a) the rate of heat loss from the steam pipe, (b) the
annual cost of this energy lost if steam is generated in a natural
gas furnace that has an efficiency of 75 percent and the price of
natural gas is $0.52/therm (1 therm = 105,500 kJ), and (c) the
thickness of fiberglass insulation (k = 0.035 W/m • °C) needed
in order to save 90 percent of the heat lost. Assume the pipe
temperature to remain constant at 150°C.
150°C
\ Q
J^C ^
Fiberglass
insulation
FIGURE P3-70
FIGURE P3-67
3-71 Consider a 2-m-high electric hot water heater that has a
diameter of 40 cm and maintains the hot water at 55°C. The
tank is located in a small room whose average temperature is
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CHAPTER 3
3 cm
27°C
Foam
insulation
40 cm
T = 55°C
2 m
FIGURE P3-71
27°C, and the heat transfer coefficients on the inner and outer
surfaces of the heater are 50 and 12 W/m 2 ■ °C, respectively.
The tank is placed in another 46-cm-diameter sheet metal tank
of negligible thickness, and the space between the two tanks is
filled with foam insulation (k = 0.03 W/m ■ °C). The thermal
resistances of the water tank and the outer thin sheet metal
shell are very small and can be neglected. The price of elec-
tricity is $0.08/kWh, and the home owner pays $280 a year for
water heating. Determine the fraction of the hot water energy
cost of this household that is due to the heat loss from the tank.
Hot water tank insulation kits consisting of 3-cm-thick fiber-
glass insulation (k = 0.035 W/m ■ °C) large enough to wrap the
entire tank are available in the market for about $30. If such an
insulation is installed on this water tank by the home owner
himself, how long will it take for this additional insulation to
pay for itself? Answers-. 17.5 percent, 1.5 years
3-72 rfigM Reconsider Problem 3-71. Using EES (or other)
b^2 software, plot the fraction of energy cost of hot
water due to the heat loss from the tank as a function of the
hot water temperature in the range of 40°C to 90°C. Discuss
the results.
3-73 Consider a cold aluminum canned drink that is initially
at a uniform temperature of 3°C. The can is 12.5 cm high and
has a diameter of 6 cm. If the combined convection/radiation
heat transfer coefficient between the can and the surrounding
air at 25 °C is 10 W/m 2 • °C, determine how long it will take for
the average temperature of the drink to rise to 10°C.
In an effort to slow down the warming of the cold drink, a
person puts the can in a perfectly fitting 1 -cm-thick cylindrical
rubber insulation (k = 0.13 W/m • °C). Now how long will it
take for the average temperature of the drink to rise to 10°C?
Assume the top of the can is not covered.
3-C
:25°C
12.5 cm
FIGURE P3-73
3-74 Repeat Problem 3-
resistance of 0.00008 m 2
insulation.
6 cm
73, assuming a thermal contact
°C/W between the can and the
3-75E Steam at 450°F is flowing through a steel pipe (k = 8.7
Btu/h • ft ■ °F) whose inner and outer diameters are 3.5 in. and
4.0 in., respectively, in an environment at 55°F. The pipe is
insulated with 2-in. -thick fiberglass insulation (k = 0.020
Btu/h • ft • °F). If the heat transfer coefficients on the inside and
the outside of the pipe are 30 and 5 Btu/h • ft 2 • °F, respectively,
determine the rate of heat loss from the steam per foot length of
the pipe. What is the error involved in neglecting the thermal
resistance of the steel pipe in calculations?
Steel pipe
Insulation
FIGURE P3-75E
3-76 Hot water at an average temperature of 90°C is flowing
through a 15-m section of a cast iron pipe (k = 52 W/m • °C)
whose inner and outer diameters are 4 cm and 4.6 cm, respec-
tively. The outer surface of the pipe, whose emissivity is 0.7, is
exposed to the cold air at 10°C in the basement, with a heat
transfer coefficient of 15 W/m 2 • °C. The heat transfer coeffi-
cient at the inner surface of the pipe is 120 W/m 2 ■ °C. Taking
the walls of the basement to be at 10°C also, determine the rate
of heat loss from the hot water. Also, determine the average
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HEAT TRANSFER
velocity of the water in the pipe if the temperature of the water
drops by 3°C as it passes through the basement.
3-77 Repeat Problem 3-76 for a pipe made of copper (k =
386 W/m ■ °C) instead of cast iron.
3-78E Steam exiting the turbine of a steam power plant at
100°F is to be condensed in a large condenser by cooling water
flowing through copper pipes (k = 223 Btu/h • ft • °F) of inner
diameter 0.4 in. and outer diameter 0.6 in. at an average
temperature of 70°F. The heat of vaporization of water at
100°F is 1037 Btu/lbm. The heat transfer coefficients are 1500
Btu/h • ft 2 • °F on the steam side and 35 Btu/h • ft 2 • °F on the
water side. Determine the length of the tube required to con-
dense steam at a rate of 120 lbm/h. Answer: 1 148 ft
Steam, 100°F
120 lbm/h
Liquid water
FIGURE P3-78E
3-79E Repeat Problem 3-78E, assuming that a 0.01 -in. -thick
layer of mineral deposit (k = 0.5 Btu/h • ft ■ °F) has formed on
the inner surface of the pipe.
3-80
Reconsider Problem 3-78E. Using EES (or
other) software, investigate the effects of the
thermal conductivity of the pipe material and the outer di-
ameter of the pipe on the length of the tube required. Let
the thermal conductivity vary from 10 Btu/h • ft • °F to 400
Btu/h • ft • °F and the outer diameter from 0.5 in. to 1.0 in. Plot
the length of the tube as functions of pipe conductivity and the
outer pipe diameter, and discuss the results.
3-81 The boiling temperature of nitrogen at atmospheric
pressure at sea level (1 atm pressure) is — 196°C. Therefore, ni-
trogen is commonly used in low -temperature scientific studies
since the temperature of liquid nitrogen in a tank open to the at-
mosphere will remain constant at — 196°C until it is depleted.
Any heat transfer to the tank will result in the evaporation of
some liquid nitrogen, which has a heat of vaporization of 198
kJ/kg and a density of 810 kg/m 3 at 1 atm.
N, vapor
Insulation
FIGURE P3-81
Consider a 3-m-diameter spherical tank that is initially filled
with liquid nitrogen at 1 atm and — 196°C. The tank is exposed
to ambient air at 15°C, with a combined convection and radia-
tion heat transfer coefficient of 35 W/m 2 ■ °C. The temperature
of the thin-shelled spherical tank is observed to be almost the
same as the temperature of the nitrogen inside. Determine
the rate of evaporation of the liquid nitrogen in the tank as a
result of the heat transfer from the ambient air if the tank is
(a) not insulated, (b) insulated with 5-cm-thick fiberglass insu-
lation (k = 0.035 W/m ■ °C), and (c) insulated with 2-cm-thick
superinsulation which has an effective thermal conductivity of
0.00005 W/m • °C.
3-82 Repeat Problem 3-81 for liquid oxygen, which has
a boiling temperature of — 183°C, a heat of vaporization of
213 kJ/kg, and a density of 1 140 kg/m 3 at 1 atm pressure.
Critical Radius of Insulation
3-83C What is the critical radius of insulation? How is it
defined for a cylindrical layer?
3-84C A pipe is insulated such that the outer radius of the
insulation is less than the critical radius. Now the insulation is
taken off. Will the rate of heat transfer from the pipe increase
or decrease for the same pipe surface temperature?
3-85C A pipe is insulated to reduce the heat loss from it.
However, measurements indicate that the rate of heat loss
has increased instead of decreasing. Can the measurements
be right?
3-86C Consider a pipe at a constant temperature whose ra-
dius is greater than the critical radius of insulation. Someone
claims that the rate of heat loss from the pipe has increased
when some insulation is added to the pipe. Is this claim valid?
3-87C Consider an insulated pipe exposed to the atmo-
sphere. Will the critical radius of insulation be greater on calm
days or on windy days? Why?
cen58933_ch03.qxd 9/10/2002 9:00 AM Page 197
3-88 A 2-mm-diameter and 1 0-m-long electric wire is tightly
wrapped with a 1-mm-thick plastic cover whose thermal con-
ductivity is k = 0.15 W/m ■ °C. Electrical measurements indi-
cate that a current of 10 A passes through the wire and there is
a voltage drop of 8 V along the wire. If the insulated wire is ex-
posed to a medium at T„ = 30°C with a heat transfer coeffi-
cient of h = 24 W/m 2 ■ °C, determine the temperature at the
interface of the wire and the plastic cover in steady operation.
Also determine if doubling the thickness of the plastic cover
will increase or decrease this interface temperature.
Electrical
wire
T. = 30°C
- Insulation
2>
■10m
FIGURE P3-88
3-89E A 0.083-in. -diameter electrical wire at 115°F is
covered by 0.02-in. -thick plastic insulation (k = 0.075
Btu/h • ft • °F). The wire is exposed to a medium at 50°F, with
a combined convection and radiation heat transfer coefficient
of 2.5 Btu/h • ft 2 • °F. Determine if the plastic insulation on the
wire will increase or decrease heat transfer from the wire.
Answer: It helps
3-90E Repeat Problem 3-89E, assuming a thermal contact
resistance of 0.001 h • ft 2 • °F/Btu at the interface of the wire
and the insulation.
3-91 A 5-mm-diameter spherical ball at 50°C is covered by a
1-mm-thick plastic insulation (k = 0.13 W/m • °C). The ball is
exposed to a medium at 15°C, with a combined convection and
radiation heat transfer coefficient of 20 W/m 2 ■ °C. Determine
if the plastic insulation on the ball will help or hurt heat trans-
fer from the ball.
FIGURE P3-91
3-92
Reconsider Problem 3-91. Using EES (or other)
software, plot the rate of heat transfer from the
ball as a function of the plastic insulation thickness in the range
of 0.5 mm to 20 mm. Discuss the results.
Heat Transfer from Finned Surfaces
3-93C What is the reason for the widespread use of fins on
surfaces?
197
CHAPTER 3
3-94C What is the difference between the fin effectiveness
and the fin efficiency?
3-95C The fins attached to a surface are determined to have
an effectiveness of 0.9. Do you think the rate of heat transfer
from the surface has increased or decreased as a result of the
addition of these fins?
3-96C Explain how the fins enhance heat transfer from a
surface. Also, explain how the addition of fins may actually
decrease heat transfer from a surface.
3-97C How does the overall effectiveness of a finned sur-
face differ from the effectiveness of a single fin?
3-98C Hot water is to be cooled as it flows through the tubes
exposed to atmospheric air. Fins are to be attached in order to
enhance heat transfer. Would you recommend attaching the
fins inside or outside the tubes? Why?
3-99C Hot air is to be cooled as it is forced to flow through
the tubes exposed to atmospheric air. Fins are to be added in
order to enhance heat transfer. Would you recommend attach-
ing the fins inside or outside the tubes? Why? When would you
recommend attaching fins both inside and outside the tubes?
3-100C Consider two finned surfaces that are identical
except that the fins on the first surface are formed by casting
or extrusion, whereas they are attached to the second surface
afterwards by welding or tight fitting. For which case do you
think the fins will provide greater enhancement in heat trans-
fer? Explain.
3-101C The heat transfer surface area of a fin is equal to the
sum of all surfaces of the fin exposed to the surrounding
medium, including the surface area of the fin tip. Under what
conditions can we neglect heat transfer from the fin tip?
3-102C Does the (a) efficiency and (b) effectiveness of a fin
increase or decrease as the fin length is increased?
3-103C Two pin fins are identical, except that the diameter
of one of them is twice the diameter of the other. For which fin
will the (a) fin effectiveness and (b) fin efficiency be higher?
Explain.
3-104C Two plate fins of constant rectangular cross section
are identical, except that the thickness of one of them is twice
the thickness of the other. For which fin will the (a) fin effec-
tiveness and (b) fin efficiency be higher? Explain.
3-105C Two finned surfaces are identical, except that the
convection heat transfer coefficient of one of them is twice that
of the other. For which finned surface will the (a) fin effective-
ness and (b) fin efficiency be higher? Explain.
3-106 Obtain a relation for the fin efficiency for a fin of con-
stant cross-sectional area A c , perimeter p, length L, and thermal
conductivity k exposed to convection to a medium at r„ with a
heat transfer coefficient h. Assume the fins are sufficiently long
so that the temperature of the fin at the tip is nearly T x . Take
the temperature of the fin at the base to be T b and neglect heat
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HEAT TRANSFER
h, T m
n{ k
I"
^ b =K
p = kD, A c =
= TdfilA
FIGURE P3-1 06
transfer from the fin tips. Simplify the relation for (a) a circu-
lar fin of diameter D and (b) rectangular fins of thickness t.
3-107 The case-to-ambient thermal resistance of a power
transistor that has a maximum power rating of 1 5 W is given to
be 25°C/W. If the case temperature of the transistor is not to
exceed 80°C, determine the power at which this transistor can
be operated safely in an environment at 40°C.
3-108 A 40-W power transistor is to be cooled by attaching
it to one of the commercially available heat sinks shown in
Table 3-A. Select a heat sink that will allow the case tempera-
ture of the transistor not to exceed 90° in the ambient air at 20°.
= 20°C
90°C
FIGURE P3-1 08
3-109 A 30-W power transistor is to be cooled by attaching
it to one of the commercially available heat sinks shown in
Table 3-A. Select a heat sink that will allow the case tempera-
ture of the transistor not to exceed 80°C in the ambient air at
35°C.
3-110 Steam in a heating system flows through tubes whose
outer diameter is 5 cm and whose walls are maintained at a
temperature of 180°C. Circular aluminum alloy 2024-T6 fins
(k = 1 86 W/m ■ °C) of outer diameter 6 cm and constant thick-
ness 1 mm are attached to the tube. The space between the fins
is 3 mm, and thus there are 250 fins per meter length of the
tube. Heat is transferred to the surrounding air at T rj -_ = 25°C,
with a heat transfer coefficient of 40 W/m 2 • °C. Determine the
increase in heat transfer from the tube per meter of its length as
a result of adding fins. Answer: 2639 W
3-111E Consider a stainless steel spoon (k = 8.7
Btu/h ■ ft • °F) partially immersed in boiling water at 200°F in
a kitchen at 75°F. The handle of the spoon has a cross section
of 0.08 in. X 0.5 in., and extends 7 in. in the air from the free
2.5 cm
= 25°C *
180°C
FIGURE P3-1 10
surface of the water. If the heat transfer coefficient at the ex-
posed surfaces of the spoon handle is 3 Btu/h ■ ft 2 • °F, deter-
mine the temperature difference across the exposed surface of
the spoon handle. State your assumptions. Answer-. 124. 6°F
Spoon
:75°F
Boiling
water
200°F
FIGURE P3-111E
VR*.
3-112E Repeat Problem 3-111 for a silver spoon (k = 247
Btu/h • ft • °F).
3-113E fu'M Reconsider Problem 3-1 HE. Using EES (or
Ei3 other) software, investigate the effects of the
thermal conductivity of the spoon material and the length of its
extension in the air on the temperature difference across the
exposed surface of the spoon handle. Let the thermal conduc-
tivity vary from 5 Btu/h ■ ft • °F to 225 Btu/h • ft ■ °F and the
length from 5 in. to 12 in. Plot the temperature difference as the
functions of thermal conductivity and length, and discuss
the results.
3-114 A 0.3-cm-thick, 12-cm-high, and 18-cm-long circuit
board houses 80 closely spaced logic chips on one side, each
dissipating 0.04 W. The board is impregnated with copper fill-
ings and has an effective thermal conductivity of 20 W/m • °C.
All the heat generated in the chips is conducted across the cir-
cuit board and is dissipated from the back side of the board
to a medium at 40°C, with a heat transfer coefficient of 50
W/m 2 • °C. (a) Determine the temperatures on the two sides
of the circuit board, (b) Now a 0.2-cm-thick, 12-cm-high, and
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CHAPTER 3
18-cm-long aluminum plate (k = 237 W/m • °C) with 864
2-cm-long aluminum pin fins of diameter 0.25 cm is attached
to the back side of the circuit board with a 0.02 -cm-thick epoxy
adhesive (k = 1.8 W/m • °C). Determine the new temperatures
on the two sides of the circuit board.
3-115 Repeat Problem 3-1 14 using a copper plate with cop-
per fins (k = 386 W/m • °C) instead of aluminum ones.
3-116 A hot surface at 100°C is to be cooled by attach-
ing 3-cm-long, 0.25-cm-diameter aluminum pin fins (k =
237 W/m • °C) to it, with a center-to-center distance of 0.6 cm.
The temperature of the surrounding medium is 30°C, and the
heat transfer coefficient on the surfaces is 35 W/m 2 • °C.
Determine the rate of heat transfer from the surface for a
1-m X 1-m section of the plate. Also determine the overall
effectiveness of the fins.
FIGURE P3-1 16
386
3-117 Repeat Problem 3-116 using copper fins (k
W/m • °C) instead of aluminum ones.
3-118 fitt'M Reconsider Problem 3-116. Using EES (or
KS other) software, investigate the effect of the cen-
ter-to-center distance of the fins on the rate of heat transfer
from the surface and the overall effectiveness of the fins. Let
the center-to-center distance vary from 0.4 cm to 2.0 cm. Plot
the rate of heat transfer and the overall effectiveness as a func-
tion of the center-to-center distance, and discuss the results.
3-119 Two 3-m-long and 0.4-cm-thick cast iron (k = 52
W/m • °C) steam pipes of outer diameter 10 cm are connected
to each other through two 1 -cm-thick flanges of outer diameter
20 cm. The steam flows inside the pipe at an average tempera-
ture of 200°C with a heat transfer coefficient of 180 W/m 2 • °C.
The outer surface of the pipe is exposed to an ambient at 12°C,
with a heat transfer coefficient of 25 W/m 2 • °C. (a) Disregard-
ing the flanges, determine the average outer surface tempera-
ture of the pipe, (b) Using this temperature for the base of the
flange and treating the flanges as the fins, determine the fin ef-
ficiency and the rate of heat transfer from the flanges, (c) What
length of pipe is the flange section equivalent to for heat trans-
fer purposes?
FIGURE P3-1 19
Heat Transfer in Common Configurations
3-120C What is a conduction shape factor? How is it related
to the thermal resistance?
3-121C What is the value of conduction shape factors in
engineering?
3-122 A 20-m-long and 8-cm-diameter hot water pipe of a
district heating system is buried in the soil 80 cm below the
ground surface. The outer surface temperature of the pipe is
60°C. Taking the surface temperature of the earth to be 5°C
and the thermal conductivity of the soil at that location to be
0.9 W/m • °C, determine the rate of heat loss from the pipe.
JL
5°C
80 cm
-60°C
m
,20 a-
FIGURE P3-1 22
3-123
Reconsider Problem 3-122. Using EES (or
other) software, plot the rate of heat loss from
the pipe as a function of the burial depth in the range of 20 cm
to 2.0 m. Discuss the results.
3-124 Hot and cold water pipes 8 m long run parallel to each
other in a thick concrete layer. The diameters of both pipes are
5 cm, and the distance between the centerlines of the pipes is
40 cm. The surface temperatures of the hot and cold pipes are
60°C and 15°C, respectively. Taking the thermal conductivity
of the concrete to be k = 0.75 W/m • °C, determine the rate of
heat transfer between the pipes. Answer: 306 W
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HEAT TRANSFER
3-125 [?(,■>! Reconsider Problem 3-124. Using EES (or
I^S other) software, plot the rate of heat transfer
between the pipes as a function of the distance between the
centerlines of the pipes in the range of 10 cm to 1 .0 m. Discuss
the results.
3-126E A row of 3-ft-long and 1 -in. -diameter used uranium
fuel rods that are still radioactive are buried in the ground par-
allel to each other with a center-to-center distance of 8 in. at a
depth 15 ft from the ground surface at a location where the
thermal conductivity of the soil is 0.6 Btu/h • ft • °F. If the sur-
face temperature of the rods and the ground are 350°F and
60°F, respectively, determine the rate of heat transfer from the
fuel rods to the atmosphere through the soil.
60°F
. •'; 1^8 in. ; »[»'8 in.r"p-8 in.^j
FIGURE P3-1 26
3-127 Hot water at an average temperature of 60°C and an
average velocity of 0.6 m/s is flowing through a 5-m section
of a thin-walled hot water pipe that has an outer diameter of
2.5 cm. The pipe passes through the center of a 14-cm-fhick
wall filled with fiberglass insulation (k = 0.035 W/m ■ °C). If
the surfaces of the wall are at 1 8°C, determine (a) the rate of
heat transfer from the pipe to the air in the rooms and (b) the
temperature drop of the hot water as it flows through this
5-m-long section of the wall. Answers: 23.5 W, 0.02°C
Wall
Hot water pipe
8°C
X
0°C
3 m
20 m: ■ f
FIGURE P3-1 28
(k = 1.5 W/m ■ °C) vertically for 3 m, and continues horizon-
tally at this depth for 20 m more before it enters the next build-
ing. The first section of the pipe is exposed to the ambient air
at 8°C, with a heat transfer coefficient of 22 W/m 2 • °C. If the
surface of the ground is covered with snow at 0°C, determine
(a) the total rate of heat loss from the hot water and (b) the
temperature drop of the hot water as it flows through this
25-m-long section of the pipe.
3-129 Consider a house with a flat roof whose outer dimen-
sions are 12 m X 12 m. The outer walls of the house are 6 m
high. The walls and the roof of the house are made of 20-cm-
thick concrete (k = 0.75 W/m • °C). The temperatures of the in-
ner and outer surfaces of the house are 15°C and 3°C,
respectively. Accounting for the effects of the edges of adjoin-
ing surfaces, determine the rate of heat loss from the house
through its walls and the roof. What is the error involved in ig-
noring the effects of the edges and corners and treating the roof
asal2mX 12m surface and the walls as 6 m X 12 m surfaces
for simplicity?
3-130 Consider a 10-m-long thick-walled concrete duct (k =
0.75 W/m • °C) of square cross-section. The outer dimensions
of the duct are 20 cm X 20 cm, and the thickness of the duct
wall is 2 cm. If the inner and outer surfaces of the duct are at
100°C and 15°C, respectively, determine the rate of heat trans-
fer through the walls of the duct. Answer: 22.9 kW
FIGURE P3-1 27
16 cm-
20 cm
FIGURE P3-1 30
3-128 Hot water at an average temperature of 80°C and an
average velocity of 1.5 m/s is flowing through a 25-m section
of a pipe that has an outer diameter of 5 cm. The pipe extends
2 m in the ambient air above the ground, dips into the ground
3-131 A 3-m-diameter spherical tank containing some radio-
active material is buried in the ground (k = 1 .4 W/m • °C). The
distance between the top surface of the tank and the ground
surface is 4 m. If the surface temperatures of the tank and the
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CHAPTER 3
ground are 140°C and 15°C, respectively, determine the rate of
heat transfer from the tank.
3-132 fitt'M Reconsider Problem 3-131. Using EES (or
k^^ other) software, plot the rate of heat transfer
from the tank as a function of the tank diameter in the range of
0.5 m to 5.0 m. Discuss the results.
3-133 Hot water at an average temperature of 85°C passes
through a row of eight parallel pipes that are 4 m long and have
an outer diameter of 3 cm, located vertically in the middle of a
concrete wall (k = 0.75 W/m ■ °C) that is 4 m high, 8 m long,
and 15 cm thick. If the surfaces of the concrete walls are
exposed to a medium at 32°C, with a heat transfer coefficient
of 12 W/m 2 • °C, determine the rate of heat loss from the hot
water and the surface temperature of the wall.
FIGURE P3-1 39
Special Topics:
Heat Transfer through the Walls and Roofs
3-134C What is the TJ-value of a wall? How does it differ
from the unit thermal resistance of the wall? How is it related
to the [/-factor?
3-135C What is effective emissivity for a plane-parallel air
space? How is it determined? How is radiation heat transfer
through the air space determined when the effective emissivity
is known?
3-136C The unit thermal resistances (/{-values) of both
40-mm and 90-mm vertical air spaces are given in Table 3-9 to
be 0.22 m 2 • °C/W, which implies that more than doubling the
thickness of air space in a wall has no effect on heat transfer
through the wall. Do you think this is a typing error? Explain.
3-137C What is a radiant barrier? What kind of materials are
suitable for use as radiant barriers? Is it worthwhile to use ra-
diant barriers in the attics of homes?
3-138C Consider a house whose attic space is ventilated ef-
fectively so that the air temperature in the attic is the same as
the ambient air temperature at all times. Will the roof still have
any effect on heat transfer through the ceiling? Explain.
3-139 Determine the summer R- value and the [/-factor of a
wood frame wall that is built around 38-mm X 140-mm wood
studs with a center-to-center distance of 400 mm. The 140-
mm-wide cavity between the studs is filled with mineral fiber
batt insulation. The inside is finished with 13-mm gypsum
wallboard and the outside with 13-mm wood fiberboard and
13-mm X 200-mm wood bevel lapped siding. The insulated
cavity constitutes 80 percent of the heat transmission area,
while the studs, headers, plates, and sills constitute 20 percent.
Answers: 3.213 m 2 • °C/W, 0.311 W/m 2 ■ °C
3-140 The 13-mm-thick wood fiberboard sheathing of the
wood stud wall in Problem 3-139 is replaced by a 25-mm-
thick rigid foam insulation. Determine the percent increase in
the R- value of the wall as a result.
3-141E Determine the winter R- value and the [/-factor of a
masonry cavity wall that is built around 4-in. -thick concrete
blocks made of lightweight aggregate. The outside is finished
with 4-in. face brick with ^-in. cement mortar between the
bricks and concrete blocks. The inside finish consists of ^-in.
gypsum wallboard separated from the concrete block by | -in.-
thick (1-in. by 3-in. nominal) vertical furring whose center-to-
center distance is 16 in. Neither side of the | -in. -thick air space
between the concrete block and the gypsum board is coated
with any reflective film. When determining the R- value of the
air space, the temperature difference across it can be taken to
be 30°F with a mean air temperature of 50°F. The air space
constitutes 80 percent of the heat transmission area, while the
vertical furring and similar structures constitute 20 percent.
FIGURE P3-1 41 E
3-142 Consider a flat ceiling that is built around 38-mm X
90-mm wood studs with a center-to-center distance of 400 mm.
The lower part of the ceiling is finished with 13-mm gypsum
wallboard, while the upper part consists of a wood subfloor
(R = 0.166 m 2 • °C/W), a 13-mm plywood, a layer of felt (R =
0.011 m 2 ■ °C/W), and linoleum (R = 0.009 m 2 ■ °C/W). Both
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HEAT TRANSFER
12 3 4 5
FIGURE P3-1 42
sides of the ceiling are exposed to still air. The air space con-
stitutes 82 percent of the heat transmission area, while the
studs and headers constitute 1 8 percent. Determine the winter
R- value and the {/-factor of the ceiling assuming the 90-mm-
wide air space between the studs (a) does not have any reflec-
tive surface, (b) has a reflective surface with s = 0.05 on one
side, and (c) has reflective surfaces with e = 0.05 on both
sides. Assume a mean temperature of 10°C and a temperature
difference of 5.6°C for the air space.
3-143 Determine the winter lvalue and the {/-factor of a
masonry cavity wall that consists of 100-mm common bricks,
a 90-mm air space, 100-mm concrete blocks made of light-
weight aggregate, 20-mm air space, and 13-mm gypsum wall-
board separated from the concrete block by 20-mm-thick
(1-in. X 3 -in. nominal) vertical furring whose center-to-center
distance is 400 mm. Neither side of the two air spaces is coated
with any reflective films. When determining the ^-value of the
air spaces, the temperature difference across them can be taken
to be 16.7°C with a mean air temperature of 10°C. The air
space constitutes 84 percent of the heat transmission area,
while the vertical furring and similar structures constitute
16 percent. Answers: 1.02 m 2 • °C/W, 0.978 W/m 2 • °C
3-144 Repeat Problem 3-143 assuming one side of both air
spaces is coated with a reflective film of e = 0.05.
3-145 Determine the winter R-vahie and the [/-factor of a
masonry wall that consists of the following layers: 100-mm
face bricks, 100-mm common bricks, 25-mm urethane rigid
foam insulation, and 13-mm gypsum wallboard.
Answers: 1.404 m 2 • °C/W, 0.712 W/m 2 • °C
3-146 The overall heat transfer coefficient (the [/-value) of a
wall under winter design conditions is U = 1.55 W/m 2 ■ °C.
Determine the [/-value of the wall under summer design
conditions.
3-147 The overall heat transfer coefficient (the [/-value) of a
wall under winter design conditions is U = 2.25 W/m 2 ■ °C.
Now a layer of 100-mm face brick is added to the outside,
leaving a 20-mm air space between the wall and the bricks. De-
termine the new [/-value of the wall. Also, determine the rate
of heat transfer through a 3-m-high, 7-m-long section of the
wall after modification when the indoor and outdoor tempera-
tures are 22°C and — 5°C, respectively.
FIGURE P3-1 47
FIGURE P3-1 43
3-148 Determine the summer and winter ^-values, in
m 2 • °C/W, of a masonry wall that consists of 100-mm face
bricks, 13-mm of cement mortar, 100-mm lightweight concrete
block, 40-mm air space, and 20-mm plasterboard.
Answers: 0.809 and 0.795 m 2 ■ °C/W
3-149E The overall heat transfer coefficient of a wall is
determined to be U = 0.09 Btu/h • ft 2 ■ °F under the conditions
of still air inside and winds of 7.5 mph outside. What will the
[/-factor be when the wind velocity outside is doubled?
Answer: 0.0907 Btu/h ■ ft 2 • °F
3-150 Two homes are identical, except that the walls of one
house consist of 200-mm lightweight concrete blocks, 20-mm
air space, and 20-mm plasterboard, while the walls of the other
house involve the standard R-2.4 m 2 • °C/W frame wall con-
struction. Which house do you think is more energy efficient?
cen58933_ch03.qxd 9/10/2002 9:00 AM Page 203
3-151 Determine the /J-value of a ceiling that consists of a
layer of 19-mm acoustical tiles whose top surface is covered
with a highly reflective aluminum foil for winter conditions.
Assume still air below and above the tiles.
Highly
reflective
foil
FIGURE P3-1 51
Review Problems
3-152E Steam is produced in the copper tubes (k = 223
Btu/h • ft • °F) of a heat exchanger at a temperature of 250°F by
another fluid condensing on the outside surfaces of the tubes at
350°F. The inner and outer diameters of the tube are 1 in. and
1.3 in., respectively. When the heat exchanger was new, the
rate of heat transfer per foot length of the tube was 2 X 10 4
Btu/h. Determine the rate of heat transfer per foot length
of the tube when a 0.01 -in. -thick layer of limestone (k =
1 .7 Btu/h ■ ft ■ °F) has formed on the inner surface of the tube
after extended use.
3-153E Repeat Problem 3-152E, assuming that a 0.01-in.-
thick limestone layer has formed on both the inner and outer
surfaces of the tube.
3-154 A 1.2-m-diameter and 6-m-long cylindrical propane
tank is initially filled with liquid propane whose density is 581
kg/m 3 . The tank is exposed to the ambient air at 30°C, with a
heat transfer coefficient of 25 W/m 2 • °C. Now a crack devel-
ops at the top of the tank and the pressure inside drops to 1 atm
while the temperature drops to — 42°C, which is the boiling
temperature of propane at 1 atm. The heat of vaporization of
Propane
T
f
ir =30°C
vapor
"\
PROPANE TANK
1.2 m
T = -42°C
P = 1 atm /
4
6 m ►
203
CHAPTER 3
propane at 1 atm is 425 kJ/kg. The propane is slowly vaporized
as a result of the heat transfer from the ambient air into the
tank, and the propane vapor escapes the tank at — 42°C through
the crack. Assuming the propane tank to be at about the same
temperature as the propane inside at all times, determine how
long it will take for the propane tank to empty if the tank is
(a) not insulated and (b) insulated with 7.5-cm-thick glass wool
insulation (k = 0.038 W/m • °C).
3-155 Hot water is flowing at an average velocity of 1 .5 m/s
through a cast iron pipe (k = 52 W/m • °C) whose inner and
outer diameters are 3 cm and 3.5 cm, respectively. The pipe
passes through a 15-m-long section of a basement whose
temperature is 15°C. If the temperature of the water drops
from 70°C to 67°C as it passes through the basement and the
heat transfer coefficient on the inner surface of the pipe is 400
W/m 2 ■ °C, determine the combined convection and radiation
heat transfer coefficient at the outer surface of the pipe.
Answer: 272.5 W/m 2 • °C
3-156 Newly formed concrete pipes are usually cured first
overnight by steam in a curing kiln maintained at a temperature
of 45°C before the pipes are cured for several days outside. The
heat and moisture to the kiln is provided by steam flowing in a
pipe whose outer diameter is 12 cm. During a plant inspection,
it was noticed that the pipe passes through a 10-m section that
is completely exposed to the ambient air before it reaches the
kiln. The temperature measurements indicate that the average
temperature of the outer surface of the steam pipe is 82°C
when the ambient temperature is 8°C. The combined convec-
tion and radiation heat transfer coefficient at the outer surface
of the pipe is estimated to be 25 W/m 2 • °C. Determine the
amount of heat lost from the steam during a 10-h curing
process that night.
Steam is supplied by a gas-fired steam generator that has
an efficiency of 80 percent, and the plant pays $0.60/therm of
natural gas (1 therm = 105,500 kJ). If the pipe is insulated and
90 percent of the heat loss is saved as a result, determine the
amount of money this facility will save a year as a result of
insulating the steam pipes. Assume that the concrete pipes are
cured 110 nights a year. State your assumptions.
8 "C
Furnace
FIGURE P3-1 54
FIGURE P3-1 56
3-157 Consider an 18-cm X 18-cm multilayer circuit board
dissipating 27 W of heat. The board consists of four layers
of 0.2-mm-thick copper (k = 386 W/m • °C) and three layers of
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HEAT TRANSFER
Copper
FIGURE P3-1 57
1.5-mm-thick epoxy glass (k = 0.26 W/m ■ °C) sandwiched
together, as shown in the figure. The circuit board is attached to
a heat sink from both ends, and the temperature of the board at
those ends is 35°C. Heat is considered to be uniformly gener-
ated in the epoxy layers of the board at a rate of 0.5 W per 1 -cm
X 18-cm epoxy laminate strip (or 1.5 W per 1-cm X 18-cm
strip of the board). Considering only a portion of the board be-
cause of symmetry, determine the magnitude and location of
the maximum temperature that occurs in the board. Assume
heat transfer from the top and bottom faces of the board to be
negligible.
3-158 The plumbing system of a house involves a 0.5-m sec-
tion of a plastic pipe {k = 0.16 W/m • °C) of inner diameter
2 cm and outer diameter 2.4 cm exposed to the ambient air.
During a cold and windy night, the ambient air temperature re-
mains at about — 5°C for a period of 14 h. The combined con-
vection and radiation heat transfer coefficient on the outer
surface of the pipe is estimated to be 40 W/m 2 • °C, and the
heat of fusion of water is 333.7 kJ/kg. Assuming the pipe to
contain stationary water initially at 0°C, determine if the water
in that section of the pipe will completely freeze that night.
Exposed
- water pipe
AIR
SOIL-
FIGURE P3-1 58
3-159 Repeat Problem 3-158 for the case of a heat transfer
coefficient of 10 W/m 2 • °C on the outer surface as a result of
putting a fence around the pipe that blocks the wind.
3-1 60E The surface temperature of a 3 -in. -diameter baked
potato is observed to drop from 300°F to 200°F in 5 minutes in
an environment at 70°F. Determine the average heat transfer
coefficient between the potato and its surroundings. Using this
heat transfer coefficient and the same surface temperature,
determine how long it will take for the potato to experience
the same temperature drop if it is wrapped completely in a
0.12-in.-thick towel (k = 0.035 Btu/h • ft ■ °F). You may use the
properties of water for potato.
3-161E Repeat Problem 3-160E assuming there is a 0.02-
in. -thick air space (k = 0.015 Btu/h • ft • °F) between the potato
and the towel.
3-162 An ice chest whose outer dimensions are 30 cm X
40 cm X 50 cm is made of 3-cm-thick Styrofoam (k = 0.033
W/m • °C). Initially, the chest is filled with 45 kg of ice at 0°C,
and the inner surface temperature of the ice chest can be taken
to be 0°C at all times. The heat of fusion of ice at 0°C is 333.7
kJ/kg, and the heat transfer coefficient between the outer
surface of the ice chest and surrounding air at 35°C is 18
W/m 2 ■ °C. Disregarding any heat transfer from the 40-cm X
50-cm base of the ice chest, determine how long it will take for
the ice in the chest to melt completely.
:35°C
^ °
a Ice chest °
~ _ 0°C °
Q o
3 cm
Styrofoam
FIGURE P3-1 62
3-163 A 4-m-high and 6-m-long wall is constructed of two
large 2-cm-thick steel plates (k = 15 W/m • °C) separated by
1-cm -thick and 20-cm-wide steel bars placed 99 cm apart. The
Steel plates
2 cm
20 cm
Fiberglass
- insulation
99 cm
1 cm
2 cm
FIGURE P3-1 63
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CHAPTER 3
remaining space between the steel plates is filled with fiber-
glass insulation (k = 0.035 W/m ■ °C). If the temperature dif-
ference between the inner and the outer surfaces of the walls is
22°C, determine the rate of heat transfer through the wall. Can
we ignore the steel bars between the plates in heat transfer
analysis since they occupy only 1 percent of the heat transfer
surface area?
3-164 A 0.2-cm-thick, 10-cm-high, and 15-cm-long circuit
board houses electronic components on one side that dissipate
a total of 15 W of heat uniformly. The board is impregnated
with conducting metal fillings and has an effective thermal
conductivity of 12 W/m • °C. All the heat generated in the com-
ponents is conducted across the circuit board and is dissipated
from the back side of the board to a medium at 37°C, with a
heat transfer coefficient of 45 W/m 2 • °C. (a) Determine the
surface temperatures on the two sides of the circuit board.
(b) Now a 0.1 -cm-thick, 10-cm-high, and 15-cm-long alumi-
num plate (k = 237 W/m • °C) with 20 0.2-cm-thick, 2-cm-
long, and 15-cm-wide aluminum fins of rectangular profile are
attached to the back side of the circuit board with a 0.03-cm-
thick epoxy adhesive (k = 1.8 W/m • °C). Determine the new
temperatures on the two sides of the circuit board.
Electronic
components
10 cm //
0.3 cm
0.2 cm
^2 mm
FIGURE P3-1 64
3-165 Repeat Problem 3-164 using a copper plate with cop-
per fins (k = 386 W/m • °C) instead of aluminum ones.
3-166 A row of 10 parallel pipes that are 5 m long and have
an outer diameter of 6 cm are used to transport steam at 150°C
through the concrete floor (k = 0.75 W/m • °C) of a 10-m X
5-m room that is maintained at 25°C. The combined con-
vection and radiation heat transfer coefficient at the floor is
12 W/m 2 • °C. If the surface temperature of the concrete floor
is not to exceed 40°C, determine how deep the steam pipes
should be buried below the surface of the concrete floor.
Room
25°C
,40°C
0OOO
oooooo
^ Steam pipes
*D = 6cm
Concrete floor
FIGURE P3-1 66
3-167 Consider two identical people each generating 60 W
of metabolic heat steadily while doing sedentary work, and dis-
sipating it by convection and perspiration. The first person
is wearing clothes made of 1-mm-thick leather (k = 0.159
W/m ■ °C) that covers half of the body while the second one is
wearing clothes made of 1-mm-thick synthetic fabric (k = 0.13
W/m ■ °C) that covers the body completely. The ambient air is
at 30°C, the heat transfer coefficient at the outer surface is
1 5 W/m 2 ■ °C, and the inner surface temperature of the clothes
can be taken to be 32°C. Treating the body of each person as a
25-cm-diameter 1.7-m-long cylinder, determine the fractions
of heat lost from each person by perspiration.
3-168 A 6-m-wide 2.8-m-high wall is constructed of one
layer of common brick (k = 0.72 W/m • °C) of thickness
20 cm, one inside layer of light-weight plaster (k = 0.36
W/m • °C) of thickness 1 cm, and one outside layer of cement
based covering (k = 1 .40 W/m • °C) of thickness 2 cm. The in-
ner surface of the wall is maintained at 23 °C while the outer
surface is exposed to outdoors at 8°C with a combined convec-
tion and radiation heat transfer coefficient of 17 W/m 2 ■ °C.
Determine the rate of heat transfer through the wall and tem-
perature drops across the plaster, brick, covering, and surface-
ambient air.
3-169 Reconsider Problem 3-1 68. It is desired to insulate the
wall in order to decrease the heat loss by 85 percent. For the
same inner surface temperature, determine the thickness of in-
sulation and the outer surface temperature if the wall is insu-
lated with (a) polyurethane foam {k = 0.025 W/m • °C) and
(b) glass fiber (k = 0.036 W/m • °C).
3-170 Cold conditioned air at 12°C is flowing inside a
1.5-cm-thick square aluminum (k = 237 W/m ■ °C) duct of
inner cross section 22 cm X 22 cm at a mass flow rate of
0.8 kg/s. The duct is exposed to air at 33°C with a combined
convection-radiation heat transfer coefficient of 8 W/m 2 • °C.
The convection heat transfer coefficient at the inner surface is
75 W/m 2 • °C. If the air temperature in the duct should not
increase by more than 1 °C determine the maximum length of
the duct.
3-171 When analyzing heat transfer through windows, it
is important to consider the frame as well as the glass area.
Consider a 2-m-wide 1.5-m-high wood-framed window with
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HEAT TRANSFER
85 percent of the area covered by 3-mm-thick single-pane glass
(k = 0.7 W/m ■ °C). The frame is 5 cm thick, and is made of
pine wood (k = 0.12 W/m • °C). The heat transfer coefficient is
7 W/m 2 • °C inside and 13 W/m 2 • °C outside. The room is
maintained at 24°C, and the temperature outdoors is 40°C. De-
termine the percent error involved in heat transfer when the
window is assumed to consist of glass only.
3-172 Steam at 235°C is flowing inside a steel pipe (k =
61 W/m • °C) whose inner and outer diameters are 10 cm and
12 cm, respectively, in an environment at 20°C. The heat trans-
fer coefficients inside and outside the pipe are 1 05 W/m 2 • °C
and 14 W/m 2 • °C, respectively. Determine (a) the thickness of
the insulation (k = 0.038 W/m ■ °C) needed to reduce the heat
loss by 95 percent and (b) the thickness of the insulation
needed to reduce the exposed surface temperature of insulated
pipe to 40°C for safety reasons.
3-173 When the transportation of natural gas in a pipeline is
not feasible for economic or other reasons, it is first liquefied at
about — 160°C, and then transported in specially insulated
tanks placed in marine ships. Consider a 6-m-diameter spheri-
cal tank that is filled with liquefied natural gas (LNG) at
— 160°C. The tank is exposed to ambient air at 18°C with a
heat transfer coefficient of 22 W/m 2 • °C. The tank is thin-
shelled and its temperature can be taken to be the same as the
LNG temperature. The tank is insulated with 5-cm-thick super
insulation that has an effective thermal conductivity of 0.00008
W/m • °C. Taking the density and the specific heat of LNG to
be 425 kg/m 3 and 3.475 kJ/kg • °C, respectively, estimate how
long it will take for the LNG temperature to rise to — 150°C.
3-174 A 15-cm X 20-cm hot surface at 85°C is to be cooled
by attaching 4-cm-long aluminum (k = 237 W/m ■ °C) fins of
2-mm X 2-mm square cross section. The temperature of sur-
rounding medium is 25°C and the heat transfer coefficient on
the surfaces can be taken to be 20 W/m 2 • °C. If it is desired to
triple the rate of heat transfer from the bare hot surface, deter-
mine the number of fins that needs to be attached.
3-175 [?(,■>! Reconsider Problem 3-174. Using EES (or
1^2 other) software, plot the number of fins as a
function of the increase in the heat loss by fins relative to no
fin case (i.e., overall effectiveness of the fins) in the range of
1.5 to 5. Discuss the results. Is it realistic to assume the heat
transfer coefficient to remain constant?
3-176 A 1 .4-m-diameter spherical steel tank filled with iced
water at 0°C is buried underground at a location where the
thermal conductivity of the soil is k = 0.55 W/m • °C. The dis-
tance between the tank center and the ground surface is 2.4 m.
For ground surface temperature of 18°C, determine the rate of
heat transfer to the iced water in the tank. What would your
answer be if the soil temperature were 1 8°C and the ground
surface were insulated?
3-177 A 0. 6-m-diameter 1.9-m-long cylindrical tank con-
taining liquefied natural gas (LNG) at — 160°C is placed at the
center of a 1 .9-m-long 1 .4-m X 1 .4-m square solid bar made of
an insulating material with k = 0.0006 W/m • °C. If the outer
surface temperature of the bar is 20°C, determine the rate of
heat transfer to the tank. Also, determine the LNG temperature
after one month. Take the density and the specific heat of LNG
to be 425 kg/m 3 and 3.475 kJ/kg ■ °C, respectively.
Design and Essay Problems
3-178 The temperature in deep space is close to absolute
zero, which presents thermal challenges for the astronauts who
do space walks. Propose a design for the clothing of the astro-
nauts that will be most suitable for the thermal environment in
space. Defend the selections in your design.
3-179 In the design of electronic components, it is very de-
sirable to attach the electronic circuitry to a substrate material
that is a very good thermal conductor but also a very effective
electrical insulator. If the high cost is not a major concern, what
material would you propose for the substrate?
3-180 Using cylindrical samples of the same material, devise
an experiment to determine the thermal contact resistance.
Cylindrical samples are available at any length, and the thermal
conductivity of the material is known.
3-181 Find out about the wall construction of the cabins of
large commercial airplanes, the range of ambient conditions
under which they operate, typical heat transfer coefficients on
the inner and outer surfaces of the wall, and the heat generation
rates inside. Determine the size of the heating and air-
conditioning system that will be able to maintain the cabin
at 20°C at all times for an airplane capable of carrying
400 people.
3-182 Repeat Problem 3-181 for a submarine with a crew of
60 people.
3-183 A house with 200-m 2 floor space is to be heated with
geothermal water flowing through pipes laid in the ground
under the floor. The walls of the house are 4 m high, and there
are 10 single-paned windows in the house that are 1.2 m wide
and 1.8 m high. The house has R-X9 (in h • ft 2 • °F/Btu) insula-
tion in the walls and R-30 on the ceiling. The floor temperature
is not to exceed 40°C. Hot geothermal water is available at
90°C, and the inner and outer diameter of the pipes to be used
are 2.4 cm and 3.0 cm. Design such a heating system for this
house in your area.
3-184 Using a timer (or watch) and a thermometer, conduct
this experiment to determine the rate of heat gain of your
refrigerator. First, make sure that the door of the refrigerator
is not opened for at least a few hours to make sure that steady
operating conditions are established. Start the timer when the
refrigerator stops running and measure the time At t it stays off
cen58933_ch03.qxd 9/10/2002 9:00 AM Page 207
before it kicks in. Then measure the time At 2 it stays on. Not-
ing that the heat removed during At 2 is equal to the heat gain of
the refrigerator during At { + At 2 and using the power con-
sumed by the refrigerator when it is running, determine the av-
erage rate of heat gain for your refrigerator, in watts. Take the
COP (coefficient of performance) of your refrigerator to be 1 .3
if it is not available.
207
CHAPTER 3
Now, clean the condenser coils of the refrigerator and re-
move any obstacles on the way of airflow through the coils By
replacing these measurements, determine the improvement in
the COP of the refrigerator.
cen58933_ch03 . qxd 9/10/2002 9:00 AM Page 208
cen58933_ch04.qxd 9/10/2002 9:12 AM Page 209
TRANSIENT HEAT
CONDUCTION
CHAPTER
The temperature of a body, in general, varies with time as well
as position. In rectangular coordinates, this variation is expressed as
T(x, y, z, t), where (x, y, z) indicates variation in the x, y, and z directions,
respectively, and t indicates variation with time. In the preceding chapter, we
considered heat conduction under steady conditions, for which the tempera-
ture of a body at any point does not change with time. This certainly simpli-
fied the analysis, especially when the temperature varied in one direction only,
and we were able to obtain analytical solutions. In this chapter, we consider
the variation of temperature with time as well as position in one- and multi-
dimensional systems.
We start this chapter with the analysis of lumped systems in which the tem-
perature of a solid varies with time but remains uniform throughout the solid
at any time. Then we consider the variation of temperature with time as well
as position for one-dimensional heat conduction problems such as those asso-
ciated with a large plane wall, a long cylinder, a sphere, and a semi-infinite
medium using transient temperature charts and analytical solutions. Finally,
we consider transient heat conduction in multidimensional systems by uti-
lizing the product solution.
CONTENTS
4-1 Lumped Systems Analysis 210
4-2 Transient Heat Conduction
in Large Plane Walls, Long
Cylinders, and Spheres
with Spatial Effects 216
4-3 Transient Heat Conduction
in Semi-Infinite Solids 228
4-4 Transient Heat Conduction in
Multidimensional Systems 231
Topic of Special Interest:
Refrigeration and
Freezing of Foods 239
cen58933_ch04.qxd 9/10/2002 9:12 AM Page 21C
210
HEAT TRANSFER
(a) Copper ball
(b) Roast beef
FIGURE 4-1
A small copper ball can be modeled
as a lumped system, but a roast
beef cannot.
SOLID BODY
m = mass
V = volume
: density
= initial temperature
T = T(t)
Q = hA s [T„-T(t)]
FIGURE 4-2
The geometry and parameters
involved in the lumped
system analysis.
4-1 ■ LUMPED SYSTEM ANALYSIS
In heat transfer analysis, some bodies are observed to behave like a "lump"
whose interior temperature remains essentially uniform at all times during a
heat transfer process. The temperature of such bodies can be taken to be a
function of time only, T(t). Heat transfer analysis that utilizes this idealization
is known as lumped system analysis, which provides great simplification
in certain classes of heat transfer problems without much sacrifice from
accuracy.
Consider a small hot copper ball coming out of an oven (Fig. 4-1). Mea-
surements indicate that the temperature of the copper ball changes with time,
but it does not change much with position at any given time. Thus the tem-
perature of the ball remains uniform at all times, and we can talk about the
temperature of the ball with no reference to a specific location.
Now let us go to the other extreme and consider a large roast in an oven. If
you have done any roasting, you must have noticed that the temperature dis-
tribution within the roast is not even close to being uniform. You can easily
verify this by taking the roast out before it is completely done and cutting it in
half. You will see that the outer parts of the roast are well done while the cen-
ter part is barely warm. Thus, lumped system analysis is not applicable in this
case. Before presenting a criterion about applicability of lumped system
analysis, we develop the formulation associated with it.
Consider a body of arbitrary shape of mass m, volume V, surface area A s ,
density p, and specific heat C p initially at a uniform temperature T { (Fig. 4-2).
At time t = 0, the body is placed into a medium at temperature r„, and heat
transfer takes place between the body and its environment, with a heat trans-
fer coefficient h. For the sake of discussion, we will assume that T m > T t , but
the analysis is equally valid for the opposite case. We assume lumped system
analysis to be applicable, so that the temperature remains uniform within the
body at all times and changes with time only, T = T(t).
During a differential time interval dt, the temperature of the body rises by a
differential amount dT. An energy balance of the solid for the time interval dt
can be expressed as
/Heat transfer into the body\
during dt
( The increase in the |
energy of the body
\ during dt j
or
hA s {T a -T)dt = mC„ dT
(4-1)
Noting that m = pV and dT = d(T — r„) since T„ = constant, Eq. 4-1 can be
rearranged as
d(T- TJ)
T-T a
hA s
pvc„
dt
(4-2)
Integrating from t = 0, at which T = T t , to any time t, at which T = T(t), gives
In
Tit) - T„
T; ~ fZ
hA s
pvc p '
(4-3)
cen58933_ch04.qxd 9/10/2002 9:12 AM Page 211
Taking the exponential of both sides and rearranging, we obtain
T(t) - T m _„,
where
pvc.
(1/s)
(4-4)
(4-5)
is a positive quantity whose dimension is (time) -1 . The reciprocal of b has
time unit (usually s), and is called the time constant. Equation 4-4 is plotted
in Fig. 4-3 for different values of b. There are two observations that can be
made from this figure and the relation above:
1. Equation 4-4 enables us to determine the temperature T(t) of a body at
time t, or alternatively, the time t required for the temperature to reach
a specified value T(t).
2. The temperature of a body approaches the ambient temperature T^
exponentially. The temperature of the body changes rapidly at the
beginning, but rather slowly later on. A large value of b indicates that
the body will approach the environment temperature in a short time.
The larger the value of the exponent b, the higher the rate of decay in
temperature. Note that b is proportional to the surface area, but inversely
proportional to the mass and the specific heat of the body. This is not
surprising since it takes longer to heat or cool a larger mass, especially
when it has a large specific heat.
Once the temperature T(t) at time t is available from Eq. A-A, the rate of con-
vection heat transfer between the body and its environment at that time can be
determined from Newton's law of cooling as
Q(t) = hA s [T(t) - rj
(W)
(4-6)
The total amount of heat transfer between the body and the surrounding
medium over the time interval t = to t is simply the change in the energy
content of the body:
Q = mC.[T(!) ~ n
(kJ)
(4-7)
The amount of heat transfer reaches its upper limit when the body reaches the
surrounding temperature T„. Therefore, the maximum heat transfer between
the body and its surroundings is (Fig. 4-4)
e„
mC p {T a - r,)
(kJ)
(4-8)
We could also obtain this equation by substituting the T(t) relation from Eq.
4-4 into the Q(t) relation in Eq. 4-6 and integrating it from t = to t — > °°.
Criteria for Lumped System Analysis
The lumped system analysis certainly provides great convenience in heat
transfer analysis, and naturally we would like to know when it is appropriate
211
CHAPTER 4
Tin
FIGURE 4-3
The temperature of a lumped
system approaches the environment
temperature as time gets larger.
1 =
h
t
— > 00
T t
T,
T a
T„
T>
r«,
T T rJ -_
r,
<
T a
r»
Q--
tsraax —
">C p (T r
-TJ
FIGURE 4-4
Heat transfer to
or from
a body
reaches its maximum value
when the body reaches
the environment temperature.
cen58933_ch04.qxd 9/10/2002 9:12 AM Page 212
212
HEAT TRANSFER
Convection
» Conduction .
SOLID
BODY
to use it. The first step in establishing a criterion for the applicability of the
lumped system analysis is to define a characteristic length as
and a Biot number Bi as
Bi
hL c
(4-9)
Bi
heat convection
heat conduction
FIGURE 4-5
The Biot number can be viewed as the
ratio of the convection at the surface
to conduction within the body.
It can also be expressed as (Fig. 4-5)
Bi
h AT Convection at the surface of the body
~klL,.~AT = ~
or
Bi
LJk
Conduction within the body
Conduction resistance within the body
\lh Convection resistance at the surface of the body
When a solid body is being heated by the hotter fluid surrounding it (such as
a potato being baked in an oven), heat is first convected to the body and
subsequently conducted within the body. The Biot number is the ratio of the
internal resistance of a body to heat conduction to its external resistance to
heat convection. Therefore, a small Biot number represents small resistance
to heat conduction, and thus small temperature gradients within the body.
Lumped system analysis assumes a uniform temperature distribution
throughout the body, which will be the case only when the thermal resistance
of the body to heat conduction (the conduction resistance) is zero. Thus,
lumped system analysis is exact when Bi = and approximate when Bi > 0.
Of course, the smaller the Bi number, the more accurate the lumped system
analysis. Then the question we must answer is, How much accuracy are we
willing to sacrifice for the convenience of the lumped system analysis?
Before answering this question, we should mention that a 20 percent
uncertainty in the convection heat transfer coefficient h in most cases is con-
sidered "normal" and "expected." Assuming h to be constant and uniform is
also an approximation of questionable validity, especially for irregular geome-
tries. Therefore, in the absence of sufficient experimental data for the specific
geometry under consideration, we cannot claim our results to be better than
±20 percent, even when Bi = 0. This being the case, introducing another
source of uncertainty in the problem will hardly have any effect on the over-
all uncertainty, provided that it is minor. It is generally accepted that lumped
system analysis is applicable if
Bi<0.1
When this criterion is satisfied, the temperatures within the body relative to
the surroundings (i.e., T — T m ) remain within 5 percent of each other even for
well-rounded geometries such as a spherical ball. Thus, when Bi < 0.1, the
variation of temperature with location within the body will be slight and can
reasonably be approximated as being uniform.
cen58933_ch04.qxd 9/10/2002 9:12 AM Page 213
213
CHAPTER 4
The first step in the application of lumped system analysis is the calculation
of the Biot number, and the assessment of the applicability of this approach.
One may still wish to use lumped system analysis even when the criterion
Bi < 0.1 is not satisfied, if high accuracy is not a major concern.
Note that the Biot number is the ratio of the convection at the surface to con-
duction within the body, and this number should be as small as possible for
lumped system analysis to be applicable. Therefore, small bodies with high
thermal conductivity are good candidates for lumped system analysis, es-
pecially when they are in a medium that is a poor conductor of heat (such as
air or another gas) and motionless. Thus, the hot small copper ball placed in
quiescent air, discussed earlier, is most likely to satisfy the criterion for
lumped system analysis (Fig. 4-6).
Some Remarks on Heat Transfer in Lumped Systems
To understand the heat transfer mechanism during the heating or cooling of a
solid by the fluid surrounding it, and the criterion for lumped system analysis,
consider this analogy (Fig. 4-7). People from the mainland are to go by boat
to an island whose entire shore is a harbor, and from the harbor to their desti-
nations on the island by bus. The overcrowding of people at the harbor de-
pends on the boat traffic to the island and the ground transportation system on
the island. If there is an excellent ground transportation system with plenty of
buses, there will be no overcrowding at the harbor, especially when the boat
traffic is light. But when the opposite is true, there will be a huge overcrowd-
ing at the harbor, creating a large difference between the populations at the
harbor and inland. The chance of overcrowding is much lower in a small is-
land with plenty of fast buses.
In heat transfer, a poor ground transportation system corresponds to poor
heat conduction in a body, and overcrowding at the harbor to the accumulation
of heat and the subsequent rise in temperature near the surface of the body
relative to its inner parts. Lumped system analysis is obviously not applicable
when there is overcrowding at the surface. Of course, we have disregarded
radiation in this analogy and thus the air traffic to the island. Like passengers
at the harbor, heat changes vehicles at the surface from convection to conduc-
tion. Noting that a surface has zero thickness and thus cannot store any energy,
heat reaching the surface of a body by convection must continue its journey
within the body by conduction.
Consider heat transfer from a hot body to its cooler surroundings. Heat will
be transferred from the body to the surrounding fluid as a result of a tempera-
ture difference. But this energy will come from the region near the surface,
and thus the temperature of the body near the surface will drop. This creates a
temperature gradient between the inner and outer regions of the body and ini-
tiates heat flow by conduction from the interior of the body toward the outer
surface.
When the convection heat transfer coefficient h and thus convection heat
transfer from the body are high, the temperature of the body near the surface
will drop quickly (Fig. 4-8). This will create a larger temperature difference
between the inner and outer regions unless the body is able to transfer heat
from the inner to the outer regions just as fast. Thus, the magnitude of the
maximum temperature difference within the body depends strongly on the
ability of a body to conduct heat toward its surface relative to the ability of
15W/m 2 -°C
V k n ° 3 ,
— -^ 4-£> = 0.02 m
1
f A s kD 2 6
Bi = jfc = j5xap2 = o,ooo 75 < p.]
* 401
FIGURE 4-6
Small bodies with high thermal
conductivities and low convection
coefficients are most likely
to satisfy the criterion for
lumped system analysis.
FIGURE 4-7
Analogy between heat transfer to a
solid and passenger traffic
to an island.
Convection
/? = 2000W/m 2 -°C
FIGURE 4-8
When the convection coefficient h is
high and k is low, large temperature
differences occur between the inner
and outer regions of a large solid.
cen58933_ch04.qxd 9/10/2002 9:12 AM Page 214
214
HEAT TRANSFER
the surrounding medium to convect this heat away from the surface. The
Biot number is a measure of the relative magnitudes of these two competing
effects.
Recall that heat conduction in a specified direction n per unit surface area is
expressed as q = —k dT/dn, where dT/dn is the temperature gradient and k is
the thermal conductivity of the solid. Thus, the temperature distribution in the
body will be uniform only when its thermal conductivity is infinite, and no
such material is known to exist. Therefore, temperature gradients and thus
temperature differences must exist within the body, no matter how small, in
order for heat conduction to take place. Of course, the temperature gradient
and the thermal conductivity are inversely proportional for a given heat flux.
Therefore, the larger the thermal conductivity, the smaller the temperature
gradient.
Thermocouple
Gas
T ,h *" ^> Junction
~D = 1 mm
T(t)
FIGURE 4-9
Schematic for Example 4-1.
EXAMPLE 4-1 Temperature Measurement by Thermocouples
The temperature of a gas stream is to be measured by a thermocouple whose
junction can be approximated as a 1-mm-diameter sphere, as shown in Fig.
4-9. The properties of the junction are k = 35 W/m • °C, p = 8500 kg/m 3 , and
C p = 320 J/kg • C C, and the convection heat transfer coefficient between the
junction and the gas is h = 210 W/m 2 • °C. Determine how long it will take for
the thermocouple to read 99 percent of the initial temperature difference.
SOLUTION The temperature of a gas stream is to be measured by a thermo-
couple. The time it takes to register 99 percent of the initial A T is to be
determined.
Assumptions 1 The junction is spherical in shape with a diameter of D =
0.001 m. 2 The thermal properties of the junction and the heat transfer coeffi-
cient are constant. 3 Radiation effects are negligible.
Properties The properties of the junction are given in the problem statement.
Analysis The characteristic length of the junction is
V_
A,
ttD 2
I
D
(0.001 m) = 1.67 X 10-
Then the Biot number becomes
hL c (210 W/m 2 ■ °C)(1.67 X 10- 4 m)
Bi = X = 35W/m.°C = 0.001 <0.1
Therefore, lumped system analysis is applicable, and the error involved in this
approximation is negligible.
In order to read 99 percent of the initial temperature difference 7" - T, M
between the junction and the gas, we must have
T(t) - 7V„
0.01
For example, when 7" = C C and T„ = 100 C C, a thermocouple is considered to
have read 99 percent of this applied temperature difference when its reading
indicates T(t) = 99°C.
cen58933_ch04.qxd 9/10/2002 9:12 AM Page 215
The value of the exponent b is
hA,
h
210 W/m 2 • °C
pC p V pC p L c (8500 kg/m 3 )(320 J/kg • °C)( 1.67 X l(T 4 m)
We now substitute these values into Eq. 4-4 and obtain
T(t) ~ T a _ kt
0.462 s"
T,
-> 0.01
(0.462 s~>
which yields
t = 10 s
Therefore, we must wait at least 10 s for the temperature of the thermocouple
junction to approach within 1 percent of the initial junction-gas temperature
difference.
Discussion Note that conduction through the wires and radiation exchange
with the surrounding surfaces will affect the result, and should be considered in
a more refined analysis.
215
CHAPTER 4
EXAMPLE 4-2 Predicting the Time of Death
A person is found dead at 5 pm in a room whose temperature is 20°C. The tem-
perature of the body is measured to be 25°C when found, and the heat trans-
fer coefficient is estimated to be ft = 8 W/m 2 • °C. Modeling the body as a
30-cm-diameter, 1.70-m-long cylinder, estimate the time of death of that per-
son (Fig. 4-10).
SOLUTION A body is found while still warm. The time of death is to be
estimated.
Assumptions 1 The body can be modeled as a 30-cm-diameter, 1.70-m-long
cylinder. 2 The thermal properties of the body and the heat transfer coefficient
are constant. 3 The radiation effects are negligible. 4 The person was healthy(l)
when he or she died with a body temperature of 37°C.
Properties The average human body is 72 percent water by mass, and thus we
can assume the body to have the properties of water at the average temperature
of (37 + 25)/2 = 31°C; k
J/kg • °C (Table A-9).
0.617 W/m • °C, p = 996 kg/m 3 , and C. = 4178
Analysis The characteristic length of the body is
9 T
TTK- L
ir(0.15m) 2 (1.7m)
L =^ =
c A s 2irr D L + 2-nrj 2ir(0.15 m)(1.7 m) + 2ir(0.15 m) 2
Then the Biot number becomes
hL c (8 W/m 2 • °C)(0.0689 m)
0.0689 m
Bi
0.617 W/m
0.89 > 0.1
FIGURE 4-10
Schematic for Example 4-2.
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HEAT TRANSFER
Therefore, lumped system analysis is not applicable. However, we can still use
it to get a "rough" estimate of the time of death. The exponent b in this case is
hA s h
8 W/m 2 ■ °C
pC p V pC p L c (996kg/m 3 )(4178J/kg • °C)(0.0689 m)
= 2.79 X 10- 5 s-'
We now substitute these values into Eq. 4-4,
T(t) - r»
which yields
25 -20
37-20
t = 43,860 s = 12.2 h
Therefore, as a rough estimate, the person died about 12 h before the body was
found, and thus the time of death is 5 am. This example demonstrates how to
obtain "ball park" values using a simple analysis.
4-2 - TRANSIENT HEAT CONDUCTION IN LARGE
PLANE WALLS, LONG CYLINDERS, AND
SPHERES WITH SPATIAL EFFECTS
In Section, 4-1, we considered bodies in which the variation of temperature
within the body was negligible; that is, bodies that remain nearly isothermal
during a process. Relatively small bodies of highly conductive materials ap-
proximate this behavior. In general, however, the temperature within a body
will change from point to point as well as with time. In this section, we con-
sider the variation of temperature with time and position in one-dimensional
problems such as those associated with a large plane wall, a long cylinder, and
a sphere.
Consider a plane wall of thickness 2L, a long cylinder of radius r , and
a sphere of radius r initially at a uniform temperature T t , as shown in Fig.
4-11. At time t = 0, each geometry is placed in a large medium that is at a
constant temperature T m and kept in that medium for t > 0. Heat transfer takes
place between these bodies and their environments by convection with a uni-
form and constant heat transfer coefficient h. Note that all three cases possess
geometric and thermal symmetry: the plane wall is symmetric about its center
plane (x = 0), the cylinder is symmetric about its centerline (r = 0), and the
sphere is symmetric about its center point (r = 0). We neglect radiation heat
transfer between these bodies and their surrounding surfaces, or incorporate
the radiation effect into the convection heat transfer coefficient h.
The variation of the temperature profile with time in the plane wall is
illustrated in Fig. 4-12. When the wall is first exposed to the surrounding
medium at T m < T t at t = 0, the entire wall is at its initial temperature T t . But
the wall temperature at and near the surfaces starts to drop as a result of heat
transfer from the wall to the surrounding medium. This creates a temperature
cen58933_ch04.qxd 9/10/2002 9:12 AM Page 217
h
Initially
T=T,
L x
(a) A large plane wall
Initially
T=T :
(b) A long cylinder
(c) A sphere
217
CHAPTER 4
FIGURE 4-1 1
Schematic of the simple
geometries in which heat
transfer is one-dimensional.
gradient in the wall and initiates heat conduction from the inner parts of the
wall toward its outer surfaces. Note that the temperature at the center of the
wall remains at T t until t = t 2 , and that the temperature profile within the wall
remains symmetric at all times about the center plane. The temperature profile
gets flatter and flatter as time passes as a result of heat transfer, and eventually
becomes uniform at T = T m . That is, the wall reaches thermal equilibrium
with its surroundings. At that point, the heat transfer stops since there is no
longer a temperature difference. Similar discussions can be given for the long
cylinder or sphere.
The formulation of the problems for the determination of the one-
dimensional transient temperature distribution T(x, t) in a wall results in a par-
tial differential equation, which can be solved using advanced mathematical
techniques. The solution, however, normally involves infinite series, which
are inconvenient and time-consuming to evaluate. Therefore, there is clear
motivation to present the solution in tabular or graphical form. However, the
solution involves the parameters x, L, t, k, a, h, T t , and T m which are too many
to make any graphical presentation of the results practical. In order to reduce
the number of parameters, we nondimensionalize the problem by defining the
following dimensionless quantities:
Dimensionless temperature:
Dimensionless distance from the center:
Dimensionless heat transfer coefficient:
Dimensionless time:
Q(x, t) --
x
T(x, t)
X
Bi
L
= hh
k
at
(Biot number)
(Fourier number)
T
\"
t=\
\\
T-__
t=^>
„ f _> CO
<
> ►■
L x
h
Initially
r„
T =
?i
h
FIGURE 4-1 2
Transient temperature profiles in a
plane wall exposed to convection
from its surfaces for T t > TL.
The nondimensionalization enables us to present the temperature in terms of
three parameters only: X, Bi, and t. This makes it practical to present the
solution in graphical form. The dimensionless quantities defined above for a
plane wall can also be used for a cylinder or sphere by replacing the space
variable x by r and the half-thickness L by the outer radius r . Note that
the characteristic length in the definition of the Biot number is taken to be the
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HEAT TRANSFER
half-thickness L for the plane wall, and the radius r for the long cylinder and
sphere instead of VIA used in lumped system analysis.
The one-dimensional transient heat conduction problem just described can
be solved exactly for any of the three geometries, but the solution involves in-
finite series, which are difficult to deal with. However, the terms in the solu-
tions converge rapidly with increasing time, and for t > 0.2, keeping the first
term and neglecting all the remaining terms in the series results in an error
under 2 percent. We are usually interested in the solution for times with
t > 0.2, and thus it is very convenient to express the solution using this one-
term approximation, given as
Plane
wall: U(JC ' ' Km "
Cylinder: 9(r, f) cy]
Sphere: 6(r, f) sph =
T(x, t)
-T„
Ti ~
T. x
T(r, t)
- r„
Ti-
r„
T(r, t) -
- r„
T s - 7*.
: A,e" x i T cos (\,x/L), t > 0.2
= A,e- x i T 7 (X 1 r/O, t > 0.2
A,e- X ' T ' ', , t>0.2
(4-10)
(4-11)
(4-12)
where the constants A, and \ l are functions of the Bi number only, and their
values are listed in Table 4-1 against the Bi number for all three geometries.
The function J is the zeroth-order Bessel function of the first kind, whose
value can be determined from Table 4-2. Noting that cos (0) = J (0) = 1 and
the limit of (sin x)lx is also 1, these relations simplify to the next ones at the
center of a plane wall, cylinder, or sphere:
Center of plane wall (x = 0):
Center of cylinder (r = 0):
Center of sphere (r = 0):
T - T
■'O, wall r r y 1
A { e-^ T
0.
T - T^
= A,e- x ' T
0, cyl r r 'Y
T n -T.
J 0, sph
A,e" x i T
(4-13)
(4-14)
(4-15)
Once the Bi number is known, the above relations can be used to determine
the temperature anywhere in the medium. The determination of the constants
A, and X, usually requires interpolation. For those who prefer reading charts
to interpolating, the relations above are plotted and the one-term approxima-
tion solutions are presented in graphical form, known as the transient temper-
ature charts. Note that the charts are sometimes difficult to read, and they are
subject to reading errors. Therefore, the relations above should be preferred to
the charts.
The transient temperature charts in Figs. 4-13, 4-14, and 4-15 for a large
plane wall, long cylinder, and sphere were presented by M. P. Heisler in 1947
and are called Heisler charts. They were supplemented in 1961 with transient
heat transfer charts by H. Grober. There are three charts associated with each
geometry: the first chart is to determine the temperature T at the center of the
geometry at a given time t. The second chart is to determine the temperature
at other locations at the same time in terms of T . The third chart is to deter-
mine the total amount of heat transfer up to the time t. These plots are valid
for t > 0.2.
cen58933_ch04.qxd 9/10/2002 9:12 AM Page 219
TABLE 4-
1
Coefficients used in the one-term approximate solution of transient one-
dimensional heat conduction in plane walls, cylinders, and spheres (Bi = hL/k
for a plane wall of thickness 2L, and Bi = hr lk\ax a cylinder or sphere of
radius r a )
Plane Wall
Cylinder
Sphere
Bi
X, A,
\j A,.
*•! A
0.01
0.02
0.04
0.06
0.08
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
2.0
3.0
4.0
5.0
6.0
7.0
8.0
9.0
10.0
20.0
30.0
40.0
50.0
100.0
0.0998
0.1410
0.1987
0.2425
0.2791
0.3111
0.4328
0.5218
0.5932
0.6533
0.7051
0.7506
0.7910
0.8274
0.8603
1.0769
1.1925
1.2646
1.3138
1.3496
1.3766
1.3978
1.4149
1.4289
1.4961
1.5202
1.5325
1.5400
1.5552
1.5708
1.0017
1.0033
1.0066
1.0098
1.0130
1.0161
1.0311
1.0450
1.0580
1.0701
1.0814
1.0918
1.1016
1.1107
1.1191
1.1785
1.2102
1.2287
1.2403
1.2479
1.2532
1.2570
1.2598
1.2620
1.2699
1.2717
1.2723
1.2727
1.2731
1.2732
0.1412
0.1995
0.2814
0.3438
0.3960
0.4417
0.6170
0.7465
0.8516
0.9408
1.0184
1.0873
1.1490
1.2048
1.2558
1.5995
1.7887
1.9081
1.9898
0490
0937
1286
1566
1795
2880
3261
3455
3572
3809
2.4048
0025
0050
0099
0148
0197
0246
0483
0712
0931
1143
1345
1539
1724
1902
2071
3384
4191
4698
5029
5253
5411
5526
5611
5677
5919
5973
5993
6002
6015
6021
0.1730
0.2445
0.3450
0.4217
0.4860
0.5423
0.7593
0.9208
0528
1656
2644
3525
4320
5044
5708
0288
2889
4556
5704
6537
7165
7654
8044
8363
9857
0372
0632
0788
1102
1416
1.0030
1.0060
1.0120
1.0179
1.0239
1.0298
1.0592
1.0880
1.1164
1.1441
1.1713
1.1978
1.2236
1.2488
1.2732
1.4793
1.6227
1.7202
1.7870
1.8338
1.8673
1.8920
1.9106
1.9249
1.9781
1.9898
1.9942
1.9962
1.9990
2.0000
219
CHAPTER 4
TABLE 4-
2
The zeroth- and first-order
functions of the first kind
Bessel
6
JJL&
M&
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
1.0000
0.9975
0.9900
0.9776
0.9604
0.9385
0.9120
0.8812
0.8463
0.8075
0.7652
0.7196
0.6711
0.6201
0.5669
0.5118
0.4554
0.3980
0.3400
0.2818
0.2239
0.1666
0.1104
0.0555
0.0025
-0.0968
-0.1850
-0.2601
-0.3202
0.0000
0.0499
0.0995
0.1483
0.1960
0.2423
0.2867
0.3290
0.3688
0.4059
0.4400
0.4709
0.4983
0.5220
0.5419
0.5579
0.5699
0.5778
0.5815
0.5812
0.5767
0.5683
0.5560
0.5399
0.5202
-0.4708
0.4097
0.3391
-0.2613
Note that the case 1/Bi = k/hL = corresponds to h — > °°, which corre-
sponds to the case of specified surface temperature T w . That is, the case in
which the surfaces of the body are suddenly brought to the temperature T„,
at t = and kept at T m at all times can be handled by setting /; to infinity
(Fig. 4-16).
The temperature of the body changes from the initial temperature T t to the
temperature of the surroundings T«, at the end of the transient heat conduction
process. Thus, the maximum amount of heat that a body can gain (or lose if
Tj > T„) is simply the change in the energy content of the body. That is,
Q n
mCJT m - r,) = pVCJT. - T : )
(kJ)
(4-16)
cen58933_ch04.qxd 9/10/2002 9:12 AM Page 22C
220
HEAT TRANSFER
Az
T.-
1.0
0.7
0.5
0.4
0.3
0.2
0.1
0.07
0.05
0.04
0.03
0.02
0.01
0.007
0.005
0.004
0.003
0.002
0.001
- 1 ' " i i i axil ' ■ fii '4iJ ' ' ' LL ' : : [ V[j~] —
k ' '-H-
■ - ~"^0$^^^^^^^^^^^^S : ^^^^^^^^^^^^^^&t=S==b=i==:^^- ' " ~^
- «/ * ~i
-Si^c'^ss>C""'^ i
S--J , . fi/ *
, ■ ■ | i , | : , | | . , | , , | I ■,..: 1 | ■ ^T j
1 ! ! ! | 1 ! ! 1 i 1 1 ! | j ii 1 ! 1 i l | ! 1 II!!
Mill! ,,,
I i ! i i i ' i 1 i i 1 : i i ! | i 1 ill
Sbw>^i ; -iP^i
i i ! ! i i i | ! i | i i i i i ] i i ! 1 1 ! mm M r-«^9n-»
, 1 1 1 11 1! im :^!ti ;|t n tta^T ifeip ! tit
PsN-JS^^""^ ^^t^T-^vV
.^^"I^zlt^U^^
5§§§fffipfc====;
i \-— -i ^yh- ri — - -i a -r,- h— -i-i-
JuTtlm-H^ mt-^---
+_T -E!-?^ >c + +- L L T3"ZdiJ -
S \ \ ^JTOH — ^-_.^
+- t-*'-tg*4-i-I-—t , -n-^>T-+4- +
W\ \±t:5jz S V - l- -n-^
m^ \ t^ T " + " N ^ ^rr- —
AV \ ^T^ L -+T-
\, \\ ^>„ -?o , u V VX
sL^5t^5st mypl\V v \i\ s s vo '
A V^ AX As A Y \
>ft VVJV as SA \ v \ \\\\\
S\ V 2 * s \ A\V^ V^\
%N tf& i \ v\ \ i\ \ W\ N
aNSF \ \ YSAVvVV
n l ? ~" ITT 1 A~ ^ \ s i^ ' '
i '
y^ V* X "" ~ L " ~ ~- N
s v u Si ? «\^ ^t s v ^ is?
fe\ _S, Vitt^WjV ^ V VAt\
\ f S, S V V '^ _L
\\S \ \m i^^\i\ \ X \ i m
\ t i \ V Y V^^Cti J i X ^ V
\aY Y uViJt.«Ai\^V V V \ Uv
A a v\ S \ s irtr i \\ \ ^
v\a 5 ffiiltuVfcA $ V \ U U
V \ $ V S V VU YYf 5 \ ^ \ S
v \ \W\\ \vfn \ ^ ^ nv
AAA Xa V - IV^Vr-lX^AV
\\\ ™\\\\Wx\ \ \ M\\
lAjAaaAJaEaAAaAa
4 6
10 14 IS
22 26 30 50
T = at/t 2
70
100 120
1 50
300 400 500 600 700
(a) Midplane temperature (from M. P. Heisler)
T-T„
Initially
T=Tj
0+ —
2L-
1.0
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.01
IT
L = t
iffi
j _|_i|
1
1
1
— 0.
6 J
—
- JT
—
— 1
[If
PL
II
Q
1.0
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
BA^y
•lYf,
; 57
"
'fl^y
04- /
t' /
/ '
tt-t
<tty
1 -t
/
/ /
tj-t 4
it a
% t
/ /
14- t
f^^
ll—t
f
/ /
tT4
k ^
t^
1 '
1 I
/ /
tirt-
-4T -J-
4T -
/ /
ill
' /
'iiA
i%4- 7
IT- -,
/
t]S o y
Oil 1 -
, r\ v
; ill! /
MJ^J-
-tt-jl
1111/
SV O'j- O'J-J o
t Mi \i
OlO'l c
id <-v__
Iff /
■ _ r -
cj r
* /
ki/ij
it t
A^/
ir
-W -t-
ty -
/J ,r
t i
/ / /
J A v
T /
K
A-t i
w /
B 1 v^
i /
4a7
1 i+ 7
I /
J^
\/
'17
t'Z
Jl / .<
itz
/ /
-Jz:^
Jffffi/ >
]>' / y
Jt --''
iz
-'M> '-
7ft' . S
— IT+P — '- — "
lit"^"
- Plate
il=S£
a^£
■:.W^^:
£;^l
;;#=•-"
0.1
1.0
1 k
10
1 00
10-
io-
i ci-
io- 2
10-' 1
Bi 2 z = h 2 at/k 2
10
10 2
10 J
10 4
Bi
hL
(b) Temperature distribution (from M. P. Heisler) (c) Heat transfer (from H. Grober et al.)
FIGURE 4-13
Transient temperature and heat transfer charts for a plane wall of thickness 2L initially at a uniform temperature T (
subjected to convection from both sides to an environment at temperature T^ with a convection coefficient of h.
where m is the mass, V is the volume, p is the density, and C p is the specific
heat of the body. Thus, Q max represents the amount of heat transfer for f — > °°.
The amount of heat transfer Q at a finite time f will obviously be less than this
cen58933_ch04.qxd 9/10/2002 9:12 AM Page 221
221
CHAPTER 4
1.0
0.7
0.5
0.4
0.3
0.2
0.1
0.07
0.05
0.04
0.03
0.02
0.01
0.007
0.005
0.004
0.003
0.002
0.001
-1 -
H
-'-
-
1 1 1 1 1
1 ijSfcJ
1 ! i ' ! ! i ! I i i ! ! i I ! i
Z-LL-l
-f-4-
J 1 _
^^- :
-i-i-i-i ! L_|_|.J j l_|_l i jI^J
__;
~~
^==
^£p§
>x
<t
-
Ttmt
VW 5 ^
t ! i m h~|~
—.
--,
■—
5=fc
£>*.
i 1' c ^ T '
35^S^
w^xloslSJ
"S-
-
-+-
sv
^»
™\lTO>sk
^
P
\J >
4>
-, [
I
V-
«/ ^
lu
M\M\vtv
\
s
4
\
<>s
s
\
\
\
N
x
x
\
w
s
S
v
<*
y
X
s
4-1
VKv
v
\
~tfo.
\
\
^
<
\
\
\
\
V
S
s
—
—
—
—
\
/ .
±r-
—
—
—
't-
-'/ft
—
—
!*n
'o n
~T
\\-
u
s
s
> t"
i
'
\
\
' :
,"
\
v
n
^
, P
■
\
\
i?
V
rv
17 s
■ \
\
■■
Q
\
s
L ,'
V
o
%,
r
\
\-
s
{
<o
*
\ \
<y
>j
3
>
+
1
?\illc
t?
%
5\
»-
■ <•
I '
o
»
\
5
\ "
c
.
L
V
i
i
\
t
\
1
\
\
T
\
\
\
T
\
\
V
.'■
\
\
J
i
\
±
\
i
\
\
u\
\
\
(\
\
, — 1
\
\
^~
\
► —
4 6 8 10
14 18 22 26 30 50 70
x = at// - ?
100 120 140 150 250 350
(a) Centerline temperature (from M. P. Heisler)
T-T a
r„-r„
Initially
T=Tj
0*-
1.0
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.01 0.1 1.0 10 100
1 _ k
Bi " hr
(£>) Temperature distribution (from M. P. Heisler)
rir
_
'in
'iltl y*'
' Mf\At
4
'WM
7 -W/~—
-$£—
— 0.
6-4
-*4
tt
I[
jt
— 0.
1
S 1
— 0.
1
L^ylind
1
(
er
1.0
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
44
'/'
l|
[4
/ /
1
' /
f
/'
TjI
/
/
l4
/ ,
_i
/ /
-m
/ /
/
/
-M
1
/
-CSfQ — o -\
" l/fll / 1/
3 fO — O 4
tiWii i
/
ii
' Tl
1 ii
a
^iW
/ I
±M
i
Q
? 1 / /
t
/ /
I
/ /
y
jfl / /
-4- -
/ /
-1
/ /
/
>1 1 /,
f
J
/ /
/
1//
tt
/
w +
/ /
t
4l
/ /
# Iff /
-t
/
-J
/ /
/ ,
/
1
'W/ '
1
i
|ll
bi
^~
W\ ^
4
III
^4
" i
tt
Cylinder
dm
^^
10- 5 10" 4 10" 3 10" 2 10" 1 1
Bi 2 T = /7 2 ar/A: 2
10
10 2
LO 3
10 4
(c) Heat transfer (from H. Grober et al.)
FIGURE 4-14
Transient temperature and heat transfer charts for a long cylinder of radius r initially at a uniform temperature T t
subjected to convection from all sides to an environment at temperature T x with a convection coefficient of h.
cen58933_ch04.qxd 9/10/2002 9:12 AM Page 222
222
HEAT TRANSFER
T; ~ T^
1.0
0.7
0.5
0.4
0.3
0.2
0.1
0.07
0.05
0.04
0.03
0.02
0.01
0.007
0.005
0.004
0.003
0.002
0.001
-------
— nt^t— — — — ^T x L-| Sphere
X-
35^ -+- o M
-+-
v.-T - #/ * Y-
-
-S "4-4=
4^<V
] a\\ O^^x \ "X^V^
sKv^vvS ^rtvjsos
-WrWS N ^yA4A\V
i\ , r^ \ A^r \\?^p< s v
~rn\rr"W'0\ %^c^ — r^i
ltd v o§^*&ra WsJv*
SAXW I5$t+ -XS,
JlV^ v\v\ V^K^K^a^A V
\\x*S i c ^ \\\
M \K \ * ^ 5\f<^- ^ $ \\ >
~ W^\ K $ SV^Sfc I'Z' \ \
n\ u Txt^ \ \ 's \n rv\^s
WW* \ \ V m K°v ■ >\ \ \ \
lu u&v? \ \ S A S
v\ iv %M%?± \ ^ \ \
ffiM V\ V A
ff^J^MsKSuw^
BVlvKSV \ OHvu\ ys
====lH^AA= S ^-E======^=^=
»B±3 ^tS5 — r vffrr^r^
HSSt*ri=^ -^ wte^
™^1TOF=
\W\ *\\SSv > \ N
JhAV P\^ \ li-O A Vl\^
■ifo\-rt \ N \ \ v- vV
4 HP V V \ Mlfe V A^ ^
a^ \ v \ \ M \\
pU \ V v v V \ \\
LpjII l-J t V\ i_i_i_L-i_L- L L ^ L V
BR V S S V VWAUWA \
\Vff3 \V vv y\w \ > v vV
lllA___\__AlIKffiKA_
jMlmLiSffiiAiLSA...
0.5 1.0 1.5
2.5 3 4 5 6 7 8 9 10 20
x = at/r?
30 40 50 100 150 200 250
(a) Midpoint temperature (from M. P. Heisler)
T-T rr
T a -T_
1.0
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.01
!i
^
Mil
i
1
t
—
1
tt
[
T [j
j .
LI
n
h
J_]
—
8- H
o
1
1+1
(l
Sphere
II
0.1
1.0
-L= A.
Bi fcr
10
100
10"' 1
Bi 2 x = h 2 at/k 2
(b) Temperature distribution (from M. P. Heisler) (c) Heat transfer (from H. Grober et al.)
FIGURE 4-15
Transient temperature and heat transfer charts for a sphere of radius r initially at a uniform temperature T t subjected to
convection from all sides to an environment at temperature T^ with a convection coefficient of h.
maximum. The ratio Q/Q max is plotted in Figures 4-13c, 4-14c, and 4-15c
against the variables Bi and h 2 at/k 2 for the large plane wall, long cylinder, and
cen58933_ch04.qxd 9/10/2002 9:12 AM Page 223
sphere, respectively. Note that once the fraction of heat transfer QIQ mzx has
been determined from these charts for the given t, the actual amount of heat
transfer by that time can be evaluated by multiplying this fraction by <2 max -
A negative sign for <2 max indicates that heat is leaving the body (Fig. 4-17).
The fraction of heat transfer can also be determined from these relations,
which are based on the one-term approximations already discussed:
Plane wall:
Cylinder:
Sphere:
Q
Q
sin A. ,
max/ wal ,
Q
■/i(Xi)
e- 1 20 |
max/ cy | A l
xl 111 A
1 -36,
O.sph
sph
M
(4-17)
(4-18)
(4-19)
The use of the Heisler/Grober charts and the one-term solutions already dis-
cussed is limited to the conditions specified at the beginning of this section:
the body is initially at a uniform temperature, the temperature of the medium
surrounding the body and the convection heat transfer coefficient are constant
and uniform, and there is no energy generation in the body.
We discussed the physical significance of the Biot number earlier and indi-
cated that it is a measure of the relative magnitudes of the two heat transfer
mechanisms: convection at the surface and conduction through the solid.
A small value of Bi indicates that the inner resistance of the body to heat con-
duction is small relative to the resistance to convection between the surface
and the fluid. As a result, the temperature distribution within the solid be-
comes fairly uniform, and lumped system analysis becomes applicable. Recall
that when Bi < 0.1, the error in assuming the temperature within the body to
be uniform is negligible.
To understand the physical significance of the Fourier number t, we ex-
press it as (Fig. 4-18)
at _ ^L-(l/L) Ar.
L 2 pC p LVt AT
The rate at which heat is conducted
across L of a body of volume I?
The rate at which heat is stored
in a body of volume L 3
(4-20)
223
CHAPTER 4
(a) Finite convection coefficient
(b) Infinite convection coefficient
FIGURE 4-16
The specified surface
temperature corresponds to the case
of convection to an environment at
T*, with a convection coefficient h
that is infinite.
Therefore, the Fourier number is a measure of heat conducted through a body
relative to heat stored. Thus, a large value of the Fourier number indicates
faster propagation of heat through a body.
Perhaps you are wondering about what constitutes an infinitely large plate
or an infinitely long cylinder. After all, nothing in this world is infinite. A plate
whose thickness is small relative to the other dimensions can be modeled as
an infinitely large plate, except very near the outer edges. But the edge effects
on large bodies are usually negligible, and thus a large plane wall such as the
wall of a house can be modeled as an infinitely large wall for heat transfer pur-
poses. Similarly, a long cylinder whose diameter is small relative to its length
can be analyzed as an infinitely long cylinder. The use of the transient tem-
perature charts and the one-term solutions is illustrated in the following
examples.
cen58933_ch04.qxd 9/10/2002 9:12 AM Page 224
224
HEAT TRANSFER
t =
t =
(a) Maximum heat transfer (t — » °°)
Q
Bi = ..
Ifat
k 2
Bi 2 x =
(Grober chart)
(b) Actual heat transfer for time t
FIGURE 4-17
The fraction of total heat transfer
Q/Qmax U P t° a specified time t is
determined using the Grober charts.
i
y
Fourier number: x :
at
L 2
*- stored
FIGURE 4-18
Fourier number at time / can be
viewed as the ratio of the rate of heat
conducted to the rate of heat stored
at that time.
EXAMPLE 4-3 Boiling Eggs
An ordinary egg can be approximated as a 5-cm-diameter sphere (Fig. 4-19).
The egg is initially at a uniform temperature of 5 C C and is dropped into boil-
ing water at 95°C. Taking the convection heat transfer coefficient to be
h = 1200 W/m 2 ■ °C, determine how long it will take for the center of the egg
to reach 70°C.
SOLUTION An egg is cooked in boiling water. The cooking time of the egg is to
be determined.
Assumptions 1 The egg is spherical in shape with a radius of r = 2.5 cm.
2 Heat conduction in the egg is one-dimensional because of thermal symmetry
about the midpoint. 3 The thermal properties of the egg and the heat transfer
coefficient are constant. 4 The Fourier number is t > 0.2 so that the one-term
approximate solutions are applicable.
Properties The water content of eggs is about 74 percent, and thus the ther-
mal conductivity and diffusivity of eggs can be approximated by those of water
at the average temperature of (5 + 70)/2 = 37.5°C; k = 0.627 W/m • °C and
a = k/pC p = 0.151 X 10- 5 m 2 /s (Table A-9).
Analysis The temperature within the egg varies with radial distance as well as
time, and the temperature at a specified location at a given time can be deter-
mined from the Heisler charts or the one-term solutions. Here we will use the
latter to demonstrate their use. The Biot number for this problem is
Bi
(1200 W/m 2 • °C)(0.025m)
0.627 W/m • °C
47.8
which is much greater than 0.1, and thus the lumped system analysis is not
applicable. The coefficients X 1 and A 1 for a sphere corresponding to this Bi are,
from Table 4-1,
3.0753,
1.9958
Substituting these and other values into Eq. 4-15 and solving for t gives
r - 7L
T; ~ T„
A,e~V
70 — 95
-> - „: = 1.9958e-' 30753 ^
5-95
-> t = 0.209
which is greater than 0.2, and thus the one-term solution is applicable with an
error of less than 2 percent. Then the cooking time is determined from the de-
finition of the Fourier number to be
tt 2 _ (0.209)(0.025 m) 2
~° r ~ 0.151 X 10- 6 m 2 /s
865 s = 14.4 min
Therefore, it will take about 15 min for the center of the egg to be heated from
5°C to 70°C.
Discussion Note that the Biot number in lumped system analysis was defined
differently as Bi = hL c /k = h{r/3)/k. However, either definition can be used in
determining the applicability of the lumped system analysis unless Bi ~ 0.1.
cen58933_ch04.qxd 9/10/2002 9:12 AM Page 225
225
CHAPTER 4
EXAMPLE 4-4 Heating of Large Brass Plates in an Oven
In a production facility, large brass plates of 4 cm thickness that are initially at
a uniform temperature of 20 C C are heated by passing them through an oven
that is maintained at 500 C C (Fig. 4-20). The plates remain in the oven for a
period of 7 min. Taking the combined convection and radiation heat transfer
coefficient to be h = 120 W/m 2 ■ °C, determine the surface temperature of the
plates when they come out of the oven.
SOLUTION Large brass plates are heated in an oven. The surface temperature
of the plates leaving the oven is to be determined.
Assumptions 1 Heat conduction in the plate is one-dimensional since the plate
is large relative to its thickness and there is thermal symmetry about the center
plane. 2 The thermal properties of the plate and the heat transfer coefficient are
constant. 3 The Fourier number is t > 0.2 so that the one-term approximate so-
lutions are applicable.
Properties The properties of brass at room temperature are k = 110 W/m • °C,
p = 8530 kg/m 3 , C p = 380 J/kg • °C, and a = 33.9 X lO" 6 m 2 /s (Table A-3).
More accurate results are obtained by using properties at average temperature.
Analysis The temperature at a specified location at a given time can be de-
termined from the Heisler charts or one-term solutions. Here we will use the
charts to demonstrate their use. Noting that the half-thickness of the plate is
L = 0.02 m, from Fig. 4-13 we have
0.46
1 k 100W/m-°C
T -
■*■ o
Bi hL (120 W/m 2 • °C)(0.02 m)
- T„
at (33.9 X 10- 6 m 2 /s)(7 X 60 s)
T t -
- 71
T L 2 (0.02 m) 2 " 5 - e \
Also,
1
k
Bi
hL
X
L
L
L
45.8
0.99
Therefore,
T - 71 T- 71 T ~ T„
T, - T„ T - 71 T t - 71
0.46 X 0.99 = 0.455
and
T = 71 + 0.455(7, - 71) = 500 + 0.455(20 - 500) = 282°C
Therefore, the surface temperature of the plates will be 282°C when they leave
the oven.
Discussion We notice that the Biot number in this case is Bi = 1/45.8 =
0.022, which is much less than 0.1. Therefore, we expect the lumped system
analysis to be applicable. This is also evident from (7~- TJ/(T - 7"J = 0.99,
which indicates that the temperatures at the center and the surface of the plate
relative to the surrounding temperature are within 1 percent of each other.
Egg X
T. = 5°C j
h =
1200W/m 2 -°C
7.
= 95°C
FIGURE 4-1 9
Schematic for Example 4-3.
T rr _ = 500°C
h = 120 W/m 2 -°C
/
2L = 4cm
Brass
plate
J, = 20°C
FIGURE 4-20
Schematic for Example 4-4.
cen58933_ch04.qxd 9/10/2002 9:12 AM Page 226
226
HEAT TRANSFER
Noting that the error involved in reading the Heisler charts is typically at least a
few percent, the lumped system analysis in this case may yield just as accurate
results with less effort.
The heat transfer surface area of the plate is 2/1, where A is the face area of
the plate (the plate transfers heat through both of its surfaces), and the volume
of the plate is V = (2L)A, where L is the half-thickness of the plate. The expo-
nent b used in the lumped system analysis is determined to be
hA s
pC p V
h(2A)
h
P C„(2LA) P C„L
120 W/m 2 • °C
(8530 kg/m 3 )(380 J/kg • °C)(0.02 m)
0.00185 s-
Then the temperature of the plate at t = 7 min = 420 s is determined from
Tit)
T(t) - 500
It yields
20 - 500
T(t ) = 279°C
-(0.00185 s - ')(420s)
which is practically identical to the result obtained above using the Heisler
charts. Therefore, we can use lumped system analysis with confidence when the
Biot number is sufficiently small.
T x = 200°C
h = 80W/m 2 -°C
Stainless steel
shaft
T : = 600°C
D = 20 cm
FIGURE 4-21
Schematic for Example 4-5.
EXAMPLE 4-5 Cooling of a Long
Stainless Steel Cylindrical Shaft
A long 20-cm-diameter cylindrical shaft made of stainless steel 304 comes out
of an oven at a uniform temperature of 600°C (Fig. 4-21). The shaft is then al-
lowed to cool slowly in an environment chamber at 200°C with an average heat
transfer coefficient of h = 80 W/m 2 • °C. Determine the temperature at the cen-
ter of the shaft 45 min after the start of the cooling process. Also, determine
the heat transfer per unit length of the shaft during this time period.
SOLUTION A long cylindrical shaft at 600°C is allowed to cool slowly. The cen-
ter temperature and the heat transfer per unit length are to be determined.
Assumptions 1 Heat conduction in the shaft is one-dimensional since it is long
and it has thermal symmetry about the centerline. 2 The thermal properties of
the shaft and the heat transfer coefficient are constant. 3 The Fourier number
is t > 0.2 so that the one-term approximate solutions are applicable.
Properties The properties of stainless steel 304 at room temperature
are k = 14.9 W/m ■ °C, p = 7900 kg/m 3 , C p = 477 J/kg • °C, and
a = 3.95 X 10~ 6 m 2 /s (Table A-3). More accurate results can be obtained by
using properties at average temperature.
Analysis The temperature within the shaft may vary with the radial distance r
as well as time, and the temperature at a specified location at a given time can
cen58933_ch04.qxd 9/10/2002 9:12 AM Page 227
be determined from the Heisler charts. Noting that the radius of the shaft is
r = 0.1 m, from Fig. 4-14 we have
1
14.9 W/m • °C
1.86
Bi hr (80 W/m 2 • °C)(0.1 m)
_ at _ (3.95 X 10- 6 m 2 /s)(45 X 60 s)
J ~Ji ~ (0.1m) 2
1.07
0.40
and
T = r„ + 0.4(7; - 7*.) = 200 + 0.4(600 - 200) = 360°C
Therefore, the center temperature of the shaft will drop from 600°C to 360°C
in 45 min.
To determine the actual heat transfer, we first need to calculate the maximum
heat that can be transferred from the cylinder, which is the sensible energy of
the cylinder relative to its environment. Taking L = 1 m,
m = pV = pirr 2 L = (7900 kg/m 3 )-jr(0.1 m) 2 (l m) = 248.2 kg
e raax = mC p (T„ - T,) = (248.2 kg)(0477 kJ/kg ■ °C)(600 - 200)°C
= 47,354 kJ
The dimensionless heat transfer ratio is determined from Fig. 4-14cfor a long
cylinder to be
Bi
1
1
1/Bi 1.86
0.537
Q
h 2 at
Bi 2 T = (0.537) 2 (1.07) = 0.309
Q n
0.62
Therefore,
Q = 0.62g„
0.62 X (47,354 kJ) = 29,360 kJ
which is the total heat transfer from the shaft during the first 45 min of
the cooling.
ALTERNATIVE SOLUTION We could also solve this problem using the one-term
solution relation instead of the transient charts. First we find the Biot number
_ hr _ (80 W/m 2 • °C)(0.1 m) _
Bl " T = 14.9 W/m- °C " °' 537
The coefficients X 1 and A x for a cylinder corresponding to this Bi are deter-
mined from Table 4-1 to be
Substituting these values into Eq. 4-14 gives
T - 71
e„
rp -j-> /\\&
1.122e-<°- 970 >-< L07) = 0.41
227
CHAPTER 4
cen58933_ch04.qxd 9/10/2002 9:12 AM Page 22E
228
HEAT TRANSFER
and thus
T
± o
= T a
f o.4i (r, -
- rj =
200 + 0.41(600 - 200) = 364°C
The value
of J^XJ for \ : =
0.97C
is determined from Table 4-2 to be 0.430.
Then the fractiona
1 heat tran
sfer is
determined from Eq. 4-18 to be
Q
*-max
= l-26
UK)
= 1 - 2 X 0.41 jJUJj = 0.636
and thus
Q =
0.636e ma ,
= 0.636 X (47,354 kJ) = 30,120 kj
Discussion
Thes
ight difference between the two results is due to the reading
error of the charts
CO \
Plane
s surface
T * \
h
n /
X
FIGURE 4-22
Schematic of a semi-infinite body.
4-3 - TRANSIENT HEAT CONDUCTION
IN SEMI-INFINITE SOLIDS
A semi-infinite solid is an idealized body that has a single plane surface and
extends to infinity in all directions, as shown in Fig. 4-22. This idealized body
is used to indicate that the temperature change in the part of the body in which
we are interested (the region close to the surface) is due to the thermal condi-
tions on a single surface. The earth, for example, can be considered to be a
semi-infinite medium in determining the variation of temperature near its sur-
face. Also, a thick wall can be modeled as a semi-infinite medium if all we are
interested in is the variation of temperature in the region near one of the sur-
faces, and the other surface is too far to have any impact on the region of in-
terest during the time of observation.
Consider a semi-infinite solid that is at a uniform temperature T t . At time
t = 0, the surface of the solid at x = is exposed to convection by a fluid at a
constant temperature T m , with a heat transfer coefficient h. This problem can
be formulated as a partial differential equation, which can be solved analyti-
cally for the transient temperature distribution T(x, t). The solution obtained is
presented in Fig. 4-23 graphically for the nondimensionalized temperature
defined as
1 - 6(x, f) = 1
T(x, t)
T(x,t)-T t
(4-21)
against the dimensionless variable x/(2\/at) for various values of the param-
eter h\faXlk.
Note that the values on the vertical axis correspond to x = 0, and thus rep-
resent the surface temperature. The curve hA/ai/k = c° corresponds to /; — > °°,
which corresponds to the case of specified temperature T m at the surface at
x = 0. That is, the case in which the surface of the semi-infinite body is sud-
denly brought to temperature T«, at t = and kept at T„ at all times can be han-
dled by setting h to infinity. The specified surface temperature case is closely
cen58933_ch04.qxd 9/10/2002 9:12 AM Page 229
1.0
0.5
0.4
0.3
0.2
0.1
0.05
-i 0.04
0.03
0.02
0.01
229
CHAPTER 4
Ambient
Ax, t)\
-
°.J
^p .
1 ^
^*0
0j \
0.25
0.5
0.75
1.0
1.25
1.5
2Var
FIGURE 4-23
Variation of temperature with position and time in a semi-infinite solid initially at T t subjected to convection to an
environment at T„ with a convection heat transfer coefficient of h (from P. J. Schneider, Ref. 10).
approximated in practice when condensation or boiling takes place on the
surface. For a finite heat transfer coefficient /;, the surface temperature
approaches the fluid temperature T m as the time t approaches infinity.
The exact solution of the transient one-dimensional heat conduction prob-
lem in a semi-infinite medium that is initially at a uniform temperature of T,
and is suddenly subjected to convection at time t = has been obtained, and
is expressed as
T(x, t)
erfc
hx ,
3 iy
h 2 at
k 2
erfc
h\/ai
2y/ai
(4-22)
where the quantity erfc (£) is the complementary error function, defined as
erfc(£) = 1 --^= f e-" 2 du
Vtt Jo
(4-23)
Despite its simple appearance, the integral that appears in the above relation
cannot be performed analytically. Therefore, it is evaluated numerically for
different values of £, and the results are listed in Table 4-3. For the special
case of h — > °o, the surface temperature T s becomes equal to the fluid temper-
ature r„, and Eq. 4-22 reduces to
T(x, f)
erfc
(4-24)
cen58933_ch04.qxd 9/10/2002 9:12 AM Page 23C
230
HEAT TRANSFER
TABLE 4-3
The complementary
error fu
iction
£
erfc (£)
£
erfc (Q
£
erfc (g)
6
erfc (Q
6
erfc (Q
e
erfc (Q
0.00
1.00000
0.38
0.5910
0.76
0.2825
1.14
0.1069
1.52
0.03159
1.90
0.00721
0.02
0.9774
0.40
0.5716
0.78
0.2700
1.16
0.10090
1.54
0.02941
1.92
0.00662
0.04
0.9549
0.42
0.5525
0.80
0.2579
1.18
0.09516
1.56
0.02737
1.94
0.00608
0.06
0.9324
0.44
0.5338
0.82
0.2462
1.20
0.08969
1.58
0.02545
1.96
0.00557
0.08
0.9099
0.46
0.5153
0.84
0.2349
1.22
0.08447
1.60
0.02365
1.98
0.00511
0.10
0.8875
0.48
0.4973
0.86
0.2239
1.24
0.07950
1.62
0.02196
2.00
0.00468
0.12
0.8652
0.50
0.4795
0.88
0.2133
1.26
0.07476
1.64
0.02038
2.10
0.00298
0.14
0.8431
0.52
0.4621
0.90
0.2031
1.28
0.07027
1.66
0.01890
2.20
0.00186
0.16
0.8210
0.54
0.4451
0.92
0.1932
1.30
0.06599
1.68
0.01751
2.30
0.00114
0.18
0.7991
0.56
0.4284
0.94
0.1837
1.32
0.06194
1.70
0.01612
2.40
0.00069
0.20
0.7773
0.58
0.4121
0.96
0.1746
1.34
0.05809
1.72
0.01500
2.50
0.00041
0.22
0.7557
0.60
0.3961
0.98
0.1658
1.36
0.05444
1.74
0.01387
2.60
0.00024
0.24
0.7343
0.62
0.3806
1.00
0.1573
1.38
0.05098
1.76
0.01281
2.70
0.00013
0.26
0.7131
0.64
0.3654
1.02
0.1492
1.40
0.04772
1.78
0.01183
2.80
0.00008
0.28
0.6921
0.66
0.3506
1.04
0.1413
1.42
0.04462
1.80
0.01091
2.90
0.00004
0.30
0.6714
0.68
0.3362
1.06
0.1339
1.44
0.04170
1.82
0.01006
3.00
0.00002
0.32
0.6509
0.70
0.3222
1.08
0.1267
1.46
0.03895
1.84
0.00926
3.20
0.00001
0.34
0.6306
0.72
0.3086
1.10
0.1198
1.48
0.03635
1.86
0.00853
3.40
0.00000
0.36
0.6107
0.74
0.2953
1.12
0.1132
1.50
0.03390
1.88
0.00784
3.60
0.00000
This solution corresponds to the case when the temperature of the exposed
surface of the medium is suddenly raised (or lowered) to T s at t = and is
maintained at that value at all times. Although the graphical solution given in
Fig. 4-23 is a plot of the exact analytical solution given by Eq. 4-23, it is sub-
ject to reading errors, and thus is of limited accuracy.
,T. = -10°C
Soil
Water pipe
r ; =i5°c
FIGURE 4-24
Schematic for Example 4-6.
EXAMPLE 4-6 Minimum Burial Depth of Water Pipes to Avoid
Freezing
In areas where the air temperature remains below 0°C for prolonged periods of
time, the freezing of water in underground pipes is a major concern. Fortu-
nately, the soil remains relatively warm during those periods, and it takes weeks
for the subfreezing temperatures to reach the water mains in the ground. Thus,
the soil effectively serves as an insulation to protect the water from subfreezing
temperatures in winter.
The ground at a particular location is covered with snow pack at -10°C for a
continuous period of three months, and the average soil properties at that loca-
tion are k = 0.4 W/m • °C and a = 0.15 X 10~ 6 m 2 /s (Fig. 4-24). Assuming an
initial uniform temperature of 15 C C for the ground, determine the minimum
burial depth to prevent the water pipes from freezing.
SOLUTION The water pipes are buried in the ground to prevent freezing. The
minimum burial depth at a particular location is to be determined.
Assumptions 1 The temperature in the soil is affected by the thermal condi-
tions at one surface only, and thus the soil can be considered to be a semi-
infinite medium with a specified surface temperature of - 10°C. 2 The thermal
properties of the soil are constant.
cen58933_ch04.qxd 9/10/2002 9:12 AM Page 231
Properties The properties of the soil are as given in the problem statement.
Analysis The temperature of the soil surrounding the pipes will be 0°C after
three months in the case of minimum burial depth. Therefore, from Fig. 4-23,
we have
hVoit
T(x,t) - T
1 ~ ' ' - = 1
(since h —> *■)
0-(-10)
H
15 - (-10)
0.6
2 Vat
0.36
We note that
t = (90 days)(24 h/day)(3600 s/h) = 7.78 X 10 6 s
and thus
x = 2i Vaf = 2 X 0.36V(0.15 X lO" 6 m 2 /s)(7.78 X 10 6 s) = 0.77 m
Therefore, the water pipes must be buried to a depth of at least 77 cm to avoid
freezing under the specified harsh winter conditions.
ALTERNATIVE SOLUTION The solution of this problem could also be deter-
mined from Eq. 4-24:
T(x,t)
erfc
2Vat,
0-15
-10 - 15
erfc
iVat
0.60
The argument that corresponds to this value of the complementary error func-
tion is determined from Table 4-3 to be £ = 0.37. Therefore,
x = 2£ vaf = 2 X 0.37V(0.15 X lO" 6 m 2 /s)(7.78 X 10 6 s) = 0.80 m
Again, the slight difference is due to the reading error of the chart.
231
CHAPTER 4
h
T(r,t)
Heat
transfer
(a) Long cylinder
4-4 - TRANSIENT HEAT CONDUCTION IN
MULTIDIMENSIONAL SYSTEMS
The transient temperature charts presented earlier can be used to determine the
temperature distribution and heat transfer in one-dimensional heat conduction
problems associated with a large plane wall, a long cylinder, a sphere, and a
semi-infinite medium. Using a superposition approach called the product
solution, these charts can also be used to construct solutions for the two-
dimensional transient heat conduction problems encountered in geometries
such as a short cylinder, a long rectangular bar, or a semi-infinite cylinder or
plate, and even three-dimensional problems associated with geometries such
as a rectangular prism or a semi-infinite rectangular bar, provided that all sur-
faces of the solid are subjected to convection to the same fluid at temperature
h
T(r,x,t)
Heat
' transfer
(b) Short cylinder (two-dimensional)
FIGURE 4-25
The temperature in a short
cylinder exposed to convection from
all surfaces varies in both the radial
and axial directions, and thus heat
is transferred in both directions.
cen58933_ch04.qxd 9/10/2002 9:12 AM Page 232
232
HEAT TRANSFER
Plane wall
^1
Long
cylinder
FIGURE 4-26
A short cylinder of radius r and
height a is the intersection of a long
cylinder of radius r and a plane wall
of thickness a.
Plane wall
Plane wall
hH
FIGURE 4-27
A long solid bar of rectangular
profile a X b is the intersection
of two plane walls of
thicknesses a and b.
T„, with the same heat transfer coefficient h, and the body involves no heat
generation (Fig. 4-25). The solution in such multidimensional geometries can
be expressed as the product of the solutions for the one-dimensional geome-
tries whose intersection is the multidimensional geometry.
Consider a short cylinder of height a and radius r initially at a uniform tem-
perature r,. There is no heat generation in the cylinder. At time t = 0, the
cylinder is subjected to convection from all surfaces to a medium at temper-
ature r„ with a heat transfer coefficient h. The temperature within the cylin-
der will change with x as well as r and time t since heat transfer will occur
from the top and bottom of the cylinder as well as its side surfaces. That is,
T = T(r, x, t) and thus this is a two-dimensional transient heat conduction
problem. When the properties are assumed to be constant, it can be shown that
the solution of this two-dimensional problem can be expressed as
T(r, x, t) - T a
short
cylinder
T(x, t)
plane
w;ill
T(r, t)
infinite
cylinder
(4-25)
That is, the solution for the two-dimensional short cylinder of height a and
radius r is equal to the product of the nondimensionalized solutions for the
one-dimensional plane wall of thickness a and the long cylinder of radius r ,
which are the two geometries whose intersection is the short cylinder, as
shown in Fig. 4-26. We generalize this as follows: the solution for a multi-
dimensional geometry is the product of the solutions of the one-dimensional
geometries whose intersection is the multidimensional body.
For convenience, the one-dimensional solutions are denoted by
0waii(*. r )
(r,f)
J c\ I
,-i„f(*, t)
(T(x, t)
- T a
[ T,~
r„
(T(r, t)
-rj
\ T,~
r- ,
(T(x, t)
- r.
plane
wall
infinite
cylinder
T, - T m
(4-26)
For example, the solution for a long solid bar whose cross section is an a X b
rectangle is the intersection of the two infinite plane walls of thicknesses
a and b, as shown in Fig. 4-27, and thus the transient temperature distribution
for this rectangular bar can be expressed as
T(x, y, t)
W-*. OQwaiiCy.
(4-27)
The proper forms of the product solutions for some other geometries are given
in Table 4—4. It is important to note that the x-coordinate is measured from the
surface in a semi-infinite solid, and from the midplane in a plane wall. The ra-
dial distance r is always measured from the centerline.
Note that the solution of a two-dimensional problem involves the product of
two one-dimensional solutions, whereas the solution of a three-dimensional
problem involves the product of three one-dimensional solutions.
A modified form of the product solution can also be used to determine
the total transient heat transfer to or from a multidimensional geometry by
using the one-dimensional values, as shown by L. S. Langston in 1982. The
cen58933_ch04.qxd 9/10/2002 9:13 AM Page 233
233
CHAPTER 4
TABLE 4-4
Multidimensional solutions expressed as products of one-dimensional solutions for bodies that are initially at a
uniform temperature 7j and exposed to convection from all surfaces to a medium at T«
9(''.0 = e cyl (n()
Infinite cylinder
Hw.') = e cy|(f ,oe s „, inf (x,0
Semi-infinite cylinder
Hx,r,t) = e cyl (r,t)() wM (x,f)
Short cylinder
Semi-infinite medium
K.v,v.O = e semWnf (x,f)e semWnf ^o
Quarter-infinite medium
B(x,y,z,t) =
9 semi-inf (' V > f )8 scmi -i„f & e s emi-inf fc f )
Corner region of a large medium
2L
e(*,o = e wall teO
Infinite plate (or plane wall)
2L
H^y.O = e wall fcOe semi . m ,Cv.O
Semi-infinite plate
Hx,y,z,t) =
^wall^Oe^.^O'.Oe^.infCZ.O
Quarter-infinite plate
i j
;-
/
/
if
ife^
/\
1
1 v
H^y.O = e wall (x,oe wdl (y,f)
Infinite rectangular bar
e(jc,y,z,f) =
e w a ii(^')e wall Cv,oe scmi , nf (z,o
Semi-infinite rectangular bar
Q(x,y,z,t) =
e „an^oe wall (.v,oe wall ( Z ,o
Rectangular parallelepiped
cen58933_ch04.qxd 9/10/2002 9:13 AM Page 234
234
HEAT TRANSFER
transient heat transfer for a two-dimensional geometry formed by the inter-
section of two one-dimensional geometries 1 and 2 is
_e
*iraa
total, 2D
_Q_
Cm.
_Q_
_Q_
(4-28)
Transient heat transfer for a three-dimensional body formed by the inter-
section of three one-dimensional bodies 1, 2, and 3 is given by
_Q_
Sc m a
_Q_
iima
_Q_
i/ma
_Q_
fcraa
_Q_
*£ma
_Q_
sima
(4-29)
The use of the product solution in transient two- and three-dimensional heat
conduction problems is illustrated in the following examples.
r„ = 25°C
h = 60W/m 2 -°C
T t = 120 D C
FIGURE 4-28
Schematic for Example 4-7.
EXAMPLE 4-7 Cooling of a Short Brass Cylinder
" A short brass cylinder of diameter D = 10 cm and height H = 12 cm is initially
I at a uniform temperature 7) = 120°C. The cylinder is now placed in atmo-
spheric air at 25°C, where heat transfer takes place by convection, with a heat
transfer coefficient of h = 60 W/m 2 ■ °C. Calculate the temperature at (a) the
center of the cylinder and (b) the center of the top surface of the cylinder
15 min after the start of the cooling.
SOLUTION A short cylinder is allowed to cool in atmospheric air. The temper-
atures at the centers of the cylinder and the top surface are to be determined.
Assumptions 1 Heat conduction in the short cylinder is two-dimensional, and
thus the temperature varies in both the axial x- and the radial r-directions. 2 The
thermal properties of the cylinder and the heat transfer coefficient are constant.
3 The Fourier number is t > 0.2 so that the one-term approximate solutions are
applicable.
Properties The properties of brass at room temperature are k= 110 W/m • °C
and a = 33.9 X 10~ 5 m 2 /s (Table A-3). More accurate results can be obtained
by using properties at average temperature.
Analysis (a) This short cylinder can physically be formed by the intersection of
a long cylinder of radius r = 5 cm and a plane wall of thickness 2L = 12 cm,
as shown in Fig. 4-28. The dimensionless temperature at the center of the
plane wall is determined from Figure 4-13a to be
at (3.39 X KT 5 m 2 /s)(900s) 1
T = ^ = r^ = 8.48
U (0.06 m) 2
J_ == _L- 110 W/m • °c
' 6wall(0,O =
T(0,t)-T a
Bi hL (60 w/m 2 . °c)(0.06 m ) .
cen58933_ch04.qxd 9/10/2002 9:13 AM Page 235
Similarly, at the center of the cylinder, we have
at (3.39 X Kr 5 m 2 /s)(900s)
r 2 (0.05 m) 2
1
_ k _ HOW/m • °C
Bi
hr o (60 W/m 2 • °C)(0.05 m)
dte
ore,
/r(o,o,o-r„\
1 'Y T 1 Ishort "wall'
cylinder
12.2
36.7
T(0, t)-T a
■ e cy[ (o,o= T _ T " = o.5
and
T(0,0,t)
(0, t ) X 6 cyl (0, t ) = 0.8 X 0.5 = 0.4
0.4(7, - r.) = 25 + 0.4(120 - 25) = 63°C
This is the temperature at the center of the short cylinder, which is also the cen-
ter of both the long cylinder and the plate.
(b) The center of the top surface of the cylinder is still at the center of the long
cylinder (r = 0), but at the outer surface of the plane wall (x = L). Therefore,
we first need to find the surface temperature of the wall. Noting that x = L =
0.06 m,
x _ 0.06 m
L ~ 0.06 m
1
_L = A = 110 W/m ■ °c
Bi hL (60 w/m 2 ■ °C)(0.06 m)
30.6
T -T x .
0.98
Then
9 wa ii(£>
T{L,t)-T^ (T(L,t)-TA(T -T,
T t -T,
Therefore,
T(L, 0, t)-T =
T 71
T- — T
X 0.8 = 0.784
Ti-Tn
shor. = 6 wall (L, f )0 cyl (0, t) = 0.784 X 0.5 = 0.392
cylinder
and
T(L, 0, t) = T a + 0.392(7; - r„) = 25 + 0.392(120 - 25) = 62.2°C
which is the temperature at the center of the top surface of the cylinder.
235
CHAPTER 4
J EXAMPLE 4-8 Heat Transfer from a Short Cylinder
Determine the total heat transfer from the short brass cylinder
, kg/m 3 , C p = 0.380 kJ/kg • °C) discussed in Example 4-7.
(P =
= 8530
cen58933_ch04.qxd 9/10/2002 9:13 AM Page 236
236
HEAT TRANSFER
SOLUTION We first determine the maximum heat that can be transferred from
the cylinder, which is the sensible energy content of the cylinder relative to its
environment:
m = pV = pirr 2 L = (8530 kg/m 3 )Tr(0.05 m) 2 (0.06 m) = 4.02 kg
g raax = mC p (T, - r„) = (4.02 kg)(0.380 kJ/kg • °C)(120 - 25)°C = 145.1 kJ
Then we determine the dimensionless heat transfer ratios for both geometries.
For the plane wall, it is determined from Fig. 4-13c to be
Bi = IM = 3^6 = °-° 327
h 2 at
Bi 2 T = (0.0327) 2 (8.48) = 0.0091
!cma
0.23
plane
Similarly, for the cylinder, we have
Bi = W = i = °- 0272
1/Bi 36.7
h 2 at
Bi 2 T = (0.0272) 2 (12.2) = 0.0090
0.47
nfinite
cylinder
Then the heat transfer ratio for the short cylinder is, from Eq. 4-28,
^ max ' short cyl V^max/j VtJmax/,
_G
= 0.23 + 0.47(1 - 0.23) = 0.592
Therefore, the total heat transfer from the cylinder during the first 15 min of
cooling is
Q = 0.592g raax = 0.592 X (145.1 kJ) = 85.9 kJ
EXAMPLE 4-9 Cooling of a Long Cylinder by Water
A semi-infinite aluminum cylinder of diameter D = 20 cm is initially at a uni-
form temperature 7", = 200 C C. The cylinder is now placed in water at 15°C
where heat transfer takes place by convection, with a heat transfer coefficient
of h = 120 W/m 2 • °C. Determine the temperature at the center of the cylinder
15 cm from the end surface 5 min after the start of the cooling.
SOLUTION A semi-infinite aluminum cylinder is cooled by water. The tem-
perature at the center of the cylinder 15 cm from the end surface is to be
determined.
Assumptions 1 Heat conduction in the semi-infinite cylinder is two-
dimensional, and thus the temperature varies in both the axial x- and the radial
r-directions. 2 The thermal properties of the cylinder and the heat transfer co-
efficient are constant. 3 The Fourier number is t > 0.2 so that the one-term
approximate solutions are applicable.
cen58933_ch04.qxd 9/10/2002 9:13 AM Page 237
Properties The properties of aluminum at room temperature are k = 237
W/m ■ °C and a = 9.71 X 10~ 6 m 2 /s (Table A-3). More accurate results can be
obtained by using properties at average temperature.
Analysis This semi-infinite cylinder can physically be formed by the inter-
section of an infinite cylinder of radius r = 10 cm and a semi-infinite medium,
as shown in Fig. 4-29.
We will solve this problem using the one-term solution relation for the cylin-
der and the analytic solution for the semi-infinite medium. First we consider the
infinitely long cylinder and evaluate the Biot number:
Bi
hr B _ (120 W/m 2 • °C)(0.1 m)
T~ 237 W/m ■ °C
0.05
The coefficients k 1 and A 1 for a cylinder corresponding to this Bi are deter-
mined from Table 4-1 to be \ x = 0.3126 and A 1 = 1.0124. The Fourier num-
ber in this case is
at _ (9.71 X 10- 5 m 2 /s)(5 X 60 s)
~7;~ (0.1 m) 2
2.91 >0.2
and thus the one-term approximation is applicable. Substituting these values
into Eq. 4-14 gives
' = B_ 1 (0,0 =A x e-
1.0124e-<°- 3126 > 2 ( 21)1 > = 0.762
The solution for the semi-infinite solid can be determined from
hx , h 2 at
1 - e s emi-inftM) = erfc
exp iy + >
erfc
hVal
at
First we determine the various quantities in parentheses
x 0.15 m
0.44
2Vaf 2V(9.71 X 10- 5 m 2 /s)(5 X 60s)
hVat _ (120 W/m 2 • °C)V(9.71 X lO" 5 m 2 /s)(300 s)
k 237 W/m
hx (120 W/m 2 • °C)(0.15m)
h 2 at (hVai
k 2
237 W/m ■ °C
2
°c
0.0759
0.086
(0.086) 2 = 0.0074
Substituting and evaluating the complementary error functions from Table 4-3,
e semHnf (Jc, t) = 1 - erfc (0.44) + exp (0.0759 + 0.0074) erfc (0.44 + 0.086)
= 1 - 0.5338 + exp (0.0833) X 0.457
= 0.963
Now we apply the product solution to get
T(jc, 0, t) - T-
semi-infinite
cylinder
6 S cmMnf(*> O9cyi(0, O = 0.963 X 0.762 = 0.734
237
CHAPTER 4
I
Tf = 200°C
I
D = 20cm
r« = i5°c
h =120W/m 2 -°C
x = 15 cm
FIGURE 4-29
Schematic for Example 4-9.
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238
HEAT TRANSFER
and
T(x,0,t) =
= T„ +
0.734(7;
-rj
= 15 + 0.734(200
" 15) =
151
°C
whic
*\ is the tem
perature at the
center
of the cylinder 15
cm from
the
exposed
botto
m surface.
5 C F
35°F
Steak
1 in
FIGURE 4-30
Schematic for Example 4-10.
EXAMPLE 4-10 Refrigerating Steaks while Avoiding Frostbite
In a meat processing plant, 1-in. -thick steaks initially at 75°F are to be cooled
in the racks of a large refrigerator that is maintained at 5°F (Fig. 4-30). The
steaks are placed close to each other, so that heat transfer from the 1-in. -thick
edges is negligible. The entire steak is to be cooled below 45°F, but its temper-
ature is not to drop below 35°F at any point during refrigeration to avoid "frost-
bite." The convection heat transfer coefficient and thus the rate of heat transfer
from the steak can be controlled by varying the speed of a circulating fan in-
side. Determine the heat transfer coefficient h that will enable us to meet both
temperature constraints while keeping the refrigeration time to a minimum. The
steak can be treated as a homogeneous layer having the properties p = 74.9
lbm/ft 3 , C p = 0.98 Btu/lbm • °F, k = 0.26 Btu/h ■ ft • °F, and a = 0.0035 ft 2 /h.
SOLUTION Steaks are to be cooled in a refrigerator maintained at 5 C F. The
heat transfer coefficient that will allow cooling the steaks below 45°F while
avoiding frostbite is to be determined.
Assumptions 1 Heat conduction through the steaks is one-dimensional since
the steaks form a large layer relative to their thickness and there is thermal sym-
metry about the center plane. 2 The thermal properties of the steaks and the
heat transfer coefficient are constant. 3 The Fourier number is t > 0.2 so that
the one-term approximate solutions are applicable.
Properties The properties of the steaks are as given in the problem statement.
Analysis The lowest temperature in the steak will occur at the surfaces and
the highest temperature at the center at a given time, since the inner part will
be the last place to be cooled. In the limiting case, the surface temperature at
x = L = 0.5 in. from the center will be 35°F, while the midplane temperature
is 45°F in an environment at 5°F. Then, from Fig. 4-136, we obtain
1.5
x _ 0.5 in. _
L ~ 0.5 in. ~
T(L,t)-T„ 35-5
' - M - 75
T -T m 45-5 U -°J
■ 1 _ k
Bi hL
which gives
1 k _ 0.26 Btu/h • ft • °F _
1.5 L 1.5(0.5/12 ft)
4.16 Btu/h
ft 2 • °F
Discussion The convection heat transfer coefficient should be kept below this
value to satisfy the constraints on the temperature of the steak during refriger-
ation. We can also meet the constraints by using a lower heat transfer coeffi-
cient, but doing so would extend the refrigeration time unnecessarily.
cen58933_ch04.qxd 9/10/2002 9:13 AM Page 239
The restrictions that are inherent in the use of Heisler charts and the one-
term solutions (or any other analytical solutions) can be lifted by using the nu-
merical methods discussed in Chapter 5.
239
CHAPTER 4
TOPIC OF SPECIAL INTEREST
Refrigeration and Freezing of Foods
Control of Microorganisms in Foods
Microorganisms such as bacteria, yeasts, molds, and viruses are widely
encountered in air, water, soil, living organisms, and unprocessed food
items, and cause off-flavors and odors, slime production, changes in the
texture and appearances, and the eventual spoilage of foods. Holding per-
ishable foods at warm temperatures is the primary cause of spoilage, and
the prevention of food spoilage and the premature degradation of quality
due to microorganisms is the largest application area of refrigeration. The
first step in controlling microorganisms is to understand what they are and
the factors that affect their transmission, growth, and destruction.
Of the various kinds of microorganisms, bacteria are the prime cause for
the spoilage of foods, especially moist foods. Dry and acidic foods create
an undesirable environment for the growth of bacteria, but not for the
growth of yeasts and molds. Molds are also encountered on moist surfaces,
cheese, and spoiled foods. Specific viruses are encountered in certain ani-
mals and humans, and poor sanitation practices such as keeping processed
foods in the same area as the uncooked ones and being careless about hand-
washing can cause the contamination of food products.
When contamination occurs, the microorganisms start to adapt to the
new environmental conditions. This initial slow or no-growth period is
called the lag phase, and the shelf life of a food item is directly propor-
tional to the length of this phase (Fig. 4-31). The adaptation period is fol-
lowed by an exponential growth period during which the population of
microorganisms can double two or more times every hour under favorable
conditions unless drastic sanitation measures are taken. The depletion of
nutrients and the accumulation of toxins slow down the growth and start
the death period.
The rate of growth of microorganisms in a food item depends on the
characteristics of the food itself such as the chemical structure, pH level,
presence of inhibitors and competing microorganisms, and water activity as
well as the environmental conditions such as the temperature and relative
humidity of the environment and the air motion (Fig. 4-32).
Microorganisms need food to grow and multiply, and their nutritional
needs are readily provided by the carbohydrates, proteins, minerals, and
vitamins in a food. Different types of microorganisms have different nu-
tritional needs, and the types of nutrients in a food determine the types of
microorganisms that may dwell on them. The preservatives added to the
*This section can be skipped without a loss of continuity.
Microorganism
population
Time
FIGURE 4-31
Typical growth curve of
microorganisms.
ENVIRONMENT
Temperature
Air motion
100
Oxygen
level
Relative
humidity
Water content
Chemical composition
Contamination level
The use of inhibitors
pH level
FIGURE 4-32
The factors that affect the rate of
growth of microorganisms.
cen58933_ch04.qxd 9/10/2002 9:13 AM Page 24C
240
HEAT TRANSFER
Rate of
growth
Temperature
FIGURE 4-33
The rate of growth of microorganisms
in a food product increases
exponentially with increasing
environmental temperature.
food may also inhibit the growth of certain microorganisms. Different
kinds of microorganisms that exist compete for the same food supply, and
thus the composition of microorganisms in a food at any time depends on
the initial make-up of the microorganisms.
All living organisms need water to grow, and microorganisms cannot
grow in foods that are not sufficiently moist. Microbiological growth in
refrigerated foods such as fresh fruits, vegetables, and meats starts at the
exposed surfaces where contamination is most likely to occur. Fresh meat
in a package left in a room will spoil quickly, as you may have noticed.
A meat carcass hung in a controlled environment, on the other hand, will
age healthily as a result of dehydration on the outer surface, which inhibits
microbiological growth there and protects the carcass.
Microorganism growth in a food item is governed by the combined ef-
fects of the characteristics of the food and the environmental factors. We
cannot do much about the characteristics of the food, but we certainly can
alter the environmental conditions to more desirable levels through heat-
ing, cooling, ventilating, humidification, dehumidification, and control of
the oxygen levels. The growth rate of microorganisms in foods is a strong
function of temperature, and temperature control is the single most effec-
tive mechanism for controlling the growth rate.
Microorganisms grow best at "warm" temperatures, usually between
20 and 60°C. The growth rate declines at high temperatures, and death
occurs at still higher temperatures, usually above 70°C for most micro-
organisms. Cooling is an effective and practical way of reducing the
growth rate of microorganisms and thus extending the shelf life of perish-
able foods. A temperature of 4°C or lower is considered to be a safe re-
frigeration temperature. Sometimes a small increase in refrigeration
temperature may cause a large increase in the growth rate, and thus a
considerable decrease in shelf life of the food (Fig. 4-33). The growth
rate of some microorganisms, for example, doubles for each 3°C rise in
temperature.
Another factor that affects microbiological growth and transmission is
the relative humidity of the environment, which is a measure of the water
content of the air. High humidity in cold rooms should be avoided since
condensation that forms on the walls and ceiling creates the proper envi-
ronment for mold growth and buildups. The drip of contaminated conden-
sate onto food products in the room poses a potential health hazard.
Different microorganisms react differently to the presence of oxygen in
the environment. Some microorganisms such as molds require oxygen for
growth, while some others cannot grow in the presence of oxygen. Some
grow best in low-oxygen environments, while others grow in environments
regardless of the amount of oxygen. Therefore, the growth of certain
microorganisms can be controlled by controlling the amount of oxygen in
the environment. For example, vacuum packaging inhibits the growth of
microorganisms that require oxygen. Also, the storage life of some fruits
can be extended by reducing the oxygen level in the storage room.
Microorganisms in food products can be controlled by (1) preventing
contamination by following strict sanitation practices, (2) inhibiting growth
by altering the environmental conditions, and (3) destroying the organisms
by heat treatment or chemicals. The best way to minimize contamination
cen58933_ch04.qxd 9/10/2002 9:13 AM Page 241
241
CHAPTER 4
in food processing areas is to use fine air filters in ventilation systems to
capture the dust particles that transport the bacteria in the air. Of course,
the filters must remain dry since microorganisms can grow in wet filters.
Also, the ventilation system must maintain a positive pressure in the food
processing areas to prevent any airborne contaminants from entering inside
by infiltration. The elimination of condensation on the walls and the ceil-
ing of the facility and the diversion of plumbing condensation drip pans of
refrigerators to the drain system are two other preventive measures against
contamination. Drip systems must be cleaned regularly to prevent micro-
biological growth in them. Also, any contact between raw and cooked food
products should be minimized, and cooked products must be stored in
rooms with positive pressures. Frozen foods must be kept at — 18°C or be-
low, and utmost care should be exercised when food products are packaged
after they are frozen to avoid contamination during packaging.
The growth of microorganisms is best controlled by keeping the temper-
ature and relative humidity of the environment in the desirable range.
Keeping the relative humidity below 60 percent, for example, prevents the
growth of all microorganisms on the surfaces. Microorganisms can be de-
stroyed by heating the food product to high temperatures (usually above
70°C), by treating them with chemicals, or by exposing them to ultraviolet
light or solar radiation.
Distinction should be made between survival and growth of micro-
organisms. A particular microorganism that may not grow at some low tem-
perature may be able to survive at that temperature for a very long time
(Fig. 4-34). Therefore, freezing is not an effective way of killing micro-
organisms. In fact, some microorganism cultures are preserved by freezing
them at very low temperatures. The rate of freezing is also an important
consideration in the refrigeration of foods since some microorganisms
adapt to low temperatures and grow at those temperatures when the cool-
ing rate is very low.
Refrigeration and Freezing of Foods
The storage life of fresh perishable foods such as meats, fish, vegetables,
and fruits can be extended by several days by storing them at temperatures
just above freezing, usually between 1 and 4°C. The storage life of foods
can be extended by several months by freezing and storing them at sub-
freezing temperatures, usually between — 18 and — 35°C, depending on the
particular food (Fig. 4-35).
Refrigeration slows down the chemical and biological processes in foods,
and the accompanying deterioration and loss of quality and nutrients.
Sweet corn, for example, may lose half of its initial sugar content in one
day at 21°C, but only 5 percent of it at 0°C. Fresh asparagus may lose
50 percent of its vitamin C content in one day at 20°C, but in 12 days at
0°C. Refrigeration also extends the shelf life of products. The first appear-
ance of unsightly yellowing of broccoli, for example, may be delayed by
three or more days by refrigeration.
Early attempts to freeze food items resulted in poor-quality products
because of the large ice crystals that formed. It was determined that the rate
of freezing has a major effect on the size of ice crystals and the quality,
texture, and nutritional and sensory properties of many foods. During slow
Z z
Frozen
FIGURE 4-34
Freezing may stop the growth of
microorganisms, but it may not
necessarily kill them.
Freezer
-18to-35°C
o
Frozen
foods
'-'
Refrigerator
1 to 4°C
ft
y
Fresh
foods
FIGURE 4-35
Recommended refrigeration and
freezing temperatures for
most perishable foods.
cen58933_ch04.qxd 9/10/2002 9:13 AM Page 242
242
HEAT TRANSFER
Temperature
End of
freezing
Time
FIGURE 4-36
Typical freezing curve of a food item.
TABLE 4-5
Thermal properties
of beef
Quantity
Typical value
Average density
1070 kg/m 3
Specific heat:
Above freezing
3.14 kJ/kg • °C
Below freezing
1.70 kJ/kg • °C
Freezing point
-2.7°C
Latent heat of fusion 249 kJ/kg
Thermal
0.41 W/m • °C
conductivity
(at 6°C)
freezing, ice crystals can grow to a large size, whereas during fast freezing
a large number of ice crystals start forming at once and are much smaller in
size. Large ice crystals are not desirable since they can puncture the walls
of the cells, causing a degradation of texture and a loss of natural juices
during thawing. A crust forms rapidly on the outer layer of the product and
seals in the juices, aromatics, and flavoring agents. The product quality is
also affected adversely by temperature fluctuations of the storage room.
The ordinary refrigeration of foods involves cooling only without any
phase change. The freezing of foods, on the other hand, involves three
stages: cooling to the freezing point (removing the sensible heat), freezing
(removing the latent heat), and further cooling to the desired subfreezing
temperature (removing the sensible heat of frozen food), as shown in Fig-
ure 4-36.
Beef Products
Meat carcasses in slaughterhouses should be cooled as fast as possible to a
uniform temperature of about 1.7°C to reduce the growth rate of micro-
organisms that may be present on carcass surfaces, and thus minimize
spoilage. The right level of temperature, humidity, and air motion should
be selected to prevent excessive shrinkage, toughening, and discoloration.
The deep body temperature of an animal is about 39°C, but this temper-
ature tends to rise a couple of degrees in the midsections after slaughter as
a result of the heat generated during the biological reactions that occur in
the cells. The temperature of the exposed surfaces, on the other hand, tends
to drop as a result of heat losses. The thickest part of the carcass is the
round, and the center of the round is the last place to cool during chilling.
Therefore, the cooling of the carcass can best be monitored by inserting a
thermometer deep into the central part of the round.
About 70 percent of the beef carcass is water, and the carcass is cooled
mostly by evaporative cooling as a result of moisture migration toward the
surface where evaporation occurs. But this shrinking translates into a loss
of salable mass that can amount to 2 percent of the total mass during an
overnight chilling. To prevent excessive loss of mass, carcasses are usually
washed or sprayed with water prior to cooling. With adequate care, spray
chilling can eliminate carcass cooling shrinkage almost entirely.
The average total mass of dressed beef, which is normally split into two
sides, is about 300 kg, and the average specific heat of the carcass is about
3.14 kJ/kg • °C (Table 4-5). The chilling room must have a capacity equal
to the daily kill of the slaughterhouse, which may be several hundred.
A beef carcass is washed before it enters the chilling room and absorbs a
large amount of water (about 3.6 kg) at its surface during the washing
process. This does not represent a net mass gain, however, since it is lost by
dripping or evaporation in the chilling room during cooling. Ideally, the
carcass does not lose or gain any net weight as it is cooled in the chilling
room. However, it does lose about 0.5 percent of the total mass in the hold-
ing room as it continues to cool. The actual product loss is determined by
first weighing the dry carcass before washing and then weighing it again
after it is cooled.
The refrigerated air temperature in the chilling room of beef carcasses
must be sufficiently high to avoid freezing and discoloration on the outer
cen58933_ch04.qxd 9/10/2002 9:13 AM Page 243
20 24 28 32 36 40 44 48 52 56 60 64
Time from start of chill, hours
72
243
CHAPTER 4
FIGURE 4-37
Typical cooling curve of a beef carcass
in the chilling and holding rooms at an
average temperature of 0°C (from
ASHRAE, Handbook: Refrigeration,
Ref. 3, Chap. 11, Fig. 2).
surfaces of the carcass. This means a long residence time for the massive
beef carcasses in the chilling room to cool to the desired temperature. Beef
carcasses are only partially cooled at the end of an overnight stay in the
chilling room. The temperature of a beef carcass drops to 1.7 to 7°C at the
surface and to about 15°C in mid parts of the round in 10 h. It takes another
day or two in the holding room maintained at 1 to 2°C to complete chilling
and temperature equalization. But hog carcasses are fully chilled during
that period because of their smaller size. The air circulation in the holding
room is kept at minimum levels to avoid excessive moisture loss and dis-
coloration. The refrigeration load of the holding room is much smaller than
that of the chilling room, and thus it requires a smaller refrigeration system.
Beef carcasses intended for distant markets are shipped the day after
slaughter in refrigerated trucks, where the rest of the cooling is done. This
practice makes it possible to deliver fresh meat long distances in a timely
manner.
The variation in temperature of the beef carcass during cooling is given
in Figure 4-37. Initially, the cooling process is dominated by sensible heat
transfer. Note that the average temperature of the carcass is reduced by
about 28°C (from 36 to 8°C) in 20 h. The cooling rate of the carcass could
be increased by lowering the refrigerated air temperature and increasing
the air velocity, but such measures also increase the risk of surface freezing.
Most meats are judged on their tenderness, and the preservation of ten-
derness is an important consideration in the refrigeration and freezing of
meats. Meat consists primarily of bundles of tiny muscle fibers bundled to-
gether inside long strings of connective tissues that hold it together. The
tenderness of a certain cut of beef depends on the location of the cut, the
age, and the activity level of the animal. Cuts from the relatively inactive
mid-backbone section of the animal such as short loins, sirloin, and prime
ribs are more tender than the cuts from the active parts such as the legs and
the neck (Fig. 4-38). The more active the animal, the more the connective
tissue, and the tougher the meat. The meat of an older animal is more fla-
vorful, however, and is preferred for stewing since the toughness of the
meat does not pose a problem for moist-heat cooking such as boiling. The
Chuck
Sirloin
Brisket Flank Round
Foreshank Short plate
FIGURE 4-38
Various cuts of beef (from National
Livestock and Meat Board).
cen58933_ch04.qxd 9/10/2002 9:13 AM Page 244
244
HEAT TRANSFER
Time in days
FIGURE 4-39
Variation of tenderness of meat stored
at 2°C with time after slaughter.
Meat freezer
Air
-40 to -30°C
2.5 to 5 m/s
FIGURE 4-40
The freezing time of meat can be
reduced considerably by using low
temperature air at high velocity.
TABLE 4-6
Storage life of frozen meat products
at different storage temperatures
(from ASHRAE Handbook:
Refrigeration, Chap. 10, Table 7)
Storage Life, Months
Temperature
Product
-12°C
-18°C-23°C
Beef
4-12
6-18 12-24
Lamb
3-8
6-16 12-18
Veal
3-4
4-14 8
Pork
2-6
4-12 8-15
Chopped
beef
3-4
4-6 8
Cooked foods
2-3
2-4
protein collagen, which is the main component of the connective tissue,
softens and dissolves in hot and moist environments and gradually trans-
forms into gelatin, and tenderizes the meat.
The old saying "one should either cook an animal immediately after
slaughter or wait at least two days" has a lot of truth in it. The biomechan-
ical reactions in the muscle continue after the slaughter until the energy
supplied to the muscle to do work diminishes. The muscle then stiffens and
goes into rigor mortis. This process begins several hours after the animal is
slaughtered and continues for 12 to 36 h until an enzymatic action sets in
and tenderizes the connective tissue, as shown in Figure 4-39. It takes
about seven days to complete tenderization naturally in storage facilities
maintained at 2°C. Electrical stimulation also causes the meat to be tender.
To avoid toughness, fresh meat should not be frozen before rigor mortis has
passed.
You have probably noticed that steaks are tender and rather tasty when
they are hot but toughen as they cool. This is because the gelatin that
formed during cooking thickens as it cools, and meat loses its tenderness.
So it is no surprise that first-class restaurants serve their steak on hot thick
plates that keep the steaks warm for a long time. Also, cooking softens the
connective tissue but toughens the tender muscle fibers. Therefore, barbe-
cuing on low heat for a long time results in a tough steak.
Variety meats intended for long-term storage must be frozen rapidly to
reduce spoilage and preserve quality. Perhaps the first thought that comes
to mind to freeze meat is to place the meat packages into the freezer and
wait. But the freezing time is too long in this case, especially for large
boxes. For example, the core temperature of a 4-cm-deep box containing
32 kg of variety meat can be as high as 16°C 24 h after it is placed into a
— 30°C freezer. The freezing time of large boxes can be shortened consid-
erably by adding some dry ice into it.
A more effective method of freezing, called quick chilling, involves the
use of lower air temperatures, —40 to — 30°C, with higher velocities of
2.5 m/s to 5 m/s over the product (Fig. 4-40). The internal temperature
should be lowered to — 4°C for products to be transferred to a storage
freezer and to — 18°C for products to be shipped immediately. The rate of
freezing depends on the package material and its insulating properties, the
thickness of the largest box, the type of meat, and the capacity of the re-
frigeration system. Note that the air temperature will rise excessively dur-
ing initial stages of freezing and increase the freezing time if the capacity
of the system is inadequate. A smaller refrigeration system will be adequate
if dry ice is to be used in packages. Shrinkage during freezing varies from
about 0.5 to 1 percent.
Although the average freezing point of lean meat can be taken to be
— 2°C with a latent heat of 249 kJ/kg, it should be remembered that freez-
ing occurs over a temperature range, with most freezing occurring between
— 1 and — 4°C. Therefore, cooling the meat through this temperature range
and removing the latent heat takes the most time during freezing.
Meat can be kept at an internal temperature of —2 to — 1°C for local use
and storage for under a week. Meat must be frozen and stored at much
lower temperatures for long-term storage. The lower the storage tempera-
ture, the longer the storage life of meat products, as shown in Table 4-6.
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CHAPTER 4
The internal temperature of carcasses entering the cooling sections
varies from 38 to 41°C for hogs and from 37 to 39°C for lambs and calves.
It takes about 15 h to cool the hogs and calves to the recommended tem-
perature of 3 to 4°C. The cooling-room temperature is maintained at — 1 to
0°C and the temperature difference between the refrigerant and the cooling
air is kept at about 6°C. Air is circulated at a rate of about 7 to 12 air
changes per hour. Lamb carcasses are cooled to an internal temperature of
1 to 2°C, which takes about 12 to 14 h, and are held at that temperature
with 85 to 90 percent relative humidity until shipped or processed. The rec-
ommended rate of air circulation is 50 to 60 air changes per hour during
the first 4 to 6 h, which is reduced to 10 to 12 changes per hour afterward.
Freezing does not seem to affect the flavor of meat much, but it affects
the quality in several ways. The rate and temperature of freezing may in-
fluence color, tenderness, and drip. Rapid freezing increases tenderness and
reduces the tissue damage and the amount of drip after thawing. Storage at
low freezing temperatures causes significant changes in animal fat. Frozen
pork experiences more undesirable changes during storage because of its
fat structure, and thus its acceptable storage period is shorter than that of
beef, veal, or lamb.
Meat storage facilities usually have a refrigerated shipping dock where
the orders are assembled and shipped out. Such docks save valuable stor-
age space from being used for shipping purposes and provide a more ac-
ceptable working environment for the employees. Packing plants that ship
whole or half carcasses in bulk quantities may not need a shipping dock; a
load-out door is often adequate for such cases.
A refrigerated shipping dock, as shown in Figure 4-41, reduces the re-
frigeration load of freezers or coolers and prevents temperature fluctua-
tions in the storage area. It is often adequate to maintain the shipping docks
at 4 to 7°C for the coolers and about 1.5°C for the freezers. The dew point
of the dock air should be below the product temperature to avoid conden-
sation on the surface of the products and loss of quality. The rate of airflow
through the loading doors and other openings is proportional to the square
root of the temperature difference, and thus reducing the temperature dif-
ference at the opening by half by keeping the shipping dock at the average
temperature reduces the rate of airflow into the dock and thus into the
freezer by 1 — \/03 = 0.3, or 30 percent. Also, the air that flows into
the freezer is already cooled to about 1 .5°C by the refrigeration unit of the
dock, which represents about 50 percent of the cooling load of the in-
coming air. Thus, the net effect of the refrigerated shipping dock is a
reduction of the infiltration load of the freezer by about 65 percent since
1 — 0.7 X 0.5 = 0.65. The net gain is equal to the difference between the
reduction of the infiltration load of the freezer and the refrigeration load of
the shipping dock. Note that the dock refrigerators operate at much higher
temperatures (1.5°C instead of about — 23°C), and thus they consume
much less power for the same amount of cooling.
Freezer
-23°C
Refrigerated
dock
1.5°C
| Sliding
door
Refrigerated
truck
FIGURE 4-41
A refrigerated truck dock for loading
frozen items to a refrigerated truck.
Poultry Products
Poultry products can be preserved by ice-chilling to 1 to 2°C or deep chill-
ing to about — 2°C for short-term storage, or by freezing them to — 18°C or
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HEAT TRANSFER
FIGURE 4-42
Air chilling causes dehydration and
thus weight loss for poultry, whereas
immersion chilling causes a weight
gain as a result of water absorption.
below for long-term storage. Poultry processing plants are completely
automated, and the small size of the birds makes continuous conveyor line
operation feasible.
The birds are first electrically stunned before cutting to prevent strug-
gling. Following a 90- to 120-s bleeding time, the birds are scalded by
immersing them into a tank of warm water, usually at 51 to 55°C, for up
to 120 s to loosen the feathers. Then the feathers are removed by feather-
picking machines, and the eviscerated carcass is washed thoroughly before
chilling. The internal temperature of the birds ranges from 24 to 35°C after
washing, depending on the temperatures of the ambient air and the wash-
ing water as well as the extent of washing.
To control the microbial growth, the USDA regulations require that poul-
try be chilled to 4°C or below in less than 4 h for carcasses of less than
1.8 kg, in less than 6 h for carcasses of 1.8 to 3.6 kg. and in less than 8 h for
carcasses more than 3.6 kg. Meeting these requirements today is not diffi-
cult since the slow air chilling is largely replaced by the rapid immersion
chilling in tanks of slush ice. Immersion chilling has the added benefit that
it not only prevents dehydration, but it causes a net absorption of water and
thus increases the mass of salable product. Cool air chilling of unpacked
poultry can cause a moisture loss of 1 to 2 percent, while water immersion
chilling can cause a moisture absorption of 4 to 15 percent (Fig. 4-42).
Water spray chilling can cause a moisture absorption of up to 4 percent.
Most water absorbed is held between the flesh and the skin and the
connective tissues in the skin. In immersion chilling, some soluble solids
are lost from the carcass to the water, but the loss has no significant effect
on flavor.
Many slush ice tank chillers today are replaced by continuous flow-type
immersion slush ice chillers. Continuous slush ice-chillers can reduce the
internal temperature of poultry from 32 to 4°C in about 30 minutes at a rate
up to 10, 000 birds per hour. Ice requirements depend on the inlet and exit
temperatures of the carcass and the water, but 0.25 kg of ice per kg of car-
cass is usually adequate. However, bacterial contamination such as salmo-
nella remains a concern with this method, and it may be necessary to
chloride the water to control contamination.
Tenderness is an important consideration for poultry products just as it is
for red meat, and preserving tenderness is an important consideration in the
cooling and freezing of poultry. Birds cooked or frozen before passing
through rigor mortis remain very tough. Natural tenderization begins soon
after slaughter and is completed within 24 h when birds are held at 4°C.
Tenderization is rapid during the first three hours and slows down there-
after. Immersion in hot water and cutting into the muscle adversely affect
tenderization. Increasing the scalding temperature or the scalding time has
been observed to increase toughness, and decreasing the scalding time has
been observed to increase tenderness. The beating action of mechanical
feather-picking machines causes considerable toughening. Therefore, it is
recommended that any cutting be done after tenderization. Cutting up the
bird into pieces before natural tenderization is completed reduces tender-
ness considerably. Therefore, it is recommended that any cutting be done
after tenderization. Rapid chilling of poultry can also have a toughening
cen58933_ch04.qxd 9/10/2002 9:13 AM Page 247
247
CHAPTER 4
effect. It is found that the tenderization process can be speeded up consid-
erably by a patented electrical stunning process.
Poultry products are highly perishable, and thus they should be kept at
the lowest possible temperature to maximize their shelf life. Studies have
shown that the populations of certain bacteria double every 36 h at — 2°C,
14 h at 0°C, 7 h at 5°C, and less than 1 h at 25°C (Fig. 4-43). Studies have
also shown that the total bacterial counts on birds held at 2°C for 14 days
are equivalent to those held at 10°C for 5 days or 24°C for 1 day. It has also
been found that birds held at — 1°C had 8 days of additional shelf life over
those held at 4°C.
The growth of microorganisms on the surfaces of the poultry causes the
development of an off-odor and bacterial slime. The higher the initial
amount of bacterial contamination, the faster the sliming occurs. Therefore,
good sanitation practices during processing such as cleaning the equipment
frequently and washing the carcasses are as important as the storage tem-
perature in extending shelf life.
Poultry must be frozen rapidly to ensure a light, pleasing appearance.
Poultry that is frozen slowly appears dark and develops large ice crystals
that damage the tissue. The ice crystals formed during rapid freezing are
small. Delaying freezing of poultry causes the ice crystals to become larger.
Rapid freezing can be accomplished by forced air at temperatures of —23
to — 40°C and velocities of 1.5 to 5 m/s in air-blast tunnel freezers. Most
poultry is frozen this way. Also, the packaged birds freeze much faster on
open shelves than they do in boxes. If poultry packages must be frozen in
boxes, then it is very desirable to leave the boxes open or to cut holes on
the boxes in the direction of airflow during freezing. For best results, the
blast tunnel should be fully loaded across its cross-section with even spac-
ing between the products to assure uniform airflow around all sides of the
packages. The freezing time of poultry as a function of refrigerated air tem-
perature is given in Figure 4-44. Thermal properties of poultry are given in
Table 4-7.
Other freezing methods for poultry include sandwiching between cold
plates, immersion into a refrigerated liquid such as glycol or calcium chlo-
ride brine, and cryogenic cooling with liquid nitrogen. Poultry can be
frozen in several hours by cold plates. Very high freezing rates can be ob-
tained by immersing the packaged birds into a low-temperature brine. The
freezing time of birds in — 29°C brine can be as low as 20 min, depending
on the size of the bird (Fig. 4-45). Also, immersion freezing produces a
very appealing light appearance, and the high rates of heat transfer make
continuous line operation feasible. It also has lower initial and maintenance
costs than forced air, but leaks into the packages through some small holes
or cracks remain a concern. The convection heat transfer coefficient is 17
W/m 2 • °C for air at -29°C and 2.5 m/s whereas it is 170 W/m 2 • °C for
sodium chloride brine at — 18°C and a velocity of 0.02 m/s. Sometimes liq-
uid nitrogen is used to crust freeze the poultry products to — 73°C. The
freezing is then completed with air in a holding room at — 23°C.
Properly packaged poultry products can be stored frozen for up to about
a year at temperatures of — 18°C or lower. The storage life drops consider-
ably at higher (but still below-freezing) temperatures. Significant changes
Storage life (days)
15 20 25
Storage temperature, °C
FIGURE 4-43
The storage life of fresh poultry
decreases exponentially with
increasing storage temperature.
7
on
3 6
o
v 5
<D -1
D
Giblets
■ Inside surface
o 1 3 ram depth
• Under skin
□ /
S 11/
o
9
o —
• —
-• — f
-84 -73 -62 -51 -40 -29 -18 -7
Air temperature, degrees Celsius
Note: Freezing time is the time required for
temperature to fall from to — 4°C. The values
are for 2.3 to 3.6 kg chickens with initial
temperature of to 2°C and with air velocity
of 2.3 to 2.8 m/s.
FIGURE 4-44
The variation of freezing time of
poultry with air temperature (from
van der Berg and Lentz, Ref. 11).
cen58933_ch04.qxd 9/10/2002 9:13 AM Page 24E
248
HEAT TRANSFER
FIGURE 4-45
The variation of temperature of the
breast of 6. 8 -kg turkeys initially at
1°C with depth during immersion
cooling at — 29°C (from van der Berg
and Lentz, Ref. 11).
Giblets
Inside surface
38 mm depth
25 mm depth
j— 13 mm depth
\\6.5 mm depth
\ Under skin
Skin surface
100 125 150
Time, min.
250
TABLE 4-7
Thermal properties of poultry
Quantity
Typical value
Average density:
Muscle
Skin
Specific heat:
Above freezing
Below freezing
Freezing point
Latent heat of fusion
1070 kg/m 3
1030 kg/m 3
2.94 kJ/kg • °C
1.55 kJ/kg • °C
-2.8°C
247 kJ/kg
Thermal conductivity: (in W/m • °C)
Breast muscle 0.502 at 20°C
1.384 at -20°C
1.506 at -40°C
Dark muscle 1.557 at -40°C
occur in flavor and juiciness when poultry is frozen for too long, and a stale
rancid odor develops. Frozen poultry may become dehydrated and experi-
ence freezer burn, which may reduce the eye appeal of the product and
cause toughening of the affected area. Dehydration and thus freezer burn
can be controlled by humidification, lowering the storage temperature, and
packaging the product in essentially impermeable film. The storage life can
be extended by packing the poultry in an oxygen-free environment. The
bacterial counts in precooked frozen products must be kept at safe levels
since bacteria may not be destroyed completely during the reheating
process at home.
Frozen poultry can be thawed in ambient air, water, refrigerator, or oven
without any significant difference in taste. Big birds like turkey should be
thawed safely by holding it in a refrigerator at 2 to 4°C for two to four days,
depending on the size of the bird. They can also be thawed by immersing
them into cool water in a large container for 4 to 6 h, or holding them in a
paper bag. Care must be exercised to keep the bird's surface cool to mini-
mize microbiological growth when thawing in air or water.
H
EXAMPLE 4-5 Chilling of Beef Carcasses in a Meat Plant
HThe chilling room of a meat plant is 18 m X 20 m X 5.5 m in size and has a
capacity of 450 beef carcasses. The power consumed by the fans and the lights
of the chilling room are 26 and 3 kW, respectively, and the room gains heat
through its envelope at a rate of 13 kW. The average mass of beef carcasses is
285 kg. The carcasses enter the chilling room at 36°C after they are washed to
facilitate evaporative cooling and are cooled to 15°C in 10 h. The water is ex-
pected to evaporate at a rate of 0.080 kg/s. The air enters the evaporator sec-
tion of the refrigeration system at 0.7°C and leaves at -2°C. The air side of
the evaporator is heavily finned, and the overall heat transfer coefficient of the
evaporator based on the air side is 20 W/m 2 • °C. Also, the average temperature
■ difference between the air and the refrigerant in the evaporator is 5.5°C.
cen58933_ch04.qxd 9/10/2002 9:13 AM Page 249
Determine (a) the refrigeration load of the chilling room, (b) the volume flow
rate of air, and (c) the heat transfer surface area of the evaporator on the air
side, assuming all the vapor and the fog in the air freezes in the evaporator.
SOLUTION The chilling room of a meat plant with a capacity of 450 beef car-
casses is considered. The cooling load, the airflow rate, and the heat transfer
area of the evaporator are to be determined.
Assumptions 1 Water evaporates at a rate of 0.080 kg/s. 2 All the moisture in
the air freezes in the evaporator.
Properties The heat of fusion and the heat of vaporization of water at 0°C are
333.7 kJ/kg and 2501 kJ/kg (Table A-9). The density and specific heat of air at
0°C are 1.292 kg/m 3 and 1.006 kJ/kg ■ °C (Table A-15). Also, the specific heat
of beef carcass is determined from the relation in Table A-7b to be
Analysis (a) A sketch of the chilling room is given in Figure 4-46. The amount
of beef mass that needs to be cooled per unit time is
m beef = (Total beef mass cooled)/(Cooling time)
= (450 carcasses)(285 kg/carcass)/(10 X 3600 s) = 3.56 kg/s
The product refrigeration load can be viewed as the energy that needs to be
removed from the beef carcass as it is cooled from 36 to 15°C at a rate of
3.56 kg/s and is determined to be
Gb
(mCAT) h
(3.56 kg/s)(3.14 kJ/kg • °C)(36 - 15)°C = 235 kW
Then the total refrigeration load of the chilling room becomes
G total, chillroom = G beef + G fan + G lights + G heat gain = 235 + 26 + 3 + 13 = 277 kW
The amount of carcass cooling due to evaporative cooling of water is
G beef, evaporative = («^)water = (0-080 kg/ S )(2490 kJ/kg) = 199 kW
which is 199/235 = 85 percent of the total product cooling load. The remain-
ing 15 percent of the heat is transferred by convection and radiation.
(£>) Heat is transferred to air at the rate determined above, and the tempera-
ture of the air rises from -2°C to 0.7°C as a result. Therefore, the mass flow
rate of air is
Ga
277 kW
(C„A7/ air ) (1.006 kJ/kg • °C)[0.7 - (-2)°C]
102.0 kg/s
Then the volume flow rate of air becomes
h ml 102 kg/s
V
Pa.r 1.292 kg/m 3
78.9 m 3 /s
(c) Normally the heat transfer load of the evaporator is the same as the refriger-
ation load. But in this case the water that enters the evaporator as a liquid is
249
CHAPTER 4
L~
H
Lights, 3 kW
13 kW
Fans, 26 kW
Evaporation
0.080 kg/s
Refrigerated
air
^tlllllllt
*-*- 0.7°C
Evaporator
f
-2°C
J
t-evap
FIGURE 4-46
Schematic for Example 4-5.
cen58933_ch04.qxd 9/10/2002 9:13 AM Page 25C
250
HEAT TRANSFER
frozen as the temperature drops to -2°C, and the evaporator must also remove
the latent heat of freezing, which is determined from
G freezing = (* ^ a .e„,)„ater = (0-080 kg/ S )(334 kj/kg) = 27 kW
Therefore, the total rate of heat removal at the evaporator is
G evaporator = G total, chill room "•" Gfreezing = 277 + 27 = 304 kW
Then the heat transfer surface area of the evaporator on the air side is deter-
mined from Q evaporator = (t//l) airside A7",
Ge
304,000 W
UAT (20 W/m 2 ■ °C)(5.5°C)
2764 m 2
Obviously, a finned surface must be used to provide such a large surface area
on the air side.
SUMMARY
In this chapter we considered the variation of temperature with
time as well as position in one- or multidimensional systems.
We first considered the lumped systems in which the tempera-
ture varies with time but remains uniform throughout the sys-
tem at any time. The temperature of a lumped body of arbitrary
shape of mass m, volume V, surface area A s , density p, and
specific heat C p initially at a uniform temperature T t that is
exposed to convection at time t = in a medium at tempera-
ture T„ with a heat transfer coefficient h is expressed as
T(t) ~ T x _ ki
where
h
hA,
pC p V pC p L c
(1/s)
is a positive quantity whose dimension is (time) -1 . This rela-
tion can be used to determine the temperature T(t) of a body at
time t or, alternately, the time t required for the temperature to
reach a specified value T{t). Once the temperature Tit) at time
t is available, the rate of convection heat transfer between the
body and its environment at that time can be determined from
Newton's law of cooling as
Q(t) = hA s [T(t)-T x ]
(W)
The total amount of heat transfer between the body and the sur-
rounding medium over the time interval t = to / is simply the
change in the energy content of the body,
Q = mCJT(t) - Ti\
(kJ)
The amount of heat transfer reaches its upper limit when the
body reaches the surrounding temperature 7^. Therefore, the
maximum heat transfer between the body and its surround-
ings is
Gma* = mC (T a - T { )
(kJ)
The error involved in lumped system analysis is negligible
when
hL c
Bi = — -<0.1
k
where Bi is the Biot number and L c = VIA S is the characteristic
length.
When the lumped system analysis is not applicable, the vari-
ation of temperature with position as well as time can be deter-
mined using the transient temperature charts given in Figs.
4-13, 4-14, 4-15, and 4-23 for a large plane wall, a long cylin-
der, a sphere, and a semi-infinite medium, respectively. These
charts are applicable for one-dimensional heat transfer in those
geometries. Therefore, their use is limited to situations in
which the body is initially at a uniform temperature, all sur-
faces are subjected to the same thermal conditions, and the
body does not involve any heat generation. These charts can
also be used to determine the total heat transfer from the body
up to a specified time /.
cen58933_ch04.qxd 9/10/2002 9:13 AM Page 251
Using a one-term approximation, the solutions of one-
dimensional transient heat conduction problems are expressed
analytically as
Plane wall:
Cylinder:
Sphere:
Q(x, f) wa n
T(x, t)
6(r, 0,
cyl
8(r, 0,
sph
2
A,e" XlT cos (XjX/L),
t>0.2
T(r, t) - T„
T - T
2
Ate-m JfartrJ,
t>0.2
T(r, t) - r«
r, - r.
, 2 sin^r/O
\ x rlr„
t>0.2
where the constants A l and \ { are functions of the Bi number
only, and their values are listed in Table 4-1 against the Bi
number for all three geometries. The error involved in one-
term solutions is less than 2 percent when t > 0.2.
Using the one-term solutions, the fractional heat transfers in
different geometries are expressed as
Plane wall:
Cylinder:
Sphere:
Q
siniax
_Q_
Q
tCmax
cy I
spli
1
1 -26,
1 -36,
sin \,
/i(Xi)
O.cyl X)
sin \, — X, cos X,
0, sph
K
The analytic solution for one-dimensional transient heat
conduction in a semi-infinite solid subjected to convection is
given by
T(x, t)
erfc
erfc
hx , h 2 at
2V^/ exp U + ^
2\fat
+
h\/ai
251
CHAPTER 4
where the quantity erfc (£) is the complementary error func-
tion. For the special case of /i — > =°, the surface temperature T s
becomes equal to the fluid temperature T m and the above equa-
tion reduces to
T(x, t)
erfc
(T s = constant)
Using a clever superposition principle called the product so-
lution these charts can also be used to construct solutions for
the two-dimensional transient heat conduction problems en-
countered in geometries such as a short cylinder, a long rectan-
gular bar, or a semi-infinite cylinder or plate, and even
three-dimensional problems associated with geometries such
as a rectangular prism or a semi-infinite rectangular bar, pro-
vided that all surfaces of the solid are subjected to convection
to the same fluid at temperature T m with the same convection
heat transfer coefficient h, and the body involves no heat
generation. The solution in such multidimensional geometries
can be expressed as the product of the solutions for the
one-dimensional geometries whose intersection is the multi-
dimensional geometry.
The total heat transfer to or from a multidimensional geom-
etry can also be determined by using the one-dimensional val-
ues. The transient heat transfer for a two-dimensional geometry
formed by the intersection of two one-dimensional geometries
1 and 2 is
_Q_
Q_
!cma
_Q_
Transient heat transfer for a three-dimensional body formed by
the intersection of three one -dimensional bodies 1 , 2, and 3 is
given by
_Q_
kl- m a
+
Jima
_Q
_Q
_Q_
*>ma
_Q_
J&ma
REFERENCES AND SUGGESTED READING
1. ASHRAE. Handbook of Fundamentals. SI version.
Atlanta, GA: American Society of Heating, Refrigerating,
and Air-Conditioning Engineers, Inc., 1993.
2. ASHRAE. Handbook of Fundamentals. SI version.
Atlanta, GA: American Society of Heating, Refrigerating,
and Air-Conditioning Engineers, Inc., 1994.
3. H. S. Carslaw and J. C. Jaeger. Conduction of Heat in
Solids. 2nd ed. London: Oxford University Press, 1959.
4. H. Grober, S. Erk, and U. Grigull. Fundamentals of Heat
Transfer. New York: McGraw-Hill, 1961.
5. M. P. Heisler. "Temperature Charts for Induction and
Constant Temperature Heating." ASME Transactions 69
(1947), pp. 227-36.
6. H. Hillman. Kitchen Science. Mount Vernon, NY:
Consumers Union, 1981.
7. F. P. Incropera and D. P. DeWitt. Introduction to Heat
Transfer. 4th ed. New York: John Wiley & Sons, 2002.
cen58933_ch04.qxd 9/10/2002 9:13 AM Page 252
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HEAT TRANSFER
8. L. S. Langston. "Heat Transfer from Multidimensional
Objects Using One-Dimensional Solutions for Heat
Loss." International Journal of Heat and Mass Transfer
25(1982), pp. 149-50.
9. M. N. Ozisik, Heat Transfer — A Basic Approach. New
York: McGraw-Hill, 1985.
10. P. J. Schneider. Conduction Heat Transfer. Reading, MA:
Addison -Wesley, 1955.
11. L. van der Berg and C. P. Lentz. "Factors Affecting
Freezing Rate and Appearance of Eviscerated Poultry
Frozen in Air." Food Technology 12 (1958).
PROBLEMS
Lumped System Analysis
4-1 C What is lumped system analysis? When is it
applicable?
4-2C Consider heat transfer between two identical hot solid
bodies and the air surrounding them. The first solid is being
cooled by a fan while the second one is allowed to cool natu-
rally. For which solid is the lumped system analysis more
likely to be applicable? Why?
4-3C Consider heat transfer between two identical hot solid
bodies and their environments. The first solid is dropped in a
large container filled with water, while the second one is al-
lowed to cool naturally in the air. For which solid is the lumped
system analysis more likely to be applicable? Why?
4-4C Consider a hot baked potato on a plate. The tempera-
ture of the potato is observed to drop by 4°C during the first
minute. Will the temperature drop during the second minute be
less than, equal to, or more than 4°C? Why?
Hot
baked
potato
FIGURE P4-4C
4-5C Consider a potato being baked in an oven that is main-
tained at a constant temperature. The temperature of the potato
is observed to rise by 5°C during the first minute. Will the tem-
perature rise during the second minute be less than, equal to, or
more than 5°C? Why?
4-6C What is the physical significance of the Biot number?
Is the Biot number more likely to be larger for highly conduct-
ing solids or poorly conducting ones?
*Problems designated by a "C" are concept questions, and
students are encouraged to answer them all. Problems designated
by an "E" are in English units, and the SI users can ignore them.
Problems with an EES-CD icon ® are solved using EES, and
complete solutions together with parametric studies are included
on the enclosed CD. Problems with a computer-EES icon H are
comprehensive in nature, and are intended to be solved with a
computer, preferably using the EES software that accompanies
this text.
4-7C Consider two identical 4-kg pieces of roast beef. The
first piece is baked as a whole, while the second is baked after
being cut into two equal pieces in the same oven. Will there be
any difference between the cooking times of the whole and cut
roasts? Why?
4-8C Consider a sphere and a cylinder of equal volume
made of copper. Both the sphere and the cylinder are initially at
the same temperature and are exposed to convection in the
same environment. Which do you think will cool faster, the
cylinder or the sphere? Why?
4-9C In what medium is the lumped system analysis more
likely to be applicable: in water or in air? Why?
4-10C For which solid is the lumped system analysis more
likely to be applicable: an actual apple or a golden apple of the
same size? Why?
4-1 1C For which kind of bodies made of the same material
is the lumped system analysis more likely to be applicable:
slender ones or well-rounded ones of the same volume? Why?
4-12 Obtain relations for the characteristic lengths of a large
plane wall of thickness 2L, a very long cylinder of radius r ,
and a sphere of radius r .
4-13 Obtain a relation for the time required for a lumped
system to reach the average temperature | (T; + TJ), where
Tj is the initial temperature and T a is the temperature of the
environment.
4-14 The temperature of a gas stream is to be measured by
a thermocouple whose junction can be approximated as a
1 .2-mm-diameter sphere. The properties of the junction are
k = 35 W/m • °C, p = 8500 kg/m 3 , and C p = 320 J/kg ■ °C, and
the heat transfer coefficient between the junction and the gas
is h = 65 W/m 2 ■ °C. Determine how long it will take for
the thermocouple to read 99 percent of the initial temperature
difference. Answer: 38.5 s
4-15E In a manufacturing facility, 2 -in. -diameter brass balls
(k = 64.1 Btu/h • ft • °F, p = 532 lbm/ft 3 , and C p = 0.092
Btu/lbm • °F) initially at 250°F are quenched in a water bath at
120°F for a period of 2 min at a rate of 120 balls per minute.
If the convection heat transfer coefficient is 42 Btu/h • ft 2 • °F,
determine (a) the temperature of the balls after quenching and
(b) the rate at which heat needs to be removed from the water
in order to keep its temperature constant at 120°F
cen58933_ch04.qxd 9/10/2002 9:13 AM Page 253
250°F
oo<
/
Brass
balls
.OOO
> 120°F r
0000000
V
Water
bath
FIGURE P4-15E
4-16E Repeat Problem 4-1 5E for aluminum balls.
4-17 To warm up some milk for a baby, a mother pours milk
into a thin-walled glass whose diameter is 6 cm. The height of
the milk in the glass is 7 cm. She then places the glass into a
large pan filled with hot water at 60°C. The milk is stirred con-
stantly, so that its temperature is uniform at all times. If the
heat transfer coefficient between the water and the glass is
120 W/m 2 • °C, determine how long it will take for the milk to
warm up from 3°C to 38°C. Take the properties of the milk
to be the same as those of water. Can the milk in this case be
treated as a lumped system? Why? Answer: 5.8 min
4-18 Repeat Problem 4-17 for the case of water also
being stirred, so that the heat transfer coefficient is doubled to
240 W/m 2 • °C.
4-1 9E During a picnic on a hot summer day, all the cold
drinks disappeared quickly, and the only available drinks were
those at the ambient temperature of 80°F. In an effort to cool a
12-fluid-oz drink in a can, which is 5 in. high and has a diame-
ter of 2.5 in., a person grabs the can and starts shaking it in the
iced water of the chest at 32°F. The temperature of the drink
can be assumed to be uniform at all times, and the heat transfer
coefficient between the iced water and the aluminum can is
30 Btu/h • ft 2 ■ °F. Using the properties of water for the drink,
estimate how long it will take for the canned drink to cool
to 45 °F
253
CHAPTER 4
4-20 Consider a 1000-W iron whose base plate is made of
0.5-cm-thick aluminum alloy 2024-T6 (p = 2770 kg/m 3 , C p =
875 J/kg • °C, a = 7.3 X 10~ 5 m 2 /s). The base plate has a sur-
face area of 0.03 m 2 . Initially, the iron is in thermal equilibrium
with the ambient air at 22°C. Taking the heat transfer
coefficient at the surface of the base plate to be 12 W/m 2 ■ °C
and assuming 85 percent of the heat generated in the resistance
wires is transferred to the plate, determine how long it will take
for the plate temperature to reach 140°C. Is it realistic to as-
sume the plate temperature to be uniform at all times?
1000 w
iron
FIGURE P4-20
4-21
Reconsider Problem 4-20. Using EES (or other)
software, investigate the effects of the heat trans-
fer coefficient and the final plate temperature on the time it will
take for the plate to reach this temperature. Let the heat trans-
fer coefficient vary from 5 W/m 2 ■ °C to 25 W/m 2 ■ °C and the
temperature from 30°C to 200°C. Plot the time as functions of
the heat transfer coefficient and the temperature, and discuss
the results.
4-22 Stainless steel ball bearings (p = 8085 kg/m 3 , k = 15.1
W/m ■ °C, C p = 0.480 kJ/kg ■ °C, and a = 3.91 X lO" 6 m 2 /s)
having a diameter of 1.2 cm are to be quenched in water. The
balls leave the oven at a uniform temperature of 900°C and are
exposed to air at 30°C for a while before they are dropped into
the water. If the temperature of the balls is not to fall below
850°C prior to quenching and the heat transfer coefficient in
the air is 125 W/m 2 • °C, determine how long they can stand in
the air before being dropped into the water. Answer: 3.7 s
Carbon steel balls (p = 7833 kg/m 3 , k = 54 W/m ■ °C,
°C, and a = 1.474 X 10~ 6 m 2 /s) 8 mm in
4-23
C p = 0.465 kJ/kg
Furnace
-,900°C
o
Air, 35°C
, Steel ball 100 o C
OOO
FIGURE P4-1 9E
FIGURE P4-23
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254
HEAT TRANSFER
diameter are annealed by heating them first to 900°C in a fur-
nace and then allowing them to cool slowly to 100°C in am-
bient air at 35°C. If the average heat transfer coefficient is
75 W/m 2 ■ °C, determine how long the annealing process will
take. If 2500 balls are to be annealed per hour, determine the
total rate of heat transfer from the balls to the ambient air.
4-24 rSpM Reconsider Problem 4-23. Using EES (or other)
b^ti software, investigate the effect of the initial tem-
perature of the balls on the annealing time and the total rate of
heat transfer. Let the temperature vary from 500°C to 1000°C.
Plot the time and the total rate of heat transfer as a function of
the initial temperature, and discuss the results.
4-25 An electronic device dissipating 30 W has a mass of
20 g, a specific heat of 850 J/kg • °C, and a surface area of
5 cm 2 . The device is lightly used, and it is on for 5 min and
then off for several hours, during which it cools to the ambient
temperature of 25°C. Taking the heat transfer coefficient to be
12 W/m 2 • °C, determine the temperature of the device at the
end of the 5 -min operating period. What would your answer be
if the device were attached to an aluminum heat sink having a
mass of 200 g and a surface area of 80 cm 2 ? Assume the device
and the heat sink to be nearly isothermal.
Transient Heat Conduction in Large Plane Walls,
Long Cylinders, and Spheres with Spatial Effects
4-26C What is an infinitely long cylinder? When is it proper
to treat an actual cylinder as being infinitely long, and when is
it not? For example, is it proper to use this model when finding
the temperatures near the bottom or top surfaces of a cylinder?
Explain.
4-27C Can the transient temperature charts in Fig. 4-13 for
a plane wall exposed to convection on both sides be used for a
plane wall with one side exposed to convection while the other
side is insulated? Explain.
4-28C Why are the transient temperature charts prepared
using nondimensionalized quantities such as the Biot and
Fourier numbers instead of the actual variables such as thermal
conductivity and time?
4-29C What is the physical significance of the Fourier num-
ber? Will the Fourier number for a specified heat transfer prob-
lem double when the time is doubled?
4-30C How can we use the transient temperature charts
when the surface temperature of the geometry is specified in-
stead of the temperature of the surrounding medium and the
convection heat transfer coefficient?
4-31 C A body at an initial temperature of T t is brought into a
medium at a constant temperature of T„. How can you deter-
mine the maximum possible amount of heat transfer between
the body and the surrounding medium?
4-32C The Biot number during a heat transfer process be-
tween a sphere and its surroundings is determined to be 0.02.
Would you use lumped system analysis or the transient tem-
perature charts when determining the midpoint temperature of
the sphere? Why?
4-33 A student calculates that the total heat transfer from
a spherical copper ball of diameter 15 cm initially at 200°C
and its environment at a constant temperature of 25°C during
the first 20 min of cooling is 4520 kJ. Is this result reason-
able? Why?
4-34 An ordinary egg can be approximated as a 5.5-cm-
diameter sphere whose properties are roughly k = 0.6 W/m •
°C and a = 0.14 X 10~ s m 2 /s. The egg is initially at a uniform
temperature of 8°C and is dropped into boiling water at 97°C.
Taking the convection heat transfer coefficient to be h = 1400
W/m 2 ■ °C, determine how long it will take for the center of the
egg to reach 70°C.
Boiling
water y
97°C
Egg
r, = 8°c
FIGURE P4-34
4-35
Reconsider Problem 4-34. Using EES (or other)
software, investigate the effect of the final center
temperature of the egg on the time it will take for the center to
reach this temperature. Let the temperature vary from 50°C
to 95°C. Plot the time versus the temperature, and discuss the
results.
4-36 In a production facility, 3-cm-thick large brass plates
(Jfc = 110 W/m • °C, p = 8530 kg/m 3 , C p = 380 J/kg • °C, and
a = 33.9 X 10~ 6 m 2 /s) that are initially at a uniform tempera-
ture of 25°C are heated by passing them through an oven main-
tained at 700°C. The plates remain in the oven for a period of
10 min. Taking the convection heat transfer coefficient to be
h = 80 W/m 2 • °C, determine the surface temperature of the
plates when they come out of the oven.
Furnace, 700°C
3 cm
Brass plate
25°C
FIGURE P4-36
cen58933_ch04.qxd 9/10/2002 9:13 AM Page 255
4-37 ScJU Reconsider Problem 4-36. Using EES (or other)
|^£^ software, investigate the effects of the tempera-
ture of the oven and the heating time on the final surface tem-
perature of the plates. Let the oven temperature vary from
500°C to 900°C and the time from 2 min to 30 min. Plot the
surface temperature as the functions of the oven temperature
and the time, and discuss the results.
4-38 A long 3 5 -cm-diameter cylindrical shaft made of stain-
less steel 304 (k = 14.9 W/m • °C, p = 7900 kg/m 3 , C p = All
J/kg • °C, and a = 3.95 X 10~ 6 m 2 /s) comes out of an oven at
a uniform temperature of 400°C. The shaft is then allowed to
cool slowly in a chamber at 150°C with an average convection
heat transfer coefficient of h = 60 W/m 2 • °C. Determine the
temperature at the center of the shaft 20 min after the start of
the cooling process. Also, determine the heat transfer per unit
length of the shaft during this time period.
Answers: 390°C, 16,015 kJ/m
4-39 rfigM Reconsider Problem 4-38. Using EES (or other)
b^S software, investigate the effect of the cooling
time on the final center temperature of the shaft and the amount
of heat transfer. Let the time vary from 5 min to 60 min. Plot
the center temperature and the heat transfer as a function of the
time, and discuss the results.
4-40E Long cylindrical AISI stainless steel rods (k = 7.74
Btu/h ■ ft • °F and a = 0.135 ft 2 /h) of 4-in. diameter are heat-
treated by drawing them at a velocity of 10 ft/min through
a 30-ft-long oven maintained at 1700°F. The heat transfer
coefficient in the oven is 20 Btu/h ■ ft 2 • °F. If the rods enter
the oven at 85°F, determine their centerline temperature when
they leave.
Oven
1700°F
10 ft/min
.
/ r
n
f
\
Stainless steel
85°F
FIGURE P4-40E
4-41 In a meat processing plant, 2-cm-thick steaks (k = 0.45
W/m • °C and a = 0.91 X 10~ 7 m 2 /s) that are initially at 25°C
are to be cooled by passing them through a refrigeration room
at — 11°C. The heat transfer coefficient on both sides of the
steaks is 9 W/m 2 • °C. If both surfaces of the steaks are to be
cooled to 2°C, determine how long the steaks should be kept in
the refrigeration room.
4-42 A long cylindrical wood log (k = 0.17 W/m ■ °C and
a = 1 .28 X 10~ 7 m 2 /s) is 10 cm in diameter and is initially at a
uniform temperature of 10°C. It is exposed to hot gases at
255
CHAPTER 4
500°C in a fireplace with a heat transfer coefficient of 13.6
W/m 2 ■ °C on the surface. If the ignition temperature of the
wood is 420°C, determine how long it will be before the log
ignites.
4-43 In Betty Crocker 's Cookbook, it is stated that it takes 2 h
45 min to roast a 3.2-kg rib initially at 4.5°C "rare" in an oven
maintained at 163°C. It is recommended that a meat ther-
mometer be used to monitor the cooking, and the rib is consid-
ered rare done when the thermometer inserted into the center of
the thickest part of the meat registers 60°C. The rib can be
treated as a homogeneous spherical object with the properties
p = 1200 kg/m 3 , C p = 4.1 kJ/kg ■ °C, k = 0.45 W/m • °C, and
a = 0.91 X 10~ 7 m 2 /s. Determine (a) the heat transfer coeffi-
cient at the surface of the rib, (b) the temperature of the outer
surface of the rib when it is done, and (c) the amount of heat
transferred to the rib. (d) Using the values obtained, predict
how long it will take to roast this rib to "medium" level, which
occurs when the innermost temperature of the rib reaches
71°C. Compare your result to the listed value of 3 h 20 min.
If the roast rib is to be set on the counter for about 15 min
before it is sliced, it is recommended that the rib be taken
out of the oven when the thermometer registers about 4°C
below the indicated value because the rib will continue cook-
ing even after it is taken out of the oven. Do you agree with this
recommendation?
Answers: {a) 156.9 W/m 2 • °C, (« 159. 5°C, (c) 1629 kJ, (d) 3.0 h
Rib
T t = 4.5°C
FIGURE P4-43
4-44 Repeat Problem 4^-3 for a roast rib that is to be "well-
done" instead of "rare." A rib is considered to be well-done
when its center temperature reaches 77°C, and the roasting in
this case takes about 4 h 15 min.
4-45 For heat transfer purposes, an egg can be considered to
be a 5. 5-cm -diameter sphere having the properties of water. An
egg that is initially at 8°C is dropped into the boiling water at
100°C. The heat transfer coefficient at the surface of the egg is
estimated to be 800 W/m 2 • °C. If the egg is considered cooked
when its center temperature reaches 60°C, determine how long
the egg should be kept in the boiling water.
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HEAT TRANSFER
4-46 Repeat Problem 4^5 for a location at 1610-m eleva-
tion such as Denver, Colorado, where the boiling temperature
of water is 94.4°C.
4-47 The author and his 6-year-old son have conducted the
following experiment to determine the thermal conductivity of
a hot dog. They first boiled water in a large pan and measured
the temperature of the boiling water to be 94°C, which is not
surprising, since they live at an elevation of about 1650 m in
Reno, Nevada. They then took a hot dog that is 12.5 cm long
and 2.2 cm in diameter and inserted a thermocouple into the
midpoint of the hot dog and another thermocouple just under
the skin. They waited until both thermocouples read 20°C,
which is the ambient temperature. They then dropped the hot
dog into boiling water and observed the changes in both tem-
peratures. Exactly 2 min after the hot dog was dropped into the
boiling water, they recorded the center and the surface temper-
atures to be 59°C and 88°C, respectively. The density of the hot
dog can be taken to be 980 kg/m 3 , which is slightly less than
the density of water, since the hot dog was observed to be float-
ing in water while being almost completely immersed. The
specific heat of a hot dog can be taken to be 3900 J/kg • °C,
which is slightly less than that of water, since a hot dog is
mostly water. Using transient temperature charts, determine
(a) the thermal diffusivity of the hot dog, (b) the thermal con-
ductivity of the hot dog, and (c) the convection heat transfer
coefficient.
Answers: (a) 2.02 x 10" 7 m 2 /s, (b) 0.771 W/m ■ °C,
(c) 467 W/m 2 ■ °C.
Refrigerator
5°F
FIGURE P4-47
4-48 Using the data and the answers given in Problem 4^-7,
determine the center and the surface temperatures of the hot
dog 4 min after the start of the cooking. Also determine the
amount of heat transferred to the hot dog.
4-49E In a chicken processing plant, whole chickens averag-
ing 5 lb each and initially at 72 °F are to be cooled in the racks
of a large refrigerator that is maintained at 5°F. The entire
chicken is to be cooled below 45 °F, but the temperature of the
chicken is not to drop below 35°F at any point during refrig-
eration. The convection heat transfer coefficient and thus the
rate of heat transfer from the chicken can be controlled by
varying the speed of a circulating fan inside. Determine the
heat transfer coefficient that will enable us to meet both tem-
perature constraints while keeping the refrigeration time to a
FIGURE P4-49E
minimum. The chicken can be treated as a homogeneous spher-
ical object having the properties p = 74.9 lbm/ft 3 , C p = 0.98
Btu/lbm ■ °F, k = 0.26 Btu/h • ft ■ °F, and a = 0.0035 ft 2 /h.
4-50 A person puts a few apples into the freezer at — 1 5°C to
cool them quickly for guests who are about to arrive. Initially,
the apples are at a uniform temperature of 20°C, and the heat
transfer coefficient on the surfaces is 8 W/m 2 • °C. Treating the
apples as 9-cm-diameter spheres and taking their properties to
be p = 840 kg/m 3 , C p = 3.81 kJ/kg • °C, k = 0.418 W/m ■ °C,
and a = 1.3 X 10~ 7 m 2 /s, determine the center and surface
temperatures of the apples in 1 h. Also, determine the amount
of heat transfer from each apple.
4-51 [ft^S Reconsider Problem 4-50. Using EES (or other)
1^2 software, investigate the effect of the initial tem-
perature of the apples on the final center and surface tem-
peratures and the amount of heat transfer. Let the initial
temperature vary from 2°C to 30°C. Plot the center tempera-
ture, the surface temperature, and the amount of heat transfer
as a function of the initial temperature, and discuss the results.
4-52 Citrus fruits are very susceptible to cold weather, and
extended exposure to subfreezing temperatures can destroy
them. Consider an 8-cm-diameter orange that is initially at
Orange
T ; = 15°C
FIGURE P4-52
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CHAPTER 4
15°C. A cold front moves in one night, and the ambient tem-
perature suddenly drops to — 6°C, with a heat transfer coeffi-
cient of 15 W/m 2 • °C. Using the properties of water for the
orange and assuming the ambient conditions to remain con-
stant for 4 h before the cold front moves out, determine if any
part of the orange will freeze that night.
4-53 An 8-cm-diameter potato (p = 1100 kg/m 3 , C p = 3900
J/kg ■ °C, k = 0.6 W/m • °C, and a = 1.4 X 10" 7 m 2 /s) that is
initially at a uniform temperature of 25°C is baked in an oven
at 170°C until a temperature sensor inserted to the center of the
potato indicates a reading of 70°C. The potato is then taken out
of the oven and wrapped in thick towels so that almost no heat
is lost from the baked potato. Assuming the heat transfer coef-
ficient in the oven to be 25 W/m 2 • °C, determine (a) how long
the potato is baked in the oven and (b) the final equilibrium
temperature of the potato after it is wrapped.
4-54 White potatoes (k = 0.50 W/m ■ °C and a = 0.13 X
10~ 6 m 2 /s) that are initially at a uniform temperature of 25°C
and have an average diameter of 6 cm are to be cooled by re-
frigerated air at 2°C flowing at a velocity of 4 m/s. The average
heat transfer coefficient between the potatoes and the air is ex-
perimentally determined to be 19 W/m 2 ■ °C. Determine how
long it will take for the center temperature of the potatoes to
drop to 6°C. Also, determine if any part of the potatoes will ex-
perience chilling injury during this process.
Air
2°C
4 m/s
FIGURE P4-54
4-55E Oranges of 2.5-in. diameter (k = 0.26 Btu/h • ft • °F
and a = 1.4 X 10~ 6 ft 2 /s) initially at a uniform temperature of
78°F are to be cooled by refrigerated air at 25°F flowing at a
velocity of 1 ft/s. The average heat transfer coefficient between
the oranges and the air is experimentally determined to be 4.6
Btu/h • ft 2 • °F. Determine how long it will take for the center
temperature of the oranges to drop to 40°F. Also, determine if
any part of the oranges will freeze during this process.
4-56 A 65-kg beef carcass {k = 0.47 W/m • °C and a =
0.13 X 10~ 6 m 2 /s) initially at a uniform temperature of 37°C is
to be cooled by refrigerated air at — 6°C flowing at a velocity
of 1.8 m/s. The average heat transfer coefficient between the
carcass and the air is 22 W/m 2 • °C. Treating the carcass as a
cylinder of diameter 24 cm and height 1 .4 m and disregarding
heat transfer from the base and top surfaces, determine how
long it will take for the center temperature of the carcass to
drop to 4°C. Also, determine if any part of the carcass will
freeze during this process. Answer: 14.0 h
Air -
Beef
-6°C -
37°C
1.8 m/s ►•
FIGURE P4-56
4-57 Layers of 23-cm-thick meat slabs (k = 0.47 W/m ■ °C
and a = 0.13 X 10~ 6 m 2 /s) initially at a uniform temperature
of 7°C are to be frozen by refrigerated air at — 30°C flowing at
a velocity of 1 .4 m/s. The average heat transfer coefficient be-
tween the meat and the air is 20 W/m 2 • °C. Assuming the size
of the meat slabs to be large relative to their thickness, deter-
mine how long it will take for the center temperature of the
slabs to drop to — 18°C. Also, determine the surface tempera-
ture of the meat slab at that time.
4-58E Layers of 6-in. -thick meat slabs (k = 0.26
Btu/h • ft • °F and a = 1.4 X 10~ 6 ft 2 /s) initially at a uniform
temperature of 50°F are cooled by refrigerated air at 23°F to a
temperature of 36°F at their center in 12 h. Estimate the aver-
age heat transfer coefficient during this cooling process.
Answer: 1.5 Btu/h • ft 2 • °F
4-59 Chickens with an average mass of 1.7 kg (k = 0.45
W/m ■ °C and a = 0.13 X 10~ 6 m 2 /s) initially at a uniform
temperature of 15°C are to be chilled in agitated brine at
— 10°C. The average heat transfer coefficient between the
chicken and the brine is determined experimentally to be
440 W/m 2 • °C. Taking the average density of the chicken to be
0.95 g/cm 3 and treating the chicken as a spherical lump, deter-
mine the center and the surface temperatures of the chicken in
2 h and 30 min. Also, determine if any part of the chicken will
freeze during this process.
FIGURE P4-59
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HEAT TRANSFER
Transient Heat Conduction in Semi-Infinite Solids
4-60C What is a semi-infinite medium? Give examples of
solid bodies that can be treated as semi-infinite mediums for
heat transfer purposes.
4-61 C Under what conditions can a plane wall be treated as
a semi-infinite medium?
4-62C Consider a hot semi-infinite solid at an initial temper-
ature of Tj that is exposed to convection to a cooler medium at
a constant temperature of T m with a heat transfer coefficient of
h. Explain how you can determine the total amount of heat
transfer from the solid up to a specified time t B .
4-63 In areas where the air temperature remains below 0°C
for prolonged periods of time, the freezing of water in under-
ground pipes is a major concern. Fortunately, the soil remains
relatively warm during those periods, and it takes weeks for the
subfreezing temperatures to reach the water mains in the
ground. Thus, the soil effectively serves as an insulation to pro-
tect the water from the freezing atmospheric temperatures in
winter.
The ground at a particular location is covered with snow
pack at — 8°C for a continuous period of 60 days, and the aver-
age soil properties at that location are k = 0.35 W/m • °C and
a = 0.15 X 80~ 6 m 2 /s. Assuming an initial uniform tempera-
ture of 8°C for the ground, determine the minimum burial
depth to prevent the water pipes from freezing.
4-64 The soil temperature in the upper layers of the earth
varies with the variations in the atmospheric conditions. Before
a cold front moves in, the earth at a location is initially at a uni-
form temperature of 10°C. Then the area is subjected to a tem-
perature of — 10°C and high winds that resulted in a convection
heat transfer coefficient of 40 W/m 2 • °C on the earth's surface
for a period of 10 h. Taking the properties of the soil at that lo-
cation to be k = 0.9 W/m ■ °C and a = 1.6 X 10~ 5 m 2 /s, deter-
mine the soil temperature at distances 0, 10, 20, and 50 cm
from the earth's surface at the end of this 10-h period.
► Winds,
" -10°C
Soil
10°C
FIGURE P4-64
4-65
Reconsider Problem 4-64. Using EES (or other)
software, plot the soil temperature as a function
of the distance from the earth's surface as the distance varies
from m to lm, and discuss the results.
4-66E The walls of a furnace are made of 1.5-ft-thick con-
crete (k = 0.64 Btu/h ■ ft • °F and a = 0.023 ft 2 /h). Initially, the
furnace and the surrounding air are in thermal equilibrium at
70°F. The furnace is then fired, and the inner surfaces of the
furnace are subjected to hot gases at 1 800°F with a very large
heat transfer coefficient. Determine how long it will take for
the temperature of the outer surface of the furnace walls to rise
to70.1°F Answer: 181 min
4-67 A thick wood slab (k = 0. 1 7 W/m • °C and a = 1 .28 X
10~ 7 m 2 /s) that is initially at a uniform temperature of 25°C is
exposed to hot gases at 550°C for a period of 5 minutes. The
heat transfer coefficient between the gases and the wood slab is
35 W/m 2 • °C. If the ignition temperature of the wood is 450°C,
determine if the wood will ignite.
4-68 A large cast iron container (k = 52 W/m • °C and a =
1.70 X 10~ 5 m 2 /s) with 5-cm-thick walls is initially at a uni-
form temperature of 0°C and is filled with ice at 0°C. Now the
outer surfaces of the container are exposed to hot water at 60°C
with a very large heat transfer coefficient. Determine how long
it will be before the ice inside the container starts melting.
Also, taking the heat transfer coefficient on the inner surface of
the container to be 250 W/m 2 • °C, determine the rate of heat
transfer to the ice through a 1.2-m-wide and 2-m-high section
of the wall when steady operating conditions are reached. As-
sume the ice starts melting when its inner surface temperature
rises to 0.1 °C.
Hot water
60°C
Cast iron
f chest
k •
U o Ice O
nop O
O Q o
Q o
° o °. Q
5 cm
FIGURE P4-68
Transient Heat Conduction in Multidimensional Systems
4-69C What is the product solution method? How is it used
to determine the transient temperature distribution in a two-
dimensional system?
4-70C How is the product solution used to determine
the variation of temperature with time and position in three-
dimensional systems?
4-71 C A short cylinder initially at a uniform temperature T :
is subjected to convection from all of its surfaces to a medium
at temperature r„. Explain how you can determine the temper-
ature of the midpoint of the cylinder at a specified time /.
4-72C Consider a short cylinder whose top and bottom sur-
faces are insulated. The cylinder is initially at a uniform tem-
perature Tj and is subjected to convection from its side surface
cen58933_ch04.qxd 9/10/2002 9:13 AM Page 259
to a medium at temperature T„ with a heat transfer coefficient
of h. Is the heat transfer in this short cylinder one- or two-
dimensional? Explain.
4-73 A short brass cylinder (p = 8530 kg/m 3 , C p = 0.389
kJ/kg • °C, k = 110 W/m • °C, and a = 3.39 X 10~ 5 m 2 /s) of
diameter D = 8 cm and height H = 15 cm is initially at a
uniform temperature of T; = 150°C. The cylinder is now
placed in atmospheric air at 20°C, where heat transfer takes
place by convection with a heat transfer coefficient of h =
40 W/m 2 • °C. Calculate (a) the center temperature of the cyl-
inder, {b) the center temperature of the top surface of the cylin-
der, and (c) the total heat transfer from the cylinder 15 min
after the start of the cooling.
15 cm
FIGURE P4-73
4-74
Reconsider Problem 4-73. Using EES (or other)
software, investigate the effect of the cooling
time on the center temperature of the cylinder, the center tem-
perature of the top surface of the cylinder, and the total heat
transfer. Let the time vary from 5 min to 60 min. Plot the cen-
ter temperature of the cylinder, the center temperature of the
top surface, and the total heat transfer as a function of the time,
and discuss the results.
4-75 A semi-infinite aluminum cylinder (k = 237 W/m • °C,
a = 9.71 X 10~ 5 m 2 /s) of diameter D = 15 cm is initially at a
uniform temperature of T, = 150°C. The cylinder is now
placed in water at 10°C, where heat transfer takes place
by convection with a heat transfer coefficient of h =
140 W/m 2 ■ °C. Determine the temperature at the center of the
cylinder 5 cm from the end surface 8 min after the start of
cooling.
4-76E A hot dog can be considered to be a cylinder 5 in.
long and 0.8 in. in diameter whose properties are p =
61.2 lbm/ft 3 , C p = 0.93 Btu/lbm ■ °F, k = 0.44 Btu/h • ft • °F,
and a = 0.0077 ft 2 /h. A hot dog initially at 40°F is dropped
into boiling water at 212°F. If the heat transfer coefficient at
the surface of the hot dog is estimated to be 120 Btu/h • ft 2 • °F,
determine the center temperature of the hot dog after 5, 10, and
15 min by treating the hot dog as (a) a finite cylinder and (b) an
infinitely long cylinder.
259
CHAPTER 4
4-77E Repeat Problem 4-76E for a location at 5300 ft
elevation such as Denver, Colorado, where the boiling temper-
ature of water is 202 °F.
4-78 A 5-cm-high rectangular ice block (k = 2.22 W/m ■ °C
and a = 0.124 X 10~ 7 m 2 /s) initially at -20°C is placed on a
table on its square base 4 cm X 4 cm in size in a room at 1 8°C.
The heat transfer coefficient on the exposed surfaces of the ice
block is 12 W/m 2 ■ °C. Disregarding any heat transfer from the
base to the table, determine how long it will be before the ice
block starts melting. Where on the ice block will the first liquid
droplets appear?
Room
FIGURE P4-78
4-79
Reconsider Problem 4-78. Using EES (or other)
software, investigate the effect of the initial tem-
perature of the ice block on the time period before the ice block
starts melting. Let the initial temperature vary from — 26°C to
— 4°C. Plot the time versus the initial temperature, and discuss
the results.
4-80 A 2-cm-high cylindrical ice block (k = 2.22 W/m • °C
and a = 0.124 X 10~ 7 m 2 /s) is placed on a table on its base of
diameter 2 cm in a room at 20°C. The heat transfer coefficient
on the exposed surfaces of the ice block is 13 W/m 2 ■ °C, and
heat transfer from the base of the ice block to the table is neg-
ligible. If the ice block is not to start melting at any point for at
least 2 h, determine what the initial temperature of the ice
block should be.
4-81 Consider a cubic block whose sides are 5 cm long and
a cylindrical block whose height and diameter are also 5 cm.
Both blocks are initially at 20°C and are made of granite (k =
2.5 W/m • °C and a = 1.15 X 10~ 6 m 2 /s). Now both blocks are
exposed to hot gases at 500°C in a furnace on all of their sur-
faces with a heat transfer coefficient of 40 W/m 2 • °C. Deter-
mine the center temperature of each geometry after 10, 20, and
60 min.
4-82 Repeat Problem 4-81 with the heat transfer coefficient
at the top and the bottom surfaces of each block being doubled
to 80 W/m 2 • °C.
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260
HEAT TRANSFER
5 cm
T, = 20°C
5 cm
Hot gases, 500°C
FIGURE P4-81
4-83 A 20-cm-long cylindrical aluminum block (p = 2702
kg/m 3 , C p = 0.896 kJ/kg ■ °C, k = 236 W/m ■ °C, and a =
9.75 X 10~ 5 m 2 /s), 15 cm in diameter, is initially at a uniform
temperature of 20°C. The block is to be heated in a furnace at
1200°C until its center temperature rises to 300°C. If the heat
transfer coefficient on all surfaces of the block is 80 W/m 2 • °C,
determine how long the block should be kept in the furnace.
Also, determine the amount of heat transfer from the aluminum
block if it is allowed to cool in the room until its temperature
drops to 20°C throughout.
4-84 Repeat Problem 4-83 for the case where the aluminum
block is inserted into the furnace on a low-conductivity mater-
ial so that the heat transfer to or from the bottom surface of the
block is negligible.
4-85 rfigM Reconsider Problem 4-83. Using EES (or other)
b^2 software, investigate the effect of the final center
temperature of the block on the heating time and the amount of
heat transfer. Let the final center temperature vary from 50°C
to 1000°C. Plot the time and the heat transfer as a function of
the final center temperature, and discuss the results.
Special Topic: Refrigeration and Freezing of Foods
4-86C What are the common kinds of microorganisms?
What undesirable changes do microorganisms cause in foods?
4-87C How does refrigeration prevent or delay the spoilage
of foods? Why does freezing extend the storage life of foods
for months?
4-88C What are the environmental factors that affect the
growth rate of microorganisms in foods?
4-89C What is the effect of cooking on the microorganisms
in foods? Why is it important that the internal temperature of a
roast in an oven be raised above 70°C?
4-90C How can the contamination of foods with micro-
organisms be prevented or minimized? How can the growth of
microorganisms in foods be retarded? How can the micro-
organisms in foods be destroyed?
4-91 C How does (a) the air motion and (b) the relative hu-
midity of the environment affect the growth of microorganisms
in foods?
4-92C The cooling of a beef carcass from 37°C to 5°C with
refrigerated air at 0°C in a chilling room takes about 48 h.
To reduce the cooling time, it is proposed to cool the carcass
with refrigerated air at -10°C. How would you evaluate this
proposal?
4-93C Consider the freezing of packaged meat in boxes with
refrigerated air. How do (a) the temperature of air, (b) the
velocity of air, (c) the capacity of the refrigeration system, and
(d) the size of the meat boxes affect the freezing time?
4-94C How does the rate of freezing affect the tenderness,
color, and the drip of meat during thawing?
4-95C It is claimed that beef can be stored for up to two
years at -23°C but no more than one year at — 12°C. Is this
claim reasonable? Explain.
4-96C What is a refrigerated shipping dock? How does it
reduce the refrigeration load of the cold storage rooms?
4-97C How does immersion chilling of poultry compare to
forced-air chilling with respect to (a) cooling time, (b) mois-
ture loss of poultry, and (c) microbial growth.
4-98C What is the proper storage temperature of frozen
poultry? What are the primary methods of freezing for poultry?
4-99C What are the factors that affect the quality of
frozen fish?
4-100 The chilling room of a meat plant is 15 m X 18 m X
5.5 m in size and has a capacity of 350 beef carcasses. The
power consumed by the fans and the lights in the chilling room
are 22 and 2 kW, respectively, and the room gains heat through
its envelope at a rate of 1 1 kW. The average mass of beef car-
casses is 280 kg. The carcasses enter the chilling room at 35°C,
after they are washed to facilitate evaporative cooling, and are
cooled to 16°C in 12 h. The air enters the chilling room at
— 2.2°C and leaves at 0.5°C. Determine (a) the refrigeration
load of the chilling room and (b) the volume flow rate of air.
The average specific heats of beef carcasses and air are 3.14
and 1 .0 kJ/kg • °C, respectively, and the density of air can be
taken to be 1 .28 kg/m 3 .
4-101 Turkeys with a water content of 64 percent that are
initially at 1 °C and have a mass of about 7 kg are to be frozen
by submerging them into brine at — 29°C. Using Figure 4^-5,
determine how long it will take to reduce the temperature of
the turkey breast at a depth of 3.8 cm to — 18°C. If the temper-
ature at a depth of 3.8 cm in the breast represents the average
^\
Brine -29°C
II
FIGURE P4-1 01
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temperature of the turkey, determine the amount of heat trans-
fer per turkey assuming (a) the entire water content of the
turkey is frozen and (b) only 90 percent of the water content of
the turkey is frozen at — 18°C. Take the specific heats of turkey
to be 2.98 and 1.65 kJ/kg ■ °C above and below the freezing
point of — 2.8°C, respectively, and the latent heat of fusion of
turkey to be 214 kJ/kg. Answers.- {a) 1753 kJ, (£>) 1617 kJ
4-102E Chickens with a water content of 74 percent, an
initial temperature of 32°F, and a mass of about 6 lbm are to be
frozen by refrigerated air at — 40°F. Using Figure 4^-4, de-
termine how long it will take to reduce the inner surface
temperature of chickens to 25 °F. What would your answer be if
the air temperature were — 80°F?
4-103 Chickens with an average mass of 2.2 kg and average
specific heat of 3.54 kJ/kg • °C are to be cooled by chilled wa-
ter that enters a continuous-flow-type immersion chiller at
0.5°C. Chickens are dropped into the chiller at a uniform tem-
perature of 15°C at a rate of 500 chickens per hour and are
cooled to an average temperature of 3°C before they are taken
out. The chiller gains heat from the surroundings at a rate of
210 kJ/min. Determine (a) the rate of heat removal from the
chicken, in kW, and {b) the mass flow rate of water, in kg/s, if
the temperature rise of water is not to exceed 2°C.
4-104 In a meat processing plant, 10-cm-thick beef slabs
(p = 1090 kg/m 3 , C p = 3.54 kJ/kg • °C, k = 0.47 W/m • °C,
and a = 0.13 X 10 6 m 2 /s) initially at 15°C are to be cooled in
the racks of a large freezer that is maintained at — 12°C. The
meat slabs are placed close to each other so that heat transfer
from the 10-cm-thick edges is negligible. The entire slab is to
be cooled below 5°C, but the temperature of the steak is not
to drop below — 1 °C anywhere during refrigeration to avoid
"frost bite." The convection heat transfer coefficient and thus
the rate of heat transfer from the steak can be controlled by
varying the speed of a circulating fan inside. Determine the
heat transfer coefficient h that will enable us to meet both tem-
perature constraints while keeping the refrigeration time to a
minimum. Answer: 9.9 W/m 2 • °C.
Aii-
Meat
-12°C M0 cm
FIGURE P4-1 04
Review Problems
4-105 Consider two 2-cm-thick large steel plates (k = 43
W/m ■ °C and a = 1.17 X 10~ 5 m 2 /s) that were put on top of
each other while wet and left outside during a cold winter night
at — 15°C. The next day, a worker needs one of the plates, but
the plates are stuck together because the freezing of the water
261
CHAPTER 4
between the two plates has bonded them together. In an effort
to melt the ice between the plates and separate them, the
worker takes a large hairdryer and blows hot air at 50°C all
over the exposed surface of the plate on the top. The convec-
tion heat transfer coefficient at the top surface is estimated to
be 40 W/m 2 • °C. Determine how long the worker must keep
blowing hot air before the two plates separate.
Answer: 482 s
4-106 Consider a curing kiln whose walls are made of
30-cm-thick concrete whose properties are k = 0.9 W/m • °C
and a = 0.23 X 10~ 5 m 2 /s. Initially, the kiln and its walls are
in equilibrium with the surroundings at 2°C. Then all the doors
are closed and the kiln is heated by steam so that the tempera-
ture of the inner surface of the walls is raised to 42°C and is
maintained at that level for 3 h. The curing kiln is then opened
and exposed to the atmospheric air after the stream flow is
turned off. If the outer surfaces of the walls of the kiln were in-
sulated, would it save any energy that day during the period
the kiln was used for curing for 3 h only, or would it make no
difference? Base your answer on calculations.
FIGURE P4-1 06
4-107 The water main in the cities must be placed at suf-
ficient depth below the earth's surface to avoid freezing during
extended periods of subfreezing temperatures. Determine the
minimum depth at which the water main must be placed at a
location where the soil is initially at 15°C and the earth's
surface temperature under the worst conditions is expected
to remain at — 10°C for a period of 75 days. Take the proper-
ties of soil at that location to be k = 0.7 W/m • °C and a =
1.4 X 10~ 5 m 2 /s. Answer: 7.05 m
4-108 A hot dog can be considered to be a 12-cm-long cyl-
inder whose diameter is 2 cm and whose properties are
p = 980 kg/m 3 , C = 3.9 kJ/kg • °C, k = 0.76 W/m ■ °C, and
FIGURE P4-1 08
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262
HEAT TRANSFER
a = 2 X 10~ 7 m 2 /s. A hot dog initially at 5°C is dropped into
boiling water at 100°C. The heat transfer coefficient at the sur-
face of the hot dog is estimated to be 600 W/m 2 • °C. If the hot
dog is considered cooked when its center temperature reaches
80°C, determine how long it will take to cook it in the boiling
water.
4-109 A long roll of 2-m-wide and 0.5-cm-thick 1 -Mn man-
ganese steel plate coming off a furnace at 820°C is to be
quenched in an oil bath (C p = 2.0 kJ/kg ■ °C) at 45°C. The
metal sheet is moving at a steady velocity of 10 m/min, and the
oil bath is 5 m long. Taking the convection heat transfer
coefficient on both sides of the plate to be 860 W/m 2 • °C, de-
termine the temperature of the sheet metal when it leaves the
oil bath. Also, determine the required rate of heat removal from
the oil to keep its temperature constant at 45°C.
Furnace
Oil bath, 45°C
FIGURE P4-1 09
4-110E In Betty Crocker 's Cookbook, it is stated that it takes
5 h to roast a 14-lb stuffed turkey initially at 40°F in an oven
maintained at 325°F. It is recommended that a meat thermome-
ter be used to monitor the cooking, and the turkey is considered
done when the thermometer inserted deep into the thickest part
of the breast or thigh without touching the bone registers
185°F. The turkey can be treated as a homogeneous spheri-
cal object with the properties p = 75 lbm/ft 3 , C p = 0.98
Btu/lbm ■ °F, k = 0.26 Btu/h • ft ■ °F, and a = 0.0035 ft 2 /h.
Assuming the tip of the thermometer is at one-third radial
distance from the center of the turkey, determine (a) the aver-
age heat transfer coefficient at the surface of the turkey, (b) the
temperature of the skin of the turkey when it is done, and
(c) the total amount of heat transferred to the turkey in the
oven. Will the reading of the thermometer be more or less than
185°F 5 min after the turkey is taken out of the oven?
4-111 (~Jb\ During a fire, the trunks of some dry oak trees (k
W = 0.17 W/m • °C and a = 1.28 X 10" 7 m 2 /s)
that are initially at a uniform temperature of 30°C are exposed
to hot gases at 520°C for a period of 5 h, with a heat transfer
coefficient of 65 W/m 2 • °C on the surface. The ignition tem-
perature of the trees is 410°C. Treating the trunks of the trees
as long cylindrical rods of diameter 20 cm, determine if these
dry trees will ignite as the fire sweeps through them.
FIGURE P4-1 1 1
4-112 We often cut a watermelon in half and put it into the
freezer to cool it quickly. But usually we forget to check on it
and end up having a watermelon with a frozen layer on the top.
To avoid this potential problem a person wants to set the timer
such that it will go off when the temperature of the exposed
surface of the watermelon drops to 3°C.
Consider a 30-cm-diameter spherical watermelon that is cut
into two equal parts and put into a freezer at — 12°C. Initially,
the entire watermelon is at a uniform temperature of 25°C, and
the heat transfer coefficient on the surfaces is 30 W/m 2 • °C.
Assuming the watermelon to have the properties of water, de-
termine how long it will take for the center of the exposed cut
surfaces of the watermelon to drop to 3°C.
FIGURE P4-1 10
Watermelon, 25°C
FIGURE P4-1 12
4-113 The thermal conductivity of a solid whose density and
specific heat are known can be determined from the relation
k = a/pC p after evaluating the thermal diffusivity a.
Consider a 2-cm-diameter cylindrical rod made of a sample
material whose density and specific heat are 3700 kg/m 3 and
920 J/kg ■ °C, respectively. The sample is initially at a uniform
temperature of 25°C. In order to measure the temperatures of
cen58933_ch04.qxd 9/10/2002 9:13 AM Page 263
r
Thermocouples
Rod «=
Boiling water center
100°C
FIGURE P4-1 13
the sample at its surface and its center, a thermocouple is
inserted to the center of the sample along the centerline, and
another thermocouple is welded into a small hole drilled on
the surface. The sample is dropped into boiling water at 100°C.
After 3 min, the surface and the center temperatures are re-
corded to be 93°C and 75°C, respectively. Determine the ther-
mal diffusivity and the thermal conductivity of the material.
4-114 In desert climates, rainfall is not a common occurrence
since the rain droplets formed in the upper layer of the atmo-
sphere often evaporate before they reach the ground. Consider
a raindrop that is initially at a temperature of 5°C and has a
diameter of 5 mm. Determine how long it will take for the
diameter of the raindrop to reduce to 3 mm as it falls through
ambient air at 18°C with a heat transfer coefficient of 400
W/m 2 ■ °C. The water temperature can be assumed to remain
constant and uniform at 5°C at all times.
4-115E Consider a plate of thickness 1 in., a long cylinder of
diameter 1 in., and a sphere of diameter 1 in., all initially at
400°F and all made of bronze (k = 15.0 Btu/h • ft • °F and a =
0.333 ft 2 /h). Now all three of these geometries are exposed to
cool air at 75°F on all of their surfaces, with a heat transfer co-
efficient of 7 Btu/h • ft 2 • °F. Determine the center temperature
of each geometry after 5, 10, and 30 min. Explain why the cen-
ter temperature of the sphere is always the lowest.
Cylinder
tin.
J
Sphere
<2L,
FIGURE P4-1 15
4-116E Repeat Problem 4-1 15E for cast iron geometries
(k = 29 Btu/h • ft ■ °F and a = 0.61 ft 2 /h).
4-117E rra| Reconsider Problem 4-1 15E. Using EES (or
b^2 other) software, plot the center temperature of
each geometry as a function of the cooling time as the time
varies fom 5 min to 60 min, and discuss the results.
263
CHAPTER 4
4-118 Engine valves (k = 48 W/m • °C, C p = 440 J/kg ■ °C,
and p = 7840 kg/m 3 ) are heated to 800°C in the heat treatment
section of a valve manufacturing facility. The valves are then
quenched in a large oil bath at an average temperature of 45 °C.
The heat transfer coefficient in the oil bath is 650 W/m 2 • °C.
The valves have a cylindrical stem with a diameter of 8 mm
and a length of 10 cm. The valve head and the stem may be as-
sumed to be of equal surface area, and the volume of the valve
head can be taken to be 80 percent of the volume of steam. De-
termine how long will it take for the valve temperature to drop
to (a) 400°C, (b) 200°C, and (c) 46°C and (d) the maximum
heat transfer from a single valve.
4-119 A watermelon initially at 35°C is to be cooled by
dropping it into a lake at 1 5°C. After 4 h and 40 min of cooling,
the center temperature of watermelon is measured to be 20°C.
Treating the watermelon as a 20-cm -diameter sphere and using
the properties k = 0.618 W/m • °C, a = 0.15 X 10~ 6 m 2 /s,
p = 995 kg/m 3 , and C p = 4.18 kJ/kg • °C, determine the aver-
age heat transfer coefficient and the surface temperature of
watermelon at the end of the cooling period.
4-120 10-cm-thick large food slabs tightly wrapped by thin
paper are to be cooled in a refrigeration room maintained
at 0°C. The heat transfer coefficient on the box surfaces is
25 W/m 2 ■ °C and the boxes are to be kept in the refrigeration
room for a period of 6 h. If the initial temperature of the boxes
is 30°C determine the center temperature of the boxes if the
boxes contain (a) margarine (k = 0.233 W/m • °C and a =
0.11 X 10- 6 m 2 /s), (b) white cake (k = 0.082 W/m • °C and
a = 0.10 X 10~ 6 m 2 /s), and (c) chocolate cake (k = 0.106
W/m ■ °C and a = 0.12 X lO" 6 m 2 /s).
4-121 A 30-cm-diameter, 3.5-m-high cylindrical column of
a house made of concrete (k = 0.79 W/m • °C, a = 5.94 X
10- 7 m 2 /s, p = 1600 kg/m 3 , and C p = 0.84 kJ/kg • °C) cooled
to 16°C during a cold night is heated again during the day by
being exposed to ambient air at an average temperature of
28°C with an average heat transfer coefficient of 14 W/m 2 ■ °C.
Determine (a) how long it will take for the column surface
temperature to rise to 27°C, (b) the amount of heat transfer
until the center temperature reaches to 28°C, and (c) the
amount of heat transfer until the surface temperature reaches
to 27°C.
4-122 Long aluminum wires of diameter 3 mm (p = 2702
kg/m 3 , C p = 0.896 kJ/kg ■ °C, k = 236 W/m • °C, and a =
9.75 X 10~ 5 m 2 /s) are extruded at a temperature of 350°C and
exposed to atmospheric air at 30°C with a heat transfer coeffi-
cient of 35 W/m 2 ■ °C. (a) Determine how long it will take for
the wire temperature to drop to 50°C. (b) If the wire is extruded
at a velocity of 10 m/min, determine how far the wire travels
after extrusion by the time its temperature drops to 50°C. What
change in the cooling process would you propose to shorten
this distance? (c) Assuming the aluminum wire leaves the ex-
trusion room at 50°C, determine the rate of heat transfer from
the wire to the extrusion room.
Answers: (a) 144 s, (b) 24 m, (c) 856 W
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HEAT TRANSFER
350°C
10 m/min
Aluminum wire
FIGURE P4-1 22
4-123 Repeat Problem 4-122 for a copper wire (p =
8950 kg/m 3 , C p = 0.383 kJ/kg • °C, k = 386 W/m • °C, and
a = 1.13 X 10- 4 m 2 /s).
4-124 Consider a brick house (k = 0.72 W/m • °C and a =
0.45 X 10~ 6 m 2 /s) whose walls are 10 m long, 3 m high, and
0.3 m thick. The heater of the house broke down one night, and
the entire house, including its walls, was observed to be 5°C
throughout in the morning. The outdoors warmed up as the day
progressed, but no change was felt in the house, which was
tightly sealed. Assuming the outer surface temperature of the
house to remain constant at 15°C, determine how long it would
take for the temperature of the inner surfaces of the walls to
riseto5.1°C.
15°C
~5°C
FIGURE P4-1 24
4-125 A 40-cm-thick brick wall (k = 0.72 W/m • °C, and a =
1.6 X 10~ 7 m 2 /s) is heated to an average temperature of 18°C
by the heating system and the solar radiation incident on it dur-
ing the day. During the night, the outer surface of the wall is ex-
posed to cold air at 2°C with an average heat transfer coefficient
of 20 W/m 2 ■ °C, determine the wall temperatures at distances
15,30, and 40 cm from the outer surface for a period of 2 hours.
4-126 Consider the engine block of a car made of cast iron
(k = 52 W/m • °C and a = 1 .7 X 10~ 5 m 2 /s). The engine can be
considered to be a rectangular block whose sides are 80 cm,
40 cm, and 40 cm. The engine is at a temperature of 150°C
when it is turned off. The engine is then exposed to atmospheric
air at 17°C with a heat transfer coefficient of 6 W/m 2 • °C. De-
termine (a) the center temperature of the top surface whose
sides are 80 cm and 40 cm and (b) the comer temperature after
45 min of cooling.
4-127 A man is found dead in a room at 16°C. The surface
temperature on his waist is measured to be 23°C and the heat
transfer coefficient is estimated to be 9 W/m 2 • °C. Modeling the
body as 28-cm diameter, 1.80-m-long cylinder, estimate how
long it has been since he died. Take the properties of the body to
be k = 0.62 W/m • °C and a = 0.15 X 10~ 6 m 2 /s, and assume
the initial temperature of the body to be 36°C.
Computer, Design, and Essay Problems
4-128 Conduct the following experiment at home to deter-
mine the combined convection and radiation heat transfer co-
efficient at the surface of an apple exposed to the room air. You
will need two thermometers and a clock.
First, weigh the apple and measure its diameter. You may
measure its volume by placing it in a large measuring cup
halfway filled with water, and measuring the change in volume
when it is completely immersed in the water. Refrigerate the
apple overnight so that it is at a uniform temperature in the
morning and measure the air temperature in the kitchen. Then
take the apple out and stick one of the thermometers to its mid-
dle and the other just under the skin. Record both temperatures
every 5 min for an hour. Using these two temperatures, calcu-
late the heat transfer coefficient for each interval and take their
average. The result is the combined convection and radiation
heat transfer coefficient for this heat transfer process. Using
your experimental data, also calculate the thermal conductivity
and thermal diffusivity of the apple and compare them to the
values given above.
4-129 Repeat Problem 4-128 using a banana instead of an
apple. The thermal properties of bananas are practically the
same as those of apples.
4-130 Conduct the following experiment to determine the
time constant for a can of soda and then predict the temperature
of the soda at different times. Leave the soda in the refrigerator
overnight. Measure the air temperature in the kitchen and the
temperature of the soda while it is still in the refrigerator by
taping the sensor of the thermometer to the outer surface of the
can. Then take the soda out and measure its temperature again
in 5 min. Using these values, calculate the exponent b. Using
this b-value, predict the temperatures of the soda in 10, 15, 20,
30, and 60 min and compare the results with the actual temper-
ature measurements. Do you think the lumped system analysis
is valid in this case?
4-131 Citrus trees are very susceptible to cold weather, and
extended exposure to subfreezing temperatures can destroy the
crop. In order to protect the trees from occasional cold fronts
with subfreezing temperatures, tree growers in Florida usually
install water sprinklers on the trees. When the temperature
drops below a certain level, the sprinklers spray water on the
trees and their fruits to protect them against the damage the
subfreezing temperatures can cause. Explain the basic mecha-
nism behind this protection measure and write an essay on how
the system works in practice.
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NUMERICAL METHODS
IN HEAT CONDUCTION
CHAPTER
So far we have mostly considered relatively simple heat conduction prob-
lems involving simple geometries with simple boundary conditions be-
cause only such simple problems can be solved analytically. But many
problems encountered in practice involve complicated geometries with com-
plex boundary conditions or variable properties and cannot be solved ana-
lytically. In such cases, sufficiently accurate approximate solutions can be
obtained by computers using a numerical method.
Analytical solution methods such as those presented in Chapter 2 are based
on solving the governing differential equation together with the boundary con-
ditions. They result in solution functions for the temperature at every point in
the medium. Numerical methods, on the other hand, are based on replacing
the differential equation by a set of n algebraic equations for the unknown
temperatures at n selected points in the medium, and the simultaneous solu-
tion of these equations results in the temperature values at those discrete
points.
There are several ways of obtaining the numerical formulation of a heat
conduction problem, such as the finite difference method, the finite element
method, the boundary element method, and the energy balance (or control
volume) method. Each method has its own advantages and disadvantages, and
each is used in practice. In this chapter we will use primarily the energy bal-
ance approach since it is based on the familiar energy balances on control vol-
umes instead of heavy mathematical formulations, and thus it gives a better
physical feel for the problem. Besides, it results in the same set of algebraic
equations as the finite difference method. In this chapter, the numerical for-
mulation and solution of heat conduction problems are demonstrated for both
steady and transient cases in various geometries.
CONTENTS
Why Numerical Methods 266
Finite Difference Formulation of
Differential Equations 269
One-Dimensional
Steady Heat Conduction 272
Two-Dimensional
Steady Heat Conduction 282
Transient Heat Conduction 291
Topic of Special Interest:
Controlling the
Numerical Error 309
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266
HEAT TRANSFER
Solution:
Q(r) = -kA
1 ' 6k v
4ty 3
dr 3
FIGURE 5-1
The analytical solution of a problem
requires solving the governing
differential equation and applying
the boundary conditions.
5-1 ■ WHY NUMERICAL METHODS?
The ready availability of high-speed computers and easy-to-use powerful soft-
ware packages has had a major impact on engineering education and practice
in recent years. Engineers in the past had to rely on analytical skills to solve
significant engineering problems, and thus they had to undergo a rigorous
training in mathematics. Today's engineers, on the other hand, have access to
a tremendous amount of computation power under their fingertips, and they
mostly need to understand the physical nature of the problem and interpret the
results. But they also need to understand how calculations are performed by
the computers to develop an awareness of the processes involved and the lim-
itations, while avoiding any possible pitfalls.
In Chapter 2 we solved various heat conduction problems in various geo-
metries in a systematic but highly mathematical manner by (1) deriving the
governing differential equation by performing an energy balance on a differ-
ential volume element, (2) expressing the boundary conditions in the proper
mathematical form, and (3) solving the differential equation and applying the
boundary conditions to determine the integration constants. This resulted in a
solution function for the temperature distribution in the medium, and the so-
lution obtained in this manner is called the analytical solution of the problem.
For example, the mathematical formulation of one-dimensional steady heat
conduction in a sphere of radius r whose outer surface is maintained at a uni-
form temperature of T, with uniform heat generation at a rate of g Q was ex-
pressed as (Fig. 5-1)
\_d_t 2 dT\ go
r 2 dr\ dr) k
dT(0)
dr
whose (analytical) solution is
and
T(r ) = T x
T{r) = T l +f k {ri-r^)
(5-1)
(5-2)
This is certainly a very desirable form of solution since the temperature at
any point within the sphere can be determined simply by substituting the
r-coordinate of the point into the analytical solution function above. The ana-
lytical solution of a problem is also referred to as the exact solution since it
satisfies the differential equation and the boundary conditions. This can be
verified by substituting the solution function into the differential equation and
the boundary conditions. Further, the rate of heat flow at any location within
the sphere or its surface can be determined by taking the derivative of the so-
lution function T(r) and substituting it into Fourier's law as
Q(r)
-kA
dT
dr
-£(4irr 2 )
gof_
~3k
4irg Q r' :
(5-3)
The analysis above did not require any mathematical sophistication beyond
the level of simple integration, and you are probably wondering why anyone
would ask for something else. After all, the solutions obtained are exact and
cen58 93 3_ch05.qxd 9/4/2002 11:41 AM Page 267
easy to use. Besides, they are instructive since they show clearly the func-
tional dependence of temperature and heat transfer on the independent vari-
able r. Well, there are several reasons for searching for alternative solution
methods.
1 Limitations
Analytical solution methods are limited to highly simplified problems in sim-
ple geometries (Fig. 5-2). The geometry must be such that its entire surface
can be described mathematically in a coordinate system by setting the vari-
ables equal to constants. That is, it must fit into a coordinate system perfectly
with nothing sticking out or in. In the case of one-dimensional heat conduc-
tion in a solid sphere of radius r , for example, the entire outer surface can be
described by r = r . Likewise, the surfaces of a finite solid cylinder of radius
r and height H can be described by r = r for the side surface and z = and
z = H for the bottom and top surfaces, respectively. Even minor complica-
tions in geometry can make an analytical solution impossible. For example, a
spherical object with an extrusion like a handle at some location is impossible
to handle analytically since the boundary conditions in this case cannot be ex-
pressed in any familiar coordinate system.
Even in simple geometries, heat transfer problems cannot be solved analyt-
ically if the thermal conditions are not sufficiently simple. For example, the
consideration of the variation of thermal conductivity with temperature, the
variation of the heat transfer coefficient over the surface, or the radiation heat
transfer on the surfaces can make it impossible to obtain an analytical solu-
tion. Therefore, analytical solutions are limited to problems that are simple or
can be simplified with reasonable approximations.
2 Better Modeling
We mentioned earlier that analytical solutions are exact solutions since they
do not involve any approximations. But this statement needs some clarifica-
tion. Distinction should be made between an actual real-world problem and
the mathematical model that is an idealized representation of it. The solutions
we get are the solutions of mathematical models, and the degree of applica-
bility of these solutions to the actual physical problems depends on the accu-
racy of the model. An "approximate" solution of a realistic model of a
physical problem is usually more accurate than the "exact" solution of a crude
mathematical model (Fig. 5-3).
When attempting to get an analytical solution to a physical problem, there
is always the tendency to oversimplify the problem to make the mathematical
model sufficiently simple to warrant an analytical solution. Therefore, it is
common practice to ignore any effects that cause mathematical complications
such as nonlinearities in the differential equation or the boundary conditions.
So it comes as no surprise that nonlinearities such as temperature dependence
of thermal conductivity and the radiation boundary conditions are seldom con-
sidered in analytical solutions. A mathematical model intended for a numeri-
cal solution is likely to represent the actual problem better. Therefore, the
numerical solution of engineering problems has now become the norm rather
than the exception even when analytical solutions are available.
T„h
267
CHAPTER 5
k = constant
No
radiation
T„h
Long
cylinder
h,T„
No
radiation
h,T„
h = constant
T„ = constant
FIGURE 5-2
Analytical solution methods are
limited to simplified problems
in simple geometries.
Exact (analytical) Approximate (numerical)
solution of model, solution of model,
but crude solution but accurate solution
of actual problem of actual problem
FIGURE 5-3
The approximate numerical solution
of a real-world problem may be more
accurate than the exact (analytical)
solution of an oversimplified
model of that problem.
cen58 93 3_ch05.qxd 9/4/2002 11:41 AM Page 268
268
HEAT TRANSFER
T(r, z)
Analytical solution:
T(r,z)-T-^ ~ J (Ar) sinhA„(L-z)
where A 's are roots of 7 n (A r n ) =
FIGURE 5-4
Some analytical solutions are very
complex and difficult to use.
3 Flexibility
Engineering problems often require extensive parametric studies to under-
stand the influence of some variables on the solution in order to choose the
right set of variables and to answer some "what-if " questions. This is an iter-
ative process that is extremely tedious and time-consuming if done by hand.
Computers and numerical methods are ideally suited for such calculations,
and a wide range of related problems can be solved by minor modifications in
the code or input variables. Today it is almost unthinkable to perform any sig-
nificant optimization studies in engineering without the power and flexibility
of computers and numerical methods.
4 Complications
Some problems can be solved analytically, but the solution procedure is so
complex and the resulting solution expressions so complicated that it is not
worth all that effort. With the exception of steady one-dimensional or transient
lumped system problems, all heat conduction problems result in partial
differential equations. Solving such equations usually requires mathematical
sophistication beyond that acquired at the undergraduate level, such as orthog-
onality, eigenvalues, Fourier and Laplace transforms, Bessel and Legendre
functions, and infinite series. In such cases, the evaluation of the solution,
which often involves double or triple summations of infinite series at a speci-
fied point, is a challenge in itself (Fig. 5-4). Therefore, even when the solu-
tions are available in some handbooks, they are intimidating enough to scare
prospective users away.
FIGURE 5-5
The ready availability of high-
powered computers with sophisticated
software packages has made
numerical solution the norm
rather than the exception.
5 Human Nature
As human beings, we like to sit back and make wishes, and we like our wishes
to come true without much effort. The invention of TV remote controls made
us feel like kings in our homes since the commands we give in our comfort-
able chairs by pressing buttons are immediately carried out by the obedient
TV sets. After all, what good is cable TV without a remote control. We cer-
tainly would love to continue being the king in our little cubicle in the engi-
neering office by solving problems at the press of a button on a computer
(until they invent a remote control for the computers, of course). Well, this
might have been a fantasy yesterday, but it is a reality today. Practically all
engineering offices today are equipped with high-powered computers with
sophisticated software packages, with impressive presentation-style colorful
output in graphical and tabular form (Fig. 5-5). Besides, the results are as
accurate as the analytical results for all practical purposes. The computers
have certainly changed the way engineering is practiced.
The discussions above should not lead you to believe that analytical solu-
tions are unnecessary and that they should be discarded from the engineering
curriculum. On the contrary, insight to the physical phenomena and engineer-
ing wisdom is gained primarily through analysis. The "feel" that engineers
develop during the analysis of simple but fundamental problems serves as
an invaluable tool when interpreting a huge pile of results obtained from a
computer when solving a complex problem. A simple analysis by hand for
a limiting case can be used to check if the results are in the proper range. Also,
cen58 93 3_ch05.qxd 9/4/2002 11:41 AM Page 269
269
CHAPTER 5
nothing can take the place of getting "ball park" results on a piece of paper
during preliminary discussions. The calculators made the basic arithmetic
operations by hand a thing of the past, but they did not eliminate the need for
instructing grade school children how to add or multiply.
In this chapter, you will learn how to formulate and solve heat transfer
problems numerically using one or more approaches. In your professional life,
you will probably solve the heat transfer problems you come across using a
professional software package, and you are highly unlikely to write your own
programs to solve such problems. (Besides, people will be highly skeptical
of the results obtained using your own program instead of using a well-
established commercial software package that has stood the test of time.) The
insight you will gain in this chapter by formulating and solving some heat
transfer problems will help you better understand the available software pack-
ages and be an informed and responsible user.
5-2 - FINITE DIFFERENCE FORMULATION
OF DIFFERENTIAL EQUATIONS
The numerical methods for solving differential equations are based on replac-
ing the differential equations by algebraic equations. In the case of the popu-
lar finite difference method, this is done by replacing the derivatives by
differences. Below we will demonstrate this with both first- and second-order
derivatives. But first we give a motivational example.
Consider a man who deposits his money in the amount of A Q = $100 in a
savings account at an annual interest rate of 18 percent, and let us try to de-
termine the amount of money he will have after one year if interest is com-
pounded continuously (or instantaneously). In the case of simple interest, the
money will earn $18 interest, and the man will have 100 + 100 X 0.18 =
$118.00 in his account after one year. But in the case of compounding, the
interest earned during a compounding period will also earn interest for the
remaining part of the year, and the year-end balance will be greater than $118.
For example, if the money is compounded twice a year, the balance will be
100 + 100 X (0.18/2) = $109 after six months, and 109 + 109 X (0.18/2) =
$118.81 at the end of the year. We could also determine the balance A di-
rectly from
A = A (l + 0" = ($100)(1 + 0.09) 2 = $118.81
(5-4)
where i is the interest rate for the compounding period and n is the number of
periods. Using the same formula, the year-end balance is determined for
monthly, daily, hourly, minutely, and even secondly compounding, and the re-
sults are given in Table 5—1.
Note that in the case of daily compounding, the year-end balance will be
$119.72, which is $1.72 more than the simple interest case. (So it is no wonder
that the credit card companies usually charge interest compounded daily when
determining the balance.) Also note that compounding at smaller time inter-
vals, even at the end of each second, does not change the result, and we sus-
pect that instantaneous compounding using "differential" time intervals dt will
give the same result. This suspicion is confirmed by obtaining the differential
TABLE 5-1
Year-end balance of a $100 account
earning interest at an annual rate
of 18 percent for various
compounding periods
Number
Compounding
of
Year-End
Period
Periods, n
Balance
1 year
1
$118.00
6 months
2
118.81
1 month
12
119.56
1 week
52
119.68
1 day
365
119.72
1 hour
8760
119.72
1 minute
525,600
119.72
1 second
31,536,000
119.72
Instantaneous
00
119.72
cen58933_ch05.qxd 9/4/2002 11:41 AM Page 27C
270
HEAT TRANSFER
equation dAldt
stitution yields
iA for the balance A, whose solution is A = A exp(it). Sub-
A = ($100)exp(0.18 X 1) = $119.72
m
f(x + Ax)
/y\^f
m
£ —1 1
'1
Ax i
Tangent line
x x + Ax x
FIGURE 5-6
The derivative of a function at a point
represents the slope of the function
at that point.
FIGURE 5-7
Schematic of the nodes and the nodal
temperatures used in the development
of the finite difference formulation
of heat transfer in a plane wall.
which is identical to the result for daily compounding. Therefore, replacing a
differential time interval dt by a finite time interval of At = 1 day gave the
same result, which leads us into believing that reasonably accurate results can
be obtained by replacing differential quantities by sufficiently small differ-
ences. Next, we develop the finite difference formulation of heat conduction
problems by replacing the derivatives in the differential equations by differ-
ences. In the following section we will do it using the energy balance method,
which does not require any knowledge of differential equations.
Derivatives are the building blocks of differential equations, and thus we
first give a brief review of derivatives. Consider a function /that depends on
x, as shown in Figure 5-6. The first derivative of fix) at a point is equivalent
to the slope of a line tangent to the curve at that point and is defined as
df(x)
dx
A/
lim - — :
a*->o Ax
lim
fix + Ax) -fix)
Ax
(5-5)
which is the ratio of the increment A/of the function to the increment Ax of the
independent variable as Ax — > 0. If we don't take the indicated limit, we will
have the following approximate relation for the derivative:
dfix) fix + Ax) - fix)
dx
Ax
(5-6)
This approximate expression of the derivative in terms of differences is the
finite difference form of the first derivative. The equation above can also be
obtained by writing the Taylor series expansion of the function / about the
point x,
fix + Ax) = fix) + Ax
dfix)
dx
1 . J 2 fix)
(5-7)
and neglecting all the terms in the expansion except the first two. The first
term neglected is proportional to Ax 2 , and thus the error involved in each step
of this approximation is also proportional to Ax 2 . However, the commutative
error involved after M steps in the direction of length L is proportional to Ax
since MAx 2 = {LI Ax) Ax 2 = LAx. Therefore, the smaller the Ax, the smaller
the error, and thus the more accurate the approximation.
Now consider steady one-dimensional heat transfer in a plane wall of thick-
ness L with heat generation. The wall is subdivided into M sections of equal
thickness Ax = LIM in the x-direction, separated by planes passing through
M + 1 points 0, 1, 2, . . . , m — 1, m, m + 1, . . . , M called nodes or nodal
points, as shown in Figure 5-7. The x-coordinate of any point m is simply
x,„ = mAx, and the temperature at that point is simply T(x m ) = T„ r
The heat conduction equation involves the second derivatives of tempera-
ture with respect to the space variables, such as d 2 T/dx 2 , and the finite differ-
ence formulation is based on replacing the second derivatives by appropriate
cen58 93 3_ch05.qxd 9/4/2002 11:41 AM Page 271
differences. But we need to start the process with first derivatives. Using
Eq. 5-6, the first derivative of temperature dTldx at the midpoints m — i and
m + ~ of the sections surrounding the node m can be expressed as
271
CHAPTER 5
dT
dx
Ax
and
dT
dx
T„, + 1
Ax
(5-8)
Noting that the second derivative is simply the derivative of the first deriva-
tive, the second derivative of temperature at node m can be expressed as
d 2 T
dx 2
dT
dx
dT
dx
Ax
" 27*
Ax 2
Ax
Ax
(5-9)
which is the finite difference representation of the second derivative at a gen-
eral internal node m. Note that the second derivative of temperature at a node
m is expressed in terms of the temperatures at node m and its two neighboring
nodes. Then the differential equation
d 2 T
dx 2 '
8_
k
(5-10)
which is the governing equation for steady one-dimensional heat transfer in a
plane wall with heat generation and constant thermal conductivity, can be ex-
pressed in the finite difference form as (Fig. 5-8)
2T,„
Ax 2
- m + 1 . 6 »;
0,
m = 1,2,3,
,M - 1
(5-11)
where g m is the rate of heat generation per unit volume at node m. If the sur-
face temperatures T and T M are specified, the application of this equation to
each of the M — \ interior nodes results in M — 1 equations for the determi-
nation of M — 1 unknown temperatures at the interior nodes. Solving these
equations simultaneously gives the temperature values at the nodes. If the
temperatures at the outer surfaces are not known, then we need to obtain two
more equations in a similar manner using the specified boundary conditions.
Then the unknown temperatures at M + 1 nodes are determined by solving
the resulting system of M + 1 equations in M + 1 unknowns simultaneously.
Note that the boundary conditions have no effect on the finite difference
formulation of interior nodes of the medium. This is not surprising since the
control volume used in the development of the formulation does not involve
any part of the boundary. You may recall that the boundary conditions had no
effect on the differential equation of heat conduction in the medium either.
The finite difference formulation above can easily be extended to two- or
three-dimensional heat transfer problems by replacing each second derivative
by a difference equation in that direction. For example, the finite difference
formulation for steady two-dimensional heat conduction in a region with
Plane wall
Differential equation:
dx 2 k
Valid at every point
Finite difference equation:
T .-27/ +7/ . g
111 - I III III + I ^jn _
Ax 2 k ~
Valid at discrete points
KaH
FIGURE 5-8
The differential equation is valid at
every point of a medium, whereas the
finite difference equation is valid at
discrete points (the nodes) only.
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272
HEAT TRANSFER
n+1
FIGURE 5-9
Finite difference mesh for two-
dimensional conduction in
rectangular coordinates.
heat generation and constant thermal conductivity can be expressed in rectan-
gular coordinates as (Fig. 5-9)
jj, IT rp
m,n m—l,n m, n + 1
~Tx 1 +
2T
Ay 2
- m, n — 1 6m, n
(5-12)
for m = 1,2,3, ... ,M —1 and n = 1, 2, 3, . . . , N — 1 at any interior node
(m, n). Note that a rectangular region that is divided into M equal subregions
in the x-direction and N equal subregions in the y-direction has a total of
(M + 1)(N + 1) nodes, and Eq. 5-12 can be used to obtain the finite differ-
ence equations at (M — 1)(N — 1) of these nodes (i.e., all nodes except those
at the boundaries).
The finite difference formulation is given above to demonstrate how differ-
ence equations are obtained from differential equations. However, we will use
the energy balance approach in the following sections to obtain the numerical
formulation because it is more intuitive and can handle boundary conditions
more easily. Besides, the energy balance approach does not require having the
differential equation before the analysis.
Plane wall
Volume
* element
of node m
°m
*^cond, left ^^^1
x *
^^^W *~cond, ri;
A general
interior node
/
L
1 2 m-1
m
m + 1
M
X
U ►!- -1
r A.v n Ax
Ax
FIGURE 5-10
The nodal points and volume
elements for the finite difference
formulation of one-dimensional
conduction in a plane wall.
5-3 - ONE-DIMENSIONAL
STEADY HEAT CONDUCTION
In this section we will develop the finite difference formulation of heat con-
duction in a plane wall using the energy balance approach and discuss how to
solve the resulting equations. The energy balance method is based on sub-
dividing the medium into a sufficient number of volume elements and then
applying an energy balance on each element. This is done by first selecting
the nodal points (or nodes) at which the temperatures are to be determined and
then forming elements (or control volumes) over the nodes by drawing lines
through the midpoints between the nodes. This way, the interior nodes remain
at the middle of the elements, and the properties at the node such as the
temperature and the rate of heat generation represent the average properties of
the element. Sometimes it is convenient to think of temperature as varying
linearly between the nodes, especially when expressing heat conduction be-
tween the elements using Fourier's law.
To demonstrate the approach, again consider steady one-dimensional heat
transfer in a plane wall of thickness L with heat generation g(x) and constant
conductivity k. The wall is now subdivided into M equal regions of thickness
Ax = LIM in the x-direction, and the divisions between the regions are
selected as the nodes. Therefore, we have M + 1 nodes labeled 0, 1, 2, ... ,
m — 1, m, m + 1, . . . , M, as shown in Figure 5-10. The x-coordinate of any
node m is simply x m = mAx, and the temperature at that point is T(x m ) = T m .
Elements are formed by drawing vertical lines through the midpoints between
the nodes. Note that all interior elements represented by interior nodes are
full-size elements (they have a thickness of Ax), whereas the two elements at
the boundaries are half-sized.
To obtain a general difference equation for the interior nodes, consider the
element represented by node m and the two neighboring nodes m — 1 and
m + 1. Assuming the heat conduction to be into the element on all surfaces,
an energy balance on the element can be expressed as
cen58 93 3_ch05.qxd 9/4/2002 11:41 AM Page 273
/Rate of heat\
conduction
at the left
surface /
Rate of heat\ /Rate of heat\
conduction
at the right
surface
generation
inside the
element
/Rate of change \
of the energy
content of
the element /
273
CHAPTER 5
or
2 cond, left ' 2
cond, right
A£„
At
(5-13)
since the energy content of a medium (or any part of it) does not change under
steady conditions and thus AE element = 0. The rate of heat generation within
the element can be expressed as
tin * element
om
AAx
(5-14)
where g m is the rate of heat generation per unit volume in W/m 3 evaluated at
node m and treated as a constant for the entire element, and A is heat transfer
area, which is simply the inner (or outer) surface area of the wall.
Recall that when temperature varies linearly, the steady rate of heat con-
duction across a plane wall of thickness L can be expressed as
e c
kA
AT
(5-15)
where AT is the temperature change across the wall and the direction of heat
transfer is from the high temperature side to the low temperature. In the case
of a plane wall with heat generation, the variation of temperature is not linear
and thus the relation above is not applicable. However, the variation of tem-
perature between the nodes can be approximated as being linear in the deter-
mination of heat conduction across a thin layer of thickness Ax between two
nodes (Fig. 5-11). Obviously the smaller the distance Ax between two nodes,
the more accurate is this approximation. (In fact, such approximations are the
reason for classifying the numerical methods as approximate solution meth-
ods. In the limiting case of Ax approaching zero, the formulation becomes ex-
act and we obtain a differential equation.) Noting that the direction of heat
transfer on both surfaces of the element is assumed to be toward the node m,
the rate of heat conduction at the left and right surfaces can be expressed as
T — T
M cond, left K/i \
and Q
cond, right
kA
Ax
(5-16)
Substituting Eqs. 5-14 and 5-16 into Eq. 5-13 gives
kA
Ax
kA-
Ax
, AAx =
(5-17)
which simplifies to
TT + T
Ax 2
m = 1,2,3,
,M~ 1
(5-18)
Linear
T ,-T f
kA-a^ *i
Ax
kA-
A.v
FIGURE 5-1 1
In finite difference formulation, the
temperature is assumed to vary
linearly between the nodes.
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274
HEAT TRANSFER
kA-
T x -T 2
Ax
g 2 AAx
-Volume
element
of node 2
T - T, T - T .
kA—. — - ~kA—. — - + e n AAx =
v kA-
T -T
2 3
Ax
Ax
Ax
T. - 271 + 71 + g 1 AAx z /k =
(a) Assuming heat transfer to be out of the
volume element at the right surface.
kA-
T x -T 2
Ax
g 2 AAx
-Volume
element
of node 2
T -71 71 - T .
kA -± — - + kA -*- — - + g^AAx =
<kA-
Ax
Ax
Ax
T l - 2T 2 + T 3 + g 2 AAx l /k =
(b) Assuming heat transfer to be into the
volume element at all surfaces.
FIGURE 5-12
The assumed direction of heat transfer
at surfaces of a volume element has
no effect on the finite difference
formulation.
which is identical to the difference equation (Eq. 5-11) obtained earlier.
Again, this equation is applicable to each of the M — 1 interior nodes, and its
application gives M — 1 equations for the determination of temperatures at
M + 1 nodes. The two additional equations needed to solve for the M + 1 un-
known nodal temperatures are obtained by applying the energy balance on the
two elements at the boundaries (unless, of course, the boundary temperatures
are specified).
You are probably thinking that if heat is conducted into the element from
both sides, as assumed in the formulation, the temperature of the medium will
have to rise and thus heat conduction cannot be steady. Perhaps a more realis-
tic approach would be to assume the heat conduction to be into the element on
the left side and out of the element on the right side. If you repeat the formu-
lation using this assumption, you will again obtain the same result since the
heat conduction term on the right side in this case will involve T m — T m + j in-
stead of T m + j — T m , which is subtracted instead of being added. Therefore,
the assumed direction of heat conduction at the surfaces of the volume ele-
ments has no effect on the formulation, as shown in Figure 5-12. (Besides, the
actual direction of heat transfer is usually not known.) However, it is conve-
nient to assume heat conduction to be into the element at all surfaces and not
worry about the sign of the conduction terms. Then all temperature differences
in conduction relations are expressed as the temperature of the neighboring
node minus the temperature of the node under consideration, and all conduc-
tion terms are added.
Boundary Conditions
Above we have developed a general relation for obtaining the finite difference
equation for each interior node of a plane wall. This relation is not applicable
to the nodes on the boundaries, however, since it requires the presence of
nodes on both sides of the node under consideration, and a boundary node
does not have a neighboring node on at least one side. Therefore, we need to
obtain the finite difference equations of boundary nodes separately. This is
best done by applying an energy balance on the volume elements of boundary
nodes.
Boundary conditions most commonly encountered in practice are the spec-
ified temperature, specified heat flux, convection, and radiation boundary
conditions, and here we develop the finite difference formulations for them
for the case of steady one-dimensional heat conduction in a plane wall of
thickness L as an example. The node number at the left surface at x = is 0,
and at the right surface at x = L it is M. Note that the width of the volume el-
ement for either boundary node is Ax/2.
The specified temperature boundary condition is the simplest boundary
condition to deal with. For one-dimensional heat transfer through a plane wall
of thickness L, the specified temperature boundary conditions on both the left
and right surfaces can be expressed as (Fig. 5-13)
7"(0) = T Q = Specified value
T{L) = T M = Specified value
(5-19)
where T and T„, are the specified temperatures at surfaces at x = and x = L,
respectively. Therefore, the specified temperature boundary conditions are
cen58 93 3_ch05.qxd 9/4/2002 11:41 AM Page 275
incorporated by simply assigning the given surface temperatures to the bound-
ary nodes. We do not need to write an energy balance in this case unless we
decide to determine the rate of heat transfer into or out of the medium after the
temperatures at the interior nodes are determined.
When other boundary conditions such as the specified heat flux, convection,
radiation, or combined convection and radiation conditions are specified at a
boundary, the finite difference equation for the node at that boundary is ob-
tained by writing an energy balance on the volume element at that boundary.
The energy balance is again expressed as
2 e
(5-20)
275
CHAPTER 5
35°C
Plane wall
82°C
L
1
2 •••
M
for heat transfer under steady conditions. Again we assume all heat transfer to
be into the volume element from all surfaces for convenience in formulation,
except for specified heat flux since its direction is already specified. Specified
heat flux is taken to be a positive quantity if into the medium and a negative
quantity if out of the medium. Then the finite difference formulation at the
node m = (at the left boundary where x = 0) of a plane wall of thickness L
during steady one-dimensional heat conduction can be expressed as (Fig.
5-14)
Qv
kA
Ax
g (AAx/2) =
(5-21)
where AAx/2 is the volume of the volume element (note that the boundary ele-
ment has half thickness), g is the rate of heat generation per unit volume (in
W/m 3 ) at x = 0, and A is the heat transfer area, which is constant for a plane
wall. Note that we have Ax in the denominator of the second term instead of
Ax/2. This is because the ratio in that term involves the temperature difference
between nodes and 1, and thus we must use the distance between those two
nodes, which is Ax.
The finite difference form of various boundary conditions can be obtained
from Eq. 5-21 by replacing Q left surface by a suitable expression. Next this is
done for various boundary conditions at the left boundary.
1. Specified Heat Flux Boundary Condition
FIGURE 5-13
Finite difference formulation of
specified temperature boundary
conditions on both surfaces
of a plane wall.
Ax
2
^- Volume element
of node
So
k i^^B T -T
surface
Ax
L
1 2 •••
x * • Ax— H
X
+ kA-
■-0
o • , Ax
FIGURE 5-14
Schematic for the finite difference
formulation of the left boundary
node of a plane wall.
(joA + kA
Ax
+ g (AAx/2) =
(5-22)
Special case: Insulated Boundary (q = 0)
T, -T,
. -^j^ + £ (AAx/2) =
(5-23)
2. Convection Boundary Condition
Tt-T
hA{T.„ - T ) + kA
Ax
+ g (AAx/2) =
(5-24)
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276
HEAT TRANSFER
FIGURE 5-15
Schematic for the finite difference
formulation of combined convection
and radiation on the left boundary
of a plane wall.
k
Medium A
k A
T , -T
A 11! - 1 1)1
» A Ax
*4,m
&B,m
Interface
Medium B
k B
V '" + Ax
m
m-\
m
m + 1
H
X
Ax
2
Ax
2
VI
T ,-T T ,-T
k.A-^ m -+k„A-^ ffi
4 Ajc b Ax
Ax
A^ =
71 2
FIGURE 5-16
Schematic for the finite difference
formulation of the interface boundary
condition for two mediums A and B
that are in perfect thermal contact.
3. Radiation Boundary Condition
eaA(T* m - T 4 ) + kA ■
Ax
MAAx/2) =
(5-25)
4. Combined Convection and Radiation Boundary Condition
(Fig. 5-15)
hA(T m - T ) + euA(T* an - T+) + kA -^j^ + g (AAx/2) =
or
,A(T x -T a ) + kA
Ax
+ g a {AAxl2) =
(5-26)
(5-27)
5. Combined Convection, Radiation, and Heat Flux Boundary
Condition
qoA + *A(r« - r ) + eaA(r s 4 urr - T 4 ) + kA -^j^ + g (AA.x/2) = (5-28)
6. Interface Boundary Condition Two different solid media A and B
are assumed to be in perfect contact, and thus at the same temperature
at the interface at node m (Fig. 5-16). Subscripts A and B indicate
properties of media A and B, respectively.
T
Ax
+ k R A
T
Ax
g A . ,„{AAxl2) + g Bt m (AAx/2) = (5-29)
In these relations, q is the specified heat flux in W/m 2 , h is the convection
coefficient, h combmed is the combined convection and radiation coefficient, r m is
the temperature of the surrounding medium, T san . is the temperature of the
surrounding surfaces, e is the emissivity of the surface, and cr is the Stefan-
Boltzman constant. The relations above can also be used for node M on the
right boundary by replacing the subscript "0" by "M" and the subscript "1" by
"M - 1".
Note that absolute temperatures must be used in radiation heat transfer
calculations, and all temperatures should be expressed in K or R when a
boundary condition involves radiation to avoid mistakes. We usually try to
avoid the radiation boundary condition even in numerical solutions since it
causes the finite difference equations to be nonlinear, which are more difficult
to solve.
Treating Insulated Boundary Nodes as Interior Nodes:
The Mirror Image Concept
One way of obtaining the finite difference formulation of a node on an insu-
lated boundary is to treat insulation as "zero" heat flux and to write an energy
balance, as done in Eq. 5-23. Another and more practical way is to treat the
node on an insulated boundary as an interior node. Conceptually this is done
cen58 93 3_ch05.qxd 9/4/2002 11:41 AM Page 277
by replacing the insulation on the boundary by a mirror and considering the
reflection of the medium as its extension (Fig. 5-17). This way the node next
to the boundary node appears on both sides of the boundary node because of
symmetry, converting it into an interior node. Then using the general formula
(Eq. 5-18) for an interior node, which involves the sum of the temperatures of
the adjoining nodes minus twice the node temperature, the finite difference
formulation of a node m = on an insulated boundary of a plane wall can be
expressed as
2T„, + T m
Sm
Ax 2
Mn
Ax 2
(5-30)
which is equivalent to Eq. 5-23 obtained by the energy balance approach.
The mirror image approach can also be used for problems that possess ther-
mal symmetry by replacing the plane of symmetry by a mirror. Alternately, we
can replace the plane of symmetry by insulation and consider only half of the
medium in the solution. The solution in the other half of the medium is sim-
ply the mirror image of the solution obtained.
277
CHAPTER 5
Insulation
Insulated
boundary
,- node
Mirror
Mirror
image
1
Equivalent
interior
, node
1
FIGURE 5-17
A node on an insulated boundary
can be treated as an interior node by
replacing the insulation by a mirror.
EXAMPLE 5-1 Steady Heat Conduction in a Large Uranium Plate
Consider a large uranium plate of thickness L = 4 cm and thermal conductivity
k = 28 W/m • °C in which heat is generated uniformly at a constant rate of
g = 5 X 10 6 W/m 3 . One side of the plate is maintained at 0°C by iced water
while the other side is subjected to convection to an environment at T x = 30°C
with a heat transfer coefficient of h = 45 W/m 2 ■ C C, as shown in Figure 5-18.
Considering a total of three equally spaced nodes in the medium, two at the
boundaries and one at the middle, estimate the exposed surface temperature of
the plate under steady conditions using the finite difference approach.
SOLUTION A uranium plate is subjected to specified temperature on one side
and convection on the other. The unknown surface temperature of the plate is
to be determined numerically using three equally spaced nodes.
Assumptions 1 Heat transfer through the wall is steady since there is no in-
dication of any change with time. 2 Heat transfer is one-dimensional since
the plate is large relative to its thickness. 3 Thermal conductivity is constant.
4 Radiation heat transfer is negligible.
Properties The thermal conductivity is given to be k = 28 W/m • °C.
Analysis The number of nodes is specified to be M = 3, and they are chosen
to be at the two surfaces of the plate and the midpoint, as shown in the figure.
Then the nodal spacing Ax becomes
Ax
M- 1
0.04 m
3 - 1
0.02 m
We number the nodes 0, 1, and 2. The temperature at node is given to be
T = C C, and the temperatures at nodes 1 and 2 are to be determined. This
problem involves only two unknown nodal temperatures, and thus we need to
have only two equations to determine them uniquely. These equations are ob-
tained by applying the finite difference method to nodes 1 and 2.
Uranium
plate
o°c
k = 28 W/m-°C
g = 5x 10 6 W/m 3
h
T„
0<
L
1
2
X
FIGURE 5-18
Schematic for Example 5—1.
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278
HEAT TRANSFER
Plate
It
Finite difference solution:
Exact solution:
T 2 = 136.0°C
FIGURE 5-19
Despite being approximate in nature,
highly accurate results can be
obtained by numerical methods.
Node 1 is an interior node, and the finite difference formulation at that node
is obtained directly from Eq. 5-18 by setting m = 1:
2T, + T 2
Ax 2
- 277] + T 2
I?
2T, - T,
gA* 2
(1)
Node 2 is a boundary node subjected to convection, and the finite difference
formulation at that node is obtained by writing an energy balance on the volume
element of thickness Ax/2 at that boundary by assuming heat transfer to be into
the medium at all sides:
hA(T a -T 2 ) + kA-
T 2
Ax
+ g\(AAxl2) =
Canceling the heat transfer area A and rearranging give
r,
1 + h -f\T 7
hAx gAx 2
k °° 2k
(2)
Equations (1) and (2) form a system of two equations in two unknowns T x and
T 2 . Substituting the given quantities and simplifying gives
2T { - T 2
- 1.0327;
71.43
-36.68
(in °C)
(in °C)
This is a system of two algebraic equations in two unknowns and can be solved
easily by the elimination method. Solving the first equation for T x and substi-
tuting into the second equation result in an equation in 7~ 2 whose solution is
T 2 = 136.1°C
This is the temperature of the surface exposed to convection, which is the
desired result. Substitution of this result into the first equation gives 7"! =
103. 8 C C, which is the temperature at the middle of the plate.
Discussion The purpose of this example is to demonstrate the use of the finite
difference method with minimal calculations, and the accuracy of the result
was not a major concern. But you might still be wondering how accurate the re-
sult obtained above is. After all, we used a mesh of only three nodes for the
entire plate, which seems to be rather crude. This problem can be solved ana-
lytically as described in Chapter 2, and the analytical (exact) solution can be
shown to be
T(x)
0.5ghL 2 /k + gL+ T^h gx 2
hL+ k X ~2k
Substituting the given quantities, the temperature of the exposed surface of the
plate at x = L = 0.04 m is determined to be 136. 0°C, which is almost identi-
cal to the result obtained here with the approximate finite difference method
(Fig. 5-19). Therefore, highly accurate results can be obtained with numerical
methods by using a limited number of nodes.
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279
CHAPTER 5
EXAMPLE 5-2 Heat Transfer from Triangular Fins
Consider an aluminum alloy fin (k = 180 W/m • °C) of triangular cross section
with length L = 5 cm, base thickness b = 1 cm, and very large width w in the
direction normal to the plane of paper, as shown in Figure 5-20. The base of
the fin is maintained at a temperature of T = 200°C. The fin is losing heat
to the surrounding medium at 7" x = 25°C with a heat transfer coefficient of
ft = 15 W/m 2 • °C. Using the finite difference method with six equally spaced
nodes along the fin in the x-direction, determine (a) the temperatures at the
nodes, (£>) the rate of heat transfer from the fin for w = 1 m, and (c) the fin
efficiency.
SOLUTION A long triangular fin attached to a surface is considered. The nodal
temperatures, the rate of heat transfer, and the fin efficiency are to be deter-
mined numerically using six equally spaced nodes.
Assumptions 1 Heat transfer is steady since there is no indication of any
change with time. 2 The temperature along the fin varies in the x direction only.
3 Thermal conductivity is constant. 4 Radiation heat transfer is negligible.
Properties The thermal conductivity is given to be k = 180 W/m ■ °C.
Analysis (a) The number of nodes in the fin is specified to be M = 6, and their
location is as shown in the figure. Then the nodal spacing Ax becomes
Ax
M- 1
0.05 m
6 - 1
0.01 m
The temperature at node is given to be T = 200°C, and the temperatures at
the remaining five nodes are to be determined. Therefore, we need to have five
equations to determine them uniquely. Nodes 1, 2, 3, and 4 are interior nodes,
and the finite difference formulation for a general interior node m is obtained
by applying an energy balance on the volume element of this node. Noting that
heat transfer is steady and there is no heat generation in the fin and assuming
heat transfer to be into the medium at all sides, the energy balance can be ex-
pressed as
2 e = o
M„
Ax
kA
right
Ax
hA conv (T^ - TJ =
Note that heat transfer areas are different for each node in this case, and using
geometrical relations, they can be expressed as
A lcft = (Height X Width) s
(Height X Width) 4
right
2w[L - (m - l/2)Ax]tan Q
2w[L - (m + l/2)Ax]tan 6
2 X Length X Width = 2w(Ax/cos 0)
Substituting,
2kw[L - (m - |)Ax]tan ■
2kw[L - (m + i)Ax]tan Q
Ax
Ax
l,n | , 2wAx,_ „ . „
— + h ^(T„ - TJ =
cos
[L-(m — )A.v]tan (
FIGURE 5-20
Schematic for Example 5-2 and the
volume element of a general
interior node of the fin.
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280
HEAT TRANSFER
Ax/2
FIGURE 5-21
Schematic of the volume element of
node 5 at the tip of a triangular fin.
Dividing each term by 2kwL tan 0/Ax gives
(.T m - 1 ~T m ) +
1 - (m - |) —
1 / _l_ K A*
1 - (m + i) y
+
h(Ax) 1
kL sin
(r. - rj = o
Note that
tan
fr/2 = 0.5 cm
L 5 cm
0.1
= tan-'O.l = 5.71°
Also, sin 5.71° = 0.0995. Then the substitution of known quantities gives
(5.5 - m)T„,
(10.00838 - 2m)T„, + (4.5 - m)T„,
-0.209
Now substituting 1, 2, 3, and 4 for m results in these finite difference equa-
tions for the interior nodes.-
m = 1
m = 2
m = 3
m = 4
-900.209
3.57*! - 6.008387 2 + 2.5T 3 = -0.209
2.5T 2 - 4.008387 3 + 1.5T 4 = -0.209
I.57/3 - 2.008387 4 + 0.57 5 = -0.209
(1)
(2)
(3)
(4)
The finite difference equation for the boundary node 5 is obtained by writing an
energy balance on the volume element of length Ax/2 at that boundary, again by
assuming heat transfer to be into the medium at all sides (Fig. 5-21):
kA„
7 4 ~7 5
Ax
+ M conv (T« - 7 5 ) =
where
2w — tan
and
A-
2w
Ax/2
cos
Canceling w in all terms and substituting the known quantities gives
7 4 - 1.008387 5 = -0.209
(5)
Equations (1) through (5) form a linear system of five algebraic equations in five
unknowns. Solving them simultaneously using an equation solver gives
7, = 198.6°C, 7 2 = 197.1°C, 7 3 = 195.7°C,
7 4 = 194.3°C, 7 5 = 192.9°C
which is the desired solution for the nodal temperatures.
(b) The total rate of heat transfer from the fin is simply the sum of the heat
transfer from each volume element to the ambient, and for w= 1 m it is deter-
mined from
cen58 93 3_ch05.qxd 9/4/2002 11:41 AM Page 281
5 5
x£ fin £j x£ element, m
in = m =
£j '^*conv, fflV m ^ «=/
Noting that the heat transfer surface area is wAx/cos 6 for the boundary nodes
and 5, and twice as large for the interior nodes 1, 2, 3, and 4, we have
G fi „ = h
h
wAx
[(T - t„) + 2(r, - rj + 2(T 2 - r„) + 2(r 3 - r„)
cos G
2(r 4 - r„) + (r 5 - r„)]
wAx
cos G
[r + 2(T, + r 2 + r 3 + r 4 ) + r 5 - iotj
(1 m)(0.01 m)
(15 W/m 2 • °C) ,- 10 [200 + 2 X 785.7 + 192.9 - 10 X 25]
cos 5.71°
258.4 W
(c) If the entire fin were at the base temperature of T = 200°C, the total rate
of heat transfer from the fin for w = 1 m would be
e„
hA„
(T - r„) = h(2wL/cos Q)(T - T m )
(15 W/m 2 • °C)[2(1 m)(0.05 m)/cos5.71°](200 - 25)°C
263.8 W
Then the fin efficiency is determined from
gfin 258.4 W
%in
e,
263.8 W
0.98
which is less than 1, as expected. We could also determine the fin efficiency in
this case from the proper fin efficiency curve in Chapter 3, which is based on
the analytical solution. We would read 0.98 for the fin efficiency, which is iden-
tical to the value determined above numerically.
281
CHAPTER 5
The finite difference formulation of steady heat conduction problems usu-
ally results in a system of N algebraic equations in N unknown nodal temper-
atures that need to be solved simultaneously. When N is small (such as 2 or 3),
we can use the elementary elimination method to eliminate all unknowns ex-
cept one and then solve for that unknown (see Example 5-1). The other un-
knowns are then determined by back substitution. When N is large, which is
usually the case, the elimination method is not practical and we need to use a
more systematic approach that can be adapted to computers.
There are numerous systematic approaches available in the literature, and
they are broadly classified as direct and iterative methods. The direct meth-
ods are based on a fixed number of well-defined steps that result in the solu-
tion in a systematic manner. The iterative methods, on the other hand, are
based on an initial guess for the solution that is refined by iteration until a
specified convergence criterion is satisfied (Fig. 5-22). The direct methods
usually require a large amount of computer memory and computation time,
Direct methods:
Solve in a systematic manner following a
series of well-defined steps.
Iterative methods:
Start with an initial guess for the solution,
and iterate until solution converges.
FIGURE 5-22
Two general categories of solution
methods for solving systems
of algebraic equations.
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HEAT TRANSFER
and they are more suitable for systems with a relatively small number of equa-
tions. The computer memory requirements for iterative methods are minimal,
and thus they are usually preferred for large systems. The convergence of it-
erative methods to the desired solution, however, may pose a problem.
y
N
n + 1
\y
Node (m, n)
n-l
\y
2
1
Ax
Ar
1 2
| « 1 • • • M x
m — 1 m + 1
FIGURE 5-23
The nodal network for the finite
difference formulation of two-
dimensional conduction in
rectangular coordinates.
n + 1
Ay
Volume
element n
m — 1, n I
Ay
n-l
-Ax-
m, n + 1
m, n
-+-
I
m, n-l
-Ax-
m + 1 , n
I
y t m - 1
m + 1
FIGURE 5-24
The volume element of a general
interior node (m, n) for two-
dimensional conduction in
rectangular coordinates.
5^ - TWO-DIMENSIONAL
STEADY HEAT CONDUCTION
In Section 5-3 we considered one-dimensional heat conduction and assumed
heat conduction in other directions to be negligible. Many heat transfer prob-
lems encountered in practice can be approximated as being one-dimensional,
but this is not always the case. Sometimes we need to consider heat transfer in
other directions as well when the variation of temperature in other directions
is significant. In this section we will consider the numerical formulation and
solution of two-dimensional steady heat conduction in rectangular coordinates
using the finite difference method. The approach presented below can be ex-
tended to three-dimensional cases.
Consider a rectangular region in which heat conduction is significant in the
x- and y-directions. Now divide the x-y plane of the region into a rectangular
mesh of nodal points spaced Ax and Ay apart in the x- and y-directions,
respectively, as shown in Figure 5-23, and consider a unit depth of Az = 1
in the z-direction. Our goal is to determine the temperatures at the nodes,
and it is convenient to number the nodes and describe their position by
the numbers instead of actual coordinates. A logical numbering scheme for
two-dimensional problems is the double subscript notation (m, n) where
m = 0,1,2, ... ,M is the node count in the x-direction and n = 0, 1, 2, . . . ,N
is the node count in the _y-direction. The coordinates of the node (m, n) are
simply x = mAx and y = nAy, and the temperature at the node (m, n) is
denoted by T nun .
Now consider a volume element of size Ax X Ay X 1 centered about a gen-
eral interior node (m, n) in a region in which heat is generated at a rate of g and
the thermal conductivity k is constant, as shown in Figure 5-24. Again
assuming the direction of heat conduction to be toward the node under
consideration at all surfaces, the energy balance on the volume element can be
expressed as
f Rate of heat conduction \
at the left, top, right,
\ and bottom surfaces /
+
I Rate of heat |
generation inside
\ the element
I Rate of change of \
the energy content
\ of the element /
or
t^cond, left ~*~ second, top ~*~ \i cond. right "*" ticond, bottom ~*~ ^eL
A£„
At
(5-31)
for the steady case. Again assuming the temperatures between the adja-
cent nodes to vary linearly and noting that the heat transfer area is
A x = Ay X 1 = Ay in the .^-direction and A y = Ax X 1 = Ax in the y-direction,
the energy balance relation above becomes
cen58 93 3_ch05.qxd 9/4/2002 11:41 AM Page 283
T — T
m—\,n m, n
kAy — + kAx — : — r + kAy
Ax Ay Ax
kAx — + g nh „ Ax Ay = (5-32)
Dividing each term by Ax X Ay and simplifying gives
(5-33)
* m— I, n ^"* m, n m+ 1, n * m, n — 1 in, n * m, n + 1 am, n
Ax~ 2 + A^ + ~k~
form= 1, 2, 3, ... ,M — 1 and n = 1, 2, 3, . . . , N — 1. This equation is iden-
tical to Eq. 5-12 obtained earlier by replacing the derivatives in the differen-
tial equation by differences for an interior node (m, n). Again a rectangular
region M equally spaced nodes in the x-direction and N equally spaced nodes
in the y-direction has a total of (M + 1)(7V + 1) nodes, and Eq. 5-33 can be
used to obtain the finite difference equations at all interior nodes.
In finite difference analysis, usually a square mesh is used for sim-
plicity (except when the magnitudes of temperature gradients in the x- and
y-directions are very different), and thus Ax and Ay are taken to be the same.
Then Ax = Ay = /, and the relation above simplifies to
g mn p
T m - i, „ + T m + (] „ + T m „ + i + T m „ _ j — 4T m ^ „ H - = (5-34)
That is, the finite difference formulation of an interior node is obtained by
adding the temperatures of the four nearest neighbors of the node, subtracting
four times the temperature of the node itself and adding the heat generation
term. It can also be expressed in this form, which is easy to remember:
'ka "r" T , op + T nght + i bottom — 4i node H - = (5-35)
When there is no heat generation in the medium, the finite difference equa-
tion for an interior node further simplifies to r node = (r left + r top + r right +
r bottom )/4, which has the interesting interpretation that the temperature of each
interior node is the arithmetic average of the temperatures of the four neigh-
boring nodes. This statement is also true for the three-dimensional problems
except that the interior nodes in that case will have six neighboring nodes in-
stead of four.
Boundary Nodes
The development of finite difference formulation of boundary nodes in two-
(or three-) dimensional problems is similar to the development in the one-
dimensional case discussed earlier. Again, the region is partitioned between
the nodes by forming volume elements around the nodes, and an energy bal-
ance is written for each boundary node. Various boundary conditions can be
handled as discussed for a plane wall, except that the volume elements
in the two-dimensional case involve heat transfer in the y-direction as well as
the x-direction. Insulated surfaces can still be viewed as "mirrors, " and the
283
CHAPTER 5
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HEAT TRANSFER
Volume element
of node 2
Boundary
subjected
/ j to convection
e,ef, + e top + Q nght + e bollom + — = o
FIGURE 5-25
The finite difference formulation of
a boundary node is obtained by
writing an energy balance
on its volume element.
Convection
y
i /', T„
1 2 /"
3 / \
t I
Av
J
/ \ Ax = Ay = I
4 i
5 !
6 7 \ t 8 9 t Vr
t
Av
1
10]
— h —
h —
12]
h —
13]
h —
14] 15
I
90°C
A'
*-Ax^
*-Ax^
-^A*^
--Ajt-
»-A*-
FIGURE 5-26
Schematic for Example 5-3 and
the nodal network (the boundaries
of volume elements of the nodes are
indicated by dashed lines).
h,T„
I
h,T a
Ti
(b) Node 2
(a) Node 1
FIGURE 5-27
Schematics for energy balances on the
volume elements of nodes 1 and 2.
mirror image concept can be used to treat nodes on insulated boundaries as in-
terior nodes.
For heat transfer under steady conditions, the basic equation to keep in mind
when writing an energy balance on a volume element is (Fig. 5-25)
S Q + SKl
(5-36)
whether the problem is one-, two-, or three-dimensional. Again we assume,
for convenience in formulation, all heat transfer to be into the volume ele-
ment from all surfaces except for specified heat flux, whose direction is al-
ready specified. This is demonstrated in Example 5-3 for various boundary
conditions.
EXAMPLE 5-3 Steady Two-Dimensional Heat Conduction
in L-Bars
Consider steady heat transfer in an L-shaped solid body whose cross section is
given in Figure 5-26. Heat transfer in the direction normal to the plane of the
paper is negligible, and thus heat transfer in the body is two-dimensional. The
thermal conductivity of the body is k = 15 W/m • °C, and heat is generated in
the body at a rate of g = 2 X 10 6 W/m 3 . The left surface of the body is insu-
lated, and the bottom surface is maintained at a uniform temperature of 90°C.
The entire top surface is subjected to convection to ambient air at 7" x = 25°C
with a convection coefficient of h = 80 W/m 2 • °C, and the right surface is sub-
jected to heat flux at a uniform rate of q R = 5000 W/m 2 . The nodal network of
the problem consists of 15 equally spaced nodes with Ax = Ay = 1.2 cm, as
shown in the figure. Five of the nodes are at the bottom surface, and thus their
temperatures are known. Obtain the finite difference equations at the remain-
ing nine nodes and determine the nodal temperatures by solving them.
SOLUTION Heat transfer in a long L-shaped solid bar with specified boundary
conditions is considered. The nine unknown nodal temperatures are to be de-
termined with the finite difference method.
Assumptions 1 Heat transfer is steady and two-dimensional, as stated. 2 Ther-
mal conductivity is constant. 3 Heat generation is uniform. 4 Radiation heat
transfer is negligible.
Properties The thermal conductivity is given to be k = 15 W/m • °C.
Analysis We observe that all nodes are boundary nodes except node 5, which
is an interior node. Therefore, we will have to rely on energy balances to obtain
the finite difference equations. But first we form the volume elements by parti-
tioning the region among the nodes equitably by drawing dashed lines between
the nodes. If we consider the volume element represented by an interior node
to be full size (i.e., Ax X Ay X 1), then the element represented by a regular
boundary node such as node 2 becomes half size (i.e., Ax X Ay/2 X 1), and
a corner node such as node 1 is quarter size (i.e., Ax/2 X Ay/2 X 1). Keeping
Eq. 5-36 in mind for the energy balance, the finite difference equations for
each of the nine nodes are obtained as follows:
(a) Node 1. The volume element of this corner node is insulated on the left and
subjected to convection at the top and to conduction at the right and bottom
surfaces. An energy balance on this element gives [Fig. 5-27a]
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285
CHAPTER 5
Ax Ay r 7 — T, A r T, — T, Ai Ay
+ 'T^-^*Tir +t TV + ^r°
Taking Ax = Ay = /, it simplifies to
hi.
SI'
2k
(£>) Node 2. The volume element of this boundary node is subjected to con-
vection at the top and to conduction at the right, bottom, and left surfaces. An
energy balance on this element gives [Fig. 5-276]
Ay 7, -T 2 T s - T 2 Ay T, - T 2 Av
hax(T^ - T 2 ) + k^-^—^ + kax^—^ + k^f^—^ + g 2 Ax— =
2 Ax Ay 2 ax 2
Taking Ax = Ay = /, it simplifies to
/ 2hl\ 2hl R2I 2
r t - (4 + ±f\ T 2 + T,+ 2T 5 = -^r_ - =p
(c) Node 3. The volume element of this corner node is subjected to convection
at the top and right surfaces and to conduction at the bottom and left surfaces.
An energy balance on this element gives [Fig. 5-28a]
^♦£W.
T 3 )
axT 6
Ay
+ k
Ay T 2 - T,
Ax Ay
Taking Ax = Ay = /, it simplifies to
, 2hl
T 3 + T 6
2hl,
Ax + ^ 3 "2 2
2k
(d) Node 4. This node is on the insulated boundary and can be treated as an
interior node by replacing the insulation by a mirror. This puts a reflected image
of node 5 to the left of node 4. Noting that Ax = Ay = /, the general interior
node relation for the steady two-dimensional case (Eq. 5-35) gives [Fig. 5-286]
T 5 + F, + T 5 + T l0 -4T 4 + ^ =
or, noting that T 10 = 90° C,
T, - 47*. + 27\
-90
&/ 2
(e) Node 5. This is an interior node, and noting that Ax = Ay = /, the finite
difference formulation of this node is obtained directly from Eq. 5-35 to be
[Fig. 5-29a]
8sl 2
T 4 + T 2 + T 6 + T u -4T 5 + — =
h,T m
Mirror
(5)
h,T~
EH
— 1 ♦
10
(a) Node 3
(b) Node 4
FIGURE 5-28
Schematics for energy balances on the
volume elements of nodes 3 and 4.
♦ 2
4 1
4 1-
— 4 1 — P,
5 1 1
ill
12
(a) Node 5
(6) Node 6
FIGURE 5-29
Schematics for energy balances on the
volume elements of nodes 5 and 6.
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286
HEAT TRANSFER
tv i
1r
15
13 J
FIGURE 5-30
Schematics for energy balances on the
volume elements of nodes 7 and 9.
or, noting that T n = 90°C,
T 2 + T 4 - AT, + T 6
-90
g S l 7
if) Node 6. The volume element of this inner corner node is subjected to con-
vection at the L-shaped exposed surface and to conduction at other surfaces.
An energy balance on this element gives [Fig. 5-296]
# + t)<-
Ay TV
kAx-
T„
, a ^5 ~~ T 6 \ x T 3 - T 6
kAy — ; h k
2 Ax Ay
3AxAy
Ax
Ay
Taking Ax = Ay = /and noting that T l2 = 90°C, it simplifies to
T 3 + 2T S - 6 +
2hl
T< + T 7
2hl
180 -^r.
k
3g 6 / 2
2fc
(g) Node 7. The volume element of this boundary node is subjected to convec-
tion at the top and to conduction at the right, bottom, and left surfaces. An en-
ergy balance on this element gives [Fig. 5-30a]
hbx(T a - r 7 )
+ k
AyT s
2 Ax
Ay T 6 - TV
kAx
T 13
Ay
Ax
Av
g 7 Ax— =
Taking Ax = Ay = /and noting that T 13 = 90°C, it simplifies to
4 + ^]:
2hl
■180 -^tv
k
k
{h) Node 8. This node is identical to Node 7, and the finite difference formu-
lation of this node can be obtained from that of Node 7 by shifting the node
numbers by 1 (i.e., replacing subscript m by m + 1). It gives
T 7
4+fr
. 180 _2« *;
k k
(/') Node 9. The volume element of this corner node is subjected to convection
at the top surface, to heat flux at the right surface, and to conduction at the
bottom and left surfaces. An energy balance on this element gives [Fig. 5-306]
, Ax
h^(T„
T 9 ) + q R
Ay , , Ax T v .
AyT %
Ay
Ax
Ax Ay
2 2
Taking Ax = Ay = /and noting that 7" 15 = 90°C, it simplifies to
T K -\2 + ^)T Q = -90
k k " 2k
cen58 93 3_ch05.qxd 9/4/2002 11:41 AM Page 287
This completes the development of finite difference formulation for this prob-
lem. Substituting the given quantities, the system of nine equations for the
determination of nine unknown nodal temperatures becomes
-2.0647*, + T 2 + T 4 =
-11.2
4.1287; + r 3 + 27* 5 =
-22.4
T 2 - 2.1287* 3 + T 6 =
-12.8
T { - 47* 4 + 27* 5 =
-109.2
T 2 + T 4 - 4T 5 + T 6 =
-109.2
2T 5 - 6.1287* 6 + T 7 =
-212.0
T 6 - 4A28T-, + T g =
-202.4
T 7 - 4.128r 8 + T 9 =
-202.4
7g - 2.0647*, =
-105.2
which is a system of nine algebraic equations with nine unknowns. Using an
equation solver, its solution is determined to be
T,=
112.1°C
T 2 =
110.8°C
7 , 3 =
106.6°C
T 4 =
109.4°C
T s =
108.1°C
T 6 =
103.2°C
T 7 =
97.3°C
T s =
96.3°C
T 9 =
97.6°C
Note that the temperature is the highest at node 1 and the lowest at node 8.
This is consistent with our expectations since node 1 is the farthest away from
the bottom surface, which is maintained at 90°C and has one side insulated,
and node 8 has the largest exposed area relative to its volume while being close
to the surface at 90°C.
287
CHAPTER 5
Irregular Boundaries
In problems with simple geometries, we can fill the entire region using simple
volume elements such as strips for a plane wall and rectangular elements for
two-dimensional conduction in a rectangular region. We can also use cylin-
drical or spherical shell elements to cover the cylindrical and spherical bodies
entirely. However, many geometries encountered in practice such as turbine
blades or engine blocks do not have simple shapes, and it is difficult to fill
such geometries having irregular boundaries with simple volume elements.
A practical way of dealing with such geometries is to replace the irregular
geometry by a series of simple volume elements, as shown in Figure 5-31.
This simple approach is often satisfactory for practical purposes, especially
when the nodes are closely spaced near the boundary. More sophisticated ap-
proaches are available for handling irregular boundaries, and they are com-
monly incorporated into the commercial software packages.
EXAMPLE 5-4 Heat Loss through Chimneys
Hot combustion gases of a furnace are flowing through a square chimney made
of concrete (k = 1.4 W/m • °C). The flow section of the chimney is 20 cm X
20 cm, and the thickness of the wall is 20 cm. The average temperature of the
Actual boundary
^- Appro
ximation
FIGURE 5-31
Approximating an irregular
boundary with a rectangular mesh.
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HEAT TRANSFER
hot gases in the chimney is T, = 300°C, and the average convection heat trans-
fer coefficient inside the chimney is h, = 70 W/m 2 • °C. The chimney is losing
heat from its outer surface to the ambient air at T = 20 C C by convection with
a heat transfer coefficient of h = 21 W/m 2 • °C and to the sky by radiation. The
emissivity of the outer surface of the wall is e = 0.9, and the effective sky tem-
perature is estimated to be 260 K. Using the finite difference method with
Ax = Ay = 10 cm and taking full advantage of symmetry, determine the
temperatures at the nodal points of a cross section and the rate of heat loss for
a 1-m-long section of the chimney.
Symmetry lines
(Equivalent to insulation)
Representative
7" k section of chimney
FIGURE 5-32
Schematic of the chimney discussed in
Example 5^4 and the nodal network
for a representative section.
h,T„
h,T w
1
\
2
V
.4
(a) Node 1 (b) Node 2
FIGURE 5-33
Schematics for energy balances on the
volume elements of nodes 1 and 2.
SOLUTION Heat transfer through a square chimney is considered. The nodal
temperatures and the rate of heat loss per unit length are to be determined with
the finite difference method.
Assumptions 1 Heat transfer is steady since there is no indication of change
with time. 2 Heat transfer through the chimney is two-dimensional since the
height of the chimney is large relative to its cross section, and thus heat con-
duction through the chimney in the axial direction is negligible. It is tempting
to simplify the problem further by considering heat transfer in each wall to be
one-dimensional, which would be the case if the walls were thin and thus the
corner effects were negligible. This assumption cannot be justified in this case
since the walls are very thick and the corner sections constitute a considerable
portion of the chimney structure. 3 Thermal conductivity is constant.
Properties The properties of chimney are given to be k = 1.4 W/m • °C and
b= 0.9.
Analysis The cross section of the chimney is given in Figure 5-32. The most
striking aspect of this problem is the apparent symmetry about the horizontal
and vertical lines passing through the midpoint of the chimney as well as the
diagonal axes, as indicated on the figure. Therefore, we need to consider only
one-eighth of the geometry in the solution whose nodal network consists of nine
equally spaced nodes.
No heat can cross a symmetry line, and thus symmetry lines can be treated
as insulated surfaces and thus "mirrors" in the finite difference formulation.
Then the nodes in the middle of the symmetry lines can be treated as interior
nodes by using mirror images. Six of the nodes are boundary nodes, so we will
have to write energy balances to obtain their finite difference formulations. First
we partition the region among the nodes equitably by drawing dashed lines be-
tween the nodes through the middle. Then the region around a node surrounded
by the boundary or the dashed lines represents the volume element of the node.
Considering a unit depth and using the energy balance approach for the bound-
ary nodes (again assuming all heat transfer into the volume element for conve-
nience) and the formula for the interior nodes, the finite difference equations
for the nine nodes are determined as follows:
(a) Node 1. On the inner boundary, subjected to convection, Figure 5-33a
Ax AyTj-r,
+hl -(T i -T0 + k T ^ r
Ax T 3 - T,
1 2 Ay
+ =
Taking Ax = Ay = /, it simplifies to
h,l\
Ti + T 2 + T,
h,l
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289
CHAPTER 5
(b)
Node 2.
On the inner boundary, subjected to convection, Figure
5-336
k Ay
K 2
-4 — - + h t == (Ti -T 2 ) + + kAx -4 — - =
Ax 2 -' Ay
Tak
ng Ax =
Ay =
= 1, it simplifies to
T l -[3 + ^JT 2 + 2T 4 =-^T,
(c)
Nodes 3
, 4, and 5. (Interior nodes, Fig. 5-34)
Node 3: T 4 + T, + T 4 + T 6 - 47/ 3 =
Node 4: T, + T 2 + T 5 + T 7 - 4T 4 =
Node 5: T 4 + T 4 + T s + T s - 47/ 5 =
(d)
Node 6
. (On
the outer boundary, subjected to convection anc
A x T 3 ~ T 6 Ay T 7 - T 6
2 Ay 2 Ax
+ h ^ (T - T 6 ) + eo- ^ (r s 4 ky - 7 6 4 ) =
radiation)
Tak
ng Ax =
Ay-
= /, it simplifies to
1 h„l\ h„l prrl
T 2 + T 3 -(2 + ^JT 6 =-^T ~^{Ti y - T*)
(e)
Node 7
(On
the outer boundary, subjected to convection anc
radiation,
Fig
5-35)
k
Ay T 6 - T 7 T 4 - T 7 Ay T, - T 7
_ . + kAx + k _ .
2 Ax Ay 2 Ax
+ h o Ax(T - T 7 ) + eo-Ax(r 4 ky - Tf) =
Tak
ng Ax =
Ay-
= /, it simplifies to
2T 4
+ T 6
/ 2hJ\ 2hJ 2eo-/ ,
- u + -^j t 7 + t s = -^t -^pca -
r 4 )
(f)
Node 8.
Same as Node 7, except shift the node numbers up by
1 (replace
4 by 5, 6 by
7, 7
by 8, and 8 by 9 in the last relation)
27/ 5
+ T 7
/ 2h„l\ 2h„l 2f(tI
r 8 4 )
(g)
Node 9
(On
the outer boundary, subjected to convection anc
radiation,
Fig
5-35)
Av r 8 - r, a r Ax
I I
(4) i Minor i
Mirror
FIGURE 5-34
Converting the boundary
nodes 3 and 5 on symmetry lines to
interior nodes by using mirror images.
Insulation
h, T„
sky
FIGURE 5-35
Schematics for energy balances on the
volume elements of nodes 7 and 9.
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HEAT TRANSFER
Temperature, °C
23 40 55 60 55 40 23
40 <
89
► ♦ ♦
152
♦
89
♦ ♦ <
•40
55 <
60 <
138 256
273
256 138
-55
• 60
• 152* 273 <
*
>273 ♦ 152 <
55 <
•55
138 21
56
273
2.
56 138
40 <
► ♦ ♦
89 138
152
♦ ♦ <
138 89
• 40
23 40 55 60 55 40 23
FIGURE 5-36
The variation of temperature
in the chimney.
Taking Ax = Ay = /, it simplifies to
1 +
h J
hj
eo7
(n
sky
This problem involves radiation, which requires the use of absolute tempera-
ture, and thus all temperatures should be expressed in Kelvin. Alternately, we
could use C C for all temperatures provided that the four temperatures in the ra-
diation terms are expressed in the form (7~+ 273) 4 . Substituting the given
quantities, the system of nine equations for the determination of nine unknown
nodal temperatures in a form suitable for use with the Gauss-Seidel iteration
method becomes
(T 2 + T 3 + 2865)/7
T 2 = (T x + 2T 4 + 2865)/8
T } = (7, + 2T 4 + T 6 )/4
T 4 = (T 2 + T 3 + T 5 + 7V)/4
T 5 = (27 4 + 27 8 )/4
T-, = (27/ 4 + T 6 + T s + 912.4 - 0.729 X 10-' r 7 4 )/7
r 8 = (2T 5 + T 7 + T g + 912.4 - 0.729 X 10-" r 8 4 )/7
T 9 = (T s + 456.2 - 0.3645 X 1Q-" T 9 4 )/2.5
which is a system of nonlinear equations. Using an equation solver, its solution
is determined to be
r,
= 545.7 K =
272.6°C
T 2
= 529.2 K =
256. 1°C
T 3
= 425.2 K =
152.1°C
T 4
= 411.2 K =
138.0°C
T 5
= 362.1 K =
89.0°C
T 6
= 332.9 K =
59.7°C
Ti
= 328.1 K =
54.9°C
Ts
= 313.1 K =
39.9°C
T 9
= 296.5 K =
23.4°C
The variation of temperature in the chimney is shown in Figure 5-36.
Note that the temperatures are highest at the inner wall (but less than
300°C) and lowest at the outer wall (but more that 260 K), as expected.
The average temperature at the outer surface of the chimney weighed by the
surface area is
_ (o.5r 6 + t 7 + r 8 + o.5r 9 )
wall, out (Q5 + 1 + 1+ Q5)
_ 0.5 X 332.9 + 328.1 + 313.1 + 0.5 X 296.5 _ ..„ ,„
— ~ — 318.6 K.
Then the rate of heat loss through the 1-m-long section of the chimney can be
determined approximately from
cen58 93 3_ch05.qxd 9/4/2002 11:42 AM Page 291
Q
chimney
"o^o (-' w
T ) + ectA (r wa n_ oul r sky )
= (21 W/m 2 • K)[4 X (0.6 m)(l m)](318.6 - 293)K
+ 0.9(5.67 X 10- 8 W/m 2 ■ K 4 )
[4 X (0.6 m)(l m)](318.6 K) 4 - (260 K) 4 ]
= 1291 + 702 = 1993 W
We could also determine the heat transfer by finding the average temperature of
the inner wall, which is (272.6 + 256.D/2 = 264.4°C, and applying Newton's
law of cooling at that surface:
G chimney = "i A (Xi ~ ^ wall, in)
= (70 W/m 2 • K)[4 X (0.2 m)(l m)](300 - 264.4)°C = 1994 W
The difference between the two results is due to the approximate nature of the
numerical analysis.
Discussion We used a relatively crude numerical model to solve this problem
to keep the complexities at a manageable level. The accuracy of the solution ob-
tained can be improved by using a finer mesh and thus a greater number of
nodes. Also, when radiation is involved, it is more accurate (but more laborious)
to determine the heat losses for each node and add them up instead of using
the average temperature.
291
CHAPTER 5
5-5 - TRANSIENT HEAT CONDUCTION
So far in this chapter we have applied the finite difference method to steady
heat transfer problems. In this section we extend the method to solve transient
problems.
We applied the finite difference method to steady problems by discretizing
the problem in the space variables and solving for temperatures at discrete
points called the nodes. The solution obtained is valid for any time since under
steady conditions the temperatures do not change with time. In transient prob-
lems, however, the temperatures change with time as well as position, and
thus the finite difference solution of transient problems requires discretization
in time in addition to discretization in space, as shown in Figure 5-37. This is
done by selecting a suitable time step At and solving for the unknown nodal
temperatures repeatedly for each A; until the solution at the desired time is ob-
tained. For example, consider a hot metal object that is taken out of the oven
at an initial temperature of T, at time t = and is allowed to cool in ambient
air. If a time step of At = 5 min is chosen, the determination of the tempera-
ture distribution in the metal piece after 3 h requires the determination of the
temperatures 3 X 60/5 = 36 times, or in 36 time steps. Therefore, the compu-
tation time of this problem will be 36 times that of a steady problem. Choos-
ing a smaller Af will increase the accuracy of the solution, but it will also
increase the computation time.
In transient problems, the superscript i is used as the index or counter
of time steps, with i = corresponding to the specified initial condition.
In the case of the hot metal piece discussed above, i = 1 corresponds to
t = 1 X Af = 5 min, i = 2 corresponds to t = 2 X At = 10 min, and a general
r
,
fi+i
J m-1
y/+l
m
m+1
h
m—s
r
in
T'
m+1
1
\At
J Ax
Ax
Ax
'
.
1 m — lmm + 1 x
FIGURE 5-37
Finite difference formulation of time-
dependent problems involves discrete
points in time as well as space.
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HEAT TRANSFER
time step i corresponds to t t = iAt. The notation T' v is used to represent the
temperature at the node m at time step i.
The formulation of transient heat conduction problems differs from that of
steady ones in that the transient problems involve an additional term repre-
senting the change in the energy content of the medium with time. This addi-
tional term appears as a first derivative of temperature with respect to time in
the differential equation, and as a change in the internal energy content during
Af in the energy balance formulation. The nodes and the volume elements in
transient problems are selected as they are in the steady case, and, again as-
suming all heat transfer is into the element for convenience, the energy bal-
ance on a volume element during a time interval At can be expressed as
Heat transferred into \
the volume element
from all of its surfaces
during Af /
/ Heat generated \
within the
volume element
during At
\
I The change in the \
energy content of
the volume element
\ during Af /
or
Af X ^ Q + At X G elcraent = A£ cl
(5-37)
where the rate of heat transfer Q normally consists of conduction terms for
interior nodes, but may involve convection, heat flux, and radiation for bound-
ary nodes.
Noting that A2i element = mCAT = pV e | ement CAT, where p is density and C is
the specific heat of the element, dividing the earlier relation by Af gives
All sides
A£„
Af
pv«
AT
element ^ a *
(5-38)
Volume element
(can be any shape)
p = density
V = volume
pV = mass
C = specific heat
A T = temperature change
AC/ = pVCAT = pVC{T<+ ' - r )
FIGURE 5-38
The change in the energy content of
the volume element of a node
during a time interval Af.
or, for any node m in the medium and its volume element,
Y i + 1 y i
' ' X? ' ^element P ^element ^
All sides
Af
(5-39)
where T l m and T,' n + ' are the temperatures of node m at times f, = iAt and t i + 1 =
(i + l)At, respectively, and T[ + l — T,' n represents the temperature change
of the node during the time interval Af between the time steps i and i + 1
(Fig. 5-38).
Note that the ratio (T,;, + ' — T' n ^)IAt is simply the finite difference approxi-
mation of the partial derivative dT/dt that appears in the differential equations
of transient problems. Therefore, we would obtain the same result for the
finite difference formulation if we followed a strict mathematical approach
instead of the energy balance approach used above. Also note that the finite
difference formulations of steady and transient problems differ by the single
term on the right side of the equal sign, and the format of that term remains the
same in all coordinate systems regardless of whether heat transfer is one-,
two-, or three-dimensional. For the special case of T]+ ' = T' m (i.e., no change
in temperature with time), the formulation reduces to that of steady case, as
expected.
cen58 93 3_ch05.qxd 9/4/2002 11:42 AM Page 293
The nodal temperatures in transient problems normally change during each
time step, and you may be wondering whether to use temperatures at the pre-
vious time step ;' or the new time step i + 1 for the terms on the left side of Eq.
5-39. Well, both are reasonable approaches and both are used in practice. The
finite difference approach is called the explicit method in the first case and
the implicit method in the second case, and they are expressed in the general
form as (Fig. 5-39)
Explicit method:
Implicit method:
^j & "*" ^element P "element ^
t^ i + I 'r
m -* ill
^j & "*" ^element P 'element ^'
At
T'/+ 1 'T'i
1 m in
At
(5-40)
(5-41)
It appears that the time derivative is expressed in forward difference form in
the explicit case and backward difference form in the implicit case. Of course,
it is also possible to mix the two fundamental formulations of Eqs. 5-40 and
5-41 and come up with more elaborate formulations, but such formulations
offer little insight and are beyond the scope of this text. Note that both for-
mulations are simply expressions between the nodal temperatures before and
after a time interval and are based on determining the new temperatures T'„f '
using the previous temperatures T] n . The explicit and implicit formulations
given here are quite general and can be used in any coordinate system re-
gardless of the dimension of heat transfer. The volume elements in multi-
dimensional cases simply have more surfaces and thus involve more terms in
the summation.
The explicit and implicit methods have their advantages and disadvantages,
and one method is not necessarily better than the other one. Next you will see
that the explicit method is easy to implement but imposes a limit on the al-
lowable time step to avoid instabilities in the solution, and the implicit method
requires the nodal temperatures to be solved simultaneously for each time step
but imposes no limit on the magnitude of the time step. We will limit the dis-
cussion to one- and two-dimensional cases to keep the complexities at a man-
ageable level, but the analysis can readily be extended to three-dimensional
cases and other coordinate systems.
Transient Heat Conduction in a Plane Wall
Consider transient one-dimensional heat conduction in a plane wall of thick-
ness L with heat generation g(x, t) that may vary with time and position and
constant conductivity k with a mesh size of Ax = L/M and nodes 0, 1, 2, ... ,
M in the x-direction, as shown in Figure 5-40. Noting that the volume ele-
ment of a general interior node m involves heat conduction from two sides and
the volume of the element is V e i ement = AAx, the transient finite difference for-
mulation for an interior node can be expressed on the basis of Eq. 5-39 as
kA
Ax
T T
*■ m , . , in
1- kA —
Ax
+ g„,AAx = pAAxC :
At
(5-42)
293
CHAPTER 5
If expressed at i + 1 : Implicit method
-py A
in in_
At
If expressed at i: Explicit method
FIGURE 5-39
The formulation of explicit and
implicit methods differs at the time
step (previous or new) at which the
heat transfer and heat generation
terms are expressed.
Plane wall
kA _m^A m
Ax
Ax
1 2
m-1
,- Volume
element
of node m
-1
l kA _m±L
Ax
Ax
m+1 M-1
M *
FIGURE 5-40
The nodal points and volume elements
for the transient finite difference
formulation of one-dimensional
conduction in a plane wall.
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294
HEAT TRANSFER
Canceling the surface area A and multiplying by Ax/k, it simplifies to
s Ajc 2 a -P-
7 1 of j_ f _i_ ^HL — ^^ A ( r ri J r\ Ti\
'«-] Z - L m ■ ~ L m + 1 ~ K n\t "
(5-43)
where a = k/pC is the thermal dijfusivity of the wall material. We now define
a dimensionless mesh Fourier number as
ocAf
Ajc 2
(5-44)
Then Eq. 5-43 reduces to
2T + T
g„Ax 2
(5-45)
Note that the left side of this equation is simply the finite difference formula-
tion of the problem for the steady case. This is not surprising since the formu-
lation must reduce to the steady case for T£~ ' = T' m . Also, we are still not
committed to explicit or implicit formulation since we did not indicate the
time step on the left side of the equation. We now obtain the explicit finite dif-
ference formulation by expressing the left side at time step i as
2T'<
T'
1 m+l
gln^X 2
(explicit)
(5-46)
This equation can be solved explicitly for the new temperature T^ +l (and thus
the name explicit method) to give
t(T,U + 7£ +1 ) + (1 -2t)71 + t
glA* 2
(5-47)
A
M(r.-r^)
Ax
2
pA Ax C ° °
y 2 m
W^kA^ °-
Ax
Ax Ax
1 2 •••
L x
FIGURE 5-41
Schematic for the explicit finite
difference formulation of the
convection condition at the left
boundary of a plane wall.
for all interior nodes m = 1, 2, 3, . . . , M — 1 in a plane wall. Expressing the
left side of Eq. 5-45 at time step i + 1 instead of i would give the implicit
finite difference formulation as
T i + 1 ITi+1 X T/+] _l_
1 m-\ LL m ~ 1 m+[ ^
Ax 2 Ti
(implicit) (5-48)
which can be rearranged as
x7;;+\-(i + 2T)7r i + T7;;,V 1 + T
g^Ax 2
+ Ti =
(5-49)
The application of either the explicit or the implicit formulation to each of the
M — 1 interior nodes gives M — 1 equations. The remaining two equations are
obtained by applying the same method to the two boundary nodes unless, of
course, the boundary temperatures are specified as constants (invariant with
time). For example, the formulation of the convection boundary condition at
the left boundary (node 0) for the explicit case can be expressed as (Fig. 5^4-1)
hA(T x - Ti) + kA ■
Ax
+ &A
Ax
. Ajc £o
PA -r- C : —
v 2 At
(5-50)
cen58 93 3_ch05.qxd 9/4/2002 11:42 AM Page 295
which simplifies to
1 - 2t - 2t
hAx
2t77
2T nr T -
gh^x 2
(5-51)
295
CHAPTER 5
Note that in the case of no heat generation and t = 0.5, the explicit
finite difference formulation for a general interior node reduces to T,'„ +1 =
(Tm-i + Tm- 1)/2, which has the interesting interpretation that the temperature
of an interior node at the new time step is simply the average of the tempera-
tures of its neighboring nodes at the previous time step.
Once the formulation (explicit or implicit) is complete and the initial condi-
tion is specified, the solution of a transient problem is obtained by marching
in time using a step size of Af as follows: select a suitable time step At and de-
termine the nodal temperatures from the initial condition. Taking the initial
temperatures as the previous solution T,'„ at t = 0, obtain the new solution T'„f '
at all nodes at time t = At using the transient finite difference relations. Now
using the solution just obtained at t = At as the previous solution T,' n , obtain
the new solution T,'„ + ' at t = 2At using the same relations. Repeat the process
until the solution at the desired time is obtained.
Stability Criterion for Explicit Method: Limitation on At
The explicit method is easy to use, but it suffers from an undesirable feature
that severely restricts its utility: the explicit method is not unconditionally sta-
ble, and the largest permissible value of the time step Af is limited by the sta-
bility criterion. If the time step Af is not sufficiently small, the solutions
obtained by the explicit method may oscillate wildly and diverge from the ac-
tual solution. To avoid such divergent oscillations in nodal temperatures, the
value of Af must be maintained below a certain upper limit established by the
stability criterion. It can be shown mathematically or by a physical argument
based on the second law of thermodynamics that the stability criterion is sat-
isfied if the coefficients of all T' ln in the T]„ +[ expressions (called the primary
coefficients) are greater than or equal to zero for all nodes m (Fig. 5-42). Of
course, all the terms involving T' m for a particular node must be grouped to-
gether before this criterion is applied.
Different equations for different nodes may result in different restrictions on
the size of the time step Af, and the criterion that is most restrictive should be
used in the solution of the problem. A practical approach is to identify the
equation with the smallest primary coefficient since it is the most restrictive
and to determine the allowable values of Af by applying the stability criterion
to that equation only. A Af value obtained this way will also satisfy the stabil-
ity criterion for all other equations in the system.
For example, in the case of transient one-dimensional heat conduction in a
plane wall with specified surface temperatures, the explicit finite difference
equations for all the nodes (which are interior nodes) are obtained from
Eq. 5-47. The coefficient of T l m in the T^ [ expression is 1 — 2t, which is
independent of the node number m, and thus the stability criterion for all
nodes in this case is 1 — 2t ^ or
Explicit formulation:
Ti + l = aJi + -
TS + ' = aJS + -
t;„ + ' = a,j;„ + ■
*m = a M*M +
Stability criterion:
a,„>0, m = 0,1,2,..
. m, .
..M
FIGURE 5-42
The stability criterion of the
explicit method requires all primary
coefficients to be positive or zero.
ctAf __, J_ /interior nodes, one-dimensional heat
Ajc 2 ~ 2 \ transfer in rectangular coordinates
(5-52)
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296
HEAT TRANSFER
When the material of the medium and thus its thermal diffusivity a is known
and the value of the mesh size Ax is specified, the largest allowable value of
the time step At can be determined from this relation. For example, in the case
of a brick wall (a = 0.45 X 10~ 6 m 2 /s) with a mesh size of Ax = 0.01 m, the
upper limit of the time step is
At:
1 Ax 2
2 «
(0.01 m) 2
2(0.45 X 10~ 6 m 2 /s)
Ills = 1.85 min
o°c
50°C
50°C
20°C
m — \ m m + \
m—\ m m + 1
Time step: i + 1
Time step: i
FIGURE 5-43
The violation of the stability criterion
in the explicit method may result in
the violation of the second law of
thermodynamics and thus
divergence of solution.
^^~ ~^^^^te- *^^m
Uranium plate
o°c
k = 28 W/m-°C
g = 5x 10 6 W/m 3
a = 12.5xl(r 6 m 2 /s
Ax Ax
h
L
0<
''
■
1
2
X
T ... . = 200°C
initial
FIGURE 5-44
Schematic for Example 5-5.
The boundary nodes involving convection and/or radiation are more re-
strictive than the interior nodes and thus require smaller time steps. Therefore,
the most restrictive boundary node should be used in the determination of the
maximum allowable time step At when a transient problem is solved with
the explicit method.
To gain a better understanding of the stability criterion, consider the explicit
finite difference formulation for an interior node of a plane wall (Eq. 5^47) for
the case of no heat generation,
^ + ' = t(7;;,_ 1 + 7;;, +1 ) + (i-2t)7^
Assume that at some time step i the temperatures 7^ , and T' m+ , are equal but
less than T*, (say, T^.j = T;„ +l = 50°C and T l m = 80°C). At the next time
step, we expect the temperature of node m to be between the two values (say,
70°C). However, if the value of t exceeds 0.5 (say, t = 1), the temperature of
node m at the next time step will be less than the temperature of the neighbor-
ing nodes (it will be 20°C), which is physically impossible and violates the
second law of thermodynamics (Fig. 5-43). Requiring the new temperature of
node m to remain above the temperature of the neighboring nodes is equiva-
lent to requiring the value of t to remain below 0.5.
The implicit method is unconditionally stable, and thus we can use any time
step we please with that method (of course, the smaller the time step, the bet-
ter the accuracy of the solution). The disadvantage of the implicit method is
that it results in a set of equations that must be solved simultaneously for each
time step. Both methods are used in practice.
EXAMPLE 5-5 Transient Heat Conduction in a Large Uranium
Plate
Consider a large uranium plate of thickness L = 4 cm, thermal conductivity k =
28 W/m ■ °C, and thermal diffusivity a = 12.5 X lO" 6 m 2 /s that is initially at
a uniform temperature of 200°C. Heat is generated uniformly in the plate at a
constant rate of g = 5 X 10 6 W/m 3 . At time t = 0, one side of the plate is
brought into contact with iced water and is maintained at 0°C at all times, while
the other side is subjected to convection to an environment at 7"^ = 30°C with
a heat transfer coefficient of h = 45 W/m 2 • °C, as shown in Figure 5-44. Con-
sidering a total of three equally spaced nodes in the medium, two at the bound-
aries and one at the middle, estimate the exposed surface temperature of the
plate 2.5 min after the start of cooling using (a) the explicit method and (b) the
implicit method.
cen58 93 3_ch05.qxd 9/4/2002 11:42 AM Page 297
SOLUTION We have solved this problem in Example 5-1 for the steady case,
and here we repeat it for the transient case to demonstrate the application of
the transient finite difference methods. Again we assume one-dimensional heat
transfer in rectangular coordinates and constant thermal conductivity. The num-
ber of nodes is specified to be M = 3, and they are chosen to be at the two sur-
faces of the plate and at the middle, as shown in the figure. Then the nodal
spacing Ax becomes
Ax
M- 1
0.04 m
3 - 1
0.02 m
We number the nodes as 0, 1, and 2. The temperature at node is given to be
T = 0°C at all times, and the temperatures at nodes 1 and 2 are to be deter-
mined. This problem involves only two unknown nodal temperatures, and thus
we need to have only two equations to determine them uniquely. These equa-
tions are obtained by applying the finite difference method to nodes 1 and 2.
(a) Node 1 is an interior node, and the explicit finite difference formulation at
that node is obtained directly from Eq. 5-47 by setting m = 1:
t(T + Tj) + (1 - 2t) 77 + t
,?i
Ax 2
(1)
Node 2 is a boundary node subjected to convection, and the finite difference
formulation at that node is obtained by writing an energy balance on the volume
element of thickness Ax/2 at that boundary by assuming heat transfer to be into
the medium at all sides (Fig. 5-45):
hA{T rj - Tj) + kA
Ax
n Ax
P A
Ax Tj +l -Tj
C
Ax
Dividing by kA/2Axand using the definitions of thermal diffusivity a = A/pCand
the dimensionless mesh Fourier number t = aAf/(Ax) 2 gives
2hAx g^Ax 2 Tj +1 -Tj
^(T„ - Tj) + 2(77 " Tj) + ^r- = "^f "
which can be solved for Tj +1 to give
T i+i
hAx\ • / • hAx fcAx 2
! _ 2t _ 2t _ n + J 2T , + 2 — T„ + —
(2)
Note that we did not use the superscript /for quantities that do not change with
time. Next we need to determine the upper limit of the time step Af from the
stability criterion, which requires the coefficient of T{ in Equation 1 and the co-
efficient of Tj in the second equation to be greater than or equal to zero. The
coefficient of Tj is smaller in this case, and thus the stability criterion for this
problem can be expressed as
1 - 2t - 2t
hAx .
1
2(1 + hAxlk)
At:
Ax 2
2a(l + hAxlk)
297
CHAPTER 5
Volume element -
of node 2
kA-
Ax
Si
T' +
hA(T- x _ - T')
Ax
2
FIGURE 5-45
Schematic for the explicit
finite difference formulation of the
convection condition at the right
boundary of a plane wall.
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HEAT TRANSFER
TABLE 5-2
The variation of the nodal
temperatures in Example 5-5 with
time obtained by the explicit
method
Node
Time
Time,
Temperature, °C
Step, /
s
T[
n
200.0
200.0
1
15
139.7
228.4
2
30
149.3
172.8
3
45
123.8
179.9
4
60
125.6
156.3
5
75
114.6
157.1
6
90
114.3
146.9
7
105
109.5
146.3
8
120
108.9
141.8
9
135
106.7
141.1
10
150
106.3
139.0
20
300
103.8
136.1
30
450
103.7
136.0
40
600
103.7
136.0
since t = aAf/(Ax) 2 . Substituting the given quantities, the maximum allowable
value of the time step is determined to be
(0.02 m) 2
Af < — - ^ = 15.5 s
2(12.5 X 10- 6 m 2 /s)[l + (45 W/m 2 • °C)(0.02 m)/28 W/m • °C]
Therefore, any time step less than 15.5 s can be used to solve this problem. For
convenience, let us choose the time step to be Af = 15 s. Then the mesh
Fourier number becomes
ctA? (12-5 X 10- 6 m 2 /s)(15s)
(A*) 2
(0.02 m) 2
0.46875 (forAf = 15 s)
Substituting this value of t and other given quantities, the explicit finite differ-
ence equations (1) and (2) developed here reduce to
77 +l
7V' +I
0.06257/ + 0.4687577 + 33.482
0.93757/ + 0.03236677 + 34.386
The initial temperature of the medium at t = and / = is given to be 200°C
throughout, and thus T° = T° = 200°C. Then the nodal temperatures at 7"/
and 77 1 at f = Af = 15 s are determined from these equations to be
0.06257/,°
0.4687577° + 33.482
0.0625 X 200 + 0.46875 X 200 + 33.482 = 139.7°C
0.93757/,° + 0.03236677° + 34.386
0.9375 X 200 + 0.032366 X 200 + 34.386 = 228.4°C
Similarly, the nodal temperatures 77/ and 7 2 2 at t = 2Af = 2 X 15 = 30 s are
determined to be
Tf = 0.06257/, 1 + 0.4687577 + 33.482
= 0.0625 X 139.7 + 0.46875 X 228.4 + 33.482
Ti = 0.93757/,' + 0.03236677 + 34.386
149.3°C
= 0.9375 X 139.7 + 0.032366 X 228.4 + 34.386 = 172.8°C
Continuing in the same manner, the temperatures at nodes 1 and 2 are de-
termined for / = 1, 2, 3, 4, 5, . . . , 50 and are given in Table 5-2. Therefore,
the temperature at the exposed boundary surface 2.5 min after the start of
cooling is
r L 25rain = 77° = 139.0°C
(b) Node 1 is an interior node, and the implicit finite difference formulation at
that node is obtained directly from Eq. 5-49 by setting m = 1:
J?n A* 2
t7/ - (1 + 2t) 77+ > + t77 +1 + t—, — + 77
(3)
Node 2 is a boundary node subjected to convection, and the implicit finite dif-
ference formulation at that node can be obtained from this formulation by ex-
pressing the left side of the equation at time step / + 1 instead of / as
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299
CHAPTER 5
2/iAx
(r»- t{ +1 ) + 2(7/+ ' - tj +1 )
&Ax 2 7<+'-7i
which can be rearranged as
2t7/ +1
1 + 2t + 2t^W' + 2x^7, ,+ t^ + 7 =
k k k
(4)
Again we did not use the superscript /or / + 1 for quantities that do not change
with time. The implicit method imposes no limit on the time step, and thus we
can choose any value we want. However, we will again choose Af = 15 s, and
thus t = 0.46875, to make a comparison with part (a) possible. Substituting
this value of t and other given quantities, the two implicit finite difference
equations developed here reduce to
-1 .93757/ +1 4
0.937577 + 1
0.4687577"
33.482 =
1 .967677 +1 + 77 + 34.386 =
Again 7? - , 2
and for / = 0, these two equations reduce to
- 1.93757/ + 0.4687577 1 + 200
33.482 =
34.386 =
The unknown nodal temperatures 7/ and 77/ at t = At -
solving these two equations simultaneously to be
15 s are determined by
168.8°C
and
TV = 199.6°C
Similarly, for / = 1, these equations reduce to
-1.93757?
0.468757 2 2 + 168.8
- 1.967677 + 199.6
33.482 =
34.386 =
The unknown nodal temperatures 11 and 7 2 2 at t = Af = 2 X 15
determined by solving these two equations simultaneously to be
30 s are
T? = 150.5°C
and
77 = 190.6°C
Continuing in this manner, the temperatures at nodes 1 and 2 are determined
for / = 2, 3, 4, 5, . . . , 40 and are listed in Table 5-3, and the temperature
at the exposed boundary surface (node 2) 2.5 min after the start of cooling is
obtained to be
r 2.5min = j\a = 143.9°C
which is close to the result obtained by the explicit method. Note that either
method could be used to obtain satisfactory results to transient problems, ex-
cept, perhaps, for the first few time steps. The implicit method is preferred
when it is desirable to use large time steps, and the explicit method is preferred
when one wishes to avoid the simultaneous solution of a system of algebraic
equations.
TABLE 5-3
The variation of the nodal
temperatures in Example 5-5 with
time obtained by the implicit
method
Node
Time
Time,
Temperature, °C
Step, /
s
T[
n
200.0
200.0
1
15
168.8
199.6
2
30
150.5
190.6
3
45
138.6
180.4
4
60
130.3
171.2
5
75
124.1
163.6
6
90
119.5
157.6
7
105
115.9
152.8
8
120
113.2
149.0
9
135
111.0
146.1
10
150
109.4
143.9
20
300
104.2
136.7
30
450
103.8
136.1
40
600
103.8
136.1
cen58933_ch05.qxd 9/4/2002 11:42 AM Page 3C
300
HEAT TRANSFER
South
FIGURE 5-46
Schematic of a Trombe wall
(Example 5-6).
TABLE 5-4
The hourly variation of monthly
average ambient temperature and
solar heat flux incident on a vertical
surface for January in Reno, Nevada
Time
Ambient
Solar
of Temperature
Radiation,
Day
°F
Btu/h ■ ft 2
7 AM-10 AM
33
114
10 AM-1 PM
43
242
1 PM-4 PM
45
178
4 PM-7 PM
37
7 PM-10 PM
32
10 PM-1 AM
27
1 AM-4 AM
26
4 AM-7 AM
25
EXAMPLE 5-6 Solar Energy Storage in Trombe Walls
Dark painted thick masonry walls called Trombe walls are commonly used on
south sides of passive solar homes to absorb solar energy, store it during the
day, and release it to the house during the night (Fig. 5-46). The idea was pro-
posed by E. L. Morse of Massachusetts in 1881 and is named after Professor
Felix Trombe of France, who used it extensively in his designs in the 1970s.
Usually a single or double layer of glazing is placed outside the wall and trans-
mits most of the solar energy while blocking heat losses from the exposed sur-
face of the wall to the outside. Also, air vents are commonly installed at the
bottom and top of the Trombe walls so that the house air enters the parallel flow
channel between the Trombe wall and the glazing, rises as it is heated, and en-
ters the room through the top vent.
Consider a house in Reno, Nevada, whose south wall consists of a 1-ft-thick
Trombe wall whose thermal conductivity is k = 0.40 Btu/h • ft • °F and whose
thermal diffusivity is a = 4.78 X 10~ s ft 2 /s. The variation of the ambient tem-
perature 7" out and the solar heat flux <j S0 | ar incident on a south-facing vertical sur-
face throughout the day for a typical day in January is given in Table 5-4 in 3-h
intervals. The Trombe wall has single glazing with an absorptivity-transmissivity
product of k = 0.77 (that is, 77 percent of the solar energy incident is ab-
sorbed by the exposed surface of the Trombe wall), and the average combined
heat transfer coefficient for heat loss from the Trombe wall to the ambient is de-
termined to be h out = 0.7 Btu/h ■ ft 2 • °F. The interior of the house is maintained
at T m = 70°F at all times, and the heat transfer coefficient at the interior sur-
face of the Trombe wall is h m =1.8 Btu/h ■ ft 2 ■ °F. Also, the vents on the
Trombe wall are kept closed, and thus the only heat transfer between the air in
the house and the Trombe wall is through the interior surface of the wall. As-
suming the temperature of the Trombe wall to vary linearly between 70°F at the
interior surface and 30°F at the exterior surface at 7 am and using the explicit
finite difference method with a uniform nodal spacing of Ax = 0.2 ft, determine
the temperature distribution along the thickness of the Trombe wall after 12,
24, 36, and 48 h. Also, determine the net amount of heat transferred to the
house from the Trombe wall during the first day and the second day. Assume the
wall is 10 ft high and 25 ft long.
SOLUTION The passive solar heating of a house through a Trombe wall is con-
sidered. The temperature distribution in the wall in 12-h intervals and the
amount of heat transfer during the first and second days are to be determined.
Assumptions 1 Heat transfer is one-dimensional since the exposed surface of
the wall is large relative to its thickness. 2 Thermal conductivity is constant.
3 The heat transfer coefficients are constant.
Properties The wall properties are given to be k = 0.40 Btu/h ■ ft • °F, a =
4.78 X 10- 6 ft 2 /s, and k = 0.77.
Analysis The nodal spacing is given to be Ax = 0.2 ft, and thus the total num-
ber of nodes along the Trombe wall is
M
A +1= _LIL
Ax 0.2 ft
1
We number the nodes as 0, 1, 2, 3, 4, and 5, with node on the interior sur-
face of the Trombe wall and node 5 on the exterior surface, as shown in Figure
5-47. Nodes 1 through 4 are interior nodes, and the explicit finite difference
formulations of these nodes are obtained directly from Eq. 5-47 to be
cen58 93 3_ch05.qxd 9/4/2002 11:42 AM Page 301
Node 1 (m = 1)
Node 2 (m = 2)
Node 3 (m = 3)
Node 4 (m = 4)
77+ ' = T(r j + Ti) + (i - 2T)r/
rj +I = T(r/' + r 3 + (l - 2t)t{
Ti +[ = T(Ti + Ti) + (1 - 2t)7^
rj +1 = T(Ti + r 5 + (i - 2j)Ti
(i)
(2)
(3)
(4)
The interior surface is subjected to convection, and thus the explicit formula-
tion of node can be obtained directly from Eq. 5-51 to be
1 -2t
h m Ax\ .
2x^^)7-,;
2t77 + 2t—, — T in
Substituting the quantities h m , Ax, k, and T m , which do not change with time,
into this equation gives
(1 - 3.80t) Ti + t(277 + 126.0)
(5)
The exterior surface of the Trombe wall is subjected to convection as well as to
heat flux. The explicit finite difference formulation at that boundary is obtained
by writing an energy balance on the volume element represented by node 5,
h out A(Ti ut - Ti) + KAq^
kA
Ti
Ax
pA — C
T i +i
Af
(5-53)
which simplifies to
j.j+1
1 - 2t - 2t ■
h m „ Ax
Ti + 2tTI + 2t -
h„„, Ax
K qj olm . Ax
2t : (5-54)
where t = aAf/Ax 2 is the dimensionless mesh Fourier number. Note that we
kept the superscript /for quantities that vary with time. Substituting the quan-
tities h out , Ax, k, and k, which do not change with time, into this equation gives
(1 - 2.70t) Ti + t(2T\ + 0.7071, + Q.770?^)
(6)
where the unit of q* | ar is Btu/h ■ ft 2 .
Next we need to determine the upper limit of the time step Af from the sta-
bility criterion since we are using the explicit method. This requires the iden-
tification of the smallest primary coefficient in the system. We know that the
boundary nodes are more restrictive than the interior nodes, and thus we exam-
ine the formulations of the boundary nodes and 5 only. The smallest and thus
the most restrictive primary coefficient in this case is the coefficient of Ti in the
formulation of node since 1 - 3.8t < 1 - 2.7t, and thus the stability cri-
terion for this problem can be expressed as
1 - 3.80t>0
aA;t
A^" :
1
3.80
Substituting the given quantities, the maximum allowable value of the time step
is determined to be
At:
(0.2 ft) 2
Ax 2
3.80a 3.80 X (4.78 X lO" 6 ft 2 /s)
2202 s
301
CHAPTER 5
A":
Trombe wall
= 0.40 Btu/h-ft-°F
V
a
= 4.78x 10- 6 ft 2 /s
70 ; F
Initial temperature
/ distribution at
,T.
in in
if 7 AM (t = 0)
/; „ T
1 out c
AX =
= 0.2 ft
•30°F
0'
*
1
2 3 4
5
L x
FIGURE 5-47
The nodal network for the Trombe
wall discussed in Example 5-6.
cen58 93 3_ch05.qxd 9/4/2002 11:42 AM Page 302
302
HEAT TRANSFER
Temperature
C F
170
— 1 st day
— 2nd day
7 pm -s
110
90
/ /l AM
1 PM
70
50
{ Initial
-"^7 AM
temperature
0.2 0.4 0.6 0.8 1 ft
Distance along the Trombe wall
FIGURE 5-48
The variation of temperatures in
the Trombe wall discussed
in Example 5-6.
Therefore, any time step less than 2202 s can be used to solve this problem.
For convenience, let us choose the time step to be Af = 900 s = 15 min. Then
the mesh Fourier number becomes
ocAf (4-78 X 10- 6 ft 2 /s)(900 s)
(Ax) 2
(0.2 ft) 2
0.10755 (for A? = 15 min)
Initially (at 7 am or f = 0), the temperature of the wall is said to vary linearly be-
tween 70 C F at node and 30°F at node 5. Noting that there are five nodal
spacings of equal length, the temperature change between two neighboring
nodes is (70 - 30)°F/5 = 8°F. Therefore, the initial nodal temperatures are
r 3 °
70°F,
46°F,
T o
TO
62°F,
38°R
54°F,
30°F
Then the nodal temperatures at t = Af = 15 min (at 7:15 am) are determined
from these equations to be
Ti = (1 - 3.80t) r ° + T(2r,°
— v,! j.oui/iq i ni^i < 126.0)
(1 - 3.80 X 0.10755) 70 + 0.10755(2 X 62 + 126.0) = 68.3° F
77
T(r ° + r 2 °) + (i - 2t) r,°
= 0.10755(70 + 54) + (1 - 2 X 0.10755)62 = 62°F
7V = t(t? + r 3 °) + (l - 2t) r 2 °
= 0.10755(62 + 46) + (1 - 2 X 0.10755)54 = 54°F
Tl = T(r 2 ° + r 4 °) + (i - 2t) r 3 °
= 0.10755(54 + 38) + (1 - 2 X 0.10755)46 = 46°F
t\ = T(r 3 ° + r 5 °) + (i - 2t) r 4 °
= 0.10755(46 + 30) + (1 - 2 X 0.10755)38 = 38°F
Ti = (1 - 2.70t) T 5 ° + t(2T 4 ° + 0.70r o ° u , + 0.770</° olar )
= (1 - 2.70 X 0.10755)30 + 0.10755(2 X 38 + 0.70 X 33 + 0.770 X 114)
= 41.4°F
Note that the inner surface temperature of the Trombe wall dropped by 1.7°F
and the outer surface temperature rose by 11.4°F during the first time step
while the temperatures at the interior nodes remained the same. This is typical
of transient problems in mediums that involve no heat generation. The nodal
temperatures at the following time steps are determined similarly with the help
of a computer. Note that the data for ambient temperature and the incident
solar radiation change every 3 hours, which corresponds to 12 time steps,
and this must be reflected in the computer program. For example, the value of
q' salar must be taken to be q' solar = 75 for / = 1-12, q' solar = 242 for / = 13-24,
<7^ ar = 178 for / = 25-36, and q' solar = for / = 37-96.
The results after 6, 12, 18, 24, 30, 36, 42, and 48 h are given in Table 5-5
and are plotted in Figure 5-48 for the first day. Note that the interior tempera-
ture of the Trombe wall drops in early morning hours, but then rises as the solar
energy absorbed by the exterior surface diffuses through the wall. The exterior
surface temperature of the Trombe wall rises from 30 to 142°F in just 6 h be-
cause of the solar energy absorbed, but then drops to 53°F by next morning as
a result of heat loss at night. Therefore, it may be worthwhile to cover the outer
surface at night to minimize the heat losses.
cen58 93 3_ch05.qxd 9/4/2002 11:42 AM Page 303
TABLE 5-5
The temperatures at the nodes of a Trombe wall at various times
Time
Step, /
Nodal Temperatures,
°F
Time
To
T,
h
h
T,
T 5
h (7 am)
70.0
62.0
54.0
46.0
38.0
30.0
6 h (1 pm)
24
65.3
61.7
61.5
69.7
94.1
142.0
12 h (7 pm)
48
71.6
74.2
80.4
88.4
91.7
82.4
18 h (1 AM)
72
73.3
75.9
77.4
76.3
71.2
61.2
24 h (7 am)
96
71.2
71.9
70.9
67.7
61.7
53.0
30 h (1 pm)
120
70.3
71.1
74.3
84.2
108.3
153.2
36 h (7 pm)
144
75.4
81.1
89.4
98.2
101.0
89.7
42 h (1 am)
168
75.8
80.7
83.5
83.0
77.4
66.2
48 h (7 am)
192
73.0
75.1
72.2
66.0
66.0
56.3
303
CHAPTER 5
The rate of heat transfer from the Trombe wall to the interior of the house dur-
ing each time step is determined from Newton's law using the average temper-
ature at the inner surface of the wall (node 0) as
Gt,
Gxrombewall Af = h iB A(T& - T m ) At = h m A[(Ti + 7T >)/2 - TJAt
Therefore, the amount of heat transfer during the first time step (/' = 1) or
during the first 15-min period is
GTron.be wall = h m A[(Tj + T °)/2 ~ TJ At
= (1.8 Btu/h • ft 2 ■ °F)(10 X 25 ft 2 )[(68.3 + 70)/2 - 70°F](0.25 h)
= -95.6 Btu
The negative sign indicates that heat is transferred to the Trombe wall from the
air in the house, which represents a heat loss. Then the total heat transfer dur-
ing a specified time period is determined by adding the heat transfer amounts
for each time step as
Q
Trombe wall
2e
Trombe wall
2 h m A[(T- + Tt l )/2 - TJ At (5-55)
where / is the total number of time intervals in the specified time period. In this
case / = 48 for 12 h, 96 for 24 h, and so on. Following the approach described
here using a computer, the amount of heat transfer between the Trombe wall
and the interior of the house is determined to be
Gt
G
Q
Q
Trombe wall
Trombe wall
Trombe wall
-17, 048 Btu after 12 h
-2483 Btu after 24 h
5610 Btu after 36 h
34, 400 Btu after 48 h
(-17, 078 Btu during the first 12 h)
(14, 565 Btu during the second 12 h)
(8093 Btu during the third 12 h)
(28, 790 Btu during the fourth 12 h)
Therefore, the house loses 2483 Btu through the Trombe wall the first day as a
result of the low start-up temperature but delivers a total of 36,883 Btu of heat
to the house the second day. It can be shown that the Trombe wall will deliver
even more heat to the house during the third day since it will start the day at a
higher average temperature.
cen58 93 3_ch05.qxd 9/4/2002 11:42 AM Page 304
n + 1
Ay
304
HEAT TRANSFER
in, n + 1
Volume
element -,
1 k
m-l,n I "••"
Ay
«-l
-+-
-Ax
m, n - 1
Ax
Two-Dimensional Transient Heat Conduction
m + 1, n
y t m - 1
m + 1
FIGURE 5-49
The volume element of a
general interior node (m, «) for two-
dimensional transient conduction
in rectangular coordinates.
Consider a rectangular region in which heat conduction is significant in the
x- and y-directions, and consider a unit depth of Az = 1 in the z-direction.
Heat may be generated in the medium at a rate of g(x, y, t), which may vary
with time and position, with the thermal conductivity k of the medium as-
sumed to be constant. Now divide the x-y-plane of the region into a rectangu-
lar mesh of nodal points spaced Ax and A_y apart in the x- and y-directions,
respectively, and consider a general interior node (m, ri) whose coordinates are
x = mAx and y = nAy, as shown in Figure 5-49. Noting that the volume ele-
ment centered about the general interior node (m, n) involves heat conduction
from four sides (right, left, top, and bottom) and the volume of the element is
Element = Ax X Ay X 1 = AxAy, the transient finite difference formulation for
a general interior node can be expressed on the basis of Eq. 5-39 as
kAy
T — T
m— \,n in, 11
Ax
T — T T — T
, , ■* m, n+ 1 7H, h , . -* m+ 1. n m, n
+ kAx - — — — + kAy -
kAx
T — T
m, n— 1 m, n
Ay
Av
g,„,„AxAy = pAxAyC
Ax
'T'i+l r r\
1 in 1 m
At
(5-56)
Taking a square mesh {Ax = Ay = I ) and dividing each term by k gives after
simplifying,
f m — 1, n m + 1, n
r + t
1 m, n + 1 -* m, n — I
AT
J :
(5-57)
where again a = k/pCis the thermal diffusivity of the material and t = a.At/1 2
is the dimensionless mesh Fourier number. It can also be expressed in terms of
the temperatures at the neighboring nodes in the following easy-to-remember
form:
, J, _|_ rp
1 1 top ' J right
+ T h ,
4r„,
,/ :
(5-58)
Again the left side of this equation is simply the finite difference formulation
of the problem for the steady case, as expected. Also, we are still not com-
mitted to explicit or implicit formulation since we did not indicate the time
step on the left side of the equation. We now obtain the explicit finite differ-
ence formulation by expressing the left side at time step i as
+ T' + T 1 + T'
1 1 top ' J right ' 1 bottom
ATI,
cSnode'
(5-59)
Expressing the left side at time step i + 1 instead of ;' would give the implicit
formulation. This equation can be solved explicitly for the new temperature
Tlodi to give
TLVc = T<T,' Bft + ri p + T' ght + Ti onom ) + (1 - 4t) 7l de + t ^t^ (5-60)
for all interior nodes (m, n) where m = 1, 2, 3, . . . , M — 1 and n = 1,2,
3, . . . , N — 1 in the medium. In the case of no heat generation and t = \, the
cen58 93 3_ch05.qxd 9/4/2002 11:42 AM Page 305
explicit finite difference formulation for a general interior node reduces to
^node = (Tift + T/op + Trfght + T{ onom )/4, which has the interpretation that the
temperature of an interior node at the new time step is simply the average
of the temperatures of its neighboring nodes at the previous time step
(Fig. 5-50).
The stability criterion that requires the coefficient of T' m in the T f ' n + ' expres-
sion to be greater than or equal to zero for all nodes is equally valid for two-
or three-dimensional cases and severely limits the size of the time step At that
can be used with the explicit method. In the case of transient two-dimensional
heat transfer in rectangular coordinates, the coefficient of T' m in the r„', + ' ex-
pression is 1 — 4t, and thus the stability criterion for all interior nodes in this
case is 1 — 4t > 0, or
aA? j_ (interior nodes, two-dimensional heat
I 2 ~4 transfer in rectangular coordinates)
(5-61)
where Ax = Ay = /. When the material of the medium and thus its thermal
diffusivity a are known and the value of the mesh size / is specified, the
largest allowable value of the time step At can be determined from the relation
above. Again the boundary nodes involving convection and/or radiation are
more restrictive than the interior nodes and thus require smaller time steps.
Therefore, the most restrictive boundary node should be used in the determi-
nation of the maximum allowable time step At when a transient problem is
solved with the explicit method.
The application of Eq. 5-60 to each of the (M — 1) X (N — 1) interior nodes
gives (M — 1) X (N — 1) equations. The remaining equations are obtained by
applying the method to the boundary nodes unless, of course, the boundary
temperatures are specified as being constant. The development of the transient
finite difference formulation of boundary nodes in two- (or three-) dimen-
sional problems is similar to the development in the one-dimensional case dis-
cussed earlier. Again the region is partitioned between the nodes by forming
volume elements around the nodes, and an energy balance is written for each
boundary node on the basis of Eq. 5-39. This is illustrated in Example 5-7.
305
CHAPTER 5
Time step i:
30°C
20°C
Node
40°C
10°C
Time step i + 1 :
yi + 1
m
25°C
Node
m
FIGURE 5-50
In the case of no heat generation
and t = i the temperature of an
interior node at the new time step is
the average of the temperatures of
its neighboring nodes at the
previous time step.
EXAMPLE 5-7 Transient Two-Dimensional Heat Conduction
in L-Bars
Consider two-dimensional transient heat transfer in an L-shaped solid body that
is initially at a uniform temperature of 90°C and whose cross section is given
in Figure 5-51. The thermal conductivity and diffusivity of the body are k =
15 W/m • °C and a = 3.2 X 10~ 6 m 2 /s, respectively, and heat is generated in
the body at a rate of g = 2 X 10 6 W/m 3 . The left surface of the body is insu-
lated, and the bottom surface is maintained at a uniform temperature of 90°C
at all times. At time f = 0, the entire top surface is subjected to convection to
ambient air at T„ = 25°C with a convection coefficient of h = 80 W/m 2 • °C,
and the right surface is subjected to heat flux at a uniform rate of q R = 5000
W/m 2 . The nodal network of the problem consists of 15 equally spaced nodes
with Ax = Ay = 1.2 cm, as shown in the figure. Five of the nodes are at the bot-
tom surface, and thus their temperatures are known. Using the explicit method,
determine the temperature at the top corner (node 3) of the body after 1, 3, 5,
10, and 60 min.
Convection
V
h, T„
1 2 /"
. 3 /\
t
J \ Ax = Ay = I
Av
h —
h —
6 7 \ 8 9 Qr
T
Av
1
+
+
ii!
h —
12!
h —
13!
h —
14| 15
)
90°C
X
-Ax^
-Ax^
-A*—
-A*—
-Ax^
FIGURE 5-51
Schematic and nodal network for
Example 5-7.
cen58 93 3_ch05.qxd 9/4/2002 11:42 AM Page 306
306
HEAT TRANSFER
h,T„
h,T m
]
X
(b) Node 2
(a) Node 1
FIGURE 5-52
Schematics for energy balances on the
volume elements of nodes 1 and 2.
SOLUTION This is a transient two-dimensional heat transfer problem in rec-
tangular coordinates, and it was solved in Example 5-3 for the steady case.
Therefore, the solution of this transient problem should approach the solution
for the steady case when the time is sufficiently large. The thermal conductiv-
ity and heat generation rate are given to be constants. We observe that all nodes
are boundary nodes except node 5, which is an interior node. Therefore, we will
have to rely on energy balances to obtain the finite difference equations. The re-
gion is partitioned among the nodes equitably as shown in the figure, and the
explicit finite difference equations are determined on the basis of the energy
balance for the transient case expressed as
2 Q' + GL
PKle
c-
At
The quantities h, T„, g, and q R do not change with time, and thus we do not
need to use the superscript /for them. Also, the energy balance expressions are
simplified using the definitions of thermal diffusivity a = k/pC and the dimen-
sionless mesh Fourier number t = aAt/P, where Ax = Ay = /.
(a) Node 1. (Boundary node subjected to convection and insulation, Fig.
5-52a)
i^X
77)
Ay Ti
Ax
+ k
AxU
Ay
Ax Ay
2 2
Ax Ay T{
H 2 2
At
Dividing by fe/4 and simplifying,
2hl
8,l 2 Tl +i -T<
t (T m - 77) + 2(T{ - T{) + 2{T\ - T[) ,
which can be solved for T{ +1 to give
1 - 4t - 2tj J 77 + 2t\T{ + Ti +^T» + |r
(b) Node 2. (Boundary node subjected to convection, Fig. 5-526)
Ay Ti - Ti Ti - Ti
hAx(T^ -T{) + k-±- — k — - + kAx -
Ax
Ay
Ay T{ - T{ Ay Ay T{ +1 - Ti
+ k T^ x - + ^-^T = ^T c ^T^
Dividing by kl2, simplifying, and solving for T^ +1 gives
1 - 4t - 2t j J Ti + t (t{ + Ti + 2Ti, + ^j- T x + ^-
cen58 93 3_ch05.qxd 9/4/2002 11:42 AM Page 307
(c) Node 3. (Boundary node subjected to convection on two sides, Fig. 5-53a)
AyTl-Ti . AxAy Ax Ay T^ 1 - 7^
Dividing by klA, simplifying, and solving for 7"3 +1 gives
At
T- + [ = I 1 - 4t - 4t^ ] Ti + 2t( Ti + T£ + 2jT^ +
hi,
2k
(d) Node 4. (On the insulated boundary, and can be treated as an interior node,
Fig. 5-536). Noting that T 10 = 90°C, Eq. 5-60 gives
U 1
(1 - 4t) T\ + t T[ + 2Tj, + 90 +
(e) Node 5. (Interior node, Fig. 5-54a). Noting that T n = 90°C, Eq. 5-60
gives
Ti, + ' = (1 - 4t) Ti + t I H + r^ + Ti + 90 +
&/ 2
(f) Node 6. (Boundary node subjected to convection on two sides, Fig. 5-546)
(Ax Ay\ Av Tj - Ti T' n - Ti k T{ - TI
Ax Ti - Tl . 3 AxAy 3 AxAy Tf - Ti
+ T^y~ + S6 ^^ = P ^^ C At
Dividing by 3k/A, simplifying, and solving for 7"g +1 gives
tw'+I
J 6
l-4 T -4x|]7
hi,
2Tj + 4Ti + 2Tj + 4 X 90 + 4 -f T w + 3 ^-
[g) Node 7. (Boundary node subjected to convection, Fig. 5-55)
Ay Ti - Tj T( 3 - Tj
hAx(T^ -Tj) + k-^- -=-; + kAx -
2 Ax
Ay U ~ Tj
Ay
Ay Ay Tj +I - Tj
2 Ax +^Ax T = pAx T C- A[
Dividing by kl2, simplifying, and solving for Tj +1 gives
1 -4T-2Ty ]
2hi , g 7 r-
Tl + Ti + 2 X 90 + —r- T„ +
307
CHAPTER 5
h, r«
r
Mirror
(5)
*,r~
t
EH
— I ♦
10
(a) Node 3
(b) Node 4
FIGURE 5-53
Schematics for energy balances on the
volume elements of nodes 3 and 4.
m
ii
12
(a) Node 5
(7?) Node 6
FIGURE 5-54
Schematics for energy balances on the
volume elements of nodes 5 and 6.
h, r„
1\ 1
13 1 ^ ^
1r
• 15
FIGURE 5-55
Schematics for energy balances on the
volume elements of nodes 7 and 9.
cen5 8 93 3_ch05.qxd 9/4/2002 11:42 AM Page 3C
308
HEAT TRANSFER
(h) Node 8. This node is identical to node 7, and the finite difference formula-
tion of this node can be obtained from that of node 7 by shifting the node num-
bers by 1 (i.e., replacing subscript m by subscript m + 1). It gives
1 - 4t - 2t
hi
n + T
2hl,
Tj + Tj + 2 X 90 + -T- 7^ +
(/) Node 9. (Boundary node subjected to convection on two sides, Fig. 5-55)
,Ax,_ _,, , . Ay A.t Tj 5 - Tj
h T (T«,-T$) + q R — k-—^-
kAyTi
Ti , . Ax^J AxAy T<'
+
2 Ax 6J 2 2
Dividing by klA, simplifying, and solving for 7"g +1 gives
p TT c "
At
T' + l
*9
hl\ I 4rI hi
i - 4t - 2t j r 9 " + 2t m + 90 + ^-+ ^r„
2fe
This completes the finite difference formulation of the problem. Next we need
to determine the upper limit of the time step Af from the stability criterion,
which requires the coefficient of T m in the 7~^ +1 expression (the primary coeffi-
cient) to be greater than or equal to zero for all nodes. The smallest primary co-
efficient in the nine equations here is the coefficient of Ti in the expression,
and thus the stability criterion for this problem can be expressed as
1 - 4t - 4t
///.
1
4(1 + hllk)
A?:
/ :
4a(l + hllk)
since t = aAf// 2 . Substituting the given quantities, the maximum allowable
value of the time step is determined to be
At:
(0.012 m) 2
4(3.2 X 10- 6 m 2 /s)[l + (80 W/m 2 • °C)(0.012 m)/(15 W/m ■ °C)]
10.6 s
Therefore, any time step less than 10.6 s can be used to solve this problem. For
convenience, let us choose the time step to be At = 10 s. Then the mesh
Fourier number becomes
aAt _ (3-2 X 10- 6 m 2 /s)(10s)
l r ~ (0.012 m) 2
0.222 (for At = 10 s)
Substituting this value of t and other given quantities, the developed transient
finite difference equations simplify to
T i+
0.08367,' + 0.444(7^ + 7] + 11.2)
0.0S36Ti + 0.222(7/ + Tj + ITi, + 22.4;
0.05527J + 0.444(7; + T b + 12.8)
0.1127] + 0.222(7/ + 27J + 109.2)
0.1127/ + 0.222(7] + 7j + 7 6 ' + 109.2)
cen58 93 3_ch05.qxd 9/4/2002 11:42 AM Page 309
0.093 17^ + 0.074(2^ + 47^' + 2Tj + 424)
0.08367V + 0.222(7,; + TJ + 202.4)
0.08367V + 0.222(Tj + T$ + 202.4)
0.08367V + 0.444(7^ + 105.2)
Using the specified initial condition as the solution at time t = (for / = 0),
sweeping through these nine equations will give the solution at intervals of
10 s. The solution at the upper corner node (node 3) is determined to be
100.2, 105.9, 106.5, 106.6, and 106. 6°C at 1, 3, 5, 10, and 60 min, re-
spectively. Note that the last three solutions are practically identical to the
solution for the steady case obtained in Example 5-3. This indicates that steady
conditions are reached in the medium after about 5 min.
309
CHAPTER 5
TOPIC OF SPECIAL INTEREST
Controlling the Numerical Error
A comparison of the numerical results with the exact results for tempera-
ture distribution in a cylinder would show that the results obtained by a nu-
merical method are approximate, and they may or may not be sufficiently
close to the exact (true) solution values. The difference between a numeri-
cal solution and the exact solution is the error involved in the numerical
solution, and it is primarily due to two sources:
• The discretization error (also called the truncation ox formulation
error), which is caused by the approximations used in the formulation
of the numerical method.
• The round-off error, which is caused by the computer's use of a
limited number of significant digits and continuously rounding (or
chopping) off the digits it cannot retain.
Below we discuss both types of errors.
Discretization Error
The discretization error involved in numerical methods is due to replacing
the derivatives by differences in each step, or the actual temperature distri-
bution between two adjacent nodes by a straight line segment.
Consider the variation of the solution of a transient heat transfer problem
with time at a specified nodal point. Both the numerical and actual (exact)
solutions coincide at the beginning of the first time step, as expected, but
the numerical solution deviates from the exact solution as the time t in-
creases. The difference between the two solutions at t = Af is due to the ap-
proximation at the first time step only and is called the local discretization
error. One would expect the situation to get worse with each step since the
second step uses the erroneous result of the first step as its starting point
and adds a second local discretization error on top of it, as shown in Figure
5-56. The accumulation of the local discretization errors continues with the
increasing number of time steps, and the total discretization error at any
Global
error
'3 Time
FIGURE 5-56
The local and global discretization
errors of the finite difference
method at the third time step
at a specified nodal point.
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310
HEAT TRANSFER
step is called the global or accumulated discretization error. Note that the
local and global discretization errors are identical for the first time step.
The global discretization error usually increases with the increasing num-
ber of steps, but the opposite may occur when the solution function
changes direction frequently, giving rise to local discretization errors of op-
posite signs, which tend to cancel each other.
To have an idea about the magnitude of the local discretization error,
consider the Taylor series expansion of the temperature at a specified nodal
point m about time t h
dT(x m , t) 1 , d 2 T(x m , t)
T(x„„ t t + At) = T(x m , f ; ) + M \ m t + i At - "J +■■■ (5-62)
The finite difference formulation of the time derivative at the same nodal
point is expressed as
dT(x„„ t t ) T(x„„ t t + At) - T(x m , t t ) n + ' - T;„
dt Ar At
(5-63)
or
dT(x m , tf)
T(x m , t, + At) = T(x m , t,) + At (5-64)
which resembles the Taylor series expansion terminated after the first two
terms. Therefore, the third and later terms in the Taylor series expansion
represent the error involved in the finite difference approximation. For a
sufficiently small time step, these terms decay rapidly as the order of de-
rivative increases, and their contributions become smaller and smaller. The
first term neglected in the Taylor series expansion is proportional to At 2 ,
and thus the local discretization error of this approximation, which is the
error involved in each step, is also proportional to At 2 .
The local discretization error is the formulation error associated with a
single step and gives an idea about the accuracy of the method used. How-
ever, the solution results obtained at every step except the first one involve
the accumulated error up to that point, and the local error alone does not
have much significance. What we really need to know is the global dis-
cretization error. At the worst case, the accumulated discretization error
after / time steps during a time period t is i(At) 2 = (f /A0(Af) 2 = t Q At,
which is proportional to At Thus, we conclude that the local discretization
error is proportional to the square of the step size At 2 while the global dis-
cretization error is proportional to the step size At itself. Therefore, the
smaller the mesh size (or the size of the time step in transient problems),
the smaller the error, and thus the more accurate is the approximation. For
example, halving the step size will reduce the global discretization error by
half. It should be clear from the discussions above that the discretization er-
ror can be minimized by decreasing the step size in space or time as much
as possible. The discretization error approaches zero as the difference
quantities such as Ax and At approach the differential quantities such as dx
and dt.
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CHAPTER 5
Round-off Error
If we had a computer that could retain an infinite number of digits for all
numbers, the difference between the exact solution and the approximate
(numerical) solution at any point would entirely be due to discretization er-
ror. But we know that every computer (or calculator) represents numbers
using a finite number of significant digits. The default value of the number
of significant digits for many computers is 7, which is referred to as single
precision. But the user may perform the calculations using 15 significant
digits for the numbers, if he or she wishes, which is referred to as double
precision. Of course, performing calculations in double precision will re-
quire more computer memory and a longer execution time.
In single precision mode with seven significant digits, a computer will
register the number 44444.666666 as 44444.67 or 44444.66, depending on
the method of rounding the computer uses. In the first case, the excess dig-
its are said to be rounded to the closest integer, whereas in the second case
they are said to be chopped off. Therefore, the numbers a = AAAAA.123A5
and b = AAAAA.12032 are equivalent for a computer that performs calcula-
tions using seven significant digits. Such a computer would give a — b =
instead of the true value 0.00313.
The error due to retaining a limited number of digits during calculations
is called the round-off error. This error is random in nature and there is no
easy and systematic way of predicting it. It depends on the number of cal-
culations, the method of rounding off, the type of computer, and even the
sequence of calculations.
In algebra you learned that a + b + c = a + c + b, which seems quite
reasonable. But this is not necessarily true for calculations performed with
a computer, as demonstrated in Figure 5-57. Note that changing the se-
quence of calculations results in an error of 30.8 percent in just two opera-
tions. Considering that any significant problem involves thousands or even
millions of such operations performed in sequence, we realize that the ac-
cumulated round-off error has the potential to cause serious error without
giving any warning signs. Experienced programmers are very much aware
of this danger, and they structure their programs to prevent any buildup of
the round-off error. For example, it is much safer to multiply a number by
10 than to add it 10 times. Also, it is much safer to start any addition
process with the smallest numbers and continue with larger numbers. This
rule is particularly important when evaluating series with a large number of
terms with alternating signs.
The round-off error is proportional to the number of computations per-
formed during the solution. In the finite difference method, the number of
calculations increases as the mesh size or the time step size decreases.
Halving the mesh or time step size, for example, will double the number of
calculations and thus the accumulated round-off error.
Controlling the Error in Numerical Methods
The total error in any result obtained by a numerical method is the sum of
the discretization error, which decreases with decreasing step size, and the
round-off error, which increases with decreasing step size, as shown in Fig-
ure 5-58. Therefore, decreasing the step size too much in order to get more
Given
a
= 1111111
b
= -1111116
c
- 0.4444432
Find:
D
= a + b + c
E
= a + c + b
Solution:
D =
7777777
- 1111116 + 0.4444432
1 + 0.4444432
1.444443
(Correct result)
E =
7777777 + 0.4444432 - 7777776
7777777
- 7777776
1.000000
(In eiTor by 30.8%)
FIGURE 5-57
A simple arithmetic operation
performed with a computer
in single precision using seven
significant digits, which results in
30.8 percent error when the order
of operation is reversed.
i Error
Optimum Step size
step size
FIGURE 5-58
As the mesh or time step size
decreases, the discretization error
decreases but the round-off
error increases.
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HEAT TRANSFER
accurate results may actually backfire and give less accurate results be-
cause of a faster increase in the round-off error. We should be careful not to
let round-off error get out of control by avoiding a large number of compu-
tations with very small numbers.
In practice, we will not know the exact solution of the problem, and thus
we will not be able to determine the magnitude of the error involved in the
numerical method. Knowing that the global discretization error is propor-
tional to the step size is not much help either since there is no easy way of
determining the value of the proportionality constant. Besides, the global
discretization error alone is meaningless without a true estimate of the
round-off error. Therefore, we recommend the following practical proce-
dures to assess the accuracy of the results obtained by a numerical method.
• Start the calculations with a reasonable mesh size Ax (and time step
size At for transient problems) based on experience. Then repeat the
calculations using a mesh size of Ax/2. If the results obtained by
halving the mesh size do not differ significantly from the results
obtained with the full mesh size, we conclude that the discretization
error is at an acceptable level. But if the difference is larger than we
can accept, then we have to repeat the calculations using a mesh size
Ax/4 or even a smaller one at regions of high temperature gradients.
We continue in this manner until halving the mesh size does not cause
any significant change in the results, which indicates that the
discretization error is reduced to an acceptable level.
• Repeat the calculations using double precision holding the mesh size
(and the size of the time step in transient problems) constant. If the
changes are not significant, we conclude that the round-off error is not
a problem. But if the changes are too large to accept, then we may try
reducing the total number of calculations by increasing the mesh size
or changing the order of computations. But if the increased mesh size
gives unacceptable discretization errors, then we may have to find a
reasonable compromise.
It should always be kept in mind that the results obtained by any numer-
ical method may not reflect any trouble spots in certain problems that re-
quire special consideration such as hot spots or areas of high temperature
gradients. The results that seem quite reasonable overall may be in consid-
erable error at certain locations. This is another reason for always repeating
the calculations at least twice with different mesh sizes before accepting
them as the solution of the problem. Most commercial software packages
have built-in routines that vary the mesh size as necessary to obtain highly
accurate solutions. But it is a good engineering practice to be aware of any
potential pitfalls of numerical methods and to examine the results obtained
with a critical eye.
SUMMARY
Analytical solution methods are limited to highly simplified
problems in simple geometries, and it is often necessary to use
a numerical method to solve real world problems with com-
plicated geometries or nonuniform thermal conditions. The
cen58 93 3_ch05.qxd 9/4/2002 11:42 AM Page 313
numerical finite difference method is based on replacing deriv-
atives by differences, and the finite difference formulation of a
heat transfer problem is obtained by selecting a sufficient num-
ber of points in the region, called the nodal points or nodes,
and writing energy balances on the volume elements centered
about the nodes.
For steady heat transfer, the energy balance on a volume el-
ement can be expressed in general as
2 Q + gv d
313
CHAPTER 5
also be used to solve a system of equations simultaneously at
the press of a button.
The finite difference formulation of transient heat conduc-
tion problems is based on an energy balance that also accounts
for the variation of the energy content of the volume element
during a time interval At. The heat transfer and heat generation
terms are expressed at the previous time step i in the explicit
method, and at the new time step ;' + 1 in the implicit method.
For a general node m, the finite difference formulations are
expressed as
whether the problem is one-, two-, or three-dimensional. For
convenience in formulation, we always assume all heat trans-
fer to be into the volume element from all surfaces toward the
node under consideration, except for specified heat flux whose
direction is already specified. The finite difference formula-
tions for a general interior node under steady conditions are ex-
pressed for some geometries as follows:
221
(Ax) 2
4. T + T
1 * right ' -* bottom
AT
One-dimensional
steady conduction
in a plane wall:
Two-
dimensional
steady ,
conduction left
in rectangular
coordinates:
where Ax is the nodal spacing for the plane wall and Ax =
Ay = I is the nodal spacing for the two-dimensional case. Insu-
lated boundaries can be viewed as mirrors in formulation, and
thus the nodes on insulated boundaries can be treated as inte-
rior nodes by using mirror images.
The finite difference formulation at node at the left bound-
ary of a plane wall for steady one-dimensional heat conduction
can be expressed as
Qu
+ kA
r,
Ax
(AAx/2) =
where AAx/2 is the volume of the volume, g is the rate of heat
generation per unit volume at x = 0, and A is the heat transfer
area. The form of the first term depends on the boundary con-
dition at x = (convection, radiation, specified heat flux, etc.).
The finite difference formulation of heat conduction prob-
lems usually results in a system of N algebraic equations in N
unknown nodal temperatures that need to be solved simultane-
ously. There are numerous systematic approaches available in
the literature. Several widely available equation solvers can
Explicit y A< ■ + r< v r
method: .,f^ U elcmenl P elcment
Implicit
method:
^j ti ' ^element P "element ^
At
At
where T' m and T'„ + ' are the temperatures of node m at times
?, = iAt and t i+l = (i + l)Af, respectively, and r,;, + 1 — T/,, rep-
resents the temperature change of the node during the time in-
terval At between the time steps i and i + 1 . The explicit and
implicit formulations given here are quite general and can be
used in any coordinate system regardless of heat transfer being
one-, two-, or three-dimensional.
The explicit formulation of a general interior node for one-
and two-dimensional heat transfer in rectangular coordinates
can be expressed as
One-
dimen- T^ '
sional case:
Two- i+l= j
dimen- nodc *■ left
t(T,U + rj, +1 ) + (i -2t)t;„ + t
£,;Ax 2
sional
case:
where
top
(i - 4t) r< ode
otAf
Ax 2
ight "■" 1 bottom
e'n I 2
is the dimensionless mesh Fourier number and a = k/pC is the
thermal diffusivity of the medium.
The implicit method is inherently stable, and any value of At
can be used with that method as the time step. The largest value
of the time step At in the explicit method is limited by the sta-
bility criterion, expressed as: the coefficients of all T' m in the
r,J, + ' expressions {called the primary coefficients) must be
greater than or equal to zero for all nodes m. The maximum
value of At is determined by applying the stability criterion to
the equation with the smallest primary coefficient since it is the
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HEAT TRANSFER
most restrictive. For problems with specified temperatures or
heat fluxes at all the boundaries, the stability criterion can be
expressed as t < | for one-dimensional problems and t <
the two-dimensional problems in rectangular coordinates.
I for
REFERENCES AND SUGGESTED READING
1. D. A. Anderson, J. C. Tannehill, and R. H. Pletcher.
Computational Fluid Mechanics and Heat Transfer. New
York: Hemisphere, 1984.
2. C. A. Brebbia. The Boundary Element Method for
Engineers. New York: Halsted Press, 1978.
3. G. E. Forsythe and W. R. Wasow. Finite Difference
Methods for Partial Differential Equations. New York:
John Wiley & Sons, 1960.
4. B. Gebhart. Heat Conduction and Mass Diffusion. New
York: McGraw-Hill, 1993.
5. K. H. Huebner and E. A. Thornton. The Finite Element
Method for Engineers. 2nd ed. New York: John Wiley &
Sons, 1982.
6. Y Jaluria and K. E. Torrance. Computational Heat
Transfer. New York: Hemisphere, 1986.
7. W. J. Minkowycz, E. M. Sparrow, G. E. Schneider, and
R. H. Pletcher. Handbook of Numerical Heat Transfer.
New York: John Wiley & Sons, 1988.
8. G. E. Myers. Analytical Methods in Conduction Heat
Transfer. New York: McGraw-Hill, 1971.
9. D. H. Norrie and G. DeVries. An Introduction to Finite
Element Analysis. New York: Academic Press, 1978.
10. M. N. Ozisik. Finite Difference Methods in Heat Transfer.
Boca Raton, FL: CRC Press, 1994.
11. S. V. Patankhar. Numerical Heat Transfer and Fluid Flow.
New York: Hemisphere, 1980.
12. T. M. Shih. Numerical Heat Transfer. New York:
Hemisphere, 1984.
PROBLEMS
Why Numerical Methods?
5-1 C What are the limitations of the analytical solution
methods?
5-2C How do numerical solution methods differ from ana-
lytical ones? What are the advantages and disadvantages of
numerical and analytical methods?
5-3C What is the basis of the energy balance method? How
does it differ from the formal finite difference method? For a
specified nodal network, will these two methods result in the
same or a different set of equations?
5-4C Consider a heat conduction problem that can be solved
both analytically, by solving the governing differential equa-
tion and applying the boundary conditions, and numerically,
by a software package available on your computer. Which
*Problems designated by a "C" are concept questions, and
students are encouraged to answer them all. Problems designated
by an "E" are in English units, and the SI users can ignore them.
Problems with an EES-CD icon d> are solved using EES, and
complete solutions together with parametric studies are included
on the enclosed CD. Problems with a computer-EES icon H are
comprehensive in nature, and are intended to be solved with a
computer, preferably using the EES software that accompanies
this text.
approach would you use to solve this problem? Explain your
reasoning.
5-5C Two engineers are to solve an actual heat transfer
problem in a manufacturing facility. Engineer A makes the nec-
essary simplifying assumptions and solves the problem
analytically, while engineer B solves it numerically using a
powerful software package. Engineer A claims he solved
the problem exactly and thus his results are better, while engi-
neer B claims that he used a more realistic model and thus his
results are better. To resolve the dispute, you are asked to solve
the problem experimentally in a lab. Which engineer do you
think the experiments will prove right? Explain.
Finite Difference Formulation of Differential Equations
5-6C Define these terms used in the finite difference formu-
lation: node, nodal network, volume element, nodal spacing,
and difference equation.
5-7 Consider three consecutive nodes n — 1, n, and n + 1 in
a plane wall. Using the finite difference form of the first deriv-
ative at the midpoints, show that the finite difference form of
the second derivative can be expressed as
A.r
cen58 93 3_ch05.qxd 9/4/2002 11:42 AM Page 315
B-l
FIGURE P5-7
5-8 The finite difference formulation of steady two-
dimensional heat conduction in a medium with heat generation
and constant thermal conductivity is given by
m — I , n
IT..
*m+l
IT 4- T
Ax 1
Ay 2
in rectangular coordinates. Modify this relation for the three-
dimensional case.
5-9 Consider steady one-dimensional heat conduction in a
plane wall with variable heat generation and constant thermal
conductivity. The nodal network of the medium consists of
nodes 0, 1,2, 3, and 4 with a uniform nodal spacing of Ax.
Using the finite difference form of the first derivative (not the
energy balance approach), obtain the finite difference formula-
tion of the boundary nodes for the case of uniform heat flux q Q
at the left boundary (node 0) and convection at the right bound-
ary (node 4) with a convection coefficient of h and an ambient
temperature of T^.
5-10 Consider steady one -dimensional heat conduction in a
plane wall with variable heat generation and constant thermal
conductivity. The nodal network of the medium consists of
315
CHAPTER 5
nodes 0, 1,2, 3, 4, and 5 with a uniform nodal spacing of Ax
Using the finite difference form of the first derivative (not the
energy balance approach), obtain the finite difference formula-
tion of the boundary nodes for the case of insulation at the left
boundary (node 0) and radiation at the right boundary (node 5)
with an emissivity of e and surrounding temperature of T smr
One-Dimensional Steady Heat Conduction
5-11C Explain how the finite difference form of a heat con-
duction problem is obtained by the energy balance method.
5-12C In the energy balance formulation of the finite differ-
ence method, it is recommended that all heat transfer at the
boundaries of the volume element be assumed to be into the
volume element even for steady heat conduction. Is this a valid
recommendation even though it seems to violate the conserva-
tion of energy principle?
5-13C How is an insulated boundary handled in the finite
difference formulation of a problem? How does a symmetry
line differ from an insulated boundary in the finite difference
formulation?
5-14C How can a node on an insulated boundary be treated
as an interior node in the finite difference formulation of a
plane wall? Explain.
5-1 5C Consider a medium in which the finite difference
formulation of a general interior node is given in its simplest
form as
L m 1 — 27,,, + -/,„,! s m
Ax 2
(a) Is heat transfer in this medium steady or transient?
(b) Is heat transfer one-, two-, or three-dimensional?
(c) Is there heat generation in the medium?
(d) Is the nodal spacing constant or variable?
(e) Is the thermal conductivity of the medium constant or
variable?
5-16 Consider steady heat conduction in a plane wall whose
left surface (node 0) is maintained at 30°C while the right sur-
face (node 8) is subjected to a heat flux of 800 W/m 2 . Express
the finite difference formulation of the boundary nodes and 8
Insulation
Ax
g(x)
Radiation
FIGURE P5-10
30°C
Ax
No heat generation
1 2 3 4 5 6 7
800-
W
h
FIGURE P5-1 6
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316
HEAT TRANSFER
for the case of no heat generation. Also obtain the finite dif-
ference formulation for the rate of heat transfer at the left
boundary.
5-17 Consider steady heat conduction in a plane wall with
variable heat generation and constant thermal conductivity.
The nodal network of the medium consists of nodes 0, 1,2, 3,
and 4 with a uniform nodal spacing of Ax. Using the energy
balance approach, obtain the finite difference formulation of
the boundary nodes for the case of uniform heat flux q at the
left boundary (node 0) and convection at the right boundary
(node 4) with a convection coefficient of h and an ambient
temperature of T^.
5-18 Consider steady one -dimensional heat conduction in a
plane wall with variable heat generation and constant thermal
conductivity. The nodal network of the medium consists of
nodes 0, 1,2, 3, 4, and 5 with a uniform nodal spacing of Ax
Using the energy balance approach, obtain the finite difference
formulation of the boundary nodes for the case of insulation at
the left boundary (node 0) and radiation at the right boundary
(node 5) with an emissivity of e and surrounding temperature
of r surr .
5-19 Consider steady one -dimensional heat conduction in a
plane wall with variable heat generation and constant thermal
conductivity. The nodal network of the medium consists of
nodes 0, 1,2, 3, 4, and 5 with a uniform nodal spacing of Ax.
The temperature at the right boundary (node 5) is specified.
Using the energy balance approach, obtain the finite difference
formulation of the boundary node on the left boundary for the
case of combined convection, radiation, and heat flux at the left
boundary with an emissivity of e, convection coefficient of h,
ambient temperature of T m surrounding temperature of T SUTI ,
and uniform heat flux of q . Also, obtain the finite difference
formulation for the rate of heat transfer at the right boundary.
gW
/
Convection
ktJH
FIGURE P5-1 9
o 1
4 5
5-20 Consider steady one -dimensional heat conduction in a
composite plane wall consisting of two layers A and B in per-
fect contact at the interface. The wall involves no heat genera-
tion. The nodal network of the medium consists of nodes 0, 1
(at the interface), and 2 with a uniform nodal spacing of Ax.
Using the energy balance approach, obtain the finite difference
formulation of this problem for the case of insulation at the left
boundary (node 0) and radiation at the right boundary (node 2)
with an emissivity of e and surrounding temperature of T smr
5-21 Consider steady one-dimensional heat conduction in a
plane wall with variable heat generation and variable thermal
conductivity. The nodal network of the medium consists of
nodes 0, 1 , and 2 with a uniform nodal spacing of Ax. Using
the energy balance approach, obtain the finite difference for-
mulation of this problem for the case of specified heat flux q
to the wall and convection at the left boundary (node 0) with a
convection coefficient of h and ambient temperature of T x , and
radiation at the right boundary (node 2) with an emissivity of e
and surrounding surface temperature of T smr
%
gw
Convection
k(T)
h ^H^
T a
Ax
'
- -
1
2
FIGURE P5-21
5-22 Consider steady one-dimensional heat conduction in a
pin fin of constant diameter D with constant thermal conduc-
tivity. The fin is losing heat by convection to the ambient air at
r„ with a heat transfer coefficient of h. The nodal network of
the fin consists of nodes (at the base), 1 (in the middle), and 2
(at the fin tip) with a uniform nodal spacing of Ax. Using the
energy balance approach, obtain the finite difference formula-
tion of this problem to determine T { and T 2 for the case of spec-
ified temperature at the fin base and negligible heat transfer at
the fin tip. All temperatures are in °C.
5-23 Consider steady one-dimensional heat conduction in a
pin fin of constant diameter D with constant thermal conduc-
tivity. The fin is losing heat by convection to the ambient air
at T„, with a convection coefficient of h, and by radiation to
the surrounding surfaces at an average temperature of T SUTT .
Convection
FIGURE P5-23
cen58 93 3_ch05.qxd 9/4/2002 11:42 AM Page 317
The nodal network of the fin consists of nodes (at the base),
1 (in the middle), and 2 (at the fin tip) with a uniform nodal
spacing of Ax. Using the energy balance approach, obtain the
finite difference formulation of this problem to determine
T x and T 2 for the case of specified temperature at the fin base
and negligible heat transfer at the fin tip. All temperatures
are in °C.
5-24 Consider a large uranium plate of thickness 5 cm and
thermal conductivity k = 28 W/m • °C in which heat is gener-
ated uniformly at a constant rate of g = 6 X 10 5 W/m 3 . One
side of the plate is insulated while the other side is subjected
to convection to an environment at 30°C with a heat transfer
coefficient of h = 60 W/m 2 • °C. Considering six equally
spaced nodes with a nodal spacing of 1 cm, (a) obtain the finite
difference formulation of this problem and (b) determine the
nodal temperatures under steady conditions by solving those
equations.
5-25 Consider an aluminum alloy fin (k = 1 80 W/m ■ °C) of
triangular cross section whose length is L = 5 cm, base thick-
ness is b = 1 cm, and width w in the direction normal to the
plane of paper is very large. The base of the fin is maintained
at a temperature of T = 180°C. The fin is losing heat by con-
vection to the ambient air at T„ = 25°C with a heat transfer
coefficient of h = 25 W/m 2 ■ °C and by radiation to the sur-
rounding surfaces at an average temperature of T sm = 290 K.
Using the finite difference method with six equally spaced
nodes along the fin in the x-direction, determine (a) the tem-
peratures at the nodes and (b) the rate of heat transfer from the
fin for w = 1 m. Take the emissivity of the fin surface to be 0.9
and assume steady one-dimensional heat transfer in the fin.
Triangular fin
FIGURE P5-25
5-26
Reconsider Problem 5-25. Using EES (or other)
software, investigate the effect of the fin base
temperature on the fin tip temperature and the rate of heat
transfer from the fin. Let the temperature at the fin base vary
317
CHAPTER 5
from 100°C to 200°C. Plot the fin tip temperature and the rate
of heat transfer as a function of the fin base temperature, and
discuss the results.
5-27 Consider a large plane wall of thickness L = 0.4 m,
thermal conductivity k = 2.3 W/m • °C, and surface area
A = 20 m 2 . The left side of the wall is maintained at a constant
temperature of 80°C, while the right side loses heat by con-
vection to the surrounding air at r„ = 15°C with a heat trans-
fer coefficient of h = 24 W/m 2 • °C. Assuming steady one-
dimensional heat transfer and taking the nodal spacing to be
10 cm, (a) obtain the finite difference formulation for all nodes,
(b) determine the nodal temperatures by solving those equa-
tions, and (c) evaluate the rate of heat transfer through the wall.
5-28 Consider the base plate of a 800-W household iron hav-
ing a thickness of L = 0.6 cm, base area of A = 160 cm 2 , and
thermal conductivity of k = 20 W/m • °C. The inner surface of
the base plate is subjected to uniform heat flux generated by
the resistance heaters inside. When steady operating conditions
are reached, the outer surface temperature of the plate is mea-
sured to be 85°C. Disregarding any heat loss through the upper
part of the iron and taking the nodal spacing to be 0.2 cm,
(a) obtain the finite difference formulation for the nodes and
(b) determine the inner surface temperature of the plate by
solving those equations. Answer: {b) 100°C
Insulation
- Resistance heater, 800 W
Base plate
Ax = 0.2 cm
85°C
160 cm 2
FIGURE P5-28
5-29 Consider a large plane wall of thickness L = 0.3 m,
thermal conductivity k = 2.5 W/m • °C, and surface area
A = 12 m 2 . The left side of the wall is subjected to a heat flux
of q = 700 W/m 2 while the temperature at that surface is mea-
sured to be T Q = 60°C. Assuming steady one -dimensional heat
transfer and taking the nodal spacing to be 6 cm, (a) obtain the
finite difference formulation for the six nodes and (b) deter-
mine the temperature of the other surface of the wall by solv-
ing those equations.
5-30E A large steel plate having a thickness of L = 5 in.,
thermal conductivity of k = 7.2 Btu/h • ft • °F, and an emissiv-
ity of e = 0.6 is lying on the ground. The exposed surface of
cen58933_ch05.qxd 9/4/2002 11:42 AM Page 31E
318
HEAT TRANSFER
FIGURE P5-30E
the plate exchanges heat by convection with the ambient air
at T a = 80°F with an average heat transfer coefficient of
h = 3.5 Btu/h • ft 2 ■ °F as well as by radiation with the open sky
at an equivalent sky temperature of r sky = 510 R. The ground
temperature below a certain depth (say, 3 ft) is not affected by
the weather conditions outside and remains fairly constant at
50°F at that location. The thermal conductivity of the soil can
be taken to be k ma = 0.49 Btu/h ■ ft ■ °F, and the steel plate can
be assumed to be in perfect contact with the ground. Assuming
steady one-dimensional heat transfer and taking the nodal
spacings to be 1 in. in the plate and 0.6 ft in the ground, (a) ob-
tain the finite difference formulation for all 1 1 nodes shown in
Figure P5-30E and (b) determine the top and bottom surface
temperatures of the plate by solving those equations.
5-31E Repeat Problem 5-30E by disregarding radiation heat
transfer from the upper surface. Answers: (£>) 78.7T, 78.4°F
5-32 Consider a stainless steel spoon (k = 15.1 W/m • C,
e = 0.6) that is partially immersed in boiling water at 95°C in
a kitchen at 25°C. The handle of the spoon has a cross section
of about 0.2 cm X 1 cm and extends 1 8 cm in the air from the
free surface of the water. The spoon loses heat by convection
to the ambient air with an average heat transfer coefficient of
h = 1 3 W/m 2 ■ °C as well as by radiation to the surrounding
surfaces at an average temperature of r surr = 295 K. Assuming
steady one -dimensional heat transfer along the spoon and tak-
ing the nodal spacing to be 3 cm, (a) obtain the finite difference
formulation for all nodes, (b) determine the temperature of the
tip of the spoon by solving those equations, and (c) determine
the rate of heat transfer from the exposed surfaces of the spoon.
5-33 Repeat Problem 5-32 using a nodal spacing of 1.5 cm.
FIGURE P5-32
5-34 fu'M Reconsider Problem 5-33. Using EES (or other)
EgS software, investigate the effects of the thermal
conductivity and the emissivity of the spoon material on the
temperature at the spoon tip and the rate of heat transfer from
the exposed surfaces of the spoon. Let the thermal conductiv-
ity vary from 10 W/m • °C to 400 W/m • °C, and the emissivity
from 0.1 to 1.0. Plot the spoon tip temperature and the heat
transfer rate as functions of thermal conductivity and emissiv-
ity, and discuss the results.
5-35 One side of a 2-m-high and 3-m-wide vertical plate
at 130°C is to be cooled by attaching aluminum fins (k =
Til W/m • °C) of rectangular profile in an environment at
35°C. The fins are 2 cm long, 0.3 cm thick, and 0.4 cm apart.
The heat transfer coefficient between the fins and the sur-
rounding air for combined convection and radiation is esti-
mated to be 30 W/m 2 • °C. Assuming steady one-dimensional
heat transfer along the fin and taking the nodal spacing to be
0.5 cm, determine (a) the finite difference formulation of this
problem, (b) the nodal temperatures along the fin by solving
these equations, (c) the rate of heat transfer from a single fin,
2 cm
FIGURE P5-3
cen58 93 3_ch05.qxd 9/4/2002 11:42 AM Page 319
319
CHAPTER 5
and (d) the rate of heat transfer from the entire finned surface
of the plate.
5-36 A hot surface at 100°C is to be cooled by attach-
ing 3-cm-long, 0.25-cm-diameter aluminum pin fins (k =
237 W/m ■ °C) with a center-to-center distance of 0.6 cm. The
temperature of the surrounding medium is 30°C, and the com-
bined heat transfer coefficient on the surfaces is 35 W/m 2 ■ °C.
Assuming steady one-dimensional heat transfer along the fin
and taking the nodal spacing to be 0.5 cm, determine (a) the fi-
nite difference formulation of this problem, (b) the nodal tem-
peratures along the fin by solving these equations, (c) the rate
of heat transfer from a single fin, and (d) the rate of heat trans-
fer from a 1-m X 1-m section of the plate.
<
100°C
0.5 cm
|
1
X
1
2 3
4
5
6
FIGURE P5-36
386
5-37 Repeat Problem 5-36 using copper fins (k
W/m • °C) instead of aluminum ones.
Answers: (b) 98.6°C, 97.5°C, 96.7°C, 96.0°C, 95.7°C, 95.5°C
5-38 Two 3-m-long and 0.4-cm-thick cast iron (k = 52
W/m • °C, e = 0.8) steam pipes of outer diameter 10 cm are
connected to each other through two 1 -cm-thick flanges of
outer diameter 20 cm, as shown in the figure. The steam flows
inside the pipe at an average temperature of 200°C with a heat
transfer coefficient of 180 W/m 2 • °C. The outer surface of the
pipe is exposed to convection with ambient air at 8°C with a
heat transfer coefficient of 25 W/m 2 • °C as well as radiation
with the surrounding surfaces at an average temperature of
r surr = 290 K. Assuming steady one -dimensional heat conduc-
tion along the flanges and taking the nodal spacing to be 1 cm
along the flange (a) obtain the finite difference formulation for
all nodes, (b) determine the temperature at the tip of the flange
by solving those equations, and (c) determine the rate of heat
transfer from the exposed surfaces of the flange.
FIGURE P5-38
5-39 Tu'M Reconsider Problem 5-38. Using EES (or other)
|££^ software, investigate the effects of the steam tem-
perature and the outer heat transfer coefficient on the flange tip
temperature and the rate of heat transfer from the exposed sur-
faces of the flange. Let the steam temperature vary from 150°C
to 300°C and the heat transfer coefficient from 15 W/m 2 ■ °C to
60 W/m 2 • °C. Plot the flange tip temperature and the heat
transfer rate as functions of steam temperature and heat trans-
fer coefficient, and discuss the results.
5-40
(a)
(b)
Using EES (or other) software, solve these sys-
tems of algebraic equations.
3*
4.v,
1 2 3
=
i i '-•X') ' X-%
= 3
1 2 3
= 2
- 2x\ + 0.5* 3
= -2
X\ X'l i X3
= 11.964
Xi \ X*2 ' X-\
= 3
Answers: (a) Xj = 2, x 2 = 3, x 3 = -1, {b) x x = 2.33, x 2 = 2.29,
x 3 = -1.62
5-41 FcT^S Using EES (or other) software, solve these sys-
tems of algebraic equations.
(a)
jX 1 \ ZX9 Xt. ' Xa
-2x,
(b)
*1
+ 2x
! X 4 ~
-3
- x 2
+ 3jc
, + X 4 =
2
3je 2
+ x 3
~ 4*4 =
-6
3jf[
+ x\
+ 2*3 =
8
x\-
f 3;t 2
+ 2*3 =
-6.293
2.v,
- V 4
x 2
+ 4*3 =
-12
cen58933_ch05.qxd 9/4/2002 11:42 AM Page 32C
320
HEAT TRANSFER
5-42 [?(,■>! Using EES (or other) software, solve these sys-
tems of algebraic equations.
(a)
4.V,
x 2 + 2x 3 + x 4
3x,
-x,
2jc
- *3
2x,
4x 4
JXa
4jc 3 — 3jc 4
(6)
2.v,
4
x 2 + 4x 2
- 2x 3
2x3 "
-X[ + x\ 4
2x 4
5x 3
8x 4
10
15
Two-Dimensional Steady Heat Conduction
5-43C Consider a medium in which the finite difference
formulation of a general interior node is given in its simplest
form as
'left ' 'top + bright
4T„,
5node'
(a) Is heat transfer in this medium steady or transient?
(b) Is heat transfer one-, two-, or three-dimensional?
(c) Is there heat generation in the medium?
(d) Is the nodal spacing constant or variable?
(e) Is the thermal conductivity of the medium constant or
variable?
5-44C Consider a medium in which the finite difference
formulation of a general interior node is given in its simplest
form as
(Tleft + 'top
' right ~*~ ' bottom )'^
(a) Is heat transfer in this medium steady or transient?
(b) Is heat transfer one-, two-, or three-dimensional?
(c) Is there heat generation in the medium?
(rf) Is the nodal spacing constant or variable?
(e) Is the thermal conductivity of the medium constant or
variable?
5-45C What is an irregular boundary? What is a practical
way of handling irregular boundary surfaces with the finite dif-
ference method?
5-46 Consider steady two-dimensional heat transfer in a long
solid body whose cross section is given in the figure. The tem-
peratures at the selected nodes and the thermal conditions at
the boundaries are as shown. The thermal conductivity of the
body is k = 45 W/m • °C, and heat is generated in the body uni-
formly at a rate of g = 6 X 10 6 W/m 3 . Using the finite differ-
ence method with a mesh size of Ax = Ay = 5.0 cm, determine
(a) the temperatures at nodes 1 , 2, and 3 and (b) the rate of heat
loss from the bottom surface through a 1-m-long section of the
body.
Insulation
200°C
5 cm
5 cm
g = 6xl
260
6 W/m 3
305
©
240
290
©
®
350°C
325
Convection
T x = 20°C, h = 50 W/m 2 ■ °C
FIGURE P5-46
5-47 Consider steady two-dimensional heat transfer in a long
solid body whose cross section is given in the figure. The mea-
sured temperatures at selected points of the outer surfaces are
as shown. The thermal conductivity of the body is k = 45
W/m ■ °C, and there is no heat generation. Using the finite dif-
ference method with a mesh size of Ax = Ay = 2.0 cm, deter-
mine the temperatures at the indicated points in the medium.
Hint: Take advantage of symmetry.
150
IS
200
IS
150°C
180-
200
180
®
©
©
©
©
©
©
©
©
?2 cm
2 cm
200
150 180 200 180 150
FIGURE P5^7
5-48 Consider steady two-dimensional heat transfer in a long
solid bar whose cross section is given in the figure. The mea-
sured temperatures at selected points of the outer surfaces are
as shown. The thermal conductivity of the body is k = 20
W/m • °C, and there is no heat generation. Using the finite dif-
ference method with a mesh size of Ax = Ay = 1.0 cm, deter-
mine the temperatures at the indicated points in the medium.
Answers: T r = 185°C, T 2 = T 3 = 7" 4 = 190°C
5-49 Starting with an energy balance on a volume element,
obtain the steady two-dimensional finite difference equation
for a general interior node in rectangular coordinates for T(x, y)
for the case of variable thermal conductivity and uniform heat
generation.
cen58 93 3_ch05.qxd 9/4/2002 11:42 AM Page 321
150
180 200°C
180-
200 A
(a)
1 00
120
140
©
©
► <
Insulation
120 120
©
©
©
©
100°C
120
140
Insulation
(b)
FIGURE P5^8
5-50 Consider steady two-dimensional heat transfer in a long
solid body whose cross section is given in the figure. The tem-
peratures at the selected nodes and the thermal conditions on
the boundaries are as shown. The thermal conductivity of the
body is k = 180 W/m • °C, and heat is generated in the body
uniformly at a rate of g = 10 7 W/m 3 . Using the finite difference
method with a mesh size of Ax = Ay = 10 cm, determine
(a) the temperatures at nodes 1, 2, 3, and 4 and (b) the rate
of heat loss from the top surface through a 1 -m-long section of
the body.
100
100
100
100°C
120
150
g = 10 7
W/m 3
©
©
©
©
•0.1 m
0.1 m
>i
120
1 50
200 200
FIGURE P5-50
200
200
5-51
Reconsider Problem 5-50. Using EES (or other)
software, investigate the effects of the thermal
conductivity and the heat generation rate on the temperatures at
nodes 1 and 3, and the rate of heat loss from the top surface.
Let the thermal conductivity vary from 10 W/m ■ °C to 400
W/m • °C and the heat generation rate from 10 5 W/m 3 to 10 8
321
CHAPTER 5
W/m 3 . Plot the temperatures at nodes 1 and 3, and the rate of
heat loss as functions of thermal conductivity and heat genera-
tion rate, and discuss the results.
5-52 Consider steady two-dimensional heat transfer in a long
solid bar whose cross section is given in the figure. The mea-
sured temperatures at selected points on the outer surfaces are
as shown. The thermal conductivity of the body is k = 20
W/m ■ °C, and there is no heat generation. Using the finite dif-
ference method with a mesh size of Ajc = Ay = 1.0 cm, deter-
mine the temperatures at the indicated points in the medium.
Hint: Take advantage of symmetry.
Answers: (b) 7i = T 4 = 143°C, T 2 =T 3 = 136°C
100°C
300
(a)
300
Insulation
100 100
100 100°C
©
®\^y©
©
1 1 cm
J-L^
200 200 200 200 200°C
(b)
FIGURE P5-52
5-53 Consider steady two-dimensional heat transfer in an
L-shaped solid body whose cross section is given in the figure.
The thermal conductivity of the body is k = 45 W/m • °C, and
heat is generated in the body at a rate of g = 5 X 10 6 W/m 3 .
Convection
h,T«,
4
5
6
7 \
— ■
1
•1.5 cm
1.5 cm
<1l
FIGURE P5-53
120°C
cen58 93 3_ch05.qxd 9/4/2002 11:43 AM Page 322
322
HEAT TRANSFER
The right surface of the body is insulated, and the bottom sur-
face is maintained at a uniform temperature of 120°C. The
entire top surface is subjected to convection with ambient air
at T^ = 30°C with a heat transfer coefficient of h = 55
W/m 2 • °C, and the left surface is subjected to heat flux at a uni-
form rate of q L = 8000 W/m 2 . The nodal network of the prob-
lem consists of 13 equally spaced nodes with Ax = Ay =
1 .5 cm. Five of the nodes are at the bottom surface and thus
their temperatures are known, (a) Obtain the finite difference
equations at the remaining eight nodes and (b) determine the
nodal temperatures by solving those equations.
5-54E Consider steady two-dimensional heat transfer in a
long solid bar of square cross section in which heat is gener-
ated uniformly at a rate of g = 0.19 X 10 5 Btu/h • ft 3 . The cross
section of the bar is 0.4 ft X 0.4 ft in size, and its thermal con-
ductivity is k = 16 Btu/h • ft • °F. All four sides of the bar are
subjected to convection with the ambient air at T«, = 70°F with
a heat transfer coefficient of h = 7.9 Btu/h ■ ft 2 • °F Using the
finite difference method with a mesh size of Ax = Ay = 0.2 ft,
determine (a) the temperatures at the nine nodes and (b) the
rate of heat loss from the bar through a 1-ft-long section.
Answer: (b) 3040 Btu/h
h, T rjc
k
4 5 6
7 8
h,T«,
h,T„
FIGURE P5-54E
h, T„
5-55 Hot combustion gases of a furnace are flowing through
a concrete chimney (k = 1.4 W/m • °C) of rectangular cross
T ,h
o o
Chimney
10 cm
10 cm
FIGURE P5-55
40 cm -
10 cm
section. The flow section of the chimney is 20 cm X 40 cm,
and the thickness of the wall is 10 cm. The average temperature
of the hot gases in the chimney is T : = 280°C, and the average
convection heat transfer coefficient inside the chimney is hj =
75 W/m 2 • °C. The chimney is losing heat from its outer surface
to the ambient air at T = 15°C by convection with a heat
transfer coefficient of h a = 18 W/m 2 • °C and to the sky by
radiation. The emissivity of the outer surface of the wall is
e = 0.9, and the effective sky temperature is estimated to be
250 K. Using the finite difference method with Ax = Ay =
10 cm and taking full advantage of symmetry, (a) obtain the
finite difference formulation of this problem for steady two-
dimensional heat transfer, (b) determine the temperatures at the
nodal points of a cross section, and (c) evaluate the rate of heat
loss for a 1 -m-long section of the chimney.
5-56 Repeat Problem 5-55 by disregarding radiation heat
transfer from the outer surfaces of the chimney.
5-57 fitt'M Reconsider Problem 5-55. Using EES (or other)
1^2 software, investigate the effects of hot-gas tem-
perature and the outer surface emissivity on the temperatures at
the outer corner of the wall and the middle of the inner surface
of the right wall, and the rate of heat loss. Let the temperature
of the hot gases vary from 200°C to 400°C and the emissivity
from 0.1 to 1.0. Plot the temperatures and the rate of heat loss
as functions of the temperature of the hot gases and the emis-
sivity, and discuss the results.
5-58 /"Tfe, Consider a long concrete dam (k = 0.6 W/m • °C,
x^sy a s = 0.7 m 2 /s) of triangular cross section whose
exposed surface is subjected to solar heat flux of q s =
800 W/m 2 and to convection and radiation to the environ-
ment at 25°C with a combined heat transfer coefficient of 30
W/m 2 • °C. The 2-m-high vertical section of the dam is sub-
jected to convection by water at 15°C with a heat transfer
coefficient of 150 W/m 2 • °C, and heat transfer through the
2-m-long base is considered to be negligible. Using the finite
difference method with a mesh size of Ax = Ay = 1 m and
assuming steady two-dimensional heat transfer, determine the
temperature of the top, middle, and bottom of the exposed sur-
face of the dam. Answers: 21.3°C, 43.2°C, 43.6°C
FIGURE P5-58
cen58 93 3_ch05.qxd 9/4/2002 11:43 AM Page 323
5-59E Consider steady two-dimensional heat transfer in a
V-grooved solid body whose cross section is given in the fig-
ure. The top surfaces of the groove are maintained at 32°F
while the bottom surface is maintained at 212°F. The side sur-
faces of the groove are insulated. Using the finite difference
method with a mesh size of Ax = Ay = 1 ft and taking advan-
tage of symmetry, determine the temperatures at the middle of
the insulated surfaces.
1 \
32°F
/ 9
2
4 \/
7 10
3
5
6
8 11
lift
1 ft
-212°F
FIGURE P5-59E
5-60
Reconsider Problem 5-59E. Using EES (or
other) software, investigate the effects of the
temperatures at the top and bottom surfaces on the temperature
in the middle of the insulated surface. Let the temperatures at
the top and bottom surfaces vary from 32°F to 212°F. Plot the
temperature in the middle of the insulated surface as functions
of the temperatures at the top and bottom surfaces, and discuss
the results.
5-61 Consider a long solid bar whose thermal conductivity is
k = 12 W/m • °C and whose cross section is given in the figure.
The top surface of the bar is maintained at 50°C while the bot-
tom surface is maintained at 120°C. The left surface is insu-
lated and the remaining three surfaces are subjected to
convection with ambient air at T x = 25°C with a heat transfer
coefficient of h = 30 W/m 2 • °C. Using the finite difference
method with a mesh size of Ax = Ay = 10 cm, (a) obtain the
finite difference formulation of this problem for steady two-
50°C
C
>10cm
10 cm
'
h, 7V
120°C
323
CHAPTER 5
dimensional heat transfer and (b) determine the unknown nodal
temperatures by solving those equations.
Answers: (b) 85.7°C, 86.4°C, 87.6°C
5-62 Consider a 5-m-long constantan block (k = 23
W/m ■ °C) 30 cm high and 50 cm wide. The block is com-
pletely submerged in iced water at 0°C that is well stirred, and
the heat transfer coefficient is so high that the temperatures on
both sides of the block can be taken to be 0°C. The bottom sur-
face of the bar is covered with a low-conductivity material so
that heat transfer through the bottom surface is negligible. The
top surface of the block is heated uniformly by a 6-kW resis-
tance heater. Using the finite difference method with a mesh
size of Ax = Ay = 10 cm and taking advantage of symmetry,
(a) obtain the finite difference formulation of this problem for
steady two-dimensional heat transfer, (b) determine the un-
known nodal temperatures by solving those equations, and
(c) determine the rate of heat transfer from the block to the iced
water.
6 kW heater
Insulation
• mmummummummu
o°c
o°c
Insulation
FIGURE P5-62
Transient Heat Conduction
5-63C How does the finite difference formulation of
a transient heat conduction problem differ from that of a
steady heat conduction problem? What does the term
pAAxC(T,'„ + ' — T/„)/At represent in the transient finite differ-
ence formulation?
5-64C What are the two basic methods of solution of tran-
sient problems based on finite differencing? How do heat
transfer terms in the energy balance formulation differ in the
two methods?
5-65C The explicit finite difference formulation of a general
interior node for transient heat conduction in a plane wall is
given by
Ax 2
FIGURE P5-61
2T' + T'
Obtain the finite difference formulation for the steady case by
simplifying the relation above.
cen58 93 3_ch05.qxd 9/4/2002 11:43 AM Page 324
324
HEAT TRANSFER
5-66C The explicit finite difference formulation of a general
interior node for transient two-dimensional heat conduction is
given by
<7y eft
+ (i ■
' - 1 tr,
4t)71
[ right ' L bottom
onode
+ T ;
Obtain the finite difference formulation for the steady case by
simplifying the relation above.
5-67C Is there any limitation on the size of the time step At
in the solution of transient heat conduction problems using
(a) the explicit method and (b) the implicit method?
5-68C Express the general stability criterion for the explicit
method of solution of transient heat conduction problems.
5-69C Consider transient one-dimensional heat conduction
in a plane wall that is to be solved by the explicit method. If
both sides of the wall are at specified temperatures, express the
stability criterion for this problem in its simplest form.
5-70C Consider transient one-dimensional heat conduction
in a plane wall that is to be solved by the explicit method.
If both sides of the wall are subjected to specified heat
flux, express the stability criterion for this problem in its sim-
plest form.
5-71 C Consider transient two-dimensional heat conduction
in a rectangular region that is to be solved by the explicit
method. If all boundaries of the region are either insulated or at
specified temperatures, express the stability criterion for this
problem in its simplest form.
5-72C The implicit method is unconditionally stable and
thus any value of time step At can be used in the solution of
transient heat conduction problems. To minimize the computa-
tion time, someone suggests using a very large value of At
since there is no danger of instability. Do you agree with this
suggestion? Explain.
5-73 Consider transient heat conduction in a plane wall
whose left surface (node 0) is maintained at 50°C while the
right surface (node 6) is subjected to a solar heat flux of 600
W/m 2 . The wall is initially at a uniform temperature of 50°C.
Express the explicit finite difference formulation of the bound-
ary nodes and 6 for the case of no heat generation. Also,
obtain the finite difference formulation for the total amount
of heat transfer at the left boundary during the first three
time steps.
5-74 Consider transient heat conduction in a plane wall with
variable heat generation and constant thermal conductivity.
The nodal network of the medium consists of nodes 0, 1,2, 3,
and 4 with a uniform nodal spacing of Ax. The wall is initially
at a specified temperature. Using the energy balance approach,
obtain the explicit finite difference formulation of the boundary
nodes for the case of uniform heat flux q Q at the left boundary
%
g(x,
k
Ax
FIGURE P5-74
(node 0) and convection at the right boundary (node 4) with a
convection coefficient of h and an ambient temperature of T a .
Do not simplify.
5-75 Repeat Problem 5-74 for the case of implicit formula-
tion.
5-76 Consider transient heat conduction in a plane wall with
variable heat generation and constant thermal conductivity.
The nodal network of the medium consists of nodes 0, 1,2,
3, 4, and 5 with a uniform nodal spacing of Ax The wall is ini-
tially at a specified temperature. Using the energy balance ap-
proach, obtain the explicit finite difference formulation of the
boundary nodes for the case of insulation at the left boundary
(node 0) and radiation at the right boundary (node 5) with an
emissivity of e and surrounding temperature of T smr .
5-77 Consider transient heat conduction in a plane wall with
variable heat generation and constant thermal conductivity.
The nodal network of the medium consists of nodes 0, 1,2, 3,
and 4 with a uniform nodal spacing of Ax. The wall is initially
at a specified temperature. The temperature at the right bound-
ary (node 4) is specified. Using the energy balance approach,
obtain the explicit finite difference formulation of the boundary
node for the case of combined convection, radiation, and heat
flux at the left boundary with an emissivity of e, convection co-
efficient of h, ambient temperature of T m surrounding temper-
ature of T sun , and uniform heat flux of q toward the wall. Also,
obtain the finite difference formulation for the total amount of
heat transfer at the right boundary for the first 20 time steps.
FIGURE P5-77
cen58 93 3_ch05.qxd 9/4/2002 11:43 AM Page 325
5-78 Starting with an energy balance on a volume element,
obtain the two-dimensional transient explicit finite difference
equation for a general interior node in rectangular coordinates
for T(x, y, t) for the case of constant thermal conductivity and
no heat generation.
5-79 Starting with an energy balance on a volume element,
obtain the two-dimensional transient implicit finite difference
equation for a general interior node in rectangular coordinates
for T(x, y, t) for the case of constant thermal conductivity and
no heat generation.
5-80 Starting with an energy balance on a disk volume ele-
ment, derive the one-dimensional transient explicit finite dif-
ference equation for a general interior node for T(z, t) in a
cylinder whose side surface is insulated for the case of constant
thermal conductivity with uniform heat generation.
5-81 Consider one-dimensional transient heat conduction in
a composite plane wall that consists of two layers A and B with
perfect contact at the interface. The wall involves no heat gen-
eration and initially is at a specified temperature. The nodal
network of the medium consists of nodes 0, 1 (at the interface),
and 2 with a uniform nodal spacing of Ax. Using the energy
balance approach, obtain the explicit finite difference formula-
tion of this problem for the case of insulation at the left bound-
ary (node 0) and radiation at the right boundary (node 2) with
an emissivity of e and surrounding temperature of T SUTI .
Insulation
Ax
r
FIGURE P5-81
Ajc
1 2
\
Interface
Radiation
5-82 Consider transient one-dimensional heat conduction in
a pin fin of constant diameter D with constant thermal conduc-
tivity. The fin is losing heat by convection to the ambient air at
T x with a heat transfer coefficient of h and by radiation to the
surrounding surfaces at an average temperature of r surr . The
nodal network of the fin consists of nodes (at the base), 1 (in
the middle), and 2 (at the fin tip) with a uniform nodal spacing
of Ax. Using the energy balance approach, obtain the explicit
finite difference formulation of this problem for the case of a
specified temperature at the fin base and negligible heat trans-
fer at the fin tip.
5-83 Repeat Problem 5-82 for the case of implicit
formulation.
325
CHAPTER 5
5-84 Consider a large uranium plate of thickness L = 8 cm,
thermal conductivity k = 28 W/m • °C, and thermal diffusivity
a = 12.5 X 10~ 6 m 2 /s that is initially at a uniform temperature
of 100°C. Heat is generated uniformly in the plate at a constant
rate of g = 10 6 W/m 3 . At time t = 0, the left side of the plate is
insulated while the other side is subjected to convection with
an environment at T„ = 20°C with a heat transfer coefficient of
h = 35 W/m 2 • °C. Using the explicit finite difference approach
with a uniform nodal spacing of Ax = 2 cm, determine (a) the
temperature distribution in the plate after 5 min and (b) how
long it will take for steady conditions to be reached in the plate.
5-85
Reconsider Problem 5-84. Using EES (or other)
software, investigate the effect of the cooling
time on the temperatures of the left and right sides of the plate.
Let the time vary from 5 min to 60 min. Plot the temperatures
at the left and right surfaces as a function of time, and discuss
the results.
5-86 Consider a house whose south wall consists of a 30-cm-
thick Trombe wall whose thermal conductivity is k = 0.70
W/m • °C and whose thermal diffusivity is a = 0.44 X 10~ 6
m 2 /s. The variations of the ambient temperature T oM and the
solar heat flux <7 S0 | ar incident on a south-facing vertical surface
throughout the day for a typical day in February are given in
the table in 3-h intervals. The Trombe wall has single glazing
with an absorptivity-transmissivity product of k = 0.76 (that
is, 76 percent of the solar energy incident is absorbed by the
exposed surface of the Trombe wall), and the average com-
bined heat transfer coefficient for heat loss from the Trombe
wall to the ambient is determined to be h out = 3.4 W/m 2 • °C.
The interior of the house is maintained at T m = 20°C at all
times, and the heat transfer coefficient at the interior surface of
the Trombe wall is h m = 9.1 W/m 2 • °C. Also, the vents on the
Trombe wall are kept closed, and thus the only heat transfer be-
tween the air in the house and the Trombe wall is through the
Sun's
rays
FIGURE P5-86
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326
HEAT TRANSFER
TABLE P5-86
The hourly variations of the monthly average ambient
temperature and solar heat flux incident on a
vertical surface
Ambient
Solar
Time of Day
Te
nperature, °C
Insolation, W/m 2
7 AM-10 AM
375
10 AM-1 PM
4
750
1 PM-4 PM
6
580
4 PM-7 PM
1
95
7 PM-10 PM
-2
10 PM-1 AM
-3
1 AM-4 AM
-4
4 AM-7 AM
4
interior surface of the wall. Assuming the temperature of the
Trombe wall to vary linearly between 20°C at the interior sur-
face and 0°C at the exterior surface at 7 am and using the ex-
plicit finite difference method with a uniform nodal spacing of
Ax = 5 cm, determine the temperature distribution along the
thickness of the Trombe wall after 6, 12, 18, 24, 30, 36, 42, and
48 hours and plot the results. Also, determine the net amount of
heat transferred to the house from the Trombe wall during the
first day if the wall is 2.8 m high and 7 m long.
5-87 Consider two-dimensional transient heat transfer in an
L-shaped solid bar that is initially at a uniform temperature
of 140°C and whose cross section is given in the figure. The
thermal conductivity and diffusivity of the body are k = 15
W/m • °C and a = 3.2 X 10~ 6 m 2 /s, respectively, and heat is
generated in the body at a rate of g = 2 X 10 7 W/m 3 . The right
surface of the body is insulated, and the bottom surface is
maintained at a uniform temperature of 140°C at all times. At
time t = 0, the entire top surface is subjected to convection
with ambient air at T^ = 25°C with a heat transfer coefficient
of h = 80 W/m 2 ■ °C, and the left surface is subjected to
uniform heat flux at a rate of q L = 8000 W/m 2 . The nodal net-
work of the problem consists of 13 equally spaced nodes with
Ax = Ay = 1.5 cm. Using the explicit method, determine the
temperature at the top corner (node 3) of the body after 2, 5,
and 30 min.
Convection
h,T a
4
5
6
7 \
■1.5cm
1.5cm
1l
FIGURE P5-87
5-88 VigM Reconsider Problem 5-87. Using EES (or other)
t£^ software, plot the temperature at the top comer as
a function of heating time varies from 2 min to 30 min, and dis-
cuss the results.
5-89 Consider a long solid bar (k = 28 W/m • °C and a =
12 X 10~ 6 m 2 /s) of square cross section that is initially at a uni-
form temperature of 20°C. The cross section of the bar is
20 cm X 20 cm in size, and heat is generated in it uniformly at
a rate of g = 8 X 10 5 W/m 3 . All four sides of the bar are sub-
jected to convection to the ambient air at T x = 30°C with
a heat transfer coefficient of h = 45 W/m 2 • °C. Using the
explicit finite difference method with a mesh size of Ax =
Ay = 10 cm, determine the centerline temperature of the bar
(a) after 10 min and (b) after steady conditions are established.
h, r„
2
g
4 5 6
>10 cm
10 cm
8
l±
h, 7L •- ■ ■• h,T m
h,T m
FIGURE P5-89
5-90E Consider a house whose windows are made of
0.375-in. -thick glass (k = 0.48 Btu/h • ft • °F and a = 4.2 X
10~ 6 ft 2 /s). Initially, the entire house, including the walls and
the windows, is at the outdoor temperature of T D = 35°F. It is
observed that the windows are fogged because the indoor tem-
perature is below the dew-point temperature of 54 °F. Now the
heater is turned on and the air temperature in the house is
raised to T, = 72°F at a rate of 2°F rise per minute. The heat
transfer coefficients at the inner and outer surfaces of the wall
can be taken to be A, = 1.2 and h = 2.6 Btu/h ■ ft 2 • °F, respec-
tively, and the outdoor temperature can be assumed to remain
constant. Using the explicit finite difference method with a
mesh size of Ax = 0.125 in., determine how long it will take
House
140°C
Window
glass
Ax
12 3 4
r
>g
0.375 in
Outdoors
FIGURE P5-90E
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327
CHAPTER 5
for the fog on the windows to clear up (i.e., for the inner sur-
face temperature of the window glass to reach 54°F).
5-91 A common annoyance in cars in winter months is the
formation of fog on the glass surfaces that blocks the view.
A practical way of solving this problem is to blow hot air or to
attach electric resistance heaters to the inner surfaces. Consider
the rear window of a car that consists of a 0.4-cm-thick glass
(k = 0.84 W/m ■ °C and a = 0.39 X lO" 6 m 2 /s). Strip heater
wires of negligible thickness are attached to the inner surface
of the glass, 4 cm apart. Each wire generates heat at a rate of
10 W/m length. Initially the entire car, including its windows,
is at the outdoor temperature of T = — 3°C. The heat transfer
coefficients at the inner and outer surfaces of the glass can be
taken to be /;, = 6 and h a = 20 W/m 2 • °C, respectively. Using
the explicit finite difference method with a mesh size of Ax =
0.2 cm along the thickness and Ay = 1 cm in the direction nor-
mal to the heater wires, determine the temperature distribution
throughout the glass 15 min after the strip heaters are turned
on. Also, determine the temperature distribution when steady
conditions are reached.
Thermal
symmetry line
Inner
surface
Outer
surface
Glass
Thermal
symmetry line
FIGURE P5-91
5-92
!►, Repeat Problem 5-9 1 using the implicit method
Kgl/ with a time step of 1 min.
5-93 The roof of a house consists of a 15-cm-thick concrete
slab (k = 1.4 W/m ■ °C and a = 0.69 X lO" 6 m 2 /s) that is 20 m
wide and 20 m long. One evening at 6 pm, the slab is observed
to be at a uniform temperature of 18°C. The average ambient
air and the night sky temperatures for the entire night are pre-
dicted to be 6°C and 260 K, respectively. The convection heat
transfer coefficients at the inner and outer surfaces of the roof
can be taken to be h t = 5 and h a = 12 W/m 2 • °C, respectively.
The house and the interior surfaces of the walls and the floor
Radiation Convection
jLd
Concrete
/ roof
mf ^Convection
W h., T.
Radiation ' '
J
y
rr
n
rr
ii
ii
15 cm
FIGURE P5-93
are maintained at a constant temperature of 20°C during the
night, and the emissivity of both surfaces of the concrete roof
is 0.9. Considering both radiation and convection heat transfers
and using the explicit finite difference method with a time step
of At = 5 min and a mesh size of Ax = 3 cm, determine the
temperatures of the inner and outer surfaces of the roof at 6 am.
Also, determine the average rate of heat transfer through the
roof during that night.
5-94 Consider a refrigerator whose outer dimensions are
1.80 m X 0.8 m X 0.7 m. The walls of the refrigerator are
constructed of 3-cm-thick urethane insulation (k = 0.026
W/m • ° C and a = 0.36 X 10~ 6 m 2 /s) sandwiched between
two layers of sheet metal with negligible thickness. The refrig-
erated space is maintained at 3°C and the average heat transfer
coefficients at the inner and outer surfaces of the wall are
6 W/m 2 • °C and 9 W/m 2 • °C, respectively. Heat transfer
through the bottom surface of the refrigerator is negligible. The
kitchen temperature remains constant at about 25°C. Initially,
the refrigerator contains 15 kg of food items at an average
specific heat of 3.6 kJ/kg • °C. Now a malfunction occurs and
the refrigerator stops running for 6 h as a result. Assuming the
Heat
25°C
FIGURE P5-94
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328
HEAT TRANSFER
temperature of the contents of the refrigerator, including the
air inside, rises uniformly during this period, predict the tem-
perature inside the refrigerator after 6 h when the repair-
man arrives. Use the explicit finite difference method with a
time step of Af = 1 min and a mesh size of Ax = 1 cm and dis-
regard corner effects (i.e., assume one -dimensional heat trans-
fer in the walls).
5-95 rSpM Reconsider Problem 5-94. Using EES (or other)
b^ti software, plot the temperature inside the refrig-
erator as a function of heating time as time varies from 1 h to
10 h, and discuss the results.
Special Topic: Controlling the Numerical Error
5-96C Why do the results obtained using a numerical
method differ from the exact results obtained analytically?
What are the causes of this difference?
5-97C What is the cause of the discretization error? How
does the global discretization error differ from the local
discretization error?
5-98C Can the global (accumulated) discretization error be
less than the local error during a step? Explain.
5-99C How is the finite difference formulation for the first
derivative related to the Taylor series expansion of the solution
function?
5-100C Explain why the local discretization error of the fi-
nite difference method is proportional to the square of the step
size. Also explain why the global discretization error is propor-
tional to the step size itself.
5-101C What causes the round-off error? What kind of
calculations are most susceptible to round-off error?
5-102C What happens to the discretization and the round-
off errors as the step size is decreased?
5-103C Suggest some practical ways of reducing the
round-off error.
5-104C What is a practical way of checking if the round-off
error has been significant in calculations?
5-105C What is a practical way of checking if the dis-
cretization error has been significant in calculations?
Review Problems
5-106 Starting with an energy balance on the volume ele-
ment, obtain the steady three-dimensional finite difference
equation for a general interior node in rectangular coordinates
for T(x, y, z) for the case of constant thermal conductivity and
uniform heat generation.
5-107 Starting with an energy balance on the volume ele-
ment, obtain the three-dimensional transient explicit finite dif-
ference equation for a general interior node in rectangular
coordinates for T(x, y, z, f) for the case of constant thermal con-
ductivity and no heat generation.
5-108 Consider steady one -dimensional heat conduction in a
plane wall with variable heat generation and constant thermal
conductivity. The nodal network of the medium consists of
nodes 0, 1,2, and 3 with a uniform nodal spacing of Ax. The
temperature at the left boundary (node 0) is specified. Using
the energy balance approach, obtain the finite difference for-
mulation of boundary node 3 at the right boundary for the case
of combined convection and radiation with an emissivity of e,
convection coefficient of h, ambient temperature of T m and
surrounding temperature of T sun . Also, obtain the finite dif-
ference formulation for the rate of heat transfer at the left
boundary.
g(x)
Ax
FIGUREP5-108
5-109 Consider one-dimensional transient heat conduction
in a plane wall with variable heat generation and variable ther-
mal conductivity. The nodal network of the medium consists of
nodes 0, 1 , and 2 with a uniform nodal spacing of Ax. Using
the energy balance approach, obtain the explicit finite differ-
ence formulation of this problem for the case of specified heat
flux q and convection at the left boundary (node 0) with a con-
vection coefficient of h and ambient temperature of T m and ra-
diation at the right boundary (node 2) with an emissivity of e
and surrounding temperature of T sun .
5-110 Repeat Problem 5-109 for the case of implicit
formulation.
5-111 Consider steady one -dimensional heat conduction in a
pin fin of constant diameter D with constant thermal conduc-
tivity. The fin is losing heat by convection with the ambient air
Convection
FIGURE P5-1 1 1
cen58 93 3_ch05.qxd 9/4/2002 11:43 AM Page 329
at T-j, (in °C) with a convection coefficient of h, and by radia-
tion to the surrounding surfaces at an average temperature of
T mTI (in K). The nodal network of the fin consists of nodes (at
the base), 1 (in the middle), and 2 (at the fin tip) with a uniform
nodal spacing of Ax. Using the energy balance approach, ob-
tain the finite difference formulation of this problem for the
case of a specified temperature at the fin base and convection
and radiation heat transfer at the fin tip.
5-112 Starting with an energy balance on the volume ele-
ment, obtain the two-dimensional transient explicit finite dif-
ference equation for a general interior node in rectangular
coordinates for T(x, y, t) for the case of constant thermal con-
ductivity and uniform heat generation.
5-113 Starting with an energy balance on a disk volume ele-
ment, derive the one-dimensional transient implicit finite dif-
ference equation for a general interior node for T(z, t) in a
cylinder whose side surface is subjected to convection with a
convection coefficient of h and an ambient temperature of 7^
for the case of constant thermal conductivity with uniform heat
generation.
5-114E The roof of a house consists of a 5-in. -thick concrete
slab (k = 0.81 Btu/h • ft • °F and a = 7.4 X 1(T 6 ft 2 /s) that is
45 ft wide and 55 ft long. One evening at 6 pm, the slab is ob-
served to be at a uniform temperature of 70°F. The ambient air
temperature is predicted to be at about 50°F from 6 pm to
10 pm, 42°F from 10 pm to 2 am, and 38°F from 2 am to 6 am,
while the night sky temperature is expected to be about 445 R
for the entire night. The convection heat transfer coefficients at
the inner and outer surfaces of the roof can be taken to be
ft, = 0.9 and h a = 2.1 Btu/h • ft 2 • °F, respectively. The house
and the interior surfaces of the walls and the floor are main-
tained at a constant temperature of 70°F during the night, and
the emissivity of both surfaces of the concrete roof is 0.9.
Considering both radiation and convection heat transfers and
using the explicit finite difference method with a mesh size of
Radiation Convection
Ji IK, T_
Concrete
/ roof
W h..T.
?
Radiation
Convection
h., T.
rr
rn
rr
ii
ii
329
CHAPTER 5
Ax = 1 in. and a time step of At = 5 min, determine the tem-
peratures of the inner and outer surfaces of the roof at 6 am.
Also, determine the average rate of heat transfer through the
roof during that night.
5-115 Solar radiation incident on a large body of clean water
(k = 0.61 W/m • °C and a = 0.15 X 10~ 6 m 2 /s) such as a lake,
a river, or a pond is mostly absorbed by water, and the amount
of absorption varies with depth. For solar radiation incident at
a 45° angle on a 1-m-deep large pond whose bottom surface is
black (zero reflectivity), for example, 2.8 percent of the solar
energy is reflected back to the atmosphere, 37.9 percent is ab-
sorbed by the bottom surface, and the remaining 59.3 percent
is absorbed by the water body. If the pond is considered to be
four layers of equal thickness (0.25 m in this case), it can be
shown that 47.3 percent of the incident solar energy is ab-
sorbed by the top layer, 6.1 percent by the upper mid layer, 3.6
percent by the lower mid layer, and 2.4 percent by the bottom
layer [for more information see Cengel and Ozisik, Solar En-
ergy, 33, no. 6 (1984), pp. 581-591]. The radiation absorbed by
the water can be treated conveniently as heat generation in the
heat transfer analysis of the pond.
Consider a large 1 -m-deep pond that is initially at a uniform
temperature of 15°C throughout. Solar energy is incident on
the pond surface at 45° at an average rate of 500 W/m 2 for a pe-
riod of 4 h. Assuming no convection currents in the water and
using the explicit finite difference method with a mesh size of
Ax = 0.25 m and a time step of At = 15 min, determine the
temperature distribution in the pond under the most favorable
conditions (i.e., no heat losses from the top or bottom surfaces
of the pond). The solar energy absorbed by the bottom surface
of the pond can be treated as a heat flux to the water at that sur-
face in this case.
Solar . „,, 2
radiation s
45
Top layer
Solar pond
Upper mid layer
Lower mid layer
Bottom layer
- Black
:
FIGURE P5-114E
FIGURE P5-1 15
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330
HEAT TRANSFER
5-116 Reconsider Problem 5-115. The absorption of solar
radiation in that case can be expressed more accurately as a
fourth-degree polynomial as
g(x) =
4,(0.859 - 3.415x + 6.704* 2 - 6.339x 3 + 2.278.x 4 ), W/m 3
where q s is the solar flux incident on the surface of the pond in
W/m 2 and x is the distance from the free surface of the pond
in m. Solve Problem 5-115 using this relation for the absorp-
tion of solar radiation.
5-117 A hot surface at 120°C is to be cooled by attaching
8 cm long, 0.8 cm in diameter aluminum pin fins (k = 237
W/m ■ °C and a = 97.1 X 10~ 6 m 2 /s) to it with a center-to-
center distance of 1 .6 cm. The temperature of the surrounding
medium is 15°C, and the heat transfer coefficient on the sur-
faces is 35 W/m 2 • °C. Initially, the fins are at a uniform tem-
perature of 30°C, and at time t = 0, the temperature of the hot
surface is raised to 120°C. Assuming one-dimensional heat
conduction along the fin and taking the nodal spacing to be
Ax = 2 cm and a time step to be At = 0.5 s, determine the
nodal temperatures after 5 min by using the explicit finite dif-
ference method. Also, determine how long it will take for
steady conditions to be reached.
Convection
h, r„
FIGURE P5-1 17
5-118E Consider a large plane wall of thickness L = 0.3 ft
and thermal conductivity k = 1.2 Btu/h • ft ■ °F in space. The
wall is covered with a material having an emissivity of
e = 0.80 and a solar absorptivity of a s = 0.45. The inner sur-
face of the wall is maintained at 520 R at all times, while the
outer surface is exposed to solar radiation that is incident at a
rate of q\ = 300 Btu/h • ft 2 . The outer surface is also losing heat
Ax
by radiation to deep space at R. Using a uniform nodal spac-
ing of Ax = 0. 1 ft, (a) obtain the finite difference formulation
for steady one-dimensional heat conduction and (b) determine
the nodal temperatures by solving those equations.
Answers: (b) 522 R, 525 R, 527 R
5-119 Frozen food items can be defrosted by simply leaving
them on the counter, but it takes too long. The process can
be speeded up considerably for flat items such as steaks by
placing them on a large piece of highly conducting metal,
called the defrosting plate, which serves as a fin. The increased
surface area enhances heat transfer and thus reduces the de-
frosting time.
Consider two 1.5-cm-thick frozen steaks at — 18°C that re-
semble a 15-cm-diameter circular object when placed next to
each other. The steaks are now placed on a 1 -cm-thick black-
anodized circular aluminum defrosting plate (k = 237
W/m • °C, a = 97.1 X 10~ 6 m 2 /s, and e = 0.90) whose outer
diameter is 30 cm. The properties of the frozen steaks are
p = 970 kg/m 3 , C p = 1.55 kJ/kg • °C, k = 1.40 W/m • °C,
a = 0.93 X 10~ 6 m 2 /s, and e = 0.95, and the heat of fusion is
h
if
187 kJ/kg. The steaks can be considered to be defrosted
when their average temperature is 0°C and all of the ice in the
steaks is melted. Initially, the defrosting plate is at the room
temperature of 20°C, and the wooden countertop it is placed on
can be treated as insulation. Also, the surrounding surfaces can
be taken to be at the same temperature as the ambient air, and
the convection heat transfer coefficient for all exposed surfaces
can be taken to be 12 W/m 2 • °C. Heat transfer from the lateral
surfaces of the steaks and the defrosting plate can be neglected.
Assuming one -dimensional heat conduction in both the steaks
and the defrosting plate and using the explicit finite difference
method, determine how long it will take to defrost the steaks.
Use four nodes with a nodal spacing of Ax = 0.5 cm for the
steaks, and three nodes with a nodal spacing of Ar = 3.75 cm
for the exposed portion of the defrosting plate. Also, use a time
step of Af = 5 s. Hint: First, determine the total amount of heat
transfer needed to defrost the steaks, and then determine how
long it will take to transfer that much heat.
Radiation
Defrosting
plate
FIGURE P5-118E
FIGURE P5-1 19
cen58 93 3_ch05.qxd 9/4/2002 11:43 AM Page 331
5-120 Repeat Problem 5-119 for a copper defrosting plate
using a time step of At = 3 s.
Design and Essay Problems
5-121 Write a two-page essay on the finite element method,
and explain why it is used in most commercial engineering
software packages. Also explain how it compares to the finite
difference method.
5-122 Numerous professional software packages are avail-
able in the market for performing heat transfer analysis, and
they are widely advertised in professional magazines such as
the Mechanical Engineering magazine published by the Amer-
ican Society of Mechanical Engineers (ASME). Your company
decides to purchase such a software package and asks you
to prepare a report on the available packages, their costs, ca-
pabilities, ease of use, and compatibility with the available
hardware, and other software as well as the reputation of the
software company, their history, financial health, customer
support, training, and future prospects, among other things.
After a preliminary investigation, select the top three packages
and prepare a full report on them.
5-123 Design a defrosting plate to speed up defrosting of flat
food items such as frozen steaks and packaged vegetables and
evaluate its performance using the finite difference method
(see Prob. 5-119). Compare your design to the defrosting
331
CHAPTER 5
plates currently available on the market. The plate must per-
form well, and it must be suitable for purchase and use as a
household utensil, durable, easy to clean, easy to manufacture,
and affordable. The frozen food is expected to be at an initial
temperature of — 1 8°C at the beginning of the thawing process
and 0°C at the end with all the ice melted. Specify the material,
shape, size, and thickness of the proposed plate. Justify your
recommendations by calculations. Take the ambient and sur-
rounding surface temperatures to be 20°C and the convection
heat transfer coefficient to be 15 W/m 2 • °C in your analysis.
For a typical case, determine the defrosting time with and
without the plate.
5-124 Design a fire-resistant safety box whose outer dimen-
sions are 0.5 m X 0.5 m X 0.5 m that will protect its com-
bustible contents from fire which may last up to 2 h. Assume
the box will be exposed to an environment at an average tem-
perature of 700°C with a combined heat transfer coefficient of
70 W/m 2 • °C and the temperature inside the box must be be-
low 150°C at the end of 2 h. The cavity of the box must be as
large as possible while meeting the design constraints, and the
insulation material selected must withstand the high tempera-
tures to which it will be exposed. Cost, durability, and strength
are also important considerations in the selection of insulation
materials.
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CHAPTER
FUNDAMENTALS OF
CONVECTION
ft.
So far, we have considered conduction, which is the mechanism of heat
transfer through a solid or a quiescent fluid. We now consider convec-
tion, which is the mechanism of heat transfer through a fluid in the
presence of bulk fluid motion.
Convection is classified as natural (ox free) and. forced convection, depend-
ing on how the fluid motion is initiated. In forced convection, the fluid is
forced to flow over a surface or in a pipe by external means such as a pump or
a fan. In natural convection, any fluid motion is caused by natural means such
as the buoyancy effect, which manifests itself as the rise of warmer fluid and
the fall of the cooler fluid. Convection is also classified as external and inter-
nal, depending on whether the fluid is forced to flow over a surface or in a
channel.
We start this chapter with a general physical description of the convection
mechanism. We then discuss the velocity and thermal boundary layers, and
laminar and turbulent flows. We continue with the discussion of the dimen-
sionless Reynolds, Prandtl, and Nusselt numbers, and their physical signifi-
cance. Next we derive the convection equations of on the basis of mass,
momentum, and energy conservation, and obtain solutions for flow over aflat
plate. We then nondimensionalize the convection equations, and obtain func-
tional forms of friction and convection coefficients. Finally, we present analo-
gies between momentum and heat transfer.
CONTENTS
6-1 Physical Mechanism on
Convection 334
6-2 Classification of Fluid
Flows 337
6-3 Velocity Boundary Layer 335
6-4 Thermal Boundary
Layer 341
6-5 Laminar and Turbulent
Flows 342
6-6 Heat and Momentum Transfer in
Turbulent Flow 343
6-7 Derivation of Differential
Convection Equations 345
6-8 Solutions of Convection
Equations for a Flat
Plate 352
6-9 Nondimensionalized
Convection Equations and
Similarity 356
6-10 Functional Forms of
Friction and Convection
Coefficients 357
6-11 Analogies between Momentum
and Heat Transfer 358
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334
HEAT TRANSFER
20°C
5 m/s
AIR
L
50°C
(a) Forced convection
(b) Free convection
AIR
C
. No convection
ii currents
H
(c) Conduction
FIGURE 6-1
Heat transfer from a hot surface to the
surrounding fluid by convection and
conduction.
-Hotplate, 110°C
Fluid
•T-
Heat transfer
through the
fluid - — -.
v Cold plate, 30°C
FIGURE 6-2
Heat transfer through a fluid
sandwiched between two parallel
plates.
6-1 - PHYSICAL MECHANISM OF CONVECTION
We mentioned earlier that there are three basic mechanisms of heat transfer:
conduction, convection, and radiation. Conduction and convection are similar
in that both mechanisms require the presence of a material medium. But they
are different in that convection requires the presence of fluid motion.
Heat transfer through a solid is always by conduction, since the molecules
of a solid remain at relatively fixed positions. Heat transfer through a liquid or
gas, however, can be by conduction or convection, depending on the presence
of any bulk fluid motion. Heat transfer through a fluid is by convection in the
presence of bulk fluid motion and by conduction in the absence of it. There-
fore, conduction in a fluid can be viewed as the limiting case of convection,
corresponding to the case of quiescent fluid (Fig. 6-1).
Convection heat transfer is complicated by the fact that it involves fluid mo-
tion as well as heat conduction. The fluid motion enhances heat transfer, since
it brings hotter and cooler chunks of fluid into contact, initiating higher rates
of conduction at a greater number of sites in a fluid. Therefore, the rate of heat
transfer through a fluid is much higher by convection than it is by conduction.
In fact, the higher the fluid velocity, the higher the rate of heat transfer.
To clarify this point further, consider steady heat transfer through a fluid
contained between two parallel plates maintained at different temperatures, as
shown in Figure 6-2. The temperatures of the fluid and the plate will be the
same at the points of contact because of the continuity of temperature. As-
suming no fluid motion, the energy of the hotter fluid molecules near the hot
plate will be transferred to the adjacent cooler fluid molecules. This energy
will then be transferred to the next layer of the cooler fluid molecules. This
energy will then be transferred to the next layer of the cooler fluid, and so on,
until it is finally transferred to the other plate. This is what happens during
conduction through a fluid. Now let us use a syringe to draw some fluid near
the hot plate and inject it near the cold plate repeatedly. You can imagine that
this will speed up the heat transfer process considerably, since some energy is
carried to the other side as a result of fluid motion.
Consider the cooling of a hot iron block with a fan blowing air over its top
surface, as shown in Figure 6-3. We know that heat will be transferred from
the hot block to the surrounding cooler air, and the block will eventually
cool. We also know that the block will cool faster if the fan is switched to a
higher speed. Replacing air by water will enhance the convection heat trans-
fer even more.
Experience shows that convection heat transfer strongly depends on the
fluid properties dynamic viscosity |x, thermal conductivity k, density p, and
specific heat C p as well as the fluid velocity T. It also depends on the geome-
try and the roughness of the solid surface, in addition to the type of fluid flow
(such as being streamlined or turbulent). Thus, we expect the convection heat
transfer relations to be rather complex because of the dependence of convec-
tion on so many variables. This is not surprising, since convection is the most
complex mechanism of heat transfer.
Despite the complexity of convection, the rate of convection heat transfer is
observed to be proportional to the temperature difference and is conveniently
expressed by Newton's law of cooling as
cen58933_ch06.qxd 9/4/2002 12:05 PM Page 335
or
where
h(T,-T„) (W/m 2 ) (6-1)
g conv = hA s (T s - r„) (W) (6-2)
h = convection heat transfer coefficient, W/m 2 • °C
A s = heat transfer surface area, m 2
T s = temperature of the surface, °C
T rjz = temperature of the fluid sufficiently far from the surface, °C
Judging from its units, the convection heat transfer coefficient h can be de-
fined as the rate of heat transfer between a solid surface and a fluid per unit
surface area per unit temperature difference.
You should not be deceived by the simple appearance of this relation, be-
cause the convection heat transfer coefficient h depends on the several of the
mentioned variables, and thus is difficult to determine.
When a fluid is forced to flow over a solid surface that is nonporous (i.e.,
impermeable to the fluid), it is observed that the fluid in motion comes to a
complete stop at the surface and assumes a zero velocity relative to the sur-
face. That is, the fluid layer in direct contact with a solid surface "sticks" to
the surface and there is no slip. In fluid flow, this phenomenon is known as the
no-slip condition, and it is due to the viscosity of the fluid (Fig. 6-4).
The no-slip condition is responsible for the development of the velocity pro-
file for flow. Because of the friction between the fluid layers, the layer that
sticks to the wall slows the adjacent fluid layer, which slows the next layer,
and so on. A consequence of the no-slip condition is that all velocity profiles
must have zero values at the points of contact between a fluid and a solid. The
only exception to the no-slip condition occurs in extremely rarified gases.
A similar phenomenon occurs for the temperature. When two bodies at dif-
ferent temperatures are brought into contact, heat transfer occurs until both
bodies assume the same temperature at the point of contact. Therefore, a fluid
and a solid surface will have the same temperature at the point of contact. This
is known as no-temperature-jump condition.
An implication of the no-slip and the no-temperature jump conditions is that
heat transfer from the solid surface to the fluid layer adjacent to the surface is
by pure conduction, since the fluid layer is motionless, and can be expressed as
~^flu
dT
dy
v=0
(W/m 2 )
335
CHAPTER 6
Velocity
profile
FIGURE 6-3
The cooling of a hot block by forced
convection.
Uniform
approach
velocity, °l
Relative
velocities
of fluid layers
Zero
velocity
at the
surface
Solid block
(6-3)
FIGURE 6-4
A fluid flowing over a stationary
surface comes to a complete stop at
the surface because of the no-slip
condition.
where T represents the temperature distribution in the fluid and (dT/dy) Y=0 is
the temperature gradient at the surface. This heat is then convected away from
the surface as a result of fluid motion. Note that convection heat transfer from
a solid surface to a fluid is merely the conduction heat transfer from the solid
surface to the fluid layer adjacent to the surface. Therefore, we can equate
Eqs. 6-1 and 6-3 for the heat flux to obtain
cen58933_ch06.qxd 9/4/2002 12:05 PM Page 336
336
HEAT TRANSFER
-knmi(dTIdy) y=0
T - T x
(W/m 2 • °C)
(6-4)
for the determination of the convection heat transfer coefficient when the tem-
perature distribution within the fluid is known.
The convection heat transfer coefficient, in general, varies along the flow
(or X-) direction. The average or mean convection heat transfer coefficient for
a surface in such cases is determined by properly averaging the local convec-
tion heat transfer coefficients over the entire surface.
Nusselt Number
In convection studies, it is common practice to nondimensionalize the gov-
erning equations and combine the variables, which group together into di-
mensionless numbers in order to reduce the number of total variables. It is also
common practice to nondimensionalize the heat transfer coefficient h with the
Nusselt number, defined as
Nu
(6-5)
&T=T^
FIGURE 6-5
Heat transfer through a fluid layer
of thickness L and temperature
difference AT.
where k is the thermal conductivity of the fluid and L c is the characteristic
length. The Nusselt number is named after Wilhelm Nusselt, who made sig-
nificant contributions to convective heat transfer in the first half of the twen-
tieth century, and it is viewed as the dimensionless convection heat transfer
coefficient.
To understand the physical significance of the Nusselt number, consider a
fluid layer of thickness L and temperature difference AT = T 2 — T u as shown
in Fig. 6-5. Heat transfer through the fluid layer will be by convection when
the fluid involves some motion and by conduction when the fluid layer is mo-
tionless. Heat flux (the rate of heat transfer per unit time per unit surface area)
in either case will be
"co
hAT
(6-6)
Blowing
on food
FIGURE 6-6
We resort to forced convection
whenever we need to increase the
rate of heat transfer.
and
Taking their ratio gives
<Zcond — *
AT
hAT hh
q cond kAT/L k
Nu
(6-7)
(6-8)
which is the Nusselt number. Therefore, the Nusselt number represents the en-
hancement of heat transfer through a fluid layer as a result of convection rel-
ative to conduction across the same fluid layer. The larger the Nusselt number,
the more effective the convection. A Nusselt number of Nu = 1 for a fluid
layer represents heat transfer across the layer by pure conduction.
We use forced convection in daily life more often than you might think
(Fig. 6-6). We resort to forced convection whenever we want to increase the
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CHAPTER 6
rate of heat transfer from a hot object. For example, we turn on the fan on hot
summer days to help our body cool more effectively. The higher the fan speed,
the better we feel. We stir our soup and blow on a hot slice of pizza to make
them cool faster. The air on windy winter days feels much colder than it actu-
ally is. The simplest solution to heating problems in electronics packaging is
to use a large enough fan.
6-2 - CLASSIFICATION OF FLUID FLOWS
Convection heat transfer is closely tied with fluid mechanics, which is the sci-
ence that deals with the behavior of fluids at rest or in motion, and the inter-
action of fluids with solids or other fluids at the boundaries. There are a wide
variety of fluid flow problems encountered in practice, and it is usually con-
venient to classify them on the basis of some common characteristics to make
it feasible to study them in groups. There are many ways to classify the fluid
flow problems, and below we present some general categories.
Viscous versus Inviscid Flow
When two fluid layers move relative to each other, a friction force develops
between them and the slower layer tries to slow down the faster layer. This in-
ternal resistance to flow is called the viscosity, which is a measure of internal
stickiness of the fluid. Viscosity is caused by cohesive forces between the
molecules in liquids, and by the molecular collisions in gases. There is no
fluid with zero viscosity, and thus all fluid flows involve viscous effects to
some degree. Flows in which the effects of viscosity are significant are called
viscous flows. The effects of viscosity are very small in some flows, and ne-
glecting those effects greatly simplifies the analysis without much loss in ac-
curacy. Such idealized flows of zero-viscosity fluids are called frictionless or
inviscid flows.
Internal versus External Flow
A fluid flow is classified as being internal and external, depending on whether
the fluid is forced to flow in a confined channel or over a surface. The flow of
an unbounded fluid over a surface such as a plate, a wire, or a pipe is exter-
nal flow. The flow in a pipe or duct is internal flow if the fluid is completely
bounded by solid surfaces. Water flow in a pipe, for example, is internal flow,
and air flow over an exposed pipe during a windy day is external flow
(Fig. 6-7). The flow of liquids in a pipe is called open-channel flow if the
pipe is partially filled with the liquid and there is a free surface. The flow of
water in rivers and irrigation ditches are examples of such flows.
Compressible versus Incompressible Flow
A fluid flow is classified as being compressible or incompressible, depending
on the density variation of the fluid during flow. The densities of liquids are
essentially constant, and thus the flow of liquids is typically incompressible.
Therefore, liquids are usually classified as incompressible substances. A pres-
sure of 210 atm, for example, will cause the density of liquid water at 1 aim to
change by just 1 percent. Gases, on the other hand, are highly compressible. A
Internal
flow
FIGURE 6-7
Internal flow of water in a pipe
and the external flow of air over
the same pipe.
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HEAT TRANSFER
pressure change of just 0.01 atm, for example, will cause a change of 1 per-
cent in the density of atmospheric air. However, gas flows can be treated as
incompressible if the density changes are under about 5 percent, which is usu-
ally the case when the flow velocity is less than 30 percent of the velocity of
sound in that gas (i.e., the Mach number of flow is less than 0.3). The veloc-
ity of sound in air at room temperature is 346 m/s. Therefore, the compress-
ibility effects of air can be neglected at speeds under 100 m/s. Note that the
flow of a gas is not necessarily a compressible flow.
Laminar versus Turbulent Flow
Some flows are smooth and orderly while others are rather chaotic. The highly
ordered fluid motion characterized by smooth streamlines is called laminar.
The flow of high-viscosity fluids such as oils at low velocities is typically
laminar. The highly disordered fluid motion that typically occurs at high ve-
locities characterized by velocity fluctuations is called turbulent. The flow of
low-viscosity fluids such as air at high velocities is typically turbulent. The
flow regime greatly influences the heat transfer rates and the required power
for pumping.
Light hot
water rising
Hot
water
Cold
water
Cold
water
Hot
t
Hot water
storage tank
(above the top
of collectors)
FIGURE 6-8
Natural circulation of water in a solar
water heater by thermo siphoning.
Natural (or Unforced) versus Forced Flow
A fluid flow is said to be natural or forced, depending on how the fluid motion
is initiated. In forced flow, a fluid is forced to flow over a surface or in a pipe
by external means such as a pump or a fan. In natural flows, any fluid motion
is due to a natural means such as the buoyancy effect, which manifests itself
as the rise of the warmer (and thus lighter) fluid and the fall of cooler (and
thus denser) fluid. This thermosiphoning effect is commonly used to replace
pumps in solar water heating systems by placing the water tank sufficiently
above the solar collectors (Fig. 6-8).
WdtCl
7*" Steady versus Unsteady (Transient) Flow
The terms steady and uniform are used frequently in engineering, and thus it
is important to have a clear understanding of their meanings. The term
steady implies no change with time. The opposite of steady is unsteady, or
transient. The term uniform, however, implies no change with location over
a specified region.
Many devices such as turbines, compressors, boilers, condensers, and heat
exchangers operate for long periods of time under the same conditions, and
they are classified as steady-flow devices. During steady flow, the fluid prop-
erties can change from point to point within a device, but at any fixed point
they remain constant.
One-, Two-, and Three-Dimensional Flows
A flow field is best characterized by the velocity distribution, and thus a flow
is said to be one-, two-, or three-dimensional if the flow velocity T varies in
one, two, or three primary dimensions, respectively. A typical fluid flow in-
volves a three-dimensional geometry and the velocity may vary in all three di-
mensions rendering the flow three-dimensional [T(x, y, z) in rectangular
or T(r, 0, z) in cylindrical coordinates]. However, the variation of velocity in
cen58933_ch06.qxd 9/4/2002 12:05 PM Page 339
certain direction can be small relative to the variation in other directions, and
can be ignored with negligible error. In such cases, the flow can be modeled
conveniently as being one- or two-dimensional, which is easier to analyze.
When the entrance effects are disregarded, fluid flow in a circular pipe is
one-dimensional since the velocity varies in the radial r direction but not in
the angular 0- or axial z-directions (Fig. 6-9). That is, the velocity profile is
the same at any axial z-location, and it is symmetric about the axis of the pipe.
Note that even in this simplest flow, the velocity cannot be uniform across the
cross section of the pipe because of the no-slip condition. However, for con-
venience in calculations, the velocity can be assumed to be constant and thus
uniform at a cross section. Fluid flow in a pipe usually approximated as one-
dimensional uniform flow.
339
CHAPTER 6
^- velocity profile
(remains unchanged)
1
Y(f)
FIGURE 6-9
One-dimensional flow in a
circular pipe.
6-3 ■ VELOCITY BOUNDARY LAYER
Consider the parallel flow of a fluid over a flat plate, as shown in Fig. 6-10.
Surfaces that are slightly contoured such as turbine blades can also be ap-
proximated as flat plates with reasonable accuracy. The x-coordinate is mea-
sured along the plate surface from the leading edge of the plate in the
direction of the flow, and y is measured from the surface in the normal direc-
tion. The fluid approaches the plate in the x-direction with a uniform upstream
velocity of T, which is practically identical to the free-stream velocity m„ over
the plate away from the surface (this would not be the case for cross flow over
blunt bodies such as a cylinder).
For the sake of discussion, we can consider the fluid to consist of adjacent
layers piled on top of each other. The velocity of the particles in the first fluid
layer adjacent to the plate becomes zero because of the no-slip condition. This
motionless layer slows down the particles of the neighboring fluid layer as a
result of friction between the particles of these two adjoining fluid layers at
different velocities. This fluid layer then slows down the molecules of the next
layer, and so on. Thus, the presence of the plate is felt up to some normal dis-
tance 8 from the plate beyond which the free-stream velocity w m remains es-
sentially unchanged. As a result, the x-component of the fluid velocity, u, will
vary from at y = to nearly u rM at y = 8 (Fig. 6-11).
Turbulent boundary
layer
j J j Turbulent
~v y layer
Buffer layer
Laminar sublayer
FIGURE 6-10
The development of the boundary layer for flow over a flat plate, and the different flow regimes.
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HEAT TRANSFER
y
0.99T
Relative
velocities
of fluid layers
- -«« a T
Zero
velocity
at the
surface
FIGURE 6-1 1
The development of a boundary layer
on a surface is due to the no-slip
condition.
Viscosity
Liquids
Gases
Temperature
FIGURE 6-12
The viscosity of liquids decreases and
the viscosity of gases increases with
temperature.
The region of the flow above the plate bounded by 8 in which the effects
of the viscous shearing forces caused by fluid viscosity are felt is called the
velocity boundary layer. The boundary layer thickness, 8, is typically de-
fined as the distance y from the surface at which u = 0.99m„.
The hypothetical line of u = 0.99m„ divides the flow over a plate into two
regions: the boundary layer region, in which the viscous effects and the ve-
locity changes are significant, and the inviscid flow region, in which the fric-
tional effects are negligible and the velocity remains essentially constant.
Surface Shear Stress
Consider the flow of a fluid over the surface of a plate. The fluid layer in con-
tact with the surface will try to drag the plate along via friction, exerting a fric-
tion force on it. Likewise, a faster fluid layer will try to drag the adjacent
slower layer and exert a friction force because of the friction between the two
layers. Friction force per unit area is called shear stress, and is denoted by t.
Experimental studies indicate that the shear stress for most fluids is propor-
tional to the velocity gradient, and the shear stress at the wall surface is as
l-i
du
dy
(N/m 2 )
(6-9)
where the constant of proportionality ix is called the dynamic viscosity of
the fluid, whose unit is kg/m • s (or equivalently, N ■ s/m 2 , or Pa • s, or poise
= 0.1 Pa • s).
The fluids that that obey the linear relationship above are called Newtonian
fluids, after Sir Isaac Newton who expressed it first in 1687. Most common
fluids such as water, air, gasoline, and oils are Newtonian fluids. Blood and
liquid plastics are examples of non-Newtonian fluids. In this text we will con-
sider Newtonian fluids only.
In fluid flow and heat transfer studies, the ratio of dynamic viscosity to den-
sity appears frequently. For convenience, this ratio is given the name kine-
matic viscosity v and is expressed as v = |x/p. Two common units of
kinematic viscosity are m 2 /s and stoke (1 stoke = 1 cm 2 /s = 0.0001 m 2 /s).
The viscosity of a fluid is a measure of its resistance to flow, and it is a
strong function of temperature. The viscosities of liquids decrease with tem-
perature, whereas the viscosities of gases increase with temperature (Fig.
6-12). The viscosities of some fluids at 20°C are listed in Table 6-1. Note that
the viscosities of different fluids differ by several orders of magnitude.
The determination of the surface shear stress t s from Eq. 6-9 is not practical
since it requires a knowledge of the flow velocity profile. A more practical ap-
proach in external flow is to relate j s to the upstream velocity T as
C,
P T 2
(N/m 2 )
(6-10)
where C f is the dimensionless friction coefficient, whose value in most cases
is determined experimentally, and p is the density of the fluid. Note that the
friction coefficient, in general, will vary with location along the surface. Once
the average friction coefficient over a given surface is available, the friction
force over the entire surface is determined from
cen58933_ch06.qxd 9/4/2002 12:05 PM Page 341
CfA,
pT 2
~2~
(N)
(6-11)
where A s is the surface area.
The friction coefficient is an important parameter in heat transfer studies
since it is directly related to the heat transfer coefficient and the power re-
quirements of the pump or fan.
6^ - THERMAL BOUNDARY LAYER
We have seen that a velocity boundary layer develops when a fluid flows over
a surface as a result of the fluid layer adjacent to the surface assuming the sur-
face velocity (i.e., zero velocity relative to the surface). Also, we defined the
velocity boundary layer as the region in which the fluid velocity varies from
zero to 0.99w ro . Likewise, a thermal boundary layer develops when a fluid at
a specified temperature flows over a surface that is at a different temperature,
as shown in Fig. 6-13.
Consider the flow of a fluid at a uniform temperature of T a over an isother-
mal flat plate at temperature T s . The fluid particles in the layer adjacent to the
surface will reach thermal equilibrium with the plate and assume the surface
temperature T s . These fluid particles will then exchange energy with the par-
ticles in the adjoining-fluid layer, and so on. As a result, a temperature profile
will develop in the flow field that ranges from T s at the surface to T^ suffi-
ciently far from the surface. The flow region over the surface in which the
temperature variation in the direction normal to the surface is significant is the
thermal boundary layer. The thickness of the thermal boundary layer 8, at
any location along the surface is defined as the distance from the surface at
which the temperature difference T — T s equals Q.99(T„ — T s ). Note that for
the special case of T s = 0, we have T = 0.99r„ at the outer edge of the ther-
mal boundary layer, which is analogous to u = 0.99m„ for the velocity bound-
ary layer.
The thickness of the thermal boundary layer increases in the flow direction,
since the effects of heat transfer are felt at greater distances from the surface
further down stream.
The convection heat transfer rate anywhere along the surface is directly re-
lated to the temperature gradient at that location. Therefore, the shape of the
temperature profile in the thermal boundary layer dictates the convection heat
transfer between a solid surface and the fluid flowing over it. In flow over a
heated (or cooled) surface, both velocity and thermal boundary layers will de-
velop simultaneously. Noting that the fluid velocity will have a strong influ-
ence on the temperature profile, the development of the velocity boundary
layer relative to the thermal boundary layer will have a strong effect on the
convection heat transfer.
341
CHAPTER 6
TABLE
3 ^^H
Dynamic
at 1 atm
stated)
viscosities of some fluids
and 20°C (unless otherwise
Fluid
Dynamic viscosity
|x, kg/m ■ s
Glycerin:
-20°C
134.0
0°C
12.1
20°C
1.49
40°C
0.27
Engine oil:
SAE 10W
0.10
SAE 10W30
0.17
SAE 30
0.29
SAE 50
0.86
Mercury
0.0015
Ethyl alcohol
0.0012
Water:
0°C
0.0018
20°C
0.0010
100°C (liquid)
0.0003
100°C (vapor)
0.000013
Blood, 37°C
0.0004
Gasoline
0.00029
Ammonia
0.00022
Air
0.000018
Hydrogen, 0°C
0.000009
Free- stream
Thermal
boundary
j layer
r s + o.99(r„-r s )
FIGURE 6-13
Thermal boundary layer on a flat plate
(the fluid is hotter than the plate
surface).
Prandtl Number
The relative thickness of the velocity and the thermal boundary layers is best
described by the dimensionless parameter Prandtl number, defined as
Molecular diffusivity of momentum v u-C p
Pr
Molecular diffusivity of heat
v
a
(6-12)
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HEAT TRANSFER
TABLE 6-2
Typical ranges of Prandtl numbers
for common fluids
Fluid
Pr
Liquid metals
Gases
Water
Light organic fluids
Oils
Glycerin
0.004-0.030
0.7-1.0
1.7-13.7
5-50
50-100,000
2000-100,000
Smoke
' j, Turbulent
flow
Laminar
flow
FIGURE 6-14
Laminar and turbulent flow regimes of
cigarette smoke.
(a) Laminar flow
-Die trace
r
(b) Turbulent flow
FIGURE 6-15
The behavior of colored fluid injected
into the flow in laminar and turbulent
flows in a tube.
It is named after Ludwig Prandtl, who introduced the concept of boundary
layer in 1904 and made significant contributions to boundary layer theory.
The Prandtl numbers of fluids range from less than 0.01 for liquid metals to
more than 100,000 for heavy oils (Table 6-2). Note that the Prandtl number is
in the order of 10 for water.
The Prandtl numbers of gases are about 1, which indicates that both momen-
tum and heat dissipate through the fluid at about the same rate. Heat diffuses
very quickly in liquid metals (Pr <§ 1) and very slowly in oils (Pr > 1) relative
to momentum. Consequently the thermal boundary layer is much thicker for
liquid metals and much thinner for oils relative to the velocity boundary layer.
6-5 - LAMINAR AND TURBULENT FLOWS
If you have been around smokers, you probably noticed that the cigarette
smoke rises in a smooth plume for the first few centimeters and then starts
fluctuating randomly in all directions as it continues its journey toward the
lungs of others (Fig. 6-14). Likewise, a careful inspection of flow in a pipe re-
veals that the fluid flow is streamlined at low velocities but turns chaotic as
the velocity is increased above a critical value, as shown in Figure 6-15. The
flow regime in the first case is said to be laminar, characterized by smooth
streamlines and highly-ordered motion, and turbulent in the second case,
where it is characterized by velocity fluctuations and highly-disordered mo-
tion. The transition from laminar to turbulent flow does not occur suddenly;
rather, it occurs over some region in which the flow fluctuates between lami-
nar and turbulent flows before it becomes fully turbulent.
We can verify the existence of these laminar, transition, and turbulent flow
regimes by injecting some dye streak into the flow in a glass tube, as the
British scientist Osborn Reynolds (1842-1912) did over a century ago. We
will observe that the dye streak will form a straight and smooth line at low ve-
locities when the flow is laminar (we may see some blurring because of mol-
ecular diffusion), will have bursts of fluctuations in the transition regime, and
will zigzag rapidly and randomly when the flow becomes fully turbulent.
These zigzags and the dispersion of the dye are indicative of the fluctuations
in the main flow and the rapid mixing of fluid particles from adjacent layers.
Typical velocity profiles in laminar and turbulent flow are also given in Fig-
ure 6-10. Note that the velocity profile is approximately parabolic in laminar
flow and becomes flatter in turbulent flow, with a sharp drop near the surface.
The turbulent boundary layer can be considered to consist of three layers. The
very thin layer next to the wall where the viscous effects are dominant is
the laminar sublayer. The velocity profile in this layer is nearly linear, and
the flow is streamlined. Next to the laminar sublayer is the buffer layer, in
which the turbulent effects are significant but not dominant of the diffusion
effects, and next to it is the turbulent layer, in which the turbulent effects
dominate.
The intense mixing of the fluid in turbulent flow as a result of rapid fluctu-
ations enhances heat and momentum transfer between fluid particles, which
increases the friction force on the surface and the convection heat transfer
rate. It also causes the boundary layer to enlarge. Both the friction and heat
transfer coefficients reach maximum values when the flow becomes fully tur-
bulent. So it will come as no surprise that a special effort is made in the design
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CHAPTER 6
of heat transfer coefficients associated with turbulent flow. The enhancement
in heat transfer in turbulent flow does not come for free, however. It may be
necessary to use a larger pump to overcome the larger friction forces accom-
panying the higher heat transfer rate.
Reynolds Number
The transition from laminar to turbulent flow depends on the surface geome-
try, surface roughness, free-stream velocity, surface temperature, and type of
fluid, among other things. After exhaustive experiments in the 1880s, Osborn
Reynolds discovered that the flow regime depends mainly on the ratio of the
inertia forces to viscous forces in the fluid. This ratio is called the Reynolds
number, which is a dimensionless quantity, and is expressed for external flow
as (Fig. 6-16)
YL C _ P TL r
Re
Inertia forces
Viscous
H
(6-13)
where T is the upstream velocity (equivalent to the free-stream velocity m„ for
a flat plate), L c is the characteristic length of the geometry, and v = jx/p is the
kinematic viscosity of the fluid. For a flat plate, the characteristic length is the
distance x from the leading edge. Note that kinematic viscosity has the unit
m 2 /s, which is identical to the unit of thermal diffusivity, and can be viewed as
viscous diffusivity or diffusivity for momentum.
At large Reynolds numbers, the inertia forces, which are proportional to the
density and the velocity of the fluid, are large relative to the viscous forces,
and thus the viscous forces cannot prevent the random and rapid fluctuations
of the fluid. At small Reynolds numbers, however, the viscous forces are large
enough to overcome the inertia forces and to keep the fluid "in line." Thus the
flow is turbulent in the first case and laminar in the second.
The Reynolds number at which the flow becomes turbulent is called the
critical Reynolds number. The value of the critical Reynolds number is dif-
ferent for different geometries. For flow over a flat plate, the generally ac-
cepted value of the critical Reynolds number is Re cr = Txjv = u^xjv =
5 X 10 5 , where x a is the distance from the leading edge of the plate at which
transition from laminar to turbulent flow occurs. The value of Re cr may
change substantially, however, depending on the level of turbulence in the free
stream.
6-6 - HEAT AND MOMENTUM TRANSFER IN
TURBULENT FLOW
Most flows encountered in engineering practice are turbulent, and thus it is
important to understand how turbulence affects wall shear stress and heat
transfer. Turbulent flow is characterized by random and rapid fluctuations of
groups of fluid particles, called eddies, throughout the boundary layer. These
fluctuations provide an additional mechanism for momentum and heat trans-
fer. In laminar flow, fluid particles flow in an orderly manner along stream-
lines, and both momentum and heat are transferred across streamlines by
molecular diffusion. In turbulent flow, the transverse motion of eddies trans-
port momentum and heat to other regions of flow before they mix with the rest
of the fluid and lose their identity, greatly enhancing momentum and heat
FIGURE 6-16
The Reynolds number can be viewed
as the ratio of the inertia forces to
viscous forces acting on a fluid
volume element.
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HEAT TRANSFER
(a) Before
turbulence
(b) After
turbulence
FIGURE 6-17
The intense mixing in turbulent flow
brings fluid particles at different
temperatures into close contact, and
thus enhances heat transfer.
X^^f^H"
Time, t
FIGURE 6-18
Fluctuations of the velocity
component u with time at a specified
location in turbulent flow.
transfer. As a result, turbulent flow is associated with much higher values of
friction and heat transfer coefficients (Fig. 6-17).
Even when the mean flow is steady, the eddying motion in turbulent flow
causes significant fluctuations in the values of velocity, temperature, pressure,
and even density (in compressible flow). Figure 6-18 shows the variation of
the instantaneous velocity component u with time at a specified location, as
can be measured with a hot-wire anemometer probe or other sensitive device.
We observe that the instantaneous values of the velocity fluctuate about a
mean value, which suggests that the velocity can be expressed as the sum of a
mean value u and a. fluctuating component u' ,
u + u'
(6-14)
This is also the case for other properties such as the velocity component v in
the y direction, and thus v = v + v' , P = P + P' , and T = T + T . The mean
value of a property at some location is determined by averaging it over a time
interval that is sufficiently large so that the net effect of fluctuations is zero.
Therefore, the time average of fluctuating components is zero, e.g., W = 0.
The magnitude of u' is usually just a few percent of u, but the high frequen-
cies of eddies (in the order of a thousand per second) makes them very effec-
tive for the transport of momentum and thermal energy. In steady turbulent
flow, the mean values of properties (indicated by an overbar) are independent
of time.
Consider the upward eddy motion of a fluid during flow over a surface. The
mass flow rate of fluid per unit area normal to flow is pv'. Noting that h = C p T
represents the energy of the fluid and T is the eddy temperature relative to the
mean value, the rate of thermal energy transport by turbulent eddies is
q t = pC p v'T'. By a similar argument on momentum transfer, the turbulent shear
stress can be shown to be t, = — pwV. Note that u'v' + even though u' =
and v' = 0, and experimental results show that u'v' is a negative quantity.
Terms such as — pu'v' are called Reynolds stresses.
The random eddy motion of groups of particles resembles the random mo-
tion of molecules in a gas — colliding with each other after traveling a certain
distance and exchanging momentum and heat in the process. Therefore, mo-
mentum and heat transport by eddies in turbulent boundary layers is analo-
gous to the molecular momentum and heat diffusion. Then turbulent wall
shear stress and turbulent heat transfer can be expressed in an analogous
manner as
-pu v
\^
dll
'dy
and
q, = pC p v'T'
-K
dT
dy
(6-15)
where u,, is called the turbulent viscosity, which accounts for momentum
transport by turbulent eddies, and k, is called the turbulent thermal conduc-
tivity, which accounts for thermal energy transport by turbulent eddies. Then
the total shear stress and total heat flux can be expressed conveniently as
(|X + |X,)
du
dy
p(v + e M )
dll
dy
and
-(*■
k,) dy
pC (a + s H )
dT
dy
(6-16)
(6-17)
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CHAPTER 6
where e M = (x,/p is the eddy diffusivity of momentum and s H = kJpC p is the
eddy diffusivity of heat.
Eddy motion and thus eddy diffusivities are much larger than their molecu-
lar counterparts in the core region of a turbulent boundary layer. The eddy mo-
tion loses its intensity close to the wall, and diminishes at the wall because of
the no-slip condition. Therefore, the velocity and temperature profiles are
nearly uniform in the core region of a turbulent boundary layer, but very steep
in the thin layer adjacent to the wall, resulting in large velocity and tempera-
ture gradients at the wall surface. So it is no surprise that the wall shear stress
and wall heat flux are much larger in turbulent flow than they are in laminar
flow (Fig. 6-19).
Note that molecular diffusivities v and a (as well as |x and k) are fluid prop-
erties, and their values can be found listed in fluid handbooks. Eddy diffusiv-
ities s M and e h (as well as (ji r and k t ), however are not fluid properties and their
values depend on flow conditions. Eddy diffusivities s M and s H decrease to-
wards the wall, becoming zero at the wall.
6-7 - DERIVATION OF DIFFERENTIAL
CONVECTION EQUATIONS*
In this section we derive the governing equations of fluid flow in the bound-
ary layers. To keep the analysis at a manageable level, we assume the flow to
be steady and two-dimensional, and the fluid to be Newtonian with constant
properties (density, viscosity, thermal conductivity, etc.).
Consider the parallel flow of a fluid over a surface. We take the flow direc-
tion along the surface to be x and the direction normal to the surface to be y,
and we choose a differential volume element of length dx, height dy, and unit
depth in the z-direction (normal to the paper) for analysis (Fig. 6-20). The
fluid flows over the surface with a uniform free-stream velocity u m but the ve-
locity within boundary layer is two-dimensional: the x-component of the ve-
locity is u, and the y-component is v. Note that u = u(x, y) and v = v(x, y) in
steady two-dimensional flow.
Next we apply three fundamental laws to this fluid element: Conservation of
mass, conservation of momentum, and conservation of energy to obtain the con-
tinuity, momentum, and energy equations for laminar flow in boundary layers.
T„ or u„
Conservation of Mass Equation
The conservation of mass principle is simply a statement that mass cannot be
created or destroyed, and all the mass must be accounted for during an analy-
sis. In steady flow, the amount of mass within the control volume remains
constant, and thus the conservation of mass can be expressed as
Rate of mass flow \
into the control volume/
Rate of mass flow \
out of the control volume/
(6-18)
*This and the upcoming sections of this chapter deal with theoretical aspects of convection,
and can be skipped and be used as a reference if desired without a loss in continuity.
du
or
3T
dy
v=0
dy
y=0
Laminar
T^ or u x
}ht
or
dT
dy
v=0
dy
y=0
Turbulent
FIGURE 6-1 9
The velocity and temperature gradients
at the wall, and thus the wall shear stress
and heat transfer rate, are much larger
for turbulent flow than they are for
laminar flow (T is shown relative to T s ).
FIGURE 6-20
Differential control volume used in the
derivation of mass balance in velocity
boundary layer in two-dimensional
flow over a surface.
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HEAT TRANSFER
Noting that mass flow rate is equal to the product of density, mean velocity,
and cross-sectional area normal to flow, the rate at which fluid enters the con-
trol volume from the left surface is pu(dy ■ 1). The rate at which the fluid
leaves the control volume from the right surface can be expressed as
p(u + ^dxj(dyl) (6-19)
Repeating this for the y direction and substituting the results into Eq. 6-18, we
obtain
pu(dy • 1) + pv(dx ■ 1) = p(u + |^ dx\dy ■ 1) + p(v + y dyXdx • 1) (6-20)
Simplifying and dividing by dx ■ dy ■ 1 gives
? + ? = (6-21)
dx dy
This is the conservation of mass relation, also known as the continuity equa-
tion, or mass balance for steady two-dimensional flow of a fluid with con-
stant density.
Conservation of Momentum Equations
The differential forms of the equations of motion in the velocity boundary
layer are obtained by applying Newton's second law of motion to a differen-
tial control volume element in the boundary layer. Newton's second law is an
expression for the conservation of momentum, and can be stated as the net
force acting on the control volume is equal to the mass times the acceleration
of the fluid element within the control volume, which is also equal to the net
rate of momentum outflow from the control volume.
The forces acting on the control volume consist of body forces that act
throughout the entire body of the control volume (such as gravity, electric, and
magnetic forces) and are proportional to the volume of the body, and surface
forces that act on the control surface (such as the pressure forces due to hy-
drostatic pressure and shear stresses due to viscous effects) and are propor-
tional to the surface area. The surface forces appear as the control volume is
isolated from its surroundings for analysis, and the effect of the detached body
is replaced by a force at that location. Note that pressure represents the com-
pressive force applied on the fluid element by the surrounding fluid, and is al-
ways directed to the surface.
We express Newton's second law of motion for the control volume as
cm \( Acceleration \ _ /Net force (body and surface)]
\in a specified direction/ I acting in that direction j
or
8m • a z = F SLrface >x + F hoi ^ x (6-23)
where the mass of the fluid element within the control volume is
8m = p(dx ■ dy • 1) (6-24)
cen58933_ch06.qxd 9/4/2002 12:05 PM Page 347
Noting that flow is steady and two-dimensional and thus u
differential of u is
u(x, y), the total
347
CHAPTER 6
du
dii
dx
dx
dlt
ay
dy
(6-25)
Then the acceleration of the fluid element in the x direction becomes
du
dt
du dx
dx dt
du dy
dy dt
du du
U h V —
dx 3y
(6-26)
You may be tempted to think that acceleration is zero in steady flow since
acceleration is the rate of change of velocity with time, and in steady flow
there is no change with time. Well, a garden hose nozzle will tell us that this
understanding is not correct. Even in steady flow and thus constant mass flow
rate, water will accelerate through the nozzle (Fig. 6-21). Steady simply
means no change with time at a specified location (and thus du/dt = 0), but
the value of a quantity may change from one location to another (and thus
duldx and du/dy may be different from zero). In the case of a nozzle, the ve-
locity of water remains constant at a specified point, but it changes from inlet
to the exit (water accelerates along the nozzle, which is the reason for attach-
ing a nozzle to the garden hose in the first place).
The forces acting on a surface are due to pressure and viscous effects. In
two-dimensional flow, the viscous stress at any point on an imaginary surface
within the fluid can be resolved into two perpendicular components: one nor-
mal to the surface called normal stress (which should not be confused with
pressure) and another along the surface called shear stress. The normal stress
is related to the velocity gradients duldx and dv/dy, that are much smaller than
du/dy, to which shear stress is related. Neglecting the normal stresses for sim-
plicity, the surface forces acting on the control volume in the x-direction will
be as shown in Fig. 6-22. Then the net surface force acting in the .x-direction
becomes
= [dy dy
dP
dx'
dxUdy 1)
9t
dy
dP
dx
(dx-dy 1)
d l u
dP
dx
{dx ■ dy ■ 1 )
(6-27)
since t = \L(duldy). Substituting Eqs. 6-21, 6-23, and 6-24 into Eq. 6-20 and
dividing by dx ■ dy ■ 1 gives
du , du
dx
|X
d 2 U
3y 2
dP
dx
(6-28)
This is the relation for the conservation of momentum in the x-direction, and
is known as the jf-momentum equation. Note that we would obtain the same
result if we used momentum flow rates for the left-hand side of this equation
instead of mass times acceleration. If there is a body force acting in the
x-direction, it can be added to the right side of the equation provided that it is
expressed per unit volume of the fluid.
In a boundary layer, the velocity component in the flow direction is much
larger than that in the normal direction, and thus u > v, and dvldx and dv/dy are
Water
FIGURE 6-21
During steady flow, a fluid may not
accelerate in time at a fixed point, but
it may accelerate in space.
dx
FIGURE 6-22
Differential control volume used in the
derivation of x-momentum equation in
velocity boundary layer in two-
dimensional flow over a surface.
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HEAT TRANSFER
1)
Velocity components:
V < u
2)
Velocity grandients:
dx dv
du du
— < —
dx dy
3)
Temperature gradients:
— < —
dx dy
FIGURE 6-23
Boundary layer approximations.
negligible. Also, u varies greatly with y in the normal direction from zero at the
wall surface to nearly the free-stream value across the relatively thin boundary
layer, while the variation of u with x along the flow is typically small. There-
fore, du/ dy > du/dx. Similarly, if the fluid and the wall are at different temper-
atures and the fluid is heated or cooled during flow, heat conduction will occur
primarily in the direction normal to the surface, and thus 3T/dy > dT/dx. That
is, the velocity and temperature gradients normal to the surface are much
greater than those along the surface. These simplifications are known as the
boundary layer approximations. These approximations greatly simplify the
analysis usually with little loss in accuracy, and make it possible to obtain ana-
lytical solutions for certain types of flow problems (Fig. 6-23).
When gravity effects and other body forces are negligible and the boundary
layer approximations are valid, applying Newton's second law of motion on
the volume element in the y-direction gives the y-momentum equation to be
dP
dy
(6-29)
That is, the variation of pressure in the direction normal to the surface is neg-
ligible, and thus P = P(x) and dP/dx = dPIdx. Then it follows that for a given
x, the pressure in the boundary layer is equal to the pressure in the free stream,
and the pressure determined by a separate analysis of fluid flow in the free
stream (which is typically easier because of the absence of viscous effects)
can readily be used in the boundary layer analysis.
The velocity components in the free stream region of a flat plate are u = u„,
= constant and v = 0. Substituting these into the x-momentum equations
(Eq. 6-28) gives dP/dx = 0. Therefore, for flow over a flat plate, the pres-
sure remains constant over the entire plate (both inside and outside the bound-
ary layer).
Conservation of Energy Equation
The energy balance for any system undergoing any process is expressed as
E in — E out = AE system , which states that the change in the energy content of a
system during a process is equal to the difference between the energy input
and the energy output. During a steady-flow process, the total energy content
of a control volume remains constant (and thus AE system = 0), and the amount
of energy entering a control volume in all forms must be equal to the amount
of energy leaving it. Then the rate form of the general energy equation reduces
for a steady -flow process to E- m — E out = 0.
Noting that energy can be transferred by heat, work, and mass only, the en-
ergy balance for a steady -flow control volume can be written explicitly as
V-^in ^out/by heat ~*~ (,-^in -^out)
by work
l-^in -^out)
out /by mass
(6-30)
h + ke +
The total energy of a flowing fluid stream per unit mass is e stre
pe where h is the enthalpy (which is the sum of internal energy and flow en-
ergy), pe = gz is the potential energy, and ke = T 2 /2 = (m 2 + v 2 )/2 is the
kinetic energy of the fluid per unit mass. The kinetic and potential energies are
usually very small relative to enthalpy, and therefore it is common practice to
neglect them (besides, it can be shown that if kinetic energy is included in the
analysis below, all the terms due to this inclusion cancel each other). We
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349
CHAPTER 6
assume the density p, specific heat C p , viscosity |x, and the thermal conductiv-
ity k of the fluid to be constant. Then the energy of the fluid per unit mass can
be expressed as e stream = h = C p T.
Energy is a scalar quantity, and thus energy interactions in all directions can
be combined in one equation. Noting that mass flow rate of the fluid entering
the control volume from the left is pu(dy ■ 1), the rate of energy transfer to the
control volume by mass in the x-direction is, from Fig. 6-24,
o(/?7£ stream ) x
V^in out/by mass, A' V '''^stream/.v
V ^^streamjjc '
t)X
- dx
djpujdy ■ l) C p T]
dx
■ dx
PC,
dx
dx
u §x + T §ji] dxdy (6 . 31)
Repeating this for the ^-direction and adding the results, the net rate of energy
transfer to the control volume by mass is determined to be
V^in ^out/
by mass
n dx
dx
u h + T T^) dxd y - p c i v |^ + T %) dxd y
" c \ u fx + v %) dxdy
(6-32)
^•heat. out. y £[
mass. out. y
^•heat in, x
mass out, v
^mass in, v
FIGURE 6-24
The energy transfers by heat and mass
flow associated with a differential
control volume in the thermal
boundary layer in steady two-
dimensional flow.
since du/dx + dv/dy = from the continuity equation.
The net rate of heat conduction to the volume element in the x-direction is
V^in ^outJ
by heat, X
Q* ~ G, +
dx
dx
~(-k(dyl)^)dx = k^dxdy (6-33)
Repeating this for the ^-direction and adding the results, the net rate of energy
transfer to the control volume by heat conduction becomes
C^in ^c
^ / &T , , , , d 2 T , , ,(d 2 T, d 2 T"
,t)byheat = * T^dx d y + k—^dxdy = k — + —
3 6x l dy 1 \dx~ dy- ,
dxdy (6-34)
Another mechanism of energy transfer to and from the fluid in the control
volume is the work done by the body and surface forces. The work done by a
body force is determined by multiplying this force by the velocity in the di-
rection of the force and the volume of the fluid element, and this work needs
to be considered only in the presence of significant gravitational, electric, or
magnetic effects. The surface forces consist of the forces due to fluid pressure
and the viscous shear stresses. The work done by pressure (the flow work) is
already accounted for in the analysis above by using enthalpy for the micro-
scopic energy of the fluid instead of internal energy. The shear stresses that re-
sult from viscous effects are usually very small, and can be neglected in many
cases. This is especially the case for applications that involve low or moderate
velocities.
Then the energy equation for the steady two-dimensional flow of a fluid
with constant properties and negligible shear stresses is obtained by substitut-
ing Eqs. 6-32 and 6-34 into 6-30 to be
"f +v
8T
dv
d 2 T
d 2 T
dy 2
(6-35)
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350
HEAT TRANSFER
which states that the net energy convected by the fluid out of the control vol-
ume is equal to the net energy transferred into the control volume by heat
conduction.
When the viscous shear stresses are not negligible, their effect is accounted
for by expressing the energy equation as
PC,
dT . dT
„, w — + v —
'i dx dy
dx 2
dy 2
IxO
(6-36)
where the viscous dissipation function <i> is obtained after a lengthy analysis
(see an advanced book such as the one by Schlichting (Ref. 9) for details) to be
<J>
duy
dx
dv_V
du
3y
dv
dx
(6-37)
Viscous dissipation may play a dominant role in high-speed flows, especially
when the viscosity of the fluid is high (like the flow of oil in journal bearings).
This manifests itself as a significant rise in fluid temperature due to the con-
version of the kinetic energy of the fluid to thermal energy. Viscous dissipa-
tion is also significant for high-speed flights of aircraft.
For the special case of a stationary fluid, u = v = and the energy equation
reduces, as expected, to the steady two-dimensional heat conduction equation,
dx 2
dy 2 '
(6-38)
Moving
plate
T=12m/s
Stationary
plate
FIGURE 6-25
Schematic for Example 6-1 .
EXAMPLE 6-1 Temperature Rise of Oil in a Journal Bearing
The flow of oil in a journal bearing can be approximated as parallel flow be-
tween two large plates with one plate moving and the other stationary. Such
flows are known as Couette flow.
Consider two large isothermal plates separated by 2-mm-thick oil film. The
upper plates moves at a constant velocity of 12 m/s, while the lower plate is sta-
tionary. Both plates are maintained at 20°C. (a) Obtain relations for the velocity
and temperature distributions in the oil. (b) Determine the maximum tempera-
ture in the oil and the heat flux from the oil to each plate (Fig. 6-25).
SOLUTION Parallel flow of oil between two plates is considered. The velocity
and temperature distributions, the maximum temperature, and the total heat
transfer rate are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Oil is an incompressible
substance with constant properties. 3 Body forces such as gravity are negligible.
4 The plates are large so that there is no variation in the z direction.
Properties The properties of oil at 20°C are (Table A-10):
k = 0.145 W/m • K and \y, = 0.800 kg/m • s = 0.800 N • s/m 2
Analysis (a) We take the x-axis to be the flow direction, and y to be the normal
direction. This is parallel flow between two plates, and thus v = 0. Then the
continuity equation (Eq. 6-21) reduces to
Continuity:
du
dx
dv
dy
du
dx
u(y)
cen58933_ch06.qxd 9/4/2002 12:05 PM Page 351
Therefore, the x-component of velocity does not change in the flow direction
(i.e., the velocity profile remains unchanged). Noting that u = u(y), v = 0, and
dP/dx = (flow is maintained by the motion of the upper plate rather than the
pressure gradient), the x-momentum equation (Eq. 6-28) reduces to
x-momentum: p u
du
dx
du
' dy
|X
d 2 U
dy 2
8P
dx
d 2 u
cly 2
This is a second-order ordinary differential equation, and integrating it twice gives
u(y) = C { y + C 2
The fluid velocities at the plate surfaces must be equal to the velocities of the
plates because of the no-slip condition. Therefore, the boundary conditions are
u(0) = and u(L) = V, and applying them gives the velocity distribution to be
u(y)
-T
Frictional heating due to viscous dissipation in this case is significant be-
cause of the high viscosity of oil and the large plate velocity. The plates are
isothermal and there is no change in the flow direction, and thus the tempera-
ture depends on y only, T = T(y). Also, u = u(y) and v = 0. Then the energy
equation with dissipation (Eqs. 6-36 and 6-37) reduce to
Energy:
,d 2 T, (du\2
,d 2 T
-^
TV
dy 2 r '\dyj dy 2 ^\L f
since du/dy = YIL. Dividing both sides by k and integrating twice give
r ^ = -£(z T ) 2 + c ^ + c ^
7" and T{L)
Applying the boundary conditions 7"(0)
ture distribution to be
T(y) =
T gives the tempera-
M-yyy f
2k \L L 2
(b) The temperature gradient is determined by differentiating T(y) with re-
spect to y,
dT
dv '
jjlT 2
~2~kL
y
The location of maximum temperature is determined by setting dT/dy = and
solving for y,
jjlT 2 -
dT
dy 2kL
y
y
Therefore, maximum temperature will occur at mid plane, which is not surpris-
ing since both plates are maintained at the same temperature. The maximum
temperature is the value of temperature at y = L/2,
\iT 2 (L/2 (L/2)'
20 +
2k
(0.8 N • s/m 2 )(12m/s) 2 / i\y
8(0.145 W/m-°C) UN-m/s
IjlT 2
8fc
119°C
351
CHAPTER 6
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HEAT TRANSFER
Heat flux at the plates is determined from the definition of heat flux,
dy y =o
(0.8N-s/m 2 )(12m/s) 2
-4^(1
k 2kL {l
2(0.002 m)
<1l
dT
'dy
ulT 2
-0)
jjlT 2
2L
\w \
IN-
mis)
-2)
jjlT 2
~~2ZT~
28,800 W/m 2
-q = 28,800 W/m 2
Therefore, heat fluxes at the two plates are equal in magnitude but opposite
in sign.
Discussion A temperature rise of 99°C confirms our suspicion that viscous dis-
sipation is very significant. Also, the heat flux is equivalent to the rate of me-
chanical energy dissipation. Therefore, mechanical energy is being converted to
thermal energy at a rate of 57.2 kW/m 2 of plate area to overcome friction in the
oil. Finally, calculations are done using oil properties at 20°C, but the oil tem-
perature turned out to be much higher. Therefore, knowing the strong depen-
dence of viscosity on temperature, calculations should be repeated using
properties at the average temperature of 70°C to improve accuracy.
u (jc, 0) = o
-v (x, 0) =
T(x,0) = T s
FIGURE 6-26
Boundary conditions for flow
over a flat plate.
6-8 - SOLUTIONS OF CONVECTION EQUATIONS FOR
A FLAT PLATE
Consider laminar flow of a fluid over aflat plate, as shown in Fig. 6-19. Sur-
faces that are slightly contoured such as turbine blades can also be approxi-
mated as flat plates with reasonable accuracy. The .r-coordinate is measured
along the plate surface from the leading edge of the plate in the direction of
the flow, and y is measured from the surface in the normal direction. The fluid
approaches the plate in the x-direction with a uniform upstream velocity,
which is equivalent to the free stream velocity u m .
When viscous dissipation is negligible, the continuity, momentum, and en-
ergy equations (Eqs. 6-21, 6-28, and 6-35) reduce for steady, incompressible,
laminar flow of a fluid with constant properties over a flat plate to
Continuity:
Momentum:
Energy:
du
dx '
du
> — -
dx
dT
dx
dv
dy
du d 2 U
v— = V
dy
dT
' dy'
with the boundary conditions (Fig. 6-26)
u(0,y)--
u(x, 0) =
U(X, co)
u m
0.
: Moo
T(0, y)-
v(x, 0) =
T(x, °°)
dy 2
d 2 T
dy 2
0, T(x, 0) = T s
-- T
(6-39)
(6-40)
(6-41)
(6-42)
When fluid properties are assumed to be constant and thus independent of tem-
perature, the first two equations can be solved separately for the velocity com-
ponents u and v. Once the velocity distribution is available, we can determine
cen58933_ch06.qxd 9/4/2002 12:05 PM Page 353
the friction coefficient and the boundary layer thickness using their definitions.
Also, knowing u and v, the temperature becomes the only unknown in the last
equation, and it can be solved for temperature distribution.
The continuity and momentum equations were first solved in 1908 by the
German engineer H. Blasius, a student of L. Prandtl. This was done by trans-
forming the two partial differential equations into a single ordinary differen-
tial equation by introducing a new independent variable, called the similarity
variable. The finding of such a variable, assuming it exists, is more of an art
than science, and it requires to have a good insight of the problem.
Noticing that the general shape of the velocity profile remains the same
along the plate, Blasius reasoned that the nondimensional velocity profile u/u„
should remain unchanged when plotted against the nondimensional distance
y/S, where 8 is the thickness of the local velocity boundary layer at a given x.
That is, although both 8 and u at a given y vary with x, the velocity u at a fixed
y/S remains constant. Blasius was also aware from the work of Stokes that 8
is proportional to y/vxluZ, and thus he defined a dimensionless similarity
variable as
r\=y
vx
353
CHAPTER 6
(6-43)
and thus ulu m = function(r|). He then introduced a stream function \\s(x, y) as
and v = — — (6-44)
so that the continuity equation (Eq. 6-39) is automatically satisfied and thus
eliminated (this can be verified easily by direct substitution). He then defined
a function /(t|) as the dependent variable as
/dn)
«!<
u^\/vxlul_
Then the velocity components become
di|( _ di|) dr\
dy dr\ dy
_d±_
dx ~
vx df In,
' v u-^ dr\ v vx
jvx df u a I v
' V «o= dx 2 y M„J
/=
d-c]
1 luoji
2 \ x \ ' dr\
TV
df
■f
(6-45)
(6-46)
(6-47)
By differentiating these u and v relations, the derivatives of the velocity com-
ponents can be shown to be
du
dx
«2 d 2 f
~2x^dr\ 2 '
du_ Ju„ d 2 f d 2 u _ uj d 3 f
dy °°\vx dr\ 2 ' dy 2 vx dr\ 3
(6-48)
Substituting these relations into the momentum equation and simplifying, we
obtain
d 3 f d 2 f
d-(Y dfy
(6-49)
which is a third-order nonlinear differential equation. Therefore, the system
of two partial differential equations is transformed into a single ordinary
cen58933_ch06.qxd 9/4/2002 12:05 PM Page 354
354
HEAT TRANSFER
TABLE 6-3
Similarity function fand its
derivatives for laminar boundary
layer along a flat plate.
11
f
df__ u_
dt] U-,_
d 2 f
dt] 2
0.332
0.5
0.042
0.166
0.331
1.0
0.166
0.330
0.323
1.5
0.370
0.487
0.303
2.0
0.650
0.630
0.267
2.5
0.996
0.751
0.217
3.0
1.397
0.846
0.161
3.5
1.838
0.913
0.108
4.0
2.306
0.956
0.064
4.5
2.790
0.980
0.034
5.0
3.283
0.992
0.016
5.5
3.781
0.997
0.007
6.0
4.280
0.999
0.002
00
00
1
differential equation by the use of a similarity variable. Using the definitions
of /and T|, the boundary conditions in terms of the similarity variables can be
expressed as
/(0) = 0,
£
0.
and
il =
£
d-q
1
(6-50)
The transformed equation with its associated boundary conditions cannot be
solved analytically, and thus an alternative solution method is necessary. The
problem was first solved by Blasius in 1908 using a power series expansion ap-
proach, and this original solution is known as the Blasius solution. The prob-
lem is later solved more accurately using different numerical approaches, and
results from such a solution are given in Table 6-3. The nondimensional ve-
locity profile can be obtained by plotting u/u„ against m,. The results obtained
by this simplified analysis are in excellent agreement with experimental results.
Recall that we defined the boundary layer thickness as the distance from the
surface for which ulu^ = 0.99. We observe from Table 6-3 that the value of m,
corresponding to u/u«, = 0.992 is m, = 5.0. Substituting r\ = 5.0 and v = 8 into
the definition of the similarity variable (Eq. 6-43) gives 5.0 = ^\ r ujvx. Then
the velocity boundary layer thickness becomes
5.0
5.0jc
\fujvx
(6-51)
since Re x = u^xlv, where x is the distance from the leading edge of the plate.
Note that the boundary layer thickness increases with increasing kinematic
viscosity v and with increasing distance from the leading edge x, but it de-
creases with increasing free-stream velocity m„. Therefore, a large free-stream
velocity will suppress the boundary layer and cause it to be thinner.
The shear stress on the wall can be determined from its definition and the
du/dy relation in Eq. 6-48:
'i vx drf „=o
i-i
dy
\LU^_
(6-52)
Substituting the value of the second derivative of /at r\ = from Table 6-3
gives
0.332k,
PIxm, _ 0.332pw;
V x VRe;
Then the local skin friction coefficient becomes
(6-53)
C,
f,x
pT 2 /2 pw|/2
0.664 Re:
(6-54)
Note that unlike the boundary layer thickness, wall shear stress and the skin
friction coefficient decrease along the plate as x~ m .
The Energy Equation
Knowing the velocity profile, we are now ready to solve the energy equation
for temperature distribution for the case of constant wall temperature T s . First
we introduce the dimensionless temperature 9 as
cen58933_ch06.qxd 9/4/2002 12:05 PM Page 355
Q(x,y)
T(x,y)
(6-55)
355
CHAPTER 6
Noting that both T s and T m are constant, substitution into the energy equation
gives
se , ae a 2 e
dx dy df
(6-56)
Temperature profiles for flow over an isothermal flat plate are similar, just like
the velocity profiles, and thus we expect a similarity solution for temperature
to exist. Further, the thickness of the thermal boundary layer is proportional to
y/vx/Un , just like the thickness of the velocity boundary layer, and thus the
similarity variable is also r\, and 9 = 0(t|). Using the chain rule and substitut-
ing the u and v expressions into the energy equation gives
4frfeari + i ,
° dt\ dt\ dx 2 V
"»v / df
rfe stj
dr\ dy
d 2 e
dr\-
dy
Simplifying and noting that Pr = via. give
d 2 Q
'dr\ 2
Pr/
clQ
dr\
(6-57)
(6-58)
with the boundary conditions 0(0) = and 0(°°) = 1. Obtaining an equation for
9 as a function of t] alone confirms that the temperature profiles are similar,
and thus a similarity solution exists. Again a closed-form solution cannot be
obtained for this boundary value problem, and it must be solved numerically.
It is interesting to note that for Pr = 1, this equation reduces to Eq. 6-49 when
is replaced by dfldi\, which is equivalent to m/m„ (see Eq. 6-46). The bound-
ary conditions for and dfldt\ are also identical. Thus we conclude that the ve-
locity and thermal boundary layers coincide, and the nondimensional velocity
and temperature profiles (m/m„ and 0) are identical for steady, incompressible,
laminar flow of a fluid with constant properties and Pr = 1 over an isothermal
flat plate (Fig. 6-27). The value of the temperature gradient at the surface
(y = or r\ = 0) in this case is, from Table 6-3, dQ/dr\ = d 2 fldrf = 0.332.
Equation 6-58 is solved for numerous values of Prandtl numbers. For
Pr > 0.6, the nondimensional temperature gradient at the surface is found to
be proportional to Pr l/3 , and is expressed as
dQ
dr\
0.332 Pr 1
T| =
The temperature gradient at the surface is
ae
or
dy
= (T„ - T s ) .
, = o sj dy
0.332 Pr 1/3 (T„
//ft
= (7*. - T s ) ^
= o dt]
jUa,
~ T s) V vr
•n = o dy
y =
(6-59)
(6-60)
'Vw
Then the local convection coefficient and Nusselt number become
q, -k(dT/dy)\ y=0
h.
0.332 Pr 1/j /t
(6-61)
Velocity or thermal
boundary layer
FIGURE 6-27
When Pr = 1, the velocity and thermal
boundary layers coincide, and the
nondimensional velocity and
temperature profiles are identical for
steady, incompressible, laminar flow
over a flat plate.
cen58933_ch06.qxd 9/4/2002 12:06 PM Page 356
356
HEAT TRANSFER
and
h x
Nu t = -^ = 0.332 Pr 1/3 Re'/ 2 Pr > 0.6 (6-62)
k
The Nu v values obtained from this relation agree well with measured values.
Solving Eq. 6-58 numerically for the temperature profile for different
Prandtl numbers, and using the definition of the thermal boundary layer, it
is determined that 8/8, = Pr 1/3 . Then the thermal boundary layer thickness
becomes
8 5.0.x:
Note that these relations are valid only for laminar flow over an isothermal flat
plate. Also, the effect of variable properties can be accounted for by evaluat-
ing all such properties at the film temperature defined as T f = (T s + T^/2.
The Blasius solution gives important insights, but its value is largely histor-
ical because of the limitations it involves. Nowadays both laminar and turbu-
lent flows over surfaces are routinely analyzed using numerical methods.
6-9 - NONDIMENSIONALIZED CONVECTION
EQUATIONS AND SIMILARITY
When viscous dissipation is negligible, the continuity, momentum, and energy
equations for steady, incompressible, laminar flow of a fluid with constant
properties are given by Eqs. 6-21, 6-28, and 6-35.
These equations and the boundary conditions can be nondimensionalized by
dividing all dependent and independent variables by relevant and meaningful
constant quantities: all lengths by a characteristic length L (which is the length
for a plate), all velocities by a reference velocity T (which is the free stream
velocity for a plate), pressure by pT 2 (which is twice the free stream dynamic
pressure for a plate), and temperature by a suitable temperature difference
(which is T m — T s for a plate). We get
* X *V * U * V $ P * 1 — l s
x =Z' y = r " = r v = t p = w and T = ^w;
where the asterisks are used to denote nondimensional variables. Introducing
these variables into Eqs. 6-21, 6-28, and 6-35 and simplifying give
(6-64)
(6-65)
(6-66)
Continuity:
Momentum:
du dv _ _
dx dy
xdu" *dM 1 d 2 u dP"
dx* dy* R e L dy* dx*
Energy:
*dT* *dr* 1 d 2 T*
" dx* V dy* ReiPrsy* 2
with the boundary
conditions
ir
(0.
y*)
= 1, u*(x*, 0) = 0, u*(x*, °°) = 1,
T
■(0
y*)
= 1, T*(x*, 0) = 0, T*(x*, oo) = 1
V*(x*, 0) = 0, (6-67)
cen58933_ch06.qxd 9/4/2002 12:06 PM Page 357
357
CHAPTER 6
where Re L = YLIv is the dimensionless Reynolds number and Pr = via is the
Prandtl number. For a given type of geometry, the solutions of problems with
the same Re and Nu numbers are similar, and thus Re and Nu numbers serve
as similarity parameters. Two physical phenomena are similar if they have the
same dimensionless forms of governing differential equations and boundary
conditions (Fig. 6-28).
A major advantage of nondimensionalizing is the significant reduction in
the number of parameters. The original problem involves 6 parameters
(L, T, Toe, T s , v, a), but the nondimensionalized problem involves just 2 para-
meters (Re L and Pr). For a given geometry, problems that have the same val-
ues for the similarity parameters have identical solutions. For example,
determining the convection heat transfer coefficient for flow over a given sur-
face will require numerical solutions or experimental investigations for sev-
eral fluids, with several sets of velocities, surface lengths, wall temperatures,
and free stream temperatures. The same information can be obtained with far
fewer investigations by grouping data into the dimensionless Re and Pr num-
bers. Another advantage of similarity parameters is that they enable us to
group the results of a large number of experiments and to report them conve-
niently in terms of such parameters (Fig. 6-29).
Re,
Water "~ <L
:>
■* —
— I, ►
T,
Re 2
Air — *<CZ
\M
L 2
If Re, = Re 2
then Cfy = C„
FIGURE 6-28
Two geometrically similar bodies have
the same value of friction coefficient
at the same Reynolds number.
6-10 - FUNCTIONAL FORMS OF FRICTION AND
CONVECTION COEFFICIENTS
The three nondimensionalized boundary layer equations (Eqs. 6-64, 6-65,
and 6-66) involve three unknown functions u*, v*, and T*, two independent
variables x* and y*, and two parameters Re L and Pr. The pressure P*(x*) de-
pends on the geometry involved (it is constant for a flat plate), and it has the
same value inside and outside the boundary layer at a specified x*. Therefore,
it can be determined separately from the free stream conditions, and dP*ldx*
in Eq. 6-65 can be treated as a known function of x*. Note that the boundary
conditions do not introduce any new parameters.
For a given geometry, the solution for u* can be expressed as
Parameters before nondimensionalizing
L,X ZL, T„v,a
Parameters after nondimensionalizing:
Re, Pr
FIGURE 6-29
The number of parameters is reduced
greatly by nondimensionalizing the
convection equations.
u * =/,(x*,y*, Re L )
(6-68)
Then the shear stress at the surface becomes
|X"
(ill
v = ()
jjlT
y =0
Mx, ReJ
(6-69)
Substituting into its definition gives the local friction coefficient,
t s |xT/L 2 * *
V^i = pW 2(x ' ReJ = R^ M*> Re '> =&*> Re ^ (6 " 70)
c,
tx
Thus we conclude that the friction coefficient for a given geometry can be ex-
pressed in terms of the Reynolds number Re and the dimensionless space vari-
able x* alone (instead of being expressed in terms of x, L, T, p, and |x). This is a
very significant finding, and shows the value of nondimensionalized equations.
Similarly, the solution of Eq. 6-66 for the dimensionless temperature T* for
a given geometry can be expressed as
cen58933_ch06.qxd 9/4/2002 12:06 PM Page 35E
358
HEAT TRANSFER
T*
= Nu
v*=o
Laminar
FIGURE 6-30
The Nusselt number is equivalent to
the dimensionless temperature
gradient at the surface.
Local Nusselt number:
Nu v = function (jt*, Re t , Pr)
Average Nusselt number:
Nu = function (Re t , Pr)
A common form of Nusselt number:
Nu = CRe'PPr"
FIGURE 6-31
For a given geometry, the average
Nusselt number is a function of
Reynolds and Prandtl numbers.
T* = g,(x*, y*, Re^ Pr) (6-71)
Using the definition of T*, the convection heat transfer coefficient becomes
h
-k(dT/dy)\ y =o
-k(T x
- T s ) 37/*
kdT
T s ~ T„
L(T S -
r„) dy*
)-*=o L dy
(6-72)
Substituting this into the Nusselt number relation gives [or alternately, we can
rearrange the relation above in dimensionless form as hLIk = (dT*/dy*)\ y * =0
and define the dimensionless group hLIk as the Nusselt number]
Nu v
hL
k
BT
dy
y =o
&(**, Re t , Pr)
(6-73)
Note that the Nusselt number is equivalent to the dimensionless temperature
gradient at the surface, and thus it is properly referred to as the dimensionless
heat transfer coefficient (Fig. 6-30). Also, the Nusselt number for a given
geometry can be expressed in terms of the Reynolds number Re, the Prandtl
number Pr, and the space variable x*, and such a relation can be used for dif-
ferent fluids flowing at different velocities over similar geometries of differ-
ent lengths.
The average friction and heat transfer coefficients are determined by inte-
grating Cf X and Nu v over the surface of the given body with respect to x* from
to 1. Integration will remove the dependence on x*, and the average friction
coefficient and Nusselt number can be expressed as
C f = / 4 (Re L ) and Nu = g 3 (Re L , Pr)
(6-74)
These relations are extremely valuable as they state that for a given geometry,
the friction coefficient can be expressed as a function of Reynolds number
alone, and the Nusselt number as a function of Reynolds and Prandtl numbers
alone (Fig. 6-31). Therefore, experimentalists can study a problem with a
minimum number of experiments, and report their friction and heat transfer
coefficient measurements conveniently in terms of Reynolds and Prandtl
numbers. For example, a friction coefficient relation obtained with air for a
given surface can also be used for water at the same Reynolds number. But it
should be kept in mind that the validity of these relations is limited by the lim-
itations on the boundary layer equations used in the analysis.
The experimental data for heat transfer is often represented with reasonable
accuracy by a simple power-law relation of the form
Nu = CRe'/'Pr"
(6-75)
where m and n are constant exponents (usually between and 1), and the
value of the constant C depends on geometry. Sometimes more complex rela-
tions are used for better accuracy.
6-1 1 - ANALOGIES BETWEEN MOMENTUM AND
HEAT TRANSFER
In forced convection analysis, we are primarily interested in the determination
of the quantities Cf (to calculate shear stress at the wall) and Nu (to calculate
heat transfer rates). Therefore, it is very desirable to have a relation between
cen58933_ch06.qxd 9/4/2002 12:06 PM Page 359
Cj and Nu so that we can calculate one when the other is available. Such rela-
tions are developed on the basis of the similarity between momentum and heat
transfers in boundary layers, and are known as Reynolds analogy and
Chilton- Colburn analogy.
Reconsider the nondimensionalized momentum and energy equations for
steady, incompressible, laminar flow of a fluid with constant properties and
negligible viscous dissipation (Eqs. 6-65 and 6-66). When Pr = 1 (which is
approximately the case for gases) and dP*/dx* = (which is the case when,
m = Moo = T = constant in the free stream, as in flow over a flat plate), these
equations simplify to
359
CHAPTER 6
Momentum:
Energy:
dx '
t dT*
dx
*d«
i av
dy Re L dy
+ v
*.df
1 d 2 T*
dy' 1 Re L dy*
(6-76)
(6-77)
which are exactly of the same form for the dimensionless velocity u* and tem-
perature T*. The boundary conditions for u* and T* are also identical. There-
fore, the functions u* and T* must be identical, and thus the first derivatives
of u* and T* at the surface must be equal to each other,
du
dy
y =o
ar
dy
y =o
Then from Eqs. 6-69, 6-70, and 6-73 we have
Re,
q,— = Nu, (Pr=l)
(6-78)
(6-79)
which is known as the Reynolds analogy (Fig. 6-32). This is an important
analogy since it allows us to determine the heat transfer coefficient for fluids
with Pr ~ 1 from a knowledge of friction coefficient which is easier to mea-
sure. Reynolds analogy is also expressed alternately as
C
where
/.*
St
St
(Pr = 1)
Nu
pcy
(6-80)
(6-81)
Profiles:
u* = T
Gradients:
du*
*> *
dy
_ dT*
/=o dy*
/=°
Analogy:
Re,
C, x — = Nu x
is the Stanton number, which is also a dimensionless heat transfer coefficient.
Reynolds analogy is of limited use because of the restrictions Pr = 1 and
dP*/dx* = on it, and it is desirable to have an analogy that is applicable over
a wide range of Pr. This is done by adding a Prandtl number correction. The
friction coefficient and Nusselt number for a flat plate are determined in Sec-
tion 6-8 to be
C fiX = 0.664 Re; 1 ' 2
and
Nil.
0.332 Pr" 3 Re!
(6-82)
Taking their ratio and rearranging give the desired relation, known as the
modified Reynolds analogy or Chilton-Colburn analogy,
FIGURE 6-32
When Pr = 1 and dP*/dx* « 0,
the nondimensional velocity and
temperature profiles become
identical, and Nu is related to C f
by Reynolds analogy.
cen58933_ch06.qxd 9/4/2002 12:06 PM Page 36C
360
HEAT TRANSFER
Re t
C f]X ^ = Nu,Pr-" 3
C,
1:1
2
pc„t
Pr 2 '
l )H
(6-83)
for 0.6 < Pr < 60. Herey H is called the Colburn j-factor. Although this rela-
tion is developed using relations for laminar flow over a flat plate (for which
dP*/dx* = 0), experimental studies show that it is also applicable approxi-
mately for turbulent flow over a surface, even in the presence of pressure gra-
dients. For laminar flow, however, the analogy is not applicable unless
dP*/dx* ~ 0. Therefore, it does not apply to laminar flow in a pipe. Analogies
between Cf and Nu that are more accurate are also developed, but they are
more complex and beyond the scope of this book. The analogies given above
can be used for both local and average quantities.
Air
20°C, 7 m/s
L = 3m
FIGURE 6-33
Schematic for Example 6-2.
■
EXAMPLE 6-2 Finding Convection Coefficient from Drag Measurement
A 2-m X 3-m flat plate is suspended in a room, and is subjected to air flow par-
allel to its surfaces along its 3-m-long side. The free stream temperature and
velocity of air are 20°C and 7 m/s. The total drag force acting on the plate is
measured to be 0.86 N. Determine the average convection heat transfer coeffi-
cient for the plate (Fig. 6-33).
SOLUTION A flat plate is subjected to air flow, and the drag force acting on it
is measured. The average convection coefficient is to be determined.
Assumptions 1 Steady operating conditions exist. 2 The edge effects are negli-
gible. 3 The local atmospheric pressure is 1 atm.
Properties The properties of air at 20°C and 1 atm are (Table A-15):
p = 1.204 kg/m 3 , C p = 1.007 kJ/kg • K, Pr = 0.7309
Analysis The flow is along the 3-m side of the plate, and thus the characteris-
tic length is L = 3 m. Both sides of the plate are exposed to air flow, and thus
the total surface area is
A,
2WL
12 m 2
2(2 m)(3 m)
For flat plates, the drag force is equivalent to friction force. The average friction
coefficient C f can be determined from Eq. 6-11,
P T 2
<f
CfA s
Solving for C f and substituting,
C f
0.86 N
/lkg • m/s 2
pA s T 2 /2 ( 1 .204 kg/m 3 )( 1 2 m 2 )(7 m/s) 2 /2 \ 1 N
0.00243
Then the average heat transfer coefficient can be determined from the modified
Reynolds analogy (Eq. 6-83) to be
h
C f pVC. 0.00243 (1.204 kg/m 3 )(7 m/s)(1007 J/kg • °C)
2 Pr 2 '
0.7309 2
12.7 W/m 2 - °C
Discussion This example shows the great utility of momentum-heat transfer
analogies in that the convection heat transfer coefficient can be determined
from a knowledge of friction coefficient, which is easier to determine.
cen58933_ch06.qxd 9/4/2002 12:06 PM Page 361
SUMMARY
361
CHAPTER 6
Convection heat transfer is expressed by Newton 's law of cool-
ing as
e conv = hAJT, - rj
where h is the convection heat transfer coefficient, T s is the sur-
face temperature, and T„ is the free-stream temperature. The
convection coefficient is also expressed as
h
-kfimd(.dT/dy) y=a
T - T x
The Nusselt number, which is the dimensionless heat transfer
coefficient, is defined as
Nu
hL„
where k is the thermal conductivity of the fluid and L c is the
characteristic length.
The highly ordered fluid motion characterized by smooth
streamlines is called laminar. The highly disordered fluid mo-
tion that typically occurs at high velocities is characterized by
velocity fluctuations is called turbulent. The random and rapid
fluctuations of groups of fluid particles, called eddies, provide
an additional mechanism for momentum and heat transfer.
The region of the flow above the plate bounded by 8 in
which the effects of the viscous shearing forces caused by fluid
viscosity are felt is called the velocity boundary layer. The
boundary layer thickness, 8, is defined as the distance from the
surface at which u = 0.99a«,. The hypothetical line of
u = 0.99^ divides the flow over a plate into the boundary
layer region in which the viscous effects and the velocity
changes are significant, and the inviscid flow region, in which
the frictional effects are negligible.
The friction force per unit area is called shear stress, and the
shear stress at the wall surface is expressed as
V-
du
By
y=o
c,
pT 2
where jjl is the dynamic viscosity, T is the upstream velocity,
and Cf is the dimensionless friction coefficient. The property
v = pVp is the kinematic viscosity. The friction force over the
entire surface is determined from
C f A
pT 2
The flow region over the surface in which the temperature
variation in the direction normal to the surface is significant is
the thermal boundary layer. The thickness of the thermal
boundary layer 8, at any location along the surface is the dis-
tance from the surface at which the temperature difference
T — T s equals 0.99(7"^ — T s ). The relative thickness of the ve-
locity and the thermal boundary layers is best described by the
dimensionless Prandtl number, defined as
Pr
Molecular diffusivity of momentum v p-C p
Molecular diffusivity of heat a k
For external flow, the dimensionless Reynolds number is ex-
pressed as
Re
Inertia forces _ °VL C _ p°VL c
Viscous forces
)X
For a flat plate, the characteristic length is the distance x from
the leading edge. The Reynolds number at which the flow be-
comes turbulent is called the critical Reynolds number. For flow
over a flat plate, its value is taken to be Re cr = Yxjv = 5 X 10 5 .
The continuity, momentum, and energy equations for steady
two-dimensional incompressible flow with constant properties
are determined from mass, momentum, and energy balances
to be
Continuity:
x-momentum:
Energy:
du
dx
P«
PC,
Bv
dy
du
dx
du
d-ll
df
dP
dx
dT , dT
dx
3yJ
d 2 T , d 2 f
Av-
ar
\x<S>
where the viscous dissipation function <£> is
<i> = 2
duV fdv\2
dx) [dyj .
+
du
dy
av\2
dx
Using the boundary layer approximations and a similarity vari-
able, these equations can be solved for parallel steady incom-
pressible flow over a flat plate, with the following results:
Velocity boundary layer thickness:
5.0
5.0a-
\/YM
C,
Local friction coefficient:
Local Nusselt number: Nu
Thermal boundary layer thickness:
•/.*
pT 2 /2
h x x
~k~ =
' Pr 1 ' 3
0.664 Re; 1 ' 2
0.332 Pr" 3 Re |/ 2
5 .Ox
The average friction coefficient and Nusselt number are ex-
pressed in functional form as
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362
HEAT TRANSFER
Cf = / 4 (Re L ) and Nu = & (Re & Pr)
The Nusselt number can be expressed by a simple power-law
relation of the form
Nu = C Ref Pr"
is the Stanton number. The analogy is extended to other Prandtl
numbers by the modified Reynolds analogy or Chilton-
Colburn analogy, expressed as
C,
Re,
%x
Nu v Pr"
where m and n are constant exponents, and the value of the
constant C depends on geometry. The Reynolds analogy relates
the convection coefficient to the friction coefficient for fluids
with Pr ~ 1 , and is expressed as
Re,
c,
f.x
St,
2
pCJ
Vr m =j H (0.6<Pr<60)
These analogies are also applicable approximately for turbu-
lent flow over a surface, even in the presence of pressure
gradients.
where
St
h
Nu
pCJV Re L Pr
REFERENCES AND SUGGESTED READING
1. H. Blasius. "The Boundary Layers in Fluids with Little
Friction (in German)." Z Math. Phys., 56, 1 (1908); pp.
1-37; English translation in National Advisory
Committee for Aeronautics Technical Memo No. 1256,
February 1950.
2. R. W. Fox and A. T. McDonald. Introduction to Fluid
Mechanics. 5th. ed. New York, Wiley, 1999.
3. J. P. Holman. Heat Transfer. 9th ed. New York: McGraw-
Hill, 2002.
4. F. P. Incropera and D. P. DeWitt. Introduction to Heat
Transfer. 4th ed. New York: John Wiley & Sons, 2002.
5. W. M. Kays and M. E. Crawford. Convective Heat and
Mass Transfer. 3rd ed. New York: McGraw-Hill, 1993.
6. F Kreith and M. S. Bonn. Principles of Heat Transfer. 6th
ed. Pacific Grove, CA: Brooks/Cole, 2001.
7. M. N. Ozisik. Heat Transfer — A Basic Approach. New
York: McGraw-Hill, 1985.
8. O. Reynolds. "On the Experimental Investigation of the
Circumstances Which Determine Whether the Motion of
Water Shall Be Direct or Sinuous, and the Law of
Resistance in Parallel Channels."' Philosophical
Transactions of the Royal Society of London \1A (1883),
pp. 935-82.
9. H. Schlichting. Boundary Layer Theory. 7th ed. New
York: McGraw-Hill, 1979.
10. G. G. Stokes. "On the Effect of the Internal Friction of
Fluids on the Motion of Pendulums." Cambridge
Philosophical Transactions, IX, 8, 1851.
11. N. V. Suryanarayana. Engineering Heat Transfer. St. Paul,
MN: West, 1995.
12. F. M. White. Heat and Mass Transfer. Reading, MA:
Addison-Wesley, 1988.
PROBLEMS
Mechanism and Types of Convection
6-1C What is forced convection? How does it differ from nat-
ural convection? Is convection caused by winds forced or nat-
ural convection?
6-2C What is external forced convection? How does it differ
from internal forced convection? Can a heat transfer system in-
volve both internal and external convection at the same time?
Give an example.
*Problems designated by a "C" are concept questions, and
students are encouraged to answer them all. Problems designated
by an "E" are in English units, and the SI users can ignore them.
Problems with an EES-CD icon ® are solved using EES, and
complete solutions together with parametric studies are included
on the enclosed CD. Problems with a computer-EES icon El are
comprehensive in nature, and are intended to be solved with a
computer, preferably using the EES software that accompanies
this text.
cen58933_ch06.qxd 9/4/2002 12:06 PM Page 363
363
CHAPTER 6
6-3C In which mode of heat transfer is the convection heat
transfer coefficient usually higher, natural convection or forced
convection? Why?
6-4C Consider a hot baked potato. Will the potato cool faster
or slower when we blow the warm air coming from our lungs
on it instead of letting it cool naturally in the cooler air in the
room? Explain.
6-5C What is the physical significance of the Nusselt number?
How is it defined?
6-6C When is heat transfer through a fluid conduction and
when is it convection? For what case is the rate of heat transfer
higher? How does the convection heat transfer coefficient dif-
fer from the thermal conductivity of a fluid?
6-7C Define incompressible flow and incompressible fluid.
Must the flow of a compressible fluid necessarily be treated as
compressible?
6-8 During air cooling of potatoes, the heat transfer coeffi-
cient for combined convection, radiation, and evaporation is
determined experimentally to be as shown:
Heat Transfer Coefficient,
Air Velocity,
m/s
W/m 2 • °C
0.66
14.0
1.00
19.1
1.36
20.2
1.73
24.4
Consider a 10-cm -diameter potato initially at 20°C with a ther-
mal conductivity of 0.49 W/m • °C. Potatoes are cooled by re-
frigerated air at 5°C at a velocity of 1 m/s. Determine the initial
rate of heat transfer from a potato, and the initial value of the
temperature gradient in the potato at the surface.
Answers: 9.0 W, -585°C/m
6-9 An average man has a body surface area of 1 .8 m 2 and a
skin temperature of 33°C. The convection heat transfer coeffi-
cient for a clothed person walking in still air is expressed as
h = 8.6T - 53 for 0.5 < T < 2 m/s, where T is the walking ve-
locity in m/s. Assuming the average surface temperature of the
clothed person to be 30°C, determine the rate of heat loss from
an average man walking in still air at 10°C by convection at
a walking velocity of (a) 0.5 m/s, (b) 1.0 m/s, (c) 1.5 m/s, and
(d) 2.0 m/s.
6-10 The convection heat transfer coefficient for a clothed
person standing in moving air is expressed as h = 14.8T - 69 for
0.15 < T < 1.5 m/s, where T is the air velocity. For a person
with a body surface area of 1.7 m 2 and an average surface
temperature of 29°C, determine the rate of heat loss from the
person in windy air at 10°C by convection for air velocities of
(a) 0.5 m/s, (b) 1.0 m/s, and (c) 1.5 m/s.
6-11 During air cooling of oranges, grapefruit, and tangelos,
the heat transfer coefficient for combined convection, radia-
tion, and evaporation for air velocities of 0. 1 1 < T < 0.33 m/s
Air
5°C
1 atm
FIGURE P6-1 1
is determined experimentally and is expressed as h = 5.05
& air Re 1/3 /D, where the diameter D is the characteristic length.
Oranges are cooled by refrigerated air at 5°C and 1 atm at a ve-
locity of 0.5 m/s. Determine (a) the initial rate of heat transfer
from a 7 -cm -diameter orange initially at 15°C with a thermal
conductivity of 0.50 W/m • °C, {b) the value of the initial tem-
perature gradient inside the orange at the surface, and (c) the
value of the Nusselt number.
Velocity and Thermal Boundary Layers
6-12C What is viscosity? What causes viscosity in liquids
and in gases? Is dynamic viscosity typically higher for a liquid
or for a gas?
6-13C What is Newtonian fluid? Is water a Newtonian
fluid?
6-14C What is the no-slip condition? What causes it?
6-15C Consider two identical small glass balls dropped into
two identical containers, one filled with water and the other
with oil. Which ball will reach the bottom of the container
first? Why?
6-16C How does the dynamic viscosity of (a) liquids and
(b) gases vary with temperature?
6-17C What fluid property is responsible for the develop-
ment of the velocity boundary layer? For what kind of fluids
will there be no velocity boundary layer on a flat plate?
6-18C What is the physical significance of the Prandtl num-
ber? Does the value of the Prandtl number depend on the type
of flow or the flow geometry? Does the Prandtl number of air
change with pressure? Does it change with temperature?
6-19C Will a thermal boundary layer develop in flow over a
surface even if both the fluid and the surface are at the same
temperature?
Laminar and Turbulent Flows
6-20C How does turbulent flow differ from laminar flow?
For which flow is the heat transfer coefficient higher?
6-21 C What is the physical significance of the Reynolds
number? How is it defined for external flow over a plate of
length L?
6-22C What does the friction coefficient represent in flow
over a flat plate? How is it related to the drag force acting on
the plate?
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364
HEAT TRANSFER
6-23C What is the physical mechanism that causes the fric-
tion factor to be higher in turbulent flow?
6-24C What is turbulent viscosity? What is it caused by?
6-25C What is turbulent thermal conductivity? What is it
caused by?
Convection Equations and Similarity Solutions
6-26C Under what conditions can a curved surface be
treated as a flat plate in fluid flow and convection analysis?
6-27C Express continuity equation for steady two-
dimensional flow with constant properties, and explain what
each term represents.
6-28C Is the acceleration of a fluid particle necessarily zero
in steady flow? Explain.
6-29C For steady two-dimensional flow, what are the
boundary layer approximations?
6-30C For what types of fluids and flows is the viscous dis-
sipation term in the energy equation likely to be significant?
6-31C For steady two-dimensional flow over an isothermal
flat plate in the x-direction, express the boundary conditions for
the velocity components u and v, and the temperature T at the
plate surface and at the edge of the boundary layer.
6-32C What is a similarity variable, and what is it used for?
For what kinds of functions can we expect a similarity solution
for a set of partial differential equations to exist?
6-33C Consider steady, laminar, two-dimensional flow over
an isothermal plate. Does the thickness of the velocity bound-
ary layer increase or decrease with (a) distance from the lead-
ing edge, (b) free-stream velocity, and (c) kinematic viscosity?
6-34C Consider steady, laminar, two-dimensional flow over
an isothermal plate. Does the wall shear stress increase, de-
crease, or remain constant with distance from the leading edge?
6-35C What are the advantages of nondimensionalizing the
convection equations?
6-36C Consider steady, laminar, two-dimensional, incom-
pressible flow with constant properties and a Prandtl number of
unity. For a given geometry, is it correct to say that both the av-
erage friction and heat transfer coefficients depend on the
Reynolds number only?
6-37 Oil flow in a journal bearing can be treated as parallel
flow between two large isothermal plates with one plate mov-
ing at a constant velocity of 12 m/s and the other stationary.
Consider such a flow with a uniform spacing of 0.7 mm be-
tween the plates. The temperatures of the upper and lower
plates are 40°C and 15°C, respectively. By simplifying and
solving the continuity, momentum, and energy equations, de-
termine (a) the velocity and temperature distributions in the oil,
(b) the maximum temperature and where it occurs, and (c) the
heat flux from the oil to each plate.
17. m/s
»(>')
FIGURE P6-37
6-38 Repeat Problem 6-37 for a spacing of 0.4 mm.
6-39 A 6-cm-diameter shaft rotates at 3000 rpm in a 20-cm-
long bearing with a uniform clearance of 0.2 mm. At steady op-
erating conditions, both the bearing and the shaft in the vicinity
of the oil gap are at 50°C, and the viscosity and thermal con-
ductivity of lubricating oil are 0.05 N • s/m 2 and 0.17 W/m • K.
By simplifying and solving the continuity, momentum, and
energy equations, determine (a) the maximum temperature of
oil, (b) the rates of heat transfer to the bearing and the shaft,
and (c) the mechanical power wasted by the viscous dissipation
in the oil. Answers: (a) 53.3°C, (b) 419 W, (c) 838 W
3000 rpm
i i
6 cm
20 cm
FIGURE P6-39
6-40 Repeat Problem 6-39 by assuming the shaft to have
reached peak temperature and thus heat transfer to the shaft to
be negligible, and the bearing surface still to be maintained
at 50°C.
6-41 Reconsider Problem 6-39. Using EES (or other) soft-
ware, investigate the effect of shaft velocity on the mechanical
power wasted by viscous dissipation. Let the shaft rotation
vary from rpm to 5000 rpm. Plot the power wasted versus the
shaft rpm, and discuss the results.
6-42 Consider a 5-cm-diameter shaft rotating at 2500 rpm in
a 10-cm-long bearing with a clearance of 0.5 mm. Determine
the power required to rotate the shaft if the fluid in the gap is
(a) air, (b) water, and (c) oil at 40°C and 1 arm.
6-43 Consider the flow of fluid between two large parallel
isothermal plates separated by a distance L. The upper plate is
moving at a constant velocity of T and maintained at tempera-
ture T while the lower plate is stationary and insulated. By
simplifying and solving the continuity, momentum, and energy
equations, obtain relations for the maximum temperature of
fluid, the location where it occurs, and heat flux at the upper
plate.
6-44 Reconsider Problem 6^-3. Using the results of this prob-
lem, obtain a relation for the volumetric heat generation rate g,
in W/m 3 . Then express the convection problem as an equivalent
cen58933_ch06.qxd 9/4/2002 12:06 PM Page 365
conduction problem in the oil layer. Verify your model by solv-
ing the conduction problem and obtaining a relation for the
maximum temperature, which should be identical to the one
obtained in the convection analysis.
6-45 A 5-cm-diameter shaft rotates at 4500 rpm in a 15-cm-
long, 8-cm-outer-diameter cast iron bearing (k = 70 W/m ■ K)
with a uniform clearance of 0.6 mm filled with lubricating oil
O = 0.03 N • s/m 2 and k = 0.14 W/m ■ K). The bearing is
cooled externally by a liquid, and its outer surface is main-
tained at 40°C. Disregarding heat conduction through the shaft
and assuming one-dimensional heat transfer, determine (a) the
rate of heat transfer to the coolant, (b) the surface temperature
of the shaft, and (c) the mechanical power wasted by the vis-
cous dissipation in oil.
4500 rpm
l l
5 cm
15 cm
8 cm
FIGURE P6-45
6-46 Repeat Problem 6^-5 for a clearance of 1 mm.
Momentum and Heat Transfer Analogies
6-47C How is Reynolds analogy expressed? What is the
value of it? What are its limitations?
6-48C How is the modified Reynolds analogy expressed?
What is the value of it? What are its limitations?
6-49 (~Jb\ A 4-m X 4-m flat plate maintained at a constant
<S$$7 temperature of 80°C is subjected to parallel flow
of air at 1 atm, 20°C, and 10 m/s. The total drag force acting
on the upper surface of the plate is measured to be 2.4 N.
Using momentum-heat transfer analogy, determine the aver-
age convection heat transfer coefficient, and the rate of heat
transfer between the upper surface of the plate and the air.
365
CHAPTER 6
6-50 A metallic airfoil of elliptical cross section has a mass of
50 kg, surface area of 12 m 2 , and a specific heat of 0.50 kJ/kg •
°C). The airfoil is subjected to air flow at 1 atm, 25°C, and 8
m/s along its 3-m-long side. The average temperature of the
airfoil is observed to drop from 160°C to 150°C within 2 min of
cooling. Assuming the surface temperature of the airfoil to be
equal to its average temperature and using momentum-heat
transfer analogy, determine the average friction coefficient of
the airfoil surface. Answer: 0.000227
6-51 Repeat Problem 6-50 for an air-flow velocity of 12 m/s.
6-52 The electrically heated 0.6-m-high and 1.8-m-long
windshield of a car is subjected to parallel winds at 1 atm, 0°C,
and 80 km/h. The electric power consumption is observed to be
50 W when the exposed surface temperature of the windshield
is 4°C. Disregarding radiation and heat transfer from the inner
surface and using the momentum-heat transfer analogy, deter-
mine drag force the wind exerts on the windshield.
6-53 Consider an airplane cruising at an altitude of 10 km
where standard atmospheric conditions are — 50°C and 26.5
kPa at a speed of 800 km/h. Each wing of the airplane can be
modeled as a 25-m X 3-m flat plate, and the friction coefficient
of the wings is 0.0016. Using the momentum-heat transfer
analogy, determine the heat transfer coefficient for the wings at
cruising conditions. Answer: 89.6 W/m 2 ■ °C
Design and Essay Problems
6-54 Design an experiment to measure the viscosity of liquids
using a vertical funnel with a cylindrical reservoir of height h
and a narrow flow section of diameter D and length L. Making
appropriate assumptions, obtain a relation for viscosity in
terms of easily measurable quantities such as density and vol-
ume flow rate.
6-55 A facility is equipped with a wind tunnel, and can mea-
sure the friction coefficient for flat surfaces and airfoils. De-
sign an experiment to determine the mean heat transfer
coefficient for a surface using friction coefficient data.
cen58933_ch06.qxd 9/4/2002 12:06 PM Page 366
cen58 93 3_ch07.qxd 9/4/2002 12:12 PM Page 367
1
1
1
•
I CHAPTER
r
i
^
EXTERNAL FORCED
CONVECTION
■ n Chapter 6 we considered the general and theoretical aspects of forced
1 convection, with emphasis on differential formulation and analytical solu-
1 tions. In this chapter we consider the practical aspects of forced convection
to or from flat or curved surfaces subjected to external flow, characterized by
the freely growing boundary layers surrounded by a free flow region that
involves no velocity and temperature gradients.
We start this chapter with an overview of external flow, with emphasis on
friction and pressure drag, flow separation, and the evaluation of average drag
and convection coefficients. We continue with parallel flow over flat plates.
In Chapter 6, we solved the boundary layer equations for steady, laminar, par-
allel flow over a flat plate, and obtained relations for the local friction coeffi-
cient and the Nusselt number. Using these relations as the starting point, we
determine the average friction coefficient and Nusselt number. We then extend
the analysis to turbulent flow over flat plates with and without an unheated
starting length.
Next we consider cross flow over cylinders and spheres, and present graphs
and empirical correlations for the drag coefficients and the Nusselt numbers,
and discuss their significance. Finally, we consider cross flow over tube banks
7
CONTENTS
7-1 Drag and Heat Transfer in
External Flow 368
7-2 Parallel Flow over Flat
Plates 371
7-3 Flow across Cylinders and
Spheres 380
7-4 Flow across Tube
Banks 389
Topic of Special Interest:
Reducing Heat Transfer
through Surfaces:
Thermal Insulation 395
in aligned and staggered configurations, and present correlations for the pres-
sure drop and the average Nusselt number for both configurations.
367
^3
cen58 93 3_ch07.qxd 9/4/2002 12:12 PM Page 368
368
HEAT TRANSFER
Wind tunnel
60 mph
FIGURE 7-1
Schematic for measuring the drag
force acting on a car in a wind tunnel.
7-1 - DRAG AND HEAT TRANSFER
IN EXTERNAL FLOW
Fluid flow over solid bodies frequently occurs in practice, and it is responsi-
ble for numerous physical phenomena such as the drag force acting on the au-
tomobiles, power lines, trees, and underwater pipelines; the lift developed by
airplane wings; upward draft of rain, snow, hail, and dust particles in high
winds; and the cooling of metal or plastic sheets, steam and hot water pipes,
and extruded wires. Therefore, developing a good understanding of external
flow and external forced convection is important in the mechanical and ther-
mal design of many engineering systems such as aircraft, automobiles, build-
ings, electronic components, and turbine blades.
The flow fields and geometries for most external flow problems are too com-
plicated to be solved analytically, and thus we have to rely on correlations
based on experimental data. The availability of high-speed computers has
made it possible to conduct series of "numerical experimentations" quickly by
solving the governing equations numerically, and to resort to the expensive and
time-consuming testing and experimentation only in the final stages of design.
In this chapter we will mostly rely on relations developed experimentally.
The velocity of the fluid relative to an immersed solid body sufficiently far
from the body (outside the boundary layer) is called the free-stream velocity,
and is denoted by «„. It is usually taken to be equal to the upstream velocity
T also called the approach velocity, which is the velocity of the approaching
fluid far ahead of the body. This idealization is nearly exact for very thin bod-
ies, such as a flat plate parallel to flow, but approximate for blunt bodies such
as a large cylinder. The fluid velocity ranges from zero at the surface (the no-
slip condition) to the free-stream value away from the surface, and the sub-
script "infinity" serves as a reminder that this is the value at a distance where
the presence of the body is not felt. The upstream velocity, in general, may
vary with location and time (e.g., the wind blowing past a building). But in the
design and analysis, the upstream velocity is usually assumed to be uniform
and steady for convenience, and this is what we will do in this chapter.
Friction and Pressure Drag
You may have seen high winds knocking down trees, power lines, and even
trailers, and have felt the strong "push" the wind exerts on your body. You ex-
perience the same feeling when you extend your arm out of the window of a
moving car. The force a flowing fluid exerts on a body in the flow direction is
called drag (Fig. 7-1)
A stationary fluid exerts only normal pressure forces on the surface of a
body immersed in it. A moving fluid, however, also exerts tangential shear
forces on the surface because of the no-slip condition caused by viscous
effects. Both of these forces, in general, have components in the direction
of flow, and thus the drag force is due to the combined effects of pressure and
wall shear forces in the flow direction. The components of the pressure
and wall shear forces in the normal direction to flow tend to move the body in
that direction, and their sum is called lift.
In general, both the skin friction (wall shear) and pressure contribute to
the drag and the lift. In the special case of a thin flat plate aligned parallel
to the flow direction, the drag force depends on the wall shear only and is
cen58 93 3_ch07.qxd 9/4/2002 12:12 PM Page 369
independent of pressure. When the flat plate is placed normal to the flow di-
rection, however, the drag force depends on the pressure only and is indepen-
dent of the wall shear since the shear stress in this case acts in the direction
normal to flow (Fig. 7-2). For slender bodies such as wings, the shear force
acts nearly parallel to the flow direction. The drag force for such slender bod-
ies is mostly due to shear forces (the skin friction).
The drag force F D depends on the density p of the fluid, the upstream ve-
locity T, and the size, shape, and orientation of the body, among other things.
The drag characteristics of a body is represented by the dimensionless drag
coefficient C D defined as
F D
Drag coefficient: C n = - — (7-1)
s jj d i p <Y2 A
where A is the frontal area (the area projected on a plane normal to the direc-
tion of flow) for blunt bodies — bodies that tends to block the flow. The frontal
area of a cylinder of diameter D and length L, for example, is A = LD. For
parallel flow over flat plates or thin airfoils, A is the surface area. The drag co-
efficient is primarily a function of the shape of the body, but it may also de-
pend on the Reynolds number and the surface roughness.
The drag force is the net force exerted by a fluid on a body in the direction
of flow due to the combined effects of wall shear and pressure forces. The part
of drag that is due directly to wall shear stress t w is called the skin friction
drag (or just friction drag) since it is caused by frictional effects, and the part
that is due directly to pressure P is called the pressure drag (also called the
form drag because of its strong dependence on the form or shape of the body).
When the friction and pressure drag coefficients are available, the total drag
coefficient is determined by simply adding them,
Cr
C 4- C
*-\D, friction ' ^D, pressure
(7-2)
The friction drag is the component of the wall shear force in the direction of
flow, and thus it depends on the orientation of the body as well as the magni-
tude of the wall shear stress t w . The friction drag is zero for a surface normal
to flow, and maximum for a surface parallel to flow since the friction drag in
this case equals the total shear force on the surface. Therefore, for parallel
flow over a flat plate, the drag coefficient is equal to the friction drag coeffi-
cient, or simply the friction coefficient (Fig. 7-3). That is,
Flat plate:
(-*n — ^-n
c,
(7-3)
Once the average friction coefficient Cy is available, the drag (or friction)
force over the surface can be determined from Eq. 7-1. In this case A is the
surface area of the plate exposed to fluid flow. When both sides of a thin
plate are subjected to flow, A becomes the total area of the top and bottom
surfaces. Note that the friction coefficient, in general, will vary with location
along the surface.
Friction drag is a strong function of viscosity, and an "idealized" fluid with
zero viscosity would produce zero friction drag since the wall shear stress
would be zero (Fig. 7-4). The pressure drag would also be zero in this case
during steady flow regardless of the shape of the body since there will be no
pressure losses. For flow in the horizontal direction, for example, the pressure
along a horizontal line will be constant (just like stationary fluids) since the
369
CHAPTER 7
High pressure
Low pressure
s Wall shear
FIGURE 7-2
Drag force acting on a flat
plate normal to flow depends on
the pressure only and is
independent of the wall shear,
which acts normal to flow.
FIGURE 7-3
For parallel flow over a flat plate,
the pressure drag is zero, and thus the
drag coefficient is equal to the friction
coefficient and the drag force is
equal to the friction force.
: if U. =
k
r
r
□ oooocTl
r®([Z>)
(H " " —
■^ML^Z^0
FIGURE 1-H
For the flow of an "idealized"
fluid with zero viscosity past a body,
both the friction drag and pressure
drag are zero regardless of the
shape of the body.
cen58 93 3_ch07.qxd 9/4/2002 12:12 PM Page 37C
370
HEAT TRANSFER
upstream velocity is constant, and thus there will be no net pressure force
acting on the body in the horizontal direction. Therefore, the total drag is zero
for the case of ideal inviscid fluid flow.
At low Reynolds numbers, most drag is due to friction drag. This is espe-
cially the case for highly streamlined bodies such as airfoils. The friction drag
is also proportional to the surface area. Therefore, bodies with a larger surface
area will experience a larger friction drag. Large commercial airplanes, for ex-
ample, reduce their total surface area and thus drag by retracting their wing
extensions when they reach the cruising altitudes to save fuel. The friction
drag coefficient is independent of surface roughness in laminar flow, but is a
strong function of surface roughness in turbulent flow due to surface rough-
ness elements protruding further into the highly viscous laminar sublayer.
The pressure drag is proportional to the difference between the pressures
acting on the front and back of the immersed body, and the frontal area.
Therefore, the pressure drag is usually dominant for blunt bodies, negligible
for streamlined bodies such as airfoils, and zero for thin flat plates parallel to
the flow.
When a fluid is forced to flow over a curved surface at sufficiently high ve-
locities, it will detach itself from the surface of the body. The low-pressure re-
gion behind the body where recirculating and back flows occur is called the
separation region. The larger the separation area is, the larger the pressure
drag will be. The effects of flow separation are felt far downstream in the form
of reduced velocity (relative to the upstream velocity). The region of flow
trailing the body where the effect of the body on velocity is felt is called the
wake (Fig. 7-5). The separated region comes to an end when the two flow
streams reattach, but the wake keeps growing behind the body until the fluid
in the wake region regains its velocity. The viscous effects are the most sig-
nificant in the boundary layer, the separated region, and the wake. The flow
outside these regions can be considered to be inviscid.
Heat Transfer
The phenomena that affect drag force also affect heat transfer, and this effect
appears in the Nusselt number. By nondimensionalizing the boundary layer
equations, it was shown in Chapter 6 that the local and average Nusselt num-
bers have the functional form
FIGURE 7-5
Separation and reattachment
during flow over a cylinder,
and the wake region.
Nu, = fix*, Re x , Pr) and Nu = /,(Re L , Pr)
(7-4a, b)
The experimental data for heat transfer is often represented conveniently
with reasonable accuracy by a simple power-law relation of the form
Nu = CRe^'Pr"
(7-5)
where m and n are constant exponents, and the value of the constant C de-
pends on geometry and flow.
The fluid temperature in the thermal boundary layer varies from T s at the
surface to about T rtJ at the outer edge of the boundary. The fluid properties also
vary with temperature, and thus with position across the boundary layer. In
order to account for the variation of the properties with temperature, the fluid
properties are usually evaluated at the so-called film temperature, defined as
cen58 93 3_ch07.qxd 9/4/2002 12:12 PM Page 371
(7-6)
which is the arithmetic average of the surface and the free-stream tempera-
tures. The fluid properties are then assumed to remain constant at those values
during the entire flow. An alternative way of accounting for the variation of
properties with temperature is to evaluate all properties at the free stream tem-
perature and to multiply the Nusselt number relation in Eq. 7-5 by (VxJPx s ) r
or (ixjfjij)''.
The local drag and convection coefficients vary along the surface as a result
of the changes in the velocity boundary layers in the flow direction. We are
usually interested in the drag force and the heat transfer rate for the entire sur-
face, which can be determined using the average friction and convection co-
efficient. Therefore, we present correlations for both local (identified with the
subscript x) and average friction and convection coefficients. When relations
for local friction and convection coefficients are available, the average fric-
tion and convection coefficients for the entire surface can be determined by
integration from
C„
l\S-
dx
and
L
h r dx
(7-7)
(7-8)
When the average drag and convection coefficients are available, the drag
force can be determined from Eq. 7-1 and the rate of heat transfer to or from
an isothermal surface can be determined from
Q = hA s (T s - TJ
(7-9)
where A, is the surface area.
371
CHAPTER 7
7-2 ■ PARALLEL FLOW OVER FLAT PLATES
Consider the parallel flow of a fluid over a flat plate of length L in the flow di-
rection, as shown in Figure 7-6. The x-coordinate is measured along the plate
surface from the leading edge in the direction of the flow. The fluid ap-
proaches the plate in the x-direction with uniform upstream velocity T and
temperature T„. The flow in the velocity boundary layer starts out as laminar,
but if the plate is sufficiently long, the flow will become turbulent at a dis-
tance x a from the leading edge where the Reynolds number reaches its critical
value for transition.
The transition from laminar to turbulent flow depends on the surface geom-
etry, surface roughness, upstream velocity, surface temperature, and the type
of fluid, among other things, and is best characterized by the Reynolds num-
ber. The Reynolds number at a distance x from the leading edge of a flat plate
is expressed as
V
^y?t
OJ Turbulent ^pj
FIGURE 7-6
Laminar and turbulent regions
of the boundary layer during
flow over a flat plate.
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372
HEAT TRANSFER
Re,
pVx = Yx
(7-10)
Note that the value of the Reynolds number varies for a flat plate along the
flow, reaching Re L = VL/v at the end of the plate.
For flow over a flat plate, transition from laminar to turbulent is usually
taken to occur at the critical Reynolds number of
Re„,
5 X 10 5
(7-11)
The value of the critical Reynolds number for a flat plate may vary from 10 5
to 3 X 10 6 , depending on the surface roughness and the turbulence level of the
free stream.
FIGURE 7-7
The average friction coefficient over
a surface is determined by integrating
the local friction coefficient over the
entire surface.
Friction Coefficient
Based on analysis, the boundary layer thickness and the local friction coeffi-
cient at location x for laminar flow over a flat plate were determined in Chap-
ter 6 to be
Laminar:
5x
Re'/ 2
and
C,
%x
0.664
ReL' 2 '
Re v < 5 X 10 5 (7-12a, b)
The corresponding relations for turbulent flow are
Turbulent: 8„
0.382jc
Re'/ 5
and C
■f.*
0.0592
Rel' 5 '
5X10 5 < Re,.
10 7
(7-13a, b)
where x is the distance from the leading edge of the plate and Re x = Vx/v is the
Reynolds number at location x. Note that Q x . is proportional to Re A 7 1/2 and thus
to x~ 111 for laminar flow. Therefore, Cf >x is supposedly infinite at the leading
edge (x = 0) and decreases by a factor of x~ 112 in the flow direction. The local
friction coefficients are higher in turbulent flow than they are in laminar flow
because of the intense mixing that occurs in the turbulent boundary layer.
Note that Q x reaches its highest values when the flow becomes fully turbu-
lent, and then decreases by a factor of x~ U5 in the flow direction.
The average friction coefficient over the entire plate is detennined by sub-
stituting the relations above into Eq. 7-7 and performing the integrations
(Fig.7-7). We get
Laminar:
C,
1.328
Re[ /2
Re L < 5 X 10 5
(7-14)
Turbulent:
C,
0.074
5 X 10 5
Re,
10 7
(7-15)
The first relation gives the average friction coefficient for the entire plate when
the flow is laminar over the entire plate. The second relation gives the average
friction coefficient for the entire plate only when the flow is turbulent over the
entire plate, or when the laminar flow region of the plate is too small relative to
the turbulent flow region (that is, x CI <S L where the length of the plate x a over
which the flow is laminar can be determined from Re cr = 5 X 10 5 = Yxjv).
cen58 93 3_ch07.qxd 9/4/2002 12:12 PM Page 373
In some cases, a flat plate is sufficiently long for the flow to become turbu-
lent, but not long enough to disregard the laminar flow region. In such cases,
the average friction coefficient over the entire plate is determined by perform-
ing the integration in Eq. 7-7 over two parts: the laminar region ^ .t < x CT
and the turbulent region x cr < x < L as
C,
K/>
^r laminar
dx
*-"/', x, turbulent "-^
(7-16)
Note that we included the transition region with the turbulent region. Again
taking the critical Reynolds number to be Re cr = 5 X 10 5 and performing the
integrations of Eq. 7-16 after substituting the indicated expressions, the aver-
age friction coefficient over the entire plate is determined to be
C,
0.074 1742
Re[' 5 Re L
5 X 10 5
Re,
10 7
(7-17)
The constants in this relation will be different for different critical Reynolds
numbers. Also, the surfaces are assumed to be smooth, and the free stream to
be turbulent free. For laminar flow, the friction coefficient depends on only
the Reynolds number, and the surface roughness has no effect. For turbulent
flow, however, surface roughness causes the friction coefficient to increase
severalfold, to the point that in fully turbulent regime the friction coefficient
is a function of surface roughness alone, and independent of the Reynolds
number (Fig. 7-8).
A curve fit of experimental data for the average friction coefficient in this
regime is given by Schlichting as
Rough surface, turbulent:
C f = 1.89 - 1.62 log :
e\-2.5
(7-18)
were s is the surface roughness, and L is the length of the plate in the flow di-
rection. In the absence of a better relation, the relation above can be used for
turbulent flow on rough surfaces for Re > 10 6 , especially when s/L > 10 4 .
Heat Transfer Coefficient
The local Nusselt number at a location x for laminar flow over a flat plate was
determined in Chapter 6 by solving the differential energy equation to be
Laminar:
Nu v
h x x
k
0.332 Re° v 5 Pr 1
Pr > 0.60
The corresponding relation for turbulent flow is
Turbulent: Nu x = -£- = 0.0296 Re° 8 Pr" 3
0.6 < Pr :
5 X 10 5 =
:60
Re v
10 7
(7-19)
(7-20)
Note that h x is proportional to Re° 5 and thus to x" 5 for laminar flow. There-
fore, h x is infinite at the leading edge (x = 0) and decreases by a factor of x~° ,5
in the flow direction. The variation of the boundary layer thickness 8 and the
friction and heat transfer coefficients along an isothermal flat plate are shown
in Figure 7-9. The local friction and heat transfer coefficients are higher in
373
CHAPTER 7
Relative
roughness,
e/L
Friction
coefficient
C,
0.0*
1 x 10- 5
1 x 10- 4
1 x 10- 3
0.0029
0.0032
0.0049
0.0084
*Smooth surface for Re = 10 7 . Others
calculated from Eq. 7-18.
FIGURE 7-8
For turbulent flow, surface roughness
may cause the friction coefficient
to increase severalfold.
FIGURE 7-9
The variation of the local friction
and heat transfer coefficients for
flow over a flat plate.
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374
HEAT TRANSFER
X turbulent
x a L *
FIGURE 7-10
Graphical representation of the
average heat transfer coefficient for a
flat plate with combined laminar and
turbulent flow.
turbulent flow than they are in laminar flow. Also, h x reaches its highest val-
ues when the flow becomes fully turbulent, and then decreases by a factor of
— T
in the flow direction, as shown in the figure.
The average Nusselt number over the entire plate is determined by substitut-
ing the relations above into Eq. 7-8 and performing the integrations. We get
Laminar:
Turbulent:
Nu
hL
0.664 Re ?- 5 Pr 1/3
Re L < 5 X 10 5
Nu
hL
0.037 Re? 8 Pr" 3
0.6 < Pr < 60
5 X 10 5 < Re, < 10 7
(7-21)
(7-22)
The first relation gives the average heat transfer coefficient for the entire plate
when the flow is laminar over the entire plate. The second relation gives the
average heat transfer coefficient for the entire plate only when the flow is tur-
bulent over the entire plate, or when the laminar flow region of the plate is too
small relative to the turbulent flow region.
In some cases, a flat plate is sufficiently long for the flow to become turbu-
lent, but not long enough to disregard the laminar flow region. In such cases,
the average heat transfer coefficient over the entire plate is determined by per-
forming the integration in Eq. 7-8 over two parts as
/;
h
x, laminar
dx
I "*, trubulent " X I
(7-23)
Again taking the critical Reynolds number to be Re cr = 5 X 10 5 and perform-
ing the integrations in Eq. 7-23 after substituting the indicated expressions, the
average Nusselt number over the entire plate is determined to be (Fig. 7-10)
Nu
hL
(0.037 Re? 8 - 871)Pr'
0.6 < Pr :
5 X 10 5 :
;60
Re L :
10 7
(7-24)
The constants in this relation will be different for different critical Reynolds
numbers.
Liquid metals such as mercury have high thermal conductivities, and are
commonly used in applications that require high heat transfer rates. However,
they have very small Prandtl numbers, and thus the thermal boundary layer
develops much faster than the velocity boundary layer. Then we can assume
the velocity in the thermal boundary layer to be constant at the free stream
value and solve the energy equation. It gives
Nu ;t = 0.565(Re v Pr) 1
Pr < 0.05
(7-25)
It is desirable to have a single correlation that applies to all fluids, including
liquid metals. By curve-fitting existing data, Churchill and Ozoe (Ref. 3) pro-
posed the following relation which is applicable for all Prandtl numbers and
is claimed to be accurate to ± 1 % ,
Nu,
h x x _ 0.3387 Pr 1/3 Re' v
~T ~ [1 + (0.0468/Pr) 2/:
(7-26)
These relations have been obtained for the case of isothermal surfaces
but could also be used approximately for the case of nonisothermal surfaces
by assuming the surface temperature to be constant at some average value.
cen58 93 3_ch07.qxd 9/4/2002 12:12 PM Page 375
Also, the surfaces are assumed to be smooth, and the free stream to be turbu-
lent free. The effect of variable properties can be accounted for by evaluating
all properties at the film temperature.
Flat Plate with Unheated Starting Length
So far we have limited our consideration to situations for which the entire
plate is heated from the leading edge. But many practical applications involve
surfaces with an unheated starting section of length £, shown in Figure 7-11,
and thus there is no heat transfer for < x < £. In such cases, the velocity
boundary layer starts to develop at the leading edge (x = 0), but the thermal
boundary layer starts to develop where heating starts (x = £).
Consider a flat plate whose heated section is maintained at a constant tem-
perature (T = T s constant for x > £). Using integral solution methods (see
Kays and Crawford, 1994), the local Nusselt numbers for both laminar and
turbulent flows are determined to be
Laminar:
Turbulent:
Nu v
Nu,
Nu
v(for£ = 0)
[i - {m m v
Nu
b((for£=0)
[1 - (£/x) 9 " ]'
0.332 Re»- 5 Pr" 3
[1 - (ilx) m V n
0.0296 Re°- 8 Pr 1/3
: [1 - (i/x) mo Y 19
(7-27)
(7-28)
for x > £. Note that for £ = 0, these Nu v relations reduce to Nu l(for ^ = 0) , which
is the Nusselt number relation for a flat plate without an unheated starting
length. Therefore, the terms in brackets in the denominator serve as correction
factors for plates with unheated starting lengths.
The determination of the average Nusselt number for the heated section of
a plate requires the integration of the local Nusselt number relations above,
which cannot be done analytically. Therefore, integrations must be done nu-
merically. The results of numerical integrations have been correlated for the
average convection coefficients [Thomas, (1977) Ref. 11] as
Laminar:
Turbulent:
h
2[1 - (£/x)™]
1 -i/L
h r= ,
, _ 5[1 - (j/xf 10 ] ,
11 ~ 4(1 -i/L) **='
(7-29)
(7-30)
375
CHAPTER 7
7'
Thermal boundary layer
Velocity boundary layer
FIGURE 7-11
Flow over a flat plate with an unheated
starting length.
The first relation gives the average convection coefficient for the entire
heated section of the plate when the flow is laminar over the entire plate. Note
that f or £ = it reduces to h L = 2h x = L , as expected. The second relation gives
the average convection coefficient for the case of turbulent flow over the en-
tire plate or when the laminar flow region is small relative to the turbulent
region.
Uniform Heat Flux
When a flat plate is subjected to uniform heat flux instead of uniform temper
ature, the local Nusselt number is given by
0.453 Re» 5 Pr" 3
0.0308 Re? 8 Pr 1 ' 3
Laminar:
Turbulent
Nu,
Nu r
(7-31)
(7-32)
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376
HEAT TRANSFER
These relations give values that are 36 percent higher for laminar flow and
4 percent higher for turbulent flow relative to the isothermal plate case. When
the plate involves an unheated starting length, the relations developed for the
uniform surface temperature case can still be used provided that Eqs. 7-31 and
7-32 are used for Nu l(for t = 0) in Eqs. 7-27 and 7-28, respectively.
When heat flux q s is prescribed, the rate of heat transfer to or from the plate
and the surface temperature at a distance x are determined from
Q = iA s
(7-33)
and
4s = h x [T s (x) - rj ->
where A, is the heat transfer surface area.
TJx) = r.
(7-34)
r„=60°c
V=2m/s
r s =20 o c
\ \\\\\\V\\\\\\\\\\\\T\\\\\\\\\\\\\\\\\\V\I
L = 5 m
FIGURE 7-12
Schematic for Example 7-1.
■
EXAMPLE 7-1 Flow of Hot Oil over a Flat Plate
Engine oil at 60°C flows over the upper surface of a 5-m-long flat plate whose
temperature is 20°C with a velocity of 2 m/s (Fig. 7-12). Determine the total
drag force and the rate of heat transfer per unit width of the entire plate.
SOLUTION Engine oil flows over a flat plate. The total drag force and the rate
of heat transfer per unit width of the plate are to be determined.
Assumptions 1 The flow is steady and incompressible. 2 The critical Reynolds
number is Re cr = 5 X 10 5 .
Properties The properties of engine oil at the film temperature of T f = {T s + 7"J/
2 = (20 + 60V2 = 40°C are (Table A-14).
p = 876 kg/m 3
k = 0.144 W/m- °C
Pr = 2870
v = 242 X 10- 6 m 2 /s
Analysis Noting that L = 5 m, the Reynolds number at the end of the plate is
(2 m/s)(5 m)
Re,
v
0.242 X 10- 5 m 2 /s
4.13 X 10 4
which is less than the critical Reynolds number. Thus we have laminar flow over
the entire plate, and the average friction coefficient is
Q
1.328 X (4.13 X 10 3 )-
0.0207
Noting that the pressure drag is zero and thus C D = C, for a flat plate, the drag
force acting on the plate per unit width becomes
P T 2
c f As ~T
181 N
0.0207 X (5 X 1 m 2 )
(876 kg/m 3 )(2 m/s) 2
1 N
1 kg ■ m/s 2
The total drag force acting on the entire plate can be determined by multiplying
the value obtained above by the width of the plate.
This force per unit width corresponds to the weight of a mass of about 18 kg.
Therefore, a person who applies an equal and opposite force to the plate to keep
cen58 93 3_ch07.qxd 9/4/2002 12:12 PM Page 377
it from moving will feel like he or she is using as much force as is necessary to
hold a 18-kg mass from dropping.
Similarly, the Nusselt number is determined using the laminar flow relations
for a flat plate,
Nu
Then,
and
hL
k
0.664 Re£- 5 Pr 1/3
0.664 X (4.13 X 10 4 ) 05 X 2870 m = 1918
Nu
0.144 W/m ■ °C
5 m
(1918) = 55.2 W/m 2 • °C
Q = hA s (T x - T s ) = (55.2 W/m 2 • °C)(5 X 1 m 2 )(60 - 20)°C = 11,040 W
Discussion Note that heat transfer is always from the higher-temperature
medium to the lower-temperature one. In this case, it is from the oil to the
plate. The heat transfer rate is per m width of the plate. The heat transfer for
the entire plate can be obtained by multiplying the value obtained by the actual
width of the plate.
377
CHAPTER 7
■ EXAMPLE 7-2 Cooling of a Hot Block by Forced Air at High
Elevation
The local atmospheric pressure in Denver, Colorado (elevation 1610 m), is
83.4 kPa. Air at this pressure and 20°C flows with a velocity of 8 m/s over a
1.5 m X 6 m flat plate whose temperature is 140°C (Fig. 7-13). Determine the
rate of heat transfer from the plate if the air flows parallel to the (a) 6-m-long
side and (b) the 1.5-m side.
SOLUTION The top surface of a hot block is to be cooled by forced air. The
rate of heat transfer is to be determined for two cases.
Assumptions 1 Steady operating conditions exist. 2 The critical Reynolds num-
ber is Re cr = 5 X 10 5 . 3 Radiation effects are negligible. 4 Air is an ideal gas.
Properties The properties k, u,, C p , and Pr of ideal gases are independent of
pressure, while the properties v and a are inversely proportional to density and
thus pressure. The properties of air at the film temperature of T f = {T s + 7" x )/2
= (140 + 20)/2 = 80°C and 1 atm pressure are (Table A-15)
0.02953 W/m • °C
2.097 X 10- 5 m 2 /s
Pr = 0.7154
The atmospheric pressure in Denver is P = (83.4 kPa)/(101.325 kPa/atm) =
0.823 atm. Then the kinematic viscosity of air in Denver becomes
v = v @lam /P = (2.097 X 10- 5 m 2 /s)/0.823 = 2.548 X lO" 5 m 2 /s
Analysis (a) When air flow is parallel to the long side, we have L = 6 m, and
the Reynolds number at the end of the plate becomes
Re,
YL
v
m/s)(6 m)
2.548 X 10~ 5 m 2 /s
1.884 X 10 6
T. = 140°C
Air
FIGURE 7-13
Schematic for Example 7-2.
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378
HEAT TRANSFER
Air
20°C
8m/s
*-
: +
140°C
m
/
m
/ .
' / ■
14,300 '\
/ H
/ ^conv
V
V
6 m
^^
(a) Flow along the long side
/
Air
//
20°C
'//,
8m/s
/
V
140°C
/
/ I
/ Q
/ ^-conv
8670 W
V
6m
(6) Flow along the short side
FIGURE 7-14
The direction of fluid flow can
have a significant effect on
convection heat transfer.
which is greater than the critical Reynolds number. Thus, we have combined
laminar and turbulent flow, and the average Nusselt number for the entire plate
is determined to be
Nu
hL
- (0.037 Re, ' 8
k L
■ 871)Pr" 3
[0.037(1.884 X 10 6 ) 08 - 871]0.7154 1/3
2687
Then
■Nu
0.02953 W/m ■ °C
1" 6 m
wL = (1.5m)(6m) = 9 m 2
(2687) = 13.2 W/m 2 • °C
and
Q
hA s (T s - r„) = (13.2 W/m 2 ■ °C)(9 m 2 )(140 - 20)°C = 1.43 X 10 4 W
Note that if we disregarded the laminar region and assumed turbulent flow over
the entire plate, we would get Nu = 3466 from Eq. 7-22, which is 29 percent
higher than the value calculated above.
(b) When air flow is along the short side, we have L = 1.5 m, and the Reynolds
number at the end of the plate becomes
T£ (8m/s)(1.5m)
Re,
2.548 X 10~ 5 m 2 /s
4.71 X 10 5
which is less than the critical Reynolds number. Thus we have laminar flow over
the entire plate, and the average Nusselt number is
Nu
Then
hL
k
0.664 Re L 05 Pr" 3 = 0.664 X (4.71 X 10 5 ) 05 X 0.7154 1 ' 3 = 408
h
: Nu
0.02953 W/m
1.5 m
(408) = 8.03 W/m 2 • °C
and
Q = hA s (T s - r„) = (8.03 W/m 2 ■ °C)(9 m 2 )(140 - 20)°C = 8670 W
which is considerably less than the heat transfer rate determined in case (a).
Discussion Note that the direction of fluid flow can have a significant effect on
convection heat transfer to or from a surface (Fig. 7-14). In this case, we can
increase the heat transfer rate by 65 percent by simply blowing the air along the
long side of the rectangular plate instead of the short side.
EXAMPLE 7-3 Cooling of Plastic Sheets by Forced Air
The forming section of a plastics plant puts out a continuous sheet of plastic
that is 4 ft wide and 0.04 in. thick at a velocity of 30 ft/min. The temperature
of the plastic sheet is 200°F when it is exposed to the surrounding air, and a
2-ft-long section of the plastic sheet is subjected to air flow at 80°F at a veloc-
ity of 10 ft/s on both sides along its surfaces normal to the direction of motion
cen58 93 3_ch07.qxd 9/4/2002 12:12 PM Page 379
379
CHAPTER 7
of the sheet, as shown in Figure 7-15. Determine (a) the rate of heat transfer
from the plastic sheet to air by forced convection and radiation and (£>) the tem-
perature of the plastic sheet at the end of the cooling section. Take the density,
specific heat, and emissivity of the plastic sheet to be p = 75 lbm/ft 3 , C p = 0.4
Btu/lbm ■ °F, and s = 0.9.
SOLUTION Plastic sheets are cooled as they leave the forming section of
a plastics plant. The rate of heat loss from the plastic sheet by convection and
radiation and the exit temperature of the plastic sheet are to be determined.
Assumptions 1 Steady operating conditions exist. 2 The critical Reynolds num-
ber is Re cr = 5 X 10 5 . 3 Air is an ideal gas. 4 The local atmospheric pressure is
1 atm. 5 The surrounding surfaces are at the temperature of the room air.
Properties The properties of the plastic sheet are given in the problem state-
ment. The properties of air at the film temperature of T f = (T s + TJ/2 = (200
+ 80)/2 = 140°F and 1 atm pressure are (Table A-15E)
k = 0.01623 Btu/h ■ ft ■ °F Pr = 0.7202
v = 0.7344 ft 2 /h = 0.204 X lO" 3 ft 2 /s
Analysis (a) We expect the temperature of the plastic sheet to drop somewhat
as it flows through the 2-ft-long cooling section, but at this point we do not
know the magnitude of that drop. Therefore, we assume the plastic sheet to be
isothermal at 200°F to get started. We will repeat the calculations if necessary
to account for the temperature drop of the plastic sheet.
Noting that L = 4 ft, the Reynolds number at the end of the air flow across
the plastic sheet is
Re,
YL_ (10ft/s)(4ft)
~ lr ~ 0.204 X 10~ 3 ft 2 /s
1.961 X 10 5
which is less than the critical Reynolds number. Thus, we have laminar flow
over the entire sheet, and the Nusselt number is determined from the laminar
flow relations for a flat plate to be
hJ
Nu = ^= = 0.664 Re" 5 Pr" 3 = 0.664 X (1.961 X 10 5 ) 05 X (0.7202)" 3 = 263.6
Then,
u k M 0.01623 Btu/h • ft • °F .. ,, „ . „ n . . „ 2 oc
h =tNu = — (263.6) = 1.07 Btu/h ■ ft- • °F
L 4 ft
A s = (2 ft)(4 ft)(2 sides) = 16 ft 2
and
2co„v = hA s (T s - T^ )
= (1.07 Btu/h • ft 2 • °F)(16 ft 2 )(200 - 80)°F
= 2054 Btu/h
e„ d = evA s (T? - T s i r )
= (0.9)(0.1714 X 10- 8 Btu/h • ft 2 ■ R 4 )(16 ft 2 )[(660 R) 4 - (540 R) 4 ]
= 2584 Btu/h
200°F
Air
80°F, 10 ft/s
30 ft/min
FIGURE 7-15
Schematic for Example 7-3.
cen5 8 93 3_ch07.qxd 9/4/2002 12:12 PM Page 3E
380
HEAT TRANSFER
Therefore, the rate of cooling of the plastic sheet by combined convection and
radiation is
e,c
2co„v + firad = 2054 + 2584 = 4638 Btu/h
(b) To find the temperature of the plastic sheet at the end of the cooling sec-
tion, we need to know the mass of the plastic rolling out per unit time (or the
mass flow rate), which is determined from
m = pA c T plastic = (75 lbm/W 4 * °"° 4 ft 3 || ^ft/:
12
60
Then, an energy balance on the cooled section of the plastic sheet yields
Q = mC p (T 2 -T0
Q
mC„
Noting that Q is a negative quantity (heat loss) for the plastic sheet and substi-
tuting, the temperature of the plastic sheet as it leaves the cooling section is
determined to be
T, = 200°F
-4638 Btu/h
1 h
(0.5 lbm/s)(0.4 Btu/lbm ■ °F) \3600 s
193.6F
Discussion The average temperature of the plastic sheet drops by about 6.4°F
as it passes through the cooling section. The calculations now can be repeated
by taking the average temperature of the plastic sheet to be 196. 8°F instead of
200°F for better accuracy, but the change in the results will be insignificant be-
cause of the small change in temperature.
Stagnation
point
Separation
point
Boundary layer
FIGURE 7-16
Typical flow patterns in
cross flow over a cylinder.
7-3 - FLOW ACROSS CYLINDERS AND SPHERES
Flow across cylinders and spheres is frequently encountered in practice. For
example, the tubes in a shell-and-tube heat exchanger involve both internal
flow through the tubes and external flow over the tubes, and both flows must
be considered in the analysis of the heat exchanger. Also, many sports such as
soccer, tennis, and golf involve flow over spherical balls.
The characteristic length for a circular cylinder or sphere is taken to be the
external diameter D. Thus, the Reynolds number is defined as Re = VD/v
where T is the uniform velocity of the fluid as it approaches the cylinder or
sphere. The critical Reynolds number for flow across a circular cylinder or
sphere is about Re cr ~ 2 X 10 5 . That is, the boundary layer remains laminar
for about Re =£ 2 X 10 5 and becomes turbulent for Re^2X 10 5 .
Cross flow over a cylinder exhibits complex flow patterns, as shown in Fig-
ure 7-16. The fluid approaching the cylinder branches out and encircles the
cylinder, forming a boundary layer that wraps around the cylinder. The fluid
particles on the midplane strike the cylinder at the stagnation point, bringing
the fluid to a complete stop and thus raising the pressure at that point. The
pressure decreases in the flow direction while the fluid velocity increases.
At very low upstream velocities (Re ^ 1), the fluid completely wraps around
the cylinder and the two arms of the fluid meet on the rear side of the cylinder
cen58 93 3_ch07.qxd 9/4/2002 12:12 PM Page 381
381
CHAPTER 7
400
200
LOO
60
40
£3
l
0.6
0.4
0.2
O.l
0.06
10-'
Smoo
h
cylindc
r
Sphere
1
»■
l
*.
-
10° 10' 10 2 10 3 10 4 10 5 10 6
Re
FIGURE 7-17
Average drag coefficient for cross flow over a smooth circular
cylinder and a smooth sphere (from Schlichting, Ref. 10).
in an orderly manner. Thus, the fluid follows the curvature of the cylinder. At
higher velocities, the fluid still hugs the cylinder on the frontal side, but it is too
fast to remain attached to the surface as it approaches the top of the cylinder.
As a result, the boundary layer detaches from the surface, forming a separation
region behind the cylinder. Flow in the wake region is characterized by random
vortex formation and pressures much lower than the stagnation point pressure.
The nature of the flow across a cylinder or sphere strongly affects the total
drag coefficient C D . Both the friction drag and the pressure drag can be sig-
nificant. The high pressure in the vicinity of the stagnation point and the low
pressure on the opposite side in the wake produce a net force on the body in
the direction of flow. The drag force is primarily due to friction drag at low
Reynolds numbers (Re < 10) and to pressure drag at high Reynolds numbers
(Re > 5000). Both effects are significant at intermediate Reynolds numbers.
The average drag coefficients C D for cross flow over a smooth single circu-
lar cylinder and a sphere are given in Figure 7-17. The curves exhibit differ-
ent behaviors in different ranges of Reynolds numbers:
• For Re ^ 1 , we have creeping flow, and the drag coefficient decreases
with increasing Reynolds number. For a sphere, it is C D = 24/Re. There is
no flow separation in this regime.
• At about Re = 10, separation starts occurring on the rear of the body with
vortex shedding starting at about Re = 90. The region of separation
increases with increasing Reynolds number up to about Re = 10 3 . At this
point, the drag is mostly (about 95 percent) due to pressure drag. The drag
coefficient continues to decrease with increasing Reynolds number in this
range of 10 < Re < 10 3 . (A decrease in the drag coefficient does not
necessarily indicate a decrease in drag. The drag force is proportional to
the square of the velocity, and the increase in velocity at higher Reynolds
numbers usually more than offsets the decrease in the drag coefficient.)
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382
HEAT TRANSFER
Laminar
boundary —
layer \^
^Kv^^r
/';/
\ -7 C -v
Separation —
(a) Laminar flow (Re < 2 X 10 5 )
Laminar Transition
boundary
layer
Turbulent
boundary
layer
Separation
(6) Turbulence occurs (Re > 2 X 10 5 )
FIGURE 7-18
Turbulence delays flow separation.
• In the moderate range of 10 3 < Re < 10 5 , the drag coefficient remains
relatively constant. This behavior is characteristic of blunt bodies. The
flow in the boundary layer is laminar in this range, but the flow in the
separated region past the cylinder or sphere is highly turbulent with a
wide turbulent wake.
• There is a sudden drop in the drag coefficient somewhere in the range of
10 5 < Re < 10 6 (usually, at about 2 X 10 5 ). This large reduction in C D is
due to the flow in the boundary layer becoming turbulent, which moves
the separation point further on the rear of the body, reducing the size of
the wake and thus the magnitude of the pressure drag. This is in contrast
to streamlined bodies, which experience an increase in the drag
coefficient (mostly due to friction drag) when the boundary layer
becomes turbulent.
Flow separation occurs at about ~ 80° (measured from the stagnation
point) when the boundary layer is laminar and at about 6 ~ 140° when it is
turbulent (Fig. 7-18). The delay of separation in turbulent flow is caused by
the rapid fluctuations of the fluid in the transverse direction, which enables the
turbulent boundary layer to travel further along the surface before separation
occurs, resulting in a narrower wake and a smaller pressure drag. In the range
of Reynolds numbers where the flow changes from laminar to turbulent, even
the drag force F D decreases as the velocity (and thus Reynolds number) in-
creases. This results in a sudden decrease in drag of a flying body and insta-
bilities in flight.
Effect of Surface Roughness
We mentioned earlier that surface roughness, in general, increases the drag
coefficient in turbulent flow. This is especially the case for streamlined bodies.
For blunt bodies such as a circular cylinder or sphere, however, an increase in
the surface roughness may actually decrease the drag coefficient, as shown in
Figure 7-19 for a sphere. This is done by tripping the flow into turbulence at
a lower Reynolds number, and thus causing the fluid to close in behind the
body, narrowing the wake and reducing pressure drag considerably. This
results in a much smaller drag coefficient and thus drag force for a rough-
surfaced cylinder or sphere in a certain range of Reynolds number compared
to a smooth one of identical size at the same velocity. At Re = 10 5 , for exam-
ple, C D = 0.1 for a rough sphere with elD = 0.0015, whereas C D = 0.5 for
a smooth one. Therefore, the drag coefficient in this case is reduced by a fac-
tor of 5 by simply roughening the surface. Note, however, that at Re = 10 6 ,
C D = 0.4 for the rough sphere while C D = 0.1 for the smooth one. Obviously,
roughening the sphere in this case will increase the drag by a factor of 4
(Fig. 7-20).
The discussion above shows that roughening the surface can be used to
great advantage in reducing drag, but it can also backfire on us if we are not
careful — specifically, if we do not operate in the right range of Reynolds num-
ber. With this consideration, golf balls are intentionally roughened to induce
turbulence at a lower Reynolds number to take advantage of the sharp drop in
the drag coefficient at the onset of turbulence in the boundary layer (the typi-
cal velocity range of golf balls is 15 to 150 m/s, and the Reynolds number
is less than 4 X 10 5 ). The critical Reynolds number of dimpled golf balls is
cen58 93 3_ch07.qxd 9/4/2002 12:12 PM Page 383
383
CHAPTER 7
*?
'<-
ii
Q
u
4xl0 6
FIGURE 7-1 9
The effect of surface roughness on the drag coefficient of a sphere (from Blevins, Ref. 1).
about 4 X 10 4 . The occurrence of turbulent flow at this Reynolds number
reduces the drag coefficient of a golf ball by half, as shown in Figure 7-19.
For a given hit, this means a longer distance for the ball. Experienced golfers
also give the ball a spin during the hit, which helps the rough ball develop a
lift and thus travel higher and further. A similar argument can be given for a
tennis ball. For a table tennis ball, however, the distances are very short, and
the balls never reach the speeds in the turbulent range. Therefore, the surfaces
of table tennis balls are made smooth.
Once the drag coefficient is available, the drag force acting on a body
in cross flow can be determined from Eq. 7-1 where A is the frontal area
(A = LD for a cylinder of length L and A = ttD 2 /4 for a sphere). It should be
kept in mind that the free-stream turbulence and disturbances by other bodies
in flow (such as flow over tube bundles) may affect the drag coefficients
significantly.
Re
Co
Smooth
surface
Rough surface,
elD= 0.0015
10 5
10 6
0.5
0.1
0.1
0.4
FIGURE 7-20
Surface roughness may increase
or decrease the drag coefficient
of a spherical object, depending on
the value of the Reynolds number.
m EXAMPLE 7-4 Drag Force Acting on a Pipe in a River
A 2.2-cm-outer-diameter pipe is to cross a river at a 30-m-wide section while
being completely immersed in water (Fig. 7-21). The average flow velocity of
water is 4 m/s and the water temperature is 15°C. Determine the drag force ex-
erted on the pipe by the river.
SOLUTION A pipe is crossing a river. The drag force that acts on the pipe is to
be determined.
Assumptions 1 The outer surface of the pipe is smooth so that Figure 7-17 can
be used to determine the drag coefficient. 2 Water flow in the river is steady.
3 The direction of water flow is normal to the pipe. 4 Turbulence in river flow is
not considered.
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384
HEAT TRANSFER
FIGURE 7-21
Schematic for Example 7-4.
800
700
600
500
400
300
200
100
40° 80° 120°
9 from stagnation point
FIGURE 7-22
Variation of the local
heat transfer coefficient along the
circumference of a circular cylinder in
cross flow of air (from Giedt, Ref. 5).
Properties The density and dynamic viscosity of water at 15°C are p =
999.1 kg/m 3 and u, = 1.138 X 1CT 3 kg/m • s (Table A-9).
Analysis Noting that D = 0.022 m, the Reynolds number for flow over the
pipe is
Re
YD
v
pYD (999.1 kg/m 3 )(4 m/s)(0.022 m)
&
1.138 X 10- 3 kg/m • s
7.73 X 10 4
The drag coefficient corresponding to this value is, from Figure 7-17, C D = 1.0.
Also, the frontal area for flow past a cylinder is A = LD. Then the drag force act-
ing on the pipe becomes
C n A
P T 2
1.0(30 X 0.022 m 2 )
(999.1 kg/m 3 )(4 m/s) 2
IN
1 kg • m/s-
5275 N
Discussion Note that this force is equivalent to the weight of a mass over 500
kg. Therefore, the drag force the river exerts on the pipe is equivalent to hanging
a total of over 500 kg in mass on the pipe supported at its ends 30 m apart. The
necessary precautions should be taken if the pipe cannot support this force.
Heat Transfer Coefficient
Flows across cylinders and spheres, in general, involve flow separation,
which is difficult to handle analytically. Therefore, such flows must be studied
experimentally or numerically. Indeed, flow across cylinders and spheres has
been studied experimentally by numerous investigators, and several empirical
correlations have been developed for the heat transfer coefficient.
The complicated flow pattern across a cylinder greatly influences heat
transfer. The variation of the local Nusselt number Nu 9 around the periphery
of a cylinder subjected to cross flow of air is given in Figure 7-22. Note that,
for all cases, the value of Nu e starts out relatively high at the stagnation point
(0 = 0°) but decreases with increasing as a result of the thickening of the
laminar boundary layer. On the two curves at the bottom corresponding to
Re = 70,800 and 101,300, Nu 9 reaches a minimum at ~ 80°, which is the
separation point in laminar flow. Then Nu e increases with increasing as a re-
sult of the intense mixing in the separated flow region (the wake). The curves
cen58 93 3_ch07.qxd 9/4/2002 12:12 PM Page 385
at the top corresponding to Re = 140,000 to 219,000 differ from the first two
curves in that they have two minima for Nu e . The sharp increase in Nu 9 at
about ~ 90° is due to the transition from laminar to turbulent flow. The later
decrease in Nu 9 is again due to the thickening of the boundary layer. Nu 9
reaches its second minimum at about 6 ~ 140°, which is the flow separation
point in turbulent flow, and increases with as a result of the intense mixing
in the turbulent wake region.
The discussions above on the local heat transfer coefficients are insightful;
however, they are of little value in heat transfer calculations since the
calculation of heat transfer requires the average heat transfer coefficient over
the entire surface. Of the several such relations available in the literature for
the average Nusselt number for cross flow over a cylinder, we present the one
proposed by Churchill and Bernstein:
5/8^4/5
Nil
cyl
ItD
k
0.3
0.62 Re 1/2 Pr 1
[1 + (0.4/Pr) 2 ' 3 ]
2/31 1/4
1 +
/ Re
1282,000
(7-35)
This relation is quite comprehensive in that it correlates available data well for
Re Pr > 0.2. The fluid properties are evaluated at the film tempera-
ture T,
f
\(Xm
+ T s ), which is the average of the free-stream and surface
temperatures.
For flow over a sphere, Whitaker recommends the following comprehen-
sive correlation:
Nu
sph
liD
k
2 + [0.4 Re" 2 + 0.06 Re 2/3 ] Pr°
1/4
(7-36)
385
CHAPTER 7
which is valid for 3.5 < Re < 80,000 and 0.7 < Pr < 380. The fluid proper-
ties in this case are evaluated at the free-stream temperature T ra , except for u^,
which is evaluated at the surface temperature T s . Although the two relations
above are considered to be quite accurate, the results obtained from them can
be off by as much as 30 percent.
The average Nusselt number for flow across cylinders can be expressed
compactly as
Nu
cyl
liD
k
C Re'" Pr"
(7-37)
where n = | and the experimentally determined constants C and m are given
in Table 7-1 for circular as well as various noncircular cylinders. The charac-
teristic length D for use in the calculation of the Reynolds and the Nusselt
numbers for different geometries is as indicated on the figure. All fluid prop-
erties are evaluated at the film temperature.
The relations for cylinders above are for single cylinders or cylinders ori-
ented such that the flow over them is not affected by the presence of others.
Also, they are applicable to smooth surfaces. Surface roughness and the free-
stream turbulence may affect the drag and heat transfer coefficients signifi-
cantly. Eq. 7-37 provides a simpler alternative to Eq. 7-35 for flow over
cylinders. However, Eq. 7-35 is more accurate, and thus should be preferred
in calculations whenever possible.
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HEAT TRANSFER
T = 110°C
Wind
Y = 8 m/s
FIGURE 7-23
Schematic for Example 7-5.
TABLE 7-1
Empirical correlations for the average Nusselt number for forced convection
over circular and noncircular cylinders in cross flow (from Zukauskas, Ref. 14,
and Jakob, Ref. 6)
Cross-section
of the cylinder
Circle
Square
Square
(tilted
45°)
Hexagon
Hexagon
(tilted
45°)
Vertical
plate
Ellipse
Fluid
Gas or
liquid
Gas
Gas
Gas
Gas
Gas
Gas
Range of Re
0.4-4
4-40
40-4000
4000-40,000
40,000-400,000
5000-100,000
5000-100,000
5000-100,000
5000-19,500
19,500-100,000
4000-15,000
2500-15,000
Nusselt number
Nu = 0.989Re 0330 Pr 1/3
Nu = 0.91 IRe 0385 Pr 1/3
Nu = 0.683Re 0466 Pr 1/3
Nu = 0.193Re 0518 Pr 1 ' 3
Nu = 0.027Re - 805 Pr 1/3
Nu = 0.102Re°
Pr 1
Nu = 0.246Re°
Pr 1
Nu = 0.153Re°
Pr 1
Nu = 0.160Re 0638 Pr 1 ' 3
Nu = 0.0385Re 0782 Pr 1/3
Nu = 0.228Re
0.731 p r l/3
Pr 1
Nu = 0.248Re 0612 Pr 1/3
EXAMPLE 7-5 Heat Loss from a Steam Pipe in Windy Air
A long 10-cm-diameter steam pipe whose external surface temperature is
110°C passes through some open area that is not protected against the winds
(Fig. 7-23). Determine the rate of heat loss from the pipe per unit of its length
cen58 93 3_ch07.qxd 9/4/2002 12:12 PM Page 387
when the air is at 1 atm pressure and 10°C and the wind is blowing across the
pipe at a velocity of 8 m/s.
SOLUTION A steam pipe is exposed to windy air. The rate of heat loss from the
steam is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Radiation effects are neg-
ligible. 3 Air is an ideal gas.
Properties The properties of air at the average film temperature of T f =
(T s + 7J/2 = (110 + 10)/2 = 60°C and 1 atm pressure are (Table A-15)
k = 0.02808 W/m • °C
v = 1.896 X 10- 5 m 2 /s
Pr = 0.7202
Analysis The Reynolds number is
(8m/s)(0.1 m)
Re
YD
v
1.896 X 10- 5 m 2 /s
The Nusselt number can be determined from
Nu
and
hD
k
0.3 +
124
h ~-
0.3
0.62 Re" 2 Pr 1 '
[1 + (0.4/Pr)
2/3-11/4
4.219 X 10 4
Re
282,000
0.62(4.219 X 10 4 )" 2 (0.7202) 1 '
[1 + (0.4/0.7202) 2 ' 3 ]" 4
1 +
4.219 X 10'
282,000
-Nu
0.02808 W/m ■ °C
(124) = 34.8W/m 2 -°C
D 0.1m
Then the rate of heat transfer from the pipe per unit of its length becomes
A s = pL = ttDL = ir(0.1 m)(l m) = 0.314 m 2
Q = hA s (T s - 7/„) = (34.8 W/m 2 • C)(0.314 m 2 )(110 - 10)°C = 1093 W
The rate of heat loss from the entire pipe can be obtained by multiplying the
value above by the length of the pipe in m.
Discussion The simpler Nusselt number relation in Table 7-1 in this case
would give Nu = 128, which is 3 percent higher than the value obtained above
using Eq. 7-35.
387
CHAPTER 7
EXAMPLE 7-6 Cooling of a Steel Ball by Forced Air
A 25-cm-diameter stainless steel ball (p = 8055 kg/m 3 , C p = 480 J/kg • °C) is
removed from the oven at a uniform temperature of 300°C (Fig. 7-24). The ball
is then subjected to the flow of air at 1 atm pressure and 25 C C with a velocity
of 3 m/s. The surface temperature of the ball eventually drops to 200°C. Deter-
mine the average convection heat transfer coefficient during this cooling
process and estimate how long the process will take.
Air
= 25°C
= 3 m/s
FIGURE 7-24
Schematic for Example 7-6.
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388
HEAT TRANSFER
SOLUTION A hot stainless steel ball is cooled by forced air. The average con-
vection heat transfer coefficient and the cooling time are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Radiation effects are neg-
ligible. 3 Air is an ideal gas. 4 The outer surface temperature of the ball is uni-
form at all times. 5 The surface temperature of the ball during cooling is
changing. Therefore, the convection heat transfer coefficient between the ball
and the air will also change. To avoid this complexity, we take the surface tem-
perature of the ball to be constant at the average temperature of (300 + 200)/2
= 250°C in the evaluation of the heat transfer coefficient and use the value ob-
tained for the entire cooling process.
Properties The dynamic viscosity of air at the average surface temperature is
M-s = l JL e25o°c = 2.76 X 10~ 5 kg/m • s. The properties of air at the free-stream
temperature of 25°C and 1 atm are (Table A-15)
k = 0.02551 W/m • °C
p. = 1.849 X 10- 5 kg/m-s
v
Pr
1.562 X 10-
0.7296
' m 2 /s
Analysis The Reynolds number is determined from
(3 m/s)(0.25 m)
Re
YD
v
1.562 X 10~ 5 m 2 /s
4.802 X 10 4
The Nusselt number is
hD
Nu
2 + [0.4 Re" 2 + 0.06 Re 2 ' 3 ] Pr°
m,
= 2 + [0.4(4.802 X 10 4 ) 1 ' 2 + 0.06(4.802 X 10 4 ) 2/3 ](0.7296) 04
/ 1.849 X 10- 5 \" 4
X \ 2.76 X 10~
= 135
Then the average convection heat transfer coefficient becomes
h = ^ Nu
0.02551 W/m ■ °C
0.25 m
(135) = 13.8 W/m 2 - °C
In order to estimate the time of cooling of the ball from 300°C to 200°C, we de-
termine the average rate of heat transfer from Newton's law of cooling by using
the average surface temperature. That is,
A s = ttD 2 = tt(0.25 m) 2 = 0.1963 m 2
g ave = hA£T, >mc - T x ) = (13.8 W/m 2 • °C)(0.1963 m 2 )(250 - 25)°C = 610 W
Next we determine the total heat transferred from the ball, which is simply the
change in the energy of the ball as it cools from 300°C to 200°C:
m = pV = p|irD 3 = (8055 kg/m 3 ) |tt(0.25 m) 3 = 65.9 kg
g,otai = mC p (T 2 - T x ) = (65.9 kg)(480 J/kg ■ °C)(300 - 200)°C = 3,163,000 J
In this calculation, we assumed that the entire ball is at 200°C, which is not
necessarily true. The inner region of the ball will probably be at a higher tem-
perature than its surface. With this assumption, the time of cooling is deter-
mined to be
cen58 93 3_ch07.qxd 9/4/2002 12:12 PM Page 389
389
CHAPTER 7
Af
Q
2a
3,163,000 J
610 J/s
5185 s = lh26min
Discussion The time of cooling could also be determined more accurately us-
ing the transient temperature charts or relations introduced in Chapter 4. But
the simplifying assumptions we made above can be justified if all we need is a
ballpark value. It will be naive to expect the time of cooling to be exactly 1 h 26
min, but, using our engineering judgment, it is realistic to expect the time of
cooling to be somewhere between one and two hours.
7^ ■ FLOW ACROSS TUBE BANKS
Cross-flow over tube banks is commonly encountered in practice in heat
transfer equipment such as the condensers and evaporators of power plants,
refrigerators, and air conditioners. In such equipment, one fluid moves
through the tubes while the other moves over the tubes in a perpendicular
direction.
In a heat exchanger that involves a tube bank, the tubes are usually placed
in a shell (and thus the name shell-and-tube heat exchanger), especially when
the fluid is a liquid, and the fluid flows through the space between the tubes
and the shell. There are numerous types of shell-and-tube heat exchangers,
some of which are considered in Chap. 13. In this section we will consider the
general aspects of flow over a tube bank, and try to develop a better and more
intuitive understanding of the performance of heat exchangers involving a
tube bank.
Flow through the tubes can be analyzed by considering flow through a sin-
gle tube, and multiplying the results by the number of tubes. This is not the
case for flow over the tubes, however, since the tubes affect the flow pattern
and turbulence level downstream, and thus heat transfer to or from them, as
shown in Figure 7-25. Therefore, when analyzing heat transfer from a tube
bank in cross flow, we must consider all the tubes in the bundle at once.
The tubes in a tube bank are usually arranged either in-line or staggered in
the direction of flow, as shown in Figure 7-26. The outer tube diameter D is
taken as the characteristic length. The arrangement of the tubes in the tube
bank is characterized by the transverse pitch S T , longitudinal pitch S L , and the
diagonal pitch S D between tube centers. The diagonal pitch is determined from
Vsi + (s T /2y-
(7-38)
As the fluid enters the tube bank, the flow area decreases from A { = S T L to
A T = (S T — D)L between the tubes, and thus flow velocity increases. In staggered
arrangement, the velocity may increase further in the diagonal region if the
tube rows are very close to each other. In tube banks, the flow characteristics
are dominated by the maximum velocitiy T max that occurs within the tube
bank rather than the approach velocity T. Therefore, the Reynolds number is
defined on the basis of maximum velocity as
Rer
.D
(7-39)
FIGURE 7-25
Flow patterns for staggered and
in-line tube banks (photos by
R. D. Willis, Ref 12).
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390
HEAT TRANSFER
Y j
1 st row 2nd row 3rd row
(a) In-line
A T =(S T -D)L
A D = (S D -D)L
(b) Staggered
FIGURE 7-26
Arrangement of the tubes in in-line
and staggered tube banks (A b A T , and
A D are flow areas at indicated
locations, and L is the length of the
tubes).
The maximum velocity is determined from the conservation of mass re-
quirement for steady incompressible flow. For in-line arrangement, the maxi-
mum velocity occurs at the minimum flow area between the tubes, and the
conservation of mass can be expressed as (see Fig. l-26a) pTA, = p T max A r
or YS T = T max (»Sy — D). Then the maximum velocity becomes
y
D
T
(7-40)
In staggered arrangement, the fluid approaching through area A { in Fig-
ure l-26b passes through area A T and then through area 2A D as it wraps
around the pipe in the next row. If 2A D > A T , maximum velocity will still oc-
cur at A T between the tubes, and thus the T max relation Eq. 7-40 can also be
used for staggered tube banks. But if 2A D < A T [or, if 2{S D - D) < (S T - £>)],
maximum velocity will occur at the diagonal cross sections, and the maximum
velocity in this case becomes
Staggered and S D < (S T + D)/2:
2(S D - D)
Y
(7-41)
since p TA, = P T m „(2A ) or YS T = 2Y max (S D - D).
The nature of flow around a tube in the first row resembles flow over a sin-
gle tube discussed in section 7-3, especially when the tubes are not too close
to each other. Therefore, each tube in a tube bank that consists of a single
transverse row can be treated as a single tube in cross-flow. The nature of flow
around a tube in the second and subsequent rows is very different, however,
because of wakes formed and the turbulence caused by the tubes upstream.
The level of turbulence, and thus the heat transfer coefficient, increases with
row number because of the combined effects of upstream rows. But there is no
significant change in turbulence level after the first few rows, and thus the
heat transfer coefficient remains constant.
Flow through tube banks is studied experimentally since it is too complex
to be treated analytically. We are primarily interested in the average heat trans-
fer coefficient for the entire tube bank, which depends on the number of tube
rows along the flow as well as the arrangement and the size of the tubes.
Several correlations, all based on experimental data, have been proposed for
the average Nusselt number for cross flow over tube banks. More recently,
Zukauskas has proposed correlations whose general form is
Nu f
hD
k
C Reg Pr"(Pr/Pr s )°
(7-42)
where the values of the constants C, m, and n depend on value Reynolds num-
ber. Such correlations are given in Table 7-2 explicitly for 0.7 < Pr < 500 and
< Re D < 2 X 10 6 . The uncertainty in the values of Nusselt number obtained
from these relations is ±15 percent. Note that all properties except Pr s are to
be evaluated at the arithmetic mean temperature of the fluid determined from
(7-43)
where T t and T e are the fluid temperatures at the inlet and the exit of the tube
bank, respectively.
cen58 93 3_ch07.qxd 9/4/2002 12:12 PM Page 391
TABLE 7-2
Nusselt number correlations for cross flow over tube banks for N > 16 and
0.7 < Pr < 500 (from Zukauskas, Ref. 15, 1987)*
Arrangement
Range of Re D
Correlation
In-line
0-100
Nu D = 0.9 Reg 4 Pr 036 (Pr/Pr s ) - 25
100-1000
Nu D = 0.52 Reg 5 Pr 036 (Pr/Pr s ) 025
1000-2 X 10 5
Nu D = 0.27 Reg 63 Pr 035 (Pr/Pr s ) 025
2 X 10 5 -2 X 10 6
Nu D = 0.033 Reg 8 Pr 04 (Pr/Pr s ) 025
Staggered
0-500
Nu D = 1.04 Reg 4 Pr a36 (Pr/Pr s ) 025
500-1000
Nu D = 0.71 Reg 5 Pr 036 (Pr/Pr s ) 025
1000-2 X 10 5
Nu D = 0.35(S T /S L ) 02 Reg 5 Pr°- 36 (Pr/Pr s ) 025
2 X 10 5 -2 X 10 6
Nu D = 0.031(S T /S L ) - 2 Reg' 8 Pr 035 (Pr/Pr s ) - 25
*AII properties except Pr s are to be evaluated at the arithmetic mean of the inlet and outlet temperatures
of the fluid (Pr. is to be evaluated at 7",).
391
CHAPTER 7
The average Nusselt number relations in Table 7-2 are for tube banks with
16 or more rows. Those relations can also be used for tube banks with N L pro-
vided that they are modified as
Nu
D,N L
FNu r
(7-44)
where F is a correction factor F whose values are given in Table 7-3. For
Re D > 1000, the correction factor is independent of Reynolds number.
Once the Nusselt number and thus the average heat transfer coefficient for
the entire tube bank is known, the heat transfer rate can be determined from
Newton's law of cooling using a suitable temperature difference AT. The first
thought that comes to mind is to use AT = T s — T m = T s — (T, + T e )/2. But
this will, in general, over predict the heat transfer rate. We will show in the
next chapter that the proper temperature difference for internal flow (flow
over tube banks is still internal flow through the shell) is the logarithmic
mean temperature difference AT ln defined as
Ar,„
(T, - T e ) - (T s - Tf)
AT, - AT
ln[(7/ s - T e )l(T s - 7V)] ln(A7;/A7/,.)
(7-45)
We will also show that the exit temperature of the fluid T e can be determined
from
TABLE 7-3
Correction factor Fto be used in Nu D Nl , = FNu D for N L < 16 and Re D > 1000
(from Zukauskas, Ref 15, 1987).
N L
1
2
3
4
5
7
10
13
In-line
0.70
0.80
0.86
0.90
0.93
0.96
0.98
0.99
Staggered
0.64
0.76
0.84
0.89
0.93
0.96
0.98
0.99
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HEAT TRANSFER
/ A s h\
T m = T,-Cr,-Tdexp[-j^-\ (7-46)
where A s = NirDL is the heat transfer surface area and m = p°V(N T S T L) is the
mass flow rate of the fluid. Here N is the total number of tubes in the bank, N T
is the number of tubes in a transverse plane, L is the length of the tubes, and
T is the velocity of the fluid just before entering the tube bank. Then the heat
transfer rate can be determined from
Q = hA s AT ]a = mC p {T e - T t ) (7-47)
The second relation is usually more convenient to use since it does not require
the calculation of AT ln .
Pressure Drop
Another quantity of interest associated with tube banks is the pressure drop
AP, which is the difference between the pressures at the inlet and the exit of
the tube bank. It is a measure of the resistance the tubes offer to flow over
them, and is expressed as
P T
AP = N L f X '—^ (7-48)
where/is the friction factor and x is the correction factor, both plotted in Fig-
ures 1-21 a and 1-21 b against the Reynolds number based on the maximum
velocity T max . The friction factor in Figure 1-21 a is for a square in-line tube
bank (S T = S L ), and the correction factor given in the insert is used to account
for the effects of deviation of rectangular in-line arrangements from square
arrangement. Similarly, the friction factor in Figure 7-27 b is for an equilateral
staggered tube bank (S T = S D ), and the correction factor is to account for the
effects of deviation from equilateral arrangement. Note that x — 1 for both
square and equilateral triangle arrangements. Also, pressure drop occurs in the
flow direction, and thus we used N L (the number of rows) in the AP relation.
The power required to move a fluid through a tube bank is proportional to
the pressure drop, and when the pressure drop is available, the pumping power
required can be determined from
W = VAP = -^= i - (7-49)
"pump ruI n v ^**#
where V = °V(N T S T L) is the volume flow rate and m = pV = p°V(N T S T L) is
the mass flow rate of the fluid through the tube bank. Note that the power re-
quired to keep a fluid flowing through the tube bank (and thus the operating
cost) is proportional to the pressure drop. Therefore, the benefits of enhancing
heat transfer in a tube bank via rearrangement should be weighed against the
cost of additional power requirements.
In this section we limited our consideration to tube banks with base surfaces
(no fins). Tube banks with finned surfaces are also commonly used in prac-
tice, especially when the fluid is a gas, and heat transfer and pressure drop cor-
relations can be found in the literature for tube banks with pin fins, plate fins,
strip fins, etc.
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CHAPTER 7
X
P L = S L ID Q-Q
P T = S T /D ®^
1 ' p - )
Ml 1
S T
10
6
X
2
1
0.6
0.2
T
-.KI--
-1 5 s
1
SC-io
-
10
S s Re Dm:v = 10- 1
-111 1 '
.
= 1.7
\^L -
2
2.0 s
(F T - mr L - i)
n ?
■■
n i
3.0
2.5
(«)
4 6 8|Ql 2 4 6 8jg2 2 4 6 8jq3 2 4 6 8JQ4 2 4 6 8jq5 2 4 6 8jq6
Re n
In-line arrangement
D.max
.6
Rp fl = 10- /
.4
II
5
/ IUJ y
-III
II-
.0
I0 2
a
J
4 6 8jq1 2 4 6 8jq2 2 4 6 8jq3 2 4 6
Re r
(fc) Staggered arrangement
10 4 2
no 3
Ho 6
D.max
FIGURE 7-27
Friction factor/and correction
factor x for tube banks (from
Zukauskas, Ref. 16, 1985).
EXAMPLE 7-7 Preheating Air by Geothermal Water in a Tube
Bank
In an industrial facility, air is to be preheated before entering a furnace by geo-
thermal water at 120°C flowing through the tubes of a tube bank located in a
duct. Air enters the duct at 20°C and 1 atm with a mean velocity of 4.5 m/s,
and flows over the tubes in normal direction. The outer diameter of the tubes is
1.5 cm, and the tubes are arranged in-line with longitudinal and transverse
pitches of S L = S T = 5 cm. There are 6 rows in the flow direction with 10 tubes
in each row, as shown in Figure 7-28. Determine the rate of heat transfer per
unit length of the tubes, and the pressure drop across the tube bank.
SOLUTION Air is heated by geothermal water in a tube bank. The rate of heat
transfer to air and the pressure drop of air are to be determined.
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HEAT TRANSFER
AIR
F= 4.5 m/s
f
T s = 120°C
T. = 20°C
-* O
o o
o o o
—
o
o o
o o o
—
o
o o
o o o
—
o
o o
o o o
—
o
o o
o o o
—
o
o o
o o o
— -
o
o o
o o o
— ►
o
o o
o o o
T o o o o o o
-k) o o o o
s,
- 5 cm D= 1.5 cm
FIGURE 7-28
Schematic for Example 7-7.
Assumptions 1 Steady operating conditions exist. 2 The surface temperature of
the tubes is equal to the temperature of geothermal water.
Properties The exit temperature of air, and thus the mean temperature, is not
known. We evaluate the air properties at the assumed mean temperature of
60°C (will be checked later) and 1 atm are Table A-15):
k = 0.02808 W/m • K, p = 1.06 kg/m 3
C p = 1 .007 kJ/kg • K, Pr = 0.7202
|x = 2.008 X 10~ 5 kg/m ■ s Pr, = Pr 87i = 0.7073
Also, the density of air at the inlet temperature of 20°C (for use in the mass flow
rate calculation at the inlet) is p x = 1.204 kg/m 3
Analysis It is given that D = 0.015 m, S L = S T = 0.05 m, and V = 4.5 m/s.
Then the maximum velocity and the Reynolds number based on the maximum
velocity become
D
•Y
0.05
0.05 - 0.015
(4.5 m/s) = 6.43 m/s
Re r
pY max D (1.06 kg/m 3 )(6.43 m/s)(0.015 m)
V
2.008 X 10~ 5 kg/m ■ s
5091
The average Nusselt number is determined using the proper relation from Table
7-2 to be
0.27(5091)°' 63 (0.7202) 036 (0.7202/0.7073)
52.2
This Nusselt number is applicable to tube banks with N L > 16. In our case, the
number of rows is N L = 6, and the corresponding correction factor from Table
7-3 is F = 0.945. Then the average Nusselt number and heat transfer coeffi-
cient for all the tubes in the tube bank become
Nu,
FNu D =
D
(0.945)(52.2) = 49.3
_ 49.3(0.02808 W/m • °C)
~ 0.015 m
92.2 W/m 2 -°C
The total number of tubes is N = N L X N T = 6 X 10 = 60. For a unit tube
length (L = 1 m), the heat transfer surface area and the mass flow rate of air
(evaluated at the inlet) are
A s = NttDL = 60tt(0.015 m)(l m) = 2.827 m 2
m = m { = p { Y{N T S T L)
= (1.204 kg/m 3 )(4.5 m/s)(10)(0.05 m)(l m) = 2.709 kg/s
Then the fluid exit temperature, the log mean temperature difference, and the
rate of heat transfer become
T e = T s - (T s - T t ) exp
mC,
(2.827 m 2 )(92.2 W/m 2 • °C)\
, l- ( 2.709kg/s) ( 1007J/k g .°C) ' =
cen58 93 3_ch07.qxd 9/4/2002 12:12 PM Page 395
Ar ln
(T, - T e ) - (T s - T { ) _ (120 - 29.11) - (120 - 20)
95.4°C
ln[(T s - T C )/(T S - Tj)] ln[(120 - 29.1 1)/(120 - 20)]
Q = hA s AT lB = (92.2 W/m 2 • °C)(2.827 m 2 )(95.4°C) = 2.49 X 10 4 W
The rate of heat transfer can also be determined in a simpler way from
Q =HAAT m = mC p (T e -T i )
= (2.709 kg/s)(1007 J/kg ■ °C)(29. 11 - 20)°C = 2.49 X 10 4 W
For this square in-line tube bank, the friction coefficient corresponding to
Re D = 5088 and S L /D = 5/1.5 = 3.33 is, from Fig. 7-27a, f = 0.16. Also,
X = 1 for the square arrangements. Then the pressure drop across the tube
bank becomes
A/> = N L f x
p^L
6(0.16)(1)
(1.06kg/m 3 )(6.43m/s) 3
IN
1 kg • m/s :
21 Pa
Discussion The arithmetic mean fluid temperature is {T, + T e )/2 = (20 +
110.9V2 = 65.4°C, which is fairly close to the assumed value of 60°C. There-
fore, there is no need to repeat calculations by reevaluating the properties at
65.4°C (it can be shown that doing so would change the results by less
than 1 percent, which is much less than the uncertainty in the equations and
the charts used).
395
CHAPTER 7
TOPIC OF SPECIAL INTEREST
Reducing Heat Transfer through Surfaces: Thermal Insulation
Thermal insulations are materials or combinations of materials that are
used primarily to provide resistance to heat flow (Fig. 7-29). You are prob-
ably familiar with several kinds of insulation available in the market. Most
insulations are heterogeneous materials made of low thermal conductivity
materials, and they involve air pockets. This is not surprising since air has
one of the lowest thermal conductivities and is readily available. The Sty-
rofoam commonly used as a packaging material for TVs, VCRs, comput-
ers, and just about anything because of its light weight is also an excellent
insulator.
Temperature difference is the driving force for heat flow, and the greater
the temperature difference, the larger the rate of heat transfer. We can slow
down the heat flow between two mediums at different temperatures by
putting "barriers" on the path of heat flow. Thermal insulations serve as
such barriers, and they play a major role in the design and manufacture of
all energy-efficient devices or systems, and they are usually the cornerstone
of energy conservation projects. A 1991 Drexel University study of the
energy-intensive U.S. industries revealed that insulation saves the U.S.
Insulation ■
Heat
loss
Heat
^>
FIGURE 7-29
Thermal insulation retards heat
transfer by acting as a barrier in the
path of heat flow.
*This section can be skipped without a loss in continuity.
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HEAT TRANSFER
FIGURE 7-30
Insulation also helps the environment
by reducing the amount of fuel burned
and the air pollutants released.
FIGURE 7-31
In cold weather, we minimize heat loss
from our bodies by putting on thick
layers of insulation (coats or furs).
industry nearly 2 billion barrels of oil per year, valued at $60 billion a year
in energy costs, and more can be saved by practicing better insulation tech-
niques and retrofitting the older industrial facilities.
Heat is generated in furnaces or heaters by burning a fuel such as coal,
oil, or natural gas or by passing electric current through a resistance heater.
Electricity is rarely used for heating purposes since its unit cost is much
higher. The heat generated is absorbed by the medium in the furnace and its
surfaces, causing a temperature rise above the ambient temperature. This
temperature difference drives heat transfer from the hot medium to the am-
bient, and insulation reduces the amount of heat loss and thus saves fuel
and money. Therefore, insulation pays for itself Trom the energy it saves.
Insulating properly requires a one-time capital investment, but its effects
are dramatic and long term. The payback period of insulation is often less
than one year. That is, the money insulation saves during the first year is
usually greater than its initial material and installation costs. On a broader
perspective, insulation also helps the environment and fights air pollution
and the greenhouse effect by reducing the amount of fuel burned and thus
the amount of C0 2 and other gases released into the atmosphere (Fig.
7-30).
Saving energy with insulation is not limited to hot surfaces. We can also
save energy and money by insulating cold surfaces (surfaces whose tem-
perature is below the ambient temperature) such as chilled water lines,
cryogenic storage tanks, refrigerated trucks, and air-conditioning ducts.
The source of "coldness" is refrigeration, which requires energy input, usu-
ally electricity. In this case, heat is transferred from the surroundings to the
cold surfaces, and the refrigeration unit must now work harder and longer
to make up for this heat gain and thus it must consume more electrical en-
ergy. A cold canned drink can be kept cold much longer by wrapping it in
a blanket. A refrigerator with well-insulated walls will consume much less
electricity than a similar refrigerator with little or no insulation. Insulating
a house will result in reduced cooling load, and thus reduced electricity
consumption for air-conditioning.
Whether we realize it or not, we have an intuitive understanding and
appreciation of thermal insulation. As babies we feel much better in our
blankies, and as children we know we should wear a sweater or coat when
going outside in cold weather (Fig. 7-31). When getting out of a pool after
swimming on a windy day, we quickly wrap in a towel to stop shivering.
Similarly, early man used animal furs to keep warm and built shelters using
mud bricks and wood. Cork was used as a roof covering for centuries. The
need for effective thermal insulation became evident with the development
of mechanical refrigeration later in the nineteenth century, and a great deal
of work was done at universities and government and private laboratories
in the 1910s and 1920s to identify and characterize thermal insulation.
Thermal insulation in the form of mud, clay, straw, rags, and wood strips
was first used in the eighteenth century on steam engines to keep workmen
from being burned by hot surfaces. As a result, boiler room temperatures
dropped and it was noticed that fuel consumption was also reduced. The re-
alization of improved engine efficiency and energy savings prompted the
search for materials with improved thermal efficiency. One of the first such
materials was mineral wool insulation, which, like many materials, was
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CHAPTER 7
discovered by accident. About 1840, an iron producer in Wales aimed a
stream of high-pressure steam at the slag flowing from a blast furnace, and
manufactured mineral wool was born. In the early 1860s, this slag wool
was a by-product of manufacturing cannons for the Civil War and quickly
found its way into many industrial uses. By 1880, builders began installing
mineral wool in houses, with one of the most notable applications being
General Grant's house. The insulation of this house was described in an ar-
ticle: "it keeps the house cool in summer and warm in winter; it prevents
the spread of fire; and it deadens the sound between floors" [Edmunds
(1989), Ref. 4]. An article published in 1887 in Scientific American detail-
ing the benefits of insulating the entire house gave a major boost to the use
of insulation in residential buildings.
The energy crisis of the 1970s had a tremendous impact on the public
awareness of energy and limited energy reserves and brought an emphasis
on energy conservation. We have also seen the development of new and
more effective insulation materials since then, and a considerable increase
in the use of insulation. Thermal insulation is used in more places than you
may be aware of. The walls of your house are probably filled with some
kind of insulation, and the roof is likely to have a thick layer of insulation.
The "thickness" of the walls of your refrigerator is due to the insulation
layer sandwiched between two layers of sheet metal (Fig. 7-32). The walls
of your range are also insulated to conserve energy, and your hot water tank
contains less water than you think because of the 2- to 4-cm-thick insula-
tion in the walls of the tank. Also, your hot water pipe may look much
thicker than the cold water pipe because of insulation.
Reasons for Insulating
If you examine the engine compartment of your car, you will notice that the
firewall between the engine and the passenger compartment as well as the
inner surface of the hood are insulated. The reason for insulating the hood
is not to conserve the waste heat from the engine but to protect people from
burning themselves by touching the hood surface, which will be too hot if
not insulated. As this example shows, the use of insulation is not limited to
energy conservation. Various reasons for using insulation can be summa-
rized as follows:
• Energy Conservation Conserving energy by reducing the rate of heat
flow is the primary reason for insulating surfaces. Insulation materials
that will perform satisfactorily in the temperature range of — 268°C to
1000°C (-450°F to 1800°F) are widely available.
• Personnel Protection and Comfort A surface that is too hot poses a
danger to people who are working in that area of accidentally touching
the hot surface and burning themselves (Fig. 7-33). To prevent this dan-
ger and to comply with the OSHA (Occupational Safety and Health
Administration) standards, the temperatures of hot surfaces should be
reduced to below 60°C (140°F) by insulating them. Also, the excessive
heat coming off the hot surfaces creates an unpleasant environment in
which to work, which adversely affects the performance or productivity
of the workers, especially in summer months.
Insulation
FIGURE 7-32
The insulation layers in the walls of a
refrigerator reduce the amount of heat
flow into the refrigerator and thus the
running time of the refrigerator, saving
electricity.
Insulation
FIGURE 7-33
The hood of the engine compartment
of a car is insulated to reduce its
temperature and to protect people
from burning themselves.
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HEAT TRANSFER
FIGURE 7-34
Insulation materials absorb vibration
and sound waves, and are used to
minimize sound transmission.
• Maintaining Process Temperature Some processes in the chemical
industry are temperature-sensitive, and it may become necessary to in-
sulate the process tanks and flow sections heavily to maintain the same
temperature throughout.
• Reducing Temperature Variation and Fluctuations The tempera-
ture in an enclosure may vary greatly between the midsection and the
edges if the enclosure is not insulated. For example, the temperature
near the walls of a poorly insulated house is much lower than the tem-
perature at the midsections. Also, the temperature in an uninsulated en-
closure will follow the temperature changes in the environment closely
and fluctuate. Insulation minimizes temperature nonuniformity in an en-
closure and slows down fluctuations.
• Condensation and Corrosion Prevention Water vapor in the air con-
denses on surfaces whose temperature is below the dew point, and the
outer surfaces of the tanks or pipes that contain a cold fluid frequently
fall below the dew-point temperature unless they have adequate insula-
tion. The liquid water on exposed surfaces of the metal tanks or pipes
may promote corrosion as well as algae growth.
• Fire Protection Damage during a fire may be minimized by keeping
valuable combustibles in a safety box that is well insulated. Insulation
may lower the rate of heat flow to such levels that the temperature in the
box never rises to unsafe levels during fire.
• Freezing Protection Prolonged exposure to subfreezing temperatures
may cause water in pipes or storage vessels to freeze and burst as a re-
sult of heat transfer from the water to the cold ambient. The bursting of
pipes as a result of freezing can cause considerable damage. Adequate
insulation will slow down the heat loss from the water and prevent
freezing during limited exposure to subfreezing temperatures. For ex-
ample, covering vegetables during a cold night will protect them from
freezing, and burying water pipes in the ground at a sufficient depth will
keep them from freezing during the entire winter. Wearing thick gloves
will protect the fingers from possible frostbite. Also, a molten metal or
plastic in a container will solidify on the inner surface if the container is
not properly insulated.
• Reducing Noise and Vibration An added benefit of thermal in-
sulation is its ability to dampen noise and vibrations (Fig. 7-34). The
insulation materials differ in their ability to reduce noise and vibration,
and the proper kind can be selected if noise reduction is an important
consideration.
There are a wide variety of insulation materials available in the market,
but most are primarily made of fiberglass, mineral wool, polyethylene,
foam, or calcium silicate. They come in various trade names such as
Ethafoam Polyethylene Foam Sheeting, Solimide Polimide Foam Sheets,
FPC Fiberglass Reinforced Silicone Foam Sheeting, Silicone Sponge Rub-
ber Sheets, fiberglass/mineral wool insulation blankets, wire-reinforced
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CHAPTER 7
mineral wool insulation, Reflect-All Insulation, granulated bulk mineral
wool insulation, cork insulation sheets, foil-faced fiberglass insulation,
blended sponge rubber sheeting, and numerous others.
Today various forms of fiberglass insulation are widely used in process
industries and heating and air-conditioning applications because of their
low cost, light weight, resiliency, and versatility. But they are not suitable
for some applications because of their low resistance to moisture and fire
and their limited maximum service temperature. Fiberglass insulations
come in various forms such as unfaced fiberglass insulation, vinyl-faced
fiberglass insulation, foil-faced fiberglass insulation, and fiberglass insula-
tion sheets. The reflective foil-faced fiberglass insulation resists vapor pen-
etration and retards radiation because of the aluminum foil on it and is
suitable for use on pipes, ducts, and other surfaces.
Mineral wool is resilient, lightweight, fibrous, wool-like, thermally
efficient, fire resistant up to 1100°C (2000°F), and forms a sound barrier.
Mineral wool insulation comes in the form of blankets, rolls, or blocks.
Calcium silicate is a solid material that is suitable for use at high tempera-
tures, but it is more expensive. Also, it needs to be cut with a saw during in-
stallation, and thus it takes longer to install and there is more waste.
Superinsulators
You may be tempted to think that the most effective way to reduce heat
transfer is to use insulating materials that are known to have very low ther-
mal conductivities such as urethane or rigid foam (k = 0.026 W/m • °C) or
fiberglass (k = 0.035 W/m • °C). After all, they are widely available, inex-
pensive, and easy to install. Looking at the thermal conductivities of mate-
rials, you may also notice that the thermal conductivity of air at room
temperature is 0.026 W/m • °C, which is lower than the conductivities of
practically all of the ordinary insulating materials. Thus you may think that
a layer of enclosed air space is as effective as any of the common insulat-
ing materials of the same thickness. Of course, heat transfer through the air
will probably be higher than what a pure conduction analysis alone would
indicate because of the natural convection currents that are likely to occur
in the air layer. Besides, air is transparent to radiation, and thus heat will
also be transferred by radiation. The thermal conductivity of air is practi-
cally independent of pressure unless the pressure is extremely high or ex-
tremely low. Therefore, we can reduce the thermal conductivity of air and
thus the conduction heat transfer through the air by evacuating the air
space. In the limiting case of absolute vacuum, the thermal conductivity
will be zero since there will be no particles in this case to "conduct" heat
from one surface to the other, and thus the conduction heat transfer will be
zero. Noting that the thermal conductivity cannot be negative, an absolute
vacuum must be the ultimate insulator, right? Well, not quite.
The purpose of insulation is to reduce "total" heat transfer from a sur-
face, not just conduction. A vacuum totally eliminates conduction but of-
fers zero resistance to radiation, whose magnitude can be comparable to
conduction or natural convection in gases (Fig. 7-35). Thus, a vacuum is
Vacuum
Radiation
FIGURE 7-35
Evacuating the space between two
surfaces completely eliminates heat
transfer by conduction or convection
but leaves the door wide open for
radiation.
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400
HEAT TRANSFER
Thin metal
sheets
J
r i
*
War
iium -*
FIGURE 7-36
Superinsulators are built by closely
packing layers of highly reflective thin
metal sheets and evacuating the space
between them.
no more effective in reducing heat transfer than sealing off one of the lanes
of a two-lane road is in reducing the flow of traffic on a one-way road.
Insulation against radiation heat transfer between two surfaces is
achieved by placing "barriers" between the two surfaces, which are highly
reflective thin metal sheets. Radiation heat transfer between two surfaces is
inversely proportional to the number of such sheets placed between the sur-
faces. Very effective insulations are obtained by using closely packed lay-
ers of highly reflective thin metal sheets such as aluminum foil (usually 25
sheets per cm) separated by fibers made of insulating material such as glass
fiber (Fig. 7-36). Further, the space between the layers is evacuated to form
a vacuum under 0.000001 aim pressure to minimize conduction or convec-
tion heat transfer through the air space between the layers. The result is an
insulating material whose apparent thermal conductivity is below 2 X 10~ 5
W/m • °C, which is one thousand times less than the conductivity of air or
any common insulating material. These specially built insulators are called
superinsulators, and they are commonly used in space applications and
cryogenics, which is the branch of heat transfer dealing with temperatures
below 100 K (— 173°C) such as those encountered in the liquefaction, stor-
age, and transportation of gases, with helium, hydrogen, nitrogen, and oxy-
gen being the most common ones.
The R-value of Insulation
The effectiveness of insulation materials is given by some manufacturers in
terms of their 7?-value, which is the thermal resistance of the material per
unit surface area. For flat insulation the i?-value is obtained by simply di-
viding the thickness of the insulation by its thermal conductivity. That is,
/J-value
L
k
(flat insulation)
(7-50)
L, ft
R-value :
FIGURE 7-37
The lvalue of an insulating material
is simply the ratio of the thickness of
the material to its thermal conductivity
in proper units.
where L is the thickness and k is the thermal conductivity of the material.
Note that doubling the thickness L doubles the lvalue of flat insulation.
Vox pipe insulation, the 7?-value is determined using the thermal resistance
relation from
lvalue
r 2 , r 2
(pipe insulation)
(7-51)
where r l is the inside radius of insulation and r 2 is the outside radius of in-
sulation. Once the 7?-value is available, the rate of heat transfer through the
insulation can be determined from
Q
Ar
R- value
X Area
(7-52)
where AT is the temperature difference across the insulation and Area is the
outer surface area for a cylinder.
In the United States, the i?-values of insulation are expressed without any
units, such as R-19 and R-30. These ^-values are obtained by dividing
the thickness of the material in feet by its thermal conductivity in the unit
Btu/h • ft • °F so that the i?-values actually have the unit h • ft 2 • °F/Btu. For
example, the lvalue of 6-in. -thick glass fiber insulation whose thermal
conductivity is 0.025 Btu/h • ft • °F is (Fig. 7-37)
cen58 93 3_ch07.qxd 9/4/2002 12:13 PM Page 401
^-value
L
k
0.5 ft
0.025 Btu/h • ft
20 h • ft 2 • °F/Btu
401
CHAPTER 7
Thus, this 6-in. -thick glass fiber insulation would be referred to as R-20 in-
sulation by the builders. The unit of i?-value is m 2 • °C/W in SI units, with
the conversion relation 1 m 2 • °C/W = 5.678 h • ft 2 • °F/Btu. Therefore, a
small 7?-value in SI corresponds to a large i?-value in English units.
Optimum Thickness of Insulation
It should be realized that insulation does not eliminate heat transfer; it
merely reduces it. The thicker the insulation, the lower the rate of heat
transfer but also the higher the cost of insulation. Therefore, there should
be an optimum thickness of insulation that corresponds to a minimum com-
bined cost of insulation and heat lost. The determination of the optimum
thickness of insulation is illustrated in Figure 7-38. Notice that the cost of
insulation increases roughly linearly with thickness while the cost of heat
loss decreases exponentially. The total cost, which is the sum of the insula-
tion cost and the lost heat cost, decreases first, reaches a minimum, and
then increases. The thickness corresponding to the minimum total cost is
the optimum thickness of insulation, and this is the recommended thickness
of insulation to be installed.
If you are mathematically inclined, you can determine the optimum thick-
ness by obtaining an expression for the total cost, which is the sum of the
expressions for the lost heat cost and insulation cost as a function of thick-
ness; differentiating the total cost expression with respect to the thickness;
and setting it equal to zero. The thickness value satisfying the resulting
equation is the optimum thickness. The cost values can be determined from
an annualized lifetime analysis or simply from the requirement that the in-
sulation pay for itself within two or three years. Note that the optimum
thickness of insulation depends on the fuel cost, and the higher the fuel
cost, the larger the optimum thickness of insulation. Considering that insu-
lation will be in service for many years and the fuel prices are likely to es-
calate, a reasonable increase in fuel prices must be assumed in calculations.
Otherwise, what is optimum insulation today will be inadequate insulation
in the years to come, and we may have to face the possibility of costly
retrofitting projects. This is what happened in the 1970s and 1980s to insu-
lations installed in the 1960s.
The discussion above on optimum thickness is valid when the type and
manufacturer of insulation are already selected, and the only thing to be
determined is the most economical thickness. But often there are several
suitable insulations for a job, and the selection process can be rather con-
fusing since each insulation can have a different thermal conductivity, dif-
ferent installation cost, and different service life. In such cases, a selection
can be made by preparing an annualized cost versus thickness chart like
Figure 7-39 for each insulation, and determining the one having the low-
est minimum cost. The insulation with the lowest annual cost is obviously
the most economical insulation, and the insulation thickness correspond-
ing to the minimum total cost is the optimum thickness. When the optimum
thickness falls between two commercially available thicknesses, it is a
good practice to be conservative and choose the thicker insulation. The
Cost
per
year
Total
cost
LostN
heat
\
— "S ""^C°^.
cost
Optimum
thickness \ !
^-— Minimum
total cost
Insulation thickness
FIGURE 7-38
Determination of the optimum
thickness of insulation on the basis of
minimum total cost.
Total cost curves
for different insulation
Insulation thickness
FIGURE 7-39
Determination of the most economical
type of insulation and its
optimum thickness.
cen58 93 3_ch07.qxd 9/4/2002 12:13 PM Page 402
402
HEAT TRANSFER
TABLE 7
Recommended insulation
thicknesses for flat hot surfaces as a
function of surface temperature
(from TIMA Energy Savings Guide)
Surface
temperature
Insulation
thickness
150°F(66°C)
2'
(5.1 cm)
250°F(121°C)
3'
(7.6 cm)
350°F(177°C)
4'
(10.2 cm)
550°F (288°C)
6'
(15.2 cm)
750°F (400°C)
9'
(22.9 cm)
950°F(510°C)
10'
(25.44 cm)
Insulation
T. r, T 2 T i T o
^^WW v ->-^WVvv v ->"^WVvv v -^AWVvv v ^
R
extra thickness will provide a little safety cushion for any possible decline
in performance over time and will help the environment by reducing the
production of greenhouse gases such as C0 2 .
The determination of the optimum thickness of insulation requires a heat
transfer and economic analysis, which can be tedious and time-consuming.
But a selection can be made in a few minutes using the tables and charts pre-
pared by TIMA (Thermal Insulation Manufacturers Association) and mem-
ber companies. The primary inputs required for using these tables or charts
are the operating and ambient temperatures, pipe diameter (in the case of
pipe insulation), and the unit fuel cost. Recommended insulation thicknesses
for hot surfaces at specified temperatures are given in Table 7-4. Recom-
mended thicknesses of pipe insulations as a function of service temperatures
are 0.5 to 1 in. for 150°F, 1 to 2 in. for 250°F, 1.5 to 3 in. for 350°F, 2 to 4.5
in. for 450°F, 2.5 to 5.5 in. for 550°F, and 3 to 6 in. for 650°F for nominal
pipe diameters of 0.5 to 36 in. The lower recommended insulation thick-
nesses are for pipes with small diameters, and the larger ones are for pipes
with large diameters.
EXAMPLE 7-8 Effect of Insulation on Surface Temperature
Hot water at T t = 120°C flows in a stainless steel pipe (k = 15 W/m • °C)
whose inner diameter is 1.6 cm and thickness is 0.2 cm. The pipe is to be
covered with adequate insulation so that the temperature of the outer surface
of the insulation does not exceed 40°C when the ambient temperature is
T = 25°C. Taking the heat transfer coefficients inside and outside the pipe to
be h, = 70 W/m z • °C and h B = 20 W/m 2 • °C, respectively, determine the
thickness of fiberglass insulation (k = 0.038 W/m • °C) that needs to be in-
stalled on the pipe.
SOLUTION A steam pipe is to be covered with enough insulation to reduce the
exposed surface temperature. The thickness of insulation that needs to be in-
stalled is to be determined.
Assumptions 1 Heat transfer is steady since there is no indication of any
change with time. 2 Heat transfer is one-dimensional since there is thermal
symmetry about the centerline and no variation in the axial direction. 3 Thermal
conductivities are constant. 4 The thermal contact resistance at the interface is
negligible.
Properties The thermal conductivities are given to be k = 15 W/m ■ C C for the
steel pipe and k = 0.038 W/m • C C for fiberglass insulation.
Analysis The thermal resistance network for this problem involves four re-
sistances in series and is given in Figure 7-40. The inner radius of the pipe is
r x = 0.8 cm and the outer radius of the pipe and thus the inner radius of the in-
sulation is r 2 = 1.0 cm. Letting r 3 represent the outer radius of the insulation,
FIGURE 7-40
Schematic for Example 7-8.
the areas of the surfaces exposed to convection for an L
the pipe become
1-m-long section of
A, = 2m-,L = 2ir(0.008 m)(l m) = 0.0503 m 2
A 3 = 2ttt 3 L = 2tit 3 (1 m) = 6.28r 3 m 2
cen58 93 3_ch07.qxd 9/4/2002 12:13 PM Page 403
Then the individual thermal resistances are determined to be
1
Ri = R
conv, 1
^1 - ^pipc
R 2 = R
1
\n(r 2 /ri)
2-nk x L 2tt(15 W/m • °C)(1 m)
In(r 3 /r 2 ) ln(r 3 /0.01)
/j.A, (70 W/m 2 • °C)(0.0503 m 2 )
ln(0.01/0.008)
= 0.284°C/W
0.0024°C/W
Rn = R
2irk 2 L 2tt(0.038 W/m ■ °C)(1 m)
4.188 ln(r 3 /0.01)°C/W
1 1 1
h c A 3 (20 W/m 2 - °C)(6.28r 3 m 2 ) 125.6r ;
°C/W
Noting that all resistances are in series, the total resistance is determined to be
^total = Ri + R t + R 2 + Ro
= [0.284 + 0.0024 + 4.188 ln(r 3 /0.01) + l/125.6r 3 ]°C/W
Then the steady rate of heat loss from the steam becomes
T i - T (120 - 125)°C
Q
R„
[0.284 + 0.0024 + 4.188 ln(r 3 /0.01) + l/125.6r 3 ]°C/W
Noting that the outer surface temperature of insulation is specified to be 40°C,
the rate of heat loss can also be expressed as
Q
T 3 -T B _ (40 - 25)°C
R a ~ (l/125.6r 3 )°C/W
1884r,
Setting the two relations above equal to each other and solving for r 3 gives
r 3 = 0.0170 m. Then the minimum thickness of fiberglass insulation required is
t = r 3 -r 2 = 0.0170 - 0.0100 = 0.0070 m = 0.70 cm
Discussion Insulating the pipe with at least 0.70-cm-thick fiberglass insu-
lation will ensure that the outer surface temperature of the pipe will be at 40°C
or below.
403
CHAPTER 7
EXAMPLE 7-9 Optimum Thickness of Insulation
During a plant visit, you notice that the outer surface of a cylindrical curing
oven is very hot, and your measurements indicate that the average temperature
of the exposed surface of the oven is 180°F when the surrounding air tempera-
ture is 75°F. You suggest to the plant manager that the oven should be insu-
lated, but the manager does not think it is worth the expense. Then you propose
to the manager to pay for the insulation yourself if he lets you keep the savings
from the fuel bill for one year. That is, if the fuel bill is $5000/yr before insula-
tion and drops to $2000/yr after insulation, you will get paid $3000. The man-
ager agrees since he has nothing to lose, and a lot to gain. Is this a smart bet
on your part?
The oven is 12 ft long and 8 ft in diameter, as shown in Figure 7-41. The
plant operates 16 h a day 365 days a year, and thus 5840 h/yr. The insulation
180°F
FIGURE 7-41
Schematic for Example 7-9.
cen58 93 3_ch07.qxd 9/4/2002 12:13 PM Page 404
404
HEAT TRANSFER
to be used is fiberglass (Aj ns = 0.024 Btu/h • ft ■ °F), whose cost is $0.70/ft 2 per
inch of thickness for materials, plus $2.00/ft 2 for labor regardless of thickness.
The combined heat transfer coefficient on the outer surface is estimated to be
h = 3.5 Btu/h • ft 2 • °F. The oven uses natural gas, whose unit cost is
$0.75/therm input (1 therm = 100,000 Btu), and the efficiency of the oven is
80 percent. Disregarding any inflation or interest, determine how much money
you will make out of this venture, if any, and the thickness of insulation (in
whole inches) that will maximize your earnings.
SOLUTION A cylindrical oven is to be insulated to reduce heat losses. The op-
timum thickness of insulation and the potential earnings are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Heat transfer through the in-
sulation is one-dimensional. 3 Thermal conductivities are constant. 4 The thermal
contact resistance at the interface is negligible. 5 The surfaces of the cylindrical
oven can be treated as plain surfaces since its diameter is greater than 3 ft.
Properties The thermal conductivity of insulation is given to be k = 0.024 Btu/
h • ft • °F.
Analysis The exposed surface area of the oven is
2A h
2-irr 2 + 2-nrL = 2tt(4 ft) 2 + 2tt(4 ft)(12 ft) = 402 ft 2
The rate of heat loss from the oven before the insulation is installed is deter-
mined from
Q = h A„(T s - T„) = (3.5 Btu/h ■ ft 2 • °F)(402 ft 2 )(180 - 75)°F = 147,700 Btu/h
Noting that the plant operates 5840 h/yr, the total amount of heat loss from the
oven per year is
Q = QAt = (147,700 Btu/h)(5840 h/yr) = 0.863 X 10 9 Btu/yr
The efficiency of the oven is given to be 80 percent. Therefore, to generate
this much heat, the oven must consume energy (in the form of natural gas) at
a rate of
G„ = G/rioven = (°- 863 X 10" Btu/yr)/0.80 = 1.079 X 10 9 Btu/yr
= 10,790 therms
since 1 therm = 100,000 Btu. Then the annual fuel cost of this oven before in-
sulation becomes
Annual cost = Q m X Unit cost
= (10,790 therm/yr)($0.75/therm) = $8093/yr
That is, the heat losses from the exposed surfaces of the oven are currently
costing the plant over $8000/yr.
When insulation is installed, the rate of heat transfer from the oven can be
determined from
Gins
+
We expect the surface temperature of the oven to increase and the heat trans-
fer coefficient to decrease somewhat when insulation is installed. We assume
cen58 93 3_ch07.qxd 9/4/2002 12:13 PM Page 405
these two effects to counteract each other. Then the relation above for 1-in.
thick insulation gives the rate of heat loss to be
A S (T S - T x ) (402 ft 2 )(180 - 75)°F
gin
1
fins J_ 1/12 ft
k,„ s K 0.024 Btu/h • ft • °F 3.5 Btu/h • ft 2
1 1,230 Btu/h
Also, the total amount of heat loss from the oven per year and the amount and
cost of energy consumption of the oven become
Q ms = Q ms At = (11,230 Btu/h)(5840 h/yr) = 0.6558 X 10 8 Btu/yr
2i„,i„ s = 2i„ s /'nove„ = (0.6558 X 10 8 Btu/yr)/0.80 = 0.820 X 10 8 Btu/yr
= 820 therms
Annual cost = Q in ins X Unit cost
= (820 therm/yr)($0.75/therm) = $615/yr
Therefore, insulating the oven by 1-in. -thick fiberglass insulation will reduce the
fuel bill by $8093 - $615 = $7362 per year. The unit cost of insulation is
given to be $2.70/ft 2 . Then the installation cost of insulation becomes
Insulation cost = (Unit cost)(Surface area) = ($2.70/ft 2 )(402 ft 2 ) = $1085
The sum of the insulation and heat loss costs is
Total cost = Insulation cost + Heat loss cost = $1085 + $615 = $1700
Then the net earnings will be
Earnings = Income - Expenses = $8093 - $1700 = $6393
To determine the thickness of insulation that maximizes your earnings, we
repeat the calculations above for 2-, 3-, 4-, and 5-in. -thick insulations, and list
the results in Table 7-5. Note that the total cost of insulation decreases first
with increasing insulation thickness, reaches a minimum, and then starts to
increase.
TABLE 7-5
The variation of total insulation cost with insulation thickness
Insulation
thickness
Heat loss,
Btu/h
Lost fuel,
therms/yr
Lost fuel
cost, $/yr
Insulation
cost, $
Total cost,
1 in.
2 in.
3 in.
4 in.
5 in.
11,230
5838
3944
2978
2392
820
426
288
217
175
615
320
216
163
131
1085
1367
1648
1930
2211
1700
1687
1864
2093
2342
We observe that the total insulation cost is a minimum at $1687 for the case of
2-in. -thick insulation. The earnings in this case are
Maximum earnings = Income — Minimum expenses
= $8093 - $1687 = $6406
405
CHAPTER 7
cen58 93 3_ch07.qxd 9/4/2002 12:13 PM Page 406
406
HEAT TRANSFER
which is not bad for a day's worth of work. The plant manager is also a big win-
ner in this venture since the heat losses will cost him only $320/yr during the
second and consequent years instead of $8093/yr. A thicker insulation could
probably be justified in this case if the cost of insulation is annualized over the
lifetime of insulation, say 20 years. Several energy conservation measures are
being marketed as explained above by several power companies and private
firms.
SUMMARY
The force a flowing fluid exerts on a body in the flow direction
is called drag. The part of drag that is due directly to wall shear
stress t w is called the skin friction drag since it is caused by
frictional effects, and the part that is due directly to pressure is
called the pressure drag or form drag because of its strong de-
pendence on the form or shape of the body.
The drag coefficient C D is a dimensionless number that rep-
resents the drag characteristics of a body, and is defined as
C r ,
Fn
i P T 2 A
where A is the frontal area for blunt bodies, and surface area
for parallel flow over flat plates or thin airfoils. For flow over
a flat plate, the Reynolds number is
Re,
pTx = Yx
[L V
Transition from laminar to turbulent occurs at the critical
Reynolds number of
Re,
P Vx c
5 X 10 5
For parallel flow over a flat plate, the local friction and con-
vection coefficients are
Laminar: C, ,. = ,„ Re,. < 5 X 10 5
/,* Re i/2
Nu v = ^- = 0.332 Re - 5 Pr" 3 Pr > 0.6
k
059?
Turbulent: C f , = ,,, , 5 X 10 5 < Re v < 10 7
/.* Re i/s
h x x 0.6 < Pr < 60
Nu Y = -^ = 0.0296 Re? 8 Pr" 3
k v 5 x 10 5 <Re v < 10 7
The average friction coefficient relations for flow over a flat
plate are:
Laminar:
Turbulent:
C,
C,
1.328
Rel' 2
0.074
Re L < 5 X 10 5
5 X 10 5 < Re L < 10 7
Combined: C f = ^^ - ^^ 5 X 10 5 < Re, < 10 7
f Re| /5 Re L L
Rough surface, turbulent: C, = 1.89 — 1.62 log
The average Nusselt number relations for flow over a flat plate
are:
hi
Laminar: Nu = -= = 0.664 Re? 5 Pr" 3 Re, < 5 X 10 5
k
Turbulent:
hL
k
Nu = ^ = 0.037 Re£ 8 Pr 1/3
Combined:
hL
k
Nu = ™ = (0.037 Re? 8 - 871) Pr" 3 ,
0.6 < Pr < 60
5 X 10 5 <Re L < 10 7
0.6 < Pr < 60
5 X 10 5 <Re L < 10 7
For isothermal surfaces with an unheated starting section of
length 4, the local Nusselt number and the average convection
coefficient relations are
Laminar:
Turbulent:
Laminar:
Turbulent:
Nu,.
Nu
i(forf = 0)
[1 - (£/*) 3/4 ]" 3 [1 - (£/x) 3 ' 4 ]
0.332 Re? 5 Pr" 3
3/411/3
Nu,.
Nu
r(fori; = 0)
0.0296 Re?.- 8 Pr 1
[1 - {l,lx) 9lw Y
[1 " &x)
9/1011/9
2[i - a/x) M ]
h= — 77T h x=L
1 - ilL
5[1 ~ (Z/x) mo
(1 " £/L)
h r= ,
cen58 93 3_ch07.qxd 9/4/2002 12:13 PM Page 407
These relations are for the case of isothermal surfaces. When a
flat plate is subjected to uniform heat flux, the local Nusselt
number is given by
Laminar:
Turbulent:
Nu, = 0.453 Re? 5 Pr 1/3
Nu = 0.0308 Re? 8 Pr 1 ' 3
The average Nusselt numbers for cross flow over a cylinder
and sphere are
hD
0.62 Re" 2 Pr 1
Nucyi k 03 + [i +(0 .4/Pr) 2/3 ] 1/4
which is valid for Re Pr > 0.2, and
hD
1 +
Re Y &
282,000/
Nu
sph
2 + [0.4 Re
1/2
0.06 Re 2/3 ]Pr°
M-
which is valid for 3.5 < Re < 80,000 and 0.7 < Pr < 380.
The fluid properties are evaluated at the film temperature
Tf = (T a + T s )/2 in the case of a cylinder, and at the free-
stream temperature T„ (except for \x s , which is evaluated at the
surface temperature T s ) in the case of a sphere.
In tube banks, the Reynolds number is based on the maxi-
mum velocity T max that is related to the approach velocity T as
In-line and Staggered with S D < (S T + D)/2:
.y
Staggered with S D < (S T + D)/2:
2(S D - D)
T"
407
CHAPTER 7
Nu,
hD
CReS'Pr'^Pr/Pr,)
where the values of the constants C, m, and n depend on value
Reynolds number. Such correlations are given in Table 7-2.
All properties except Pr s are to be evaluated at the arithmetic
mean of the inlet and outlet temperatures of the fluid defined as
T m = (T, + T e )l2.
The average Nusselt number for tube banks with less than
16 rows is expressed as
Nu AWi = FNu fl
where F is the correction factor whose values are given in
Table 7-3. The heat transfer rate to or from a tube bank is de-
termined from
Q = hA s AT ln = mC p (T e -T i )
where A7"| n is the logarithmic mean temperature difference de-
fined as
A7,„
(T,
(T s - Ti) AT e
at
IntCr, - T.y(T, - T,)] \n(AT e IAT,)
and the exit temperature of the fluid T e is
T e = T s ~ (T, - T { ) exp
AJx_
mC,
where A s = NttDL is the heat transfer surface area and m =
p°V(N T S T L) is the mass flow rate of the fluid. The pressure
drop AP for a tube bank is expressed as
p¥ 2
AP = N L fx ■
1-2
max
where S T the transverse pitch and S D is the diagonal pitch. The
average Nusselt number for cross flow over tube banks is ex-
pressed as
where/ is the friction factor and x is the correction factor, both
given in Figs. 7-27.
REFERENCES AND SUGGESTED READING
1. R. D. Blevin. Applied Fluid Dynamics Handbook.
New York: Van Nostrand Reinhold, 1984.
2. S. W. Churchill and M. Bernstein. "A Correlating
Equation for Forced Convection from Gases and Liquids
to a Circular Cylinder in Cross Flow." Journal of Heat
Transfer 99 (1977), pp. 300-306.
3. S. W. Churchill and H. Ozoe. "Correlations for Laminar
Forced Convection in Flow over an Isothermal Flat Plate
and in Developing and Fully Developed Flow in an
Isothermal Tube."' Journal of Heat Transfer 95 (Feb.
1973), pp. 78-84.
4. W. M. Edmunds. "Residential Insulation." ASTM
Standardization News (Jan. 1989), pp. 36-39.
5. W. H. Giedt. "Investigation of Variation of Point Unit-
Heat Transfer Coefficient around a Cylinder Normal to an
Air Stream." Transactions oftheASME 71 (1949),
pp. 375-381.
6. M. Jakob. Heat Transfer. Vol. 1. New York: John Wiley &
Sons, 1949.
7. W. M. Kays and M. E. Crawford. Convective Heat and
Mass Transfer. 3rd ed. New York: McGraw-Hill, 1993.
8. F. Kreith and M. S. Bohn. Principles of Heat Transfer, 6th
ed. Pacific Grove, CA: Brooks/Cole, 2001.
9. H. Schlichting. Boundary Layer Theory, 7th ed. New
York, McGraw-Hill, 1979.
cen58933_ch07.qxd 9/4/2002 12:13 PM Page 4C
408
HEAT TRANSFER
10. N. V. Suryanarayana. Engineering Heat Transfer. St. Paul,
MN: West, 1995.
11. W. C. Thomas. "Note on the Heat Transfer Equation for
Forced Convection Flow over a Flat Plate with and
Unheated Starting Length." Mechanical Engineering
News, 9, no.l (1977), p. 361.
12. R. D. Willis. "Photographic Study of Fluid Flow Between
Banks of Tubes." Engineering (1934), pp. 423—425.
13. A. Zukauskas, "Convection Heat Transfer in Cross Flow."
In Advances in Heat Transfer, J. P. Hartnett and T. F.
Irvine, Jr., Eds. New York: Academic Press, 1972, Vol. 8,
pp. 93-106.
14. A. Zukauskas. "Heat Transfer from Tubes in Cross Flow."
In Advances in Heat Transfer, ed. J. P. Hartnett and T. F.
Irvine, Jr. Vol. 8. New York: Academic Press, 1972.
15. A. Zukauskas. "Heat Transfer from Tubes in Cross Flow."
In Handbook of Single Phase Convective Heat Transfer,
Eds. S. Kakac, R. K. Shah, and Win Aung. New York:
Wiley Interscience, 1987.
16. A. Zukauskas and R. Ulinskas, "Efficiency Parameters for
Heat Transfer in Tube Banks." Heat Transfer Engineering
no. 2(1985), pp. 19-25.
PROBLEMS
Drag Force and Heat Transfer in External Flow
7-1 C What is the difference between the upstream velocity
and the free-stream velocity? For what types of flow are these
two velocities equal to each other?
7-2C What is the difference between streamlined and blunt
bodies? Is a tennis ball a streamlined or blunt body?
7-3C What is drag? What causes it? Why do we usually try
to minimize it?
7-4C What is lift? What causes it? Does wall shear con-
tribute to the lift?
7-5C During flow over a given body, the drag force, the up-
stream velocity, and the fluid density are measured. Explain
how you would detennine the drag coefficient. What area
would you use in calculations?
7-6C Define frontal area of a body subjected to external
flow. When is it appropriate to use the frontal area in drag and
lift calculations?
7-7C What is the difference between skin friction drag and
pressure drag? Which is usually more significant for slender
bodies such as airfoils?
7-8C What is the effect of surface roughness on the friction
drag coefficient in laminar and turbulent flows?
7-9C What is the effect of streamlining on (a) friction drag
and (b) pressure drag? Does the total drag acting on a body
necessarily decrease as a result of streamlining? Explain.
*Problems designated by a "C" are concept questions, and
students are encouraged to answer them all. Problems designated
by an "E" are in English units, and the SI users can ignore them.
Problems with an EES-CD icon ® are solved using EES, and
complete solutions together with parametric studies are included
on the enclosed CD. Problems with a computer-EES icon H are
comprehensive in nature, and are intended to be solved with a
computer, preferably using the EES software that accompanies
this text.
7-10C What is flow separation? What causes it? What is the
effect of flow separation on the drag coefficient?
Flow Over Flat Plates
7-11C What does the friction coefficient represent in flow
over a flat plate? How is it related to the drag force acting on
the plate?
7-12C Consider laminar flow over a flat plate. Will the fric-
tion coefficient change with distance from the leading edge?
How about the heat transfer coefficient?
7-13C How are the average friction and heat transfer coeffi-
cients determined in flow over a flat plate?
7-14 Engine oil at 80°C flows over a 6-m-long flat plate
whose temperature is 30°C with a velocity of 3 m/s. Determine
the total drag force and the rate of heat transfer over the entire
plate per unit width.
7-15 The local atmospheric pressure in Denver, Colorado
(elevation 1610 m), is 83.4 kPa. Air at this pressure and
at 30°C flows with a velocity of 6 m/s over a 2.5-m X 8-m flat
plate whose temperature is 120°C. Determine the rate of
heat transfer from the plate if the air flows parallel to the
(a) 8-m-long side and (b) the 2.5-m side.
7-16 During a cold winter day, wind at 55 km/h is blowing
parallel to a 4-m-high and 10-m-long wall of a house. If the air
outside is at 5°C and the surface temperature of the wall is
Attic
space
FIGURE P7-1 6
cen58 93 3_ch07.qxd 9/4/2002 12:13 PM Page 409
12°C, determine the rate of heat loss from that wall by convec-
tion. What would your answer be if the wind velocity was dou-
bled? Answers: 9081 W, 16,200 W
7-17 [?(,■)! Reconsider Problem 7-16. Using EES (or other)
k^^ software, investigate the effects of wind velocity
and outside air temperature on the rate of heat loss from the
wall by convection. Let the wind velocity vary from 10 km/h to
80 km/h and the outside air temperature from 0°C to 10°C. Plot
the rate of heat loss as a function of the wind velocity and of
the outside temperature, and discuss the results.
7-18E Air at 60°F flows over a 1 0-ft-long flat plate at 7 ft/s.
Determine the local friction and heat transfer coefficients at in-
tervals of 1 ft, and plot the results against the distance from the
leading edge.
7-19E r^| Reconsider Problem 7-18. Using EES (or
)£ZS other) software, evaluate the local friction and
heat transfer coefficients along the plate at intervals of 0.1 ft,
and plot them against the distance from the leading edge.
7-20 Consider a hot automotive engine, which can be ap-
proximated as a 0.5-m-high, 0.40-m-wide, and 0.8-m-long rec-
tangular block. The bottom surface of the block is at a
temperature of 80°C and has an emissivity of 0.95. The ambi-
ent air is at 20°C, and the road surface is at 25°C. Determine
the rate of heat transfer from the bottom surface of the engine
block by convection and radiation as the car travels at a veloc-
ity of 80 km/h. Assume the flow to be turbulent over the entire
surface because of the constant agitation of the engine block.
7-21 The forming section of a plastics plant puts out a con-
tinuous sheet of plastic that is 1.2 m wide and 2 mm thick at a
rate of 15 m/min. The temperature of the plastic sheet is 90°C
when it is exposed to the surrounding air, and the sheet is sub-
jected to air flow at 30°C at a velocity of 3 m/s on both sides
along its surfaces normal to the direction of motion of the
sheet. The width of the air cooling section is such that a fixed
point on the plastic sheet passes through that section in 2 s. De-
termine the rate of heat transfer from the plastic sheet to the air.
Air
30°C, 3 m/s
409
CHAPTER 7
7-22 The top surface of the passenger car of a train moving
at a velocity of 70 km/h is 2.8 m wide and 8 m long. The top
surface is absorbing solar radiation at a rate of 200 W/m 2 , and
the temperature of the ambient air is 30°C. Assuming the roof
of the car to be perfectly insulated and the radiation heat ex-
change with the surroundings to be small relative to convec-
tion, determine the equilibrium temperature of the top surface
of the car. Answer: 35. 1°C
Air
30°C
200 W/m 2
////
70 km/h
1 5 m/min
FIGURE P7-21
FIGURE P7-22
7-23 Reconsider Problem 7-22. Using EES (or other) soft-
ware, investigate the effects of the train velocity and the rate
of absorption of solar radiation on the equilibrium tempera-
ture of the top surface of the car. Let the train velocity vary
from 10 km/h to 120 km/h and the rate of solar absorption from
100 W/m 2 to 500 W/m 2 . Plot the equilibrium temperature as
functions of train velocity and solar radiation absorption rate,
and discuss the results.
7-24 A 15-cm X 15-cm circuit board dissipating 15 W of
power uniformly is cooled by air, which approaches the circuit
board at 20°C with a velocity of 5 m/s. Disregarding any heat
transfer from the back surface of the board, determine the sur-
face temperature of the electronic components (a) at the lead-
ing edge and {b) at the end of the board. Assume the flow to be
turbulent since the electronic components are expected to act
as turbulators.
7-25 Consider laminar flow of a fluid over a flat plate main-
tained at a constant temperature. Now the free-stream velocity
of the fluid is doubled. Determine the change in the drag force
on the plate and rate of heat transfer between the fluid and the
plate. Assume the flow to remain laminar.
7-26E Consider a refrigeration truck traveling at 55 mph at
a location where the air temperature is 80°F. The refrigerated
compartment of the truck can be considered to be a 9-ft-wide,
8-ft-high, and 20-ft-long rectangular box. The refrigeration
system of the truck can provide 3 tons of refrigeration (i.e., it
can remove heat at a rate of 600 Btu/min). The outer surface
of the truck is coated with a low-emissivity material, and thus
radiation heat transfer is very small. Determine the average
temperature of the outer surface of the refrigeration compart-
ment of the truck if the refrigeration system is observed to be
cen58 93 3_ch07.qxd 9/4/2002 12:13 PM Page 41C
410
HEAT TRANSFER
20 ft
Air, 80°F
V = 55 mph
FIGURE P7-26E
operating at half the capacity. Assume the air flow over the en-
tire outer surface to be turbulent and the heat transfer coeffi-
cient at the front and rear surfaces to be equal to that on side
surfaces.
7-27 Solar radiation is incident on the glass cover of a solar
collector at a rate of 700 W/m 2 . The glass transmits 88 percent
of the incident radiation and has an emissivity of 0.90. The en-
tire hot water needs of a family in summer can be met by two
collectors 1.2 m high and 1 m wide. The two collectors are at-
tached to each other on one side so that they appear like a sin-
gle collector 1 .2 m X 2 m in size. The temperature of the glass
cover is measured to be 35°C on a day when the surrounding
air temperature is 25°C and the wind is blowing at 30 km/h.
The effective sky temperature for radiation exchange between
the glass cover and the open sky is — 40°C. Water enters the
tubes attached to the absorber plate at a rate of 1 kg/min. As-
suming the back surface of the absorber plate to be heavily in-
sulated and the only heat loss to occur through the glass cover,
determine (a) the total rate of heat loss from the collector,
(b) the collector efficiency, which is the ratio of the amount of
heat transferred to the water to the solar energy incident on the
collector, and (c) the temperature rise of water as it flows
through the collector.
r sky = -40°c
FIGURE P7-27
7-28 A transformer that is 10 cm long, 6.2 cm wide, and
5 cm high is to be cooled by attaching a 10 cm X 6.2 cm wide
polished aluminum heat sink (emissivity = 0.03) to its top sur-
face. The heat sink has seven fins, which are 5 mm high, 2 mm
thick, and 10 cm long. A fan blows air at 25°C parallel to the
passages between the fins. The heat sink is to dissipate 20 W of
heat and the base temperature of the heat sink is not to exceed
60°C. Assuming the fins and the base plate to be nearly isother-
mal and the radiation heat transfer to be negligible, determine
the minimum free-stream velocity the fan needs to supply to
avoid overheating.
Air
25°C
\\\\\
FIGURE P7-28
7-29 Repeat Problem 7-28 assuming the heat sink to be
black-anodized and thus to have an effective emissivity of
0.90. Note that in radiation calculations the base area (10 cm X
6.2 cm) is to be used, not the total surface area.
7-30 An array of power transistors, dissipating 6 W of power
each, are to be cooled by mounting them on a 25-cm X 25-cm
square aluminum plate and blowing air at 35°C over the plate
with a fan at a velocity of 4 m/s. The average temperature of
the plate is not to exceed 65°C. Assuming the heat transfer
from the back side of the plate to be negligible and disregard-
ing radiation, determine the number of transistors that can be
placed on this plate.
Aluminum
plate
Power
transistor, 6 W
35°C "
Air -
4 m/s -
25 cm
65°C
25 cm
FIGURE P7-30
cen58 93 3_ch07.qxd 9/4/2002 12:13 PM Page 411
7-31 Repeat Problem 7-30 for a location at an elevation of
1610 m where the atmospheric pressure is 83.4 kPa.
Answer: 4
7-32 Air at 25°C and 1 atm is flowing over a long flat plate
with a velocity of 8 m/s. Determine the distance from the lead-
ing edge of the plate where the flow becomes turbulent, and the
thickness of the boundary layer at that location.
7-33 Repeat Problem 7-32 for water.
7-34 The weight of a thin flat plate 50 cm X 50 cm in size is
balanced by a counterweight that has a mass of 2 kg, as shown
in the figure. Now a fan is turned on, and air at 1 atm and 25°C
flows downward over both surfaces of the plate with a free-
stream velocity of 10 m/s. Determine the mass of the counter-
weight that needs to be added in order to balance the plate in
this case.
Air
25°C, 10 m/s
50 cm
FIGURE P7-34
Flow across Cylinders and Spheres
7-35C Consider laminar flow of air across a hot circular
cylinder. At what point on the cylinder will the heat transfer be
highest? What would your answer be if the flow were turbulent?
7-36C In flow over cylinders, why does the drag coefficient
suddenly drop when the flow becomes turbulent? Isn't turbu-
lence supposed to increase the drag coefficient instead of
decreasing it?
7-37C In flow over blunt bodies such as a cylinder, how
does the pressure drag differ from the friction drag?
7-38C Why is flow separation in flow over cylinders de-
layed in turbulent flow?
7-39 A long 8-cm-diameter steam pipe whose external sur-
face temperature is 90°C passes through some open area that is
not protected against the winds. Determine the rate of heat loss
from the pipe per unit of its length when the air is at 1 atm pres-
sure and 7°C and the wind is blowing across the pipe at a ve-
locity of 50 km/h.
7-40 A stainless steel ball (p = 8055 kg/m 3 , C p = 480 J/kg •
°C) of diameter D = 15 cm is removed from the oven at a uni-
form temperature of 350°C. The ball is then subjected to the flow
411
CHAPTER 7
of air at 1 atm pressure and 30°C with a velocity of 6 m/s. The
surface temperature of the ball eventually drops to 250°C. Deter-
mine the average convection heat transfer coefficient during this
cooling process and estimate how long this process has taken.
7-41 TccM Reconsider Problem 7^40. Using EES (or other)
ggSa software, investigate the effect of air velocity on
the average convection heat transfer coefficient and the cooling
time. Let the air velocity vary from 1 m/s to 10 m/s. Plot the
heat transfer coefficient and the cooling time as a function of
air velocity, and discuss the results.
7-42E A person extends his uncovered arms into the windy
air outside at 54°F and 20 mph in order to feel nature closely.
Initially, the skin temperature of the arm is 86°F. Treating the
arm as a 2-ft-long and 3 -in. -diameter cylinder, determine the
rate of heat loss from the arm.
Air
54°F, 20 mph
FIGURE P7-12E
7-43E
Reconsider Problem 7^12E. Using EES (or
other) software, investigate the effects of air
temperature and wind velocity on the rate of heat loss from the
arm. Let the air temperature vary from 20°F to 80°F and the
wind velocity from 10 mph to 40 mph. Plot the rate of heat loss
as a function of air temperature and of wind velocity, and dis-
cuss the results.
7-44 An average person generates heat at a rate of 84 W
while resting. Assuming one-quarter of this heat is lost from
the head and disregarding radiation, determine the average sur-
face temperature of the head when it is not covered and is sub-
jected to winds at 10°C and 35 km/h. The head can be
approximated as a 30-cm-diameter sphere. Answer: 12.7°C
7-45 Consider the flow of a fluid across a cylinder main-
tained at a constant temperature. Now the free-stream velocity
of the fluid is doubled. Determine the change in the drag force
on the cylinder and the rate of heat transfer between the fluid
and the cylinder.
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412
HEAT TRANSFER
7-46 A 6-mm-diameter electrical transmission line carries an
electric current of 50 A and has a resistance of 0.002 ohm per
meter length. Determine the surface temperature of the wire
during a windy day when the air temperature is 10°C and the
wind is blowing across the transmission line at 40 km/h.
Wind, 40 km/h
10 ° C /////
^^
"^
^T
^7T
Transmission
lines
FIGURE P7-46
7-47
Reconsider Problem l-\&. Using EES (or other)
software, investigate the effect of the wind veloc-
ity on the surface temperature of the wire. Let the wind veloc-
ity vary from 10 km/h to 80 km/h. Plot the surface temperature
as a function of wind velocity, and discuss the results.
7-48 A heating system is to be designed to keep the wings of
an aircraft cruising at a velocity of 900 km/h above freezing
temperatures during flight at 12,200-m altitude where the stan-
dard atmospheric conditions are — 55.4°C and 18.8 kPa. Ap-
proximating the wing as a cylinder of elliptical cross section
whose minor axis is 30 cm and disregarding radiation, determine
the average convection heat transfer coefficient on the wing sur-
face and the average rate of heat transfer per unit surface area.
7-49 fl&\ A long aluminum wire of diameter 3 mm is
xgsy extruded at a temperature of 370°C. The wire
is subjected to cross air flow at 30°C at a velocity of 6 m/s.
Determine the rate of heat transfer from the wire to the air per
meter length when it is first exposed to the air.
370°C
3 mm
Aluminum
wire
J 30°C
6 m/s
FIGURE P7-49
7-50E Consider a person who is trying to keep cool on a hot
summer day by turning a fan on and exposing his entire body
to air flow. The air temperature is 85°F and the fan is blowing
air at a velocity of 6 ft/s. If the person is doing light work and
generating sensible heat at a rate of 300 Btu/h, determine the
average temperature of the outer surface (skin or clothing) of
the person. The average human body can be treated as a 1-ft-
diameter cylinder with an exposed surface area of 18 ft 2 . Dis-
regard any heat transfer by radiation. What would your answer
be if the air velocity were doubled? Answers: 95. 1°F, 9 1 .6°F
85°F
6 ft/s
Btu/h
FIGURE P7-50E
7-51 An incandescent lightbulb is an inexpensive but highly
inefficient device that converts electrical energy into light. It
converts about 10 percent of the electrical energy it consumes
into light while converting the remaining 90 percent into heat.
(A fluorescent lightbulb will give the same amount of light
while consuming only one-fourth of the electrical energy, and
it will last 10 times longer than an incandescent lightbulb.) The
glass bulb of the lamp heats up very quickly as a result of ab-
sorbing all that heat and dissipating it to the surroundings by
convection and radiation.
Consider a 10-cm -diameter 100-W lightbulb cooled by a fan
that blows air at 25°C to the bulb at a velocity of 2 m/s. The
surrounding surfaces are also at 25°C, and the emissivity of the
glass is 0.9. Assuming 10 percent of the energy passes through
the glass bulb as light with negligible absorption and the rest of
the energy is absorbed and dissipated by the bulb itself, deter-
mine the equilibrium temperature of the glass bulb.
Air .
25°C ■
2m/s"
£ = 0.9
FIGURE P7-51
Light, 10 W
7-52 During a plant visit, it was noticed that a 12-m-long
section of a 10-cm-diameter steam pipe is completely exposed
to the ambient air. The temperature measurements indicate that
cen58 93 3_ch07.qxd 9/4/2002 12:13 PM Page 413
the average temperature of the outer surface of the steam pipe
is 75°C when the ambient temperature is 5°C. There are also
light winds in the area at 10 km/h. The emissivity of the outer
surface of the pipe is 0.8, and the average temperature of the
surfaces surrounding the pipe, including the sky, is estimated to
be 0°C. Determine the amount of heat lost from the steam dur-
ing a 10-h-long work day.
Steam is supplied by a gas-fired steam generator that has an
efficiency of 80 percent, and the plant pays $0.54/therm of nat-
ural gas (1 therm = 105,500 kJ). If the pipe is insulated and 90
percent of the heat loss is saved, determine the amount of
money this facility will save a year as a result of insulating the
steam pipes. Assume the plant operates every day of the year
for 10 h. State your assumptions.
:0°C
e = 0i
'75°C
10 cm
Steam pipe
5°C
10 km/h
FIGURE P7-52
7-53 Reconsider Problem 7-52. There seems to be some un-
certainty about the average temperature of the surfaces sur-
rounding the pipe used in radiation calculations, and you are
asked to determine if it makes any significant difference in
overall heat transfer. Repeat the calculations for average sur-
rounding and surface temperatures of — 20°C and 25°C, re-
spectively, and determine the change in the values obtained.
7-54E A 12-ft-long, 1.5-kW electrical resistance wire is
made of 0.1 -in. -diameter stainless steel (k = 8.7 Btu/h • ft ■ °F).
The resistance wire operates in an environment at 85°F. Deter-
mine the surface temperature of the wire if it is cooled by a fan
blowing air at a velocity of 20 ft/s.
FIGURE P7-54E
7-55 The components of an electronic system are located in
a 1.5-m-long horizontal duct whose cross section is 20 cm X
20 cm. The components in the duct are not allowed to come
into direct contact with cooling air, and thus are cooled by air
413
CHAPTER 7
at 30°C flowing over the duct with a velocity of 200 m/min.
If the surface temperature of the duct is not to exceed 65°C, de-
termine the total power rating of the electronic devices that can
be mounted into the duct. Answer: 640 W
30°C
200 m/min
65 °C
20 cm
FIGURE P7-55
7-56 Repeat Problem 7-55 for a location at 4000-m altitude
where the atmospheric pressure is 61 .66 kPa.
7-57 A 0.4-W cylindrical electronic component with diame-
ter 0.3 cm and length 1 .8 cm and mounted on a circuit board is
cooled by air flowing across it at a velocity of 150 m/min. If
the air temperature is 40°C, determine the surface temperature
of the component.
7-58 Consider a 50-cm-diameter and 95-cm-long hot water
tank. The tank is placed on the roof of a house. The water in-
side the tank is heated to 80°C by a flat-plate solar collector
during the day. The tank is then exposed to windy air at 18°C
with an average velocity of 40 km/h during the night. Estimate
the temperature of the tank after a 45-mm period. Assume the
tank surface to be at the same temperature as the water inside,
and the heat transfer coefficient on the top and bottom surfaces
to be the same as that on the side surface.
7-59 Tu'M Reconsider Problem 7-58. Using EES (or other)
|££^ software, plot the temperature of the tank as a
function of the cooling time as the time varies from 30 mm to
5 h, and discuss the results.
7-60 A 1 .8-m-diameter spherical tank of negligible thickness
contains iced water at 0°C. Air at 25°C flows over the tank with
a velocity of 7 m/s. Determine the rate of heat transfer to the
tank and the rate at which ice melts. The heat of fusion of wa-
ter at 0°C is 333.7 kJ/kg.
7-61 A 10-cm-diameter, 30-cm-high cylindrical bottle con-
tains cold water at 3°C. The bottle is placed in windy air at
27°C. The water temperature is measured to be 11°C after 45
minutes of cooling. Disregarding radiation effects and heat
transfer from the top and bottom surfaces, estimate the average
wind velocity.
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414
HEAT TRANSFER
Flow across Tube Banks
7-62C In flow across tube banks, why is the Reynolds num-
ber based on the maximum velocity instead of the uniform ap-
proach velocity?
7-63C In flow across tube banks, how does the heat transfer
coefficient vary with the row number in the flow direction?
How does it vary with in the transverse direction for a given
row number?
7-64 Combustion air in a manufacturing facility is to be pre-
heated before entering a furnace by hot water at 90°C flowing
through the tubes of a tube bank located in a duct. Air enters
the duct at 15°C and 1 atm with a mean velocity of 3.8 m/s, and
flows over the tubes in normal direction. The outer diameter
of the tubes is 2.1 cm, and the tubes are arranged in-line with
longitudinal and transverse pitches of S L = S T = 5 cm. There
are eight rows in the flow direction with eight tubes in each
row. Determine the rate of heat transfer per unit length of the
tubes, and the pressure drop across the tube bank.
7-65 Repeat Problem 7-64 for staggered arrangement with
S L = S T = 5 cm.
7-66 Air is to be heated by passing it over a bank of
3-m-long tubes inside which steam is condensing at 100°C. Air
approaches the tube bank in the normal direction at 20°C and
1 atm with a mean velocity of 5.2 m/s. The outer diameter of
the tubes is 1 .6 cm, and the tubes are arranged staggered with
longitudinal and transverse pitches of S L = S T = 4 cm. There
are 20 rows in the flow direction with 10 tubes in each row.
Determine (a) the rate of heat transfer, (b) and pressure drop
across the tube bank, and (c) the rate of condensation of steam
inside the tubes.
7-67 Repeat Problem 7-66 for in-line arrangement with
S L = S T = 5 cm.
7-68 Exhaust gases at 1 atm and 300°C are used to preheat
water in an industrial facility by passing them over a bank of
tubes through which water is flowing at a rate of 6 kg/s. The
mean tube wall temperature is 80°C. Exhaust gases approach
the tube bank in normal direction at 4.5 m/s. The outer diame-
ter of the tubes is 2.1 cm, and the tubes are arranged in-line
with longitudinal and transverse pitches of S L = S T = 8 cm.
There are 16 rows in the flow direction with eight tubes in each
row. Using the properties of air for exhaust gases, determine
(a) the rate of heat transfer per unit length of tubes, (b) and
pressure drop across the tube bank, and (c) the temperature rise
of water flowing through the tubes per unit length of tubes.
7-69 Water at 15°C is to be heated to 65°C by passing it over
a bundle of 4-m-long 1 -cm-diameter resistance heater rods
maintained at 90°C. Water approaches the heater rod bundle in
normal direction at a mean velocity of 0.8 m/s. The rods arc
arranged in-line with longitudinal and transverse pitches of
S L = A cm and S T = 3 cm. Determine the number of tube rows
N L in the flow direction needed to achieve the indicated tem-
perature rise.
Water
15°C
0.8m/s i s i= 4cm
FIGURE P7-69
4
D= 1 cm
T. = 90°C
r
7-70 Air is to be cooled in the evaporator section of a re-
frigerator by passing it over a bank of 0.8-cm-outer-diameter
and 0.4-m-long tubes inside which the refrigerant is evaporat-
ing at — 20°C. Air approaches the tube bank in the normal di-
rection at 0°C and 1 atm with a mean velocity of 4 m/s. The
tubes are arranged in-line with longitudinal and transverse
pitches of S L = S T = 1.5 cm. There are 30 rows in the flow di-
rection with 15 tubes in each row. Determine (a) the refriger-
ation capacity of this system and (b) and pressure drop across
the tube bank.
o°c
1 atm
4 m/s
Air
0.8 cm
FIGURE P7-70
Refrigerant, — 20°C
7-71 Repeat Problem 7-70 by solving it for staggered
arrangement with S L = S T = 1.5 cm, and compare the perfor-
mance of the evaporator for the in-line and staggered arrange-
ments.
7-72 A tube bank consists of 300 tubes at a distance of 6 cm
between the centerlines of any two adjacent tubes. Air ap-
proaches the tube bank in the normal direction at 40°C and
1 atm with a mean velocity of 7 m/s. There are 20 rows in the
flow direction with 1 5 tubes in each row with an average sur-
face temperature of 140°C. For an outer tube diameter of 2 cm,
determine the average heat transfer coefficient.
Special Topic: Thermal Insulation
7-73C What is thermal insulation? How does a thermal insu-
lator differ in purpose from an electrical insulator and from a
sound insulator?
7-74C Does insulating cold surfaces save energy? Explain.
cen58 93 3_ch07.qxd 9/4/2002 12:13 PM Page 415
7-75C What is the /?-value of insulation? How is it deter-
mined? Will doubling the thickness of flat insulation double its
fl-value?
7-76C How does the R-va\ue of an insulation differ from its
thermal resistance?
7-77C Why is the thermal conductivity of superinsulation
orders of magnitude lower than the thermal conductivities of
ordinary insulations?
7-78C Someone suggests that one function of hair is to insu-
late the head. Do you agree with this suggestion?
7-79C Name five different reasons for using insulation in in-
dustrial facilities.
7-80C What is optimum thickness of insulation? How is it
determined?
7-81 What is the thickness of flat R-S (in SI units) insulation
whose thermal conductivity is 0.04 W/m • °C?
7-82E What is the thickness of flat R-20 (in English units)
insulation whose thermal conductivity is 0.02 Btu/h ■ ft • °F?
7-83 Hot water at 110°C flows in a cast iron pipe (k = 52
W/m • °C) whose inner radius is 2.0 cm and thickness is 0.3
cm. The pipe is to be covered with adequate insulation so that
the temperature of the outer surface of the insulation does not
exceed 30°C when the ambient temperature is 22°C. Taking
the heat transfer coefficients inside and outside the pipe to be
h, = 80 W/m 2 ■ °C and h B = 22 W/m 2 ■ °C, respectively, deter-
mine the thickness of fiber glass insulation (k = 0.038 W/m ■
°C) that needs to be installed on the pipe.
Answer: 1.32 cm
7-84 [JJ^l Reconsider Problem 7-83. Using EES (or other)
b^2 software, plot the thickness of the insulation as a
function of the maximum temperature of the outer surface of
insulation in the range of 24°C to 48°C. Discuss the results.
7-85 f~J&\ Consider a furnace whose average outer surface
^Ms temperature is measured to be 90°C when the av-
erage surrounding air temperature is 27°C. The furnace is 6 m
long and 3 m in diameter. The plant operates 80 h per week for
52 weeks per year. You are to insulate the furnace using fiber-
glass insulation (k ms = 0.038 W/m ■ °C) whose cost is S10/m 2
per cm of thickness for materials, plus $30/m 2 for labor regard-
less of thickness. The combined heat transfer coefficient on the
outer surface is estimated to be h = 30 W/m 2 • °C. The furnace
uses natural gas whose unit cost is $0.50/therm input (1 therm =
105,500 kJ), and the efficiency of the furnace is 78 percent. The
management is willing to authorize the installation of the thick-
est insulation (in whole cm) that will pay for itself (materials and
labor) in one year. That is, the total cost of insulation should be
roughly equal to the drop in the fuel cost of the furnace for one
year. Determine the thickness of insulation to be used and the
money saved per year. Assume the surface temperature of the
furnace and the heat transfer coefficient are to remain constant.
Answer: 14 cm
415
CHAPTER 7
7-85 Repeat Problem 7-85 for an outer surface temperature
of 75 °C for the furnace.
7-87E Steam at 400°F is flowing through a steel pipe (k = 8.7
Btu/h ■ ft • °F) whose inner and outer diameters are 3.5 in. and
4.0 in., respectively, in an environment at 60°F. The pipe is insu-
lated with 1 -in. -thick fiberglass insulation (k = 0.020 Btu/h • ft ■
°F), and the heat transfer coefficients on the inside and the out-
side of the pipe are 30 Btu/h • ft 2 • °F and 5 Btu/h ■ ft 2 • °F, re-
spectively. It is proposed to add another 1 -in. -thick layer of
fiberglass insulation on top of the existing one to reduce the heat
losses further and to save energy and money. The total cost of
new insulation is $7 per ft length of the pipe, and the net fuel cost
of energy in the steam is $0.01 per 1000 Btu (therefore, each
1000 Btu reduction in the heat loss will save the plant $0.01).
The policy of the plant is to implement energy conservation
measures that pay for themselves within two years. Assuming
continuous operation (8760 h/year), determine if the proposed
additional insulation is justified.
7-88 The plumbing system of a plant involves a section of a
plastic pipe (k = 0.16 W/m • °C) of inner diameter 6 cm and
outer diameter 6.6 cm exposed to the ambient air. You are to
insulate the pipe with adequate weather-jacketed fiberglass
insulation (k = 0.035 W/m • °C) to prevent freezing of water in
the pipe. The plant is closed for the weekends for a period of
60 h, and the water in the pipe remains still during that period.
The ambient temperature in the area gets as low as — 10°C in
winter, and the high winds can cause heat transfer coefficients
as high as 30 W/m 2 ■ °C. Also, the water temperature in the
pipe can be as cold as 15°C, and water starts freezing when its
temperature drops to 0°C. Disregarding the convection resis-
tance inside the pipe, determine the thickness of insulation that
will protect the water from freezing under worst conditions.
7-89 Repeat Problem 7-88 assuming 20 percent of the water
in the pipe is allowed to freeze without jeopardizing safety.
Answer: 27.9 cm
Review Problems
7-90 Consider a house that is maintained at 22°C at all times.
The walls of the house have R-3.38 insulation in SI units (i.e.,
an Llk value or a thermal resistance of 3.38 m 2 ■ °C/W). During
a cold winter night, the outside air temperature is 4°C and wind
at 50 km/h is blowing parallel to a 3-m-high and 8-m-long wall
of the house. If the heat transfer coefficient on the interior
surface of the wall is 8 W/m 2 • °C, determine the rate of heat
loss from that wall of the house. Draw the thermal resistance
network and disregard radiation heat transfer.
Answer: 122 W
7-91 An automotive engine can be approximated as a 0.4-m-
high, 0.60-m-wide, and 0.7-m-long rectangular block. The
bottom surface of the block is at a temperature of 75°C and has
an emissivity of 0.92. The ambient air is at 5°C, and the road
surface is at 10°C. Determine the rate of heat transfer from the
bottom surface of the engine block by convection and radiation
cen58 93 3_ch07.qxd 9/4/2002 12:13 PM Page 416
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HEAT TRANSFER
as the car travels at a velocity of 60 km/h. Assume the flow to
be turbulent over the entire surface because of the constant
agitation of the engine block. How will the heat transfer be
affected when a 2-mm-thick gunk (k = 3 W/m ■ °C) has
formed at the bottom surface as a result of the dirt and oil
collected at that surface over time? Assume the metal
temperature under the gunk still to be 75°C.
Engine
block
Air
60 km/h
5°C
Gunk
FIGURE P7-91
7-92E The passenger compartment of a minivan traveling at
60 mph can be modeled as a 3.2-ft-high, 6-ft-wide, and 11-ft-
long rectangular box whose walls have an insulating value
of R-3 (i.e., a wall thickness-to-thermal conductivity ratio of
3 h • ft 2 • °F/Btu). The interior of a minivan is maintained at an
average temperature of 70°F during a trip at night while the
outside air temperature is 90°F. The average heat transfer
coefficient on the interior surfaces of the van is 1 .2 Btu/h • ft 2 ■
°F. The air flow over the exterior surfaces can be assumed to be
turbulent because of the intense vibrations involved, and the
heat transfer coefficient on the front and back surfaces can be
taken to be equal to that on the top surface. Disregarding any
heat gain or loss by radiation, determine the rate of heat
transfer from the ambient air to the van.
Air
60 mph
90°F
FIGURE P7-92E
7-93 Consider a house that is maintained at a constant
temperature of 22°C. One of the walls of the house has three
single-pane glass windows that are 1.5 m high and 1.2 m long.
The glass (k = 0.78 W/m • °C) is 0.5 cm thick, and the heat
transfer coefficient on the inner surface of the glass is
8 W/m 2 • C. Now winds at 60 km/h start to blow parallel to the
surface of this wall. If the air temperature outside is — 2°C,
determine the rate of heat loss through the windows of this
wall. Assume radiation heat transfer to be negligible.
7-94 Consider a person who is trying to keep cool on a hot
summer day by turning a fan on and exposing his body to air
flow. The air temperature is 32°C, and the fan is blowing air at
a velocity of 5 m/s. The surrounding surfaces are at 40°C, and
the emissivity of the person can be taken to be 0.9. If the
person is doing light work and generating sensible heat at a rate
of 90 W, determine the average temperature of the outer
surface (skin or clothing) of the person. The average human
body can be treated as a 30-cm-diameter cylinder with an
exposed surface area of 1 .7 m 2 . Answer: 36.2°C
7-95 Four power transistors, each dissipating 12 W, are
mounted on a thin vertical aluminum plate (k = 237 W/m • °C)
22 cm X 22 cm in size. The heat generated by the transistors is
to be dissipated by both surfaces of the plate to the surrounding
air at 20°C, which is blown over the plate by a fan at a velocity
of 250 m/min. The entire plate can be assumed to be nearly
isothermal, and the exposed surface area of the transistor can
be taken to be equal to its base area. Determine the temperature
of the aluminum plate.
7-96 A 3-m-internal-diameter spherical tank made of 1 -cm-
thick stainless steel (k = 15 W/m • °C) is used to store iced
water at 0°C. The tank is located outdoors at 30°C and is
subjected to winds at 25 km/h. Assuming the entire steel tank
to be at 0°C and thus its thermal resistance to be negligible,
determine (a) the rate of heat transfer to the iced water in the
tank and (b) the amount of ice at 0°C that melts during a 24-h
period. The heat of fusion of water at atmospheric pressure is
h jf = 333.7 kJ/kg. Disregard any heat transfer by radiation.
T = 30°C
25 km/h
1 cm
FIGURE P7-96
7-97 Repeat Problem 7-96, assuming the inner surface of
the tank to be at 0°C but by taking the thermal resistance of the
tank and heat transfer by radiation into consideration. Assume
the average surrounding surface temperature for radiation
exchange to be 15°C and the outer surface of the tank to have
an emissivity of 0.9. Answers: (a) 9630 W, (b) 2493 kg
7-98E A transistor with a height of 0.25 in. and a diameter of
0.22 in. is mounted on a circuit board. The transistor is cooled
by air flowing over it at a velocity of 500 ft/min. If the air
temperature is 120°F and the transistor case temperature is not
to exceed 1 80°F, determine the amount of power this transistor
can dissipate safely.
cen58 93 3_ch07.qxd 9/4/2002 12:13 PM Page 417
Air, 500 ft/min
120°F
lllllll
Power
transistor
T. < 180°F
0.22 in.
0.25 in.
FIGURE P7-98E
7-99 The roof of a house consists of a 15-cm-thick concrete
slab (k = 2 W/m ■ °C) 15 m wide and 20 m long. The
convection heat transfer coefficient on the inner surface of the
roof is 5 W/m 2 • °C. On a clear winter night, the ambient air is
reported to be at 10°C, while the night sky temperature is 100
K. The house and the interior surfaces of the wall are
maintained at a constant temperature of 20°C. The emissivity
of both surfaces of the concrete roof is 0.9. Considering both
radiation and convection heat transfer, determine the rate of
heat transfer through the roof when wind at 60 km/h is blowing
over the roof.
If the house is heated by a furnace burning natural gas with
an efficiency of 85 percent, and the price of natural gas is
$0.60/therm (1 therm = 105,500 kJ of energy content),
determine the money lost through the roof that night during a
14-h period. Answers: 28 kW, $9.44
r air =io°c
Concrete
m i n r0 °f
60 km/h 20 m
15 cm
FIGURE P7-99
7-100 Steam at 250°C flows in a stainless steel pipe (k = 15
W/m • °C) whose inner and outer diameters are 4 cm and 4.6
cm, respectively. The pipe is covered with 3.5-cm-thick glass
wool insulation (k = 0.038 W/m • °C) whose outer surface has
an emissivity of 0.3. Heat is lost to the surrounding air and
surfaces at 3°C by convection and radiation. Taking the heat
transfer coefficient inside the pipe to be 80 W/m 2 • °C,
determine the rate of heat loss from the steam per unit length of
the pipe when air is flowing across the pipe at 4 m/s.
417
CHAPTER 7
7-101 The boiling temperature of nitrogen at atmospheric
pressure at sea level (1 atm pressure) is — 196°C. Therefore,
nitrogen is commonly used in low-temperature scientific
studies, since the temperature of liquid nitrogen in a tank open
to the atmosphere will remain constant at — 196°C until it is
depleted. Any heat transfer to the tank will result in the
evaporation of some liquid nitrogen, which has a heat of
vaporization of 198 kJ/kg and a density of 810 kg/m 3 at 1 atm.
Consider a 4-m-diameter spherical tank that is initially filled
with liquid nitrogen at 1 atm and — 196°C. The tank is exposed
to 20°C ambient air and 40 km/h winds. The temperature of the
thin-shelled spherical tank is observed to be almost the same as
the temperature of the nitrogen inside. Disregarding any radia-
tion heat exchange, determine the rate of evaporation of the
liquid nitrogen in the tank as a result of heat transfer from the
ambient air if the tank is (a) not insulated, (b) insulated with
5-cm-thick fiberglass insulation (k = 0.035 W/m • °C), and
(c) insulated with 2-cm-thick superinsulation that has an effec-
tive thermal conductivity of 0.00005 W/m • °C.
N 2 vapor
40 km/h
Insulation
FIGURE P7-1 01
7-102 Repeat Problem 7-101 for liquid oxygen, which has a
boiling temperature of — 183°C, a heat of vaporization of 213
kJ/kg, and a density of 1140 kg/m 3 at 1 atm pressure.
7-103 A 0.3 -cm-thick, 12-cm-high, and 18-cm-long circuit
board houses 80 closely spaced logic chips on one side, each
dissipating 0.06 W. The board is impregnated with copper
fillings and has an effective thermal conductivity of 16
W/m ■ °C. All the heat generated in the chips is conducted
across the circuit board and is dissipated from the back side of
the board to the ambient air at 30°C, which is forced to flow
over the surface by a fan at a free-stream velocity of 400
m/min. Determine the temperatures on the two sides of the
circuit board.
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HEAT TRANSFER
7-104E fife. It is we ll known that cold air feels much
xjgy colder in windy weather than what the ther-
mometer reading indicates because of the "chilling effect" of
the wind. This effect is due to the increase in the convection
heat transfer coefficient with increasing air velocities. The
equivalent windchill temperature in °F is given by (1993
ASHRAE Handbook of Fundamentals, Atlanta, GA, p. 8.15)
T =914
'1-4 - r amb ie„t) (0.475 - 0.0203T + 0.304VT)
where T is the wind velocity in mph and T^^^ is the ambient
air temperature in °F in calm air, which is taken to be air with
light winds at speeds up to 4 mph. The constant 91.4°F in the
above equation is the mean skin temperature of a resting person
in a comfortable environment. Windy air at a temperature r amb i e nt
and velocity T will feel as cold as calm air at a temperature
^cquiv The equation above is valid for winds up to 43 mph.
Winds at higher velocities produce little additional chilling
effect. Determine the equivalent wind chill temperature of an
environment at 10°F at wind speeds of 10, 20, 30, and 40 mph.
Exposed flesh can freeze within one minute at a temperature
below — 25°F in calm weather. Does a person need to be
concerned about this possibility in any of the cases above?
Winds
40°F
35 mph
FIGURE P7-104E
7-105E
Reconsider Problem 7-104E. Using EES (or
other) software, plot the equivalent wind chill
temperatures in °F as a function of wind velocity in the range
of 4 mph to 100 mph for ambient temperatures of 20°F, 40°F
and 60°F Discuss the results.
Design and Essay Problems
7-106 On average, superinsulated homes use just 15 percent
of the fuel required to heat the same size conventional home
built before the energy crisis in the 1970s. Write an essay on
superinsulated homes, and identify the features that make them
so energy efficient as well as the problems associated with
them. Do you think superinsulated homes will be economically
attractive in your area?
7-107 Conduct this experiment to determine the heat loss co-
efficient of your house or apartment in W/°C or But/h • °F First
make sure that the conditions in the house are steady and the
house is at the set temperature of the thermostat. Use an out-
door thermometer to monitor outdoor temperature. One
evening, using a watch or timer, determine how long the heater
was on during a 3-h period and the average outdoor tempera-
ture during that period. Then using the heat output rating of
your heater, determine the amount of heat supplied. Also, esti-
mate the amount of heat generation in the house during that pe-
riod by noting the number of people, the total wattage of lights
that were on, and the heat generated by the appliances and
equipment. Using that information, calculate the average rate
of heat loss from the house and the heat loss coefficient.
7-108 The decision of whether to invest in an energy-saving
measure is made on the basis of the length of time for it to pay
for itself in projected energy (and thus cost) savings. The easi-
est way to reach a decision is to calculate the simple payback
period by simply dividing the installed cost of the measure by
the annual cost savings and comparing it to the lifetime of the
installation. This approach is adequate for short payback peri-
ods (less than 5 years) in stable economies with low interest
rates (under 10 percent) since the error involved is no larger
than the uncertainties. However, if the payback period is long,
it may be necessary to consider the interest rate if the money is
to be borrowed, or the rate of return if the money is invested
elsewhere instead of the energy conservation measure. For ex-
ample, a simple payback period of five years corresponds to
5.0, 6.12, 6.64, 7.27, 8.09, 9.919, 10.84, and 13.91 for an inter-
est rate (or return on investment) of 0, 6, 8, 10, 12, 14, 16, and
18 percent, respectively. Finding out the proper relations from
engineering economics books, determine the payback periods
for the interest rates given above corresponding to simple pay-
back periods of 1 through 10 years.
7-109 Obtain information on frostbite and the conditions
under which it occurs. Using the relation in Problem 7-1 04E,
prepare a table that shows how long people can stay in cold and
windy weather for specified temperatures and wind speeds
before the exposed flesh is in danger of experiencing frostbite.
7-110 Write an article on forced convection cooling with air,
helium, water, and a dielectric liquid. Discuss the advantages
and disadvantages of each fluid in heat transfer. Explain the
circumstances under which a certain fluid will be most suitable
for the cooling job.
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CHAPTER
INTERNAL FORCED
CONVECTION
t
Liquid or gas flow through pipes or ducts is commonly used in heating and
cooling applications. The fluid in such applications is forced to flow by a
fan or pump through a tube that is sufficiently long to accomplish the
desired heat transfer. In this chapter we will pay particular attention to the de-
termination of the friction factor and convection coefficient since they are di-
rectly related to the pressure drop and heat transfer rate, respectively. These
quantities are then used to determine the pumping power requirement and the
required tube length.
There is a fundamental difference between external and internal flows. In
external flow, considered in Chapter 7, the fluid has a free surface, and thus
the boundary layer over the surface is free to grow indefinitely. In internal
flow, however, the fluid is completely confined by the inner surfaces of the
tube, and thus there is a limit on how much the boundary layer can grow.
We start this chapter with a general physical description of internal flow,
and the mean velocity and mean temperature. We continue with the discussion
of the hydrodynamic and thermal entry lengths, developing flow, and fully
developed flow. We then obtain the velocity and temperature profiles for
fully developed laminar flow, and develop relations for the friction factor and
Nusselt number. Finally we present empirical relations for developing
and fully developed flows, and demonstrate their use.
CONTENTS
8-1 Introduction 420
8-2 Mean Velocity and
Mean Temperature 420
8-3 The Entrance Region 423
8-4 General Thermal Analysis 426
8-5 Laminar Flow in Tubes 431
8-6 Turbulent Flow in Tubes 441
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HEAT TRANSFER
Circular pipe
Rectangular
duct
FIGURE 8-1
Circular pipes can withstand large
pressure differences between the
inside and the outside without
undergoing any distortion, but
the noncircular pipes cannot.
8-1 - INTRODUCTION
You have probably noticed that most fluids, especially liquids, are transported
in circular pipes. This is because pipes with a circular cross section can with-
stand large pressure differences between the inside and the outside without
undergoing any distortion. Noncircular pipes are usually used in applications
such as the heating and cooling systems of buildings where the pressure dif-
ference is relatively small and the manufacturing and installation costs are
lower (Fig. 8-1). For a fixed surface area, the circular tube gives the most heat
transfer for the least pressure drop, which explains the overwhelming popu-
larity of circular tubes in heat transfer equipment.
The terms pipe, duct, tube, and conduit are usually used interchangeably for
flow sections. In general, flow sections of circular cross section are referred to
as pipes (especially when the fluid is a liquid), and the flow sections of non-
circular cross section as ducts (especially when the fluid is a gas). Small di-
ameter pipes are usually referred to as tubes. Given this uncertainty, we will
use more descriptive phrases (such as a circular pipe or a rectangular duct)
whenever necessary to avoid any misunderstandings.
Although the theory of fluid flow is reasonably well understood, theoretical
solutions are obtained only for a few simple cases such as the fully developed
laminar flow in a circular pipe. Therefore, we must rely on the experimental
results and the empirical relations obtained for most fluid flow problems
rather than closed form analytical solutions. Noting that the experimental re-
sults are obtained under carefully controlled laboratory conditions, and that no
two systems are exactly alike, we must not be so naive as to view the results
obtained as "exact." An error of 10 percent (or more) in friction or convection
coefficient calculated using the relations in this chapter is the "norm" rather
than the "exception."
Perhaps we should mention that the friction between the fluid layers in a
tube may cause a slight rise in fluid temperature as a result of mechanical en-
ergy being converted to thermal energy. But this frictional heating is too small
to warrant any consideration in calculations, and thus is disregarded. For ex-
ample, in the absence of any heat transfer, no noticeable difference will be de-
tected between the inlet and exit temperatures of a fluid flowing in a tube. The
primary consequence of friction in fluid flow is pressure drop. Thus, it is rea-
sonable to assume that any temperature change in the fluid is due to heat
transfer. But frictional heating must be considered for flows that involve
highly viscous fluids with large velocity gradients.
In most practical applications, the flow of a fluid through a pipe or duct can
be approximated to be one-dimensional, and thus the properties can be as-
sumed to vary in one direction only (the direction of flow). As a result, all
properties are uniform at any cross section normal to the flow direction, and
the properties are assumed to have bulk average values over the cross section.
But the values of the properties at a cross section may change with time unless
the flow is steady.
8-2 - MEAN VELOCITY AND MEAN TEMPERATURE
In external flow, the free-stream velocity served as a convenient reference
velocity for use in the evaluation of the Reynolds number and the friction
cen58933_ch08.qxd 9/4/2002 11:29 AM Page 421
coefficient. In internal flow, there is no free stream and thus we need an alter-
native. The fluid velocity in a tube changes from zero at the surface because
of the no-slip condition, to a maximum at the tube center. Therefore, it is con-
venient to work with an average or mean velocity T„„ which remains con-
stant for incompressible flow when the cross sectional area of the tube is
constant.
The mean velocity in actual heating and cooling applications may change
somewhat because of the changes in density with temperature. But, in prac-
tice, we evaluate the fluid properties at some average temperature and treat
them as constants. The convenience in working with constant properties usu-
ally more than justifies the slight loss in accuracy.
The value of the mean velocity T„, in a tube is determined from the require-
ment that the conservation of mass principle be satisfied (Fig. 8-2). That is,
pT m A c
[ pV(r,x)dA c
J A
(8-1)
421
CHAPTER 8
(a) Actual
■y
(b) Idealized
FIGURE 8-2
Actual and idealized velocity profiles
for flow in a tube (the mass flow rate
of the fluid is the same for both cases).
where m is the mass flow rate, p is the density, A c is the cross sectional area,
and Y(r, x) is the velocity profile. Then the mean velocity for incompressible
flow in a circular tube of radius R can be expressed as
T
[ pY(r,x)dA c \ R pY{r,x)2Ttrdr
Ja, Jo
pA c
pirR 2
2 f R
x)rdr
(8-2)
Therefore, when we know the mass flow rate or the velocity profile, the mean
velocity can be determined easily.
When a fluid is heated or cooled as it flows through a tube, the temperature
of the fluid at any cross section changes from T s at the surface of the wall to
some maximum (or minimum in the case of heating) at the tube center. In
fluid flow it is convenient to work with an average or mean temperature T m
that remains uniform at a cross section. Unlike the mean velocity, the mean
temperature T m will change in the flow direction whenever the fluid is heated
or cooled.
The value of the mean temperature T m is determined from the requirement
that the conservation of energy principle be satisfied. That is, the energy trans-
ported by the fluid through a cross section in actual flow must be equal to the
energy that would be transported through the same cross section if the fluid
were at a constant temperature T m . This can be expressed mathematically as
(Fig. 8-3)
mC p T„,
f C p Thm = f P C,
Jm J A,
TTdA„
(8-3)
where C p is the specific heat of the fluid. Note that the product rhC p T„, at any
cross section along the tube represents the energy flow with the fluid at that
cross section. Then the mean temperature of a fluid with constant density and
specific heat flowing in a circular pipe of radius R can be expressed as
f C p Thm f
Jm JO
CT{pY2ixrdr)
mC„
pT m (TTR 2 )C„
T(r, x) T(r, x) rdr
(8-4)
(a) Actual
(b) Idealized
FIGURE 8-3
Actual and idealized temperature
profiles for flow in a tube (the rate at
which energy is transported with the
fluid is the same for both cases).
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HEAT TRANSFER
Note that the mean temperature T m of a fluid changes during heating or cool-
ing. Also, the fluid properties in internal flow are usually evaluated at the bulk
mean fluid temperature, which is the arithmetic average of the mean temper-
atures at the inlet and the exit. That is, T b = (T m , + T e )/2.
Circular tube:
i-Sg*-
Square duct:
4a 2
D,,
4a
Rectangular duct:
D,
4d/)
lab
a + b
2(a + b)
FIGURE 8-4
The hydraulic diameter D h = AA C Ip
is defined such that it reduces to
ordinary diameter for circular tubes.
Laminar Turbulent
^rn^h-
Die trace ' ^-Pipe wall
FIGURE 8-5
In the transitional flow region of
2300 < Re < 4000, the flow
switches between laminar
and turbulent randomly.
Laminar and Turbulent Flow In Tubes
Flow in a tube can be laminar or turbulent, depending on the flow conditions.
Fluid flow is streamlined and thus laminar at low velocities, but turns turbu-
lent as the velocity is increased beyond a critical value. Transition from lami-
nar to turbulent flow does not occur suddenly; rather, it occurs over some
range of velocity where the flow fluctuates between laminar and turbulent
flows before it becomes fully turbulent. Most pipe flows encountered in prac-
tice are turbulent. Laminar flow is encountered when highly viscous fluids
such as oils flow in small diameter tubes or narrow passages.
For flow in a circular tube, the Reynolds number is defined as
Re
P y„,D_T m D
JJL V
(8-5)
where T„, is the mean fluid velocity, D is the diameter of the tube, and v =
|x/p is the kinematic viscosity of the fluid.
For flow through noncircular tubes, the Reynolds number as well as the
Nusselt number and the friction factor are based on the hydraulic diameter
D h defined as (Fig. 8-4)
D„
AA C
(8-6)
where A c is the cross sectional area of the tube and p is its perimeter. The
hydraulic diameter is defined such that it reduces to ordinary diameter D for
circular tubes since
Circular tubes:
D„
AA C
~P~
4tt£> 2 /4
ttD
D
It certainly is desirable to have precise values of Reynolds numbers for
laminar, transitional, and turbulent flows, but this is not the case in practice.
This is because the transition from laminar to turbulent flow also depends on
the degree of disturbance of the flow by surface roughness, pipe vibrations,
and the fluctuations in the flow. Under most practical conditions, the flow in a
tube is laminar for Re < 2300, turbulent for Re > 10,000, and transitional in
between. That is,
Re < 2300
2300 < Re < 10,000
Re > 10,000
laminar flow
transitional flow
turbulent flow
In transitional flow, the flow switches between laminar and turbulent
randomly (Fig. 8-5). It should be kept in mind that laminar flow can be
maintained at much higher Reynolds numbers in very smooth pipes by
avoiding flow disturbances and tube vibrations. In such carefully controlled
cen58933_ch08.qxd 9/4/2002 11:29 AM Page 423
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CHAPTER 8
experiments, laminar flow has been maintained at Reynolds numbers of up to
100,000.
8-3 - THE ENTRANCE REGION
Consider a fluid entering a circular tube at a uniform velocity. As in external
flow, the fluid particles in the layer in contact with the surface of the tube will
come to a complete stop. This layer will also cause the fluid particles in the
adjacent layers to slow down gradually as a result of friction. To make up for
this velocity reduction, the velocity of the fluid at the midsection of the tube
will have to increase to keep the mass flow rate through the tube constant. As
a result, a velocity boundary layer develops along the tube. The thickness of
this boundary layer increases in the flow direction until the boundary layer
reaches the tube center and thus fills the entire tube, as shown in Figure 8-6.
The region from the tube inlet to the point at which the boundary layer
merges at the centerline is called the hydrodynamic entrance region, and the
length of this region is called the hydrodynamic entry length L h . Flow in the
entrance region is called hydro dynamic ally developing flow since this is the
region where the velocity profile develops. The region beyond the entrance re-
gion in which the velocity profile is fully developed and remains unchanged
is called the hydrodynamically fully developed region. The velocity profile
in the fully developed region is parabolic in laminar flow and somewhat flat-
ter in turbulent flow due to eddy motion in radial direction.
Now consider a fluid at a uniform temperature entering a circular tube
whose surface is maintained at a different temperature. This time, the fluid
particles in the layer in contact with the surface of the tube will assume the
surface temperature. This will initiate convection heat transfer in the tube and
the development of a thermal boundary layer along the tube. The thickness of
this boundary layer also increases in the flow direction until the boundary
layer reaches the tube center and thus fills the entire tube, as shown in
Figure 8-7.
The region of flow over which the thermal boundary layer develops and
reaches the tube center is called the thermal entrance region, and the length
of this region is called the thermal entry length L t . Flow in the thermal en-
trance region is called thermally developing flow since this is the region where
the temperature profile develops. The region beyond the thermal entrance re-
gion in which the dimensionless temperature profile expressed as (T s — T)l
(T s — T m ) remains unchanged is called the thermally fully developed region.
The region in which the flow is both hydrodynamically and thermally devel-
oped and thus both the velocity and dimensionless temperature profiles re-
main unchanged is called fully developed flow. That is,
-Velocity boundary layer
-Velocity profile
t r
/
s
^
^*V
~-
Vi
\
— »
./
V
^
fc/
— ►*
Hyd
rodynamic
entrai
ice region
— Hydr
odynamically
fully developed region
FIGURE 8-6
The development of the
velocity boundary layer in a
tube. (The developed mean velocity
profile will be parabolic in laminar
flow, as shown, but somewhat
blunt in turbulent flow.)
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HEAT TRANSFER
FIGURE 8-7
The development of the
thermal boundary layer in a tube.
(The fluid in the tube is being cooled.)
• Thermal
boundary layer
entrance region
Thermally -
fully developed region
Hydro dynamic ally fully developed:
Thermally fully developed:
3T(r, x)
dx
-> T = V(r)
d_
dx
T s (x)-T(r,x)
TJx) - TJx)
(8-7)
(8-8)
The friction factor is related to the shear stress at the surface, which is re-
lated to the slope of the velocity profile at the surface. Noting that the veloc-
ity profile remains unchanged in the hydrodynamically fully developed
region, the friction factor also remains constant in that region. A similar argu-
ment can be given for the heat transfer coefficient in the thermally fully de-
veloped region.
In a thermally fully developed region, the derivative of (T s — T)/(T S — T„,)
with respect to x is zero by definition, and thus (T s — T)/(T S — T m ) is indepen-
dent of x. Then the derivative of (T s — T)/(T S — T ni ) with respect r must also
be independent of x. That is,
d_
dr\T,
Surface heat flux can be expressed as
4s = h x (T s - T m )
(dT/dr)\ r=R
dr
*m
k(dT/dr)\ r=R
(8-9)
(8-10)
which, from Eq. 8-9, is independent of x. Thus we conclude that in the ther-
mally fully developed region of a tube, the local convection coefficient is con-
stant (does not vary with x). Therefore, both the friction and convection
coefficients remain constant in the fully developed region of a tube.
Note that the temperature profile in the thermally fully developed region
may vary with x in the flow direction. That is, unlike the velocity profile, the
temperature profile can be different at different cross sections of the tube in
the developed region, and it usually is. However, the dimensionless tempera-
ture profile defined above remains unchanged in the thermally developed re-
gion when the temperature or heat flux at the tube surface remains constant.
During laminar flow in a tube, the magnitude of the dimensionless Prandtl
number Pr is a measure of the relative growth of the velocity and thermal
boundary layers. For fluids with Pr ~ 1, such as gases, the two boundary lay-
ers essentially coincide with each other. For fluids with Pr > 1, such as oils,
the velocity boundary layer outgrows the thermal boundary layer. As a result,
cen58933_ch08.qxd 9/4/2002 11:29 AM Page 425
425
CHAPTER 8
the hydrodynamic entry length is smaller than the thermal entry length. The
opposite is true for fluids with Pr -4 1 such as liquid metals.
Consider a fluid that is being heated (or cooled) in a tube as it flows through
it. The friction factor and the heat transfer coefficient are highest at the tube
inlet where the thickness of the boundary layers is zero, and decrease gradu-
ally to the fully developed values, as shown in Figure 8-8. Therefore, the
pressure drop and heat flux are higher in the entrance regions of a tube, and
the effect of the entrance region is always to enhance the average friction and
heat transfer coefficients for the entire tube. This enhancement can be signif-
icant for short tubes but negligible for long ones.
Entry Lengths
The hydrodynamic entry length is usually taken to be the distance from the
tube entrance where the friction coefficient reaches within about 2 percent of
the fully developed value. In laminar flow, the hydrodynamic and thermal
entry lengths are given approximately as [see Kays and Crawford (1993),
Ref. 13, and Shah and Bhatti (1987), Ref. 25]
PrL,
(8-11)
(8-12)
For Re = 20, the hydrodynamic entry length is about the size of the diameter,
but increases linearly with the velocity. In the limiting case of Re = 2300, the
hydrodynamic entry length is 115D.
In turbulent flow, the intense mixing during random fluctuations usually
overshadows the effects of momentum and heat diffusion, and therefore the
hydrodynamic and thermal entry lengths are of about the same size and inde-
pendent of the Prandtl number. Also, the friction factor and the heat transfer
coefficient remain constant in fully developed laminar or turbulent flow since
the velocity and normalized temperature profiles do not vary in the flow di-
rection. The hydrodynamic entry length for turbulent flow can be determined
from [see Bhatti and Shah (1987), Ref. 1, and Zhi-qing (1982), Ref. 31]
- Thermal boundary layer
-Velocity boundary layer
FIGURE 8-8
Variation of the friction
factor and the convection
heat transfer coefficient in the flow
direction for flow in a tube (Pr > 1).
1.359 Re 1
(8-13)
The hydrodynamic entry length is much shorter in turbulent flow, as expected,
and its dependence on the Reynolds number is weaker. It is IID at Re =
10,000, and increases to 43D at Re = 10 5 . In practice, it is generally agreed
that the entrance effects are confined within a tube length of 10 diameters, and
the hydrodynamic and thermal entry lengths are approximately taken to be
10D
(8-14)
The variation of local Nusselt number along a tube in turbulent flow for
both uniform surface temperature and uniform surface heat flux is given in
Figure 8-9 for the range of Reynolds numbers encountered in heat transfer
equipment. We make these important observations from this figure:
• The Nusselt numbers and thus the convection heat transfer coefficients
are much higher in the entrance region.
cen58933_ch08.qxd 9/4/2002 11:29 AM Page 426
426
HEAT TRANSFER
FIGURE 8-9
Variation of local Nusselt number
along a tube in turbulent flow for both
uniform surface temperature and
uniform surface heat flux
[Deissler(1953), Ref. 4].
Z
800
700
600
500
400
300
200
100
1 1
1 1 1 1 1
I
-D — ►
I
_
-\\
V
\\
V
V
\\
\\
^^
■=-=.
- Nu v H (q s = constant) I
Re =2 X 10 5
105
6 X 10 4
VWT
3 X 10 4
-
v^
1A4
1 1 1 1 1 1 1 1 1
10
x/D
12
14
16
18
20
• The Nusselt number reaches a constant value at a distance of less than
10 diameters, and thus the flow can be assumed to be fully developed for
x > 10D.
• The Nusselt numbers for the uniform surface temperature and uniform
surface heat flux conditions are identical in the fully developed regions,
and nearly identical in the entrance regions. Therefore, Nusselt number
is insensitive to the type of thermal boundary condition, and the turbulent
flow correlations can be used for either type of boundary condition.
Precise correlations for the friction and heat transfer coefficients for the en-
trance regions are available in the literature. However, the tubes used in prac-
tice in forced convection are usually several times the length of either entrance
region, and thus the flow through the tubes is often assumed to be fully devel-
oped for the entire length of the tube. This simplistic approach gives reason-
able results for long tubes and conservative results for short ones.
mC p T :
I • 1 T
I I ',
I I
I I
I I
I I
mC p T e
Energy balance:
Q=mC p (T e -Ti)
FIGURE 8-10
The heat transfer to a fluid flowing in
a tube is equal to the increase in
the energy of the fluid.
S-A - GENERAL THERMAL ANALYSIS
You will recall that in the absence of any work interactions (such as electric
resistance heating), the conservation of energy equation for the steady flow of
a fluid in a tube can be expressed as (Fig. 8-10)
Q =mCST e -n
(W)
(8-15)
where T, and T e are the mean fluid temperatures at the inlet and exit of the
tube, respectively, and Q is the rate of heat transfer to or from the fluid. Note
that the temperature of a fluid flowing in a tube remains constant in the ab-
sence of any energy interactions through the wall of the tube.
The thermal conditions at the surface can usually be approximated with
reasonable accuracy to be constant surface temperature (T s = constant) or
constant surface heat flux (q s = constant). For example, the constant surface
cen58933_ch08.qxd 9/4/2002 11:29 AM Page 427
temperature condition is realized when a phase change process such as boil-
ing or condensation occurs at the outer surface of a tube. The constant surface
heat flux condition is realized when the tube is subjected to radiation or elec-
tric resistance heating uniformly from all directions.
Surface heat flux is expressed as
4s = h,(T s - T m )
(W/m 2 )
(8-16)
where h x is the local heat transfer coefficient and T s and T m are the surface and
the mean fluid temperatures at that location. Note that the mean fluid temper-
ature T m of a fluid flowing in a tube must change during heating or cooling.
Therefore, when h x = h = constant, the surface temperature T s must change
when q 5 = constant, and the surface heat flux q 5 must change when T s = con-
stant. Thus we may have either T s = constant or q s = constant at the surface
of a tube, but not both. Next we consider convection heat transfer for these
two common cases.
Constant Surface Heat Flux (q s = constant)
In the case of q s = constant, the rate of heat transfer can also be expressed as
Q=q,A. = ihC,(T m -Td (W)
Then the mean fluid temperature at the tube exit becomes
T " = Ti + ~rhC„
(8-17)
(8-18)
Note that the mean fluid temperature increases linearly in the flow direction
in the case of constant surface heat flux, since the surface area increases lin-
early in the flow direction (A s is equal to the perimeter, which is constant,
times the tube length).
The surface temperature in the case of constant surface heat flux q s can be
determined from
4s = h(T s ~ T m )
_> 7=7- +
<ls
h
(8-19)
In the fully developed region, the surface temperature T s will also increase lin-
early in the flow direction since h is constant and thus T s — T m = constant
(Fig. 8-11). Of course this is true when the fluid properties remain constant
during flow.
The slope of the mean fluid temperature T m on a T-x diagram can be deter-
mined by applying the steady-flow energy balance to a tube slice of thickness
dx shown in Figure 8-12. It gives
mC p dT m = q s (pdx)
dT,„
q s p
dx mC„
constant
(8-20)
where p is the perimeter of the tube.
Noting that both q s and h are constants, the differentiation of Eq. 8-19 with
respect to x gives
dT m = dT s
dx dx
(8-21)
427
CHAPTER 8
Entrance , Fully developed
region
region
q = constant
t t t I t t t t t t t t t t
FIGURE 8-1 1
Variation of the tube surface
and the mean fluid temperatures
along the tube for the case of
constant surface heat flux.
5Q=h{T s -TJdA
T m ! ! T m + dT m
mCJ m ± ±mCJT m + dTJ
p m
I I
I I
I I
I I
L
dx
FIGURE 8-1 2
Energy interactions for a differential
control volume in a tube.
cen58933_ch08.qxd 9/4/2002 11:29 AM Page 42E
428
HEAT TRANSFER
Also, the requirement that the dimensionless temperature profile remains
unchanged in the fully developed region gives
d_
dx
since T,
(ST,
\dx
dT
dx
dx
dT,
dx
(8-22)
constant. Combining Eqs. 8-20, 8-21, and 8-22 gives
§T = dT 1
dx dx
dx
mC„
constant
(8-23)
T(r)
m
V"^ ^
t t t Xl t t t t x 2 t t
FIGURE 8-13
The shape of the temperature
profile remains unchanged in the
fully developed region of a tube
subjected to constant surface heat flux.
Then we conclude that in fully developed flow in a tube subjected to constant
surface heat flux, the temperature gradient is independent of x and thus the
shape of the temperature profile does not change along the tube (Fig. 8-13).
For a circular tube, p = 2ir7? and rh = pT m A c = pT m (trR 2 ), and Eq. 8-23
becomes
Circular tube:
dT dT s dT m
2*
constant
(8-24)
dx dx dx pY m C p R
where T„, is the mean velocity of the fluid.
Constant Surface Temperature (T s = constant)
From Newton's law of cooling, the rate of heat transfer to or from a fluid
flowing in a tube can be expressed as
Q = M s Ar ave = hA s (T s - TJ,
(W)
(8-25)
where h is the average convection heat transfer coefficient, A s is the heat trans-
fer surface area (it is equal to ttZ)L for a circular pipe of length L), and Ar ave
is some appropriate average temperature difference between the fluid and the
surface. Below we discuss two suitable ways of expressing Ar ave .
In the constant surface temperature (T s = constant) case, AT ave can be
expressed approximately by the arithmetic mean temperature difference
AT^as
AT; + A 7; (T, - T;) + (T s - T e ) T, + T e
A7 « A7 = — ' = — — = T ■
tJX ave iJX am r\ r* *- s r*
= T S - T b (8-26)
where T b = (T t + T e )/2 is the bulk mean fluid temperature, which is the arith-
metic average of the mean fluid temperatures at the inlet and the exit of
the tube.
Note that the arithmetic mean temperature difference Ar am is simply the av-
erage of the temperature differences between the surface and the fluid at the
inlet and the exit of the tube. Inherent in this definition is the assumption that
the mean fluid temperature varies linearly along the tube, which is hardly ever
the case when T s = constant. This simple approximation often gives accept-
able results, but not always. Therefore, we need a better way to evaluate AT ave .
Consider the heating of a fluid in a tube of constant cross section whose
inner surface is maintained at a constant temperature of T y We know that the
cen58933_ch08.qxd 9/4/2002 11:29 AM Page 429
mean temperature of the fluid T m will increase in the flow direction as a result
of heat transfer. The energy balance on a differential control volume shown in
Figure 8-12 gives
m C p dT m
h(T s - T m )dA s
(8-27)
429
CHAPTER 8
That is, the increase in the energy of the fluid (represented by an increase in
its mean temperature by dT m ) is equal to the heat transferred to the fluid from
the tube surface by convection. Noting that the differential surface area is
dA s = pdx, where p is the perimeter of the tube, and that dT m = —d(T s — T m ),
since T 5 is constant, the relation above can be rearranged as
d(T s - T m )
hp
riiC„
dx
(8-28)
Integrating from x = (tube inlet where T m
T„, = T e ) gives
T,) to x = L (tube exit where
In-
mC„
(8-29)
where A s = pL is the surface area of the tube and h is the constant average
convection heat transfer coefficient. Taking the exponential of both sides and
solving for T e gives the following relation which is very useful for the deter-
mination of the mean fluid temperature at the tube exit:
T e = T s - (T, - T,) e xp(-hAJrhC p )
(8-30)
This relation can also be used to determine the mean fluid temperature T m {x)
at any x by replacing A s = pL by px.
Note that the temperature difference between the fluid and the surface de-
cays exponentially in the flow direction, and the rate of decay depends on the
magnitude of the exponent hA x lmC p , as shown in Figure 8-14. This di-
mensionless parameter is called the number of transfer units, denoted by
NTU, and is a measure of the effectiveness of the heat transfer systems. For
NUT > 5, the exit temperature of the fluid becomes almost equal to the sur-
face temperature, T e ~ T s (Fig. 8-15). Noting that the fluid temperature can
approach the surface temperature but cannot cross it, an NTU of about 5 indi-
cates that the limit is reached for heat transfer, and the heat transfer will not in-
crease no matter how much we extend the length of the tube. A small value of
NTU, on the other hand, indicates more opportunities for heat transfer, and the
heat transfer will continue increasing as the tube length is increased. A large
NTU and thus a large heat transfer surface area (which means a large tube)
may be desirable from a heat transfer point of view, but it may be unaccept-
able from an economic point of view. The selection of heat transfer equipment
usually reflects a compromise between heat transfer performance and cost.
Solving Eq. 8-29 for m C p gives
riiC,,
hA,
ln[(7 s - T e )/(T, ~ Td]
(8-31)
- T s = constant
FIGURE 8-14
The variation of the mean fluid
temperature along the tube for the
case of constant temperature.
100°C
m,C„
~A„h
NTU = hA s 1
mC p
T e ,°C
0.01
20.8
0.05
23.9
0.10
27.6
0.50
51.5
1.00
70.6
5.00
99.5
10.00
100.0
FIGURE 8-15
An NTU greater than 5 indicates that
the fluid flowing in a tube will reach
the surface temperature at the exit
regardless of the inlet temperature.
cen58933_ch08.qxd 9/4/2002 11:29 AM Page 43C
430
HEAT TRANSFER
Substituting this into Eq. 8-17, we obtain
Q = hAAT la
(8-32)
where
Ar,„
AT, - AT,
ln[(T s - T„)/(T S - T,)] \MATJAT,)
(8-33)
is the logarithmic mean temperature difference. Note that AT t = T s — T t
and AT e = T s — T e are the temperature differences between the surface and
the fluid at the inlet and the exit of the tube, respectively. This Ar ln relation
appears to be prone to misuse, but it is practically fail-safe, since using T t in
place of T e and vice versa in the numerator and/or the denominator will, at
most, affect the sign, not the magnitude. Also, it can be used for both heating
(T s > Tj and T e ) and cooling (T, < T t and T e ) of a fluid in a tube.
The logarithmic mean temperature difference Ar ln is obtained by tracing the
actual temperature profile of the fluid along the tube, and is an exact repre-
sentation of the average temperature difference between the fluid and the sur-
face. It truly reflects the exponential decay of the local temperature difference.
When AT e differs from AT t by no more than 40 percent, the error in using the
arithmetic mean temperature difference is less than 1 percent. But the error in-
creases to undesirable levels when AT e differs from AT, by greater amounts.
Therefore, we should always use the logarithmic mean temperature difference
when determining the convection heat transfer in a tube whose surface is
maintained at a constant temperature T s .
Steam
T. = 120°
Water ff\
i 5 =c4M
0.3kg/s \J
J15°C
D = 2.5 cm
FIGURE 8-
Schematic
16
for Example !
EXAMPLE 8-1 Heating of Water in a Tube by Steam
Water enters a 2.5-cm-internal-diameter thin copper tube of a heat exchanger
at 15°C at a rate of 0.3 kg/s, and is heated by steam condensing outside at
120°C. If the average heat transfer coefficient is 800 W/m 2 • C, determine the
length of the tube required in order to heat the water to 1 15 C C (Fig. 8-16).
SOLUTION Water is heated by steam in a circular tube. The tube length
required to heat the water to a specified temperature is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Fluid properties are con-
stant. 3 The convection heat transfer coefficient is constant. 4 The conduction
resistance of copper tube is negligible so that the inner surface temperature of
the tube is equal to the condensation temperature of steam.
Properties The specific heat of water at the bulk mean temperature of
(15 + 115)/2 = 65°C is 4187 J/kg • °C. The heat of condensation of steam at
120°C is 2203 kJ/kg (Table A-9).
Analysis Knowing the inlet and exit temperatures of water, the rate of heat
transfer is determined to be
Q = mC p (T e - T t ) = (0.3 kg/s)(4.187 kJ/kg • °C)(115°C - 15°C) = 125.6 kW
The logarithmic mean temperature difference is
cen58933_ch08.qxd 9/4/2002 11:29 AM Page 431
Mi = T S -T t = 120°C - 15°C = 105°C
AT e - ATj 5 - 105
ln MATJAT;) ln(5/105)
The heat transfer surface area is
Q 125.6 kW
Q = hA s AT ln » A,
hAT t „ (0.8 kW/m 2 ■ °C)(32.85°C)
4.78 m 2
Then the required length of tube becomes
t A * 4.78 m 2 „
ttD tt(0.025 m)
Discussion The bulk mean temperature of water during this heating process
is 65°C, and thus the arithmetic mean temperature difference is A7" am =
120 - 65 = 55°C. Using A7" am instead of AT,„ would give L = 36 m, which is
grossly in error. This shows the importance of using the logarithmic mean tem-
perature in calculations.
431
CHAPTER 8
8-5 - LAMINAR FLOW IN TUBES
We mentioned earlier that flow in tubes is laminar for Re < 2300, and that the
flow is fully developed if the tube is sufficiently long (relative to the entry
length) so that the entrance effects are negligible. In this section we consider
the steady laminar flow of an incompressible fluid with constant properties in
the fully developed region of a straight circular tube. We obtain the momen-
tum and energy equations by applying momentum and energy balances to a
differential volume element, and obtain the velocity and temperature profiles
by solving them. Then we will use them to obtain relations for the friction fac-
tor and the Nusselt number. An important aspect of the analysis below is that
it is one of the few available for viscous flow and forced convection.
In fully developed laminar flow, each fluid particle moves at a constant
axial velocity along a streamline and the velocity profile V(r) remains un-
changed in the flow direction. There is no motion in the radial direction, and
thus the velocity component v in the direction normal to flow is everywhere
zero. There is no acceleration since the flow is steady.
Now consider a ring-shaped differential volume element of radius r, thick-
ness dr, and length dx oriented coaxially with the tube, as shown in Figure
8-17. The pressure force acting on a submerged plane surface is the product
of the pressure at the centroid of the surface and the surface area.
The volume element involves only pressure and viscous effects, and thus
the pressure and shear forces must balance each other. A force balance on the
volume element in the flow direction gives
x r + dr
f x
I 1
"'Vl') \R,
dr
z_
—
A t '
t
» ) »
J
— w—
^^^S '' max
(2irrdrP\ - {2-nrdrP\
(2irrdxr) r — (lurdxi),. .
(8-34)
FIGURE 8-17
Free body diagram of a cylindrical
fluid element of radius r, thickness dr,
and length dx oriented coaxially with a
horizontal tube in fully developed
steady flow.
cen58933_ch08.qxd 9/4/2002 11:29 AM Page 432
432
HEAT TRANSFER
which indicates that in fully developed flow in a tube, the viscous and pres-
sure forces balance each other. Dividing by 2irdrdx and rearranging,
dx
</T). r+rfr - (n) r
dr
(8-35)
Taking the limit as dr, dx — > gives
dP , d(n)
dx
dr
(8-36)
Substituting t = — \x(dYldr) and rearranging gives the desired equation,
^d_ dT
r drV dr
dP
dx
(8-37)
The quantity dYldr is negative in tube flow, and the negative sign is included
to obtain positive values for t. (Or, dYldr = —dYldy since y = R — r.) The
left side of this equation is a function of r and the right side is a function of x.
The equality must hold for any value of r and x, and an equality of the form
f(r) = g(x) can happen only if both f(r) and g(x) are equal to constants. Thus
we conclude that dPIdx = constant. This can be verified by writing a force
balance on a volume element of radius R and thickness dx (a slice of the tube),
which gives dPIdx = —2tJR. Here t s is constant since the viscosity and
the velocity profile are constants in the fully developed region. Therefore,
dPIdx = constant.
Equation 8-37 can be solved by rearranging and integrating it twice to give
T(r)
_\_!dP
4jjl \dx
CAnr + C?
(8-38)
The velocity profile T(r) is obtained by applying the boundary conditions
dY/dr = at r = (because of symmetry about the centerline) and T = at
r = R (the no-slip condition at the tube surface). We get
T(r)
R
teP'||
r_
R 2
(8-39)
Therefore, the velocity profile in fully developed laminar flow in a tube is
parabolic with a maximum at the centerline and minimum at the tube surface.
Also, the axial velocity Tis positive for any r, and thus the axial pressure gra-
dient dPIdx must be negative (i.e., pressure must decrease in the flow direc-
tion because of viscous effects).
The mean velocity is determined from its definition by substituting
Eq. 8-39 into Eq. 8-2, and performing the integration. It gives
2_
R 2
Yrdr
-2 f R F?_(dP
R 2 J 4ix\dx
r_
R 2
rdr
I?_(dP
8|jl \dx
(8-40)
Combining the last two equations, the velocity profile is obtained to be
T(r) = 2T J 1
(8-41)
cen58933_ch08.qxd 9/4/2002 11:29 AM Page 433
This is a convenient form for the velocity profile since T„, can be determined
easily from the flow rate information.
The maximum velocity occurs at the centerline, and is determined from
Eq. 8-39 by substituting r = 0,
T = 2T
max m
Therefore, the mean velocity is one-half of the maximum velocity.
(8-42)
Pressure Drop
A quantity of interest in the analysis of tube flow is the pressure drop AP since
it is directly related to the power requirements of the fan or pump to maintain
flow. We note that dPIdx = constant, and integrate it from x = where the
pressure is P, to x = L where the pressure is P 2 . We get
dP
dx
Pi
AP
" L
(8-43)
433
CHAPTER 8
Note that in fluid mechanics, the pressure drop AP is a positive quantity, and
is defined as AP = P, — P 2 . Substituting Eq. 8-43 into the T m expression in
Eq. 8-40, the pressure drop can be expressed as
Laminar flow:
AP
8|jlL c
32|jlL c
R 1
D 2
(8-44)
In practice, it is found convenient to express the pressure drop for all types of
internal flows (laminar or turbulent flows, circular or noncircular tubes,
smooth or rough surfaces) as (Fig. 8-18)
AP=/
l v>v;
D 2
(8-45)
where the dimensionless quantity / is the friction factor (also called the
Darcy friction factor after French engineer Henry Darcy, 1803-1858, who
first studied experimentally the effects of roughness on tube resistance). It
should not be confused with the friction coefficient Cf (also called the Fanning
friction factor), which is defined as Cj = T/pT^/2) = //4.
Equation 8-45 gives the pressure drop for a flow section of length L pro-
vided that (1) the flow section is horizontal so that there are no hydrostatic or
gravity effects, (2) the flow section does not involve any work devices such as
a pump or a turbine since they change the fluid pressure, and (3) the cross sec-
tional area of the flow section is constant and thus the mean flow velocity is
constant.
Setting Eqs. 8-44 and 8-45 equal to each other and solving for /gives the
friction factor for the fully developed laminar flow in a circular tube to be
Circular tube, laminar:
f-
64p,
64
Re
(8-46)
This equation shows that in laminar flow, the friction factor is a function of
the Reynolds number only and is independent of the roughness of the tube
FIGURE 8-18
The relation for pressure
drop is one of the most general
relations in fluid mechanics, and it is
valid for laminar or turbulent flows,
circular or noncircular pipes, and
smooth or rough surfaces.
cen58933_ch08.qxd 9/4/2002 11:29 AM Page 434
434
HEAT TRANSFER
surface. Once the pressure drop is available, the required pumping power is
determined from
W = VAP
"pump ful
(8-47)
where V is the volume flow rate of flow, which is expressed as
\> =
APR 2 „, ttR 4 AP ttD 4 AP
,A r = — — — ttR-
8|xL
8|xL 128jjlL
(8-48)
%u mp = 16hp
J.
%u„, P = lhp
FIGURE 8-1 9
The pumping power requirement for
a laminar flow piping system can
be reduced by a factor of 16 by
doubling the pipe diameter.
, 6, + dr
FIGURE 8-20
The differential volume element
used in the derivation of
energy balance relation.
This equation is known as the Poiseuille's Law, and this flow is called the
Hagen-Poiseuille flow in honor of the works of G. Hagen (1797-1839) and
J. Poiseuille (1799-1869) on the subject. Note from Eq. 8-48 that for a speci-
fied flow rate, the pressure drop and thus the required pumping power is pro-
portional to the length of the tube and the viscosity of the fluid, but it is
inversely proportional to the fourth power of the radius (or diameter) of the
tube. Therefore, the pumping power requirement for a piping system can be
reduced by a factor of 16 by doubling the tube diameter (Fig. 8-19). Of course
the benefits of the reduction in the energy costs must be weighed against the
increased cost of construction due to using a larger diameter tube.
The pressure drop is caused by viscosity, and it is directly related to the wall
shear stress. For the ideal inviscid flow, the pressure drop is zero since there
are no viscous effects. Again, Eq. 8-47 is valid for both laminar and turbulent
flows in circular and noncircular tubes.
Temperature Profile and the Nusselt Number
In the analysis above, we have obtained the velocity profile for fully devel-
oped flow in a circular tube from a momentum balance applied on a volume
element, determined the friction factor and the pressure drop. Below we ob-
tain the energy equation by applying the energy balance to a differential vol-
ume element, and solve it to obtain the temperature profile for the constant
surface temperature and the constant surface heat flux cases.
Reconsider steady laminar flow of a fluid in a circular tube of radius R. The
fluid properties p, k, and C p are constant, and the work done by viscous
stresses is negligible. The fluid flows along the x-axis with velocity T. The
flow is fully developed so that Tis independent of a: and thus T = T(r). Not-
ing that energy is transferred by mass in the .r-direction, and by conduction in
the r-direction (heat conduction in the x-direction is assumed to be negligible),
the steady-flow energy balance for a cylindrical shell element of thickness dr
and length dx can be expressed as (Fig. 8-20)
mC p T x
mC p T x
Iz+Qr-Qr
(8-49)
where m = pVA c
after rearranging,
pVX2irrdr). Substituting and dividing by 2^rdrdx gives,
PQT
T x+C
1 Qr-
Q,
dx
lurdx
dr
(8-50)
cen58933_ch08.qxd 9/4/2002 11:29 AM Page 435
or
y
dT
dx
1
dQ
2pC urdx dr
(8-51)
435
CHAPTER 8
where we used the definition of derivative. But
dT
d A = ±
dr dr
-klurdx
dr
ZTTKCIX — \r —
dr\ dr
(8-52)
Substituting and using a = k/pC„ gives
r dT = ad_ (dT
dx r dr\ dr
(8-53)
which states that the rate of net energy transfer to the control volume by mass
flow is equal to the net rate of heat conduction in the radial direction.
Constant Surface Heat Flux
For fully developed flow in a circular pipe subjected to constant surface heat
flux, we have, from Eq. 8-24,
dT = dT s = dT n = 2q s
dx dx dx pV m CR
constant
(8-54)
If heat conduction in the .i-direction were considered in the derivation of
Eq. 8-53, it would give an additional term ad 2 T/dx 2 , which would be equal to
zero since dT/dx = constant and thus T = T(r). Therefore, the assumption that
there is no axial heat conduction is satisfied exactly in this case.
Substituting Eq. 8-54 and the relation for velocity profile (Eq. 8-41) into
Eq. 8-53 gives
1*
kR
R 2
1 d dT
r dr\ dr
(8-55)
which is a second-order ordinary differential equation. Its general solution is
obtained by separating the variables and integrating twice to be
kR
AR-
C,r+ C 9
(8-56)
The desired solution to the problem is obtained by applying the boundary
conditions dT/dx = at r = (because of symmetry) and T = T s at r = R.
We get
(8-57)
T =T-^—\ : 7 - L -: +
k \4 R 2 AR*
cen58933_ch08.qxd 9/4/2002 11:29 AM Page 436
436
HEAT TRANSFER
The bulk mean temperature T m is determined by substituting the velocity and
temperature profile relations (Eqs. 8-41 and 8-57) into Eq. 8-4 and perform-
ing the integration. It gives
\\4sR
24 k
(8-58)
Combining this relation with q s
h
h{T 5
24 k
IIR
4§A
11D
TJ gives
k
4.36
D
(8-59)
or
Circular tube, laminar {q x = constant):
Nu
hD
4.36
(8-60)
D-
Nu = 3.66
■v^l
n\ y
i i majt
C)
jy y
Fully developed
laminar flow
FIGURE 8-21
In laminar flow in a tube with
constant surface temperature,
both the friction factor and
the heat transfer coefficient
remain constant in the fully
developed region.
Therefore, for fully developed laminar flow in a circular tube subjected to
constant surface heat flux, the Nusselt number is a constant. There is no de-
pendence on the Reynolds or the Prandtl numbers.
Constant Surface Temperature
A similar analysis can be performed for fully developed laminar flow in a cir-
cular tube for the case of constant surface temperature T s . The solution proce-
dure in this case is more complex as it requires iterations, but the Nusselt
number relation obtained is equally simple (Fig. 8-21):
Circular tube, laminar (T s = constant):
Nu
hD
k
3.66
(8-61)
The thermal conductivity k for use in the Nu relations above should be evalu-
ated at the bulk mean fluid temperature, which is the arithmetic average of the
mean fluid temperatures at the inlet and the exit of the tube. For laminar flow,
the effect of surface roughness on the friction factor and the heat transfer co-
efficient is negligible.
Laminar Flow in Noncircular Tubes
The friction factor /and the Nusselt number relations are given in Table 8-1
for fully developed laminar flow in tubes of various cross sections. The
Reynolds and Nusselt numbers for flow in these tubes are based on the hy-
draulic diameter D h = 4A c /p, where A c is the cross sectional area of the tube
and p is its perimeter. Once the Nusselt number is available, the convection
heat transfer coefficient is determined from h = kNu/D h .
Developing Laminar Flow in the Entrance Region
For a circular tube of length L subjected to constant surface temperature, the
average Nusselt number for the thermal entrance region can be determined
from (Edwards et al., 1979)
Entry region, laminar:
Nu = 3.66
0.065 (DIL) Re Pr
1 + 0.04[(D/L) Re Pr] 2
(8-62)
cen58933_ch08.qxd 9/4/2002 11:29 AM Page 437
TABLE 8-1
Nusselt number and friction factor for fully developed laminar flow in tubes of
various cross sections (D h = AA c lp, Re = V m D h /v, and Nu = hD h lk)
Tube Geometry
Circle
Rectangle
Ellipse
Triangle
alb
or9°
alb
1
2
3
4
6
8
alb
1
2
4
8
16
10°
30°
60°
90°
120°
Nusselt Number
7\ = Const.
3.66
2.98
3.39
3.96
4.44
5.14
5.60
7.54
3.66
3.74
3.79
3.72
3.65
1.61
2.26
2.47
2.34
2.00
q s = Const.
4.36
3.61
4.12
4.79
5.33
6.05
6.49
8.24
4.36
4.56
4.88
5.09
5.18
2.45
2.91
3.11
2.98
2.68
Friction Factor
f
64.00/Re
56.92/Re
62.20/Re
68.36/Re
72.92/Re
78.80/Re
82.32/Re
96.00/Re
64.00/Re
67.28/Re
72.96/Re
76.60/Re
78.16/Re
50.80/Re
52.28/Re
53.32/Re
52.60/Re
50.96/Re
437
CHAPTER 8
Note that the average Nusselt number is larger at the entrance region, as ex-
pected, and it approaches asymptotically to the fully developed value of 3.66
as L — > oo. This relation assumes that the flow is hydrodynamically developed
when the fluid enters the heating section, but it can also be used approxi-
mately for flow developing hydrodynamically.
When the difference between the surface and the fluid temperatures is large,
it may be necessary to account for the variation of viscosity with temperature.
The average Nusselt number for developing laminar flow in a circular tube in
that case can be determined from [Sieder and Tate (1936), Ref. 26]
Nu =1.86i
(Tfl)
(8-63)
All properties are evaluated at the bulk mean fluid temperature, except for u^,
which is evaluated at the surface temperature.
cen58933_ch08.qxd 9/4/2002 11:29 AM Page 43E
438
HEAT TRANSFER
The average Nusselt number for the thermal entrance region of flow
between isothermal parallel plates of length L is expressed as (Edwards
et al., 1979)
Entry region, laminar:
Nu = 7.54+
0.03 (D A /L) Re Pr
1 + 0.016[(D ; ,/L)RePr] 2
(8-64)
where D h is the hydraulic diameter, which is twice the spacing of the plates.
This relation can be used for Re < 2800.
-3 ft/s 0.15 in.
30 ft
FIGURE 8-22
Schematic for Example 8-2.
EXAMPLE 8-2 Pressure Drop in a Pipe
Water at 40°F (p = 62.42 lbm/ft 3 and p. = 3.74 Ibm/ft ■ h) is flowing in a 0.15-
in. -diameter 30-ft-long pipe steadily at an average velocity of 3 ft/s (Fig. 8-22).
Determine the pressure drop and the pumping power requirement to overcome
this pressure drop.
SOLUTION The average flow velocity in a pipe is given. The pressure drop and
the required pumping power are to be determined.
Assumptions 1 The flow is steady and incompressible. 2 The entrance effects
are negligible, and thus the flow is fully developed. 3 The pipe involves no com-
ponents such as bends, valves, and connectors.
Properties The density and dynamic viscosity of water are given to be p =
62.42 lbm/ft 3 and p, = 3.74 Ibm/ft • h = 0.00104 Ibm/ft • s.
Analysis First we need to determine the flow regime. The Reynolds number is
Re = P r - D = (62.42 lbm/ft 3 )(3ft/s)(0.12/12 ft) / 3 600 g \ = lgQ3
|X
3.74 lbm/ft • h
\ lh
which is less than 2300. Therefore, the flow is laminar. Then the friction factor
and the pressure drop become
/ =
64 64
Re 1803
0.0355
1 lbf
LP^ 30 ft (62.42 lbm/ft 3 )(3 ft/s) 2
f D 2 0.0355 Q12/12 ft 2 \32.1741bm-ft/s 2
= 930 lbf/ft 2 = 6.46 psi
The volume flow rate and the pumping power requirements are
V= V m A c = T m (iTD 2 /4) = (3 ft/s)[ir(0.12/12 ft) 2 /4] = 0.000236 ft 3 /s
1 W
J¥ pump = VAP = (0.000236 ft 3 /s)(930 lbf/ft 2 )
0.737 lbf ■ ft/s
0.30 W
Therefore, mechanical power input in the amount of 0.30 W is needed to over-
come the frictional losses in the flow due to viscosity.
cen58933_ch08.qxd 9/4/2002 11:29 AM Page 439
EXAMPLE 8-3 Flow of Oil in a Pipeline through a Lake
Consider the flow of oil at 20°C in a 30-cm-diameter pipeline at an average
velocity of 2 m/s (Fig. 8-23). A 200-m-long section of the pipeline passes
through icy waters of a lake at 0°C. Measurements indicate that the surface
temperature of the pipe is very nearly 0°C. Disregarding the thermal resistance
of the pipe material, determine (a) the temperature of the oil when the pipe
leaves the lake, (b) the rate of heat transfer from the oil, and (c) the pumping
power required to overcome the pressure losses and to maintain the flow of the
oil in the pipe.
SOLUTION Oil flows in a pipeline that passes through icy waters of a lake at
0°C. The exit temperature of the oil, the rate of heat loss, and the pumping
power needed to overcome pressure losses are to be determined.
Assumptions 1 Steady operating conditions exist. 2 The surface temperature of
the pipe is very nearly 0°C. 3 The thermal resistance of the pipe is negligible.
4 The inner surfaces of the pipeline are smooth. 5 The flow is hydrodynamically
developed when the pipeline reaches the lake.
Properties We do not know the exit temperature of the oil, and thus we cannot
determine the bulk mean temperature, which is the temperature at which the
properties of oil are to be evaluated. The mean temperature of the oil at the
inlet is 20°C, and we expect this temperature to drop somewhat as a result
of heat loss to the icy waters of the lake. We evaluate the properties of the oil
at the inlet temperature, but we will repeat the calculations, if necessary,
using properties at the evaluated bulk mean temperature. At 20°C we read
(Table A-14)
p = 888 kg/m 3
v = 901 X 10- 6 m 2 /s
k = 0.145 W/m
°C
C p = 1880 J/kg • °C
Pr = 10,400
Analysis (a) The Reynolds number is
° ir m D h (2 m/s)(0.3 m)
Re
v
901 X 10- 6 m 2 /s
666
which is less than the critical Reynolds number of 2300. Therefore, the flow is
laminar, and the thermal entry length in this case is roughly
L, « 0.05 Re Pr D = 0.05 X 666 X 10,400 X (0.3 m) = 104,000 m
which is much greater than the total length of the pipe. This is typical of fluids
with high Prandtl numbers. Therefore, we assume thermally developing flow
and determine the Nusselt number from
Nu
hD
k
3.66 +
37.3
3.66
0.065 (D/L) Re Pr
1 + 0.04[(D/L)RePr] 2/3
0.065(0.3/200) X 666 X 10,400
1 + 0.04[(0.3/200) X 666 X 10,400] 2
20°C
439
CHAPTER 8
\" - _ Icy lake, 0^C -" -~J
Oil 1
► D = 0.3m
2 m/s l
jC
\^_^>o°c_^^/
FIGURE 8-23
Schematic for Example 8-3.
cen58933_ch08.qxd 9/4/2002 11:29 AM Page 44C
440
HEAT TRANSFER
Note that this Nusselt number is considerably higher than the fully developed
value of 3.66. Then,
h = A Nu = 0A * 5Wfm (37.3) = is.o W/m 2 • °C
D 0.3 m
Also,
A s = pL = ttDL = tt(0.3 m)(200 m) = 188.5 m 2
m = pA ( .T„, = (888 kg/m 3 )[iir(0.3 m) 2 ](2 m/s) = 125.5 kg/s
Next we determine the exit temperature of oil from
T e =T s - (J, - T f ) exp {-hAJrhCp)
(18.0 W/m 2 • °C)(188.5m 2 )
0°C - [(0 - 20)°C] exp
19.71°C
(125.5 kg/s)( 1 880 J/kg ■ °C)
Thus, the mean temperature of oil drops by a mere 0.29°C as it crosses the
lake. This makes the bulk mean oil temperature 19.86°C, which is practically
identical to the inlet temperature of 20°C. Therefore, we do not need to re-
evaluate the properties.
(b) The logarithmic mean temperature difference and the rate of heat loss from
the oil are
AT,.
20 - 19.71
In
In
- 19.71
0-20
19.85°C
Q = hA s AT la = (18.0 W/m 2 • °C)(188.5 m 2 )(-19.85°C) = -6.74 X 10 4
Therefore, the oil will lose heat at a rate of 67.4 kW as it flows through the pipe
in the icy waters of the lake. Note that A7^ n is identical to the arithmetic mean
temperature in this case, since A 7", ~ A7" e .
(c) The laminar flow of oil is hydrodynamically developed. Therefore, the friction
factor can be determined from
/=£=« 0.0961
J Re 666
Then the pressure drop in the pipe and the required pumping power become
n
AP=f
0.0961
200 m (888 kg/m 3 )(2 m/s) 2
W„„
D 2 """" 0.3 m 2
rhAP (125.5 kg/s)(l. 14 X 10 5 N/m 2 )
1.14 X 10 5 N/m 2
888 kg/m 3
16.1 kW
Discussion We will need a 16.1-kW pump just to overcome the friction in the
pipe as the oil flows in the 200-m-long pipe through the lake.
cen58933_ch08.qxd 9/4/2002 11:29 AM Page 441
8-6 - TURBULENT FLOW IN TUBES
We mentioned earlier that flow in smooth tubes is fully turbulent for Re >
10,000. Turbulent flow is commonly utilized in practice because of the higher
heat transfer coefficients associated with it. Most correlations for the friction
and heat transfer coefficients in turbulent flow are based on experimental
studies because of the difficulty in dealing with turbulent flow theoretically.
For smooth tubes, the friction factor in turbulent flow can be determined
from the explicit first Petukhov equation [Petukhov (1970), Ref. 21] given as
Smooth tubes: f = (0.790 In Re - 1 .64)~ 2 10 4 < Re < 10 6 (8-65)
The Nusselt number in turbulent flow is related to the friction factor through
the Chilton-Colburn analogy expressed as
Nu = 0.125/RePr" 3 (8-66)
Once the friction factor is available, this equation can be used conveniently to
evaluate the Nusselt number for both smooth and rough tubes.
For fully developed turbulent flow in smooth tubes, a simple relation for the
Nusselt number can be obtained by substituting the simple power law relation
/= 0.184 Re~ - 2 for the friction factor into Eq. 8-66. It gives
which is known as the Colburn equation. The accuracy of this equation can be
improved by modifying it as
Nu = 0.023 Re 08 Pr" (8-68)
where n = 0.4 for heating and 0.3 for cooling of the fluid flowing through
the tube. This equation is known as the Dittus-Boelter equation [Dittus and
Boelter (1930), Ref. 6] and it is preferred to the Colburn equation.
The fluid properties are evaluated at the bulk mean fluid temperature T b =
(7, + T e )/2. When the temperature difference between the fluid and the wall is
very large, it may be necessary to use a correction factor to account for the dif-
ferent viscosities near the wall and at the tube center.
The Nusselt number relations above are fairly simple, but they may give
errors as large as 25 percent. This error can be reduced considerably to less
than 10 percent by using more complex but accurate relations such as the sec-
ond Petukhov equation expressed as
= (//8)RePr /rj.5 < Pr < 2000
1.07 + 12.7(//8) 05 (Pr 2 ' 3 - 1) \10 4 < Re < 5 X 10
The accuracy of this relation at lower Reynolds numbers is improved by mod-
ifying it as [Gnielinski (1976), Ref. 8]
(//8)(Re-1000)Pr /0.5 < Pr < 2000 \
1 + 12.7(//8f 5 (Pr 2/3 - 1) \3 X 10 3 < Re < 5 X 10 6 j l8 " 70)
441
CHAPTER 8
cen58933_ch08.qxd 9/4/2002 11:29 AM Page 442
442
HEAT TRANSFER
Relative
Friction
Roughness,
Factor,
e/L
f
0.0*
0.0119
0.00001
0.0119
0.0001
0.0134
0.0005
0.0172
0.001
0.0199
0.005
0.0305
0.01
0.0380
0.05
0.0716
*Smooth surface. All va
ues are for Re = 10 6 ,
and are calculated from
Eq
8-73.
FIGURE 8-24
The friction factor is
minimum for a smooth pipe
and increases with roughness.
TABLE 8-2
Standard sizes for Schedule 40
steel pipes
Nominal
Size, in.
Actual Inside
Diameter, in.
where the friction factor/can be determined from an appropriate relation such
as the first Petukhov equation. Gnielinski's equation should be preferred
in calculations. Again properties should be evaluated at the bulk mean fluid
temperature.
The relations above are not very sensitive to the thermal conditions at the
tube surfaces and can be used for both T s = constant and q s = constant cases.
Despite their simplicity, the correlations already presented give sufficiently
accurate results for most engineering purposes. They can also be used to ob-
tain rough estimates of the friction factor and the heat transfer coefficients in
the transition region 2300 < Re < 10,000, especially when the Reynolds
number is closer to 10,000 than it is to 2300.
The relations given so far do not apply to liquid metals because of their
very low Prandtl numbers. For liquid metals (0.004 < Pr < 0.01), the fol-
lowing relations are recommended by Sleicher and Rouse (1975, Ref. 27) for
10 4 < Re < 10 6 :
Liquid metals, T s = constant:
Liquid metals, q s = constant:
Nu = 4.8 + 0.0156 Re 085 Pr?- 93
Nu = 6.3 + 0.0167 Re 085 Pr" 93
(8-71)
(8-72)
where the subscript s indicates that the Prandtl number is to be evaluated at
the surface temperature.
Rough Surfaces
Any irregularity or roughness on the surface disturbs the laminar sublayer,
and affects the flow. Therefore, unlike laminar flow, the friction factor and
the convection coefficient in turbulent flow are strong functions of surface
roughness.
The friction factor in fully developed turbulent flow depends on the
Reynolds number and the relative roughness elD. In 1939, C. F. Colebrook
(Ref. 3) combined all the friction factor data for transition and turbulent flow
in smooth as well as rough pipes into the following implicit relation known as
the Colebrook equation.
1
Vf
-2.0 Ioe
sID
2.51
3.7 Re y/f,
(turbulent flow)
(8-73)
%
0.269
Va
0.364
%
0.493
y?
0.622
%
0.824
1
1.049
VA
1.610
2
2.067
2 1 /
2.469
3
3.068
5
5.047
10
10.02
In 1944, L. F Moody (Ref. 17) plotted this formula into the famous Moody
chart given in the Appendix. It presents the friction factors for pipe flow as a
function of the Reynolds number and e/D over a wide range. For smooth
tubes, the agreement between the Petukhov and Colebrook equations is very
good. The friction factor is minimum for a smooth pipe (but still not zero be-
cause of the no-slip condition), and increases with roughness (Fig. 8-24).
Although the Moody chart is developed for circular pipes, it can also be
used for noncircular pipes by replacing the diameter by the hydraulic diame-
ter. At very large Reynolds numbers (to the right of the dashed line on the
chart) the friction factor curves corresponding to specified relative roughness
curves are nearly horizontal, and thus the friction factors are independent of
the Reynolds number. In calculations, we should make sure that we use the in-
ternal diameter of the pipe, which may be different than the nominal diameter.
For example, the internal diameter of a steel pipe whose nominal diameter is
1 in. is 1.049 in. (Table 8-2).
cen58933_ch08.qxd 9/4/2002 11:29 AM Page 443
Commercially available pipes differ from those used in the experiments in
that the roughness of pipes in the market is not uniform, and it is difficult to
give a precise description of it. Equivalent roughness values for some com-
mercial pipes are given in Table 8-3, as well as on the Moody chart. But it
should be kept in mind that these values are for new pipes, and the relative
roughness of pipes may increase with use as a result of corrosion, scale
buildup, and precipitation. As a result, the friction factor may increase by a
factor of 5 to 10. Actual operating conditions must be considered in the design
of piping systems. Also, the Moody chart and its equivalent Colebrook equa-
tion involve several uncertainties (the roughness size, experimental error,
curve fitting of data, etc.), and thus the results obtained should not be treated
as "exact." It is usually considered to be accurate to ± 15 percent over the en-
tire range in the figure.
The Colebrook equation is implicit in /, and thus the determination of the
friction factor requires tedious iteration unless an equation solver is used.
An approximate explicit relation for / is given by S. E. Haaland in 1983
(Ref. 9) as
1
Vf
■1.8 log
63_
Re
s/dV
3.7/
(8-74)
The results obtained from this relation are within 2 percent of those obtained
from Colebrook equation, and we recommend using this relation rather than
the Moody chart to avoid reading errors.
In turbulent flow, wall roughness increases the heat transfer coefficient h by
a factor of 2 or more [Dipprey and Sabersky (1963), Ref. 5]. The convection
heat transfer coefficient for rough tubes can be calculated approximately from
the Nusselt number relations such as Eq. 8-70 by using the friction factor
determined from the Moody chart or the Colebrook equation. However, this
approach is not very accurate since there is no further increase in Ir with /for
/> 4/smooth [Norris (1970), Ref. 20] and correlations developed specifically for
rough tubes should be used when more accuracy is desired.
Developing Turbulent Flow in the Entrance Region
The entry lengths for turbulent flow are typically short, often just 10 tube
diameters long, and thus the Nusselt number determined for fully developed
turbulent flow can be used approximately for the entire tube. This simple ap-
proach gives reasonable results for pressure drop and heat transfer for long
tubes and conservative results for short ones. Correlations for the friction and
heat transfer coefficients for the entrance regions are available in the literature
for better accuracy.
Turbulent Flow in Noncircular Tubes
The velocity and temperature profiles in turbulent flow are nearly straight
lines in the core region, and any significant velocity and temperature gradients
occur in the viscous sublayer (Fig. 8-25). Despite the small thickness of
laminar sublayer (usually much less than 1 percent of the pipe diameter), the
characteristics of the flow in this layer are very important since they set the
stage for flow in the rest of the pipe. Therefore, pressure drop and heat trans-
fer characteristics of turbulent flow in tubes are dominated by the very thin
443
CHAPTER 8
TABLE 8-3
Equivalent roi
ghness values
for
new commercial pipes*
Roughness, e
Material
ft
mm
Glass, plastic
(smooth)
Concrete
0.003-0.03
0.9-9
Wood stave
0.0016
0.5
Rubber,
smoothed
0.000033
0.01
Copper or
brass tubing
0.000005
0.0015
Cast iron
0.00085
0.26
Galvanized
iron
0.0005
0.15
Wrought iron
0.00015
0.046
Stainless steel 0.000007
0.002
Commercial
steel
0.00015
0.045
*The uncertainty in these values can be as much
as ±60 percent.
V
^^»-s
p^-V(r)
T Turbulent layer
g, ) 1 Overlap layer
Laminar sublayer
FIGURE 8-25
In turbulent flow, the velocity
profile is nearly a straight line in the
core region, and any significant
velocity gradients occur in the
viscous sublayer.
cen58933_ch08.qxd 9/4/2002 11:29 AM Page 444
444
HEAT TRANSFER
Tube
' Annulus
FIGURE 8-26
A double-tube heat exchanger that
consists of two concentric tubes.
viscous sublayer next to the wall surface, and the shape of the core region is
not of much significance. Consequently, the turbulent flow relations given
above for circular tubes can also be used for noncircular tubes with reasonable
accuracy by replacing the diameter D in the evaluation of the Reynolds num-
ber by the hydraulic diameter D h = 4A c /p.
Flow through Tube Annulus
Some simple heat transfer equipments consist of two concentric tubes, and are
properly called double-tube heat exchangers (Fig. 8-26). In such devices, one
fluid flows through the tube while the other flows through the annular space.
The governing differential equations for both flows are identical. Therefore,
steady laminar flow through an annulus can be studied analytically by using
suitable boundary conditions.
Consider a concentric annulus of inner diameter D, and outer diameter D .
The hydraulic diameter of annulus is
D„
AA C _ 4tt(DI - D?)/4
T~ tt(D + D,)
D„ ~ D,
(8-75)
TABLE 8-4
Nusselt number for fully developed
laminar flow in an annulus with
one surface isothermal and the
other adiabatic (Kays and Perkins,
Ref. 14)
D,ID
Nu,
Nu
—
3.66
0.05
17.46
4.06
0.10
11.56
4.11
0.25
7.37
4.23
0.50
5.74
4.43
1.00
4.86
4.86
(a) Finned surface
Roughness
(b) Roughened surface
FIGURE 8-27
Tube surfaces are often roughened,
corrugated, ox finned in order to
enhance convection heat transfer.
Annular flow is associated with two Nusselt numbers — Nu, on the inner
tube surface and Nu on the outer tube surface — since it may involve heat
transfer on both surfaces. The Nusselt numbers for fully developed laminar
flow with one surface isothermal and the other adiabatic are given in
Table 8-4. When Nusselt numbers are known, the convection coefficients for
the inner and the outer surfaces are determined from
Nu,
h;D,,
and
Nu„
h„D h
(8-76)
For fully developed turbulent flow, the inner and outer convection coeffi-
cients are approximately equal to each other, and the tube annulus can be
treated as a noncircular duct with a hydraulic diameter of D h = D — D t . The
Nusselt number in this case can be determined from a suitable turbulent flow
relation such as the Gnielinski equation. To improve the accuracy of Nusselt
numbers obtained from these relations for annular flow, Petukhov and Roizen
(1964, Ref. 22) recommend multiplying them by the following correction fac-
tors when one of the tube walls is adiabatic and heat transfer is through the
other wall:
F,
= 0.86
(1)
-0.16
F„
= 0.86
(£)
-0.16
(outer wall adiabatic)
(inner wall adiabatic)
(8-77)
(8-78)
Heat Transfer Enhancement
Tubes with rough surfaces have much higher heat transfer coefficients than
tubes with smooth surfaces. Therefore, tube surfaces are often intention-
ally roughened, corrugated, or finned in order to enhance the convection
heat transfer coefficient and thus the convection heat transfer rate (Fig. 8-27).
Heat transfer in turbulent flow in a tube has been increased by as much as
cen58933_ch08.qxd 9/4/2002 11:29 AM Page 445
400 percent by roughening the surface. Roughening the surface, of course,
also increases the friction factor and thus the power requirement for the pump
or the fan.
The convection heat transfer coefficient can also be increased by inducing
pulsating flow by pulse generators, by inducing swirl by inserting a twisted
tape into the tube, or by inducing secondary flows by coiling the tube.
445
CHAPTER 8
EXAMPLE 8-4 Pressure Drop in a Water Pipe
Water at 60°F (p = 62.36 lbm/ft 3 and jjl = 2.713 Ibm/ft ■ h) is flowing steadily
in a 2-in. -diameter horizontal pipe made of stainless steel at a rate of 0.2 ft 3 /s
(Fig. 8-28). Determine the pressure drop and the required pumping power in-
put for flow through a 200-ft-long section of the pipe.
SOLUTION The flow rate through a specified water pipe is given. The pressure
drop and the pumping power requirements are to be determined.
Assumptions 1 The flow is steady and incompressible. 2 The entrance effects
are negligible, and thus the flow is fully developed. 3 The pipe involves no com-
ponents such as bends, valves, and connectors. 4 The piping section involves
no work devices such as a pump or a turbine.
Properties The density and dynamic viscosity of water are given by p = 62.36
lbm/ft 3 and ia = 2.713 Ibm/ft ■ h = 0.0007536 Ibm/ft • s, respectively.
Analysis First we calculate the mean velocity and the Reynolds number to
determine the flow regime:
T
Re
V
0.2 ft 3 /s
9.17 ft/s
A c ttD 2 /4 Tr(2/12ft) 2 /4
pVD _ (62.36 lbm/ft 3 )(9.17 ft/s) (2/1 2 ft) /3600 s
H* 2.713 lbm/ft ■ h [ lh
126,400
which is greater than 10,000. Therefore, the flow is turbulent. The relative
roughness of the pipe is
s/D
0.000007 ft
2/12 ft
0.000042
The friction factor corresponding to this relative roughness and the Reynolds
number can simply be determined from the Moody chart. To avoid the reading
error, we determine it from the Colebrook equation:
1
-2.0 log
2.51
I BID
v 3.7 ' Rcvfi
1
-2.0 los
0.000042
3.7
2.51
126,400 \Zf,
Using an equation solver or an iterative scheme, the friction factor is deter-
mined to be f= 0.0174. Then the pressure drop and the required power input
become
AP=f
0.0174
200 ft (62.36 lbm/ft 3 )(9.17 ft/s) 2
W
D 2 2/12 ft
1700 lbf/ft 2 = 11.8 psi
VAP = (0.2 ft 3 /s)(1700 lbf/ft 2 )
2
1 W
1 lbf
32.2 lbm • ft/s 2
0.737 lbf • ft/s
461 W
-t
0.2 ft 3 /s
water
2 in.
200 ft H
FIGURE 8-28
Schematic for Example 8^1.
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446
HEAT TRANSFER
Therefore, power input in the amount of 461 W is needed to overcome the fric-
tional losses in the pipe.
Discussion The friction factor also could be determined easily from the explicit
Haaland relation. It would give f = 0.0172, which is sufficiently close to
0.0174. Also, the friction factor corresponding to e = in this case is 0.0171,
which indicates that stainless steel pipes can be assumed to be smooth with
negligible error.
q = constant
1111)1111
FIGURE 8-
Schematic
29
for Example 8-5.
EXAMPLE 8-5 Heating of Water by Resistance Heaters in a Tube
Water is to be heated from 15°C to 65°C as it flows through a 3-cm-internal-
diameter 5-m-long tube (Fig. 8-29). The tube is equipped with an electric re-
sistance heater that provides uniform heating throughout the surface of the
tube. The outer surface of the heater is well insulated, so that in steady opera-
tion all the heat generated in the heater is transferred to the water in the tube.
If the system is to provide hot water at a rate of 10 L/min, determine the power
rating of the resistance heater. Also, estimate the inner surface temperature of
the pipe at the exit.
SOLUTION Water is to be heated in a tube equipped with an electric resis-
tance heater on its surface. The power rating of the heater and the inner surface
temperature are to be determined.
Assumptions 1 Steady flow conditions exist. 2 The surface heat flux is uniform.
3 The inner surfaces of the tube are smooth.
Properties The properties of water at the bulk mean temperature of T b =
{T, + T e )l2 = (15 + 65)/2 = 40°C are (Table A-9).
p = 992.1 kg/m 3
C v = 4179J/kg
°C
k = 0.631 W/m • °C
Pr = 4.32
v = uVp = 0.658 X 10~ 6 nr/s
Analysis The cross sectional and heat transfer surface areas are
A c = i-TTD 2 = i-ir(0.03 m) 2 = 7.069 X 10~ 4 m 2
A s = pL = ttDL = tt(0.03 m)(5 m) = 0.471 m 2
The volume flow rate of water is given as V = 10 L/min = 0.01 m 3 /min. Then
the mass flow rate becomes
m = pV= (992.1 kg/m 3 )(0.01 rnVmin) = 9.921 kg/min = 0.1654 kg/s
To heat the water at this mass flow rate from 15°C to 65°C, heat must be sup-
plied to the water at a rate of
Q =riiC p {T e -T i )
= (0.1654 kg/s)(4.179 kJ/kg • °C)(65
= 34.6 kJ/s = 34.6 kW
15)°C
cen58933_ch08.qxd 9/4/2002 11:29 AM Page 447
All of this energy must come from the resistance heater. Therefore, the power
rating of the heater must be 34.6 kW.
The surface temperature T s of the tube at any location can be determined
from
KT S - T m )
?1
h
where h is the heat transfer coefficient and T m is the mean temperature of the
fluid at that location. The surface heat flux is constant in this case, and its
value can be determined from
. Q 34.6 kW ™ /lfilrW/ 2
q. = -7- = = 73.46 kW/m 2
ls A s 0.471 m 2
To determine the heat transfer coefficient, we first need to find the mean ve-
locity of water and the Reynolds number:
T
V = 0.010 mVmin
A c 7.069 X 10- 4 m 2
14.15 m/min = 0.236 m/s
Y„,D (0.236 m/s)(0.03 m)
Re = — t. — = — ; — = 10,760
v 0.658 X 10- 6 m 2 /s
which is greater than 10,000. Therefore, the flow is turbulent and the entry
length is roughly
which is much shorter than the total length of the pipe. Therefore, we can as-
sume fully developed turbulent flow in the entire pipe and determine the Nus-
selt number from
Nu = -y = 0.023 Re - 8 Pr - 4 = 0.023(10,760) 08 (4.34) 04 = 69.5
Then,
h = A Nu = 0-631 W/m • °C = M62 w/m2 _ oc
D 0.03 m
and the surface temperature of the pipe at the exit becomes
T,
q s 73,460 W/m 2
T,„ + t- = 65°C + ; = 115°C
'" h 1462 W/m 2 ■ °C
Discussion Note that the inner surface temperature of the pipe will be 50°C
higher than the mean water temperature at the pipe exit. This temperature dif-
ference of 50°C between the water and the surface will remain constant
throughout the fully developed flow region.
447
CHAPTER 8
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448
HEAT TRANSFER
T =60°C
FIGURE 8-30
Schematic for Example 8-6.
J EXAMPLE 8-6
m
Heat Loss from the Ducts of a Heating System
Hot air at atmospheric pressure and 80°C enters an 8-m-long uninsulated
square duct of cross section 0.2 m X 0.2 m that passes through the attic of a
house at a rate of 0.15 m 3 /s (Fig. 8-30). The duct is observed to be nearly
isothermal at 60°C. Determine the exit temperature of the air and the rate of
heat loss from the duct to the attic space.
SOLUTION Heat loss from uninsulated square ducts of a heating system in
the attic is considered. The exit temperature and the rate of heat loss are to be
determined.
Assumptions 1 Steady operating conditions exist. 2 The inner surfaces of the
duct are smooth. 3 Air is an ideal gas.
Properties We do not know the exit temperature of the air in the duct, and thus
we cannot determine the bulk mean temperature of air, which is the tempera-
ture at which the properties are to be determined. The temperature of air at the
inlet is 80°C and we expect this temperature to drop somewhat as a result of
heat loss through the duct whose surface is at 60°C. At 80°C and 1 atm we
read (Table A-15)
p =
= 0.9994 kg/m 3
c P -
= 1008 J/kg
°C
k~-
= 0.02953 W/m • °C
Pr =
= 0.7154
v -
= 2.097 X 10- 5 m 2 /s
Analysis The characteristic length (which is the hydraulic diameter), the mean
velocity, and the Reynolds number in this case are
4A C 4«2
D,, = —pr = ~, — = a = 0.2 m
" P Aa
V 0.15 m 3 /s
3.75 m/s
Re
A c (0.2 m) 2
°V m D h (3.75 m/s)(0.2 m)
2.097 X 10" 5 m 2 /s
35,765
which is greater than 10,000. Therefore, the flow is turbulent and the entry
lengths in this case are roughly
L h ~ L, ~ 10D = 10 X 0.2 m = 2 m
which is much shorter than the total length of the duct. Therefore, we can
assume fully developed turbulent flow in the entire duct and determine the
Nusselt number from
hD
Nu = — - = 0.023 Re 08 Pr 03 = 0.023(35,765) - 8 (0.7154) - 3 = 91.4
cen58933_ch08.qxd 9/4/2002 11:29 AM Page 449
Then,
h
D,
Nu
0.02953 W/m ■ °C
0.2 m
(91.4) = 13.5 W/m 2 • °C
A s = pL = AaL = 4 X (0.2 m)(8 m) = 6.4 m 2
m =pV= (1.009 kg/m 3 )(0.15 m 3 /s) = 0.151 kg/s
Next, we determine the exit temperature of air from
T e = T s - (T s - r ; ) exp (-hA s /mC p )
= 60°C - [(60 - 80)°C] exp
= 71.3°C
(13.5 W/m 2 ■ °C)(6.4 m 2 )
(0.151 kg/s)(1008 J/kg • °C)
Then the logarithmic mean temperature difference and the rate of heat loss
from the air become
Ar,„
80 - 71.3
In
In
60 - 71.3
15.2°C
T - T 60 - 80
Q = hA s A7 ln = (13.5 W/m 2 ■ °C)(6.4 m 2 )(-15.2°C) = -1313 W
Therefore, air will lose heat at a rate of 1313 W as it flows through the duct in
the attic.
Discussion The average fluid temperature is (80 + 71.3)/2 = 75.7°C, which
is sufficiently close to 80°C at which we evaluated the properties of air. There-
fore, it is not necessary to re-evaluate the properties at this temperature and to
repeat the calculations.
449
CHAPTER 8
SUMMARY
Internal flow is characterized by the fluid being completely
confined by the inner surfaces of the tube. The mean velocity
and mean temperature for a circular tube of radius R are ex-
pressed as
fR
2 Jo V(r '
x)rdr and T m
2 fR
YTrdr
The Reynolds number for internal flow and the hydraulic di-
ameter are defined as
Re
,D
and D,,
4A„
The flow in a tube is laminar for Re < 2300, turbulent for
Re > 10,000, and transitional in between.
The length of the region from the tube inlet to the point at
which the boundary layer merges at the centerline is the hydro-
dynamic entry length L h , The region beyond the entrance
region in which the velocity profile is fully developed is the
hydro dynamic ally fully developed region. The length of the re-
gion of flow over which the thermal boundary layer develops
and reaches the tube center is the thermal entry length L,. The
region in which the flow is both hydrodynamically and ther-
mally developed is the fully developed flow region. The entry
lengths are given by
10D
For q s = constant, the rate of heat transfer is expressed as
Q = <i s A s = mC p (T e - T,)
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450
HEAT TRANSFER
For T, = constant, we have
Ar ln
Q =hA s AT in = mC p (T e -Td
T e = T s - (T s - 7V)exp(-M s /mC ? )
T-, - T e AT e - AT,
ln[(r, - T e )l(T s - T,)] MATJAT,)
The pressure drop and required pumping power for a volume
flow rate of Vare
L PV„
kp = — — - an d W = VAP
For fully developed laminar flow in a circular pipe, we have:
T(r) = 2T„,(l-^)=T raax |:
R 2
64p, 64
P W„, Re
APRl _n2 _ TR 4 AP = TtR*AP
8|jlL 128jjlL
V=T A
avc r 8|xL
nR 2
hD
Circular tube, laminar (q s = constant): Nu = — = 4.36
hD
Circular tube, laminar (T s = constant): Nu = — = 3.66
For fully developed turbulent flow with smooth surfaces,
we have
/= (0.790 In Re - 1.64) - 2 10 4 < Re < 10 6
Nu = 0.125/RePr" 3
0.7<Pr< 160\
Re > 10,000 J
Nu = 0.023 Re 08 Pr" with n = 0.4 for heating and 0.3 for
cooling of fluid
(//8)(Re - 1000) Pr / .5 < Pr < 2000
Nu -
Nu = 0.023 Re 08 Pr 1
1 + 12.7(//8) 05 (Pr 2/3 - 1) \3 X 10 3 < Re < 5 X 10 6
The fluid properties are evaluated at the bulk mean fluid
temperature T b = (T, + T e )l2. For liquid metal flow in the
range of 10 4 < Re < 10 6 we have:
T s = constant:
q s = constant:
Nu = 4.8 + 0.0156 Re 085 Pr?" 53
Nu = 6.3 + 0.0167 Re 085 Pr? 93
For fully developed turbulent flow with rough surfaces, the
friction factor/ is determined from the Moody chart or
1 . _ . eID , 2.51
^ = " 2 - 01O8 l^ + R^,
-1.8 log
6^9 IsID
Re I 3.7
For a concentric annulus, the hydraulic diameter is D h
D — Dj, and the Nusselt numbers are expressed as
For developing laminar flow in the entrance region with con-
stant surface temperature, we have
Nu,
and Nu„
Circular tube: Nu = 3.66
0.065(D/L) Re Pr
1 + 0.04[(D/L) Re Pr] 2
13 /,, \0.I4
/RePr/)\ l/3 /jjiA
Circular tube: Nu = 1 .86 I — I
Parallel plates: Nu = 7.54
0.03(D ; ,/L)RePr
1 + O.Oiet^/^RePr] 2
where the values for the Nusselt numbers are given in
Table 8^1.
REFERENCES AND SUGGESTED READING
1. M. S. Bhatti and R. K. Shah. "Turbulent and Transition
Flow Convective Heat Transfer in Ducts." In Handbook
of Single-Phase Convective Heat Transfer, ed. S. Kakaf,
R. K. Shah, and W. Aung. New York: Wiley
Interscience, 1987.
2. A. P. Colburn. Transactions oftheAIChE 26 (1933),
p. 174.
3. C. F Colebrook. "Turbulent flow in Pipes, with Particular
Reference to the Transition between the Smooth and
cen58933_ch08.qxd 9/4/2002 11:29 AM Page 451
Rough Pipe Laws." Journal of the Institute of Civil
Engineers London. 11 (1939), pp. 133-156.
4. R. G. Deissler. "Analysis of Turbulent Heat Transfer and
Flow in the Entrance Regions of Smooth Passages." 1953.
Referred to in Handbook of Single-Phase Convective
Heat Transfer, ed. S. Kakac, R. K. Shah, and W. Aung.
New York: Wiley Interscience, 1987.
5. D. F. Dipprey and D. H. Sabersky. "Heat and Momentum
Transfer in Smooth and Rough Tubes at Various Prandtl
Numbers." International Journal of Heat Mass Transfer 6
(1963), pp. 329-353.
6. F. W. Dittus and L. M. K. Boelter. University of
California Publications on Engineering 2 (1930), p. 433.
7. D. K. Edwards, V. E. Denny, and A. F. Mills. Transfer
Processes. 2nd ed. Washington, DC: Hemisphere, 1979.
8. V. Gnielinski. "New Equations for Heat and Mass
Transfer in Turbulent Pipe and Channel Flow."
International Chemical Engineering 16 (1976),
pp. 359-368.
9. S. E. Haaland. "Simple and Explicit Formulas for the
Friction Factor in Turbulent Pipe Flow." Journal of Fluids
Engineering (March 1983), pp. 89-90.
10. J. P. Holman. Heat Transfer. 8th ed. New York:
McGraw-Hill, 1997.
11. F. P. Incropera and D. P. DeWitt. Introduction to Heat
Transfer. 3rd ed. New York: John Wiley & Sons, 1996.
12. S. Kaka§, R. K. Shah, and W. Aung, eds. Handbook of
Single-Phase Convective Heat Transfer. New York:
Wiley Interscience, 1987.
13. W. M. Kays and M. E. Crawford. Convective Heat and
Mass Transfer. 3rd ed. New York: McGraw-Hill, 1993.
14. W. M. Kays and H. C. Perkins. Chapter 7. In Handbook of
Heat Transfer, ed. W. M. Rohsenow and J. P. Hartnett.
New York: McGraw-Hill, 1972.
15. F. Kreith and M. S. Bohn. Principles of Heat Transfer.
6th ed. Pacific Grove, CA: Brooks/Cole, 2001.
16. A. F Mills. Basic Heat and Mass Transfer. 2nd ed.
Upper Saddle River, NJ: Prentice Hall, 1999.
17. L. F. Moody. "Friction Factors for Pipe Flows."
Transactions of the ASME 66 (1944), pp. 671-684.
18. M. Molki and E. M. Sparrow. "An Empirical Correlation
for the Average Heat Transfer Coefficient in Circular
Tubes." Journal of Heat Transfer 108 (1986),
pp. 482-484.
19. B. R. Munson, D. F Young, and T Okiishi. Fundamentals
of Fluid Mechanics. 4th ed. New York: Wiley, 2002.
20. R. H. Norris. "Some Simple Approximate Heat Transfer
Correlations for Turbulent Flow in Ducts with Rough
451
CHAPTER 8
Surfaces." In Augmentation of Convective Heat Transfer,
ed. A. E. Bergles and R. L. Webb. New York:
ASME, 1970.
21. B. S. Petukhov. "Heat Transfer and Friction in Turbulent
Pipe Flow with Variable Physical Properties." In
Advances in Heat Transfer, ed. T. F Irvine and J. P.
Hartnett, Vol. 6. New York: Academic Press, 1970.
22. B. S. Petukhov and L. I. Roizen. "Generalized
Relationships for Heat Transfer in a Turbulent Flow of
a Gas in Tubes of Annular Section." High Temperature
(USSR) 2 (1964), pp. 65-68.
23. O. Reynolds. "On the Experimental Investigation of the
Circumstances Which Determine Whether the Motion
of Water Shall Be Direct or Sinuous, and the Law of
Resistance in Parallel Channels." Philosophical
Transactions of the Royal Society of London \1A (1883),
pp. 935-982.
24. H. Schlichting. Boundary Layer Theory. 7th ed.
New York: McGraw-Hill, 1979.
25. R. K. Shah and M. S. Bhatti. "Laminar Convective
Heat Transfer in Ducts." In Handbook of Single-Phase
Convective Heat Transfer, ed. S. Kakac, R. K. Shah, and
W. Aung. New York: Wiley Interscience, 1987.
26. E. N. Sieder and G. E. Tate. "Heat Transfer and Pressure
Drop of Liquids in Tubes." Industrial Engineering
Chemistry 28 (1936), pp. 1429-1435.
27. C. A. Sleicher and M. W. Rouse. "A Convenient
Correlation for Heat Transfer to Constant and Variable
Property Fluids in Turbulent Pipe Flow." International
Journal of Heat Mass Transfer 18 (1975), pp. 1429-1435.
28. N. V. Suryanarayana. Engineering Heat Transfer. St. Paul,
MN: West, 1995.
29. F M. White. Heat and Mass Transfer. Reading, MA:
Addison-Wesley, 1988.
30. S. Whitaker. "Forced Convection Heat Transfer
Correlations for Flow in Pipes, Past Flat Plates, Single
Cylinders, and for Flow in Packed Beds and Tube
Bundles." AIChE Journal 18 (1972), pp. 361-371.
31. W. Zhi-qing. "Study on Correction Coefficients of
Laminar and Turbulent Entrance Region Effects in
Round Pipes." Applied Mathematical Mechanics 3
(1982), p. 433.
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HEAT TRANSFER
PROBLEMS
General Flow Analysis
8-1C Why are liquids usually transported in circular pipes?
8-2C Show that the Reynolds number for flow in a circular
tube of diameter D can be expressed as Re = Aml{ r uD\h).
8-3C Which fluid at room temperature requires a larger
pump to move at a specified velocity in a given tube: water or
engine oil? Why?
8-4C What is the generally accepted value of the Reynolds
number above which the flow in smooth pipes is turbulent?
8-5C What is hydraulic diameter? How is it defined? What
is it equal to for a circular tube of diameter?
8-6C How is the hydrodynamic entry length defined for flow
in a tube? Is the entry length longer in laminar or turbulent
flow?
8-7C Consider laminar flow in a circular tube. Will the
friction factor be higher near the inlet of the tube or near the
exit? Why? What would your response be if the flow were
turbulent?
8-8C How does surface roughness affect the pressure drop in
a tube if the flow is turbulent? What would your response be if
the flow were laminar?
8-9C How does the friction factor /vary along the flow di-
rection in the fully developed region in (a) laminar flow and
(b) turbulent flow?
8-10C What fluid property is responsible for the develop-
ment of the velocity boundary layer? For what kinds of fluids
will there be no velocity boundary layer in a pipe?
8-11 C What is the physical significance of the number of
transfer units NTU = hAlrh C p ? What do small and large NTU
values tell about a heat transfer system?
8-12C What does the logarithmic mean temperature differ-
ence represent for flow in a tube whose surface temperature is
constant? Why do we use the logarithmic mean temperature
instead of the arithmetic mean temperature?
8-13C How is the thermal entry length defined for flow in a
tube? In what region is the flow in a tube fully developed?
*Problems designated by a "C" are concept questions, and
students are encouraged to answer them all. Problems designated
by an "E" are in English units, and the SI users can ignore them.
Problems with an EES-CD icon ® are solved using EES, and
complete solutions together with parametric studies are included
on the enclosed CD. Problems with a computer-EES icon H are
comprehensive in nature, and are intended to be solved with a
computer, preferably using the EES software that accompanies
this text.
8-14C Consider laminar forced convection in a circular tube.
Will the heat flux be higher near the inlet of the tube or near the
exit? Why?
8-1 5C Consider turbulent forced convection in a circular
tube. Will the heat flux be higher near the inlet of the tube or
near the exit? Why?
8-16C In the fully developed region of flow in a circular
tube, will the velocity profile change in the flow direction?
How about the temperature profile?
8-17C Consider the flow of oil in a tube. How will the
hydrodynamic and thermal entry lengths compare if the flow is
laminar? How would they compare if the flow were turbulent?
8-18C Consider the flow of mercury (a liquid metal) in a
tube. How will the hydrodynamic and thermal entry lengths
compare if the flow is laminar? How would they compare if the
flow were turbulent?
8-19C What do the mean velocity Y m and the mean tem-
perature T m represent in flow through circular tubes of constant
diameter?
8-20C Consider fluid flow in a tube whose surface tempera-
ture remains constant. What is the appropriate temperature dif-
ference for use in Newton's law of cooling with an average
heat transfer coefficient?
8-21 Air enters a 20-cm-diameter 12-m-long underwater
duct at 50°C and 1 atm at a mean velocity of 7 m/s, and is
cooled by the water outside. If the average heat transfer coeffi-
cient is 85 W/m 2 • °C and the tube temperature is nearly equal
to the water temperature of 5°C, determine the exit temperature
of air and the rate of heat transfer.
8-22 Cooling water available at 10°C is used to condense
steam at 30°C in the condenser of a power plant at a rate of
0.15 kg/s by circulating the cooling water through a bank of
5-m-long 1 .2-cm-internal-diameter thin copper tubes. Water
enters the tubes at a mean velocity of 4 m/s, and leaves at a
temperature of 24°C. The tubes are nearly isothermal at 30°C.
Determine the average heat transfer coefficient between the
water and the tubes, and the number of tubes needed to achieve
the indicated heat transfer rate in the condenser.
8-23 Repeat Problem 8-22 for steam condensing at a rate of
0.60 kg/s.
8-24 Combustion gases passing through a 3-cm-internal-
diameter circular tube are used to vaporize waste water at at-
mospheric pressure. Hot gases enter the tube at 115 kPa and
250°C at a mean velocity of 5 m/s, and leave at 150°C. If the
average heat transfer coefficient is 120 W/m 2 • °C and the in-
ner surface temperature of the tube is 1 10°C, determine (a) the
tube length and (b) the rate of evaporation of water.
8-25 Repeat Problem 8-24 for a heat transfer coefficient of
60 W/m 2 • °C.
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Laminar and Turbulent Flow in Tubes
8-26C How is the friction factor for flow in a tube related to
the pressure drop? How is the pressure drop related to the
pumping power requirement for a given mass flow rate?
8-27C Someone claims that the shear stress at the center of
a circular pipe during fully developed laminar flow is zero.
Do you agree with this claim? Explain.
8-28C Someone claims that in fully developed turbulent
flow in a tube, the shear stress is a maximum at the tube sur-
face. Do you agree with this claim? Explain.
8-29C Consider fully developed flow in a circular pipe with
negligible entrance effects. If the length of the pipe is doubled,
the pressure drop will (a) double, (b) more than double, (c) less
than double, (d) reduce by half, or (e) remain constant.
8-30C Someone claims that the volume flow rate in a circu-
lar pipe with laminar flow can be determined by measuring the
velocity at the centerline in the fully developed region, multi-
plying it by the cross sectional area, and dividing the result by
2. Do you agree? Explain.
8-31C Someone claims that the average velocity in a circu-
lar pipe in fully developed laminar flow can be determined by
simply measuring the velocity at R/2 (midway between the
wall surface and the centerline). Do you agree? Explain.
8-32C Consider fully developed laminar flow in a circular
pipe. If the diameter of the pipe is reduced by half while the
flow rate and the pipe length are held constant, the pressure
drop will (a) double, (b) triple, (c) quadruple, (rf) increase by a
factor of 8, or (e) increase by a factor of 16.
8-33C Consider fully developed laminar flow in a circular
pipe. If the viscosity of the fluid is reduced by half by heating
while the flow rate is held constant, how will the pressure drop
change?
8-34C How does surface roughness affect the heat transfer
in a tube if the fluid flow is turbulent? What would your re-
sponse be if the flow in the tube were laminar?
8-35 Water at 15°C (p = 999.1 kg/m 3 and p. = 1.138 X 1(T 3
kg/m ■ s) is flowing in a 4-cm-diameter and 30-m long hori-
zontal pipe made of stainless steel steadily at a rate of 5 L/s.
Determine (a) the pressure drop and (b) the pumping power
requirement to overcome this pressure drop.
5 L/s
4 cm
30 ra
FIGURE P8-35
8-36 In fully developed laminar flow in a circular pipe, the
velocity at R/2 (midway between the wall surface and the cen-
453
CHAPTER 8
terline) is measured to be 6 m/s. Determine the velocity at the
center of the pipe. Answer: 8 m/s
8-37 The velocity profile in fully developed laminar flow in
a circular pipe of inner radius R = 2 cm, in m/s, is given by
°V(r) = 4(1 — r 2 /R 2 ). Determine the mean and maximum ve-
locities in the pipe, and the volume flow rate.
FIGURE P8-37
8-38 Repeat Problem 8-37 for a pipe of inner radius 5 cm.
8-39 Water at 10°C (p = 999.7 kg/m 3 and jjl = 1 .307 X 1(T 3
kg/m • s) is flowing in a 0.20-cm-diameter 15-m-long pipe
steadily at an average velocity of 1 .2 m/s. Determine (a) the
pressure drop and (b) the pumping power requirement to over-
come this pressure drop.
Answers: (a) 188 kPa,(W 0.71 W
8-40 Water is to be heated from 10°C to 80°C as it flows
through a 2-cm-internal-diameter, 7-m-long tube. The tube is
equipped with an electric resistance heater, which provides
uniform heating throughout the surface of the tube. The outer
surface of the heater is well insulated, so that in steady op-
eration all the heat generated in the heater is transferred to
the water in the tube. If the system is to provide hot water at a
rate of 8 L/min, determine the power rating of the resistance
heater. Also, estimate the inner surface temperature of the pipe
at the exit.
8-41 Hot air at atmospheric pressure and 85°C enters a
10-m-long uninsulated square duct of cross section 0.15 m X
0.15 m that passes through the attic of a house at a rate of
0.10 m 3 /s. The duct is observed to be nearly isothermal at
70°C. Determine the exit temperature of the air and the rate of
heat loss from the duct to the air space in the attic.
Answers: 75.7°C, 941 W
Attic
space
FIGURE P8^1
8-42
Reconsider Problem 8^41. Using EES (or other)
software, investigate the effect of the volume
flow rate of air on the exit temperature of air and the rate of
heat loss. Let the flow rate vary from 0.05 m 3 /s to 0.15 m 3 /s.
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HEAT TRANSFER
Plot the exit temperature and the rate of heat loss as a function
of flow rate, and discuss the results.
8-43 Consider an air solar collector that is 1 m wide and 5 m
long and has a constant spacing of 3 cm between the glass
cover and the collector plate. Air enters the collector at 30°C at
a rate of 0.15 m 3 /s through the 1-m-wide edge and flows along
the 5-m-long passage way. If the average temperatures of the
glass cover and the collector plate are 20°C and 60°C, respec-
tively, determine (a) the net rate of heat transfer to the air in the
collector and (b) the temperature rise of air as it flows through
the collector.
cover
20°C
Insulation
Collector
plate, 60°C
FIGURE P8-43
8-44 Consider the flow of oil at 10°C in a 40-cm-diameter
pipeline at an average velocity of 0.5 m/s. A 300-m-long sec-
tion of the pipeline passes through icy waters of a lake at 0°C.
Measurements indicate that the surface temperature of the pipe
is very nearly 0°C. Disregarding the thermal resistance of the
pipe material, determine (a) the temperature of the oil when the
pipe leaves the lake, (b) the rate of heat transfer from the oil,
and (c) the pumping power required to overcome the pressure
losses and to maintain the flow oil in the pipe.
8—45 Consider laminar flow of a fluid through a square chan-
nel maintained at a constant temperature. Now the mean veloc-
ity of the fluid is doubled. Determine the change in the
pressure drop and the change in the rate of heat transfer be-
tween the fluid and the walls of the channel. Assume the flow
regime remains unchanged.
8-46 Repeat Problem 8^15 for turbulent flow.
8-47E The hot water needs of a household are to be met by
heating water at 55°F to 200°F by a parabolic solar collector at
a rate of 4 lbm/s. Water flows through a 1.25-in. -diameter thin
aluminum tube whose outer surface is blackanodized in order
to maximize its solar absorption ability. The centerline of the
tube coincides with the focal line of the collector, and a glass
Parabolic
solar collector
Glass tube
Water tube
FIGURE P8-47E
sleeve is placed outside the tube to minimize the heat losses. If
solar energy is transferred to water at a net rate of 350 Btu/h
per ft length of the tube, determine the required length of the
parabolic collector to meet the hot water requirements of this
house. Also, determine the surface temperature of the tube at
the exit.
8-48 A 15-cm X 20-cm printed circuit board whose compo-
nents are not allowed to come into direct contact with air for
reliability reasons is to be cooled by passing cool air through a
20-cm-long channel of rectangular cross section 0.2 cm X 14
cm drilled into the board. The heat generated by the electronic
components is conducted across the thin layer of the board to
the channel, where it is removed by air that enters the channel
at 15°C. The heat flux at the top surface of the channel can be
considered to be uniform, and heat transfer through other sur-
faces is negligible. If the velocity of the air at the inlet of the
channel is not to exceed 4 m/s and the surface temperature of
the channel is to remain under 50°C, determine the maximum
total power of the electronic components that can safely be
mounted on this circuit board.
Air channel
0.2 cm X 14 cm
FIGURE P8-48
Electronic
components
8-49 Repeat Problem 8^-8 by replacing air with helium,
which has six times the thermal conductivity of air.
8-50 rSi'M Reconsider Problem 8^48. Using EES (or other)
Ki3 software, investigate the effects of air velocity at
the inlet of the channel and the maximum surface temperature
on the maximum total power dissipation of electronic compo-
nents. Let the air velocity vary from 1 m/s to 10 m/s and the
surface temperature from 30°C to 90°C. Plot the power dissi-
pation as functions of air velocity and surface temperature, and
discuss the results.
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8-51 Air enters a 7-m-long section of a rectangular duct of
cross section 15 cm X 20 cm at 50°C at an average velocity of
7 m/s. If the walls of the duct are maintained at 10°C, deter-
mine (a) the outlet temperature of the air, (b) the rate of heat
transfer from the air, and (c) the fan power needed to overcome
the pressure losses in this section of the duct.
Answers: (a) 32.8°C, lb) 3674 W, (c) 4.2 W
8-52 [tj^l Reconsider Problem 8-51. Using EES (or other)
b^2 software, investigate the effect of air velocity on
the exit temperature of air, the rate of heat transfer, and the fan
power. Let the air velocity vary from 1 m/s to 10 m/s. Plot the
exit temperature, the rate of heat transfer, and the fan power as
a function of the air velocity, and discuss the results.
8-53 Hot air at 60°C leaving the furnace of a house enters a
12-m-long section of a sheet metal duct of rectangular cross
section 20 cm X 20 cm at an average velocity of 4 m/s. The
thermal resistance of the duct is negligible, and the outer sur-
face of the duct, whose emissivity is 0.3, is exposed to the cold
air at 10°C in the basement, with a convection heat transfer co-
efficient of 10 W/m 2 • °C. Taking the walls of the basement to
be at 10°C also, determine (a) the temperature at which the hot
air will leave the basement and (b) the rate of heat loss from the
hot air in the duct to the basement.
10°C
h = 10 W/m 2 -°C
Hot air Air duct
60°C 20 cm X 20 cm
4 m/s E = 0.3
FIGURE P8-53
8-54
Reconsider Problem 8-53. Using EES (or other)
software, investigate the effects of air velocity
and the surface emissivity on the exit temperature of air and the
rate of heat loss. Let the air velocity vary from 1 m/s to 10 m/s
and the emissivity from 0.1 to 1.0. Plot the exit temperature
and the rate of heat loss as functions of air velocity and emis-
sivity, and discuss the results.
8-55 The components of an electronic system dissipating
90 W are located in a 1-m-long horizontal duct whose cross
section is 16 cm X 16 cm. The components in the duct are
cooled by forced air, which enters at 32 °C at a rate of
0.65 m 3 /min. Assuming 85 percent of the heat generated inside
is transferred to air flowing through the duct and the remaining
15 percent is lost through the outer surfaces of the duct, deter-
455
CHAPTER 8
mine (a) the exit temperature of air and (b) the highest compo-
nent surface temperature in the duct.
8-56 Repeat Problem 8-55 for a circular horizontal duct of
15 -cm diameter.
8-57 Consider a hollow-core printed circuit board 12 cm
high and 18 cm long, dissipating a total of 20 W. The width of
the air gap in the middle of the PCB is 0.25 cm. The cooling air
enters the 12-cm-wide core at 32°C at a rate of 0.8 L/s. Assum-
ing the heat generated to be uniformly distributed over the two
side surfaces of the PCB, determine (a) the temperature at
which the air leaves the hollow core and (b) the highest tem-
perature on the inner surface of the core.
Answers: (a) 54.0°C, (b) 72.8°C
8-58 Repeat Problem 8-57 for a hollow-core PCB dissipat-
ing 35 W.
8-59E Water at 54 °F is heated by passing it through 0.75-in.-
internal-diameter thin-walled copper tubes. Heat is supplied to
the water by steam that condenses outside the copper tubes at
250°F. If water is to be heated to 140°F at a rate of 0.7 lbm/s,
determine (a) the length of the copper tube that needs to be
used and (b) the pumping power required to overcome pressure
losses. Assume the entire copper tube to be at the steam tem-
perature of 250°F.
8-60 A computer cooled by a fan contains eight PCBs, each
dissipating 10 W of power. The height of the PCBs is 12 cm
and the length is 18 cm. The clearance between the tips of the
components on the PCB and the back surface of the adjacent
PCB is 0.3 cm. The cooling air is supplied by a 10-W fan
mounted at the inlet. If the temperature rise of air as it flows
through the case of the computer is not to exceed 10°C, deter-
mine (a) the flow rate of the air that the fan needs to deliver,
(b) the fraction of the temperature rise of air that is due to the
heat generated by the fan and its motor, and (c) the highest
allowable inlet air temperature if the surface temperature of the
18 cm
Aii-
inlet
FIGURE P8-60
PCB, 10W
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HEAT TRANSFER
components is not to exceed 70°C anywhere in the system. Use
air properties at 25 °C.
Review Problems
8-61 A geothermal district heating system involves the trans-
port of geothermal water at 110°C from a geothermal well to a
city at about the same elevation for a distance of 1 2 km at a rate
of 1.5 m 3 /s in 60-cm-diameter stainless steel pipes. The fluid
pressures at the wellhead and the arrival point in the city are to
be the same. The minor losses are negligible because of the
large length-to-diameter ratio and the relatively small number
of components that cause minor losses, (a) Assuming the
pump-motor efficiency to be 65 percent, determine the electric
power consumption of the system for pumping, (b) Determine
the daily cost of power consumption of the system if the unit
cost of electricity is $0.06/kWh. (c) The temperature of geo-
thermal water is estimated to drop 0.5°C during this long flow.
Determine if the frictional heating during flow can make up for
this drop in temperature.
8-62 Repeat Problem 8-61 for cast iron pipes of the same
diameter.
8-63 The velocity profile in fully developed laminar flow in
a circular pipe, in m/s, is given by °V(r) = 6(1 — lOOr 2 ) where
r is the radial distance from the centerline of the pipe in m. De-
termine (a) the radius of the pipe, (b) the mean velocity
through the pipe, and (c) the maximum velocity in the pipe.
8-64E The velocity profile in fully developed laminar flow
of water at 40°F in a 80-ft-long horizontal circular pipe, in ft/s,
is given by T(r) = 0.8(1 — 625r 2 ) where r is the radial dis-
tance from the centerline of the pipe in ft. Determine (a) the
volume flow rate of water through the pipe, (b) the pressure
drop across the pipe, and (c) the useful pumping power re-
quired to overcome this pressure drop.
8-65 The compressed air requirements of a manufacturing
facility are met by a 150-hp compressor located in a room that
is maintained at 20°C. In order to minimize the compressor
work, the intake port of the compressor is connected to the
outside through an 1 1 -m-long, 20-cm-diameter duct made of
thin aluminum sheet. The compressor takes in air at a rate of
0.27 m 3 /s at the outdoor conditions of 10°C and 95 kPa. Dis-
regarding the thermal resistance of the duct and taking the heat
transfer coefficient on the outer surface of the duct to be 10
W/m 2 • °C, determine (a) the power used by the compressor to
overcome the pressure drop in this duct, (b) the rate of heat
transfer to the incoming cooler air, and (c) the temperature rise
of air as it flows through the duct.
8-66 A house built on a riverside is to be cooled in summer
by utilizing the cool water of the river, which flows at an aver-
age temperature of 15°C. A 15-m-long section of a circular
duct of 20-cm diameter passes through the water. Air enters the
underwater section of the duct at 25°C at a velocity of 3 m/s.
Assuming the surface of the duct to be at the temperature of the
Ail, 0.27 m 3 /s
10°C, 95 kPa
FIGURE P8-65
water, determine the outlet temperature of air as it leaves the
underwater portion of the duct. Also, for an overall fan effi-
ciency of 55 percent, determine the fan power input needed to
overcome the flow resistance in this section of the duct.
Air
25°C, 3 m/s
J_
15°C
Air
River, 15°C
FIGURE P8-66
8-67 Repeat Problem 8-66 assuming that a 0.15-mm-thick
layer of mineral deposit {k = 3 W/m • °C) formed on the inner
surface of the pipe.
8-68E (Jt\ The exhaust gases of an automotive engine
<s£$) leave the combustion chamber and enter a
8-ft-long and 3. 5 -in. -diameter thin-walled steel exhaust pipe at
800°F and 15.5 psia at a rate of 0.2 lbm/s. The surrounding
ambient air is at a temperature of 80°F, and the heat transfer
coefficient on the outer surface of the exhaust pipe is
3 Btu/h • ft 2 • °F. Assuming the exhaust gases to have the prop-
erties of air, determine (a) the velocity of the exhaust gases at
the inlet of the exhaust pipe and (b) the temperature at which
the exhaust gases will leave the pipe and enter the air.
8-69 Hot water at 90°C enters a 15-m section of a cast iron
pipe (k = 52 W/m • °C) whose inner and outer diameters are 4
and 4.6 cm, respectively, at an average velocity of 0.8 m/s. The
outer surface of the pipe, whose emissivity is 0.7, is exposed to
the cold air at 10°C in a basement, with a convection heat
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= 0.7
Hot
water
90°C
0.8 m/s
15m
FIGURE P8-69
transfer coefficient of 15 W/m 2 • °C. Taking the walls of the
basement to be at 10°C also, determine (a) the rate of heat loss
from the water and (b) the temperature at which the water
leaves the basement.
8-70 Repeat Problem 8-69 for a pipe made of copper (k =
386 W/m ■ °C) instead of cast iron.
8-71 D. B. Tuckerman and R. F. Pease of Stanford Univer-
sity demonstrated in the early 1980s that integrated circuits can
be cooled very effectively by fabricating a series of micro-
scopic channels 0.3 mm high and 0.05 mm wide in the back of
the substrate and covering them with a plate to confine the
fluid flow within the channels. They were able to dissipate
790 W of power generated in a 1 -cm 2 silicon chip at a junction-
to-ambient temperature difference of 71°C using water as the
coolant flowing at a rate of 0.01 L/s through 100 such channels
under a 1-cm X 1-cm silicon chip. Heat is transferred primar-
ily through the base area of the channel, and it was found that
the increased surface area and thus the fin effect are of lesser
importance. Disregarding the entrance effects and ignoring any
heat transfer from the side and cover surfaces, determine
(a) the temperature rise of water as it flows through the micro-
channels and (b) the average surface temperature of the base of
the microchannels for a power dissipation of 50 W. Assume the
water enters the channels at 20°C.
J \ ~~ Microscopic
0.05 mm \ channels
Electronic
circuits
on this side
FIGURE P8-71
457
CHAPTER 8
8-72 Liquid-cooled systems have high heat transfer coeffi-
cients associated with them, but they have the inherent disad-
vantage that they present potential leakage problems.
Therefore, air is proposed to be used as the microchannel
coolant. Repeat Problem 8-71 using air as the cooling fluid in-
stead of water, entering at a rate of 0.5 L/s.
8-73 Hot exhaust gases leaving a stationary diesel engine at
450°C enter a 15 -cm-diameter pipe at an average velocity of
3.6 m/s. The surface temperature of the pipe is 180°C. Deter-
mine the pipe length if the exhaust gases are to leave the pipe
at 250°C after transferring heat to water in a heat recovery unit.
Use properties of air for exhaust gases.
8-74 Geothermal steam at 165°C condenses in the shell side
of a heat exchanger over the tubes through which water flows.
Water enters the 4-cm-diameter, 14-m-long tubes at 20°C at a
rate of 0.8 kg/s. Determine the exit temperature of water and
the rate of condensation of geothermal steam.
8-75 Cold air at 5°C enters a 12-cm-diameter 20-m-long
isothermal pipe at a velocity of 2.5 m/s and leaves at 19°C.
Estimate the surface temperature of the pipe.
8-76 Oil at 10°C is to be heated by saturated steam at 1 atm
in a double-pipe heat exchanger to a temperature of 30°C. The
inner and outer diameters of the annular space are 3 cm and
5 cm, respectively, and oil enters at with a mean velocity of 0.8
m/s. The inner tube may be assumed to be isothermal at 100°C,
and the outer tube is well insulated. Assuming fully developed
flow for oil, determine the tube length required to heat the oil
to the indicated temperature. In reality, will you need a shorter
or longer tube? Explain.
Design and Essay Problems
8-77 Electronic boxes such as computers are commonly
cooled by a fan. Write an essay on forced air cooling of elec-
tronic boxes and on the selection of the fan for electronic
devices.
8-78 Design a heat exchanger to pasteurize milk by steam in
a dairy plant. Milk is to flow through a bank of 1 .2-cm internal
diameter tubes while steam condenses outside the tubes at
1 atm. Milk is to enter the tubes at 4°C, and it is to be heated to
72°C at a rate of 15 L/s. Making reasonable assumptions, you
are to specify the tube length and the number of tubes, and the
pump for the heat exchanger.
8-79 A desktop computer is to be cooled by a fan. The elec-
tronic components of the computer consume 80 W of power
under full-load conditions. The computer is to operate in envi-
ronments at temperatures up to 50°C and at elevations up to
3000 m where the atmospheric pressure is 70.12 kPa. The exit
temperature of air is not to exceed 60°C to meet the reliability
requirements. Also, the average velocity of air is not to exceed
120 m/min at the exit of the computer case, where the fan is in-
stalled to keep the noise level down. Specify the flow rate of
the fan that needs to be installed and the diameter of the casing
of the fan.
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NATURAL CONVECTION
CHAPTER
In Chapters 7 and 8, we considered heat transfer by forced convection,
where a fluid was forced to move over a surface or in a tube by external
means such as a pump or a fan. In this chapter, we consider natural con-
vection, where any fluid motion occurs by natural means such as buoyancy.
The fluid motion in forced convection is quite noticeable, since a fan or a
pump can transfer enough momentum to the fluid to move it in a certain di-
rection. The fluid motion in natural convection, however, is often not notice-
able because of the low velocities involved.
The convection heat transfer coefficient is a strong function of velocity: the
higher the velocity, the higher the convection heat transfer coefficient. The
fluid velocities associated with natural convection are low, typically less than
1 m/s. Therefore, the heat transfer coefficients encountered in natural convec-
tion are usually much lower than those encountered in forced convection. Yet
several types of heat transfer equipment are designed to operate under natural
convection conditions instead of forced convection, because natural convec-
tion does not require the use of a fluid mover.
We start this chapter with a discussion of the physical mechanism of natural
convection and the Grashof number. We then present the correlations to eval-
uate heat transfer by natural convection for various geometries, including
finned surfaces and enclosures. Finally, we discuss simultaneous forced and
natural convection.
CONTENTS
9-1 Physical Mechanism of Natural
Convection 460
9-2 Equation of Motion and the
Grashof Number 463
9-3 Natural Convection over
Surfaces 466
9-4 Natural Convection from Finned
Surfaces and PCBs 473
9-5 Natural Convection inside
Enclosures 477
9-6 Combined Natural and Forced
Convection 486
Topic of Special Interest:
Heat Transfer
Through Windows 489
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460
HEAT TRANSFER
Cool
FIGURE 9-1
The cooling of a boiled egg in a cooler
environment by natural convection.
FIGURE 9-2
The warming up of a cold drink
in a warmer environment by
natural convection.
9-1 - PHYSICAL MECHANISM OF NATURAL
CONVECTION
Many familiar heat transfer applications involve natural convection as the pri-
mary mechanism of heat transfer. Some examples are cooling of electronic
equipment such as power transistors, TVs, and VCRs; heat transfer from elec-
tric baseboard heaters or steam radiators; heat transfer from the refrigeration
coils and power transmission lines; and heat transfer from the bodies of ani-
mals and human beings. Natural convection in gases is usually accompanied
by radiation of comparable magnitude except for low-emissivity surfaces.
We know that a hot boiled egg (or a hot baked potato) on a plate eventually
cools to the surrounding air temperature (Fig. 9-1). The egg is cooled by
transferring heat by convection to the air and by radiation to the surrounding
surfaces. Disregarding heat transfer by radiation, the physical mechanism of
cooling a hot egg (or any hot object) in a cooler environment can be explained
as follows:
As soon as the hot egg is exposed to cooler air, the temperature of the outer
surface of the egg shell will drop somewhat, and the temperature of the air ad-
jacent to the shell will rise as a result of heat conduction from the shell to the
air. Consequently, the egg will soon be surrounded by a thin layer of warmer
air, and heat will then be transferred from this warmer layer to the outer lay-
ers of air. The cooling process in this case would be rather slow since the egg
would always be blanketed by warm air, and it would have no direct contact
with the cooler air farther away. We may not notice any air motion in the
vicinity of the egg, but careful measurements indicate otherwise.
The temperature of the air adjacent to the egg is higher, and thus its density
is lower, since at constant pressure the density of a gas is inversely propor-
tional to its temperature. Thus, we have a situation in which some low-density
or "light" gas is surrounded by a high-density or "heavy" gas, and the natural
laws dictate that the light gas rise. This is no different than the oil in a vine-
gar-and-oil salad dressing rising to the top (since p oiI < p V i ne gar)- This phe-
nomenon is characterized incorrectly by the phrase "heat rises," which is
understood to mean heated air rises. The space vacated by the warmer air in
the vicinity of the egg is replaced by the cooler air nearby, and the presence of
cooler air in the vicinity of the egg speeds up the cooling process. The rise
of warmer air and the flow of cooler air into its place continues until the egg
is cooled to the temperature of the surrounding air. The motion that results
from the continual replacement of the heated air in the vicinity of the egg by
the cooler air nearby is called a natural convection current, and the heat
transfer that is enhanced as a result of this natural convection current is called
natural convection heat transfer. Note that in the absence of natural con-
vection currents, heat transfer from the egg to the air surrounding it would be
by conduction only, and the rate of heat transfer from the egg would be much
lower.
Natural convection is just as effective in the heating of cold surfaces in a
warmer environment as it is in the cooling of hot surfaces in a cooler envi-
ronment, as shown in Figure 9-2. Note that the direction of fluid motion is
reversed in this case.
In a gravitational field, there is a net force that pushes upward a light fluid
placed in a heavier fluid. The upward force exerted by a fluid on a body
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CHAPTER 9
completely or partially immersed in it is called the buoyancy force. The mag-
nitude of the buoyancy force is equal to the weight of the fluid displaced by
the body. That is,
buoyancy rfluid 5 'body
(9-1)
where p fluid is the average density of the fluid (not the body), g is the gravita-
tional acceleration, and V body is the volume of the portion of the body im-
mersed in the fluid (for bodies completely immersed in the fluid, it is the total
volume of the body). In the absence of other forces, the net vertical force
acting on a body is the difference between the weight of the body and the
buoyancy force. That is,
W- Fu
buoyancy
Pbody 8 'body ~~ Pfluid c? 'body
(Pbody ~~ Pfluid) 8 'body
(9-2)
Note that this force is proportional to the difference in the densities of the fluid
and the body immersed in it. Thus, a body immersed in a fluid will experience
a "weight loss" in an amount equal to the weight of the fluid it displaces. This
is known as Archimedes' principle.
To have a better understanding of the buoyancy effect, consider an egg
dropped into water. If the average density of the egg is greater than the density
of water (a sign of freshness), the egg will settle at the bottom of the container.
Otherwise, it will rise to the top. When the density of the egg equals the
density of water, the egg will settle somewhere in the water while remaining
completely immersed, acting like a "weightless object" in space. This occurs
when the upward buoyancy force acting on the egg equals the weight of the
egg, which acts downward.
The buoyancy effect has far-reaching implications in life. For one thing,
without buoyancy, heat transfer between a hot (or cold) surface and the fluid
surrounding it would be by conduction instead of by natural convection. The
natural convection currents encountered in the oceans, lakes, and the atmos-
phere owe their existence to buoyancy. Also, light boats as well as heavy war-
ships made of steel float on water because of buoyancy (Fig. 9-3). Ships are
designed on the basis of the principle that the entire weight of a ship and its
contents is equal to the weight of the water that the submerged volume of the
ship can contain. The "chimney effect" that induces the upward flow of hot
combustion gases through a chimney is also due to the buoyancy effect, and
the upward force acting on the gases in the chimney is proportional to the dif-
ference between the densities of the hot gases in the chimney and the cooler
air outside. Note that there is no gravity in space, and thus there can be no nat-
ural convection heat transfer in a spacecraft, even if the spacecraft is filled
with atmospheric air.
In heat transfer studies, the primary variable is temperature, and it is desir-
able to express the net buoyancy force (Eq. 9-2) in terms of temperature dif-
ferences. But this requires expressing the density difference in terms of a
temperature difference, which requires a knowledge of a property that repre-
sents the variation of the density of a fluid with temperature at constant pres-
sure. The property that provides that information is the volume expansion
coefficient (3, defined as (Fig. 9-4)
(W--
FIGURE 9-3
It is the buoyancy force that
keeps the ships afloat in water
= ^buoyancy for floating objects).
20°C
100 kPa
lkg
(a) A substance with a large P
20°C
100 kPa
lkg
\dTlp
1
r —
* i
21°C
100 kPa
lkg
(b) A substance with a small P
FIGURE 9-4
The coefficient of volume expansion
is a measure of the change in volume
of a substance with temperature
at constant pressure.
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HEAT TRANSFER
p = Ki), =- K§), (1/K) (9_3)
In natural convection studies, the condition of the fluid sufficiently far from
the hot or cold surface is indicated by the subscript "infinity" to serve as a re-
minder that this is the value at a distance where the presence of the surface is
not felt. In such cases, the volume expansion coefficient can be expressed ap-
proximately by replacing differential quantities by differences as
1 Ap i p-^ - p
P ~ ~P AT = ~P T - T ^ at constant ^> (9 " 4)
or
p„ - p = p0(7 - r„) (at constant P) (9-5)
where p„ is the density and T m is the temperature of the quiescent fluid away
from the surface.
We can show easily that the volume expansion coefficient (3 of an ideal gas
(P = pRT) at a temperature 7Ms equivalent to the inverse of the temperature:
0ide alg as = ^ (1/K) (9-6)
where Tis the absolute temperature. Note that a large value of (3 for a fluid
means a large change in density with temperature, and that the product (3 AT
represents the fraction of volume change of a fluid that corresponds to a tem-
perature change Arat constant pressure. Also note that the buoyancy force is
proportional to the density difference, which is proportional to the temperature
difference at constant pressure. Therefore, the larger the temperature differ-
ence between the fluid adjacent to a hot (or cold) surface and the fluid away
from it, the larger the buoyancy force and the stronger the natural convection
currents, and thus the higher the heat transfer rate.
The magnitude of the natural convection heat transfer between a surface and
a fluid is directly related to the//ow rate of the fluid. The higher the flow rate,
the higher the heat transfer rate. In fact, it is the very high flow rates that in-
crease the heat transfer coefficient by orders of magnitude when forced con-
vection is used. In natural convection, no blowers are used, and therefore the
flow rate cannot be controlled externally. The flow rate in this case is estab-
lished by the dynamic balance of buoyancy and friction.
As we have discussed earlier, the buoyancy force is caused by the density dif-
ference between the heated (or cooled) fluid adjacent to the surface and the
fluid surrounding it, and is proportional to this density difference and the vol-
ume occupied by the warmer fluid. It is also well known that whenever two
bodies in contact (solid-solid, solid-fluid, or fluid-fluid) move relative to each
other, a friction force develops at the contact surface in the direction opposite to
that of the motion. This opposing force slows down the fluid and thus reduces
the flow rate of the fluid. Under steady conditions, the air flow rate driven by
buoyancy is established at the point where these two effects balance each other.
The friction force increases as more and more solid surfaces are introduced, se-
riously disrupting the fluid flow and heat transfer. For that reason, heat sinks
with closely spaced fins are not suitable for natural convection cooling.
Most heat transfer correlations in natural convection are based on ex-
perimental measurements. The instrument often used in natural convection
cen58933_ch09.qxd 9/4/2002 12:25 PM Page 463
463
CHAPTER 9
experiments is the Mach—Zehnder interferometer, which gives a plot
of isotherms in the fluid in the vicinity of a surface. The operation principle of
interferometers is based on the fact that at low pressure, the lines of constant
temperature for a gas correspond to the lines of constant density, and that the
index of refraction of a gas is a function of its density. Therefore, the degree
of refraction of light at some point in a gas is a measure of the tempera-
ture gradient at that point. An interferometer produces a map of interference
fringes, which can be interpreted as lines of constant temperature as shown
in Figure 9-5. The smooth and parallel lines in (a) indicate that the flow is
laminar, whereas the eddies and irregularities in (b) indicate that the flow is
turbulent. Note that the lines are closest near the surface, indicating a higher
temperature gradient.
9-2 - EQUATION OF MOTION AND THE GRASHOF
NUMBER
In this section we derive the equation of motion that governs the natural con-
vection flow in laminar boundary layer. The conservation of mass and energy
equations derived in Chapter 6 for forced convection are also applicable for
natural convection, but the momentum equation needs to be modified to in-
corporate buoyancy.
Consider a vertical hot flat plate immersed in a quiescent fluid body. We as-
sume the natural convection flow to be steady, laminar, and two-dimensional,
and the fluid to be Newtonian with constant properties, including density, with
one exception: the density difference p — p^, is to be considered since it is this
density difference between the inside and the outside of the boundary layer
that gives rise to buoyancy force and sustains flow. (This is known as the
Boussinesq approximation.) We take the upward direction along the plate to
be x, and the direction normal to surface to be y, as shown in Figure 9-6.
Therefore, gravity acts in the — .i-direction. Noting that the flow is steady and
two-dimensional, the x- and y-components of velocity within boundary layer
are u = u(x, y) and v = v(x, y), respectively.
The velocity and temperature profiles for natural convection over a vertical
hot plate are also shown in Figure 9-6. Note that as in forced convection, the
thickness of the boundary layer increases in the flow direction. Unlike forced
convection, however, the fluid velocity is zero at the outer edge of the veloc-
ity boundary layer as well as at the surface of the plate. This is expected since
the fluid beyond the boundary layer is motionless. Thus, the fluid velocity in-
creases with distance from the surface, reaches a maximum, and gradually de-
creases to zero at a distance sufficiently far from the surface. At the surface,
the fluid temperature is equal to the plate temperature, and gradually de-
creases to the temperature of the surrounding fluid at a distance sufficiently
far from the surface, as shown in the figure. In the case of cold surfaces, the
shape of the velocity and temperature profiles remains the same but their di-
rection is reversed.
Consider a differential volume element of height dx, length dy, and unit
depth in the z-direction (normal to the paper) for analysis. The forces acting
on this volume element are shown in Figure 9-7. Newton's second law of mo-
tion for this control volume can be expressed as
(a) Laminar flow (b) Turbulent flow
FIGURE 9-5
Isotherms in natural convection over
a hot plate in air.
« = o
Stationary
fluid
atJoo
FIGURE 9-6
Typical velocity and temperature
profiles for natural convection flow
over a hot vertical plate at temperature
T s inserted in a fluid at temperature 7^.
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464
HEAT TRANSFER
F + »*
FIGURE 9-7
Forces acting on a differential
control volume in the natural
convection boundary layer
over a vertical flat plate.
8m
(9-7)
where 8m = p(dx ■ dy ■ 1) is the mass of the fluid element within the control
volume. The acceleration in the x-direction is obtained by taking the total dif-
ferential of u(x, y), which is du = (du/dx)dx + (du/dy)dy, and dividing it by dt.
We get
du _ du dx du dy
dt ~ dx dt dy dt
du , du
dx dy
(9-8)
The forces acting on the differential volume element in the vertical direction
are the pressure forces acting on the top and bottom surfaces, the shear
stresses acting on the side surfaces (the normal stresses acting on the top and
bottom surfaces are small and are disregarded), and the force of gravity act-
ing on the entire volume element. Then the net surface force acting in the
.r-direction becomes
= [% dy
dP
dx
dx \(dy ■ 1) — pg(dx ■ dy ■ 1)
d~u
dy-
dP
dx
(9-9)
pgKdx-dy 1)
since t = |x(dw/dy). Substituting Eqs. 9-8 and 9-9 into Eq. 9-7 and dividing by
p • dx • dy • 1 gives the conservation of momentum in the x-direction as
du
du
P|"ta + V 3y
|X
d 2 u dP
dy 2 dx
PS
(9-10)
The jc-momentum equation in the quiescent fluid outside the boundary layer
can be obtained from the relation above as a special case by setting u = 0. It
gives
dP r _
dx
Pcog
(9-11)
which is simply the relation for the variation of hydrostatic pressure in a qui-
escent fluid with height, as expected. Also, noting that v <^ u in the boundary
layer and thus dv/dx ~ dv/dy ~ 0, and that there are no body forces (including
gravity) in the _y-direction, the force balance in that direction gives dP/dy = 0.
That is, the variation of pressure in the direction normal to the surface is neg-
ligible, and for a given x the pressure in the boundary layer is equal to the
pressure in the quiescent fluid. Therefore, P = P(x) = P ro (x) and dP/dx =
dPJdx = -p„g. Substituting into Eq. 9-10,
P "
du
dx
du
< —
dy
d 2 u
P- T^ + (P*
dy-
P)8
(9-12)
The last term represents the net upward force per unit volume of the fluid (the
difference between the buoyant force and the fluid weight). This is the force
that initiates and sustains convection currents.
From Eq. 9-5, we have p ra — p = pfi(T - T m ). Substituting it into the
last equation and dividing both sides by p gives the desired form of the
.r-momentum equation,
cen58933_ch09.qxd 9/4/2002 12:25 PM Page 465
dll
dx
+ V
du
dy
V + * p(r
T.)
(9-13)
This is the equation that governs the fluid motion in the boundary layer due
to the effect of buoyancy. Note that the momentum equation involves the
temperature, and thus the momentum and energy equations must be solved
simultaneously.
The set of three partial differential equations (the continuity, momentum,
and the energy equations) that govern natural convection flow over vertical
isothermal plates can be reduced to a set of two ordinary nonlinear differential
equations by the introduction of a similarity variable. But the resulting equa-
tions must still be solved numerically [Ostrach (1953), Ref. 27]. Interested
reader is referred to advanced books on the topic for detailed discussions [e.g.,
Kays and Crawford (1993), Ref. 23].
The Grashof Number
The governing equations of natural convection and the boundary conditions
can be nondimensionalized by dividing all dependent and independent vari-
ables by suitable constant quantities: all lengths by a characteristic length L c ,
all velocities by an arbitrary reference velocity T (which, from the definition
of Reynolds number, is taken to be T = Re L v/L c ), and temperature by a suit-
able temperature difference (which is taken to be T s — T„) as
* X $ " * U * V , ~~*. I ^ cc
x - j- y - J- u ~y v -f
and
465
CHAPTER 9
where asterisks are used to denote nondimensional variables. Substituting
them into the momentum equation and simplifying give
du
dx '
du
g$(T s - T„)Ll
Re 2 , Re L dy* 1
(9-14)
The dimensionless parameter in the brackets represents the natural convection
effects, and is called the Grashof number Gr L ,
gfi(T, - TJL)
Gr L = -. (9-15)
v 1
where
g = gravitational acceleration, m/s 2
(3 = coefficient of volume expansion, 1/K ((3 = 1/rfor ideal gases)
T s = temperature of the surface, °C
T a = temperature of the fluid sufficiently far from the surface, °C
L c = characteristic length of the geometry, m
v = kinematic viscosity of the fluid, m 2 /s
We mentioned in the preceding chapters that the flow regime in forced con-
vection is governed by the dimensionless Reynolds number, which represents
the ratio of inertial forces to viscous forces acting on the fluid. The flow
regime in natural convection is governed by the dimensionless Grashof num-
ber, which represents the ratio of the buoyancy force to the viscous force act-
ing on the fluid (Fig. 9-8).
Cold
fluid
Buoyancy
force
FIGURE 9-8
The Grashof number Gr is a measure
of the relative magnitudes of the
buoyancy force and the opposing
viscous force acting on the fluid.
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HEAT TRANSFER
The role played by the Reynolds number in forced convection is played by
the Grashof number in natural convection. As such, the Grashof number pro-
vides the main criterion in determining whether the fluid flow is laminar or
turbulent in natural convection. For vertical plates, for example, the critical
Grashof number is observed to be about 10 9 . Therefore, the flow regime on a
vertical plate becomes turbulent at Grashof numbers greater than 10 9 .
When a surface is subjected to external flow, the problem involves both nat-
ural and forced convection. The relative importance of each mode of heat
transfer is determined by the value of the coefficient Gr L /Re^: Natural con-
vection effects are negligible if Gr L /Ref <§ 1, free convection dominates and
the forced convection effects are negligible if Gr L /Re^ > 1, and both effects
are significant and must be considered if Gr L /Re^ ~ 1.
FIGURE 9-9
Natural convection heat transfer
correlations are usually expressed in
terms of the Rayleigh number raised
to a constant n multiplied by another
constant C, both of which are
determined experimentally.
9-3 - NATURAL CONVECTION OVER SURFACES
Natural convection heat transfer on a surface depends on the geometry of the
surface as well as its orientation. It also depends on the variation of tempera-
ture on the surface and the thermophysical properties of the fluid involved.
Although we understand the mechanism of natural convection well, the
complexities of fluid motion make it very difficult to obtain simple analytical
relations for heat transfer by solving the governing equations of motion and
energy. Some analytical solutions exist for natural convection, but such solu-
tions lack generality since they are obtained for simple geometries under some
simplifying assumptions. Therefore, with the exception of some simple cases,
heat transfer relations in natural convection are based on experimental studies.
Of the numerous such correlations of varying complexity and claimed accu-
racy available in the literature for any given geometry, we present here the
ones that are best known and widely used.
The simple empirical correlations for the average Nusselt number Nu in nat-
ural convection are of the form (Fig. 9-9)
Nu
hh,.
C(Gr L Pr)" = C Ra£
(9-16)
where Ra L is the Rayleigh number, which is the product of the Grashof and
Prandtl numbers:
g$(T s - TJI? C
Ra L = Gr L Pr = . Pr
(9-17)
The values of the constants C and n depend on the geometry of the surface and
the flow regime, which is characterized by the range of the Rayleigh number.
The value of n is usually i for laminar flow and i for turbulent flow. The value
of the constant C is normally less than 1 .
Simple relations for the average Nusselt number for various geometries are
given in Table 9-1, together with sketches of the geometries. Also given in
this table are the characteristic lengths of the geometries and the ranges of
Rayleigh number in which the relation is applicable. All fluid properties are to
be evaluated at the film temperature ZV = -(7*^ + TJ).
When the average Nusselt number and thus the average convection coef-
ficient is known, the rate of heat transfer by natural convection from a solid
cen58933_ch09.qxd 9/4/2002 12:25 PM Page 467
surface at a uniform temperature T s to the surrounding fluid is expressed by
Newton's law of cooling as
e conv = hA s {T s - r„) (W) (9-18)
where A s is the heat transfer surface area and h is the average heat transfer co-
efficient on the surface.
Vertical Plates (T s = constant)
For a vertical flat plate, the characteristic length is the plate height L. In Table
9-1 we give three relations for the average Nusselt number for an isothermal
vertical plate. The first two relations are very simple. Despite its complexity,
we suggest using the third one (Eq. 9-21) recommended by Churchill and Chu
(1975, Ref. 13) since it is applicable over the entire range of Rayleigh number.
This relation is most accurate in the range of 10 -1 < Ra L < 10 9 .
Vertical Plates (q s = constant)
In the case of constant surface heat flux, the rate of heat transfer is known (it
is simply Q = q s A s ), but the surface temperature T s is not. In fact, T s in-
creases with height along the plate. It turns out that the Nusselt number rela-
tions for the constant surface temperature and constant surface heat flux cases
are nearly identical [Churchill and Chu (1975), Ref. 13]. Therefore, the rela-
tions for isothermal plates can also be used for plates subjected to uniform
heat flux, provided that the plate midpoint temperature T L/2 is used for T s in
the evaluation of the film temperature, Rayleigh number, and the Nusselt
number. Noting that h = q s /(T L/2 — TJ), the average Nusselt number in this
case can be expressed as
Nu
hL
k
<1 S L
«T L
r.)
(9-27)
467
CHAPTER 9
The midpoint temperature T L/2 is determined by iteration so that the Nusselt
numbers determined from Eqs. 9-21 and 9-27 match.
Vertical Cylinders
An outer surface of a vertical cylinder can be treated as a vertical plate when
the diameter of the cylinder is sufficiently large so that the curvature effects
are negligible. This condition is satisfied if
D
35L
(9-28)
When this criteria is met, the relations for vertical plates can also be used for
vertical cylinders. Nusselt number relations for slender cylinders that do not
meet this criteria are available in the literature [e.g., Cebeci (1974), Ref. 8].
Inclined Plates
Consider an inclined hot plate that makes an angle 8 from the vertical, as
shown in Figure 9-10, in a cooler environment. The net force F = g(p m — p)
(the difference between the buoyancy and gravity) acting on a unit volume of
the fluid in the boundary layer is always in the vertical direction. In the case
FIGURE 9-10
Natural convection flows on the
upper and lower surfaces of
an inclined hot plate.
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HEAT TRANSFER
TABLE 9-1
Empirical correlations for the average Nusselt number for natural convection over surfaces
Geometry
Characteristic
length L c
Range of Ra
Nu
Vertical plate
10 4 -10 9
10 9 -10 13
Entire range
Nu = 0.59Ral' 4
Nu = 0.1 RaP
Nu
0.825
0.387RaJ"
[1 + (0.492/Pr) 9 ' 16 ] 8 ' 27
(complex but more accurate)
(9-19)
(9-20)
(9-21)
Inclined plate
Use vertical plate equations for the upper
surface of a cold plate and the lower
surface of a hot plate
Replace g by gcosd
for
Ra < 10 9
Horiontal plate
(Surface area A and perimeter p)
(a) Upper surface of a hot plate
(or lower surface of a cold plate)
Hot surface 1
(b) Lower surface of a hot plate
(or upper surface of a cold plate)
A.lp
Hot surface
10 4 -10 7
10 7 -10 u
Nu = 0.54Rai' 4
Nu = 0.15Ra[' 3
lQS-lO 1
Nu = 0.27R3 1 / 4
(9-22)
(9-23)
(9-24)
Vertical cylinder
A vertical cylinder can be treated as a
vertical plate when
D>
35/.
Grl' 4
Horizontal cylinder
T
D
_L_
RanfSlO 1
Nu
0.6 +■
0.387Ra# 6
[1 + (0.559/Pr) 9 ' 16 ] 8 ' 27
(9-25)
Sphere
Ra D < 10 11
(Pr >0.7)
Nu
[1 + (0.469/Pr) 9
(9-26)
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CHAPTER 9
of inclined plate, this force can be resolved into two components: F y = F cos
9 parallel to the plate that drives the flow along the plate, and F y = F sin 6
normal to the plate. Noting that the force that drives the motion is reduced, we
expect the convection currents to be weaker, and the rate of heat transfer to be
lower relative to the vertical plate case.
The experiments confirm what we suspect for the lower surface of a hot
plate, but the opposite is observed on the upper surface. The reason for this cu-
rious behavior for the upper surface is that the force component F y initiates
upward motion in addition to the parallel motion along the plate, and thus the
boundary layer breaks up and forms plumes, as shown in the figure. As a re-
sult, the thickness of the boundary layer and thus the resistance to heat trans-
fer decreases, and the rate of heat transfer increases relative to the vertical
orientation.
In the case of a cold plate in a warmer environment, the opposite occurs as
expected: The boundary layer on the upper surface remains intact with weaker
boundary layer flow and thus lower rate of heat transfer, and the boundary
layer on the lower surface breaks apart (the colder fluid falls down) and thus
enhances heat transfer.
When the boundary layer remains intact (the lower surface of a hot plate or
the upper surface of a cold plate), the Nusselt number can be determined from
the vertical plate relations provided that g in the Rayleigh number relation is
replaced by g cos 9 for < 60°. Nusselt number relations for the other two
surfaces (the upper surface of a hot plate or the lower surface of a cold plate)
are available in the literature [e.g., Fujiii and Imura (1972), Ref. 18].
Horizontal Plates
The rate of heat transfer to or from a horizontal surface depends on whether
the surface is facing upward or downward. For a hot surface in a cooler envi-
ronment, the net force acts upward, forcing the heated fluid to rise. If the hot
surface is facing upward, the heated fluid rises freely, inducing strong natural
convection currents and thus effective heat transfer, as shown in Figure 9-11.
But if the hot surface is facing downward, the plate will block the heated fluid
that tends to rise (except near the edges), impeding heat transfer. The opposite
is true for a cold plate in a warmer environment since the net force (weight
minus buoyancy force) in this case acts downward, and the cooled fluid near
the plate tends to descend.
The average Nusselt number for horizontal surfaces can be determined from
the simple power-law relations given in Table 9-1. The characteristic length
for horizontal surfaces is calculated from
P
(9-29)
where A s is the surface area and p is the perimeter. Note that L c = a/4 for a
horizontal square surface of length a, and DIA for a horizontal circular surface
of diameter D.
Horizontal Cylinders and Spheres
The boundary layer over a hot horizontal cylinder start to develop at the bot-
tom, increasing in thickness along the circumference, and forming a rising
"^ r^ fh m
Natural /
convection
currents
Hot
plate
FIGURE 9-11
Natural convection flows on the
upper and lower surfaces of
a horizontal hot plate.
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HEAT TRANSFER
Boundary
layer flow
FIGURE 9-12
Natural convection flow over a
horizontal hot cylinder.
r„ = 2o°c
,70°C
D=8cm
J
6 m
FIGURE 9-13
Schematic for Example 9-1 .
plume at the top, as shown in Figure 9-12. Therefore, the local Nusselt num-
ber is highest at the bottom, and lowest at the top of the cylinder when the
boundary layer flow remains laminar. The opposite is true in the case of a cold
horizontal cylinder in a warmer medium, and the boundary layer in this case
starts to develop at the top of the cylinder and ending with a descending plume
at the bottom.
The average Nusselt number over the entire surface can be determined from
Eq. 9-26 [Churchill and Chu (1975), Ref. 13] for an isothermal horizontal
cylinder, and from Eq. 9-27 for an isothermal sphere [Churchill (1983), Ref.
11] both given in Table 9-1.
EXAMPLE 9-1 Heat Loss from Hot Water Pipes
A 6-m-long section of an 8-cm-diameter horizontal hot water pipe shown in Fig-
ure 9-13 passes through a large room whose temperature is 20°C. If the outer
surface temperature of the pipe is 70°C, determine the rate of heat loss from
the pipe by natural convection.
SOLUTION A horizontal hot water pipe passes through a large room. The rate
of heat loss from the pipe by natural convection is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas. 3 The
local atmospheric pressure is 1 atm.
Properties The properties of air at the film temperature of T f = (T s + TJ/2 =
(70 + 20)/2 = 45°C and 1 atm are (Table A-15)
k = 0.02699 W/m • °C
v = 1.749 X 10- 5 m 2 /s
Pr = 0.7241
1 1
P
'/
318K
Analysis The characteristic length in this case is the outer diameter of the
pipe, L c = D = 0.08 m. Then the Rayleigh number becomes
Ra r
g&(T, - TJD 3
Pr
(9.81 m/s 2 )[l/(318 K)](70 - 20 K)(0.08 m) 3
(1.749 X 10- 5 m 2 /s) 2
(0.7241) = 1.869 X 10 6
The natural convection Nusselt number in this case can be determined from
Eq. 9-25 to be
Nu
Then,
0.6 +
17.40
0.387 Ra
D
[1 + (0.559/Pr) 9 " 6 ] 8
0.6 +
0.387(1869 X 10 6 ) 1/6
[1 + (0.559/0.7241) 9 " 6 ] 8
h = ^ Nu
0.02699 W/m • °C
0.08 m
A s = ttDL = tt(0.08 m)(6 m) = 1.508 m 2
(17.40) = 5.869 W/m • °C
and
Q = hA s (T s - r„) = (5.869 W/m 2 • °C)(1.508 m 2 )(70 - 20)°C = 443 W
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CHAPTER 9
Therefore, the pipe will lose heat to the air in the room at a rate of 443 W by
natural convection.
Discussion The pipe will lose heat to the surroundings by radiation as well as by
natural convection. Assuming the outer surface of the pipe to be black (emissiv-
ity e = 1) and the inner surfaces of the walls of the room to be at room temper-
ature, the radiation heat transfer is determined to be (Fig. 9-14)
Grad = ^Al'i ~~ *suir)
= (l)(1.508m 2 )(5.67 X 1(T 8 W/m 2 • K 4 )[(70 + 273 K) 4 - (20 + 273 K) 4 ]
= 553 W
which is larger than natural convection. The emissivity of a real surface is less
than 1, and thus the radiation heat transfer for a real surface will be less. But
radiation will still be significant for most systems cooled by natural convection.
Therefore, a radiation analysis should normally accompany a natural convection
analysis unless the emissivity of the surface is low.
/
r m =2o°c
e na ,co„v=443W
T =70°C
"
e ra d,„, a x = 553W
FIGURE 9-14
Radiation heat transfer is usually
comparable to natural convection in
magnitude and should be considered in
heat transfer analysis.
EXAMPLE 9-2 Cooling of a Plate in Different Orientations
Consider a 0.6-m X 0.6-m thin square plate in a room at 30°C. One side of the
plate is maintained at a temperature of 90°C, while the other side is insulated,
as shown in Figure 9-15. Determine the rate of heat transfer from the plate by
natural convection if the plate is (a) vertical, (b) horizontal with hot surface fac-
ing up, and (c) horizontal with hot surface facing down.
SOLUTION A hot plate with an insulated back is considered. The rate of heat
loss by natural convection is to be determined for different orientations.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas. 3 The
local atmospheric pressure is 1 atm.
Properties The properties of air at the film temperature of T f = (T s + TJ/2 =
(90 + 30)/2 = 60°C and 1 atm are (Table A-15)
= 0.02808 W/m • C Pr
a. 896 X 10- 5 m 2 /s B
0.7202
1 1
'/
333 K
Analysis (a) Vertical. The characteristic length in this case is the height of the
plate, which is L = 0.6 m. The Rayleigh number is
Ra L
g $(T s - r„)L 3
Pr
(9.81 m/s 2 )[l/(333 K)](90 - 30 K)(0.6 m) 3
(1.896 X 10- 5 m 2 /s) 2
(0.722) = 7.656 X 10 8
Then the natural convection Nusselt number can be determined from Eq. 9-21
to be
Nu
0.825 +
0.387 Ra[ /6
0.825 +
[1 + (0.492/Pr) 9 ' 16 ] 8 ' 27
0.387(7.656 X 10 8 ) 1
1 + (0.492/0.7202) 9 ' 16 ;
113.4
L=0.6m/
90°C
T„ = 30°C
(a) Vertical
\1/
(c) Hot surface facing down
FIGURE 9-15
Schematic for Example 9-2.
cen58933_ch09.qxd 9/4/2002 12:25 PM Page 472
472
HEAT TRANSFER
Note that the simpler relation Eq. 9-19 would give Nu = 0.59 Ra L 1/4 = 98.14,
which is 13 percent lower. Then,
h
L
V-
Nu
= 0.02808 W/m ■ °C
0.6 m
(0.6 m) 2 = 0.36 m 2
(113.4) = 5.306 W/m 2 • °C
and
Q = hAJT s - r„) = (5.306 W/m 2 • °C)(0.36 m 2 )(90 - 30)°C = 115 W
(b) Horizontal with hot surface facing up. The characteristic length and the
Rayleigh number in this case are
4
Ra L
A s L 2 L 0.6 m
P AL 4 4
g$(T s - TJL* n
lr
0.15 m
(9.81 m/s 2 )[l/(333 K)](90 - 30 K)(0.15 m) 3
(0.7202) = 1.196 X 10 7
(1.896 X 10- 5 m 2 /s) 2
The natural convection Nusselt number can be determined from Eq. 9-22 to be
Nu = 0.54 Ra[' 4 = 0.54(1.196 X 10 7 ) 1 ' 4 = 31.76
Then,
1}
Nu
0.0280 W/m ■ °C
0.15 m
(0.6 m) 2 = 0.36 m 2
(31.76) = 5.946 W/m 2 • "C
and
Q = hA s (T s - r ra ) = (5.946 W/m 2 • °C)(0.36 m 2 )(90 - 30)°C = 128 W
(c) Horizontal with hot surface facing down. The characteristic length, the heat
transfer surface area, and the Rayleigh number in this case are the same as
those determined in (b). But the natural convection Nusselt number is to be de-
termined from Eq. 9-24,
Nu
0.27(1.196 X 10 7 ) 1
15.86
Then,
and
Lr
Nu
0.02808 W/m • °C
0.15 m
(15.86) = 2.973 W/m 2 • °C
Q = hA s (T s - TJ = (2.973 W/m 2 ■ °C)(0.36 m 2 )(90 - 30)°C = 64.2 W
Note that the natural convection heat transfer is the lowest in the case of the
hot surface facing down. This is not surprising, since the hot air is "trapped"
under the plate in this case and cannot get away from the plate easily. As a re-
sult, the cooler air in the vicinity of the plate will have difficulty reaching the
plate, which results in a reduced rate of heat transfer.
Discussion The plate will lose heat to the surroundings by radiation as well as
by natural convection. Assuming the surface of the plate to be black (emissivity
cen58933_ch09.qxd 9/4/2002 12:25 PM Page 473
e=l) and the inner surfaces of the walls of the room to be at room tempera-
ture, the radiation heat transfer in this case is determined to be
2 rad = eA s <j(T s 4 - KJ
= (l)(0.36m 2 )(5.67 X 10"
= 182 W
; W/m 2 ■ K 4 )[(90 + 273 K) 4 - (30 + 273 K) 4 ]
which is larger than that for natural convection heat transfer for each case.
Therefore, radiation can be significant and needs to be considered in surfaces
cooled by natural convection.
473
CHAPTER 9
9-4 - NATURAL CONVECTION FROM FINNED
SURFACES AND PCBs
Natural convection flow through a channel formed by two parallel plates as
shown in Figure 9-16 is commonly encountered in practice. When the plates
are hot (T s > T^), the ambient fluid at T„ enters the channel from the lower
end, rises as it is heated under the effect of buoyancy, and the heated fluid
leaves the channel from the upper end. The plates could be the fins of a finned
heat sink, or the PCBs (printed circuit boards) of an electronic device. The
plates can be approximated as being isothermal (T s = constant) in the first
case, and isoflux (q s = constant) in the second case.
Boundary layers start to develop at the lower ends of opposing surfaces, and
eventually merge at the midplane if the plates are vertical and sufficiently
long. In this case, we will have fully developed channel flow after the merger
of the boundary layers, and the natural convection flow is analyzed as channel
flow. But when the plates are short or the spacing is large, the boundary lay-
ers of opposing surfaces never reach each other, and the natural convection
flow on a surface is not affected by the presence of the opposing surface. In
that case, the problem should be analyzed as natural convection from two in-
dependent plates in a quiescent medium, using the relations given for surfaces,
rather than natural convection flow through a channel.
Natural Convection Cooling of Finned Surfaces
{T s = constant)
Finned surfaces of various shapes, called heat sinks, are frequently used in the
cooling of electronic devices. Energy dissipated by these devices is transferred
to the heat sinks by conduction and from the heat sinks to the ambient air by
natural or forced convection, depending on the power dissipation require-
ments. Natural convection is the preferred mode of heat transfer since it in-
volves no moving parts, like the electronic components themselves. However,
in the natural convection mode, the components are more likely to run at a
higher temperature and thus undermine reliability. A properly selected heat
sink may considerably lower the operation temperature of the components and
thus reduce the risk of failure.
Natural convection from vertical finned surfaces of rectangular shape has
been the subject of numerous studies, mostly experimental. Bar-Cohen and
Fully
developed
flow
Isothermal
plate at T,
Boundary
layer
FIGURE 9-16
Natural convection flow through a
channel between two isothermal
vertical plates.
cen58933_ch09.qxd 9/4/2002 12:25 PM Page 474
474
HEAT TRANSFER
&
la)
(b)
FIGURE 9-17
Heat sinks with (a) widely spaced and
(b) closely packed fins (courtesy of
Vemaline Products).
Quiescent
air, T
FIGURE 9-18
Various dimensions of a finned surface
oriented vertically.
Rohsenow (1984, Ref. 5) have compiled the available data under various
boundary conditions, and developed correlations for the Nusselt number and
optimum spacing. The characteristic length for vertical parallel plates used as
fins is usually taken to be the spacing between adjacent fins S, although the fin
height L could also be used. The Rayleigh number is expressed as
Ra,
g$(T s
Pr
and
Ra,
g$(T s - T„)L :
Pr = Ra ?
(9-30)
The recommended relation for the average Nusselt number for vertical
isothermal parallel plates is
constant:
Nu
hS_
k
576
2.873
(Ra s S/L) 2 {Rn s S/L)°
(9-31)
A question that often arises in the selection of a heat sink is whether to se-
lect one with closely packed fins or widely spaced fins for a given base area
(Fig. 9-17). A heat sink with closely packed fins will have greater surface area
for heat transfer but a smaller heat transfer coefficient because of the extra
resistance the additional fins introduce to fluid flow through the interfin
passages. A heat sink with widely spaced fins, on the other hand, will have a
higher heat transfer coefficient but a smaller surface area. Therefore, there
must be an optimum spacing that maximizes the natural convection heat trans-
fer from the heat sink for a given base area WL, where W and L are the width
and height of the base of the heat sink, respectively, as shown in Figure 9-18.
When the fins are essentially isothermal and the fin thickness t is small rela-
tive to the fin spacing S, the optimum fin spacing for a vertical heat sink is de-
termined by Bar-Cohen and Rohsenow to be
constant:
S ~ = 2J14 £
2.714;
It can be shown by combining the three equations above that when S
the Nusselt number is a constant and its value is 1.307,
Nu
hS n
1.307
(9-32)
J opt>
(9-33)
The rate of heat transfer by natural convection from the fins can be deter-
mined from
Q = h(2nLH){T s - rj
(9-34)
where n = W/(S + t) ~ WIS is the number of fins on the heat sink and T s is the
surface temperature of the fins. All fluid properties are to be evaluated at the
average temperature 7 , ave = (T s + T„)I2.
Natural Convection Cooling of Vertical PCBs
(q s = constant)
Arrays of printed circuit boards used in electronic systems can often be mod-
eled as parallel plates subjected to uniform heat flux q s (Fig. 9-19). The plate
temperature in this case increases with height, reaching a maximum at the
cen58933_ch09.qxd 9/4/2002 12:25 PM Page 475
upper edge of the board. The modified Rayleigh number for uniform heat flux
on both plates is expressed as
gV4sS 4
r&:
kv 2
Pl-
(9-35)
The Nusselt number at the upper edge of the plate where maximum tempera-
ture occurs is determined from [Bar-Cohen and Rohsenow (1984), Ref. 5]
Nu,
48
2.51
IRa sS/L (Ra*5VLf 4 J
(9-36)
The optimum fin spacing for the case of uniform heat flux on both plates is
given as
constant:
2.12
s 4 lY 2
Ra'
The total rate of heat transfer from the plates is
Q = q s A s = q s (2nLH)
(9-37)
(9-38)
where n = W/(S + t) ~ WIS is the number of plates. The critical surface tem-
perature T L occurs at the upper edge of the plates, and it can be determined
from
q s = h L (T L - rj (9-39)
All fluid properties are to be evaluated at the average temperature r ave =
(T L + rj/2.
Mass Flow Rate through the Space between Plates
As we mentioned earlier, the magnitude of the natural convection heat trans-
fer is directly related to the mass flow rate of the fluid, which is established by
the dynamic balance of two opposing effects: buoyancy and friction.
The fins of a heat sink introduce both effects: inducing extra buoyancy as a
result of the elevated temperature of the fin surfaces and slowing down the
fluid by acting as an added obstacle on the flow path. As a result, increasing
the number of fins on a heat sink can either enhance or reduce natural con-
vection, depending on which effect is dominant. The buoyancy-driven fluid
flow rate is established at the point where these two effects balance each
other. The friction force increases as more and more solid surfaces are intro-
duced, seriously disrupting fluid flow and heat transfer. Under some condi-
tions, the increase in friction may more than offset the increase in buoyancy.
This in turn will tend to reduce the flow rate and thus the heat transfer. For that
reason, heat sinks with closely spaced fills are not suitable for natural convec-
tion cooling.
When the heat sink involves closely spaced fins, the narrow channels
formed tend to block or "suffocate" the fluid, especially when the heat sink is
long. As a result, the blocking action produced overwhelms the extra buoy-
ancy and downgrades the heat transfer characteristics of the heat sink. Then,
at a fixed power setting, the heat sink runs at a higher temperature relative to
the no-shroud case. When the heat sink involves widely spaced fins, the
475
CHAPTER 9
FIGURE 9-19
Arrays of vertical printed circuit
boards (PCBs) cooled by natural
convection.
cen58933_ch09.qxd 9/4/2002 12:25 PM Page 476
476
HEAT TRANSFER
shroud does not introduce a significant increase in resistance to flow, and the
buoyancy effects dominate. As a result, heat transfer by natural convection
may improve, and at a fixed power level the heat sink may run at a lower tem-
perature.
When extended surfaces such as fins are used to enhance natural convection
heat transfer between a solid and a fluid, the flow rate of the fluid in the vicin-
ity of the solid adjusts itself to incorporate the changes in buoyancy and fric-
tion. It is obvious that this enhancement technique will work to advantage
only when the increase in buoyancy is greater than the additional friction in-
troduced. One does not need to be concerned with pressure drop or pumping
power when studying natural convection since no pumps or blowers are used
in this case. Therefore, an enhancement technique in natural convection is
evaluated on heat transfer performance alone.
The failure rate of an electronic component increases almost exponentially
with operating temperature. The cooler the electronic device operates, the
more reliable it is. A rule of thumb is that the semiconductor failure rate is
halved for each 10°C reduction injunction operating temperature. The desire
to lower the operating temperature without having to resort to forced convec-
tion has motivated researchers to investigate enhancement techniques for nat-
ural convection. Sparrow and Prakash (Ref. 31) have demonstrated that, under
certain conditions, the use of discrete plates in lieu of continuous plates of the
same surface area increases heat transfer considerably. In other experimental
work, using transistors as the heat source, £engel and Zing (Ref. 9) have
demonstrated that temperature recorded on the transistor case dropped by as
much as 30°C when a shroud was used, as opposed to the corresponding no-
shroud case.
t= l mm -
FIGURE 9-20
Schematic for Example 9-3.
EXAMPLE 9-3 Optimum Fin Spacing of a Heat Sink
A 12-cm-wide and 18-cm-high vertical hot surface in 30°C air is to be cooled by
a heat sink with equally spaced fins of rectangular profile (Fig. 9-20). The fins
are 0.1 cm thick and 18 cm long in the vertical direction and have a height of
2.4 cm from the base. Determine the optimum fin spacing and the rate of heat
transfer by natural convection from the heat sink if the base temperature is 80°C.
SOLUTION A heat sink with equally spaced rectangular fins is to be used to
cool a hot surface. The optimum fin spacing and the rate of heat transfer are to
be determined.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas. 3 The
atmospheric pressure at that location is 1 atm. 4 The thickness t of the fins is
very small relative to the fin spacing S so that Eq. 9-32 for optimum fin spac-
ing is applicable. 5 All fin surfaces are isothermal at base temperature.
Properties The properties of air at the film temperature of T f = (T s + TJ/2 =
(80 + 30)/2 = 55°C and 1 atm pressure are (Table A-15)
k = 0.02772 W/m • °C
v = 1.846 X 10- 5 m 2 /s
Pr
P
0.7215
1/7}= 1/328 K
Analysis We take the characteristic length to be the length of the fins in the
vertical direction (since we do not know the fin spacing). Then the Rayleigh
number becomes
cen58933_ch09.qxd 9/4/2002 12:25 PM Page 477
Ra L = Pr
v l
(981 m/s 2 )[l/(328 K)](80 - 30 K)(0.18 m) 3
(1.846 X 10- 5 m 2 /s) 2
The optimum fin spacing is determined from Eq. 7-32 to be
L „„,, 0.8 m
(0.7215) = 1.846 X 10 7
2.714:
2.714
(1.846 X 10 7 )
7.45 X 10" 3 m = 7.45 mm
which is about seven times the thickness of the fins. Therefore, the assumption
of negligible fin thickness in this case is acceptable. The number of fins and
the heat transfer coefficient for this optimum fin spacing case are
W
0.12 m
15 fins
S + t (0.00745 + 0.0001) m
The convection coefficient for this optimum in spacing case is, from Eq. 9-33,
k , „„„ 0.02772 W/m-°C
h = Nu
opt c
1.307-
0.00745 m
0.2012 W/m 2 • °C
Then the rate of natural convection heat transfer becomes
Q = hA s (T s - rj = h(2nLH)(T s - TJ
= (0.2012 W/m 2 • °C)[2 X 15(0.18 m)(0.024 m)](80 - 30)°C = 1.30 W
Therefore, this heat sink can dissipate heat by natural convection at a rate of
1.30 W.
9-5 - NATURAL CONVECTION INSIDE ENCLOSURES
A considerable portion of heat loss from a typical residence occurs through
the windows. We certainly would insulate the windows, if we could, in order
to conserve energy. The problem is finding an insulating material that is trans-
parent. An examination of the thermal conductivities of the insulting materi-
als reveals that air is a better insulator than most common insulating
materials. Besides, it is transparent. Therefore, it makes sense to insulate the
windows with a layer of air. Of course, we need to use another sheet of glass
to trap the air. The result is an enclosure, which is known as a double-pane
window in this case. Other examples of enclosures include wall cavities, solar
collectors, and cryogenic chambers involving concentric cylinders or spheres.
Enclosures are frequently encountered in practice, and heat transfer through
them is of practical interest. Heat transfer in enclosed spaces is complicated
by the fact that the fluid in the enclosure, in general, does not remain station-
ary. In a vertical enclosure, the fluid adjacent to the hotter surface rises and the
fluid adjacent to the cooler one falls, setting off a rotationary motion within
the enclosure that enhances heat transfer through the enclosure. Typical flow
patterns in vertical and horizontal rectangular enclosures are shown in Figures
9-21 and 9-22.
477
CHAPTER 9
Cold
surface
Q-+-
ir
M
Hot
surface
Velocity
profile
FIGURE 9-21
Convective currents in a vertical
rectangular enclosure.
Light fluid
-Hot
(No fluid motion)
Heavy fluid
(a) Hot plate at the top
s- Heavy fluid
-Cold
Cold
Mit \\i \i tf ^
Light fluid
\
Hot
(b) Hot plate at the bottom
FIGURE 9-22
Convective currents in a horizontal
enclosure with (a) hot plate at the top
and (b) hot plate at the bottom.
cen58933_ch09.qxd 9/4/2002 12:25 PM Page 47E
478
HEAT TRANSFER
The characteristics of heat transfer through a horizontal enclosure depend
on whether the hotter plate is at the top or at the bottom, as shown in Fig-
ure 9-22. When the hotter plate is at the top, no convection currents will de-
velop in the enclosure, since the lighter fluid will always be on top of the
heavier fluid. Heat transfer in this case will be by pure conduction, and we
will have Nu = 1. When the hotter plate is at the bottom, the heavier fluid will
be on top of the lighter fluid, and there will be a tendency for the lighter fluid
to topple the heavier fluid and rise to the top, where it will come in contact
with the cooler plate and cool down. Until that happens, however, the heat
transfer is still by pure conduction and Nu = 1. When Ra > 1708, the buoy-
ant force overcomes the fluid resistance and initiates natural convection cur-
rents, which are observed to be in the form of hexagonal cells called Benard
cells. For Ra > 3 X 10 5 , the cells break down and the fluid motion becomes
turbulent.
The Rayleigh number for an enclosure is determined from
*0(r, - t 2 )l] i
Ra,
■Pr
(9-40)
where the characteristic length L c is the distance between the hot and cold sur-
faces, and Ty and T 2 are the temperatures of the hot and cold surfaces, respec-
tively. All fluid properties are to be evaluated at the average fluid temperature
r ave = (r, + r 2 )/2.
Effective Thermal Conductivity
When the Nusselt number is known, the rate of heat transfer through the en-
closure can be determined from
Q = hA s (T, - T 2 ) = kNuA s
(9-41)
Hot
Nu = 3
Cold Hot k eff =3k Cold
(No
motion)
Q = 10W
e = 3ow
Pure
conduction
Natural
convection
FIGURE 9-23
A Nusselt number of 3 for an
enclosure indicates that heat transfer
through the enclosure by natural
convection is three times that by pure
conduction.
since h = kNu/L. The rate of steady heat conduction across a layer of thick-
ness L c , area A s , and thermal conductivity k is expressed as
Q cond — kA s -
T, -T,
(9-42)
where T { and T 2 are the temperatures on the two sides of the layer. A compar-
ison of this relation with Eq. 9-41 reveals that the convection heat transfer in
an enclosure is analogous to heat conduction across the fluid layer in the en-
closure provided that the thermal conductivity k is replaced by £Nu. That is,
the fluid in an enclosure behaves like a fluid whose thermal conductivity is
kNu as a result of convection currents. Therefore, the quantity kNu is called
the effective thermal conductivity of the enclosure. That is,
£Nu
(9-43)
Note that for the special case of Nu = 1, the effective thermal conductivity of
the enclosure becomes equal to the conductivity of the fluid. This is expected
since this case corresponds to pure conduction (Fig. 9-23).
Natural convection heat transfer in enclosed spaces has been the subject
of many experimental and numerical studies, and numerous correlations for
the Nusselt number exist. Simple power-law type relations in the form of
cen58933_ch09.qxd 9/4/2002 12:25 PM Page 479
Nu = CRa£, where C and n are constants, are sufficiently accurate, but they
are usually applicable to a narrow range of Prandtl and Rayleigh numbers
and aspect ratios. The relations that are more comprehensive are naturally
more complex. Next we present some widely used relations for various types
of enclosures.
Horizontal Rectangular Enclosures
We need no Nusselt number relations for the case of the hotter plate being at
the top, since there will be no convection currents in this case and heat trans-
fer will be downward by conduction (Nu = 1). When the hotter plate is at the
bottom, however, significant convection currents set in for Ra L > 1708, and
the rate of heat transfer increases (Fig. 9-24).
For horizontal enclosures that contain air, Jakob (1949, Ref. 22) recom-
mends the following simple correlations
Nu = 0.195Ra| /4
Nu = 0.068Ra| /3
10 4 < Rai < 4 X 10 5
4 X 10 5 < Ra L < 10 7
(9-44)
(9-45)
These relations can also be used for other gases with 0.5 < Pr < 2. Using wa-
ter, silicone oil, and mercury in their experiments, Globe and Dropkin (1959)
obtained this correlation for horizontal enclosures heated from below,
Nu
0.069Ra/° Pr 0074
3 X 10 5 < Ra, < 7 X 10 9
(9-46)
Based on experiments with air, Hollands et al (1976, Ref. 19) recommend this
correlation for horizontal enclosures,
Nu = 1 + 1.44
1
1708
Ra,
18
1
Ra L < 10 8
(9-47)
The notation [ ] + indicates that if the quantity in the bracket is negative, it should
be set equal to zero. This relation also correlates data well for liquids with mod-
erate Prandtl numbers for Ra L < 10 5 , and thus it can also be used for water.
479
CHAPTER 9
FIGURE 9-24
A horizontal rectangular enclosure
with isothermal surfaces.
Inclined Rectangular Enclosures
Air spaces between two inclined parallel plates are commonly encountered in
flat-plate solar collectors (between the glass cover and the absorber plate) and
the double-pane skylights on inclined roofs. Heat transfer through an inclined
enclosure depends on the aspect ratio HIL as well as the tilt angle from the
horizontal (Fig. 9-25).
For large aspect ratios {HIL > 12), this equation [Hollands et al., 1976, Ref.
19] correlates experimental data extremely well for tilt angles up to 70°,
Nu = 1 + 1.44
1
1708
Ra L cos
1708(sin 1.86) 1
Ra L cos 6
(Ra L cose)'
If
1
(9-48)
for Ra L < 10 5 , < < 70°, and HIL > 12. Again any quantity in [ ] + should
be set equal to zero if it is negative. This is to ensure that Nu = 1 for Ra L cos
< 1708. Note that this relation reduces to Eq. 9-47 for horizontal enclosures
for = 0°, as expected.
FIGURE 9-25
An inclined rectangular enclosure
with isothermal surfaces.
cen58933_ch09.qxd 9/4/2002 12:25 PM Page 4E
480
HEAT TRANSFER
TABLE 9-2
Critical angles for inclined
rectangular enclosures
Aspect ratio,
Critical angle,
H/L
9cr
1
25°
3
53°
6
60°
12
67°
> 12
70°
yk
J
T,>T 7
FIGURE 9-26
A vertical rectangular enclosure with
isothermal surfaces.
Outer cylinder
atT„
Inner cylinder
atT,
FIGURE 9-27
Two concentric horizontal isothermal
cylinders.
For enclosures with smaller aspect ratios (H/L < 12), the next correlation
can be used provided that the tilt angle is less than the critical value cr listed
in Table 9-2 [Catton (1978), Ref. 7]
Nu = Nu„
/Nu 6=s
'\Nu 6=(r
(sine cr )
e/(4e„>
o°< e<
(9-49)
For tilt angles greater than the critical value (0 cr < < 90°), the Nusselt
number can be obtained by multiplying the Nusselt number for a vertical en-
closure by (sin 0) 1/4 [Ayyaswamy and Catton (1973), Ref. 3],
Nu = Nu e = 90 .(sine)" 4
3 cr < G < 90°, any H/L
(9-50)
For enclosures tilted more than 90°, the recommended relation is [Arnold et
al., (1974), Ref. 2]
Nu = 1 + (Nu e = ,
l)sin6 90° <e< 180°, any ///L
(9-51)
More recent but more complex correlations are also available in the literature
[e.g., and ElSherbiny et al. (1982), Ref. 17].
Vertical Rectangular Enclosures
For vertical enclosures (Fig. 9-26), Catton (1978, Ref. 7) recommends these
two correlations due to Berkovsky and Polevikov (1977, Ref. 6),
Nu = 0.18i
Nu = 0.22
Pr „ Y 29
0.2 + Pr Kai
Pr
0.2 + Pr
Ra,
H
KH/L<2
any Prandtl number (9-52)
Ra L Pr/(0.2 + Pr) > 10 3
2<H/L< 10
1/4
any Prandtl number (9-53)
For vertical enclosures with larger aspect ratios, the following correlations can
be used [MacGregor and Emery (1969), Ref. 26]
Nu
Nu
//V°
10 < H/L < 40
1 < Pr < 2 X 10 4 (9-54)
10 4 <Ra L < 10 7
1 <H/L< 40
1 < Pr < 20 (9-55)
10 6 < Ra L < 10"
Again all fluid properties are to be evaluated at the average temperature
(r, + r 2 )/2.
Concentric Cylinders
Consider two long concentric horizontal cylinders maintained at uniform but
different temperatures of T t and T , as shown in Figure 9-27. The diameters of
the inner and outer cylinders are D t and D , respectively, and the characteris-
tic length is the spacing between the cylinders, L c = (D — D t )/2. The rate of
heat transfer through the annular space between the natural convection unit is
expressed as
cen58933_ch09.qxd 9/4/2002 12:25 PM Page 481
Q
2-nkr
ln(D /D,
■(T,-T )
(W/m)
(9-56)
The recommended relation for effective thermal conductivity is [Raithby and
Hollands (1975), Ref. 28]
0.3861
Pr
0.861 + Pr
OfVRaJ 1
where the geometric factor for concentric cylinders F cyl is
[ln(D /A)] 4
cyl
4 3 0Dr 3/5 + D-^f
(9-57)
(9-58)
The k eff relation in Eq. 9-57 is applicable for 0.70 < Pr < 6000 and 10 2 <
F cy iRa L < 10 7 . For F cyi Ra L < 100, natural convection currents are negligible
and thus k eS = k. Note that k cB cannot be less than k, and thus we should set
k eB = k if k eB /k < 1 . The fluid properties are evaluated at the average temper-
ature of (T { + T )/2.
Concentric Spheres
For concentric isothermal spheres, the rate of heat transfer through the gap
between the spheres by natural convection is expressed as (Fig. 9-28)
2 = fceffirl
D,D,
(T t ~ T .
(W)
(9-59)
where L c = (D — D t )/2 is the characteristic length. The recommended rela-
tion for effective thermal conductivity is [Raithby and Hollands (1975), Ref. 28]
°- 74 l 0.861 V Pr
(^phRaJ 1
where the geometric factor for concentric spheres F sph is
sph
(D,D D ) 4 (Dr 7/5 + D-^f
(9-60)
(9-61)
The k eS relation in Eq. 9-60 is applicable for 0.70 < Pr < 4200 and 10 2 <
f sph Ra L < 10 4 . If k eS /k < 1, we should set k eS = k.
Combined Natural Convection and Radiation
Gases are nearly transparent to radiation, and thus heat transfer through a gas
layer is by simultaneous convection (or conduction, if the gas is quiescent)
and radiation. Natural convection heat transfer coefficients are typically very
low compared to those for forced convection. Therefore, radiation is usually
disregarded in forced convection problems, but it must be considered in nat-
ural convection problems that involve a gas. This is especially the case for
surfaces with high emissivities. For example, about half of the heat transfer
through the air space of a double pane window is by radiation. The total
rate of heat transfer is determined by adding the convection and radiation
components,
481
CHAPTER 9
A. T;
FIGURE 9-28
Two concentric isothermal spheres.
fit
2c
e,
(9-62)
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482
HEAT TRANSFER
Radiation heat transfer from a surface at temperature T s surrounded by surfaces
at a temperature T smr (both in absolute temperature unit K) is determined from
G rad = svA s {T* - T s l n ) (W)
(9-63)
where e is the emissivity of the surface, A s is the surface area, and cr = 5.67 X
10~ 8 W/m 2 • K 4 is the Stefan-Boltzmann constant.
When the end effects are negligible, radiation heat transfer between two
large parallel plates at absolute temperatures T [ and T 2 is expressed as (see
Chapter 12 for details)
Qr
vA s {T\ - T\)
1/e, + 1/e,
1
; <M,(rf
71)
(W)
(9-64)
where e, and e 2 are the emissivities of the plates, and s effectiv
emissivity defined as
1
1/e, + 1/e, - 1
is the effective
(9-65)
The emissivity of an ordinary glass surface, for example, is 0.84. Therefore,
the effective emissivity of two parallel glass surfaces facing each other is 0.72.
Radiation heat transfer between concentric cylinders and spheres is discussed
in Chapter 12.
Note that in some cases the temperature of the surrounding medium may be
below the surface temperature (T«, < T s ), while the temperature of the sur-
rounding surfaces is above the surface temperature (T mn > T s ). In such cases,
convection and radiation heat transfers are subtracted from each other instead
of being added since they are in opposite directions. Also, for a metal surface,
the radiation effect can be reduced to negligible levels by polishing the surface
and thus lowering the surface emissivity to a value near zero.
Glass -
fl = 0.8m
Air
-L = 2 cm
- Glass
FIGURE 9-29
Schematic for Example 9-A.
EXAMPLE 9-4 Heat Loss through a Double-Pane Window
The vertical 0.8-m-high, 2-m-wide double-pane window shown in Fig. 9-29
consists of two sheets of glass separated by a 2-cm air gap at atmospheric pres-
sure. If the glass surface temperatures across the air gap are measured to be
12°C and 2°C, determine the rate of heat transfer through the window.
SOLUTION Two glasses of a double-pane window are maintained at specified
temperatures. The rate of heat transfer through the window is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas. 3 Radi-
ation heat transfer is not considered.
Properties The properties of air at the average temperature of 7" ave = {7i +
7" 2 )/2 = (12 + 2)/2 = 7°C and 1 atm pressure are (Table A- 15)
k = 0.02416 W/m- °C
v = 1.399 X 10- 5 m 2 /s
Pr = 0.7344
P
1
Analysis We have a rectangular enclosure filled with air. The characteristic
length in this case is the distance between the two glasses, L = 0.02 m. Then
the Rayleigh number becomes
cen58933_ch09.qxd 9/4/2002 12:25 PM Page 483
Ra L
*P(r, - T 2 )L?
(9.81 m/s 2 )[l/(280 K)](12 - 2 K)(0.02 m) 3
(1.399 X 10- 5 m 2 /s) 2
(0.7344) = 1.051 X 10 4
The aspect ratio of the geometry is HIL = 0.8/0.02 = 40. Then the Nusselt
number in this case can be determined from Eq. 9-54 to be
1/4 p r 0.012 ££ '
Nu = 0.42Ra[ /4 Pr
0.42(1.051 X 10 4 )" 4 (0.7344)'
0.02
1.401
Then,
and
A s = H X W = (0.8 m)(2 m) = 1.6 m 2
7*1 ~T 2
Q = hA s (T t -T 2 ) = kNuA s -
„ (12 - 2)°C
= (0.02416 W/m- °C)(1.401)(1.6 m 2 ) = 27.1 W
Therefore, heat will be lost through the window at a rate of 27.1 W.
Discussion Recall that a Nusselt number of Nu = 1 for an enclosure corre-
sponds to pure conduction heat transfer through the enclosure. The air in the
enclosure in this case remains still, and no natural convection currents occur in
the enclosure. The Nusselt number in our case is 1.32, which indicates that
heat transfer through the enclosure is 1.32 times that by pure conduction. The
increase in heat transfer is due to the natural convection currents that develop
in the enclosure.
EXAMPLE 9-5 Heat Transfer through a Spherical Enclosure
The two concentric spheres of diameters D, = 20 cm and D = 30 cm shown in
Fig. 9-30 are separated by air at 1 atm pressure. The surface temperatures of
the two spheres enclosing the air are 7", = 320 K and T = 280 K, respectively.
Determine the rate of heat transfer from the inner sphere to the outer sphere by
natural convection.
SOLUTION Two surfaces of a spherical enclosure are maintained at specified
temperatures. The rate of heat transfer through the enclosure is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas. 3 Radi-
ation heat transfer is not considered.
Properties The properties of air at the average temperature of 7" ave = ( 7", + T )I2
= (320 + 280)/2 = 300 K = 27°C and 1 atm pressure are (Table A- 15)
k = 0.02566 W/m • °C
v = 1.580 X 10- 5 m 2 /s
Pr = 0.7290
1 1
P
483
CHAPTER 9
280 K
5 cm
FIGURE 9-30
Schematic for Example 9-5.
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484
HEAT TRANSFER
Glass cover
Aluminum tube
Water
FIGURE 9-31
Schematic for Example 9-6.
Analysis We have a spherical enclosure filled with air. The characteristic
length in this case is the distance between the two spheres,
L c = (D - D ; )/2 = (0.3 - 0.2)/2 = 0.05 m
The Rayleigh number is
gpcr,. - r g )L 3
Ra L = ; Pr
_ (9.81 m/s 2 )[l/(300 K)](320 - 280 K)(0.05 m) 3
(1.58 X 10- 5 m 2 /s) 2
The effective thermal conductivity is
(0.729) = 4.776 X 10 5
sph
(AA)W
0.05 m
^ = °- 74 Ho.861
[(0.2 m)(0.3 m)] 4 [(0.2 trr 7 ' 5 + (0.3 m)- 7 ' 5 ] 5
Pr
0.005229
Pj. / v* sph
0.74(0.02566 W/m ■ °C)
0F sph Ra L )" 4
0.729
(0.005229 X 4.776 X 10 5 ) 1
0.861 + 0.729
= 0.1 104 W/m- °C
Then the rate of heat transfer between the spheres becomes
/AAA
Q = k M-Tr) {Ti ~ To)
/(0.2 m)(0.3 m)\
= (0.1104 W/m • °C)tt I- — t-^z -1(320 - 280)K = 16.7 W
Therefore, heat will be lost from the inner sphere to the outer one at a rate of
16.7 W.
Discussion Note that the air in the spherical enclosure will act like a station-
ary fluid whose thermal conductivity is k eff /k = 0.1104/0.02566 = 4.3 times
that of air as a result of natural convection currents. Also, radiation heat trans-
fer between spheres is usually very significant, and should be considered in a
complete analysis.
EXAMPLE 9-6 Heating Water in a Tube by Solar Energy
A solar collector consists of a horizontal aluminum tube having an outer di-
ameter of 2 in. enclosed in a concentric thin glass tube of 4-in. -diameter (Fig.
9-31). Water is heated as it flows through the tube, and the annular space be-
tween the aluminum and the glass tubes is filled with air at 1 atm pressure.
The pump circulating the water fails during a clear day, and the water tem-
perature in the tube starts rising. The aluminum tube absorbs solar radiation
at a rate of 30 Btu/h per foot length, and the temperature of the ambient air
outside is 70°F. Disregarding any heat loss by radiation, determine the tem-
perature of the aluminum tube when steady operation is established (i.e.,
when the rate of heat loss from the tube equals the amount of solar energy
gained by the tube).
cen58933_ch09.qxd 9/4/2002 12:25 PM Page 485
SOLUTION The circulating pump of a solar collector that consists of a hori-
zontal tube and its glass cover fails. The equilibrium temperature of the tube is
to be determined.
Assumptions 1 Steady operating conditions exist. 2 The tube and its cover are
isothermal. 3 Air is an ideal gas. 4 Heat loss by radiation is negligible.
Properties The properties of air should be evaluated at the average tempera-
ture. But we do not know the exit temperature of the air in the duct, and thus
we cannot determine the bulk fluid and glass cover temperatures at this point,
and thus we cannot evaluate the average temperatures. Therefore, we will as-
sume the glass temperature to be HOT, and use properties at an anticipated
average temperature of (70 + 110)/2 = 90T (Table A-15E),
k = 0.01505 Btu/h • ft • °F
v = 0.6310 ftVh = 1.753 X 10~ 4 ft 2 /s
Pr = 0.7275
1
Analysis We have a horizontal cylindrical enclosure filled with air at 1 atm
pressure. The problem involves heat transfer from the aluminum tube to the
glass cover and from the outer surface of the glass cover to the surrounding am-
bient air. When steady operation is reached, these two heat transfer rates must
equal the rate of heat gain. That is,
Stube-glass = Q glass-ambient = fisolargain = 30 Btu/h (per foot Of tllbe)
The heat transfer surface area of the glass cover is
A n A
(ttD L) = tt(4/1 2 ft)( 1 ft) = 1 .047 ft 2 (per foot of tube)
To determine the Rayleigh number, we need to know the surface temperature of
the glass, which is not available. Therefore, it is clear that the solution will re-
quire a trial-and-error approach. Assuming the glass cover temperature to be
100°F, the Rayleigh number, the Nusselt number, the convection heat transfer
coefficient, and the rate of natural convection heat transfer from the glass cover
to the ambient air are determined to be
Rar
g$(T s - rjDj
Pr
(32.2 ft/s 2 )[l/(550 R)](l 10 - 70 R)(4/12 ft) 3
(1.753 X 10- 4 ft 2 /s) 2
(0.7275) = 2.054 X 10 6
Nu
0.6 +
0.387 RaK 6
[1 + (0.559/Pr) 9 " 6 ] 8
0.6 +
0.387(2.054 X 10 6 )" 6
[1 + (0.559/0.7275) 9 ' 16 ] 8
17.89
K
D,
Nu
0.0150 Btu/h- ft -°F
4/12 ft
(17.89) = 0.8075 Btu/h • ft 2 ■ °F
Q = h A (T - T m ) = (0.8075 Btu/h ■ ft 2 ■ °F)( 1.047 ft 2 )(110 - 70)°F
= 33.8 Btu/h
which is more than 30 Btu/h. Therefore, the assumed temperature of 110°F for
the glass cover is high. Repeating the calculations with lower temperatures, the
glass cover temperature corresponding to 30 Btu/h is determined to be 106°F.
485
CHAPTER 9
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486
HEAT TRANSFER
The temperature of the aluminum tube is determined in a similar manner us-
ing the natural convection relations for two horizontal concentric cylinders. The
characteristic length in this case is the distance between the two cylinders,
which is
L c = (D - D;)/2 = (4 - 2)/2 = 1 in.
1/12 ft
We start the calculations by assuming the tube temperature to be 200°F, and
thus an average temperature of (106 + 200)/2 = 154°F = 614 R. This gives
Ra L = Pr
v-
_ (32.2 ft/s 2 )[l/614 R)](200 - 106 R)(l/12 ft) 3
(2.117 X 10- 4 ft 2 /s) 2
The effective thermal conductivity is
[ln(D /Z?,)] 4
cyl L 3 (£>r 3/5 + D D - 3 ' 5 ) 5
[ln(4/2)] 4
(0.7184) = 4.579 X 10 4
(1/12 ft) 3 [(2/12 ft)- 3 ' 5 + (4/12 ft)- 3 ' 5 ] 5
0.1466
6 Ha86lVpr
(F cyl Ra L ) 1
0.386(0.01653 Btu/h -ft- °F)
0.04743 Btu/h ■ ft ■ T
0.7184
0.861 + 0.7184
(0.1466 X 4.579 X 10 4 ) 1
Then the rate of heat transfer between the cylinders becomes
Q
2ttL,
(T,
In(ZVA)
2tt(0.04743 Btu/h • ft • °F)
ln(4/2)
(200- 106)°F = 40.4 Btu/h
which is more than 30 Btu/h. Therefore, the assumed temperature of 200T for
the tube is high. By trying other values, the tube temperature corresponding to
30 Btu/h is determined to be 180°F. Therefore, the tube will reach an equilib-
rium temperature of 180T when the pump fails.
Discussion Note that we have not considered heat loss by radiation in the cal-
culations, and thus the tube temperature determined above is probably too
high. This problem is considered again in Chapter 12 by accounting for the ef-
fect of radiation heat transfer.
9-6 - COMBINED NATURAL AND FORCED
CONVECTION
The presence of a temperature gradient in a fluid in a gravity field always
gives rise to natural convection currents, and thus heat transfer by natural
convection. Therefore, forced convection is always accompanied by natural
convection.
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487
CHAPTER 9
We mentioned earlier that the convection heat transfer coefficient, natural or
forced, is a strong function of the fluid velocity. Heat transfer coefficients
encountered in forced convection are typically much higher than those en-
countered in natural convection because of the higher fluid velocities associ-
ated with forced convection. As a result, we tend to ignore natural convection
in heat transfer analyses that involve forced convection, although we recog-
nize that natural convection always accompanies forced convection. The error
involved in ignoring natural convection is negligible at high velocities but
may be considerable at low velocities associated with forced convection.
Therefore, it is desirable to have a criterion to assess the relative magnitude of
natural convection in the presence of forced convection.
For a given fluid, it is observed that the parameter Gr/Re 2 represents the im-
portance of natural convection relative to forced convection. This is not
surprising since the convection heat transfer coefficient is a strong function of
the Reynolds number Re in forced convection and the Grashof number Gr in
natural convection.
A plot of the nondimensionalized heat transfer coefficient for combined nat-
ural and forced convection on a vertical plate is given in Fig. 9-32 for differ-
ent fluids. We note from this figure that natural convection is negligible when
Gr/Re 2 < 0.1, forced convection is negligible when Gr/Re 2 > 10, and neither
is negligible when 0.1 < Gr/Re 2 < 10. Therefore, both natural and forced
convection must be considered in heat transfer calculations when the Gr and
Re 2 are of the same order of magnitude (one is within a factor of 10 times the
other). Note that forced convection is small relative to natural convection only
in the rare case of extremely low forced flow velocities.
Natural convection may help or hurt forced convection heat transfer, de-
pending on the relative directions of buoyancy-induced and the forced con-
vection motions (Fig. 9-33):
10
1.0
0.1
0.01
© Experiment
_ Approximate solution
Pure forced convection
_ — ■•— Pure free convection —
-••
100
— „.-
VB-rrS?~"
—
10
- Pr = 0.72 (air)
nm
—:■
■-•■
-
0.01 —
Pr = 0.003
0.01
0.1 1.0
Gr V /Re v 2
10
FIGURE 9-32
Variation of the local Nusselt number
NU X for combined natural and forced
convection from a hot isothermal
vertical plate (from Lloyd and
Sparrow, Ref. 25).
Hot plate
Buoyant
flow
(a) Assisting flow
Cold plate
Buoyant
flow
Buoyant
flow
tttt
Forced
flow
(b) Opposing flow
(c) Transverse flow
FIGURE 9-33
Natural convection can enhance or inhibit heat transfer, depending on the relative
directions of buoyancy-induced motion and the forced convection motion.
cen58933_ch09.qxd 9/4/2002 12:25 PM Page 4E
488
HEAT TRANSFER
1. In assisting flow, the buoyant motion is in the same direction as the
forced motion. Therefore, natural convection assists forced convection
and enhances heat transfer. An example is upward forced flow over a
hot surface.
2. In opposing flow, the buoyant motion is in the opposite direction to the
forced motion. Therefore, natural convection resists forced convection
and decreases heat transfer. An example is upward forced flow over a
cold surface.
3. In transverse flow, the buoyant motion is perpendicular to the forced
motion. Transverse flow enhances fluid mixing and thus enhances heat
transfer. An example is horizontal forced flow over a hot or cold
cylinder or sphere.
When determining heat transfer under combined natural and forced con-
vection conditions, it is tempting to add the contributions of natural and forced
convection in assisting flows and to subtract them in opposing flows. How-
ever, the evidence indicates differently. A review of experimental data sug-
gests a correlation of the form
Nu combined = (NuLced ± NiC ural )"" (9-41)
where Nu forced and Nu natural are determined from the correlations for pure
forced and pure natural convection, respectively. The plus sign is for assisting
and transverse flows and the minus sign is for opposing flows. The value of
the exponent n varies between 3 and 4, depending on the geometry involved.
It is observed that n = 3 correlates experimental data for vertical surfaces
well. Larger values of n are better suited for horizontal surfaces.
A question that frequently arises in the cooling of heat-generating equip-
ment such as electronic components is whether to use a fan (or a pump if the
cooling medium is a liquid) — that is, whether to utilize natural ox forced con-
vection in the cooling of the equipment. The answer depends on the maximum
allowable operating temperature. Recall that the convection heat transfer rate
from a surface at temperature T s in a medium at T„ is given by
where h is the convection heat transfer coefficient and A s is the surface area.
Note that for a fixed value of power dissipation and surface area, h and T s are
inversely proportional. Therefore, the device will operate at a higher temper-
ature when h is low (typical of natural convection) and at a lower temperature
when h is high (typical of forced convection).
Natural convection is the preferred mode of heat transfer since no blowers
or pumps are needed and thus all the problems associated with these, such as
noise, vibration, power consumption, and malfunctioning, are avoided. Nat-
ural convection is adequate for cooling low-power-output devices, especially
when they are attached to extended surfaces such as heat sinks. For high-
power-output devices, however, we have no choice but to use a blower or a
pump to keep the operating temperature below the maximum allowable level.
For very-high-power-output devices, even forced convection may not be suf-
ficient to keep the surface temperature at the desirable levels. In such cases,
we may have to use boiling and condensation to take advantage of the very
high heat transfer coefficients associated with phase change processes.
cen58933_ch09.qxd 9/4/2002 12:25 PM Page 489
TOPIC OF SCPECIAL INTEREST'
Heat Transfer Through Windows
489
CHAPTER 9
Windows are glazed apertures in the building envelope that typically con-
sist of single or multiple glazing (glass or plastic), framing, and shading. In
a building envelope, windows offer the least resistance to heat flow. In a
typical house, about one-third of the total heat loss in winter occurs through
the windows. Also, most air infiltration occurs at the edges of the windows.
The solar heat gain through the windows is responsible for much of the
cooling load in summer. The net effect of a window on the heat balance of
a building depends on the characteristics and orientation of the window as
well as the solar and weather data. Workmanship is very important in the
construction and installation of windows to provide effective sealing
around the edges while allowing them to be opened and closed easily.
Despite being so undesirable from an energy conservation point of view,
windows are an essential part of any building envelope since they enhance
the appearance of the building, allow daylight and solar heat to come in,
and allow people to view and observe outside without leaving their home.
For low-rise buildings, windows also provide easy exit areas during emer-
gencies such as fire. Important considerations in the selection of windows
are thermal comfort and energy conservation. A window should have a
good light transmittance while providing effective resistance to heat flow.
The lighting requirements of a building can be minimized by maximizing
the use of natural daylight. Heat loss in winter through the windows can be
minimized by using airtight double- or triple-pane windows with spectrally
selective films or coatings, and letting in as much solar radiation as possi-
ble. Heat gain and thus cooling load in summer can be minimized by using
effective internal or external shading on the windows.
Even in the absence of solar radiation and air infiltration, heat transfer
through the windows is more complicated than it appears to be. This is be-
cause the structure and properties of the frame are quite different than the
glazing. As a result, heat transfer through the frame and the edge section of
the glazing adjacent to the frame is two-dimensional. Therefore, it is cus-
tomary to consider the window in three regions when analyzing heat transfer
through it: (1) the center-of- glass, (2) the edge-of- glass, and (3) the. frame re-
gions, as shown in Figure 9-34. Then the total rate of heat transfer through
the window is determined by adding the heat transfer through each region as
x£ window \*L center x£ e
Qfr*
where
U m
\^ center ^c
II A
L -' edge -'-'-edge
77 A \/A
w frame z - l frame/'-'- 1 windov>
(9-67)
(9-68)
is the I/-factor or the overall heat transfer coefficient of the window;
^window i s the window area; A center , A edge , and A frame are the areas of the
Frame
Edge of glass
Center of glass
*This section can be skipped without a loss of continuity.
FIGURE 9-34
The three regions of a window
considered in heat transfer analysis.
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490
HEAT TRANSFER
Glass
J\ inside
*■ — wv^-
k
D
glass
L_
k
R.
outside T
w — *
K
FIGURE 9-35
The thermal resistance network for
heat transfer through a single glass.
center, edge, and frame sections of the window, respectively; and t/ center ,
[/ edge , and C/ frame are the heat transfer coefficients for the center, edge, and
frame sections of the window. Note that A window = A center + A edge + A frame ,
and the overall [/-factor of the window is determined from the area-
weighed [/-factors of each region of the window. Also, the inverse of the
[/-factor is the 7?-value, which is the unit thermal resistance of the window
(thermal resistance for a unit area).
Consider steady one-dimensional heat transfer through a single-pane glass
of thickness L and thermal conductivity k. The thermal resistance network
of this problem consists of surface resistances on the inner and outer sur-
faces and the conduction resistance of the glass in series, as shown in Figure
9-35, and the total resistance on a unit area basis can be expressed as
R«
R„
_i_ p _i_ p
//,
If h
^glass ll o
(9-69)
Using common values of 3 mm for the thickness and 0.92 W/m • °C for the
thermal conductivity of the glass and the winter design values of 8.29 and
34.0 W/m 2 • °C for the inner and outer surface heat transfer coefficients, the
thermal resistance of the glass is determined to be
R
1
0.003 m
1
8.29 W/m 2 • °C
= 0.121 + 0.003 4
0.92 W/m • °C 34.0 W/m 2
0.029 = 0.153 m 2 -°C/W
Note that the ratio of the glass resistance to the total resistance is
^giass 0.003 m 2 • °C/W
R ma , 0.153 m 2 • °C/W
2.0%
J 1 . inside
♦! — w^
ll;
Ail-
space
space
1
space
Glass
l outside T
w — »°
_}_
h
FIGURE 9-36
The thermal resistance network
for heat transfer through the center
section of a double-pane window
(the resistances of the glasses
are neglected).
That is, the glass layer itself contributes about 2 percent of the total ther-
mal resistance of the window, which is negligible. The situation would
not be much different if we used acrylic, whose thermal conductivity is
0. 19 W/m • °C, instead of glass. Therefore, we cannot reduce the heat trans-
fer through the window effectively by simply increasing the thickness of
the glass. But we can reduce it by trapping still air between two layers of
glass. The result is a double-pane window, which has become the norm in
window construction.
The thermal conductivity of air at room temperature is k ak = 0.025
W/m • °C, which is one-thirtieth that of glass. Therefore, the thermal resis-
tance of 1 -cm-thick still air is equivalent to the thermal resistance of a 30-
cm-thick glass layer. Disregarding the thermal resistances of glass layers,
the thermal resistance and [/-factor of a double -pane window can be ex-
pressed as (Fig. 9-36)
1
U.
double-pane {center region)
h: h„
1
(9-70)
where h.
space
^rad, space + ' 7 conv, space i s trle combined radiation and convec-
tion heat transfer coefficient of the space trapped between the two glass
layers.
Roughly half of the heat transfer through the air space of a double-pane
window is by radiation and the other half is by conduction (or convection,
cen58933_ch09.qxd 9/4/2002 12:26 PM Page 491
if there is any air motion). Therefore, there are two ways to minimize /z space
and thus the rate of heat transfer through a double-pane window:
1. Minimize radiation heat transfer through the air space. This can be
done by reducing the emissivity of glass surfaces by coating them with
low-emissivity (or "low-e" for short) material. Recall that the effective
emissivity of two parallel plates of emissivities e { and e 2 is given by
1
1/e, + 1/e,
1
(9-71)
The emissivity of an ordinary glass surface is 0.84. Therefore, the effective
emissivity of two parallel glass surfaces facing each other is 0.72. But
when the glass surfaces are coated with a film that has an emissivity of 0. 1 ,
the effective emissivity reduces to 0.05, which is one-fourteenth of 0.72.
Then for the same surface temperatures, radiation heat transfer will also go
down by a factor of 14. Even if only one of the surfaces is coated, the over-
all emissivity reduces to 0.1, which is the emissivity of the coating. Thus it
is no surprise that about one -fourth of all windows sold for residences have
a low-e coating. The heat transfer coefficient h space for the air space trapped
between the two vertical parallel glass layers is given in Table 9-3 for
13-mm- (^-in.) and 6-mm- (l-in.) thick air spaces for various effective
emissivities and temperature differences.
It can be shown that coating just one of the two parallel surfaces facing
each other by a material of emissivity e reduces the effective emissivity
nearly to e. Therefore, it is usually more economical to coat only one of the
facing surfaces. Note from Figure 9-37 that coating one of the interior sur-
faces of a double -pane window with a material having an emissivity of 0.1
491
CHAPTER 9
The heat transfer coefficient h spaze for the air space trapped between the
two vertical parallel glass layers for 13-mm- and 6-mm-thick air spaces
(from Building Materials and Structures, Report 151, U.S. Dept. of
Commerce).
(a) A
r space thick
ness =
13 mm
(MA
r space thick
ness =
6 mm
AT",
°C
hspace, W/rr
2. °C
t
T
AT",
°C
hs^ce, W/rr
2. °C J
T
^effective
^effective
°c
0.72
0.4
0.2
0.1
°C
0.72
0.4
0.2
0.1
5
5.3
3.8
2.9
2.4
5
7.2
5.7
4.8
4.3
15
5.3
3.8
2.9
2.4
50
7.2
5.7
4.8
4.3
30
5.5
4.0
3.1
2.6
10
5
7.7
6.0
5.0
4.5
10
5
5.7
4.1
3.0
2.5
10
50
7.7
6.1
5.0
4.5
10
10
15
30
5.7
6.0
4.1
4.3
3.1
3.3
2.5
2.7
30
30
5
50
8.8
8.8
6.8
6.8
5.5
5.5
4.9
4.9
30
30
30
5
15
30
5.7
5.7
6.0
4.6
4.7
4.9
3.4
3.4
3.6
2.7
2.8
3.0
50
50
5
50
10.0
10.0
7.5
7.5
6.0
6.0
5.2
5.2
'Multiply by 0.176 to convert to Btu/h • ft 2 ■ °F.
cen58933_ch09.qxd 9/4/2002 12:26 PM Page 492
4.5
4
3.5
00 ^
-2 B
i s
5 £
3 -
2.5
1.5
1
0.5
492
HEAT TRANSFER
-
Gas fill in gap
-
Air
□ Argon
7 vV
A Krypton
aO^Nx r z = °- 84
Outer ^^— 6— A — A — A-r-^" - ""
glass Inner
0. 10 on surface 2 or 3
:$
<S
glass
-
9)
(4~) Double-pane
JT glazing
-
, I
10 15
Gap width, mm
20
25
4.5
4
3.5
-2 B
?!
? -
b o
c I
3 -
2.5
<?
$
©
Triple-pane
glazing
Outer
glass
Inner
glass
Gas fill in gap
o Air
□ Argon
A Krypton
1.5
1 -
0.5
5
(b) Triple-pane window
- e = 0.10 on surfaces
2 or 3 and 4 or 5
10 15
Gap width, mm
20
25
(a) Double-pane window
FIGURE 9-37
The variation of the [/-factor for the center section of double- and triple-pane windows with uniform spacing between the
panes (from ASHRAE Handbook of Fundamentals, Ref. 1, Chap. 27, Fig. 1).
reduces the rate of heat transfer through the center section of the window
by half.
2. Minimize conduction heat transfer through air space. This can be done
by increasing the distance d between the two glasses. However, this cannot
be done indefinitely since increasing the spacing beyond a critical value
initiates convection currents in the enclosed air space, which increases the
heat transfer coefficient and thus defeats the purpose. Besides, increasing
the spacing also increases the thickness of the necessary framing and the
cost of the window. Experimental studies have shown that when the spac-
ing d is less than about 13 mm, there is no convection, and heat transfer
through the air is by conduction. But as the spacing is increased further,
convection currents appear in the air space, and the increase in heat trans-
fer coefficient offsets any benefit obtained by the thicker air layer. As a re-
sult, the heat transfer coefficient remains nearly constant, as shown in
Figure 9-37. Therefore, it makes no sense to use an air space thicker than
13 mm in a double-pane window unless a thin polyester film is used to di-
vide the air space into two halves to suppress convection currents. The film
provides added insulation without adding much to the weight or cost of the
double-pane window. The thermal resistance of the window can be in-
creased further by using triple- or quadruple-pane windows whenever it is
economical to do so. Note that using a triple-pane window instead of a dou-
ble-pane reduces the rate of heat transfer through the center section of the
window by about one-third.
cen58933_ch09.qxd 9/4/2002 12:26 PM Page 493
493
CHAPTER 9
Another way of reducing conduction heat transfer through a double -pane
window is to use a less-conducting fluid such as argon or krypton to fill the
gap between the glasses instead of air. The gap in this case needs to be well
sealed to prevent the gas from leaking outside. Of course, another alterna-
tive is to evacuate the gap between the glasses completely, but it is not
practical to do so.
Edge-of-Glass [/-Factor of a Window
The glasses in double- and triple-pane windows are kept apart from each
other at a uniform distance by spacers made of metals or insulators like
aluminum, fiberglass, wood, and butyl. Continuous spacer strips are placed
around the glass perimeter to provide an edge seal as well as uniform spac-
ing. However, the spacers also serve as undesirable "thermal bridges"
between the glasses, which are at different temperatures, and this short-
circuiting may increase heat transfer through the window considerably.
Heat transfer in the edge region of a window is two-dimensional, and lab
measurements indicate that the edge effects are limited to a 6.5-cm-wide
band around the perimeter of the glass.
The [/-factor for the edge region of a window is given in Figure 9-38
relative to the [/-factor for the center region of the window. The curve
would be a straight diagonal line if the two [/-values were equal to each
other. Note that this is almost the case for insulating spacers such as wood
and fiberglass. But the [/-factor for the edge region can be twice that of the
center region for conducting spacers such as those made of aluminum. Val-
ues for steel spacers fall between the two curves for metallic and insulating
spacers. The edge effect is not applicable to single -pane windows.
Frame [/-Factor
The framing of a window consists of the entire window except the glazing.
Heat transfer through the framing is difficult to determine because of the
different window configurations, different sizes, different constructions,
and different combination of materials used in the frame construction. The
type of glazing such as single pane, double pane, and triple pane affects the
thickness of the framing and thus heat transfer through the frame. Most
frames are made of wood, aluminum, vinyl, or fiberglass. However, using a
combination of these materials (such as aluminum-clad wood and vinyl-
clad aluminum) is also common to improve appearance and durability.
Aluminum is a popular framing material because it is inexpensive,
durable, and easy to manufacture, and does not rot or absorb water like
wood. However, from a heat transfer point of view, it is the least desirable
framing material because of its high thermal conductivity. It will come as
no surprise that the [/-factor of solid aluminum frames is the highest, and
thus a window with aluminum framing will lose much more heat than a
comparable window with wood or vinyl framing. Heat transfer through the
aluminum framing members can be reduced by using plastic inserts be-
tween components to serve as thermal barriers. The thickness of these in-
serts greatly affects heat transfer through the frame. For aluminum frames
without the plastic strips, the primary resistance to heat transfer is due to
the interior surface heat transfer coefficient. The [/-factors for various
1 1
Spacer type
i4
Metallic
P 4
S
Insulating
&
fj
Ideal
sy
- 3
^
!-i
ss
% i
•
• /
* /
s /
o
U i
OX) 1
T3
W
12 3 4 5
Center-of-glass U-factor, W/m 2 -K
FIGURE 9-38
The edge-of-glass [/-factor relative to
the center-of-glass [/-factor for
windows with various spacers (from
ASHRAE Handbook of Fundamentals,
Ref. 1, Chap. 27, Fig. 2).
cen58933_ch09.qxd 9/4/2002 12:26 PM Page 494
494
HEAT TRANSFER
TABLE 9-4
Representative frame [/-factors
for fixed vertical windows (from
ASHRAE Handbook of
Fundamentals, Ref. 1, Chap. 27,
Table 2)
[/-factor,
Frame material W/m 2 ■ °C*
Aluminum:
Single glazing (3 mm) 10.1
Double glazing (18 mm) 10.1
Triple glazing (33 mm) 10.1
Wood or vinyl:
Single glazing (3 mm) 2.9
Double glazing (18 mm) 2.8
Triple glazing (33 mm) 2.7
•Multiply by 0.176 to convert to Btu/h • ft 2 ■ °F
TABLE 9-5
Combined convection and
radiation heat transfer coefficient
hj at the inner surface of a vertical
glass under still air conditions
(in W/m 2 • °C)*
T
°C
Glass emissivity, s g
°C
0.05
0.20
0.84
20
17
2.6
3.5
7.1
20
15
2.9
3.8
7.3
20
10
3.4
4.2
7.7
20
5
3.7
4.5
7.9
20
4.0
4.8
8.1
20
-5
4.2
5.0
8.2
20
-10
4.4
5.1
8.3
frames are listed in Table 9-4 as a function of spacer materials and the
glazing unit thicknesses. Note that the [/-factor of metal framing and thus
the rate of heat transfer through a metal window frame is more than three
times that of a wood or vinyl window frame.
Interior and Exterior Surface Heat
Transfer Coefficients
Heat transfer through a window is also affected by the convection and ra-
diation heat transfer coefficients between the glass surfaces and surround-
ings. The effects of convection and radiation on the inner and outer
surfaces of glazings are usually combined into the combined convection
and radiation heat transfer coefficients hj and h , respectively. Under still
air conditions, the combined heat transfer coefficient at the inner surface of
a vertical window can be determined from
h,
1.77(7; - r,)°
sMT* - Tt)
(W/m 2 ■ °C)
(9-72)
where T g = glass temperature in K, T { = indoor air temperature in K, e g =
emissivity of the inner surface of the glass exposed to the room (taken to be
0.84 for uncoated glass), and a = 5.67 X 1(T 8 W/m 2 • K 4 is the Stefan-
Boltzmann constant. Here the temperature of the interior surfaces facing
the window is assumed to be equal to the indoor air temperature. This as-
sumption is reasonable when the window faces mostly interior walls, but it
becomes questionable when the window is exposed to heated or cooled
surfaces or to other windows. The commonly used value of h t for peak load
calculation is
hi = 8.29 W/m 2 ■ °C = 1.46 Btu/h • ft 2 • °F
(winter and summer)
•Multiply by 0.176 to convert to Btu/h • ft 2 ■ °F.
which corresponds to the winter design conditions of T { = 22°C and
T g = — 7°C for uncoated glass with e g = 0.84. But the same value of h t can
also be used for summer design conditions as it corresponds to summer con-
ditions of Tj = 24°C and T g = 32°C. The values of h t for various tempera-
tures and glass emissivities are given in Table 9-5. The commonly used
values of h for peak load calculations are the same as those used for outer
wall surfaces (34.0 W/m 2 • °C for winter and 22.7 W/m 2 • °C for summer).
Overall [/-Factor of Windows
The overall [/-factors for various kinds of windows and skylights are eval-
uated using computer simulations and laboratory testing for winter design
conditions; representative values are given in Table 9-6. Test data may pro-
vide more accurate information for specific products and should be pre-
ferred when available. However, the values listed in the table can be used
to obtain satisfactory results under various conditions in the absence of
product-specific data. The [/-factor of a fenestration product that differs
considerably from the ones in the table can be determined by (1) determin-
ing the fractions of the area that are frame, center-of-glass, and edge-of-
glass (assuming a 65-mm-wide band around the perimeter of each glazing),
cen58933_ch09.qxd 9/4/2002 12:26 PM Page 495
495
CHAPTER 9
TABLE 9-6
Overall f-factors (heat transfer coefficients) for various windows and skylights in W/m 2
(from ASHRAE Handbook of Fundamentals, Ref. 1, Chap. 27, Table 5)
Glass section
Aluminum frame
(without thermal
Type — >
Frame width — >
Spacer type ->
(glazing) only
break)
Wood or vinyl frame
Center- Edge-of-
of -glass glass
Fixed
Double
door
Sloped
skylight
Fixed
Double
door
Sloped
skylight
(Not applicable)
32 mm
(liin.)
53 mm
(2 in.)
19 mm
(| in.)
41 mm
(If in.)
88 mm
(3^ in.)
23 mm
(|in.)
Metal Insul.
All
All
All
Metal Insul. Metal Insul. Metal Insul.
6.30
—
6.63
7.16
9.88
5.93
—
5.57
—
7.57
—
5.28
—
5.69
6.27
8.86
5.02
—
4.77
—
6.57
—
5.79
—
6.16
6.71
9.94
5.48
—
5.17
—
7.63
—
3.71
3.34
3.90
4.55
6.70
3.26
3.16
3.20
3.09
4.37
4.22
3.40
2.91
3.51
4.18
6.65
2.88
2.76
2.86
2.74
4.32
4.17
3.52
3.07
3.66
4.32
6.47
3.03
2.91
2.98
2.87
4.14
3.97
3.28
2.76
3.36
4.04
6.47
2.74
2.61
2.73
2.60
4.14
3.97
Glazing Type
Single Glazing
3 mm (| in.) glass 6.30
6.4 mm (i in.) acrylic 5.28
3 mm (| in.) acrylic 5.79
Double Glazing (no coating)
6.4 mm air space 3.24
12.7 mm air space 2.78
6.4 mm argon space 2.95
12.7 mm argon space 2.61
Double Glazing [e = 0.1, coating on one of the surfaces of air space (surface 2 or 3, counting from the outside toward
inside)]
6.4 mm air space 2.44
12.7 mm air space 1.82
6.4 mm argon space 1.99
12.7 mm argon space 1.53
Triple Glazing (no coating)
6.4 mm air space 2.16
12.7 mm air space 1.76
6.4 mm argon space 1.93
12.7 mm argon space 1.65
Triple Glazing [e = 0.1, coating on one of the surfaces of air spaces (surfaces 3 and 5, counting from the outside toward
inside)]
6.4 mm air space
12.7 mm air space
6.4 mm argon space
12.7 mm argon space
3.16
2.60
3.21
3.89
6.04
2.59
2.46
2.60
2.47
3.73
3.53
2.71
2.06
2.67
3.37
6.04
2.06
1.92
2.13
1.99
3.73
3.53
2.83
2.21
2.82
3.52
5.62
2.21
2.07
2.26
2.12
3.32
3.09
2.49
1.83
2.42
3.14
5.71
1.82
1.67
1.91
1.78
3.41
3.19
2.96
2.35
2.97
3.66
5.81
2.34
2.18
2.36
2.21
3.48
3.24
2.67
2.02
2.62
3.33
5.67
2.01
1.84
2.07
1.91
3.34
3.09
2.79
2.16
2.77
3.47
5.57
2.15
1.99
2.19
2.04
3.25
3.00
2.58
1.92
2.52
3.23
5.53
1.91
1.74
1.98
1.82
3.20
2.95
1.53
2.49
1.83
2.42
3.14
5.24
1.81
1.64
1.89
1.73
2.92
2.66
0.97
2.05
1.38
1.92
2.66
5.10
1.33
1.15
1.46
1.30
2.78
2.52
1.19
2.23
1.56
2.12
2.85
4.90
1.52
1.35
1.64
1.47
2.59
2.33
0.80
1.92
1.25
1.77
2.51
4.86
1.18
1.01
1.33
1.17
2.55
2.28
Notes:
(1) Multiply by 0.176 to obtain U-factors in Btu/h • ft 2 • °F.
(2) The ^/-factors in this table include the effects of surface heat transfer coefficients and are based on winter conditions of -18°C outdoor
air and 21°C indoor air temperature, with 24 km/h (15 mph) winds outdoors and zero solar flux. Small changes in indoor and outdoor tem-
peratures will not affect the overall U-factors much. Windows are assumed to be vertical, and the skylights are tilted 20° from the horizontal
with upward heat flow. Insulation spacers are wood, fiberglass, or butyl. Edge-of-glass effects are assumed to extend the 65-mm band around
perimeter of each glazing. The product sizes are 1.2 m x 1.8 m for fixed windows, 1.8 m x 2.0 m for double-door windows, and 1.2 m x 0.6
m for the skylights, but the values given can also be used for products of similar sizes. All data are based on 3-mm (|-in.) glass unless noted
otherwise.
cen58933_ch09.qxd 9/4/2002 12:26 PM Page 496
496
HEAT TRANSFER
(2) determining the [/-factors for each section (the center-of-glass and
edge-of-glass [/-factors can be taken from the first two columns of Table
9-6 and the frame [/-factor can be taken from Table 9-5 or other sources),
and (3) multiplying the area fractions and the [/-factors for each section
and adding them up (or from Eq. 9-68 for [/ window ).
Glazed wall systems can be treated as fixed windows. Also, the data for
double-door windows can be used for single-glass doors. Several observa-
tions can be made from the data in the table:
1. Skylight [/-factors are considerably greater than those of vertical
windows. This is because the skylight area, including the curb, can be
13 to 240 percent greater than the rough opening area. The slope of
the skylight also has some effect.
2. The [/-factor of multiple-glazed units can be reduced considerably
by filling cavities with argon gas instead of dry air. The
performance of C0 2 -filled units is similar to those filled with
argon. The [/-factor can be reduced even further by filling the
glazing cavities with krypton gas.
3. Coating the glazing surfaces with low-e (low-emissivity) films
reduces the [/-factor significantly. For multiple-glazed units, it is
adequate to coat one of the two surfaces facing each other.
4. The thicker the air space in multiple-glazed units, the lower the
[/-factor, for a thickness of up to 13 mm (| in.) of air space. For a
specified number of glazings, the window with thicker air layers will
have a lower [/-factor. For a specified overall thickness of glazing,
the higher the number of glazings, the lower the [/-factor. Therefore,
a triple-pane window with air spaces of 6.4 mm (two such air spaces)
will have a lower [/-value than a double-pane window with an air
space of 12.7 mm.
5. Wood or vinyl frame windows have a considerably lower [/-value
than comparable metal-frame windows. Therefore, wood or vinyl
frame windows are called for in energy-efficient designs.
Glass
e = 0.84
FIGURE 9-39
Schematic of Example 9-7.
EXAMPLE 9-7
{/-Factor for Center-of-Glass Section of Windows
Determine the [/-factor for the center-of-glass section of a double-pane window
with a 6-mm air space for winter design conditions (Fig. 9-39). The glazings
are made of clear glass that has an emissivity of 0.84. Take the average air
space temperature at design conditions to be 0°C.
SOLUTION The U-factor for the center-of-glass section of a double-pane win-
dow is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Heat transfer through the
window is one-dimensional. 3 The thermal resistance of glass sheets is negligible.
Properties The emissivity of clear glass is 0.84.
Analysis Disregarding the thermal resistance of glass sheets, which are small,
the [/-factor for the center region of a double-pane window is determined from
1
£/,,
1
1
cen58933_ch09.qxd 9/4/2002 12:26 PM Page 497
where ft,-, h space , and ft are the heat transfer coefficients at the inner surface of
the window, the air space between the glass layers, and the outer surface of the
window, respectively. The values of h, and h for winter design conditions were
given earlier to be ft, = 8.29 W/m 2 ■ °C and ft = 34.0 W/m 2 ■ °C. The effective
emissivity of the air space of the double-pane window is
1
1
1/e, + 1/e, - 1 1/0.84 + 1/0.84 - 1
0.72
For this value of emissivity and an average air space temperature of 0°C,
we read ft.
Therefore,
7.2 W/m 2 • °C from Table 9-3 for 6-mm-thick air space.
f/„
8.29 7.2 34.0
U r ,
3.46 W/m 2 • °C
Discussion The center-of-glass ^-factor value of 3.24 W/m 2 • °C in Table 9-6
(fourth row and second column) is obtained by using a standard value of h =
29 W/m 2 • °C (instead of 34.0 W/m 2 • °C) and ft s|
erage air space temperature of - 15 C C.
6.5 W/m 2 • °C at an av-
497
CHAPTER 9
EXAMPLE 9-8 Heat Loss through Aluminum Framed Windows
A fixed aluminum-framed window with glass glazing is being considered for an
opening that is 4 ft high and 6 ft wide in the wall of a house that is maintained
at 72°F (Fig. 9-40). Determine the rate of heat loss through the window and
the inner surface temperature of the window glass facing the room when the
outdoor air temperature is 15°F if the window is selected to be (a) |-in. single
glazing, (fa) double glazing with an air space of i in., and (c) low-e-coated triple
glazing with an air space of | in.
SOLUTION The rate of heat loss through an aluminum framed window and the
inner surface temperature are to be determined from the cases of single-pane,
double-pane, and low-e triple-pane windows.
Assumptions 1 Steady operating conditions exist. 2 Heat transfer through the
window is one-dimensional. 3 Thermal properties of the windows and the heat
transfer coefficients are constant.
Properties The U-factors of the windows are given in Table 9-6.
Analysis The rate of heat transfer through the window can be determined from
e= V A IT — T 1
window overall windowV-* i L o)
where 7" and 7" arethe indoor and outdoor air temperatures, respectively; t/ OTera n
is the U-factor (the overall heat transfer coefficient) of the window; and /4 lvindow
is the window area, which is determined to be
A,
Height X Width = (4 ft)(6 ft) = 24 ft 2
The L/-factors for the three cases can be determined directly from Table 9-6 to
be 6.63, 3.51, and 1.92 W/m 2 ■ °C, respectively, to be multiplied by the factor
0.176 to convert them to Btu/h ■ ft 2 • °F. Also, the inner surface temperature of
the window glass can be determined from Newton's law
4 ft
6 ft
-Aluminum
- Glazing
frame
(a) Single
(b) Double
(c) Low-e triple
FIGURE 9-40
Schematic for
Example 9-8.
cen58933_ch09.qxd 9/4/2002 12:26 PM Page 49 E
498
HEAT TRANSFER
«^ window "/^window K-*- i
s)
x£ window
h-A
where h, is the heat transfer coefficient on the inner surface of the window,
which is determined from Table 9-5 to be h, = 8.3 W/m 2 • °C = 1.46 Btu/h ■
ft 2 • °F. Then the rate of heat loss and the interior glass temperature for each
case are determined as follows:
(a) Single glazing:
Gwindow = (6.63 X 0.176 Btu/h ■ ft 2 • °F)(24 ft 2 )(72 - 15)°F = 1596 Btu/h
Sd window
h-A
72°F
1596 Btu/h
(1.46 Btu/h • ft 2 ■ °F)(24ft 2 )
26.5°F
(b) Double glazing (| in. air space):
Gwindow = (3-51 X 0.176 Btu/h • ft 2 ■ °F)(24 ft 2 )(72 - 15)°F = 845 Btu/h
e,
72°F
845 Btu/h
g ' Mwi„dow (1.46 Btu/h • ft 2 • °F)(24 ft 2 )
(c) Triple glazing (| in. air space, low-e coated):
Swmdow = (1-92 X 0.176 Btu/h • ft 2 • °F)(24 ft 2 )(72 - 15)°F = 462 Btu/h
47.9°F
x£ window
72°F
462 Btu/h
(1.46 Btu/h • ft 2 • °F)(24ft 2 )
58.8°F
Therefore, heat loss through the window will be reduced by 47 percent in the
case of double glazing and by 71 percent in the case of triple glazing relative to
the single-glazing case. Also, in the case of single glazing, the low inner-glass
surface temperature will cause considerable discomfort in the occupants be-
cause of the excessive heat loss from the body by radiation. It is raised from
26.5°F, which is below freezing, to 47.9°F in the case of double glazing and to
58.8°F in the case of triple glazing.
Edge Center
Frame of glass of glass
1.8 m
- 0.94 m —\ \— 0.94 m
2 m
FIGURE 9-41
Schematic for Example 9-9.
EXAMPLE 9-9 {/-Factor of a Double-Door Window
Determine the overall L/-factor for a double-door-type, wood-framed double-
pane window with metal spacers, and compare your result to the value listed in
Table 9-6. The overall dimensions of the window are 1.80 m X 2.00 m, and
the dimensions of each glazing are 1.72 m X 0.94 m (Fig. 9-41).
SOLUTION The overall U-factor for a double-door type window is to be deter-
mined, and the result is to be compared to the tabulated value.
Assumptions 1 Steady operating conditions exist. 2 Heat transfer through the
window is one-dimensional.
Properties The (7-factors for the various sections of windows are given in
Tables 9-4 and 9-6.
Analysis The areas of the window, the glazing, and the frame are
^window = Height X Width = (1.8 m)(2.0 m) = 3.60 m 2
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CHAPTER 9
^glazing
A,
1 Mrame
2 X (Height X Width) = 2(1.72 m)(0.94 m) = 3.23 m 2
^window ~~ ^glazing = 3.60 — 3.23 = 0.37 m-
The edge-of-glass region consists of a 6.5-cm-wide band around the perimeter
of the glazings, and the areas of the center and edge sections of the glazing are
determined to be
A,.
2 X (Height X Width) = 2(1.72 - 0.13 m)(0.94 - 0.13 m) = 2.58 m 2
*edge
A„
3.23 - 2.58 = 0.65 m 2
The L/-factor for the frame section is determined from Table 9-4 to be
'-'frame = 2.8 W/m 2 • °C. The ^/-factors for the center and edge sections are de-
termined from Table 9-6 (fifth row, second and third columns) to be U center =
3.24 W/m 2 -°C and U,
tire window becomes
edge
3.71 W/m 2 • °C. Then the overall U-factor of the en-
^window \^ center ^center ^edge -^edge ^ frame ^frame/'-'^window
= (3.24 X 2.58 + 3.71 X 0.65 + 2.8 X 0.37)/3.60
= 3.28 W/m 2 • °C
The overall L/-factor listed in Table 9-6 for the specified type of window is
3.20 W/m 2 • °C, which is sufficiently close to the value obtained above.
SUMMARY
In this chapter, we have considered natural convection heat
transfer where any fluid motion occurs by natural means such
as buoyancy. The volume expansion coefficient of a substance
represents the variation of the density of that substance with
temperature at constant pressure, and for an ideal gas, it is ex-
pressed as (3 = l/T, where T is the absolute temperature in
KorR.
The flow regime in natural convection is governed by a di-
mensionless number called the Grashof number, which repre-
sents the ratio of the buoyancy force to the viscous force acting
on the fluid and is expressed as
Gr,
g${T s - Tjm
where L c is the characteristic length, which is the height L for
a vertical plate and the diameter D for a horizontal cylinder.
The correlations for the Nusselt number Nu = hL c Ik in natural
convection are expressed in terms of the Rayleigh number
defined as
Ra, = Gr, Pr
Pr
Nusselt number relations for various surfaces are given in
Table 9-1 . All fluid properties are evaluated at the film tempera-
ture of Tf = UT S + T rjL ). The outer surface of a vertical cylinder
can be treated as a vertical plate when the curvature effects are
negligible. The characteristic length for a horizontal surface is
L c = AJp, where A s is the surface area and p is the perimeter.
The average Nusselt number for vertical isothermal parallel
plates of spacing S and height L is given as
Nu
hS
k
576
2.873
_(Ra s SVL) 2 (Ra s 5/L) 05 _
The optimum fin spacing for a vertical heat sink and the Nus-
selt number for optimally spaced fins is
V = 2.714
S 3 LW- 5
hS„
2.714
Ra 0.25
and Nu
1.307
In a horizontal rectangular enclosure with the hotter plate at
the top, heat transfer is by pure conduction and Nu = 1 . When
the hotter plate is at the bottom, the Nusselt is
Nu = 1 + 1.44
I
1708
Ra,
Ra[«
IS
1
Ra L < 10 8
The notation [ ] + indicates that if the quantity in the bracket is
negative, it should be set equal to zero. For vertical horizontal
enclosures, the Nusselt number can be determined from
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HEAT TRANSFER
Nu = 0.18
Nu = 0.22
Pr p Y 29
0.2 + Pr N
Pr
0.2 + Pr
Ra L
0.28m
1<H/L<2
any Prandtl number
Ra L Pr/(0.2 + Pr) > 10 3
2<H/L< 10
1/4
any Prandtl number
az.'
For aspect ratios greater than 10, Eqs. 9-54 and 9-55 should be
used. For inclined enclosures, Eqs. 9-48 through 9-51 should
be used.
For concentric horizontal cylinders, the rate of heat transfer
through the annular space between the cylinders by natural
convection per unit length is
Q
2TSk F
m(ZVA)
{T, - T a )
where
and
0.3861
Pr
0.861 + Pr
(F cyl Ra L y
cyl
[ln(D /D,)] 4
For a spherical enclosure, the rate of heat transfer through
the space between the spheres by natural convection is ex-
pressed as
Q =Krfn
D,n
(T; ~ T )
where
k °' 74 io^6TTPr
L c = (D - D,)I2
U
sph
{D t D )XDi™ + Z)-™)
-7/5\5
The quantity feNu is called the effective thermal conductivity of
the enclosure, since a fluid in an enclosure behaves like a qui-
escent fluid whose thermal conductivity is £Nu as a result of
convection currents. The fluid properties are evaluated at the
average temperature of {T t + T B )I2.
For a given fluid, the parameter Gr/Re 2 represents the im-
portance of natural convection relative to forced convection.
Natural convection is negligible when Gr/Re 2 < 0.1, forced
convection is negligible when Gr/Re 2 > 10, and neither is neg-
ligible when 0.1 < Gr/Re 2 < 10.
REFERENCES AND SUGGESTED READING
1. American Society of Heating, Refrigeration, and Air
Conditioning Engineers. Handbook of Fundamentals.
Atlanta: ASHRAE, 1993.
2. J. N Arnold, I. Catton, and D. K. Edwards. "Experimental
Investigation of Natural Convection in Inclined
Rectangular Region of Differing Aspects Ratios." ASME
Paper No. 75-HT-62, 1975.
3. P. S. Ayyaswamy and I. Catton. "The Boundary-Layer
Regime for Natural Convection in a Differently Heated
Tilted Rectangular Cavity." Journal of Heat Transfer 95
(1973), p. 543.
4. A. Bar-Cohen. "Fin Thickness for an Optimized Natural
Convection Array of Rectangular Fins." Journal of Heat
Transfer 101 (1979), pp. 564-566.
5. A. Bar-Cohen and W. M. Rohsenow. "Thermally
Optimum Spacing of Vertical Natural Convection Cooled
Parallel Plates." Journal of Heat Transfer 106 (1984),
p. 116.
6. B. M. Berkovsky and V. K. Polevikov. "Numerical Study
of Problems on High-Intensive Free Convection." In
Heat Transfer and Turbulent Buoyant Convection, ed.
D. B. Spalding and N. Afgan, pp. 443^145. Washington,
DC: Hemisphere, 1977.
7. 1. Catton. "Natural Convection in Enclosures."
Proceedings of Sixth International Heat Transfer
Conference, Toronto, Canada, 1978, Vol. 6, pp. 13-31.
8. T. Cebeci. "Laminar Free Convection Heat Transfer from
the Outer Surface of a Vertical Slender Circular
Cylinder." Proceedings Fifth International Heat Transfer
Conference paper NCI.4, 1974 pp. 15-19.
9. Y. A. Cengel and P. T. L. Zing. "Enhancement of Natural
Convection Heat Transfer from Heat Sinks by
Shrouding." Proceedings of ASME/JSME Thermal
Engineering Conference, Honolulu, HA, 1987, Vol. 3,
pp. 451-475.
10. S. W. Churchill. "A Comprehensive Correlating Equation
for Laminar Assisting Forced and Free Convection."
AIChE Journal 23 (1977), pp. 10-16.
11. S. W. Churchill. "Free Convection Around Immersed
Bodies." In Heat Exchanger Design Handbook, ed.
E. U. Schliinder, Section 2.5.7. New York: Hemisphere,
1983.
cen58933_ch09.qxd 9/4/2002 12:26 PM Page 501
12. S. W. Churchill. "Combined Free and Forced Convection
around Immersed Bodies." In Heat Exchanger Design
Handbook, Section 2.5.9. New York: Hemisphere
Publishing, 1986.
13. S. W. Churchill and H. H. S. Chu. "Correlating Equations
for Laminar and Turbulent Free Convection from a
Horizontal Cylinder.'" International Journal of Heat Mass
Transfer 18 (\915), p. 1049.
14. S. W. Churchill and H. H. S. Chu. "Correlating Equations
for Laminar and Turbulent Free Convection from a
Vertical Plate." International Journal of Heat Mass
Transfer 18 (\97 5), p. 1323.
15. E. R. G. Eckert and E. Soehngen. "Studies on Heat
Transfer in Laminar Free Convection with Zehnder-Mach
Interferometer." USAF Technical Report 5747, December
1948.
16. E. R. G. Eckert and E. Soehngen. "Interferometric Studies
on the Stability and Transition to Turbulence of a Free
Convection Boundary Layer." Proceedings of General
Discussion, Heat Transfer ASME-IME, London, 1951.
17. S. M. ElSherbiny, G. D. Raithby, and K. G. T Hollands.
"Heat Transfer by Natural Convection Across Vertical and
Inclined Air Layers. Journal of Heat Transfer 104 (1982),
pp. 96-102.
18. T Fujiii and H. Imura. "Natural Convection Heat Transfer
from a Plate with Arbitrary Inclination." International
Journal of Heat Mass Transfer 15 (1972), p. 755.
19. K. G. T. Hollands, T. E. Unny, G. D. Raithby, and L.
Konicek. "Free Convective Heat Transfer Across
Inclined Air Layers." Journal of Heat Transfer 98
(1976), pp. 189-193.
20. J. P. Holman. Heat Transfer. 7th ed. New York: McGraw-
Hill, 1990.
21. F. P. Incropera and D. P. DeWitt. Introduction to Heat
Transfer. 3rd ed. New York: John Wiley & Sons, 1996.
22. M. Jakob. Heat Transfer. New York: Wiley, 1949.
501
CHAPTER 9
23. W. M. Kays and M. E. Crawford. Convective Heat and
Mass Transfer. 3rd ed. New York: McGraw-Hill, 1993.
24. F Kreith and M. S. Bohn. Principles of Heat Transfer. 6th
ed. Pacific Grove, CA: Brooks/Cole, 2001.
25. J. R. Lloyd and E. M. Sparrow. "Combined Forced and
Free Convection Flow on Vertical Surfaces."
International Journal of Heat Mass Transfer 13 (1970),
p. 434.
26. R. K. Macgregor and A. P. Emery. "Free Convection
Through Vertical Plane Layers: Moderate and High
Prandtl Number Fluids." Journal of Heat Transfer 91
(1969), p. 391.
27. S. Ostrach. "An Analysis of Laminar Free Convection
Flow and Heat Transfer About a Flat Plate Parallel to the
Direction of the Generating Body Force." National
Advisory Committee for Aeronautics, Report 1111, 1953.
28. G. D. Raithby and K. G. T. Hollands. "A General Method
of Obtaining Approximate Solutions to Laminar and
Turbulent Free Convection Problems." In Advances in
Heat Transfer, ed. F. Irvine and J. P. Hartnett, Vol. II, pp.
265-315. New York: Academic Press, 1975.
29. E. M. Sparrow and J. L. Gregg. "Laminar Free
Convection from a Vertical Flat Plate." Transactions of
the ASME 18 (1956), p. 438.
30. E. M. Sparrow and J. L. Gregg. "Laminar Free
Convection Heat Transfer from the Outer Surface of a
Vertical Circular Cylinder." ASME 78 (1956), p. 1823.
31. E. M. Sparrow and C. Prakash. "Enhancement of Natural
Convection Heat Transfer by a Staggered Array of
Vertical Plates." Journal of Heat Transfer 102 (1980), pp.
215-220.
32. E. M. Sparrow and S. B. Vemuri. "Natural
Convection/Radiation Heat Transfer from Highly
Populated Pin Fin Arrays." Journal of Heat Transfer 107
(1985), pp. 190-197.
PROBLEMS
Physical Mechanism of Natural Convection
9-1C What is natural convection? How does it differ from
forced convection? What force causes natural convection
currents?
9-2C In which mode of heat transfer is the convection heat
transfer coefficient usually higher, natural convection or forced
convection? Why?
9-3C Consider a hot boiled egg in a spacecraft that is filled
with air at atmospheric pressure and temperature at all times.
Will the egg cool faster or slower when the spacecraft is in
space instead of on the ground? Explain.
*Problems designated by a "C" are concept questions, and
students are encouraged to answer them all. Problems designated
by an "E" are in English units, and the SI users can ignore them.
Problems with an EES-CD icon ® are solved using EES, and
complete solutions together with parametric studies are included
on the enclosed CD. Problems with a computer-EES icon H are
comprehensive in nature, and are intended to be solved with a
computer, preferably using the EES software that accompanies
this text.
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HEAT TRANSFER
9-4C What is buoyancy force? Compare the relative magni-
tudes of the buoyancy force acting on a body immersed in
these mediums: (a) air, (b) water, (c) mercury, and (cl) an evac-
uated chamber.
9-5C When will the hull of a ship sink in water deeper: when
the ship is sailing in fresh water or in sea water? Why?
9-6C A person weighs himself on a waterproof spring scale
placed at the bottom of a 1-m-deep swimming pool. Will the
person weigh more or less in water? Why?
9-7C Consider two fluids, one with a large coefficient of
volume expansion and the other with a small one. In what fluid
will a hot surface initiate stronger natural convection currents?
Why? Assume the viscosity of the fluids to be the same.
9-8C Consider a fluid whose volume does not change with
temperature at constant pressure. What can you say about nat-
ural convection heat transfer in this medium?
9-9C What do the lines on an interferometer photograph rep-
resent? What do closely packed lines on the same photograph
represent?
9-10C Physically, what does the Grashof number represent?
How does the Grashof number differ from the Reynolds
number?
9-11 Show that the volume expansion coefficient of an ideal
gas is (3 = 1/r, where 7" is the absolute temperature.
Natural Convection over Surfaces
9-12C How does the Rayleigh number differ from the
Grashof number?
9-13C Under what conditions can the outer surface of a ver-
tical cylinder be treated as a vertical plate in natural convection
calculations?
9-14C Will a hot horizontal plate whose back side is insu-
lated cool faster or slower when its hot surface is facing down
instead of up?
9-15C Consider laminar natural convection from a vertical
hot plate. Will the heat flux be higher at the top or at the bottom
of the plate? Why?
9-16 A 1 0-m-long section of a 6-cm-diameter horizontal hot
water pipe passes through a large room whose temperature is
22°C. If the temperature and the emissivity of the outer sur-
face of the pipe are 65°C and 0.8, respectively, determine the
rate of heat loss from the pipe by (a) natural convection and
(b) radiation.
9-17 Consider a wall-mounted power transistor that dissi-
pates 0.18 W of power in an environment at 35°C. The transis-
tor is 0.45 cm long and has a diameter of 0.4 cm. The
emissivity of the outer surface of the transistor is 0.1, and the
average temperature of the surrounding surfaces is 25°C. Dis-
regarding any heat transfer from the base surface, determine
the surface temperature of the transistor. Use air properties at
100°C. Answer: 183°C
35°C
Power
transistor
0.18W
e = 0.1
0.4 cm
— 0.45 cm
IGURE P9-17
9-18 Tu'M Reconsider Problem 9-17. Using EES (or other)
b^2 software, investigate the effect of ambient tem-
perature on the surface temperature of the transistor. Let the en-
vironment temperature vary from 10°C to 40°C and assume
that the surrounding surfaces are 10°C colder than the environ-
ment temperature. Plot the surface temperature of the transistor
versus the environment temperature, and discuss the results.
9-19E Consider a 2-ft X 2-ft thin square plate in a room at
75°F. One side of the plate is maintained at a temperature of
130°F, while the other side is insulated. Determine the rate of
heat transfer from the plate by natural convection if the plate
is (a) vertical, (b) horizontal with hot surface facing up, and
(c) horizontal with hot surface facing down.
9-20E PSa| Reconsider Problem 9-19E. Using EES (or
I^S other) software, plot the rate of natural convec-
tion heat transfer for different orientations of the plate as a
function of the plate temperature as the temperature varies
from 80°F to 180°F, and discuss the results.
9-21 A 400-W cylindrical resistance heater is 1 m long and
0.5 cm in diameter. The resistance wire is placed horizontally
in a fluid at 20°C. Determine the outer surface temperature of
the resistance wire in steady operation if the fluid is (a) air and
(b) water. Ignore any heat transfer by radiation. Use properties
at 500°C for air and 40°C for water.
9-22 Water is boiling in a 12-cm-deep pan with an outer di-
ameter of 25 cm that is placed on top of a stove. The ambient
air and the surrounding surfaces are at a temperature of 25°C,
and the emissivity of the outer surface of the pan is 0.95. As-
suming the entire pan to be at an average temperature of 98°C,
determine the rate of heat loss from the cylindrical side surface
of the pan to the surroundings by (a) natural convection and
(b) radiation, (c) If water is boiling at a rate of 2 kg/h at 100°C,
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determine the ratio of the heat lost from the side surfaces of the
pan to that by the evaporation of water. The heat of
vaporization of water at 100°C is 2257 kJ/kg.
Answers: 46.2 W, 56.1 W, 0.082
Vapor
2kg/h
25°C
98°C
e = 0.95
FIGURE P9-22
503
CHAPTER 9
radiation. An estimate is obtained from a local insulation con-
tractor, who proposes to do the insulation job for $350, includ-
ing materials and labor. Would you support this proposal? How
long will it take for the insulation to pay for itself from the en-
ergy it saves?
9-26 Consider a 15-cm X 20-cm printed circuit board
(PCB) that has electronic components on one side. The board
is placed in a room at 20°C. The heat loss from the back sur-
face of the board is negligible. If the circuit board is dissipat-
ing 8 W of power in steady operation, determine the average
temperature of the hot surface of the board, assuming the
board is (a) vertical, (b) horizontal with hot surface facing up,
and (c) horizontal with hot surface facing down. Take the
emissivity of the surface of the board to be 0.8 and assume the
surrounding surfaces to be at the same temperature as the air
in the room. Answers: (a) 46.6°C, (b) 42.6°C, (c) 50.7°C
9-23 Repeat Problem 9-22 for a pan whose outer surface is
polished and has an emissivity of 0.1.
9-24 In a plant that manufactures canned aerosol paints, the
cans are temperature-tested in water baths at 55°C before they
are shipped to ensure that they will withstand temperatures up
to 55°C during transportation and shelving. The cans, moving
on a conveyor, enter the open hot water bath, which is 0.5 m
deep, 1 m wide, and 3.5 m long, and move slowly in the hot
water toward the other end. Some of the cans fail the test and
explode in the water bath. The water container is made of
sheet metal, and the entire container is at about the same tem-
perature as the hot water. The emissivity of the outer surface
of the container is 0.7. If the temperature of the surrounding
air and surfaces is 20°C, determine the rate of heat loss from
the four side surfaces of the container (disregard the top sur-
face, which is open).
The water is heated electrically by resistance heaters, and the
cost of electricity is $0.085/kWh. If the plant operates 24 h a
day 365 days a year and thus 8760 h a year, determine the an-
nual cost of the heat losses from the container for this facility.
FIGURE P9-24
9-25 Reconsider Problem 9-24. In order to reduce the heat-
ing cost of the hot water, it is proposed to insulate the side and
bottom surfaces of the container with 5-cm-thick fiberglass in-
sulation (k = 0.035 W/m • °C) and to wrap the insulation with
aluminum foil (8 = 0.1) in order to minimize the heat loss by
FIGURE P9-26
9-27 flfipM Reconsider Problem 9-26. Using EES (or other)
1^2 software, investigate the effects of the room tem-
perature and the emissivity of the board on the temperature
of the hot surface of the board for different orientations of the
board. Let the room temperature vary from 5°C to 35°C and
the emissivity from 0.1 to 1.0. Plot the hot surface temperature
for different orientations of the board as the functions of the
room temperature and the emissivity, and discuss the results.
9-28 f7&\ A manufacturer makes absorber plates that are
xgty 1.2 m X 0.8 m in size for use in solar collec-
tors. The back side of the plate is heavily insulated, while its
front surface is coated with black chrome, which has an ab-
sorptivity of 0.87 for solar radiation and an emissivity of
0.09. Consider such a plate placed horizontally outdoors in
calm air at 25°C. Solar radiation is incident on the plate at a
rate of 700 W/m 2 . Taking the effective sky temperature to be
10°C, determine the equilibrium temperature of the absorber
plate. What would your answer be if the absorber plate is
made of ordinary aluminum plate that has a solar absorptiv-
ity of 0.28 and an emissivity of 0.07?
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HEAT TRANSFER
Solar radiation
700 W/m 2
\ \ \ \ \ \ \
Absorber plate
/ a s = 0.87
£ = 0.09
Insulation
FIGURE P9-28
9-29 Repeat Problem 9-28 for an aluminum plate painted
flat black (solar absorptivity 0.98 and emissivity 0.98) and
also for a plate painted white (solar absorptivity 0.26 and
emissivity 0.90).
9-30 The following experiment is conducted to determine the
natural convection heat transfer coefficient for a horizontal
cylinder that is 80 cm long and 2 cm in diameter. A 80-cm-long
resistance heater is placed along the centerline of the cylinder,
and the surfaces of the cylinder are polished to minimize the ra-
diation effect. The two circular side surfaces of the cylinder are
well insulated. The resistance heater is turned on, and the power
dissipation is maintained constant at 40 W. If the average surface
temperature of the cylinder is measured to be 120°C in the 20°C
room air when steady operation is reached, determine the natural
convection heat transfer coefficient. If the emissivity of the outer
surface of the cylinder is 0.1 and a 5 percent error is acceptable,
do you think we need to do any correction for the radiation ef-
fect? Assume the surrounding surfaces to be at 20°C also.
120°C
Insulated
9-31
Resistance
heater
40 W
Insulated
FIGURE P9-30
Thick fluids such as asphalt and waxes and the pipes in
which they flow are often heated in order to reduce the viscos-
ity of the fluids and thus to reduce the pumping costs. Consider
the flow of such a fluid through a 100-m-long pipe of outer di-
ameter 30 cm in calm ambient air at 0°C. The pipe is heated
electrically, and a thermostat keeps the outer surface tempera-
ture of the pipe constant at 25°C. The emissivity of the outer
surface of the pipe is 0.8, and the effective sky temperature is
— 30°C, Determine the power rating of the electric resistance
heater, in kW, that needs to be used. Also, determine the cost of
electricity associated with heating the pipe during a 10-h pe-
riod under the above conditions if the price of electricity is
$0.09/kWh. Answers: 29.1 kW, $26.2
sky "
-30°C
Asphalt
FIGURE P9-31
9-32 Reconsider Problem 9-31. To reduce the heating cost
of the pipe, it is proposed to insulate it with sufficiently thick
fiberglass insulation (k = 0.035 W/m • °C) wrapped with alu-
minum foil (8 = 0.1) to cut down the heat losses by 85 percent.
Assuming the pipe temperature to remain constant at 25°C, de-
termine the thickness of the insulation that needs to be used.
How much money will the insulation save during this 10-h
period? Answers: 1.3 cm, $22.3
9-33E Consider an industrial furnace that resembles a 13-ft-
long horizontal cylindrical enclosure 8 ft in diameter whose
end surfaces are well insulated. The furnace burns natural gas
at a rate of 48 therms/h (1 therm = 100,000 Btu). The combus-
tion efficiency of the furnace is 82 percent (i.e., 18 percent of
the chemical energy of the fuel is lost through the flue gases as
a result of incomplete combustion and the flue gases leaving
the furnace at high temperature). If the heat loss from the outer
surfaces of the furnace by natural convection and radiation is
not to exceed 1 percent of the heat generated inside, determine
the highest allowable surface temperature of the furnace. As-
sume the air and wall surface temperature of the room to be
75 °F, and take the emissivity of the outer surface of the furnace
to be 0.85. If the cost of natural gas is $0.65/therm and the fur-
nace operates 2800 h per year, determine the annual cost of this
heat loss to the plant.
FIGURE P9-33
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CHAPTER 9
9-34 Consider a 1.2-m-high and 2-m-wide glass window
with a thickness of 6 mm, thermal conductivity k = 0.78
W/m • °C, and emissivity e = 0.9. The room and the walls that
face the window are maintained at 25°C, and the average tem-
perature of the inner surface of the window is measured to be
5°C. If the temperature of the outdoors is — 5°C, determine
(a) the convection heat transfer coefficient on the inner sur-
face of the window, (b) the rate of total heat transfer through
the window, and (c) the combined natural convection and
radiation heat transfer coefficient on the outer surface of the
window. Is it reasonable to neglect the thermal resistance of
the glass in this case?
Wall
Room
25°C
1.2 m
5°C
e = 0.9
- Glass
-5°C
FIGURE P9-34
9-35 A 3-mm-diameter and 12-m-long electric wire is
tightly wrapped with a 1 .5-mm-thick plastic cover whose ther-
mal conductivity and emissivity are k = 0.15 W/m • °C and
8 = 0.9. Electrical measurements indicate that a current of
10 A passes through the wire and there is a voltage drop of 8 V
along the wire. If the insulated wire is exposed to calm atmo-
spheric air at T a = 30°C, determine the temperature at the
interface of the wire and the plastic cover in steady operation.
Take the surrounding surfaces to be at about the same temper-
ature as the air.
9-36 During a visit to a plastic sheeting plant, it was ob-
served that a 60-m-long section of a 2-in. nominal (6.03-cm
outer-diameter) steam pipe extended from one end of the plant
to the other with no insulation on it. The temperature measure-
ments at several locations revealed that the average tempera-
ture of the exposed surfaces of the steam pipe was 170°C,
while the temperature of the surrounding air was 20°C. The
outer surface of the pipe appeared to be oxidized, and its emis-
sivity can be taken to be 0.7. Taking the temperature of the sur-
rounding surfaces to be 20°C also, determine the rate of heat
loss from the steam pipe.
Steam is generated in a gas furnace that has an efficiency of
78 percent, and the plant pays $0,538 per therm (1 therm =
105,500 kJ) of natural gas. The plant operates 24 h a day 365
days a year, and thus 8760 h a year. Determine the annual cost
of the heat losses from the steam pipe for this facility.
Steam
FIGURE P9-36
9-37 [T(^N Reconsider Problem 9-36. Using EES (or other)
1^2 software, investigate the effect of the surface
temperature of the steam pipe on the rate of heat loss from the
pipe and the annual cost of this heat loss. Let the surface tem-
perature vary from 100°C to 200°C. Plot the rate of heat loss
and the annual cost as a function of the surface temperature,
and discuss the results.
9-38 Reconsider Problem 9-36. In order to reduce heat
losses, it is proposed to insulate the steam pipe with 5-cm-thick
fiberglass insulation (k = 0.038 W/m • °C) and to wrap it with
aluminum foil (e = 0.1) in order to minimize the radiation
losses. Also, an estimate is obtained from a local insulation
contractor, who proposed to do the insulation job for $750, in-
cluding materials and labor. Would you support this proposal?
How long will it take for the insulation to pay for itself from
the energy it saves? Assume the temperature of the steam pipe
to remain constant at 170°C.
9-39 A 30-cm X 30-cm circuit board that contains 121
square chips on one side is to be cooled by combined natural
convection and radiation by mounting it on a vertical surface in
a room at 25°C. Each chip dissipates 0.05 W of power, and the
emissivity of the chip surfaces is 0.7. Assuming the heat trans-
fer from the back side of the circuit board to be negligible, and
the temperature of the surrounding surfaces to be the same as
the air temperature of the room, determine the surface temper-
ature of the chips. Answer: 33.4°C
9-40 Repeat Prob. 9-35 assuming the circuit board to be po-
sitioned horizontally with (a) chips facing up and (b) chips fac-
ing down.
9-41 The side surfaces of a 2-m-high cubic industrial furnace
burning natural gas are not insulated, and the temperature at the
outer surface of this section is measured to be 1 10°C. The tem-
perature of the furnace room, including its surfaces, is 30°C,
and the emissivity of the outer surface of the furnace is 0.7. It
is proposed that this section of the furnace wall be insulated
with glass wool insulation (k = 0.038 W/m ■ °C) wrapped by a
reflective sheet (s = 0.2) in order to reduce the heat loss by 90
percent. Assuming the outer surface temperature of the metal
section still remains at about 110°C, determine the thickness of
the insulation that needs to be used.
The furnace operates continuously throughout the year and
has an efficiency of 78 percent. The price of the natural gas is
cen58933_ch09.qxd 9/4/2002 12:26 PM Page 506
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HEAT TRANSFER
$0.55/therm (1 therm = 105,500 kJ of energy content). If the
installation of the insulation will cost $550 for materials and la-
bor, determine how long it will take for the insulation to pay
for itself from the energy it saves.
Hot
gases
30°C
Furnace
110°C
e = 0.7
2 m
FIGURE P9^1
2m
2 m
9-42 A 1 .5-m-diameter, 5-m-long cylindrical propane tank is
initially filled with liquid propane, whose density is 581 kg/m 3 .
The tank is exposed to the ambient air at 25°C in calm weather.
The outer surface of the tank is polished so that the radiation
heat transfer is negligible. Now a crack develops at the top of
the tank, and the pressure inside drops to 1 atm while the tem-
perature drops to — 42°C, which is the boiling temperature of
propane at 1 atm. The heat of vaporization of propane at 1 atm
is 425 kJ/kg. The propane is slowly vaporized as a result of the
heat transfer from the ambient air into the tank, and the
propane vapor escapes the tank at — 42°C through the crack.
Assuming the propane tank to be at about the same temperature
as the propane inside at all times, determine how long it will
take for the tank to empty if it is not insulated.
25°C
Propane
vapor
Propane tank
-42°C
FIGURE P9^2
4m
9-43E An average person generates heat at a rate of 287
Btu/h while resting in a room at 77°F. Assuming one-quarter of
this heat is lost from the head and taking the emissivity of the
skin to be 0.9, determine the average surface temperature of the
head when it is not covered. The head can be approximated as
a 12-in. -diameter sphere, and the interior surfaces of the room
can be assumed to be at the room temperature.
9-44 An incandescent lightbulb is an inexpensive but highly
inefficient device that converts electrical energy into light. It
converts about 10 percent of the electrical energy it consumes
into light while converting the remaining 90 percent into heat.
The glass bulb of the lamp heats up very quickly as a result of
absorbing all that heat and dissipating it to the surroundings by
convection and radiation. Consider an 8-cm-diameter 60-W
light bulb in a room at 25°C. The emissivity of the glass is 0.9.
Assuming that 10 percent of the energy passes through the
glass bulb as light with negligible absorption and the rest of the
energy is absorbed and dissipated by the bulb itself by natural
convection and radiation, determine the equilibrium tempera-
ture of the glass bulb. Assume the interior surfaces of the room
to be at room temperature. Answer: 169°C
Light, 6 W
FIGURE P9-44
9-45 A 40-cm-diameter, 110-cm-high cylindrical hot water
tank is located in the bathroom of a house maintained at 20°C.
The surface temperature of the tank is measured to be 44°C and
its emissivity is 0.4. Taking the surrounding surface tempera-
ture to be also 20°C, determine the rate of heat loss from all
surfaces of the tank by natural convection and radiation.
9-46 A 28-cm-high, 18-cm-long, and 18-cm-wide rectangu-
lar container suspended in a room at 24°C is initially filled with
cold water at 2°C. The surface temperature of the container is
observed to be nearly the same as the water temperature inside.
The emissivity of the container surface is 0.6, and the temper-
ature of the surrounding surfaces is about the same as the air
temperature. Determine the water temperature in the container
after 3 h, and the average rate of heat transfer to the water. As-
sume the heat transfer coefficient on the top and bottom sur-
faces to be the same as that on the side surfaces.
9-47 ru$% Reconsider Problem 9^-6. Using EES (or other)
|££g software, plot the water temperature in the con-
tainer as a function of the heating time as the time varies from
30 min to 10 h, and discuss the results.
9-48 A room is to be heated by a coal-burning stove, which
is a cylindrical cavity with an outer diameter of 32 cm and a
height of 70 cm. The rate of heat loss from the room is esti-
mated to be 1.2 kW when the air temperature in the room is
maintained constant at 24°C. The emissivity of the stove sur-
face is 0.85 and the average temperature of the surrounding
cen58933_ch09.qxd 9/4/2002 12:26 PM Page 507
wall surfaces is 17°C. Determine the surface temperature of the
stove. Neglect the transfer from the bottom surface and take the
heat transfer coefficient at the top surface to be the same as that
on the side surface.
The heating value of the coal is 30,000 kJ/kg, and the
combustion efficiency is 65 percent. Determine the amount of
coal burned a day if the stove operates 14 h a day.
9-49 The water in a 40-L tank is to be heated from 15°C to
45°C by a 6-cm-diameter spherical heater whose surface tem-
perature is maintained at 85 °C. Determine how long the heater
should be kept on.
Natural Convection from Finned Surfaces and PCBs
9-50C Why are finned surfaces frequently used in practice?
Why are the finned surfaces referred to as heat sinks in the
electronics industry?
9-51 C Why are heat sinks with closely packed fins not suit-
able for natural convection heat transfer, although they increase
the heat transfer surface area more?
9-52C Consider a heat sink with optimum fin spacing. Ex-
plain how heat transfer from this heat sink will be affected by
(a) removing some of the fins on the heat sink and (b) doubling
the number of fins on the heat sink by reducing the fin spacing.
The base area of the heat sink remains unchanged at all times.
9-53 Aluminum heat sinks of rectangular profile are com-
monly used to cool electronic components. Consider a
7.62-cm-long and 9.68-cm-wide commercially available heat
sink whose cross section and dimensions are as shown in Fig-
ure P9-53. The heat sink is oriented vertically and is used to
cool a power transistor that can dissipate up to 125 W of power.
The back surface of the heat sink is insulated. The surfaces of
the heat sink are untreated, and thus they have a low emissivity
(under 0.1). Therefore, radiation heat transfer from the heat
sink can be neglected. During an experiment conducted in
room air at 22°C, the base temperature of the heat sink was
measured to be 120°C when the power dissipation of the tran-
sistor was 1 5 W. Assuming the entire heat sink to be at the base
temperature, determine the average natural convection heat
transfer coefficient for this case. Answer: 7.1 W/m2 • °C
3.17 cm 1.45 cm
Transistor --r— >i r "i
UJL^UUI
-Heat sink
9.68 cm
1.52 cm
0.48 cm
FIGURE P9-53
9-54 Reconsider the heat sink in Problem 9-53. In order to
enhance heat transfer, a shroud (a thin rectangular metal plate)
whose surface area is equal to the base area of the heat sink is
placed very close to the tips of the fins such that the interfin
507
CHAPTER 9
spaces are converted into rectangular channels. The base tem-
perature of the heat sink in this case was measured to be
108°C. Noting that the shroud loses heat to the ambient air
from both sides, determine the average natural convection heat
transfer coefficient in this shrouded case. (For complete details,
see Cengel and Zing, Ref. 9).
Heat sink
J
Airflow "
L/
7.
r
52
11
Shroud
FIGURE P9-54
9-55E A 6-in.-wide and 8-in.-high vertical hot surface in
78°F air is to be cooled by a heat sink with equally spaced fins
of rectangular profile. The fins are 0.08 in. thick and 8 in. long
in the vertical direction and have a height of 1 .2 in. from the
base. Determine the optimum fin spacing and the rate of heat
transfer by natural convection from the heat sink if the base
temperature is 180°F
9-56E [J^l Reconsider Problem 9-55E. Using EES (or
1^2 other) software, investigate the effect of the
length of the fins in the vertical direction on the optimum fin
spacing and the rate of heat transfer by natural convection. Let
the fin length vary from 2 in. to 10 in. Plot the optimum fin
spacing and the rate of convection heat transfer as a function of
the fin length, and discuss the results.
9-57 A 1 2 . 1 -cm -wide and 1 8-cm-high vertical hot surface in
25 °C air is to be cooled by a heat sink with equally spaced fins
of rectangular profile. The fins are 0.1 cm thick and 18 cm long
in the vertical direction. Determine the optimum fin height and
the rate of heat transfer by natural convection from the heat
sink if the base temperature is 65°C.
Natural Convection inside Enclosures
9-58C The upper and lower compartments of a well-
insulated container are separated by two parallel sheets of glass
with an air space between them. One of the compartments is to
be filled with a hot fluid and the other with a cold fluid. If it is
desired that heat transfer between the two compartments
be minimal, would you recommend putting the hot fluid into
the upper or the lower compartment of the container? Why?
9-59C Someone claims that the air space in a double-pane
window enhances the heat transfer from a house because of
the natural convection currents that occur in the air space and
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HEAT TRANSFER
recommends that the double -pane window be replaced by a
single sheet of glass whose thickness is equal to the sum of the
thicknesses of the two glasses of the double-pane window to
save energy. Do you agree with this claim?
9-60C Consider a double-pane window consisting of two
glass sheets separated by a 1 -cm-wide air space. Someone sug-
gests inserting a thin vinyl sheet in the middle of the two
glasses to form two 0.5-cm-wide compartments in the window
in order to reduce natural convection heat transfer through the
window. From a heat transfer point of view, would you be in
favor of this idea to reduce heat losses through the window?
9-61 C What does the effective conductivity of an enclosure
represent? How is the ratio of the effective conductivity to ther-
mal conductivity related to the Nusselt number?
9-62 Show that the thermal resistance of a rectangular enclo-
sure can be expressed as R = §/(Ak Nu), where k is the thermal
conductivity of the fluid in the enclosure.
9-63E A vertical 4-ft-high and 6-ft-wide double-pane win-
dow consists of two sheets of glass separated by a 1-in. air gap
at atmospheric pressure. If the glass surface temperatures
across the air gap are measured to be 65°F and 40°F, determine
the rate of heat transfer through the window by (a) natural con-
vection and (b) radiation. Also, determine the /J-value of insu-
lation of this window such that multiplying the inverse of the
^-value by the surface area and the temperature difference
gives the total rate of heat transfer through the window. The ef-
fective emissivity for use in radiation calculations between two
large parallel glass plates can be taken to be 0.82.
4ft
Frame
FIGURE P9-63E
9-64E
Reconsider Problem 9-63E. Using EES (or
other) software, investigate the effect of the air
gap thickness on the rates of heat transfer by natural con-
vection and radiation, and the R-wahxe, of insulation. Let the air
gap thickness vary from 0.2 in. to 2.0 in. Plot the rates of heat
transfer by natural convection and radiation, and the R-va\ue.
of insulation as a function of the air gap thickness, and discuss
the results.
9-65 Two concentric spheres of diameters 15 cm and 25 cm
are separated by air at 1 arm pressure. The surface temperatures
of the two spheres enclosing the air are T x = 350 K and T 2 =
275 K, respectively. Determine the rate of heat transfer from
the inner sphere to the outer sphere by natural convection.
9-66 Tu'M Reconsider Problem 9-65. Using EES (or other)
Ei3 software, plot the rate of natural convection heat
transfer as a function of the hot surface temperature of the
sphere as the temperature varies from 300 K to 500 K, and dis-
cuss the results.
9-67 Flat-plate solar collectors are often tilted up toward the
sun in order to intercept a greater amount of direct solar radia-
tion. The tilt angle from the horizontal also affects the rate of
heat loss from the collector. Consider a 2-m-high and 3-m-
wide solar collector that is tilted at an angle from the hori-
zontal. The back side of the absorber is heavily insulated. The
absorber plate and the glass cover, which are spaced 2.5 cm
from each other, are maintained at temperatures of 80°C and
40°C, respectively. Determine the rate of heat loss from the ab-
sorber plate by natural convection for = 0°, 20°, and 90°.
Absorber
plate
Air space
Insulation
FIGURE P9-67
9-68 A simple solar collector is built by placing a 5-cm-
diameter clear plastic tube around a garden hose whose outer
diameter is 1 .6 cm. The hose is painted black to maximize solar
absorption, and some plastic rings are used to keep the spacing
between the hose and the clear plastic cover constant. During a
clear day, the temperature of the hose is measured to be 65°C,
Solar
radiation
WW
26°C
a
2T
:r ( f
- Clear plastic tube
V/ V
Spacer
Garden hose
65°C
FIGURE P9-68
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CHAPTER 9
while the ambient air temperature is 26°C. Determine the rate
of heat loss from the water in the hose per meter of its length
by natural convection. Also, discuss how the performance of
this solar collector can be improved. Answer: 8.2 W
9-69 TMt Reconsider Problem 9-68. Using EES (or other)
k^^ software, plot the rate of heat loss from the water
by natural convection as a function of the ambient air tempera-
ture as the temperature varies from 4°C to 40°C, and discuss
the results.
9-70 A vertical 1.3-m-high, 2.8-m-wide double-pane win-
dow consists of two layers of glass separated by a 2.2-cm air
gap at atmospheric pressure. The room temperature is 26°C
while the inner glass temperature is 18°C. Disregarding radia-
tion heat transfer, determine the temperature of the outer glass
layer and the rate of heat loss through the window by natural
convection.
9-71 Consider two concentric horizontal cylinders of diame-
ters 55 cm and 65 cm, and length 125 cm. The surfaces of the
inner and outer cylinders are maintained at 46°C and 74°C, re-
spectively. Determine the rate of heat transfer between the
cylinders by natural convection if the annular space is filled
with (a) water and (b) air.
Combined Natural and Forced Convection
9-72C When is natural convection negligible and when is it
not negligible in forced convection heat transfer?
9-73C Under what conditions does natural convection en-
hance forced convection, and under what conditions does it
hurt forced convection?
9-74C When neither natural nor forced convection is negli-
gible, is it correct to calculate each independently and add them
to determine the total convection heat transfer?
9-75 Consider a 5-m-long vertical plate at 85°C in air at
30°C. Determine the forced motion velocity above which nat-
ural convection heat transfer from this plate is negligible.
Answer: 9.04 m/s
9-76 Reconsider Problem 9-75. Using EES (or other) soft-
ware, plot the forced motion velocity above which natural con-
vection heat transfer is negligible as a function of the plate
temperature as the temperature varies from 50°C to 150°C, and
discuss the results.
9-77 Consider a 5-m-long vertical plate at 60°C in water at
25 °C. Determine the forced motion velocity above which nat-
ural convection heat transfer from this plate is negligible. Take
= 0.0004 K ' for water.
9-78 In a production facility, thin square plates 2 m X 2 m in
size coming out of the oven at 270°C are cooled by blowing
ambient air at 30°C horizontally parallel to their surfaces. De-
termine the air velocity above which the natural convection ef-
fects on heat transfer are less than 10 percent and thus are
negligible.
270°C
FIGURE P9-78
9-79 A 12-cm-high and 20-cm-wide circuit board houses
100 closely spaced logic chips on its surface, each dissipating
0.05 W. The board is cooled by a fan that blows air over the hot
surface of the board at 35°C at a velocity of 0.5 m/s. The heat
transfer from the back surface of the board is negligible. De-
termine the average temperature on the surface of the circuit
board assuming the air flows vertically upwards along the 12-
cm-long side by (a) ignoring natural convection and (b) con-
sidering the contribution of natural convection. Disregard any
heat transfer by radiation.
Special Topic: Heat Transfer through Windows
9-80C Why are the windows considered in three regions
when analyzing heat transfer through them? Name those re-
gions and explain how the overall U- value of the window is de-
termined when the heat transfer coefficients for all three
regions are known.
9-81 C Consider three similar double-pane windows with air
gap widths of 5, 10, and 20 mm. For which case will the heat
transfer through the window will be a minimum?
9-82C In an ordinary double -pane window, about half of the
heat transfer is by radiation. Describe a practical way of reduc-
ing the radiation component of heat transfer.
9-83C Consider a double-pane window whose air space
width is 20 mm. Now a thin polyester film is used to divide the
air space into two 10-mm-wide layers. How will the film affect
(a) convection and (b) radiation heat transfer through the win-
dow?
9-84C Consider a double -pane window whose air space is
flashed and filled with argon gas. How will replacing the air in
the gap by argon affect (a) convection and (b) radiation heat
transfer through the window?
9-85C Is the heat transfer rate through the glazing of a dou-
ble-pane window higher at the center or edge section of the
glass area? Explain.
9-86C How do the relative magnitudes of {/-factors of win-
dows with aluminum, wood, and vinyl frames compare? As-
sume the windows are identical except for the frames.
9-87 Determine the [/-factor for the center-of -glass section
of a double-pane window with a 13 -mm air space for winter
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HEAT TRANSFER
design conditions. The glazings are made of clear glass having
an emissivity of 0.84. Take the average air space temperature at
design conditions to be 10°C and the temperature difference
across the air space to be 15°C.
9-88 A double -door wood-framed window with glass glaz-
ing and metal spacers is being considered for an opening that is
1 .2 m high and 1.8 m wide in the wall of a house maintained at
20°C. Determine the rate of heat loss through the window and
the inner surface temperature of the window glass facing the
room when the outdoor air temperature is — 8°C if the window
is selected to be (a) 3-mm single glazing, (b) double glazing
with an air space of 1 3 mm, and (c) low-e-coated triple glazing
with an air space of 13 mm.
Double-door
window
Wood frame
Glass
Glass
FIGURE P9-88
9-89 Determine the overall [/-factor for a double -door-type
wood-framed double -pane window with 13-mm air space and
metal spacers, and compare your result to the value listed in
Table 9-6. The overall dimensions of the window are 2.00 m X
2.40 m, and the dimensions of each glazing are 1.92 m X
1.14m.
9-90 Consider a house in Atlanta, Georgia, that is maintained
at 22°C and has a total of 20 m 2 of window area. The windows
are double-door-type with wood frames and metal spacers. The
glazing consists of two layers of glass with 12.7 mm of air
space with one of the inner surfaces coated with reflective film.
The winter average temperature of Atlanta is 11.3°C. Determine
the average rate of heat loss through the windows in winter.
Answer: 456 W
9-91E Consider an ordinary house with R- 13 walls (walls
that have an R- value of 13 h • ft 2 • °F/Btu). Compare this to the
R- value of the common double -door windows that are double
pane with ~ in. of air space and have aluminum frames. If the
windows occupy only 20 percent of the wall area, determine if
more heat is lost through the windows or through the remain-
ing 80 percent of the wall area. Disregard infiltration losses.
9-92 The overall {/-factor of a fixed wood-framed window
with double glazing is given by the manufacturer to be U =
2.76 W/m 2 • °C under the conditions of still air inside and
winds of 12 km/h outside. What will the [/-factor be when the
wind velocity outside is doubled? Answer: 2.88 W/m 2 • °C
9-93 The owner of an older house in Wichita, Kansas, is con-
sidering replacing the existing double-door type wood-framed
single -pane windows with vinyl-framed double -pane windows
with an air space of 6.4 mm. The new windows are of double-
door type with metal spacers. The house is maintained at 22°C
at all times, but heating is needed only when the outdoor tem-
perature drops below 1 8°C because of the internal heat gain
from people, lights, appliances, and the sun. The average win-
ter temperature of Wichita is 7.1 °C, and the house is heated by
electric resistance heaters. If the unit cost of electricity is
$0.07/kWh and the total window area of the house is 12 m 2 , de-
termine how much money the new windows will save the
home owner per month in winter.
Single pane
Double pane
FIGURE P9-93
Review Problems
9-94E A 0.1 -W small cylindrical resistor mounted on a
lower part of a vertical circuit board is 0.3 in. long and has a di-
ameter of 0.2 in. The view of the resistor is largely blocked by
another circuit board facing it, and the heat transfer through the
connecting wires is negligible. The air is free to flow through
the large parallel flow passages between the boards as a result
of natural convection currents. If the air temperature at the
vicinity of the resistor is 120°F, determine the approximate sur-
face temperature of the resistor. Answer: 212°F
FIGURE P9-94E
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CHAPTER 9
9-95 An ice chest whose outer dimensions are 30 cm X
40 cm X 40 cm is made of 3-cm-thick styrofoam (k = 0.033
W/m ■ °C). Initially, the chest is filled with 30 kg of ice at 0°C,
and the inner surface temperature of the ice chest can be taken
to be 0°C at all times. The heat of fusion of water at 0°C is
333.7 kJ/kg, and the surrounding ambient air is at 20°C. Dis-
regarding any heat transfer from the 40 cm X 40 cm base of
the ice chest, determine how long it will take for the ice in the
chest to melt completely if the ice chest is subjected to (a) calm
air and (b) winds at 50 km/h. Assume the heat transfer coeffi-
cient on the front, back, and top surfaces to be the same as that
on the side surfaces.
9-96 An electronic box that consumes 1 80 W of power is
cooled by a fan blowing air into the box enclosure. The dimen-
sions of the electronic box are 15 cm X 50 cm X 50 cm, and all
surfaces of the box are exposed to the ambient except the base
surface. Temperature measurements indicate that the box is at
an average temperature of 32°C when the ambient temperature
and the temperature of the surrounding walls are 25°C. If the
emissivity of the outer surface of the box is 0.85, determine the
fraction of the heat lost from the outer surfaces of the elec-
tronic box.
15 cm
50 cm
FIGURE P9-96
9-97 A 6-m-internal-diameter spherical tank made of 1.5-
cm-thick stainless steel (k = 15 W/m • °C) is used to store iced
water at 0°C in a room at 20°C. The walls of the room are also
at 20°C. The outer surface of the tank is black (emissivity
e = 1 ), and heat transfer between the outer surface of the tank
and the surroundings is by natural convection and radiation.
Assuming the entire steel tank to be at 0°C and thus the ther-
mal resistance of the tank to be negligible, determine (a) the
rate of heat transfer to the iced water in the tank and (b) the
amount of ice at 0°C that melts during a 24-h period.
Answers: (a) 15.4 kW, (b) 3988 kg
9-98 Consider a 1.2-m-high and 2-m-wide double-pane
window consisting of two 3-mm-thick layers of glass (k =
0.78 W/m • °C) separated by a 3-cm-wide air space. De-
termine the steady rate of heat transfer through this window
and the temperature of its inner surface for a day during
which the room is maintained at 20°C while the temperature
of the outdoors is 0°C. Take the heat transfer coefficients
on the inner and outer surfaces of the window to be h.
10
W/m 2 ■ °C and h 2 = 25 W/m 2 • °C and disregard any heat
transfer by radiation.
9-99 An electric resistance space heater is designed such that
it resembles a rectangular box 50 cm high, 80 cm long, and
Electric heater
Heating element
FIGURE P9-99
15 cm wide filled with 45 kg of oil. The heater is to be placed
against a wall, and thus heat transfer from its back surface is
negligible for safety considerations. The surface temperature of
the heater is not to exceed 45°C in a room at 25°C. Disregard-
ing heat transfer from the bottom and top surfaces of the heater
in anticipation that the top surface will be used as a shelf, de-
termine the power rating of the heater in W. Take the emissiv-
ity of the outer surface of the heater to be 0.8 and the average
temperature of the ceiling and wall surfaces to be the same as
the room air temperature.
Also, determine how long it will take for the heater to reach
steady operation when it is first turned on (i.e., for the oil tem-
perature to rise from 25°C to 45°C). State your assumptions in
the calculations.
9-100 Skylights or "roof windows" are commonly used in
homes and manufacturing facilities since they let natural light
in during day time and thus reduce the lighting costs.
However, they offer little resistance to heat transfer, and
large amounts of energy are lost through them in winter un-
less they are equipped with a motorized insulating cover that
can be used in cold weather and at nights to reduce heat
losses. Consider a 1-m-wide and 2.5-m-long horizontal
skylight on the roof of a house that is kept at 20°C. The glaz-
ing of the skylight is made of a single layer of 0.5-cm-thick
r sky = -30°c
-10°C
Skylight
e = 0.9
FIGURE P9-1 00
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HEAT TRANSFER
glass (k = 0.78 W/m • °C and e = 0.9). Determine the rate of
heat loss through the skylight when the air temperature out-
side is — 10°C and the effective sky temperature is — 30°C.
Compare your result with the rate of heat loss through an
equivalent surface area of the roof that has a common R-5.34
construction in SI units (i.e., a thickness-to-effective-
thermal-conductivity ratio of 5.34 m 2 • °C/W).
9-101 A solar collector consists of a horizontal copper tube
of outer diameter 5 cm enclosed in a concentric thin glass tube
of 9 cm diameter. Water is heated as it flows through the tube,
and the annular space between the copper and glass tube is
filled with air at 1 atm pressure. During a clear day, the tem-
peratures of the tube surface and the glass cover are measured
to be 60°C and 32°C, respectively. Determine the rate of heat
loss from the collector by natural convection per meter length
of the tube. Answer: 17.4 W
9 cm
Glass cover
FIGURE P9-101
9-102 A solar collector consists of a horizontal aluminum
tube of outer diameter 4 cm enclosed in a concentric thin
glass tube of 7 cm diameter. Water is heated as it flows
through the aluminum tube, and the annular space between
the aluminum and glass tubes is filled with air at 1 atm pres-
sure. The pump circulating the water fails during a clear day,
and the water temperature in the tube starts rising. The alu-
minum tube absorbs solar radiation at a rate of 20 W per me-
ter length, and the temperature of the ambient air outside is
30°C. Approximating the surfaces of the tube and the glass
cover as being black (emissivity e = 1) in radiation cal-
culations and taking the effective sky temperature to be 20°C,
determine the temperature of the aluminum tube when equi-
librium is established (i.e., when the net heat loss from the
tube by convection and radiation equals the amount of solar
energy absorbed by the tube).
9-103E The components of an electronic system dissipating
1 80 W are located in a 4-ft-long horizontal duct whose cross-
section is 6 in. X 6 in. The components in the duct are cooled
by forced air, which enters at 85 °F at a rate of 22 cfm and
leaves at 100°F. The surfaces of the sheet metal duct are not
painted, and thus radiation heat transfer from the outer surfaces
is negligible. If the ambient air temperature is 80°F, determine
(a) the heat transfer from the outer surfaces of the duct to the
ambient air by natural convection and (b) the average temper-
ature of the duct.
,100°F
FIGURE P9-103E
9-104E Repeat Problem 9-1 03E for a circular horizontal
duct of diameter 4 in.
9-105E Repeat Problem 9-103E assuming the fan fails and
thus the entire heat generated inside the duct must be rejected
to the ambient air by natural convection through the outer sur-
faces of the duct.
9-106 Consider a cold aluminum canned drink that is ini-
tially at a uniform temperature of 5°C. The can is 12.5 cm high
and has a diameter of 6 cm. The emissivity of the outer surface
of the can is 0.6. Disregarding any heat transfer from the bot-
tom surface of the can, determine how long it will take for the
average temperature of the drink to rise to 7°C if the surround-
ing air and surfaces are at 25°C. Answer-. 12.1 min
9-107 Consider a 2-m-high electric hot water heater that has
a diameter of 40 cm and maintains the hot water at 60°C. The
tank is located in a small room at 20°C whose walls and the
ceiling are at about the same temperature. The tank is placed in
a 46-cm-diameter sheet metal shell of negligible thickness, and
the space between the tank and the shell is filled with foam in-
sulation. The average temperature and emissivity of the outer
surface of the shell are 40°C and 0.7, respectively. The price of
3 cm
20°C
Foam
insulation
,'U I ; (
40 cm
T,„ = 60°C
2 m
FIGURE P9-1 07
cen58933_ch09.qxd 9/4/2002 12:26 PM Page 513
electricity is $0.08/kWh. Hot water tank insulation kits large
enough to wrap the entire tank are available on the market for
about $30. If such an insulation is installed on this water tank
by the home owner himself, how long will it take for this addi-
tional insulation to pay for itself? Disregard any heat loss from
the top and bottom surfaces, and assume the insulation to re-
duce the heat losses by 80 percent.
9-108 During a plant visit, it was observed that a 1 .5-m-high
and 1 -m-wide section of the vertical front section of a natural
gas furnace wall was too hot to touch. The temperature mea-
surements on the surface revealed that the average temperature
of the exposed hot surface was 1 10°C, while the temperature of
the surrounding air was 25 °C. The surface appeared to be oxi-
dized, and its emissivity can be taken to be 0.7. Taking the tem-
perature of the surrounding surfaces to be 25°C also, determine
the rate of heat loss from this furnace.
The furnace has an efficiency of 79 percent, and the plant
pays $0.75 per therm of natural gas. If the plant operates 10 h a
day, 310 days a year, and thus 3100 h a year, determine the an-
nual cost of the heat loss from this vertical hot surface on the
front section of the furnace wall.
FIGURE P9-1 08
9-109 A group of 25 power transistors, dissipating 1.5 W
each, are to be cooled by attaching them to a black-anodized
square aluminum plate and mounting the plate on the wall of a
room at 30°C. The emissivity of the transistor and the plate sur-
faces is 0.9. Assuming the heat transfer from the back side of
the plate to be negligible and the temperature of the surrounding
surfaces to be the same as the air temperature of the room, de-
termine the size of the plate if the average surface temperature
of the plate is not to exceed 50°C. Answer-. 43 cm 3 43 cm
9-110 Repeat Problem 9-109 assuming the plate to be posi-
tioned horizontally with (a) transistors facing up and (b) tran-
sistors facing down.
9-111E Hot water is flowing at an average velocity of 4 ft/s
through a cast iron pipe (k = 30 Btu/h • ft • °F) whose inner and
outer diameters are 1.0 in. and 1.2 in., respectively. The pipe
passes through a 50-ft-long section of a basement whose tem-
perature is 60°F. The emissivity of the outer surface of the pipe
Black-anodized
aluminum
plate
513
CHAPTER 9
Power
transistor, 1.5 W
FIGURE P9-1 09
is 0.5, and the walls of the basement are also at about 60°F. If
the inlet temperature of the water is 150°F and the heat transfer
coefficient on the inner surface of the pipe is 30 Btu/h ■ ft 2 • °F,
determine the temperature drop of water as it passes through
the basement.
9-112 Consider a flat-plate solar collector placed horizon-
tally on the flat roof of a house. The collector is 1.5 m wide
and 6 m long, and the average temperature of the exposed sur-
face of the collector is 42°C. Determine the rate of heat loss
from the collector by natural convection during a calm day
when the ambient air temperature is 15°C. Also, determine the
heat loss by radiation by taking the emissivity of the collector
surface to be 0.9 and the effective sky temperature to be
-30°C. Answers: 1295 W, 2921 W
9-113 Solar radiation is incident on the glass cover of a solar
collector at a rate of 650 W/m 2 . The glass transmits 88 percent
of the incident radiation and has an emissivity of 0.90. The hot
water needs of a family in summer can be met completely by a
FIGURE P9-1 13
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HEAT TRANSFER
collector 1.5 m high and 2 m wide, and tilted 40° from the hor-
izontal. The temperature of the glass cover is measured to be
40°C on a calm day when the surrounding air temperature is
20°C. The effective sky temperature for radiation exchange be-
tween the glass cover and the open sky is — 40°C. Water enters
the tubes attached to the absorber plate at a rate of 1 kg/min.
Assuming the back surface of the absorber plate to be heavily
insulated and the only heat loss occurs through the glass cover,
determine (a) the total rate of heat loss from the collector, (b)
the collector efficiency, which is the ratio of the amount of heat
transferred to the water to the solar energy incident on the col-
lector, and (c) the temperature rise of water as it flows through
the collector.
Design and Essay Problems
9-114 Write a computer program to evaluate the variation of
temperature with time of thin square metal plates that are re-
moved from an oven at a specified temperature and placed ver-
tically in a large room. The thickness, the size, the initial
temperature, the emissivity, and the thermophysical properties
of the plate as well as the room temperature are to be specified
by the user. The program should evaluate the temperature of
the plate at specified intervals and tabulate the results against
time. The computer should list the assumptions made during
calculations before printing the results.
For each step or time interval, assume the surface tempera-
ture to be constant and evaluate the heat loss during that time
interval and the temperature drop of the plate as a result of this
heat loss. This gives the temperature of the plate at the end of a
time interval, which is to serve as the initial temperature of the
plate for the beginning of the next time interval.
Try your program for 0.2-cm-thick vertical copper plates of
40 cm X 40 cm in size initially at 300°C cooled in a room at
25°C. Take the surface emissivity to be 0.9. Use a time interval
of 1 s in calculations, but print the results at 1 0-s intervals for a
total cooling period of 15 min.
9-115 Write a computer program to optimize the spacing be-
tween the two glasses of a double -pane window. Assume the
spacing is filled with dry air at atmospheric pressure. The pro-
gram should evaluate the recommended practical value of the
spacing to minimize the heat losses and list it when the size of
the window (the height and the width) and the temperatures of
the two glasses are specified.
9-116 Contact a manufacturer of aluminum heat sinks and
obtain their product catalog for cooling electronic components
by natural convection and radiation. Write an essay on how to
select a suitable heat sink for an electronic component when its
maximum power dissipation and maximum allowable surface
temperature are specified.
9-117 The top surfaces of practically all flat-plate solar col-
lectors are covered with glass in order to reduce the heat losses
from the absorber plate underneath. Although the glass cover
reflects or absorbs about 15 percent of the incident solar radia-
tion, it saves much more from the potential heat losses from the
absorber plate, and thus it is considered to be an essential part
of a well-designed solar collector. Inspired by the energy effi-
ciency of double-pane windows, someone proposes to use
double glazing on solar collectors instead of a single glass. In-
vestigate if this is a good idea for the town in which you live.
Use local weather data and base your conclusion on heat trans-
fer analysis and economic considerations.
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1
1
1
•
I CHAPTER
r
i
■
BOILING AND
CONDENSATION
■ ■ M e know from thermodynamics that when the temperature of a liquid
lfll at a specified pressure is raised to the saturation temperature T iM at
W W that pressure, boiling occurs. Likewise, when the temperature of a
vapor is lowered to T s . dt , condensation occurs. In this chapter we study the
rates of heat transfer during such liquid-to-vapor and vapor-to-liquid phase
transformations.
Although boiling and condensation exhibit some unique features, they are
considered to be forms of convection heat transfer since they involve fluid
motion (such as the rise of the bubbles to the top and the flow of condensate
to the bottom). Boiling and condensation differ from other forms of convec-
tion in that they depend on the latent heat of vaporization ht g of the fluid and
the surface tension u at the liquid-vapor interface, in addition to the proper-
ties of the fluid in each phase. Noting that under equilibrium conditions the
temperature remains constant during a phase-change process at a fixed pres-
sure, large amounts of heat (due to the large latent heat of vaporization re-
leased or absorbed) can be transferred during boiling and condensation
essentially at constant temperature. In practice, however, it is necessary to
maintain some difference between the surface temperature T s and T sat for ef-
_.
CONTENTS
10-1 Boiling Heat Transfer 516
10-2 Pool Boiling 518
10-3 Flow Boiling 530
10-4 Condensation
Heat Transfer 532
10-5 Film Condensation 532
10-6 Film Condensation Inside Hori-
zontal Tubes 545
10-7 Dropwise Condensation 545
Topic of Special Interest:
Heat Pipes 546
fective heat transfer. Heat transfer coefficients h associated with boiling and
condensation are typically much higher than those encountered in other forms
of convection processes that involve a single phase.
We start this chapter with a discussion of the boiling curve and the modes of
pool boiling such as free convection boiling, nucleate boiling, and film boil-
ing. We then discuss boiling in the presence of forced convection. In the
second part of this chapter, we describe the physical mechanism of film con-
densation and discuss condensation heat transfer in several geometrical
arrangements and orientations. Finally, we introduce dropwise condensation
and discuss ways of maintaining it.
515
^3
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HEAT TRANSFER
Evaporation
Ail-
Boiling
Heating
FIGURE 10-1
A liquid-to-vapor phase change
process is called evaporation if it
occurs at a liquid-vapor interface
and boiling if it occurs at a
solid-liquid interface.
P = 1 atm
Water Bubbles 11Q o C
T sa,= l 00 ° C / (
X
Heating element
<x$
FIGURE 10-2
Boiling occurs when a liquid is
brought into contact with a surface
at a temperature above the saturation
temperature of the liquid.
10-1 - BOILING HEAT TRANSFER
Many familiar engineering applications involve condensation and boiling heat
transfer. In a household refrigerator, for example, the refrigerant absorbs heat
from the refrigerated space by boiling in the evaporator section and rejects
heat to the kitchen air by condensing in the condenser section (the long coils
behind the refrigerator). Also, in steam power plants, heat is transferred to the
steam in the boiler where water is vaporized, and the waste heat is rejected
from the steam in the condenser where the steam is condensed. Some elec-
tronic components are cooled by boiling by immersing them in a fluid with an
appropriate boiling temperature.
Boiling is a liquid-to-vapor phase change process just like evaporation, but
there are significant differences between the two. Evaporation occurs at the
liquid-vapor interface when the vapor pressure is less than the saturation
pressure of the liquid at a given temperature. Water in a lake at 20°C, for
example, will evaporate to air at 20°C and 60 percent relative humidity since
the saturation pressure of water at 20°C is 2.3 kPa and the vapor pressure of
air at 20°C and 60 percent relative humidity is 1.4 kPa (evaporation rates are
determined in Chapter 14). Other examples of evaporation are the drying of
clothes, fruits, and vegetables; the evaporation of sweat to cool the human
body; and the rejection of waste heat in wet cooling towers. Note that evapo-
ration involves no bubble formation or bubble motion (Fig. 10-1).
Boiling, on the other hand, occurs at the solid-liquid interface when a liq-
uid is brought into contact with a surface maintained at a temperature T s suf-
ficiently above the saturation temperature T m of the liquid (Fig. 10-2). At 1
atm, for example, liquid water in contact with a solid surface at 110°C will
boil since the saturation temperature of water at 1 atm is 100°C. The boiling
process is characterized by the rapid formation of vapor bubbles at the
solid-liquid interface that detach from the surface when they reach a certain
size and attempt to rise to the free surface of the liquid. When cooking, we do
not say water is boiling until we see the bubbles rising to the top. Boiling is a
complicated phenomenon because of the large number of variables involved
in the process and the complex fluid motion patterns caused by the bubble for-
mation and growth.
As a form of convection heat transfer, the boiling heat flux from a solid
surface to the fluid is expressed from Newton's law of cooling as
(fboilina
HT S - T S J = htf es
(W/m 2 )
(10-1)
where AT.,
r sat is called the excess temperature, which represents
the excess of the surface above the saturation temperature of the fluid.
In the preceding chapters we considered forced and free convection heat
transfer involving a single phase of a fluid. The analysis of such convection
processes involves the thermophysical properties p, |x, k, and C p of the fluid.
The analysis of boiling heat transfer involves these properties of the liquid
(indicated by the subscript /) or vapor (indicated by the subscript v) as well as
the properties h fg (the latent heat of vaporization) and a (the surface tension).
The hf g represents the energy absorbed as a unit mass of liquid vaporizes
at a specified temperature or pressure and is the primary quantity of energy
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CHAPTER 10
transferred during boiling heat transfer. The hf g values of water at various tem-
peratures are given in Table A-9.
Bubbles owe their existence to the surface-tension cr at the liquid-vapor in-
terface due to the attraction force on molecules at the interface toward the liq-
uid phase. The surface tension decreases with increasing temperature and
becomes zero at the critical temperature. This explains why no bubbles are
formed during boiling at supercritical pressures and temperatures. Surface
tension has the unit N/m.
The boiling processes in practice do not occur under equilibrium conditions,
and normally the bubbles are not in thermodynamic equilibrium with the sur-
rounding liquid. That is, the temperature and pressure of the vapor in a bubble
are usually different than those of the liquid. The pressure difference between
the liquid and the vapor is balanced by the surface tension at the interface. The
temperature difference between the vapor in a bubble and the surrounding liq-
uid is the driving force for heat transfer between the two phases. When the liq-
uid is at a lower temperature than the bubble, heat will be transferred from the
bubble into the liquid, causing some of the vapor inside the bubble to con-
dense and the bubble to collapse eventually. When the liquid is at a higher
temperature than the bubble, heat will be transferred from the liquid to the
bubble, causing the bubble to grow and rise to the top under the influence of
buoyancy.
Boiling is classified as pool boiling or flow boiling, depending on the pres-
ence of bulk fluid motion (Fig. 10-3). Boiling is called pool boiling in the ab-
sence of bulk fluid flow and flow boiling {ox forced convection boiling) in the
presence of it. In pool boiling, the fluid is stationary, and any motion of the
fluid is due to natural convection currents and the motion of the bubbles un-
der the influence of buoyancy. The boiling of water in a pan on top of a stove
is an example of pool boiling. Pool boiling of a fluid can also be achieved by
placing a heating coil in the fluid. In flow boiling, the fluid is forced to move
in a heated pipe or over a surface by external means such as a pump. There-
fore, flow boiling is always accompanied by other convection effects.
Pool and flow boiling are further classified as subcooled boiling or satu-
rated boiling, depending on the bulk liquid temperature (Fig. 10-4). Boiling
is said to be subcooled (or local) when the temperature of the main body of
the liquid is below the saturation temperature r sat (i.e., the bulk of the liquid is
subcooled) and saturated (or bulk) when the temperature of the liquid is
equal to r sat (i.e., the bulk of the liquid is saturated). At the early stages of boil-
ing, the bubbles are confined to a narrow region near the hot surface. This is
because the liquid adjacent to the hot surface vaporizes as a result of being
heated above its saturation temperature. But these bubbles disappear soon af-
ter they move away from the hot surface as a result of heat transfer from the
bubbles to the cooler liquid surrounding them. This happens when the bulk of
the liquid is at a lower temperature than the saturation temperature. The bub-
bles serve as "energy movers" from the hot surface into the liquid body by ab-
sorbing heat from the hot surface and releasing it into the liquid as they
condense and collapse. Boiling in this case is confined to a region in the lo-
cality of the hot surface and is appropriately called local or subcooled boiling.
When the entire liquid body reaches the saturation temperature, the bubbles
start rising to the top. We can see bubbles throughout the bulk of the liquid,
r') ><0
It
Heating
(a) Pool boiling
t *
Heating
(b) Flow boiling
FIGURE 10-3
Classification of boiling on the basis
of the presence of bulk fluid motion.
P = 1 atm P = 1 atm
Subcooled 80°C
water
107°C
II
Saturated 100°C
water
107°C
II
llllll 1 Bubble J {J H I I
Heating Heating
(a) Subcooled boiling (b) Saturated boiling
FIGURE 10-4
Classification of boiling
on the basis of the presence of
bulk liquid temperature.
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518
HEAT TRANSFER
and boiling in this case is called the bulk or saturated boiling. Next, we con-
sider different boiling regimes in detail.
M^'J Mm)
100°C
-103°C
100°C
-110°C
Heating
(a) Natural convection
boiling
-Vapor pockets
Heating
(b) Nucleate boiling
<
>
/- Vapor mm
)i \ )
100°C
-180°C
Jl
100°C
400°C
Heating
(c) Transition boiling
FIGURE 10-5
Different boiling rej
in pool boiling.
Heating
(d) Film boiling
10-2 - POOL BOILING
So far we presented some general discussions on boiling. Now we turn our
attention to the physical mechanisms involved in pool boiling, that is, the
boiling of stationary fluids. In pool boiling, the fluid is not forced to flow
by a mover such as a pump, and any motion of the fluid is due to natural
convection currents and the motion of the bubbles under the influence of
buoyancy.
As a familiar example of pool boiling, consider the boiling of tap water in a
pan on top of a stove. The water will initially be at about 15°C, far below the
saturation temperature of 100°C at standard atmospheric pressure. At the early
stages of boiling, you will not notice anything significant except some bubbles
that stick to the surface of the pan. These bubbles are caused by the release of
air molecules dissolved in liquid water and should not be confused with vapor
bubbles. As the water temperature rises, you will notice chunks of liquid wa-
ter rolling up and down as a result of natural convection currents, followed by
the first vapor bubbles forming at the bottom surface of the pan. These bub-
bles get smaller as they detach from the surface and start rising, and eventu-
ally collapse in the cooler water above. This is subcooled boiling since the
bulk of the liquid water has not reached saturation temperature yet. The inten-
sity of bubble formation increases as the water temperature rises further, and
you will notice waves of vapor bubbles coming from the bottom and rising to
the top when the water temperature reaches the saturation temperature (100°C
at standard atmospheric conditions). This full scale boiling is the saturated
boiling.
Boiling Regimes and the Boiling Curve
Boiling is probably the most familiar form of heat transfer, yet it remains to be
the least understood form. After hundreds of papers written on the subject, we
still do not fully understand the process of bubble formation and we must still
rely on empirical or semi-empirical relations to predict the rate of boiling heat
transfer.
The pioneering work on boiling was done in 1934 by S. Nukiyama, who
used electrically heated nichrome and platinum wires immersed in liquids in
his experiments. Nukiyama noticed that boiling takes different forms, de-
pending on the value of the excess temperature Ar excess . Four different boiling
regimes are observed: natural convection boiling, nucleate boiling, transition
boiling, and film boiling (Fig. 10-5). These regimes are illustrated on the boil-
ing curve in Figure 10-6, which is a plot of boiling heat flux versus the ex-
cess temperature. Although the boiling curve given in this figure is for water,
the general shape of the boiling curve remains the same for different fluids.
The specific shape of the curve depends on the fluid-heating surface mate-
rial combination and the fluid pressure, but it is practically independent of
the geometry of the heating surface. We will describe each boiling regime
in detail.
cen58933_chl0.qxd 9/4/2002 12:38 PM Page 519
Natural convection
boiling
Nucleate
boiling
Transition
boiling
i r
h H
Bubbles
collapse
in the
liquid
Maximum
(critical)
Film
boiling
519
CHAPTER 10
FIGURE 10-6
Typical boiling curve for water
at 1 atm pressure.
Natural Convection Boiling (to Point A on the Boiling Curve)
We learned in thermodynamics that a pure substance at a specified pressure
starts boiling when it reaches the saturation temperature at that pressure. But
in practice we do not see any bubbles forming on the heating surface until the
liquid is heated a few degrees above the saturation temperature (about 2 to
6°C for water). Therefore, the liquid is slightly superheated in this case
(a metastable condition) and evaporates when it rises to the free surface.
The fluid motion in this mode of boiling is governed by natural convection
currents, and heat transfer from the heating surface to the fluid is by natural
convection.
Nucleate Boiling (between Points A and C)
The first bubbles start forming at point A of the boiling curve at various pref-
erential sites on the heating surface. The bubbles form at an increasing rate at
an increasing number of nucleation sites as we move along the boiling curve
toward point C.
The nucleate boiling regime can be separated into two distinct regions. In
region A-B, isolated bubbles are formed at various preferential nucleation
sites on the heated surface. But these bubbles are dissipated in the liquid
shortly after they separate from the surface. The space vacated by the rising
bubbles is filled by the liquid in the vicinity of the heater surface, and the
process is repeated. The stirring and agitation caused by the entrainment of the
liquid to the heater surface is primarily responsible for the increased heat
transfer coefficient and heat flux in this region of nucleate boiling.
In region B-C, the heater temperature is further increased, and bubbles form
at such great rates at such a large number of nucleation sites that they form
numerous continuous columns of vapor in the liquid. These bubbles move all
the way up to the free surface, where they break up and release their vapor
content. The large heat fluxes obtainable in this region are caused by the com-
bined effect of liquid entrainment and evaporation.
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HEAT TRANSFER
At large values of Ar excess , the rate of evaporation at the heater surface
reaches such high values that a large fraction of the heater surface is covered
by bubbles, making it difficult for the liquid to reach the heater surface and
wet it. Consequently, the heat flux increases at a lower rate with increasing
Ar excess , and reaches a maximum at point C. The heat flux at this point is
called the critical (or maximum) heat flux, q max . For water, the critical heat
flux exceeds 1 MW/m 2 .
Nucleate boiling is the most desirable boiling regime in practice because
high heat transfer rates can be achieved in this regime with relatively small
values of Ar excess , typically under 30°C for water. The photographs in Figure
10-7 show the nature of bubble formation and bubble motion associated with
nucleate, transition, and film boiling.
Transition Boiling (between Points Cand Don the
Boiling Curve)
As the heater temperature and thus the AT excess is increased past point C, the
heat flux decreases, as shown in Figure 10-6. This is because a large fraction
of the heater surface is covered by a vapor film, which acts as an insulation
due to the low thermal conductivity of the vapor relative to that of the liquid.
In the transition boiling regime, both nucleate and film boiling partially occur.
Nucleate boiling at point C is completely replaced by film boiling at point D.
Operation in the transition boiling regime, which is also called the unstable
film boiling regime, is avoided in practice. For water, transition boiling occurs
over the excess temperature range from about 30°C to about 120°C.
Sudden drop
in temperature
Sudden jump
in temperature
^--Bypassed
\ part of the
^ boiling //
* curve//
10
at
100 1000
: T - T ., °C
FIGURE 10-8
The actual boiling curve obtained
with heated platinum wire in water
as the heat flux is increased and
then decreased.
Film Boiling (beyond Point D)
In this region the heater surface is completely covered by a continuous stable
vapor film. Point D, where the heat flux reaches a minimum, is called the
Leidenfrost point, in honor of J. C. Leidenfrost, who observed in 1756 that
liquid droplets on a very hot surface jump around and slowly boil away. The
presence of a vapor film between the heater surface and the liquid is responsi-
ble for the low heat transfer rates in the film boiling region. The heat transfer
rate increases with increasing excess temperature as a result of heat transfer
from the heated surface to the liquid through the vapor film by radiation,
which becomes significant at high temperatures.
A typical boiling process will not follow the boiling curve beyond point C,
as Nukiyama has observed during his experiments. Nukiyama noticed, with
surprise, that when the power applied to the nichrome wire immersed in wa-
ter exceeded ^ax even slightly, the wire temperature increased suddenly to the
melting point of the wire and burnout occurred beyond his control. When he
repeated the experiments with platinum wire, which has a much higher melt-
ing point, he was able to avoid burnout and maintain heat fluxes higher than
4nax- When he gradually reduced power, he obtained the cooling curve shown
in Figure 10-8 with a sudden drop in excess temperature when q min is reached.
Note that the boiling process cannot follow the transition boiling part of the
boiling curve past point C unless the power applied is reduced suddenly.
The burnout phenomenon in boiling can be explained as follows: In order to
move beyond point C where q max occurs, we must increase the heater surface
temperature T s . To increase T s , however, we must increase the heat flux. But
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CHAPTER 10
(a)
(b)
(c)
FIGURE 10-7
Various boiling regimes during
boiling of methanol on a horizontal
1 -cm-diameter steam -heated
copper tube: (a) nucleate boiling,
(b) transition boiling, and (c) film
boiling (from J. W. Westwater and
J. G. Santangelo, University of
Illinois at Champaign-Urbana).
the fluid cannot receive this increased energy at an excess temperature just be-
yond point C. Therefore, the heater surface ends up absorbing the increased
energy, causing the heater surface temperature T s to rise. But the fluid can re-
ceive even less energy at this increased excess temperature, causing the heater
surface temperature T s to rise even further. This continues until the surface
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HEAT TRANSFER
1
W
nr
r °c
melting
FIGURE 10-9
An attempt to increase the boiling heat
flux beyond the critical value often
causes the temperature of the heating
element to jump suddenly to a value
that is above the melting point,
resulting in burnout.
temperature reaches a point at which it no longer rises and the heat supplied
can be transferred to the fluid steadily. This is point E on the boiling curve,
which corresponds to very high surface temperatures. Therefore, any attempt
to increase the heat flux beyond q max will cause the operation point on the boil-
ing curve to jump suddenly from point C to point E. However, surface tem-
perature that corresponds to point E is beyond the melting point of most heater
materials, and burnout occurs. Therefore, point C on the boiling curve is also
called the burnout point, and the heat flux at this point the burnout heat flux
(Fig. 10-9).
Most boiling heat transfer equipment in practice operate slightly below q m!lx
to avoid any disastrous burnout. However, in cryogenic applications involving
fluids with very low boiling points such as oxygen and nitrogen, point E usu-
ally falls below the melting point of the heater materials, and steady film boil-
ing can be used in those cases without any danger of burnout.
Heat Transfer Correlations in Pool Boiling
Boiling regimes discussed above differ considerably in their character, and
thus different heat transfer relations need to be used for different boiling
regimes. In the natural convection boiling regime, boiling is governed by nat-
ural convection currents, and heat transfer rates in this case can be determined
accurately using natural convection relations presented in Chapter 9.
Nucleate Boiling
In the nucleate boiling regime, the rate of heat transfer strongly depends on
the nature of nucleation (the number of active nucleation sites on the surface,
the rate of bubble formation at each site, etc.), which is difficult to predict.
The type and the condition of the heated surface also affect the heat transfer.
These complications made it difficult to develop theoretical relations for heat
transfer in the nucleate boiling regime, and people had to rely on relations
based on experimental data. The most widely used correlation for the rate of
heat transfer in the nucleate boiling regime was proposed in 1952 by
Rohsenow, and expressed as
Enucleate P'/ ^
'ft
g(Pl ~ Pv)
^p\l s * sat/
C sf h fg Pfl
(10-2)
where
1 nucleate
P</
Pi
P,
(T
C p i
T s
T
1 sat
Csf
Pr,
nucleate boiling heat flux, W/m 2
viscosity of the liquid, kg/m • s
enthalpy of vaporization, J/kg
gravitational acceleration, m/s 2
density of the liquid, kg/m 3
density of the vapor, kg/m 3
surface tension of liquid-vapor interface, N/m
specific heat of the liquid, J/kg • °C
surface temperature of the heater, °C
saturation temperature of the fluid, °C
experimental constant that depends on surface-fluid combination
Prandtl number of the liquid
experimental constant that depends on the fluid
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CHAPTER 10
It can be shown easily that using property values in the specified units in
the Rohsenow equation produces the desired unit W/m 2 for the boiling heat
flux, thus saving one from having to go through tedious unit manipulations
(Fig. 10-10).
The surface tension at the vapor-liquid interface is given in Table 10-1 for
water, and Table 10-2 for some other fluids. Experimentally determined val-
ues of the constant C^are given in Table 10-3 for various fluid-surface com-
binations. These values can be used for any geometry since it is found that the
rate of heat transfer during nucleate boiling is essentially independent of the
geometry and orientation of the heated surface. The fluid properties in Eq.
10-2 are to be evaluated at the saturation temperature r sat .
The condition of the heater surface greatly affects heat transfer, and the
Rohsenow equation given above is applicable to clean and relatively smooth
surfaces. The results obtained using the Rohsenow equation can be in error by
± 100% for the heat transfer rate for a given excess temperature and by ±30%
for the excess temperature for a given heat transfer rate. Therefore, care
should be exercised in the interpretation of the results.
Recall from thermodynamics that the enthalpy of vaporization h fg of a pure
substance decreases with increasing pressure (or temperature) and reaches
zero at the critical point. Noting that h fg appears in the denominator of the
Rohsenow equation, we should see a significant rise in the rate of heat trans-
fer at high pressures during nucleate boiling.
Peak Heat Flux
In the design of boiling heat transfer equipment, it is extremely important for
the designer to have a knowledge of the maximum heat flux in order to avoid
the danger of burnout. The maximum (or critical) heat flux in nucleate pool
boiling was determined theoretically by S. S. Kutateladze in Russia in 1948
and N. Zuber in the United States in 1958 using quite different approaches,
and is expressed as (Fig. 10-11)
C„ h f [cTgpUp, ~ p v )V
(10-3)
where C cr is a constant whose value depends on the heater geometry. Exhaus-
tive experimental studies by Lienhard and his coworkers indicated that the
value of C cr is about 0.15. Specific values of C cr for different heater geome-
tries are listed in Table 10-4. Note that the heaters are classified as being large
or small based on the value of the parameter L*.
Equation 10-3 will give the maximum heat flux in W/m 2 if the properties
are used in the units specified earlier in their descriptions following Eq. 10-2.
The maximum heat flux is independent of the fluid-heating surface combina-
tion, as well as the viscosity, thermal conductivity, and the specific heat of the
liquid.
Note that p v increases but u and h fg decrease with increasing pressure, and
thus the change in q max with pressure depends on which effect dominates. The
experimental studies of Cichelli and Bonilla indicate that q max increases with
pressure up to about one-third of the critical pressure, and then starts to de-
crease and becomes zero at the critical pressure. Also note that q max is propor-
tional to hf g , and large maximum heat fluxes can be obtained using fluids with
a large enthalpy of vaporization, such as water.
H^)fe)
X
s 2 m 3
N
\ m 1
1/2
/ J °c\
kg- °c
J
\ k § 1
3
-££)"<*
= W/m 2
FIGURE 10-10
Equation 10-2 gives the
boiling heat flux in W/m 2 when
the quantities are expressed in the
units specified in their descriptions.
TABLE 10-1
Surface tension of liquid-vapor
interface for water
r, °C
a, N/m*
0.0757
20
0.0727
40
0.0696
60
0.0662
80
0.0627
100
0.0589
120
0.0550
140
0.0509
160
0.0466
180
0.0422
200
0,0377
220
0.0331
240
0.0284
260
0.0237
280
0.0190
300
0.0144
320
0.0099
340
0.0056
360
0.0019
374
0.0
♦Multiply by 0.06852 to convert to Ibf/ft or by
2.2046 to convert to lbm/s 2 .
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HEAT TRANSFER
TABLE 10-2
Surface tension of some fluids (from
Suryanarayana, Ref. 26; originally
based on data from Jasper, Ref. 14)
Substance
and Temp.
Surface Tension,
Range
a, N/m* (Tin °C)
Ammonia, -75 to -40°C:
0.0264 + 0.0002237
Benzene, 10 to 80°C:
0.0315- 0.0001297
Butane, -70 to-20°C:
0.0149- 0.0001217
Carbon dioxide, -30 to -20°C:
0.0043- 0.000160 7"
Ethyl alcohol, 10 to 70°C:
0.0241 - 0.000083 7"
Mercury, 5 to 200°C:
0.4906- 0.0002057
Methyl alcohol, 10 to 60°C:
0.0240- 0.0000777
Pentane, 10 to 30°C:
0.0183- 0.0001107
Propane, -90 to-10°C:
0.0092- 0.0000877
•Multiply by 0.06852 to convert to Ibf/ft or by
2.2046 to convert to lbm/s 2 .
TABLE 10-3
Values of the coefficient C sf and n for various fluid-surface combinations
Fluid-Heating Surface Combination C sf
n
Water-copper (polished)
Water-copper (scored)
Water-stainless steel (mechanically polished)
Water-stainless steel (ground and polished)
Water-stainless steel (teflon pitted)
Water-stainless steel (chemically etched)
Water-brass
Water-nickel
Water-platinum
n-Pentane-copper (polished)
n-Pentane-chromium
Benzene-chromium
Ethyl alcohol-chromium
Carbon tetrachloride-copper
Isopropanol-copper
0.0130
1.0
0.0068
1.0
0.0130
1.0
0.0060
1.0
0.0058
1.0
0.0130
1.0
0.0060
1.0
0.0060
1.0
0.0130
1.0
0.0154
1.7
0.0150
1.7
0.1010
1.7
0.0027
1.7
0.0130
1.7
0.0025
1.7
Minimum Heat Flux
Minimum heat flux, which occurs at the Leidenfrost point, is of practical in-
terest since it represents the lower limit for the heat flux in the film boiling
regime. Using the stability theory, Zuber derived the following expression for
the minimum heat flux for a large horizontal plate,
<7n
0.09 Pv h fg
vg(Pi ~ Pv)
(p< + Pv)
(10-4)
where the constant 0.09 was determined by Berenson in 1961. He replaced the
theoretically determined value of ^ by 0.09 to match the experimental data
better. Still, the relation above can be in error by 50 percent or more.
TABLE 10-4
Values of the coefficient C„for use in
Eq. 10-3 for maximum heat flux
(dimensionless parameter L* = Llgip,
- Pv )/<tF 2 )
Charac.
Dimension
Heater Geometry C cr
of Heater, L Range of L*
Large horizontal flat heater 0.149 Width or diameter
Small horizontal flat heater 1 18.9K 1 Width or diameter
Large horizontal cylinder 0.12 Radius
Small horizontal cylinder 0.12Z.*~ 025 Radius
Large sphere 0.11 Radius
Small sphere
0.227/.*
Radius
L* > 27
< L* < 20
L* > 1.2
0.15 < L* < 1.2
L* > 4.26
0.15 < L* < 4.26
% =o7[g(p, - p„)/W er ]
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CHAPTER 10
Film Boiling
Using an analysis similar to Nusselt's theory on filmwise condensation pre-
sented in the next section, Bromley developed a theory for the prediction of
heat flux for stable film boiling on the outside of a horizontal cylinder. The
heat flux for film boiling on a horizontal cylinder or sphere of diameter D is
given by
<7film ~~ Qi
gkl p v (p, - p v )[h f + 0.4C (T s - 7 sat )
^v D(T S - r sat )
(T s -T mt ) (10-5)
where k v is the thermal conductivity of the vapor in W/m • °C and
„ J0.62 for horizontal cylinders
fllm ~ (0.67 for spheres
Other properties are as listed before in connection with Eq. 10-2. We used a
modified latent heat of vaporization in Eq. 10-5 to account for the heat trans-
fer associated with the superheating of the vapor.
The vapor properties are to be evaluated at the film temperature, given as
Tj = (T s + T sM )/2, which is the average temperature of the vapor film. The
liquid properties and h fg are to be evaluated at the saturation temperature at the
specified pressure. Again, this relation will give the film boiling heat flux in
W/m 2 if the properties are used in the units specified earlier in their descrip-
tions following Eq. 10-2.
At high surface temperatures (typically above 300°C), heat transfer across
the vapor film by radiation becomes significant and needs to be considered
(Fig. 10-12). Treating the vapor film as a transparent medium sandwiched be-
tween two large parallel plates and approximating the liquid as a blackbody,
radiation heat transfer can be determined from
q mi = 8ct (7/ 4 - r 4 ,)
(10-6)
where e is the emissivity of the heating surface and a = 5.67 X
10~ 8 W/m 2 • K 4 is the Stefan-Boltzman constant. Note that the temperature in
this case must be expressed in K, not °C, and that surface tension and the
Stefan-Boltzman constant share the same symbol.
You may be tempted to simply add the convection and radiation heat trans-
fers to determine the total heat transfer during film boiling. However, these
two mechanisms of heat transfer adversely affect each other, causing the total
heat transfer to be less than their sum. For example, the radiation heat transfer
from the surface to the liquid enhances the rate of evaporation, and thus
the thickness of the vapor film, which impedes convection heat transfer. For
4ad < '/film' Bromley determined that the relation
T -T ,
s sat
FIGURE 10-11
Different relations are
used to determine the heat
flux in different boiling regimes.
P = 1 atm
<7filmboili„g J 1 J 1 J 1 J 1 J 1 J ^ad
Heating
FIGURE 10-12
At high heater surface temperatures,
radiation heat transfer becomes
significant during film boiling.
<7total ~~ <7filn
(10-7)
correlates experimental data well.
Operation in the transition boiling regime is normally avoided in the design
of heat transfer equipment, and thus no major attempt has been made to de-
velop general correlations for boiling heat transfer in this regime.
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HEAT TRANSFER
FIGURE 10-13
The cavities on a rough surface act as
nucleation sites and enhance
boiling heat transfer.
P = 1 atm
A f.U )/)
Heating
FIGURE 10-15
Schematic for Example 10-1.
Note that the gravitational acceleration g, whose value is approximately
9.81 m/s 2 at sea level, appears in all of the relations above for boiling heat
transfer. The effects of low and high gravity (as encountered in aerospace
applications and turbomachinery) are studied experimentally. The studies
confirm that the critical heat flux and heat flux in film boiling are proportional
to g 114 . However, they indicate that heat flux in nucleate boiling is practically
independent of gravity g, instead of being proportional to g 112 , as dictated by
Eq. 10-2.
Enhancement of Heat Transfer in Pool Boiling
The pool boiling heat transfer relations given above apply to smooth surfaces.
Below we will discuss some methods to enhance heat transfer in pool boiling.
We pointed out earlier that the rate of heat transfer in the nucleate boiling
regime strongly depends on the number of active nucleation sites on the sur-
face, and the rate of bubble formation at each site. Therefore, any modifica-
tion that will enhance nucleation on the heating surface will also enhance heat
transfer in nucleate boiling. It is observed that irregularities on the heating
surface, including roughness and dirt, serve as additional nucleation sites dur-
ing boiling, as shown in Figure 10-13. For example, the first bubbles in a pan
filled with water are most likely to form at the scratches at the bottom surface.
These scratches act like "nests" for the bubbles to form and thus increase the
rate of bubble formation. Berensen has shown that heat flux in the nucleate
boiling regime can be increased by a factor of 10 by roughening the heating
surface. However, these high heat transfer rates cannot be sustained for long
since the effect of surface roughness is observed to decay with time, and the
heat flux to drop eventually to values encountered on smooth surfaces. The ef-
fect of surface roughness is negligible on the critical heat flux and the heat
flux in film boiling.
Surfaces that provide enhanced heat transfer in nucleate boiling perma-
nently are being manufactured and are available in the market. Enhancement
in nucleation and thus heat transfer in such special surfaces is achieved either
by coating the surface with a thin layer (much less than 1 mm) of very porous
material or by forming cavities on the surface mechanically to facilitate con-
tinuous vapor formation. Such surfaces are reported to enhance heat transfer
in the nucleate boiling regime by a factor of up to 10, and the critical heat flux
by a factor of 3. The enhancement provided by one such material prepared by
machine roughening, the thermoexcel-E, is shown in Figure 10-14. The use
of finned surfaces is also known to enhance nucleate boiling heat transfer and
the critical heat flux.
Boiling heat transfer can also be enhanced by other techniques such as me-
chanical agitation and surface vibration. These techniques are not practical,
however, because of the complications involved.
EXAMPLE 10-1 Nucleate Boiling of Water in a Pan
Water is to be boiled at atmospheric pressure in a mechanically polished stain-
less steel pan placed on top of a heating unit, as shown in Figure 10-15. The
inner surface of the bottom of the pan is maintained at 108°C. If the diameter
of the bottom of the pan is 30 cm, determine (a) the rate of heat transfer to the
water and (b) the rate of evaporation of water.
cen58933_chl0.qxd 9/4/2002 12:38 PM Page 527
10 5
sat
= o°c
$/
$/
J
V
7/
§7
zt /
10 4
1 '
527
CHAPTER 10
Vapor
Tunnel
0.5 1 2 5 10
(T-TJCC)
FIGURE 10-14
The enhancement of boiling heat
transfer in Freon-12 by a mechanically
roughened surface, thermoexcel-E.
SOLUTION Water is boiled at 1 atm pressure on a stainless steel surface. The
rate of heat transfer to the water and the rate of evaporation of water are to be
determined.
Assumptions 1 Steady operating conditions exist. 2 Heat losses from the heater
and the pan are negligible.
Properties The properties of water at the saturation temperature of 100°C are
a = 0.0589 N/m (Table 10-1) and, from Table A-9,
p, = 957.9 kg/m 3
p,. = 0.6 kg/m 3
Pr, = 1.75
h f g
2257.0 X 10 3 J/kg
0.282 X 10- 3 kg • m/s
C p i = 4217 J/kg- °C
Also, C sf = 0.0130 and n = 1.0 for the boiling of water on a mechanically pol-
ished stainless steel surface (Table 10-3). Note that we expressed the proper-
ties in units specified under Eq. 10-2 in connection with their definitions in
order to avoid unit manipulations.
Analysis (a) The excess temperature in this case is A 7" = T s — 7" sat =
108 - 100 = 8°C which is relatively low (less than 30°C). Therefore, nucleate
boiling will occur. The heat flux in this case can be determined from the
Rohsenow relation to be
Mis
g(Pl ~ Pv)
t-'pl \* s -*sat)
L QA ft Pr'/ j
(0.282 X 10- 3 )(2257 X 10 3 )
4217(108 - 100)
9.81 X (957.9 - 0.6)
0.0589
\0.0130(2257 X 10 3 )1.75
7.20 X 10 4 W/m 2
The surface area of the bottom of the pan is
A = ttD 2 /4 = tt(0.3 m) 2 /4 = 0.07069 m 2
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HEAT TRANSFER
Then
the rate of heat transfer during nucleate boi
ling becomes
*~ boiling
= Aq mdeate = (0.07069 m 2 )(7.20 X
10 4 W/m 2 ) =
5093 W
(b) The rate of
evaporation of water is determined from
*"eva
Q boiling 5093 J/s
= 2.26 X 10"
3 kg/s
h fg 2257 X 10 3 J/kg
That
is, water
in the pan will boil at a rate of more than 2 grams per second.
P = 1 atm
FIGURE 10-16
Schematic for Example 10-2.
EXAMPLE 10-2 Peak Heat Flux in Nucleate Boiling
Water in a tank is to be boiled at sea level by a 1-cm-diameter nickel plated
steel heating element equipped with electrical resistance wires inside, as shown
in Figure 10-16. Determine the maximum heat flux that can be attained in the
nucleate boiling regime and the surface temperature of the heater surface in
that case.
SOLUTION Water is boiled at 1 atm pressure on a nickel plated steel sur-
face. The maximum (critical) heat flux and the surface temperature are to be
determined.
Assumptions 1 Steady operating conditions exist. 2 Heat losses from the boiler
are negligible.
Properties The properties of water at the saturation temperature of 100°C are
a = 0.0589 N/m (Table 10-1) and, from Table A-9,
Pi
= 957.9 kg/m 3
h h = 2257 X 10 3 J/kg
Pv
= 0.6 kg/m 3
jjl, = 0.282 X 10~ 3 kg • m/s
Pr,
= 1.75
C i = 4217 J/kg- °C
Also, C sf = 0.0060 and n = 1.0 for the boiling of water on a nickel plated sur-
face (Table 10-3). Note that we expressed the properties in units specified
under Eqs. 10-2 and 10-3 in connection with their definitions in order to avoid
unit manipulations.
Analysis The heating element in this case can be considered to be a short
cylinder whose characteristic dimension is its radius. That is, L = r = 0.005 m.
The dimensionless parameter L* and the constant C cr are determined from
Table 10-4 to be
L*
g(Pl ~ Pv)
/ (9.81)(957.8 - 0.6)\" 2
5) l O0589 >
which corresponds to C cr = 0.12.
Then the maximum or critical heat flux is determined from Eq. 10-3 to be
?max = C cr h fg [vgpl (p, - p v )]" 4
= 0.12(2257 X 10 3 )[0.0589 X 9.8 X (0.6) 2 (957.9 - 0.6)]" 4
= 1.02 X 10 6 W/m 2
cen58933_chl0.qxd 9/4/2002 12:38 PM Page 529
The Rohsenow relation, which gives the nucleate boiling heat flux for a spec-
ified surface temperature, can also be used to determine the surface tempera-
ture when the heat flux is given. Substituting the maximum heat flux into Eq.
10-2 together with other properties gives
M/<
g(Pi ~ Pv)
^pl \* s * sat/
C sf h fs W{
1,017,200 = (0.282 X 10- 3 )(2257 X 10 3 )
4217(7; - 100)
9.81(957.9 - 0.6)
0.0589
_0.0130(2257 X 10 3 ) 1.75_
T, = 119°C
Discussion Note that heat fluxes on the order of 1 MW/m 2 can be obtained in
nucleate boiling with a temperature difference of less than 20°C.
529
CHAPTER 10
EXAMPLE 10-3 Film Boiling of Water on a Heating Element
Water is boiled at atmospheric pressure by a horizontal polished copper heating
element of diameter D = 5 mm and emissivity e = 0.05 immersed in water, as
shown in Figure 10-17. If the surface temperature of the heating wire is
350°C, determine the rate of heat transfer from the wire to the water per unit
length of the wire.
SOLUTION Water is boiled at 1 atm by a horizontal polished copper heating
element. The rate of heat transfer to the water per unit length of the heater is to
be determined.
Assumptions 1 Steady operating conditions exist. 2 Heat losses from the boiler
are negligible.
Properties The properties of water at the saturation temperature of 100°C are
h fg = 2257 X 10 3 J/kg and p, = 957.9 kg/m 3 (Table A-9). The properties of va-
por at the film temperature of T,= (7" sat + T S )I2 = (100 + 350)/2 = 225°C =
498 K (which is sufficiently close to 500 K) are, from Table A-16,
p v = 0.441 kg/m 3
C pv = 1911 J/kg • °C
jjl v = 1.73 X 10- 5 kg/m-s
k v = 0.0357 W/m • °C
Note that we expressed the properties in units that will cancel each other in
boiling heat transfer relations. Also note that we used vapor properties at 1 atm
pressure from Table A-16 instead of the properties of saturated vapor from Table
A-9 at 250°C since the latter are at the saturation pressure of 4.0 MPa.
Analysis The excess temperature in this case is A 7" = T s — F sat =
350 - 100 = 250°C, which is much larger than 30°C for water. Therefore, film
boiling will occur. The film boiling heat flux in this case can be determined from
Eq. 10-5 to be
P = 1 atm
{ \ \\ S > ) { >
100°C
i^^v^v^v^ W
r
Heating
element
Vapor
film
FIGURE 10-17
Schematic for Example 10-3.
cen58933_chl0.qxd 9/4/2002 12:38 PM Page 53C
530
HEAT TRANSFER
High
velocity
x N Low
velocity
Nucleate pool
boiling regime
AT
ex
FIGURE 10-18
The effect of forced convection on
external flow boiling for different
flow velocities.
<7fiin
0.62
0.62
gkl Pv (p, - 9v )[h fg + 0AC pv (T, - TJ)]
V- v D(T S - r sat )
9.81(0.0357) 3 (0.441)(957.9 - 0.441)
[(2257 X 10 3 + 0.4 X 1977(250)]
(T s ~ r sat )
X 250
(1.73 X 10- 5 )(5 X 10- 3 )(250)
= 5.93 X 10 4 W/m 2
The radiation heat flux is determined from Eq. 10-6 to be
?md = eo- (r 4 - r 4 t )
= (0.05)(5.67 X lO" 8 W/m 2 ■ K 4 )[(250 + 273 K) 4 - (100 + 273 K) 4 ]
= 157 W/m 2
Note that heat transfer by radiation is negligible in this case because of the low
emissivity of the surface and the relatively low surface temperature of the heat-
ing element. Then the total heat flux becomes (Eq. 10-7)
<7total — <7filn
5.93 X 10 4 + 7 X 157
4
5.94 X 10 4 W/m 2
Finally, the rate of heat transfer from the heating element to the water is deter-
mined by multiplying the heat flux by the heat transfer surface area,
2 total = Motai = i^DL)q lotA
= (it X 0.005 m X 1 m)(5.94 X 10 4 W/m 2 )
= 933W
Discussion Note that the 5-mm-diameter copper heating element will consume
about 1 kW of electric power per unit length in steady operation in the film boil-
ing regime. This energy is transferred to the water through the vapor film that
forms around the wire.
10-3 - FLOW BOILING
The pool boiling we considered so far involves a pool of seemingly motion-
less liquid, with vapor bubbles rising to the top as a result of buoyancy effects.
In flow boiling, the fluid is forced to move by an external source such as a
pump as it undergoes a phase-change process. The boiling in this case exhibits
the combined effects of convection and pool boiling. The flow boiling is also
classified as either external and internal flow boiling depending on whether
the fluid is forced to flow over a heated surface or inside a heated tube.
External flow boiling over a plate or cylinder is similar to pool boiling, but
the added motion increases both the nucleate boiling heat flux and the critical
heat flux considerably, as shown in Figure 10-18. Note that the higher the ve-
locity, the higher the nucleate boiling heat flux and the critical heat flux. In ex-
periments with water, critical heat flux values as high as 35 MW/m 2 have been
obtained (compare this to the pool boiling value of 1.3 MW/m 2 at 1 arm pres-
sure) by increasing the fluid velocity.
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531
CHAPTER 10
Internal flow boiling is much more complicated in nature because there is
no free surface for the vapor to escape, and thus both the liquid and the vapor
are forced to flow together. The two-phase flow in a tube exhibits different
flow boiling regimes, depending on the relative amounts of the liquid and the
vapor phases. This complicates the analysis even further.
The different stages encountered in flow boiling in a heated tube are illus-
trated in Figure 10-19 together with the variation of the heat transfer coeffi-
cient along the tube. Initially, the liquid is subcooled and heat transfer to the
liquid is by forced convection. Then bubbles start forming on the inner sur-
faces of the tube, and the detached bubbles are drafted into the mainstream.
This gives the fluid flow a bubbly appearance, and thus the name bubbly flow
regime. As the fluid is heated further, the bubbles grow in size and eventually
coalesce into slugs of vapor. Up to half of the volume in the tube in this slug-
flow regime is occupied by vapor. After a while the core of the flow consists
of vapor only, and the liquid is confined only in the annular space between the
vapor core and the tube walls. This is the annular-flow regime, and very high
heat transfer coefficients are realized in this regime. As the heating continues,
the annular liquid layer gets thinner and thinner, and eventually dry spots start
to appear on the inner surfaces of the tube. The appearance of dry spots is ac-
companied by a sharp decrease in the heat transfer coefficient. This transition
regime continues until the inner surface of the tube is completely dry. Any liq-
uid at this moment is in the form of droplets suspended in the vapor core,
which resembles a mist, and we have a mist-flow regime until all the liquid
droplets are vaporized. At the end of the mist-flow regime we have saturated
vapor, which becomes superheated with any further heat transfer.
Note that the tube contains a liquid before the bubbly flow regime and a
vapor after the mist-flow regime. Heat transfer in those two cases can be
determined using the appropriate relations for single-phase convection heat
transfer. Many correlations are proposed for the determination of heat transfer
Low
Forced convection
Mist flow
Transition flow
Annular flow
Slug flow
Bubbly flow
Forced convection
Coefficient of heat transfer
FIGURE 10-19
Different flow regimes
encountered in flow boiling
in a tube under forced convection.
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532
HEAT TRANSFER
Droplets
Liquid film
(a) Film
condensation
(£>) Dropwise
condensation
FIGURE 10-20
When a vapor is exposed to a
surface at a temperature below r sat ,
condensation in the form of a liquid
film or individual droplets occurs
on the surface.
Cold
plate
FIGURE 10-21
Film condensation on a vertical plate.
in the two-phase flow (bubbly flow, slug-flow, annular-flow, and mist-flow)
cases, but they are beyond the scope of this introductory text. A crude estimate
for heat flux in flow boiling can be obtained by simply adding the forced con-
vection and pool boiling heat fluxes.
1(ML - CONDENSATION HEAT TRANSFER
Condensation occurs when the temperature of a vapor is reduced below its
saturation temperature r sat . This is usually done by bringing the vapor into
contact with a solid surface whose temperature T s is below the saturation tem-
perature r sat of the vapor. But condensation can also occur on the free surface
of a liquid or even in a gas when the temperature of the liquid or the gas to
which the vapor is exposed is below r sat . In the latter case, the liquid droplets
suspended in the gas form a fog. In this chapter, we will consider condensa-
tion on solid surfaces only.
Two distinct forms of condensation are observed: film condensation and
dropwise condensation. In film condensation, the condensate wets the sur-
face and forms a liquid film on the surface that slides down under the influ-
ence of gravity. The thickness of the liquid film increases in the flow direction
as more vapor condenses on the film. This is how condensation normally oc-
curs in practice. In dropwise condensation, the condensed vapor forms
droplets on the surface instead of a continuous film, and the surface is covered
by countless droplets of varying diameters (Fig. 10-20).
In film condensation, the surface is blanketed by a liquid film of increasing
thickness, and this "liquid wall" between solid surface and the vapor serves as
a resistance to heat transfer. The heat of vaporization h fg released as the vapor
condenses must pass through this resistance before it can reach the solid sur-
face and be transferred to the medium on the other side. In dropwise conden-
sation, however, the droplets slide down when they reach a certain size,
clearing the surface and exposing it to vapor. There is no liquid film in this
case to resist heat transfer. As a result, heat transfer rates that are more than
10 times larger than those associated with film condensation can be achieved
with dropwise condensation. Therefore, dropwise condensation is the pre-
ferred mode of condensation in heat transfer applications, and people have
long tried to achieve sustained dropwise condensation by using various vapor
additives and surface coatings. These attempts have not been very successful,
however, since the dropwise condensation achieved did not last long and con-
verted to film condensation after some time. Therefore, it is common practice
to be conservative and assume film condensation in the design of heat trans-
fer equipment.
10-5 - FILM CONDENSATION
We now consider film condensation on a vertical plate, as shown in Figure
10-21. The liquid film starts forming at the top of the plate and flows down-
ward under the influence of gravity. The thickness of the film 8 increases in
the flow direction x because of continued condensation at the liquid-vapor in-
terface. Heat in the amount hf g (the latent heat of vaporization) is released dur-
ing condensation and is transferred through the film to the plate surface at
temperature T s . Note that T s must be below the saturation temperature r sat of
the vapor for condensation to occur.
cen58933_chl0.qxd 9/4/2002 12:38 PM Page 533
Typical velocity and temperature profiles of the condensate are also given
in Figure 10-21. Note that the velocity of the condensate at the wall is zero be-
cause of the "no-slip" condition and reaches a maximum at the liquid-vapor
interface. The temperature of the condensate is T sM at the interface and de-
creases gradually to T s at the wall.
As was the case in forced convection involving a single phase, heat transfer
in condensation also depends on whether the condensate flow is laminar or
turbulent. Again the criterion for the flow regime is provided by the Reynolds
number, which is defined as
Re
D h p l Y l _ 4A c p l V l _ 4p l V l b _ 4rh
V-i PV-i Pi P&i
(10-8)
533
CHAPTER 10
where
D h = AA C Ip = 48 = hydraulic diameter of the condensate flow, m
,4
wetted perimeter of the condensate, m
pb = wetted perimeter X film thickness, m 2 , cross-sectional area of the
condensate flow at the lowest part of the flow
p, = density of the liquid, kg/m 3
jjl/ = viscosity of the liquid, kg/m ■ s
T = average velocity of the condensate at the lowest part of the flow, m/s
til = p, T; A c = mass flow rate of the condensate at the lowest part, kg/s
The evaluation of the hydraulic diameter D h for some common geometries is
illustrated in Figure 10-22. Note that the hydraulic diameter is again defined
such that it reduces to the ordinary diameter for flow in a circular tube, as was
done in Chapter 8 for internal flow, and it is equivalent to 4 times the thick-
ness of the condensate film at the location where the hydraulic diameter is
evaluated. That is, D h = 48.
The latent heat of vaporization hr g is the heat released as a unit mass of
vapor condenses, and it normally represents the heat transfer per unit mass of
condensate formed during condensation. However, the condensate in an actual
(a) Vertical plate
(c) Horizontal cylinder
FIGURE 10-22
The wetted perimeter p, the
condensate cross-sectional area A c
and the hydraulic diameter D h for
some common geometries.
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HEAT TRANSFER
condensation process is cooled further to some average temperature between
T sat and T s , releasing more heat in the process. Therefore, the actual heat trans-
fer will be larger. Rohsenow showed in 1956 that the cooling of the liquid be-
low the saturation temperature can be accounted for by replacing h fg by the
modified latent heat of vaporization h%, defined as
In
h fe + 0.68C P/ (T sat - T s )
(10-9a)
where C pi is the specific heat of the liquid at the average film temperature.
We can have a similar argument for vapor that enters the condenser as
superheated vapor at a temperature T v instead of as saturated vapor. In this
case the vapor must be cooled first to T m before it can condense, and this heat
must be transferred to the wall as well. The amount of heat released as a unit
mass of superheated vapor at a temperature T v is cooled to T m is simply
C pv (T v — r sat ), where C pv is the specific heat of the vapor at the average tem-
perature of (T v + T sM )/2. The modified latent heat of vaporization in this case
becomes
it r.
hf, + 0.68C„, (T snt - T 5 ) + C pv (T v - TJ)
(10-9b)
With these considerations, the rate of heat transfer can be expressed as
Gconden = M,(L " TJ = W.h% (10-10)
where A s is the heat transfer area (the surface area on which condensation oc-
curs). Solving for m from the equation above and substituting it into Eq. 10-8
gives yet another relation for the Reynolds number,
Re
4fico„de„ 4A J / ! (r sal - T s )
PVI h %
p\x,hi
(10-11)
Re =
Re = 30
Laminar
(wave-free)
Laminar
(wavy)
— Re= 1800
Turbulent
FIGURE 10-23
Flow regimes during film
condensation on a vertical plate.
This relation is convenient to use to determine the Reynolds number when the
condensation heat transfer coefficient or the rate of heat transfer is known.
The temperature of the liquid film varies from r sat on the liquid-vapor in-
terface to T s at the wall surface. Therefore, the properties of the liquid should
be evaluated at the film temperature T f = (T sat + T s )/2, which is approximately
the average temperature of the liquid. The h fg , however, should be evaluated
at r sat since it is not affected by the subcooling of the liquid.
Flow Regimes
The Reynolds number for condensation on the outer surfaces of vertical tubes
or plates increases in the flow direction due to the increase of the liquid film
thickness 8. The flow of liquid film exhibits different regimes, depending on
the value of the Reynolds number. It is observed that the outer surface of the
liquid film remains smooth and wave-free for about Re < 30, as shown in Fig-
ure 10-23, and thus the flow is clearly laminar. Ripples or waves appear on
the free surface of the condensate flow as the Reynolds number increases, and
the condensate flow becomes fully turbulent at about Re ~ 1800. The con-
densate flow is called wavy-laminar in the range of 450 < Re < 1800 and
turbulent for Re > 1800. However, some disagreement exists about the value
of Re at which the flow becomes wavy-laminar or turbulent.
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535
CHAPTER 10
Heat Transfer Correlations for Film Condensation
Below we discuss relations for the average heat transfer coefficient /; for the
case of laminar film condensation for various geometries.
1 Vertical Plates
Consider a vertical plate of height L and width b maintained at a constant tem-
perature T s that is exposed to vapor at the saturation temperature T sat . The
downward direction is taken as the positive x-direction with the origin placed
at the top of the plate where condensation initiates, as shown in Figure 10-24.
The surface temperature is below the saturation temperature (T s < !T sat ) and
thus the vapor condenses on the surface. The liquid film flows downward un-
der the influence of gravity. The film thickness 8 and thus the mass flow rate
of the condensate increases with x as a result of continued condensation on the
existing film. Then heat transfer from the vapor to the plate must occur
through the film, which offers resistance to heat transfer. Obviously the
thicker the film, the larger its thermal resistance and thus the lower the rate of
heat transfer.
The analytical relation for the heat transfer coefficient in film condensation
on a vertical plate described above was first developed by Nusselt in 1916
under the following simplifying assumptions:
1. Both the plate and the vapor are maintained at constant temperatures of
T s and r sat , respectively, and the temperature across the liquid film varies
linearly.
2. Heat transfer across the liquid film is by pure conduction (no convection
currents in the liquid film).
3. The velocity of the vapor is low (or zero) so that it exerts no drag on the
condensate (no viscous shear on the liquid-vapor interface).
4. The flow of the condensate is laminar and the properties of the liquid
are constant.
5. The acceleration of the condensate layer is negligible.
Then Newton's second law of motion for the volume element shown in Figure
10-24 in the vertical x-direction can be written as
Shear force
du
V ■■
aty:
Buoyancy force
p v g(S-y)(bdx)
Idealized
velocity
profile
No vapor drag
Idealized
temperature
profile
T ,
sat
- Linear
FIGURE 10-24
The volume element of condensate
on a vertical plate considered
in Nusselt's analysis.
2>,
since the acceleration of the fluid is zero. Noting that the only force acting
downward is the weight of the liquid element, and the forces acting upward
are the viscous shear (or fluid friction) force at the left and the buoyancy
force, the force balance on the volume element becomes
downward 4-
ii-d T
Weight = Viscous shear force + Buoyancy force
P/£(8 - y){bdx) = \L,— (bdx) + p v g(8 - y)(bdx)
Canceling the plate width b and solving for du/dy gives
du g(Pi - Pv)g(S - y)
dy
|X,
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HEAT TRANSFER
Integrating from y = where u = (because of the no-slip boundary condi-
tion) to y = y where u = u(y) gives
"W = — ^ — bs-2
(10-12)
The mass flow rate of the condensate at a location x, where the boundary layer
thickness is 8, is determined from
m{x) = p,uiy)dA = p,uiy)bdy
JA Jy =
(10-13)
Substituting the u(y) relation from Equation 10-12 into Eq. 10-13 gives
gbpiiPi ~ p„)8 3
mix)
3p,
(10-14)
whose derivative with respect to x is
dm gbp,i Pl - p v )8 2 rf§
dx
Vi
dx
(10-15)
which represents the rate of condensation of vapor over a vertical distance dx.
The rate of heat transfer from the vapor to the plate through the liquid film is
simply equal to the heat released as the vapor is condensed and is expressed as
dQ = h f dm = k, ibdx) ■
dm K b T sz
dx h f ,
(10-16)
Equating Eqs. 10-15 and 10-16 for drhldx to each other and separating the
variables give
8 3 dh = W IK sa, ,; dx
gPi (P/ ~ Pv)hf,
(10-17)
Integrating from x = where 8 = (the top of the plate) to x = x where
8 = 8(.x), the liquid film thickness at any location x is determined to be
8(jc)
4p J/ fc,(7 , sat - T s )x
- gPl (Pi ~ Pv)hn
(10-18)
The heat transfer rate from the vapor to the plate at a location x can be
expressed as
4x = h AT sM ~ T s ) = k,
* sat * s
h
" 8(x)
(10-19)
Substituting the 8(x) expression from Eq. 10-18, the local heat transfer coef-
ficient h r is determined to be
gPii.Pi- Pv)hf g kf
L 4\x, (r sat - T„)x J
(10-20)
cen58933_chl0.qxd 9/4/2002 12:38 PM Page 537
The average heat transfer coefficient over the entire plate is determined from
its definition by substituting the h x relation and performing the integration.
It gives
h = A„,
i/>
dx
h r=l
0.943
gPi (Pi
Pv)hf s kf
P-i (T„t ~ T S )L
(10-21)
Equation 10-21, which is obtained with the simplifying assumptions stated
earlier, provides good insight on the functional dependence of the condensa-
tion heat transfer coefficient. However, it is observed to underpredict heat
transfer because it does not take into account the effects of the nonlinear tem-
perature profile in the liquid film and the cooling of the liquid below the satu-
ration temperature. Both of these effects can be accounted for by replacing hj g
by hf g given by Eq. 10-9. With this modification, the average heat transfer
coefficient for laminar film condensation over a vertical flat plate of height L
is determined to be
gPi(Pi
Pv)h% kf
- P-i (T sal ~ T S )L _
(W/m 2 • °C), < Re < 30 (10-22)
537
CHAPTER 10
where
g = gravitational acceleration, m/s 2
p,, p y = densities of the liquid and vapor, respectively, kg/m 3
\l, = viscosity of the liquid, kg/m • s
hf g = h fg + 0.68C p/ (7* sat — T s ) = modified latent heat of vaporization, J/kg
k, = thermal conductivity of the liquid, W/m • °C
L = height of the vertical plate, m
T s = surface temperature of the plate, °C
J' sat = saturation temperature of the condensing fluid, °C
At a given temperature, p„ <S p, and thus p, — p v ~ p, except near the critical
point of the substance. Using this approximation and substituting Eqs. 10-14
ki
and 10-18 at x = L into Eq. 10-8 by noting that 8 A . = L = and
K
K =l
\K = l (Eqs. 10-19 and 10-21) give
4gP/(P/- p v )8 3
Re
3p, 2
4gP7( k,
4,?
Then the heat transfer coefficient /i v
3vj \3/z vert /4
in terms of Re becomes
(10-23)
1.47fc,Re
< Re < 30
Pv^Pl
(10-24)
The results obtained from the theoretical relations above are in excellent
agreement with the experimental results. It can be shown easily that using
property values in Eqs. 10-22 and 10-24 in the specified units gives the con-
densation heat transfer coefficient in W/m 2 • °C, thus saving one from having
cen58933_chl0.qxd 9/4/2002 12:38 PM Page 53E
538
HEAT TRANSFER
h„n =
/mkgkg j / w y\'«
s- m 3 m 3 kg \m • °C/
kg =o
\ ws ■ c ' ra
ml W 3 J "
. s m 6 m 3 • °C 3 °C
' W 4 \" 4
W 8 • °c 4 !
= W/m 2 • °C
FIGURE 10-25
Equation 10-22 gives the
condensation heat transfer coefficient
in W/m 2 • °C when the quantities are
expressed in the units specified
in their descriptions.
to go through tedious unit manipulations each time (Fig. 10-25). This is also
true for the equations below. All properties of the liquid are to be evaluated at
the film temperature T f = (T sat + T s )/2. The h fg and p v are to be evaluated at
the saturation temperature T sat .
Wavy Laminar Flow on Vertical Plates
At Reynolds numbers greater than about 30, it is observed that waves form at
the liquid-vapor interface although the flow in liquid film remains laminar.
The flow in this case is said to be wavy laminar. The waves at the liquid-
vapor interface tend to increase heat transfer. But the waves also complicate
the analysis and make it very difficult to obtain analytical solutions. There-
fore, we have to rely on experimental studies. The increase in heat transfer due
to the wave effect is, on average, about 20 percent, but it can exceed 50 per-
cent. The exact amount of enhancement depends on the Reynolds number.
Based on his experimental studies, Kutateladze (1963, Ref. 15) recommended
the following relation for the average heat transfer coefficient in wavy lami-
nar condensate flow for pv <§ p, and 30 < Re < 1800,
K
Rek,
1.08 Re 1
£
2W
30 < Re < 1800
Pv<Pl
(10-25)
A simpler alternative to the relation above proposed by Kutateladze (1963,
Ref. 15) is
h = 8 T?p°" h
"vert, wavy u.o ivc " V ert (smooth)
(10-26)
which relates the heat transfer coefficient in wavy laminar flow to that in
wave-free laminar flow. McAdams (1954, Ref. 2) went even further and
suggested accounting for the increase in heat transfer in the wavy region by
simply increasing the heat transfer coefficient determined from Eq. 10-22 for
the laminar case by 20 percent. Holman (1990) suggested using Eq. 10-22
for the wavy region also, with the understanding that this is a conservative
approach that provides a safety margin in thermal design. In this book we will
use Eq. 10-25.
A relation for the Reynolds number in the wavy laminar region can be
determined by substituting the h relation in Eq. 10-25 into the Re relation in
Eq. 10-11 and simplifying. It yields
Re„
4.81 +
3.70 Lk,(T.
M-/ h f>
\vi
. P, < Pi
(10-27)
Turbulent Flow on Vertical Plates
At a Reynolds number of about 1800, the condensate flow becomes turbulent.
Several empirical relations of varying degrees of complexity are proposed for
the heat transfer coefficient for turbulent flow. Again assuming p„ <§ p ; for
simplicity, Labuntsov (1957, Ref. 17) proposed the following relation for the
turbulent flow of condensate on vertical plates:
K
ReA-,
8750 + 58 Pr-° 5 (Re
253) W
Re > 1800
Pv -^ Pi
(10-28)
cen58933_chl0.qxd 9/4/2002 12:38 PM Page 539
1.0
0.1
X^-Eq. 10-24
yS 5
%.
^^^3
Eq. 10-25
^^"T,
i.
Wave-free
Eq. 10-28
10
30
100
1000 1800
10.000
Re
539
CHAPTER 10
FIGURE 10-26
Nondimensionalized heat transfer
coefficients for the wave-free laminar,
wavy laminar, and turbulent flow
of condensate on vertical plates.
The physical properties of the condensate are again to be evaluated at the film
temperature 7} = (T sat + T s )/2. The Re relation in this case is obtained by sub-
stituting the h relation above into the Re relation in Eq. 10-11, which gives
Re
vert, turbulent
0.0690 Lk,Pr 0S (T S!it -T s )(g
\L,h
'is
151 Pr°
253
(10-29)
Nondimensionalized heat transfer coefficients for the wave-free laminar,
wavy laminar, and turbulent flow of condensate on vertical plates are plotted
in Figure 10-26.
2 Inclined Plates
Equation 10-12 was developed for vertical plates, but it can also be used for
laminar film condensation on the upper surfaces of plates that are inclined by
an angle from the vertical, by replacing g in that equation by g cos 6 (Fig.
10-27). This approximation gives satisfactory results especially for ^ 60°.
Note that the condensation heat transfer coefficients on vertical and inclined
plates are related to each other by
K
/i vcrt (cos 0)'
(laminar)
(10-30)
Equation 10-30 is developed for laminar flow of condensate, but it can also
be used for wavy laminar flows as an approximation.
3 Vertical Tubes
Equation 10-22 for vertical plates can also be used to calculate the average
heat transfer coefficient for laminar film condensation on the outer surfaces of
vertical tubes provided that the tube diameter is large relative to the thickness
of the liquid film.
4 Horizontal Tubes and Spheres
Nusselt's analysis of film condensation on vertical plates can also be extended
to horizontal tubes and spheres. The average heat transfer coefficient for film
condensation on the outer surfaces of a horizontal tube is determined to be
FIGURE 10-27
Film condensation on
an inclined plate.
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540
HEAT TRANSFER
0.729
gPiiPi- P v )h%kf
MT^t ~ T S )D
(W/m 2 ■ °C)
(10-31)
where D is the diameter of the horizontal tube. Equation 10-31 can easily be
modified for a sphere by replacing the constant 0.729 by 0.815.
A comparison of the heat transfer coefficient relations for a vertical tube of
height L and a horizontal tube of diameter D yields
1.29
D
(10-32)
Setting /jyerticai = ^horizontal gives L = 1.29 4 D = 2.77 D, which implies that for
a tube whose length is 2.77 times its diameter, the average heat transfer coef-
ficient for laminar film condensation will be the same whether the tube is po-
sitioned horizontally or vertically. For L > 2.77 D, the heat transfer coefficient
will be higher in the horizontal position. Considering that the length of a tube
in any practical application is several times its diameter, it is common practice
to place the tubes in a condenser horizontally to maximize the condensation
heat transfer coefficient on the outer surfaces of the tubes.
J)
'/J
u
/J ^\
FIGURE 10-28
Film condensation on a
vertical tier of horizontal tubes.
5 Horizontal Tube Banks
Horizontal tubes stacked on top of each other as shown in Figure 10-28 are
commonly used in condenser design. The average thickness of the liquid film
at the lower tubes is much larger as a result of condensate falling on top of
them from the tubes directly above. Therefore, the average heat transfer coef-
ficient at the lower tubes in such arrangements is smaller. Assuming the con-
densate from the tubes above to the ones below drain smoothly, the average
film condensation heat transfer coefficient for all tubes in a vertical tier can be
expressed as
0.729
gPi(Pi~ Pv)h%k]
»,(T mt -T s )ND
1
a/1/4 h or i z < 1 tube
(10-33)
Note that Eq. 10-33 can be obtained from the heat transfer coefficient relation
for a horizontal tube by replacing D in that relation by ND. This relation does
not account for the increase in heat transfer due to the ripple formation and
turbulence caused during drainage, and thus generally yields conservative
results.
Effect of Vapor Velocity
In the analysis above we assumed the vapor velocity to be small and thus the
vapor drag exerted on the liquid film to be negligible, which is usually the
case. However, when the vapor velocity is high, the vapor will "pull" the liq-
uid at the interface along since the vapor velocity at the interface must drop to
the value of the liquid velocity. If the vapor flows downward (i.e., in the same
direction as the liquid), this additional force will increase the average velocity
of the liquid and thus decrease the film thickness. This, in turn, will decrease
the thermal resistance of the liquid film and thus increase heat transfer.
Upward vapor flow has the opposite effects: the vapor exerts a force on the
cen58933_chl0.qxd 9/4/2002 12:38 PM Page 541
liquid in the opposite direction to flow, thickens the liquid film, and thus
decreases heat transfer. Condensation in the presence of high vapor flow is
studied [e.g., Shekriladze and Gomelauri (1966), Ref. 23] and heat transfer re-
lations are obtained, but a detailed analysis of this topic is beyond the scope of
this introductory text.
541
CHAPTER 10
The Presence of Noncondensable Gases in Condensers
Most condensers used in steam power plants operate at pressures well below
the atmospheric pressure (usually under 0.1 aim) to maximize cycle thermal
efficiency, and operation at such low pressures raises the possibility of air (a
noncondensable gas) leaking into the condensers. Experimental studies show
that the presence of noncondensable gases in the vapor has a detrimental ef-
fect on condensation heat transfer. Even small amounts of a noncondensable
gas in the vapor cause significant drops in heat transfer coefficient during con-
densation. For example, the presence of less than 1 percent (by mass) of air in
steam can reduce the condensation heat transfer coefficient by more than half.
Therefore, it is common practice to periodically vent out the noncondensable
gases that accumulate in the condensers to ensure proper operation.
The drastic reduction in the condensation heat transfer coefficient in the
presence of a noncondensable gas can be explained as follows: When the va-
por mixed with a noncondensable gas condenses, only the noncondensable
gas remains in the vicinity of the surface (Fig. 10-29). This gas layer acts as a
barrier between the vapor and the surface, and makes it difficult for the vapor
to reach the surface. The vapor now must diffuse through the noncondensable
gas first before reaching the surface, and this reduces the effectiveness of the
condensation process.
Experimental studies show that heat transfer in the presence of a noncon-
densable gas strongly depends on the nature of the vapor flow and the flow
velocity. As you would expect, a high flow velocity is more likely to remove
the stagnant noncondensable gas from the vicinity of the surface, and thus im-
prove heat transfer.
Vapor + Noncondensable gas
.4-
Cold
surface"^
Condensate
Noncondensable gas
* Vapor
FIGURE 10-29
The presence of a noncondensable
gas in a vapor prevents the vapor
molecules from reaching the cold
surface easily, and thus impedes
condensation heat transfer.
EXAMPLE W-4
Condensation of Steam on a Vertical Plate
Saturated steam at atmospheric pressure condenses on a 2-m-high and 3-m-
wide vertical plate that is maintained at 80°C by circulating cooling water
through the other side (Fig. 10-30). Determine (a) the rate of heat transfer by
condensation to the plate and (b) the rate at which the condensate drips off the
plate at the bottom.
SOLUTION Saturated steam at 1 atm condenses on a vertical plate. The rates
of heat transfer and condensation are to be determined.
Assumptions 1 Steady operating conditions exist. 2 The plate is isothermal.
3 The condensate flow is wavy-laminar over the entire plate (will be verified).
4 The density of vapor is much smaller than the density of liquid, p v < p,.
Properties The properties of water at the saturation temperature of 100°C are
h fg = 2257 X 10 3 J/kg and p„ = 0.60 kg/m 3 . The properties of liquid water
at the film temperature of T f = (7" sat + T S )I2 = (100 + 80)/2 = 90°C are
(Table A-9)
2 m
Condensate
FIGURE 10-30
Schematic for Example 10-4.
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HEAT TRANSFER
Steam
100°C
FIGURE 10-31
Schematic for Example 10-5.
p, = 965.3 kg/m 3
jjl, = 0.315 X 10- 3 kg/m-s
v, = |x,/p, = 0.326 X 10~ 6 m 2 /s
C pl = 4206 J/kg • °C
k, = 0.675 W/m ■ °C
Analysis (a) The modified latent heat of vaporization is
h% = h fg + 0.68C,,, (T SM - T s )
= 2257 X 10 3 J/kg + 0.68 X (4206 J/kg • °C)(100 - 80)°C
= 2314 X 10 3 J/kg
For wavy-laminar flow, the Reynolds number is determined from Eq. 10-27
to be
Re = Re
vertical, wavy
4.81
3.70^(7^ -T s )(g
4.81 +
p., hj g \v,
3.70(3 m)(0.675 W/m ■ °C)(100 - 90)°C
X
(0.315 X 10- 3 kg/m • s)(2314 X 10 3 J/kg)
9.81 m/s 2
(0.326 X 10- 6 m 2 /s) 2
1287
which is between 30 and 1800, and thus our assumption of wavy laminar flow
is verified. Then the condensation heat transfer coefficient is determined from
Eq. 10-25 to be
h = h
Rek,
vertical, wavy
1.08 Re 1 - 22 - 5.
1287 X (0.675 W/m • °C) I 9.8I m/s 2
1.08(1287)'- 22 - 5.2 \(0.326 X lO" 6 m 2 /s) 2
5848 W/m 2 • °C
The heat transfer surface area of the plate is A s = W X L = (3 m)(2 m) = 6 m 2 .
Then the rate of heat transfer during this condensation process becomes
Q = hA s (T sM - T s ) = (5848 W/m 2 • °C)(6 m 2 )(100 - 80)°C = 7.02 X 10 s W
(fa) The rate of condensation of steam is determined from
Q 7.02 X 10 5 J/s
lit.
2314 X 10 3 J/kg
0.303 kg/s
That is, steam will condense on the surface at a rate of 303 grams per second.
EXAMPLE 10-5
Condensation of Steam on a Tilted Plate
What would your answer be to the preceding example problem if the plate were
tilted 30° from the vertical, as shown in Figure 10-31?
cen58933_chl0.qxd 9/4/2002 12:38 PM Page 543
SOLUTION (a) The heat transfer coefficient in this case can be determined
from the vertical plate relation by replacing g by g cos 0. But we will use
Eq. 10-30 instead since we already know the value for the vertical plate from
the preceding example:
h = h„
h yat (cos 9) 1/4 = (5848 W/m 2 • °C)(cos 30°) 1/4 = 5641 W/m 2 • °C
The heat transfer surface area of the plate is still 6 m 2 . Then the rate of con-
densation heat transfer in the tilted plate case becomes
Q = hA s {T sM - T s ) = (5641 W/m 2 • °C)(6 m 2 )(100 - 80)°C = 6.77 X 10 s W
(b) The rate of condensation of steam is again determined from
Q 6.77 X 10 5 J/s
1 condensation
h
■fg 2314 X 10 3 J/kg
0.293 kg/s
Discussion Note that the rate of condensation decreased by about 3.6 percent
when the plate is tilted.
543
CHAPTER 10
EXAMPLE 10-6
Condensation of Steam on Horizontal Tubes
The condenser of a steam power plant operates at a pressure of 7.38 kPa.
Steam at this pressure condenses on the outer surfaces of horizontal pipes
through which cooling water circulates. The outer diameter of the pipes is 3 cm,
and the outer surfaces of the pipes are maintained at 30°C (Fig. 10-32). De-
termine (a) the rate of heat transfer to the cooling water circulating in the pipes
and (b) the rate of condensation of steam per unit length of a horizontal pipe.
SOLUTION Saturated steam at a pressure of 7.38 kPa (Table A-9) condenses
on a horizontal tube at 30°C. The rates of heat transfer and condensation are to
be determined.
Assumptions 1 Steady operating conditions exist. 2 The tube is isothermal.
Properties The properties of water at the saturation temperature of 40°C
corresponding to 7.38 kPa are h fg = 2407 X 10 3 J/kg and p„ = 0.05 kg/m 3 .
The properties of liquid water at the film temperature of T f
(40 + 30)/2 = 35°C are (Table A-9)
(7"sat + T S )I2
p, = 994 kg/m 3
p., = 0.720 X 10" 3 kg/m • s
C,
pi
4178 J/kg- °C
0.623 W/m • °C
Analysis (a) The modified latent heat of vaporization is
h% = h fg + 0.68C;, (T m - T s )
= 2407 X 10 3 J/kg + 0.68 X (4178 J/kg • °C)(40
= 2435 X 10 3 J/kg
30)°C
Noting that p v < p, (since 0.05 < 994), the heat transfer coefficient for con-
densation on a single horizontal tube is determined from Eq. 10-31 to be
c
Steam, 40°C , 30°C
■Cooling water
}
FIGURE 10-32
Schematic for Example 10-6.
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HEAT TRANSFER
h = h
n "honzonti
0.729
gPi(Pi~ Pv)h}gkf
^(T sat ~ T s ) D
0.729
gpjh}-kj
l^i (T sat ~ T s ) D
0.729
(9.81 m/s 2 )(994 kg/m 3 ) 2 (2435 X 10 3 J/kg)(0.623 W/m ■ °C) 3
(0.720 X 10- 3 kg/m -s)(40 - 30)°C(0.03 m)
9292 W/m 2 • °C
The heat transfer surface area of the pipe per unit of its length is A s = ttDL =
tt(0.03 m)(l m) = 0.09425 m 2 . Then the rate of heat transfer during this con-
densation process becomes
Q = hA s (T sat - T s ) = (9292 W/m 2 • °C)(0.09425 m 2 )(40 - 30)°C = 8758 W
(b) The rate of condensation of steam is
8578 J/s
' condensation
Q_ =
h% 2435 X 10 3 J/kg
0.00360 kg/s
Therefore, steam will condense on the horizontal tube at a rate of 3.6 g/s or
12.9 kg/h per meter of its length.
Condensate
flow
FIGURE 10-33
Schematic for Example 10-7.
EXAMPLE 10-7
Condensation of Steam on Horizontal Tube Banks
Repeat the proceeding example problem for the case of 12 horizontal tubes
arranged in a rectangular array of 3 tubes high and 4 tubes wide, as shown in
Figure 10-33.
SOLUTION (a) Condensation heat transfer on a tube is not influenced by the
presence of other tubes in its neighborhood unless the condensate from other
tubes drips on it. In our case, the horizontal tubes are arranged in four vertical
tiers, each tier consisting of 3 tubes. The average heat transfer coefficient for a
vertical tier of N horizontal tubes is related to the one for a single horizontal
tube by Eq. 10-33 and is determined to be
h
1
horiz, N tubes
A'
1
1/4 "horiz, 1 tube
(9292 W/m 2 ■ °C) = 7060 W/m 2 ■ °C
Each vertical tier consists of 3 tubes, and thus the heat transfer coefficient de-
termined above is valid for each of the four tiers. In other words, this value can
be taken to be the average heat transfer coefficient for all 12 tubes.
The surface area for all 12 tubes per unit length of the tubes is
A s = Ntotai ttDL = 12tt(0.03 m)(l m) = 1.1310 m 2
Then the rate of heat transfer during this condensation process becomes
Q = hA s (T sal - T s ) = (7060 W/m 2 • °C)(1.131 m 2 )(40 - 30)°C = 79,850 W
(fa) The rate of condensation of steam is again determined from
cen58933_chl0.qxd 9/4/2002 12:38 PM Page 545
Q
79,850 J/s
h% 2435 X 10 3 J/kg
0.0328 kg/s
Therefore, steam will condense on the horizontal pipes at a rate of 32.8 g/s per
meter length of the tubes.
545
CHAPTER 10
10-6 - FILM CONDENSATION
INSIDE HORIZONTAL TUBES
So far we have discussed film condensation on the outer surfaces of tubes and
other geometries, which is characterized by negligible vapor velocity and the
unrestricted flow of the condensate. Most condensation processes encountered
in refrigeration and air-conditioning applications, however, involve condensa-
tion on the inner surfaces of horizontal or vertical tubes. Heat transfer analy-
sis of condensation inside tubes is complicated by the fact that it is strongly
influenced by the vapor velocity and the rate of liquid accumulation on the
walls of the tubes (Fig. 10-34).
For low vapor velocities, Chato recommends this expression for
condensation
0.555
gPl(Pl~ Pv)*J
- l^iC'sat T s )
3
T.)
(10-34)
Liquid
Vapor
for
Re.
pJ»Q
< 35,000
(10-35)
where the Reynolds number of the vapor is to be evaluated at the tube inlet
conditions using the internal tube diameter as the characteristic length. Heat
transfer coefficient correlations for higher vapor velocities are given by
Rohsenow.
^ Tube
FIGURE 10-34
Condensate flow in a horizontal tube
with large vapor velocities.
10-7 - DR0PWISE CONDENSATION
Dropwise condensation, characterized by countless droplets of varying diam-
eters on the condensing surface instead of a continuous liquid film, is one of
the most effective mechanisms of heat transfer, and extremely large heat trans-
fer coefficients can be achieved with this mechanism (Fig. 10-35).
In dropwise condensation, the small droplets that form at the nucleation
sites on the surface grow as a result of continued condensation, coalesce into
large droplets, and slide down when they reach a certain size, clearing the sur-
face and exposing it to vapor. There is no liquid film in this case to resist heat
transfer. As a result, with dropwise condensation, heat transfer coefficients can
be achieved that are more than 10 times larger than those associated with film
condensation. Large heat transfer coefficients enable designers to achieve a
specified heat transfer rate with a smaller surface area, and thus a smaller (and
m
FIGURE 10-35
Dropwise condensation of steam on a
vertical surface (from Hampson
and Ozisik, Ref. 11).
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HEAT TRANSFER
less expensive) condenser. Therefore, dropwise condensation is the preferred
mode of condensation in heat transfer applications.
The challenge in dropwise condensation is not to achieve it, but rather, to
sustain it for prolonged periods of time. Dropwise condensation is achieved
by adding a promoting chemical into the vapor, treating the surface with a
promoter chemical, or coating the surface with a polymer such as teflon or a
noble metal such as gold, silver, rhodium, palladium, or platinum. The pro-
moters used include various waxes and fatty acids such as oleic, stearic, and
linoic acids. They lose their effectiveness after a while, however, because of
fouling, oxidation, and the removal of the promoter from the surface. It is pos-
sible to sustain dropwise condensation for over a year by the combined effects
of surface coating and periodic injection of the promoter into the vapor. How-
ever, any gain in heat transfer must be weighed against the cost associated
with sustaining dropwise condensation.
Dropwise condensation has been studied experimentally for a number of
surface-fluid combinations. Of these, the studies on the condensation of steam
on copper surfaces has attracted the most attention because of their wide-
spread use in steam power plants. P. Griffith (1983) recommends these simple
correlations for dropwise condensation of steam on copper surfaces:
h
dropwise
51,104 + 2044r sat 22°C < r sat < 100°C (10-36)
255,310 ' T SM > 100°C (10-37)
where T sat is in °C and the heat transfer coefficient /? dropwise is in W/m 2 • °C.
The very high heat transfer coefficients achievable with dropwise conden-
sation are of little significance if the material of the condensing surface is not
a good conductor like copper or if the thermal resistance on the other side of
the surface is too large. In steady operation, heat transfer from one medium to
another depends on the sum of the thermal resistances on the path of heat
flow, and a large thermal resistance may overshadow all others and dominate
the heat transfer process. In such cases, improving the accuracy of a small re-
sistance (such as one due to condensation or boiling) makes hardly any dif-
ference in overall heat transfer calculations.
TOPIC OF SPECIAL INTEREST
Heat Pipes
A heat pipe is a simple device with no moving parts that can transfer large
quantities of heat over fairly large distances essentially at a constant tem-
perature without requiring any power input. A heat pipe is basically a
sealed slender tube containing a wick structure lined on the inner surface
and a small amount of fluid such as water at the saturated state, as shown
in Figure 10-36. It is composed of three sections: the evaporator section at
one end, where heat is absorbed and the fluid is vaporized; a condenser
section at the other end, where the vapor is condensed and heat is rejected;
and the adiabatic section in between, where the vapor and the liquid phases
of the fluid flow in opposite directions through the core and the wick,
*This section can be skipped without a loss in continuity.
cen58933_chl0.qxd 9/4/2002 12:38 PM Page 547
Tube wall
Wick
Liquid
flow
547
CHAPTER 10
Copper
tube
Wick
(liquid flow
passage)
Cross- section
of a heat pipe
Evaporation
Adiabatic
section
Condenser
section
FIGURE 10-36
Schematic and operation of a heat pipe.
respectively, to complete the cycle with no significant heat transfer be-
tween the fluid and the surrounding medium.
The type of fluid and the operating pressure inside the heat pipe depend
on the operating temperature of the heat pipe. For example, the critical-
and triple-point temperatures of water are 0.0 1°C and 374. 1°C, respec-
tively. Therefore, water can undergo a liquid-to-vapor or vapor-to-liquid
phase change process in this temperature range only, and thus it will not be
a suitable fluid for applications involving temperatures beyond this range.
Furthermore, water will undergo a phase-change process at a specified tem-
perature only if its pressure equals the saturation pressure at that tempera-
ture. For example, if a heat pipe with water as the working fluid is designed
to remove heat at 70°C, the pressure inside the heat pipe must be main-
tained at 31.2 kPa, which is the boiling pressure of water at this tem-
perature. Note that this value is well below the atmospheric pressure of
101 kPa, and thus the heat pipe will operate in a vacuum environment in
this case. If the pressure inside is maintained at the local atmospheric pres-
sure instead, heat transfer would result in an increase in the temperature of
the water instead of evaporation.
Although water is a suitable fluid to use in the moderate temperature
range encountered in electronic equipment, several other fluids can be used
in the construction of heat pipes to enable them to be used in cryogenic as
well as high-temperature applications. The suitable temperature ranges for
some common heat pipe fluids are given in Table 10-5. Note that the over-
all temperature range extends from almost absolute zero for cryogenic flu-
ids such as helium to over 1600°C for liquid metals such as lithium. The
ultimate temperature limits for a fluid are the triple- and critical-point tem-
peratures. However, a narrower temperature range is used in practice to
avoid the extreme pressures and low heats of vaporization that occur near
the critical point. Other desirable characteristics of the candidate fluids are
having a high surface tension to enhance the capillary effect and being
compatible with the wick material, as well as being readily available,
chemically stable, nontoxic, and inexpensive.
The concept of heat pipe was originally conceived by R. S. Gaugler of
the General Motors Corporation, who filed a patent application for it in
TABLE 10-5
Suitable temperature ranges for
some fluids used in heat pipes
Temperature
Fluid Range, °C
Helium
-271 to -268
Hydrogen
-259 to -240
Neon
-248 to -230
Nitrogen
-210 to -150
Methane
-182 to -82
Ammonia
-78 to -130
Water
5 to 230
Mercury
200 to 500
Cesium
400 to 1000
Sodium
500 to 1200
Lithium
850 to 1600
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HEAT TRANSFER
1942. However, it did not receive much attention until 1962, when it was
suggested for use in space applications. Since then, heat pipes have found
a wide range of applications, including the cooling of electronic equipment.
The Operation of a Heat Pipe
The operation of a heat pipe is based on the following physical principles:
• At a specified pressure, a liquid will vaporize or a vapor will condense
at a certain temperature, called the saturation temperature. Thus,
fixing the pressure inside a heat pipe fixes the temperature at which
phase change will occur.
• At a specified pressure or temperature, the amount of heat absorbed as
a unit mass of liquid vaporizes is equal to the amount of heat rejected
as that vapor condenses.
• The capillary pressure developed in a wick will move a liquid
in the wick even against the gravitational field as a result of the
capillary effect.
• A fluid in a channel flows in the direction of decreasing pressure.
Initially, the wick of the heat pipe is saturated with liquid and the core
section is filled with vapor. When the evaporator end of the heat pipe is
brought into contact with a hot surface or is placed into a hot environment,
heat will transfer into the heat pipe. Being at a saturated state, the liquid in
the evaporator end of the heat pipe will vaporize as a result of this heat
transfer, causing the vapor pressure there to rise. This resulting pressure
difference drives the vapor through the core of the heat pipe from the evap-
orator toward the condenser section. The condenser end of the heat pipe is
in a cooler environment, and thus its surface is slightly cooler. The vapor
that comes into contact with this cooler surface condenses, releasing the
heat a vaporization, which is rejected to the surrounding medium. The liq-
uid then returns to the evaporator end of the heat pipe through the wick as
a result of capillary action in the wick, completing the cycle. As a result,
heat is absorbed at one end of the heat pipe and is rejected at the other end,
with the fluid inside serving as a transport medium for heat.
The boiling and condensation processes are associated with extremely
high heat transfer coefficients, and thus it is natural to expect the heat pipe
to be a very effective heat transfer device, since its operation is based on al-
ternative boiling and condensation of the working fluid. Indeed, heat pipes
have effective conductivities several hundred times that of copper or silver.
That is, replacing a copper bar between two mediums at different tempera-
tures by a heat pipe of equal size can increase the rate of heat transfer be-
tween those two mediums by several hundred times. A simple heat pipe
with water as the working fluid has an effective thermal conductivity of the
order of 100,000 W/m • °C compared with about 400 W/m • °C for copper.
For a heat pipe, it is not unusual to have an effective conductivity
of 400,000 W/m • °C, which is 1000 times that of copper. A 15-cm-long,
0.6-cm-diameter horizontal cylindrical heat pipe with water inside, for
example, can transfer heat at a rate of 300 W. Therefore, heat pipes are
preferred in some critical applications, despite their high initial cost.
cen58933_chl0.qxd 9/4/2002 12:38 PM Page 549
There is a small pressure difference between the evaporator and
condenser ends, and thus a small temperature difference between the two
ends of the heat pipe. This temperature difference is usually between 1°C
and 5°C.
The Construction of a Heat Pipe
The wick of a heat pipe provides the means for the return of the liquid to
the evaporator. Therefore, the structure of the wick has a strong effect on
the performance of a heat pipe, and the design and construction of the wick
are the most critical aspects of the manufacturing process.
The wicks are often made of porous ceramic or woven stainless wire
mesh. They can also be made together with the tube by extruding axial
grooves along its inner surface, but this approach presents manufacturing
difficulties.
The performance of a wick depends on its structure. The characteristics
of a wick can be changed by changing the size and the number of the pores
per unit volume and the continuity of the passageway. Liquid motion in the
wick depends on the dynamic balance between two opposing effects: the
capillary pressure, which creates the suction effect to draw the liquid, and
the internal resistance to flow as a result of friction between the mesh sur-
faces and the liquid. A small pore size increases the capillary action, since
the capillary pressure is inversely proportional to the effective capillary ra-
dius of the mesh. But decreasing the pore size and thus the capillary radius
also increases the friction force opposing the motion. Therefore, the core
size of the mesh should be reduced so long as the increase in capillary force
is greater than the increase in the friction force.
Note that the optimum pore size will be different for different fluids and
different orientations of the heat pipe. An improperly designed wick will
result in an inadequate liquid supply and eventual failure of the heat pipe.
Capillary action permits the heat pipe to operate in any orientation in a
gravity field. However, the performance of a heat pipe will be best when
the capillary and gravity forces act in the same direction (evaporator end
down) and will be worst when these two forces act in opposite directions
(evaporator end up). Gravity does not affect the capillary force when the
heat pipe is in the horizontal position. The heat removal capacity of a hor-
izontal heat pipe can be doubled by installing it vertically with evaporator
end down so that gravity helps the capillary action. In the opposite case,
vertical orientation with evaporator end up, the performance declines con-
siderably relative the horizontal case since the capillary force in this case
must work against the gravity force.
Most heat pipes are cylindrical in shape. However, they can be manufac-
tured in a variety of shapes involving 90° bends, S -turns, or spirals. They
can also be made as a flat layer with a thickness of about 0.3 cm. Flat heat
pipes are very suitable for cooling high-power-output (say, 50 W or
greater) PCBs. In this case, flat heat pipes are attached directly to the back
surface of the PCB, and they absorb and transfer the heat to the edges.
Cooling fins are usually attached to the condenser end of the heat pipe to
improve its effectiveness and to eliminate a bottleneck in the path of heat
flow from the components to the environment when the ultimate heat sink
is the ambient air.
549
CHAPTER 10
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HEAT TRANSFER
FIGURE 10-37
Variation of the heat removal
capacity of a heat pipe with tilt angle
from the horizontal when the liquid
flows in the wick against gravity
(from Steinberg, Ref. 18).
Evaporator end
100
c
Condenser end
TABLE 10-6
Typical heat removal capacity of
various heat pipes
Outside
Heat
Diameter,
Length,
Removal
cm (in.)
cm (in.)
Rate, W
0.64(1)
15.2(6)
300
30.5(12)
175
45.7(18)
150
0.95(f)
15.2(6)
500
30.5(12)
375
45.7(18)
350
1.27(i)
15.2(6)
700
30.5(12)
575
45.7(18)
550
180W.
/ Heat
fc^ Pipe
Ar = 3°C
0.6 cm
L = 30 cm
FIGURE 10-38
Schematic for Example 10-8.
^180W
>
The decline in the performance of a 122-cm-long water heat pipe with
the tilt angle from the horizontal is shown in Figure 10-37 for heat pipes
with coarse, medium, and fine wicks. Note that for the horizontal case, the
heat pipe with a coarse wick performs best, but the performance drops off
sharply as the evaporator end is raised from the horizontal. The heat pipe
with a fine wick does not perform as well in the horizontal position but
maintains its level of performance greatly at tilted positions. It is clear from
this figure that heat pipes that work against gravity must be equipped with
fine wicks. The heat removal capacities of various heat pipes are given in
Table 10-6.
A major concern about the performance of a heat pipe is degradation
with time. Some heat pipes have failed within just a few months after they
are put into operation. The major cause of degradation appears to be con-
tamination that occurs during the sealing of the ends of the heat pipe tube
and affects the vapor pressure. This form of contamination has been mini-
mized by electron beam welding in clean rooms. Contamination of the
wick prior to installation in the tube is another cause of degradation. Clean-
liness of the wick is essential for its reliable operation for a long time. Heat
pipes usually undergo extensive testing and quality control process before
they are put into actual use.
An important consideration in the design of heat pipes is the compatibil-
ity of the materials used for the tube, wick, and fluid. Otherwise, reaction
between the incompatible materials produces noncondensable gases, which
degrade the performance of the heat pipe. For example, the reaction be-
tween stainless steel and water in some early heat pipes generated hydro-
gen gas, which destroyed the heat pipe.
EXAMPLE 10-8 Replacing a Heat Pipe by a Copper Rod
A 30-cm-long cylindrical heat pipe having a diameter of 0.6 cm is dissipating
heat at a rate of 180 W, with a temperature difference of 3°C across the heat
pipe, as shown in Figure 10-38. If we were to use a 30-cm-long copper rod in-
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CHAPTER 10
stead to remove heat at the same rate, determine the diameter and the mass of
the copper rod that needs to be installed.
SOLUTION A cylindrical heat pipe dissipates heat at a specified rate. The
diameter and mass of a copper rod that can conduct heat at the same rate are
to be determined.
Assumptions Steady operating conditions exist.
Properties The properties of copper at room temperature are p = 8950 kg/m 3
and k= 386 W/m • °C.
Analysis The rate of heat transfer Q through the copper rod can be ex-
pressed as
Q=kA
AT
where k is the thermal conductivity, L is the length, and AT is the temperature
difference across the copper bar. Solving for the cross-sectional area A and
substituting the specified values gives
. _ L a 0.3 m
~ kAT ^ ~ (386 W/m ■ °C)(3°C)
(180 W) = 0.04663 m 2 = 466.3 cm 2
Then the diameter and the mass of the copper rod become
1
A = 7TtD 2
4
D = V4A/tt = V4(466.3 cm 2 )/ir = 24.4 cm
m = pV= pAL = (8590 kg/m 3 )(0.04663 m 2 )(0.3 m) = 125.2 kg
Therefore, the diameter of the copper rod needs to be almost 25 times that of
the heat pipe to transfer heat at the same rate. Also, the rod would have a mass
of 125.2 kg, which is impossible for an average person to lift.
SUMMARY
Boiling occurs when a liquid is in contact with a surface main-
tained at a temperature T s sufficiently above the saturation tem-
perature r sat of the liquid. Boiling is classified as pool boiling
or flow boiling depending on the presence of bulk fluid motion.
Boiling is called pool boiling in the absence of bulk fluid flow
and/tow boiling (ox forced convection boiling) in its presence.
Pool and flow boiling are further classified as subcooled boil-
ing and saturated boiling depending on the bulk liquid temper-
ature. Boiling is said to be subcooled (or local) when the
temperature of the main body of the liquid is below the satura-
tion temperature T sM and saturated (or bulk) when the temper-
ature of the liquid is equal to T m . Boiling exhibits different
regimes depending on the value of the excess temperature
Ar cxccss . Four different boiling regimes are observed: natural
convection boiling, nucleate boiling, transition boiling, and
film boiling. These regimes are illustrated on the boiling curve.
The rate of evaporation and the rate of heat transfer in nucleate
boiling increase with increasing Ar cxcess and reach a maximum
at some point. The heat flux at this point is called the critical
(or maximum) heat flux, q max . The rate of heat transfer in nu-
cleate pool boiling is determined from
M,
'Is
g(Pl - Pv)
Cpl (T s J sat)
L Q/ Vs Pr/' j
The maximum (or critical) heat flux in nucleate pool boiling is
determined from
<? raa x = C a h fg [ugp 2 v (p, - p v )]" 4
where the value of the constant C cr is about 0.15. The minimum
heat flux is given by
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HEAT TRANSFER
0.09 Pv h fg
Vg(Pl ~ Pv)
L (p, - p„)" J
The heat flux for stable film boiling on the outside of a hori-
zontal cylinder or sphere of diameter D is given by
<7filn
gkl p v (P/ - p v )[h f!! + 0AC„ V (T s - r sal )
Qilm
p v D(T S - r sat )
where the constant C fllm = 0.62 for horizontal cylinders and
0.67 for spheres, and k v is the thermal conductivity of the
vapor. The vapor properties are to be evaluated at the film tem-
perature T f = (T SBt + T s )/2, which is the average temperature
of the vapor film. The liquid properties and h fg are to be evalu-
ated at the saturation temperature at the specified pressure.
Two distinct forms of condensation are observed in nature:
film condensation and dropwise condensation. In film conden-
sation, the condensate wets the surface and forms a liquid film
on the surface that slides down under the influence of gravity.
In dropwise condensation, the condensed vapor forms count-
less droplets of varying diameters on the surface instead of a
continuous film.
The Reynolds number for the condensate flow is defined as
Re
D hPl r, 4A P ,r,
\x,
PP-i
Aril
PP-i
and
Re
42co„de„ 4A s /j(7 sa , - T s )
p\x,,h%
p\x,,h%
where h% is the modified latent heat of vaporization, defined as
h% = h fg + 0.68C,, (T SM - T s )
and represents heat transfer during condensation per unit
mass of condensate. Here C pl is the specific heat of the liquid in
J/kg • °C.
Using some simplifying assumptions, the average heat
transfer coefficient for film condensation on a vertical plate of
height L is determined to be
gPi (Pi ~ Pv) h% kf
_ \x, {T s - T sM )L _
(W/m 2 • °C)
All properties of the liquid are to be evaluated at the film tem-
perature Tf = (T sat + T s )/2. The h fg and p v are to be evaluated at
r sat . Condensate flow is smooth and wave-free laminar for
about Re < 30, wavy-laminar in the range of 30 < Re < 1800,
and fully turbulent for Re > 1800. Heat transfer coefficients in
the wavy-laminar and turbulent flow regions are determined
from
Re/t,
Ret
30 < Re < 1800
Pv^Pl
L/3
"vert, turbulent g^Q + 5g p r _ . 5 (Re o. 75 _ 253) ^2 j >
Re > 1800
Pv^Pl
Equations for vertical plates can also be used for laminar
film condensation on the upper surfaces of the plates that are
inclined by an angle G from the vertical, by replacing g in that
equation by g cos 0. Vertical plate equations can also be used to
calculate the average heat transfer coefficient for laminar film
condensation on the outer surfaces of vertical tubes provided
that the tube diameter is large relative to the thickness of the
liquid film.
The average heat transfer coefficient for film condensation
on the outer surfaces of a horizontal tube is determined to be
0.729
£P/(P/- P v )h%k]
P-i (T s ~ T sat )D
(W/m 2 • °C)
where D is the diameter of the horizontal tube. This relation
can easily be modified for a sphere by replacing the constant
0.729 by 0.815. It can also be used for N horizontal tubes
stacked on top of each other by replacing D in the denominator
by ND.
For low vapor velocities, film condensation heat transfer
inside horizontal tubes can be determined from
0.555
gPi(Pi~ P v )kj
and
Re„
P-i (T^ - T s )
p v r v p
"fe "■" o CdI v^si
< 35,000
T s )
where the Reynolds number of the vapor is to be evaluated at
the tube inlet conditions using the internal tube diameter as the
characteristic length. Finally, the heat transfer coefficient for
dropwise condensation of steam on copper surfaces is given by
h
dropwise
51,104 + 2044 7/ sa
255,310
22°C < 7/ sat < 100°C
T,„ > 100°C
where r sat is in °C and the heat transfer coefficient h,
W/m 2 • °C.
dropwise -
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REFERENCES AND SUGGESTED READING
553
CHAPTER 10
1. N. Arai, T. Fukushima, A. Arai, T. Nakajima, K. Fujie,
and Y. Nakayama. "Heat Transfer Tubes Enhancing
Boiling and Condensation in Heat Exchangers of a
Refrigeration Machine." ASHRAE Journal 83 (1977),
p. 58.
2. P. J. Berensen. "Film Boiling Heat Transfer for a
Horizontal Surface." Journal of Heat Transfer 83 (1961),
pp. 351-358.
3. P. J. Berensen. "Experiments in Pool Boiling Heat
Transfer." International Journal of Heat Mass Transfer 5
(1962), pp. 985-999.
4. L. A. Bromley. "Heat Transfer in Stable Film Boiling."
Chemical Engineering Prog. 46 (1950), pp. 221-227.
5. J. C. Chato. "Laminar Condensation inside Horizontal
and Inclined Tubes." ASHRAE Journal 4 (1962), p. 52.
6. S. W. Chi. Heat Theory and Practice. Washington, D.C.:
Hemisphere, 1976.
7. M. T. Cichelli and C. F Bonilla. "Heat Transfer to Liquids
Boiling under Pressure." Transactions ofAIChE 41
(1945), pp. 755-787.
8. R. A. Colclaser, D. A. Neaman, and C. F Hawkins.
Electronic Circuit Analysis. New York: John Wiley &
Sons, 1984.
9. J. W. Dally. Packaging of Electronic Systems. New York:
McGraw-Hill, 1960.
10. P. Griffith. "Dropwise Condensation." In Heat Exchanger
Design Handbook, ed. E. U. Schlunder, Vol 2, Ch. 2.6.5.
New York: Hemisphere, 1983.
11. H. Hampson and N. Ozisik. "An Investigation into the
Condensation of Steam." Proceedings of the Institute of
Mechanical Engineers, London IB (1952), pp. 282-294.
12. J. P. Holman. Heat Transfer. 8th ed. New York:
McGraw-Hill, 1997.
13. F P. Incropera and D. P. DeWitt. Introduction to Heat
Transfer. 4th ed. New York: John Wiley & Sons, 2002.
14. J. J. Jasper. "The Surface Tension of Pure Liquid
Compounds." Journal of Physical and Chemical
Reference Data 1, No. 4 (1972), pp. 841-1009.
15. R. Kemp. "The Heat Pipe — A New Tune on an Old Pipe."
Electronics and Power (August 9, 1973), p. 326.
16. S. S. Kutateladze. Fundamentals of Heat Transfer. New
York: Academic Press, 1963.
17. S. S. Kutateladze. "On the Transition to Film Boiling
under Natural Convection." Kotloturbostroenie 3 (1948),
p. 48.
18. D. A. Labuntsov. "Heat Transfer in Film Condensation of
Pure Steam on Vertical Surfaces and Horizontal Tubes."
Teploenergetika 4 (1957), pp. 72-80.
19. J. H. Lienhard and V K. Dhir. "Extended Hydrodynamic
Theory of the Peak and Minimum Pool Boiling Heat
Fluxes." NASA Report, NASA-CR-2270, July 1973.
20. J. H. Lienhard and V K. Dhir. "Hydrodynamic Prediction
of Peak Pool Boiling Heat Fluxes from Finite Bodies."
Journal of Heat Transfer 95 (1973), pp. 152-158.
21. W. H. McAdams. Heat Transmission. 3rd ed. New York:
McGraw-Hill, 1954.
22. W. M. Rohsenow. "A Method of Correlating Heat
Transfer Data for Surface Boiling of Liquids." ASME
Transactions 74 (1952), pp. 969-975.
23. D. S. Steinberg. Cooling Techniques for Electronic
Equipment. New York: John Wiley & Sons, 1980.
24. W. M. Rohsenow. "Film Condensation." In Handbook of
Heat Transfer, ed. W. M. Rohsenow and J. P. Hartnett, Ch.
12A. New York: McGraw-Hill, 1973.
25. 1. G. Shekriladze, I. G. Gomelauri, and V I. Gomelauri.
"Theoretical Study of Laminar Film Condensation of
Flowing Vapor." International Journal of Heat Mass
Transfer 9 (1966), pp. 591-592.
26. N. V Suryanarayana. Engineering Heat Transfer. St. Paul,
MN: West Publishing, 1995.
27. J. W. Westwater and J. G. Santangelo. Industrial
Engineering Chemistry 47 (1955), p. 1605.
28. N. Zuber. "On the Stability of Boiling Heat Transfer."
ASME Transactions 80 (1958), pp. 711-720.
PROBLEMS
Boiling Heat Transfer
10-1C What is boiling? What mechanisms are responsible
for the very high heat transfer coefficients in nucleate boiling?
10-2C Does the amount of heat absorbed as 1 kg of saturated
liquid water boils at 100°C have to be equal to the amount of
heat released as 1 kg of saturated water vapor condenses at
100°C?
*Problems designated by a "C" are concept questions, and
students are encouraged to answer them all. Problems designated
by an "E" are in English units, and the SI users can ignore them.
Problems with an EES-CD icon ® are solved using EES, and
complete solutions together with parametric studies are included
on the enclosed CD. Problems with a computer-EES icon El are
comprehensive in nature, and are intended to be solved with a
computer, preferably using the EES software that accompanies
this text.
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HEAT TRANSFER
10-3C What is the difference between evaporation and
boiling?
10-4C What is the difference between pool boiling and flow
boiling?
10-5C What is the difference between subcooled and satu-
rated boiling?
10-6C Draw the boiling curve and identify the different boil-
ing regimes. Also, explain the characteristics of each regime.
10-7C How does film boiling differ from nucleate boiling?
Is the boiling heat flux necessarily higher in the stable film
boiling regime than it is in the nucleate boiling regime?
10-8C Draw the boiling curve and identify the burnout point
on the curve. Explain how burnout is caused. Why is the
burnout point avoided in the design of boilers?
10-9C Discuss some methods of enhancing pool boiling heat
transfer permanently.
10-10C Name the different boiling regimes in the order they
occur in a vertical tube during flow boiling.
10-11 Water is to be boiled at atmospheric pressure in a me-
chanically polished steel pan placed on top of a heating unit.
The inner surface of the bottom of the pan is maintained at
1 1 0°C. If the diameter of the bottom of the pan is 25 cm, deter-
mine (a) the rate of heat transfer to the water and (b) the rate of
evaporation.
1 atm
no°c
FIGURE P1 0-11
10-12 Water is to be boiled at atmospheric pressure on a
3-cm-diameter mechanically polished steel heater. Determine
the maximum heat flux that can be attained in the nucleate
boiling regime and the surface temperature of the heater sur-
face in that case.
10-13 rS^| Reconsider Problem 10-12. Using EES (or
1^2 other) software, investigate the effect of local
atmospheric pressure on the maximum heat flux and the tem-
perature difference T s — 7* sat . Let the atmospheric pressure vary
from 70 kPa and 101.3 kPa. Plot the maximum heat flux and
the temperature difference as a function of the atmospheric
pressure, and discuss the results.
10-14E Water is boiled at atmospheric pressure by a hori-
zontal polished copper heating element of diameter D = 0.5 in.
and emissivity e = 0.08 immersed in water. If the surface tem-
perature of the heating element is 788°F, determine the rate of
heat transfer to the water per unit length of the heating element.
Answer: 2465 Btu/h
10-15E Repeat Problem 10-14E for a heating element tem-
perature of 988°F.
10-16 Water is to be boiled at sea level in a 30-cm-diameter
mechanically polished AISI 304 stainless steel pan placed on
top of a 3-kW electric burner. If 60 percent of the heat gener-
ated by the burner is transferred to the water during boiling,
determine the temperature of the inner surface of the bottom
of the pan. Also, determine the temperature difference between
the inner and outer surfaces of the bottom of the pan if it is
6 mm thick.
1 atm
FIGURE P1 0-1 6
10-17 Repeat Problem 10-16 for a location at an elevation of
1500 m where the atmospheric pressure is 84.5 kPa and thus
the boiling temperature of water is 95°C.
Answers: 100. 9°C, 10.3°C
10-18 Water is boiled at sea level in a coffee maker equipped
with a 20-cm long 0.4-cm-diameter immersion-type electric
1 atm
FIGURE P1 0-1 8
cen58933_chl0.qxd 9/4/2002 12:38 PM Page 555
heating element made of mechanically polished stainless steel.
The coffee maker initially contains 1 L of water at 18°C. Once
boiling starts, it is observed that half of the water in the coffee
maker evaporates in 25 min. Determine the power rating of the
electric heating element immersed in water and the surface
temperature of the heating element. Also determine how long it
will take for this heater to raise the temperature of 1 L of cold
water from 18°C to the boiling temperature.
10-19 Repeat Problem 10-18 for a copper heating element.
10-20 A 65-cm-long, 2-cm-diameter brass heating element is
to be used to boil water at 120°C. If the surface temperature of
the heating element is not to exceed 125°C, determine the
highest rate of steam production in the boiler, in kg/h.
Answer: 19.4 kg/h
10-21 To understand the burnout phenomenon, boiling ex-
periments are conducted in water at atmospheric pressure using
an electrically heated 30-cm-long, 3-mm-diameter nickel-
plated horizontal wire. Determine (a) the critical heat flux and
(b) the increase in the temperature of the wire as the operating
point jumps from the nucleate boiling to the film boiling
regime at the critical heat flux. Take the emissivity of the wire
to be 0.5.
10-22 PE^| Reconsider Problem 10-21. Using EES (or
k^S other) software, investigate the effects of the
local atmospheric pressure and the emissivity of the wire on
the critical heat flux and the temperature rise of wire. Let the
atmospheric pressure vary from 70 kPa and 101.3 kPa and
the emissivity from 0.1 to 1 .0. Plot the critical heat flux and the
temperature rise as functions of the atmospheric pressure and
the emissivity, and discuss the results.
10-23 Water is boiled at 1 atm pressure in a 20-cm-internal-
diameter teflon-pitted stainless steel pan on an electric range. If
it is observed that the water level in the pan drops by 10 cm in
30 min, determine the inner surface temperature of the pan.
Answer: 111.5°C
10-24 Repeat Problem 10-23 for a polished copper pan.
10-25 In a gas-fired boiler, water is boiled at 150°C by hot
gases flowing through 50-m-long, 5 -cm-outer-diameter me-
chanically polished stainless steel pipes submerged in water. If
the outer surface temperature of the pipes is 165°C, determine
(a) the rate of heat transfer from the hot gases to water, (b) the
rate of evaporation, (c) the ratio of the critical heat flux to the
present heat flux, and {d) the surface temperature of the pipe at
which critical heat flux occurs.
Answers: (a) 10,865 kW, (b) 5.139 kg/s, (c) 1.34, (d) 166. 5°C
555
CHAPTER 10
Vent
10-26
160°C.
Repeat Problem 10-25 for a boiling temperature of
10-27E Water is boiled at 250°F by a 2-ft-long and 0.5-in.-
diameter nickel-plated electric heating element maintained
at 280°F. Determine (a) the boiling heat transfer coefficient,
Hot gases
FIGURE P1 0-25
(b) the electric power consumed by the heating element, and
(c) the rate of evaporation of water.
10-28E Repeat Problem 10-27E for a platinum-plated heat-
ing element.
10-29E FE^| Reconsider Problem 10-27E. Using EES (or
l^ti other) software, investigate the effect of sur-
face temperature of the heating element on the boiling heat
transfer coefficient, the electric power, and the rate of evapora-
tion of water. Let the surface temperature vary from 260°F to
300°F. Plot the boiling heat transfer coefficient, the electric
power consumption, and the rate of evaporation of water as a
function of the surface temperature, and discuss the results.
10-30 Cold water enters a steam generator at 15°C and
leaves as saturated steam at 100°C. Determine the fraction of
heat used to preheat the liquid water from 15°C to the satura-
tion temperature of 100°C in the steam generator.
Answer: 13.6 percent
10-31 Cold water enters a steam generator at 20°C and
leaves as saturated steam at the boiler pressure. At what pres-
sure will the amount of heat needed to preheat the water to sat-
uration temperature be equal to the heat needed to vaporize the
liquid at the boiler pressure?
10-32 rfi£M Reconsider Problem 10-31. Using EES (or
kS other) software, plot the boiler pressure as a
function of the cold water temperature as the
temperature varies from 0°C to 30°C, and discuss the results.
10-33 A 50-cm-long, 2-mm-diameter electric resistance
wire submerged in water is used to determine the boiling
heat transfer coefficient in water at 1 atm experimentally. The
wire temperature is measured to be 130°C when a wattmeter
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556
HEAT TRANSFER
(Q| 3.8 kW
1 atm
130°C
FIGURE P 10-33
indicates the electric power consumed to be 3.8 kW. Using
Newton's law of cooling, determine the boiling heat transfer
coefficient.
Condensation Heat Transfer
10-34C What is condensation? How does it occur?
10-35C What is the difference between film and dropwise
condensation? Which is a more effective mechanism of heat
transfer?
10-36C In condensate flow, how is the wetted perimeter
defined? How does wetted perimeter differ from ordinary
perimeter?
10-37C What is the modified latent heat of vaporization?
For what is it used? How does it differ from the ordinary latent
heat of vaporization?
10-38C Consider film condensation on a vertical plate. Will
the heat flux be higher at the top or at the bottom of the plate?
Why?
10-39C Consider film condensation on the outer surfaces of
a tube whose length is 10 times its diameter. For which ori-
entation of the tube will the heat transfer rate be the highest:
horizontal or vertical? Explain. Disregard the base and top sur-
faces of the tube.
10-40C Consider film condensation on the outer surfaces of
four long tubes. For which orientation of the tubes will the con-
densation heat transfer coefficient be the highest: (a) vertical,
(b) horizontal side by side, (c) horizontal but in a vertical tier
(directly on top of each other), or (d) a horizontal stack of two
tubes high and two tubes wide?
10-41C How does the presence of a noncondensable gas in a
vapor influence the condensation heat transfer?
10-42 The Reynolds number for condensate flow is defined
as Re = 4rii/p\Xi, where p is the wetted perimeter. Obtain sim-
plified relations for the Reynolds number by expressing p and
m by their equivalence for the following geometries: (a) a ver-
tical plate of height L and width w, (b) a. tilted plate of height L
and width w inclined at an angle G from the vertical, (c) a ver-
tical cylinder of length L and diameter D, (d) a horizontal
cylinder of length L and diameter D, and (e) a sphere of di-
ameter D.
10-43 Consider film condensation on the outer surfaces of N
horizontal tubes arranged in a vertical tier. For what value of N
will the average heat transfer coefficient for the entire stack of
tubes be equal to half of what it is for a single horizontal tube?
Answer: 16
10-44 Saturated steam at 1 atm condenses on a 3-m-high and
5-m-wide vertical plate that is maintained at 90°C by circulat-
ing cooling water through the other side. Determine (a) the rate
of heat transfer by condensation to the plate, and (b) the rate at
which the condensate drips off the plate at the bottom.
Answers: (a) 942 kW, (b) 0.412 kg/s
3 m
FIGURE P 10-44
10-45 Repeat Problem 10^14 for the case of the plate being
tilted 60° from the vertical.
10-46 Saturated steam at 30°C condenses on the outside of a
4-cm-outer-diameter, 2-m-long vertical tube. The temperature
of the tube is maintained at 20°C by the cooling water. Deter-
mine (a) the rate of heat transfer from the steam to the cooling
water, (b) the rate of condensation of steam, and (c) the ap-
proximate thickness of the liquid film at the bottom of the tube.
Steam
30°C
Condensate
20°C ■
L = 2m
FIGURE P1 0-46
10-47E Saturated steam at 95°F is condensed on the outer
surfaces of an array of horizontal pipes through which cooling
water circulates. The outer diameter of the pipes is 1 in. and the
outer surfaces of the pipes are maintained at 85°F Determine
(a) the rate of heat transfer to the cooling water circulating in
the pipes and (b) the rate of condensation of steam per unit
length of a single horizontal pipe.
cen58933_chl0.qxd 9/4/2002 12:38 PM Page 557
10-48E Repeat Problem 10-47E for the case of 32 horizon-
tal pipes arranged in a rectangular array of 4 pipes high and
8 pipes wide.
10-49 Saturated steam at 55°C is to be condensed at a rate of
10 kg/h on the outside of a 3-cm-outer-diameter vertical tube
whose surface is maintained at 45°C by the cooling water. De-
termine the tube length required.
10-50 Repeat Problem 10—49 for a horizontal tube.
Answer: 0.70 m
10-51 Saturated steam at 100°C condenses on a 2-m X 2-m
plate that is tilted 40° from the vertical. The plate is maintained
at 80°C by cooling it from the other side. Determine (a) the av-
erage heat transfer coefficient over the entire plate and (b) the
rate at which the condensate drips off the plate at the bottom.
10-52 fitt'M Reconsider Problem 10-51. Using EES (or
k££^ other) software, investigate the effects of plate
temperature and the angle of the plate from the vertical on the
average heat transfer coefficient and the rate at which the con-
densate drips off. Let the plate temperature vary from 40°C to
90°C and the plate angle from 0° to 60°. Plot the heat transfer
coefficient and the rate at which the condensate drips off as the
functions of the plate temperature and the tilt angle, and dis-
cuss the results.
10-53 Saturated ammonia vapor at 10°C condenses on the
outside of a 2-cm-outer-diameter, 8-m-long horizontal tube
whose outer surface is maintained at — 10°C. Determine (a) the
rate of heat transfer from the ammonia and (b) the rate of con-
densation of ammonia.
10-54 The condenser of a steam power plant operates at a
pressure of 4.25 kPa. The condenser consists of 100 horizontal
tubes arranged in a 10 X 10 square array. The tubes are 8 m
FIGURE P 10-54
557
CHAPTER 10
long and have an outer diameter of 3 cm. If the tube surfaces
are at 20°C, determine (a) the rate of heat transfer from the
steam to the cooling water and (b) the rate of condensation of
steam in the condenser. Answers: (a) 3678 kW, (£>) 1.496 kg/s
10-55 VLgM Reconsider Problem 10-54. Using EES (or
I^S other) software, investigate the effect of the
condenser pressure on the rate of heat transfer and the rate of
condensation of the steam. Let the condenser pressure vary
from 3 kPa to 15 kPa. Plot the rate of heat transfer and the rate
of condensation of the steam as a function of the condenser
pressure, and discuss the results.
10-56 A large heat exchanger has several columns of tubes,
with 20 tubes in each column. The outer diameter of the tubes
is 1.5 cm. Saturated steam at 50°C condenses on the outer sur-
faces of the tubes, which are maintained at 20°C. Determine
(a) the average heat transfer coefficient and (b) the rate of con-
densation of steam per m length of a column.
10-57 Saturated refrigerant- 134a vapor at 30°C is to be con-
densed in a 5-m-long, 1 -cm-diameter horizontal tube that is
maintained at a temperature of 20°C. If the refrigerant enters
the tube at a rate of 2.5 kg/min, determine the fraction of the
refrigerant that will have condensed at the end of the tube.
10-58 Repeat Problem 10-57 for a tube length of 8 m.
Answer: 17.2 percent
10-59 Tu'M Reconsider Problem 10-57. Using EES (or
k^^ other) software, plot the fraction of the refrig-
erant condensed at the end of the tube as a function of the tem-
perature of the saturated R-134a vapor as the temperature
varies from 25°C to 50°C, and discuss the results.
Special Topic: Heat Pipes
10-60C What is a heat pipe? How does it operate? Does it
have any moving parts?
10-61 C A heat pipe with water as the working fluid is said to
have an effective thermal conductivity of 100,000 W/m ■ °C,
which is more than 100,000 times the conductivity of water.
How can this happen?
10-62C What is the effect of a small amount of noncondens-
able gas such as air on the performance of a heat pipe?
10-63C Why do water-based heat pipes used in the cooling
of electronic equipment operate below atmospheric pressure?
10-64C What happens when the wick of a heat pipe is too
coarse or too fine?
10-65C Does the orientation of a heat pipe affect its perfor-
mance? Does it matter if the evaporator end of the heat pipe is
up or down? Explain.
10-66C How can the liquid in a heat pipe move up against
gravity without a pump? For heat pipes that work against grav-
ity, is it better to have coarse or fine wicks? Why?
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HEAT TRANSFER
10-67C What are the important considerations in the design
and manufacture of heat pipes?
10-68C What is the major cause for the premature degrada-
tion of the perfo nuance of some heat pipes?
10-69 A 40-cm-long cylindrical heat pipe having a diameter
of 0.5 cm is dissipating heat at a rate of 150 W, with a tem-
perature difference of 4°C across the heat pipe. If we were
to use a 40-cm-long copper rod (k = 386 W/m • °C and p =
8950 kg/m 3 ) instead to remove heat at the same rate, determine
the diameter and the mass of the copper rod that needs to be in-
stalled.
10-70 Repeat Problem 10-69 for an aluminum rod instead of
copper.
10-71E A plate that supports 10 power transistors, each dis-
sipating 35 W, is to be cooled with 1 -ft-long heat pipes having
a diameter of 1 in. Using Table 10-6, determine how many
pipes need to be attached to this plate. Answer: 2
Heat sink
Heat
pipe
Transistor
FIGURE P1 0-71 E
Review Problems
10-72 Steam at 40°C condenses on the outside of a 3-cm di-
ameter thin horizontal copper tube by cooling water that enters
the tube at 25°C at an average velocity of 2 m/s and leaves at
35°C. Determine the rate of condensation of steam, the average
overall heat transfer coefficient between the steam and the
cooling water, and the tube length.
Steam
40°C
FIGURE P1 0-72
10-73 Saturated ammonia vapor at 25°C condenses on the
outside of a 2-m-long, 3.2-cm-outer-diameter vertical tube
maintained at 15°C. Determine (a) the average heat transfer
coefficient, (b) the rate of heat transfer, and (c) the rate of con-
densation of ammonia.
10-74 Saturated isobutane vapor in a binary geothermal
power plant is to be condensed outside an array of eight hori-
zontal tubes. Determine the ratio of the condensation rate for
the cases of the tubes being arranged in a horizontal tier versus
in a vertical tier of horizontal tubes. Answer: 1.68
10-75E The condenser of a steam power plant operates at a
pressure of 0.95 psia. The condenser consists of 144 horizontal
tubes arranged in a 12 X 12 square array. The tubes are 15 ft
long and have an outer diameter of 1 .2 in. If the outer surfaces
of the tubes are maintained at 80°F, determine (a) the rate of
heat transfer from the steam to the cooling water and (b) the
rate of condensation of steam in the condenser.
10-76E Repeat Problem 10-75E for a tube diameter of 2 in.
10-77 Water is boiled at 100°C electrically by a 80-cm-long,
2-mm-diameter horizontal resistance wire made of chemically
etched stainless steel. Determine (a) the rate of heat transfer to
the water and the rate of evaporation of water if the tempera-
ture of the wire is 1 15°C and (b) the maximum rate of evapo-
ration in the nucleate boiling regime.
Answers: (a) 2387 W, 3.81 kg/h, (b) 1280 kW/m 2
- Steam
fl
Water
100°C
FIGURE P1 0-77
10-78E Saturated steam at 100°F is condensed on a 6-ft-high
vertical plate that is maintained at 80°F. Determine the rate of
heat transfer from the steam to the plate and the rate of con-
densation per foot width of the plate.
10-79 Saturated refrigerant- 134a vapor at 35 C C is to be con-
densed on the outer surface of a 7-m-long, 1.5-cm-diameter
horizontal tube that is maintained at a temperature of 25°C.
Determine the rate at which the refrigerant will condense, in
kg/min.
10-80 Repeat Problem 10-79 for a tube diameter of 3 cm.
10-81 Saturated steam at 270.1 kPa condenses inside a hori-
zontal, 6-m-long, 3 -cm-internal-diameter pipe whose surface is
maintained at 110°C. Assuming low vapor velocity, determine
cen58933_chl0.qxd 9/4/2002 12:38 PM Page 559
the average heat transfer coefficient and the rate of condensa-
tion of the steam inside the pipe.
Answers: 3345 W/m 2 • °C, 0.0174 kg/s
10-82 f~J&\ A 1.5-cm-diameter silver sphere initially at
xg$7 30°C is suspended in a room filled with satu-
rated steam at 100°C. Using the lumped system analysis, de-
termine how long it will take for the temperature of the ball to
rise to 50°C. Also, determine the amount of steam that con-
denses during this process and verify that the lumped system
analysis is applicable.
10-83 Repeat Problem 10-82 for a 3-cm-diameter cop-
per ball.
10-84 You have probably noticed that water vapor that con-
denses on a canned drink slides down, clearing the surface for
further condensation. Therefore, condensation in this case can
be considered to be dropwise. Determine the condensation heat
transfer coefficient on a cold canned drink at 5°C that is placed
in a large container filled with saturated steam at 95°C.
Steam
95°C
FIGURE P 10-84
10-85 A resistance heater made of 2-mm-diameter nickel
wire is used to heat water at 1 atm pressure. Determine the
highest temperature at which this heater can operate safely
without the danger of burning out. Answer: 109. 6°C
Computer, Design, and Essay Problems
10-86 Design the condenser of a steam power plant that has
a thermal efficiency of 40 percent and generates 10 MW of net
electric power. Steam enters the condenser as saturated vapor
at 10 kPa, and it is to be condensed outside horizontal tubes
through which cooling water from a nearby river flows. The
temperature rise of the cooling water is limited to 8°C, and the
velocity of the cooling water in the pipes is limited to 6 m/s to
keep the pressure drop at an acceptable level. Specify the pipe
diameter, total pipe length, and the arrangement of the pipes to
minimize the condenser volume.
10-87 The refrigerant in a household refrigerator is con-
densed as it flows through the coil that is typically placed
559
CHAPTER 10
behind the refrigerator. Heat transfer from the outer surface of
the coil to the surroundings is by natural convection and radia-
tion. Obtaining information about the operating conditions of
the refrigerator, including the pressures and temperatures of the
refrigerant at the inlet and the exit of the coil, show that the coil
is selected properly, and determine the safety margin in the
selection.
10-88 Water-cooled steam condensers are commonly used in
steam power plants. Obtain information about water-cooled
steam condensers by doing a literature search on the topic and
also by contacting some condenser manufacturers. In a report,
describe the various types, the way they are designed, the lim-
itation on each type, and the selection criteria.
10-89 Steam boilers have long been used to provide process
heat as well as to generate power. Write an essay on the history
of steam boilers and the evolution of modern supercritical
steam power plants. What was the role of the American Society
of Mechanical Engineers in this development?
10-90 The technology for power generation using geother-
mal energy is well established, and numerous geothermal
power plants throughout the world are currently generating
electricity economically. Binary geothermal plants utilize a
volatile secondary fluid such as isobutane, n-pentane, and
R-114 in a closed loop. Consider a binary geothermal plant
with R-114 as the working fluid that is flowing at a rate of
600 kg/s. The R-114 is vaporized in a boiler at 115°C by the
geothermal fluid that enters at 165°C and is condensed at 30°C
outside the tubes by cooling water that enters the tubes at 18°C.
Design the condenser of this binary plant.
Specify (a) the length, diameter, and number of tubes and
their arrangement in the condenser, (b) the mass flow rate of
cooling water, and (c) the flow rate of make-up water needed if
a cooling tower is used to reject the waste heat from the cool-
ing water. The liquid velocity is to remain under 6 m/s and the
length of the tubes is limited to 8 m.
10-91 A manufacturing facility requires saturated steam at
120°C at a rate of 1.2 kg/min. Design an electric steam boiler
for this purpose under these constraints:
• The boiler will be in cylindrical shape with a height-to-
diameter ratio of 1 .5. The boiler can be horizontal or
vertical.
Boiler
FIGURE P1 0-91
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560
HEAT TRANSFER
• The boiler will operate in the nucleate boiling regime,
and the design heat flux will not exceed 60 percent of the
critical heat flux to provide an adequate safety margin.
• A commercially available plug-in type electrical heating
element made of mechanically polished stainless steel will
be used. The diameter of the heater cannot be between
0.5 cm and 3 cm.
• Half of the volume of the boiler should be occupied by
steam, and the boiler should be large enough to hold
enough water for 2 h supply of steam. Also, the boiler
will be well insulated.
You are to specify the following: (1) The height and inner di-
ameter of the tank, (2) the length, diameter, power rating, and
surface temperature of the electric heating element, (3) the
maximum rate of steam production during short periods of
overload conditions, and how it can be accomplished.
10-92 Repeat Problem 10-91 for a boiler that produces
steam at 150°C at a rate of 2.5 kg/min.
10-93 Conduct this experiment to determine the boiling heat
transfer coefficient. You will need a portable immersion-type
electric heating element, an indoor-outdoor thermometer, and
metal glue (all can be purchased for about $15 in a hardware
store). You will also need a piece of string and a ruler to calcu-
late the surface area of the heater. First, boil water in a pan us-
ing the heating element and measure the temperature of the
boiling water away from the heating element. Based on your
reading, estimate the elevation of your location, and compare it
to the actual value. Then glue the tip of the thermocouple wire
of the thermometer to the midsection of the heater surface. The
temperature reading in this case will give the surface tempera-
ture of the heater. Assuming the rated power of the heater to be
the actual power consumption during heating (you can check
this by measuring the electric current and voltage), calculate
the heat transfer coefficients from Newton's law of cooling.
FIGURE P1 0-93
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1
1
1
•
I CHAPTER
r
i
■
FUNDAMENTALS OF
THERMAL RADIATION
^^ o far, we have considered the conduction and convection modes of heat
^^^ transfer, which are related to the nature of the materials involved and
^^ the presence of fluid motion, among other things. We now turn our at-
tention to the third mechanism of heat transfer: radiation, which is character-
istically different from the other two.
We start this chapter with a discussion of electromagnetic waves and the
electromagnetic spectrum, with particular emphasis on thermal radiation.
Then we introduce the idealized blackbody, blackbody radiation, and black-
body radiation function, together with the Stefan-Bo Itzmann law, Planck's
law, and Wien's displacement law.
Radiation is emitted by every point on a plane surface in all directions into
the hemisphere above the surface. The quantity that describes the magnitude
of radiation emitted or incident in a specified direction in space is the ra-
diation intensity. Various radiation fluxes such as emissive power, irradiation,
and radiosity are expressed in terms of intensity. This is followed by a discus-
sion of radiative properties of materials such as emissivity, absorptivity, re-
flectivity, and transmissivity and their dependence on wavelength, direction,
_.
CONTENTS
11-1 Introduction 562
11-2 Thermal Radiation 563
11-3 Blackbody Radiation 565
11-4 Radiation Intensity 571
11-5 Radiative Properties 577
11-6 Atmospheric and Solar
Radiation 586
Topic of Special Interest:
Solar Heat Gain Through
Windows 590
and temperature.
The greenhouse effect is presented as an example of the consequences of the
wavelength dependence of radiation properties. The last section is devoted to
the discussions of atmospheric and solar radiation because of their importance.
561
^3
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562
HEAT TRANSFER
FIGURE 11-1
A hot object in a vacuum chamber
loses heat by radiation only.
FIGURE 11-2
Unlike conduction and convection,
heat transfer by radiation can occur
between two bodies, even when they
are separated by a medium colder than
both of them.
11-1 - INTRODUCTION
Consider a hot object that is suspended in an evacuated chamber whose walls
are at room temperature (Fig. 11-1). The hot object will eventually cool down
and reach thermal equilibrium with its surroundings. That is, it will lose heat
until its temperature reaches the temperature of the walls of the chamber. Heat
transfer between the object and the chamber could not have taken place by
conduction or convection, because these two mechanisms cannot occur in a
vacuum. Therefore, heat transfer must have occurred through another mecha-
nism that involves the emission of the internal energy of the object. This
mechanism is radiation.
Radiation differs from the other two heat transfer mechanisms in that it does
not require the presence of a material medium to take place. In fact, energy
transfer by radiation is fastest (at the speed of light) and it suffers no attenua-
tion in a vacuum. Also, radiation transfer occurs in solids as well as liquids
and gases. In most practical applications, all three modes of heat transfer oc-
cur concurrently at varying degrees. But heat transfer through an evacuated
space can occur only by radiation. For example, the energy of the sun reaches
the earth by radiation.
You will recall that heat transfer by conduction or convection takes place in
the direction of decreasing temperature; that is, from a high-temperature
medium to a lower-temperature one. It is interesting that radiation heat trans-
fer can occur between two bodies separated by a medium colder than both
bodies (Fig. 11-2). For example, solar radiation reaches the surface of the
earth after passing through cold air layers at high altitudes. Also, the radiation-
absorbing surfaces inside a greenhouse reach high temperatures even when its
plastic or glass cover remains relatively cool.
The theoretical foundation of radiation was established in 1864 by physicist
James Clerk Maxwell, who postulated that accelerated charges or changing
electric currents give rise to electric and magnetic fields. These rapidly moving
fields are called electromagnetic waves or electromagnetic radiation, and
they represent the energy emitted by matter as a result of the changes in the
electronic configurations of the atoms or molecules. In 1887, Heinrich Hertz
experimentally demonstrated the existence of such waves. Electromagnetic
waves transport energy just like other waves, and all electromagnetic waves
travel at the speed of light in a vacuum, which is C = 2.9979 X 10 8 m/s. Elec-
tromagnetic waves are characterized by their frequency v or wavelength \.
These two properties in a medium are related by
(11-1)
where c is the speed of propagation of a wave in that medium. The speed
of propagation in a medium is related to the speed of light in a vacuum by
c = c /n, where n is the index of refraction of that medium. The refractive index
is essentially unity for air and most gases, about 1.5 for glass, and about 1.33 for
water. The commonly used unit of wavelength is the micrometer (|xm) or mi-
cron, where 1 |xm = 10~ 6 m. Unlike the wavelength and the speed of propaga-
tion, the frequency of an electromagnetic wave depends only on the source and
is independent of the medium through which the wave travels. The frequency
(the number of oscillations per second) of an electromagnetic wave can range
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563
CHAPTER 11
from less than a million Hz to a septillion Hz or higher, depending on the
source. Note from Eq. 11-1 that the wavelength and the frequency of electro-
magnetic radiation are inversely proportional.
It has proven useful to view electromagnetic radiation as the propagation of
a collection of discrete packets of energy called photons or quanta, as pro-
posed by Max Planck in 1900 in conjunction with his quantum theory. In this
view, each photon of frequency v is considered to have an energy of
hv
lic_
(11-2)
where h = 6.6256 X 10~ 34 J • s is Planck's constant. Note from the second
part of Eq. 11-2 that the energy of a photon is inversely proportional to its
wavelength. Therefore, shorter-wavelength radiation possesses larger photon
energies. It is no wonder that we try to avoid very-short-wavelength radiation
such as gamma rays and X-rays since they are highly destructive.
11-2 - THERMAL RADIATION
Although all electromagnetic waves have the same general features, waves of
different wavelength differ significantly in their behavior. The electromag-
netic radiation encountered in practice covers a wide range of wavelengths,
varying from less than 10~ 10 |xm for cosmic rays to more than 10 10 |xm for
electrical power waves. The electromagnetic spectrum also includes gamma
rays, X-rays, ultraviolet radiation, visible light, infrared radiation, thermal ra-
diation, microwaves, and radio waves, as shown in Figure 11-3.
Different types of electromagnetic radiation are produced through various
mechanisms. For example, gamma rays are produced by nuclear reactions,
X-rays by the bombardment of metals with high-energy electrons, microwaves
by special types of electron tubes such as klystrons and magnetrons, and radio
waves by the excitation of some crystals or by the flow of alternating current
through electric conductors.
The short-wavelength gamma rays and X-rays are primarily of concern to
nuclear engineers, while the long-wavelength microwaves and radio waves
are of concern to electrical engineers. The type of electromagnetic radiation
that is pertinent to heat transfer is the thermal radiation emitted as a result
of energy transitions of molecules, atoms, and electrons of a substance. Tem-
perature is a measure of the strength of these activities at the microscopic
level, and the rate of thermal radiation emission increases with increasing
temperature. Thermal radiation is continuously emitted by all matter whose
temperature is above absolute zero. That is, everything around us such as
walls, furniture, and our friends constantly emits (and absorbs) radiation
(Fig. 11-4). Thermal radiation is also defined as the portion of the electro-
magnetic spectrum that extends from about 0.1 to 100 |xm, since the radiation
emitted by bodies due to their temperature falls almost entirely into this wave-
length range. Thus, thermal radiation includes the entire visible and infrared
(IR) radiation as well as a portion of the ultraviolet (UV) radiation.
What we call light is simply the visible portion of the electromagnetic spec-
trum that lies between 0.40 and 0.76 |xm. Light is characteristically no differ-
ent than other electromagnetic radiation, except that it happens to trigger the
X, pm
10 10
10"
10 8
10N
10 6
10 5
If) 4
10 3
10-
10
1
io- 1
io- 2
IO- 3
io- 4
IO- 5
io- 6
io- 7
io- 8
10-9
Electrical
power waves
Radio and
TV waves
Microwaves
I
Thermal Infrared
radiation I
1 j Visible
I Ultraviolet
X-rays
y-rays
Cosmic
rays
FIGURE 11-3
The electromagnetic wave spectrum.
FIGURE 11^
Everything around us constantly
emits thermal radiation.
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HEAT TRANSFER
TABLE 11-1
The wave
length
ranges of
different
colors
Color
Wavelength band
Violet
0.40-0.44 |xm
Blue
0.44-0.49 |xm
Green
0.49-0.54 |xm
Yellow
0.54-0.60 |xm
Orange
0.60-0.67 |xm
Red
0.63-0.76 |xm
r^r
FIGURE 11-5
Food is heated or cooked in a
microwave oven by absorbing the
electromagnetic radiation energy
generated by the magnetron of the
oven.
sensation of seeing in the human eye. Light, or the visible spectrum, consists
of narrow bands of color from violet (0.40-0.44 (Jim) to red (0.63-0.76 |xm),
as shown in Table 11-1.
A body that emits some radiation in the visible range is called a light source.
The sun is obviously our primary light source. The electromagnetic radiation
emitted by the sun is known as solar radiation, and nearly all of it falls into
the wavelength band 0.3-3 (xm. Almost half of solar radiation is light (i.e., it
falls into the visible range), with the remaining being ultraviolet and infrared.
The radiation emitted by bodies at room temperature falls into the infrared
region of the spectrum, which extends from 0.76 to 100 |xm. Bodies start
emitting noticeable visible radiation at temperatures above 800 K. The tung-
sten filament of a lightbulb must be heated to temperatures above 2000 K be-
fore it can emit any significant amount of radiation in the visible range.
The ultraviolet radiation includes the low-wavelength end of the thermal ra-
diation spectrum and lies between the wavelengths 0.01 and 0.40 |xm. Ultra-
violet rays are to be avoided since they can kill microorganisms and cause
serious damage to humans and other living organisms. About 12 percent of so-
lar radiation is in the ultraviolet range, and it would be devastating if it were
to reach the surface of the earth. Fortunately, the ozone (0 3 ) layer in the at-
mosphere acts as a protective blanket and absorbs most of this ultraviolet radi-
ation. The ultraviolet rays that remain in sunlight are still sufficient to cause
serious sunburns to sun worshippers, and prolonged exposure to direct sunlight
is the leading cause of skin cancer, which can be lethal. Recent discoveries of
"holes" in the ozone layer have prompted the international community to ban
the use of ozone-destroying chemicals such as the refrigerant Freon-12 in or-
der to save the earth. Ultraviolet radiation is also produced artificially in fluo-
rescent lamps for use in medicine as a bacteria killer and in tanning parlors as
an artificial tanner. The connection between skin cancer and ultraviolet rays
has caused dermatologists to issue strong warnings against its use for tanning.
Microwave ovens utilize electromagnetic radiation in the microwave
region of the spectrum generated by microwave tubes called magnetrons.
Microwaves in the range of 10 2 — 10 5 (Jim are very suitable for use in cooking
since they are reflected by metals, transmitted by glass and plastics, and
absorbed by food (especially water) molecules. Thus, the electric energy con-
verted to radiation in a microwave oven eventually becomes part of the inter-
nal energy of the food. The fast and efficient cooking of microwave ovens has
made them as one of the essential appliances in modern kitchens (Fig. 11-5).
Radars and cordless telephones also use electromagnetic radiation in the mi-
crowave region. The wavelength of the electromagnetic waves used in radio
and TV broadcasting usually ranges between 1 and 1000 m in the radio wave
region of the spectrum.
In heat transfer studies, we are interested in the energy emitted by bodies
because of their temperature only. Therefore, we will limit our consideration
to thermal radiation, which we will simply call radiation. The relations
developed below are restricted to thermal radiation only and may not be
applicable to other forms of electromagnetic radiation.
The electrons, atoms, and molecules of all solids, liquids, and gases above
absolute zero temperature are constantly in motion, and thus radiation is con-
stantly emitted, as well as being absorbed or transmitted throughout the entire
volume of matter. That is, radiation is a volumetric phenomenon. However,
cen58933_chll.qxd 9/9/2002 9:38 AM Page 565
for opaque (nontransparent) solids such as metals, wood, and rocks, radiation
is considered to be a surface phenomenon, since the radiation emitted by the
interior regions can never reach the surface, and the radiation incident on such
bodies is usually absorbed within a few microns from the surface (Fig. 11-6).
Note that the radiation characteristics of surfaces can be changed completely
by applying thin layers of coatings on them.
11-3 - BLACKBODY RADIATION
A body at a temperature above absolute zero emits radiation in all directions
over a wide range of wavelengths. The amount of radiation energy emitted
from a surface at a given wavelength depends on the material of the body and
the condition of its surface as well as the surface temperature. Therefore, dif-
ferent bodies may emit different amounts of radiation per unit surface area,
even when they are at the same temperature. Thus, it is natural to be curious
about the maximum amount of radiation that can be emitted by a surface at a
given temperature. Satisfying this curiosity requires the definition of an ideal-
ized body, called a blackbody, to serve as a standard against which the radia-
tive properties of real surfaces may be compared.
A blackbody is defined as a perfect emitter and absorber of radiation. At a
specified temperature and wavelength, no surface can emit more energy than a
blackbody. A blackbody absorbs all incident radiation, regardless of wavelength
and direction. Also, a blackbody emits radiation energy uniformly in all direc-
tions per unit area normal to direction of emission. (Fig. 1 1-7). That is, a black-
body is a diffuse emitter. The term diffuse means "independent of direction."
The radiation energy emitted by a blackbody per unit time and per unit
surface area was determined experimentally by Joseph Stefan in 1879 and
expressed as
E b (T) = ar 4
(W/m 2 )
(11-3)
where cr = 5.67 X 10~ 8 W/m 2 • K 4 is the Stefan-Bo Itzmann constant and 7 is
the absolute temperature of the surface in K. This relation was theoretically
verified in 1884 by Ludwig Boltzmann. Equation 11-3 is known as the
Stefan-Boltzmann law and E b is called the blackbody emissive power. Note
that the emission of thermal radiation is proportional to the fourth power of
the absolute temperature.
Although a blackbody would appear black to the eye, a distinction should
be made between the idealized blackbody and an ordinary black surface. Any
surface that absorbs light (the visible portion of radiation) would appear black
to the eye, and a surface that reflects it completely would appear white. Con-
sidering that visible radiation occupies a very narrow band of the spectrum
from 0.4 to 0.76 (xm, we cannot make any judgments about the blackness of a
surface on the basis of visual observations. For example, snow and white paint
reflect light and thus appear white. But they are essentially black for infrared
radiation since they strongly absorb long-wavelength radiation. Surfaces
coated with lampblack paint approach idealized blackbody behavior.
Another type of body that closely resembles a blackbody is a large cavity
with a small opening, as shown in Figure 11-8. Radiation coming in through
the opening of area A will undergo multiple reflections, and thus it will have
several chances to be absorbed by the interior surfaces of the cavity before any
565
CHAPTER 11
Radiation
emitted
Gas or
vacuum
FIGURE 11-6
Radiation in opaque solids is
considered a surface phenomenon
since the radiation emitted only by the
molecules at the surface can escape
the solid.
Uniform
Nonuniform
Blackbody
Real body
FIGURE 11-7
A blackbody is said to be a diffuse
emitter since it emits radiation energy
uniformly in all directions.
FIGURE 11-8
A large isothermal cavity at
temperature T with a small opening of
area A closely resembles a blackbody
of surface area A at the same
temperature.
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HEAT TRANSFER
part of it can possibly escape. Also, if the surface of the cavity is isothermal at
temperature T, the radiation emitted by the interior surfaces will stream
through the opening after undergoing multiple reflections, and thus it will
have a diffuse nature. Therefore, the cavity will act as a perfect absorber and
perfect emitter, and the opening will resemble a blackbody of surface area A
at temperature T, regardless of the actual radiative properties of the cavity.
The Stefan-Boltzmann law in Eq. 11-3 gives the total blackbody emissive
power E b , which is the sum of the radiation emitted over all wavelengths.
Sometimes we need to know the spectral blackbody emissive power, which
is the amount of radiation energy emitted by a blackbody at an absolute tem-
perature T per unit time, per unit surface area, and per unit wavelength about
the wavelength X. For example, we are more interested in the amount of radi-
ation an incandescent light bulb emits in the visible wavelength spectrum than
we are in the total amount emitted.
The relation for the spectral blackbody emissive power E bk was developed
by Max Planck in 1901 in conjunction with his famous quantum theory. This
relation is known as Planck's law and is expressed as
E bk (K T) = — (W/m 2 • jxm) (1 1-4)
Xlexp (C 2 /XT) - 1]
where
C, = 2ir/!C5 = 3.742 X 10 8 W • jjinrVm 2
C 2 = hc /k = 1.439 X 10 4 |xm ■ K
Also, T is the absolute temperature of the surface, \ is the wavelength of the
radiation emitted, and k = 1.38065 X 10 23 J/K is Boltzmann's constant. This
relation is valid for a surface in a vacuum or a gas. For other mediums, it
needs to be modified by replacing C x by C t /n 2 , where n is the index of refrac-
tion of the medium. Note that the term spectral indicates dependence on
wavelength.
The variation of the spectral blackbody emissive power with wavelength is
plotted in Figure 11-9 for selected temperatures. Several observations can be
made from this figure:
1. The emitted radiation is a continuous function of wavelength. At any
specified temperature, it increases with wavelength, reaches a peak, and
then decreases with increasing wavelength.
2. At any wavelength, the amount of emitted radiation increases with
increasing temperature.
3. As temperature increases, the curves shift to the left to the shorter-
wavelength region. Consequently, a larger fraction of the radiation is
emitted at shorter wavelengths at higher temperatures.
4. The radiation emitted by the sun, which is considered to be a blackbody
at 5780 K (or roughly at 5800 K), reaches its peak in the visible region
of the spectrum. Therefore, the sun is in tune with our eyes. On the other
hand, surfaces at T < 800 K emit almost entirely in the infrared region
and thus are not visible to the eye unless they reflect light coming from
other sources.
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CHAPTER 11
> OS
Visible light region
Locus of
maximum power:
X,7/=2898u.m-K
100 K
0.1 1 10
Wavelength X, urn
1000
FIGURE 11-9
The variation of the blackbody
emissive power with wavelength for
several temperatures.
As the temperature increases, the peak of the curve in Figure 11-9 shifts
toward shorter wavelengths. The wavelength at which the peak occurs for a
specified temperature is given by Wien's displacement law as
(xr)„
2897.8 |xm • K
(11-5)
This relation was originally developed by Willy Wien in 1894 using classical
thermodynamics, but it can also be obtained by differentiating Eq. 11-4 with
respect to \ while holding T constant and setting the result equal to zero.
A plot of Wien's displacement law, which is the locus of the peaks of the
radiation emission curves, is also given in Figure 11-9.
The peak of the solar radiation, for example, occurs at A. = 2897.8/
5780 = 0.50 |xm, which is near the middle of the visible range. The peak of
the radiation emitted by a surface at room temperature (r = 298 K) occurs at
9.72 |xm, which is well into the infrared region of the spectrum.
An electrical resistance heater starts radiating heat soon after it is plugged
in, and we can feel the emitted radiation energy by holding our hands facing
the heater. But this radiation is entirely in the infrared region and thus cannot
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HEAT TRANSFER
Incident
light
Reflected
FIGURE 11-10
A surface that reflects red while
absorbing the remaining parts of the
incident light appears red to the eye.
E«A T)
FIGURE 11-11
On an E b}C -k chart, the area under
a curve for a given temperature
represents the total radiation energy
emitted by a blackbody at that
temperature.
be sensed by our eyes. The heater would appear dull red when its temperature
reaches about 1000 K, since it will start emitting a detectable amount (about
1 W/m 2 • (Jim) of visible red radiation at that temperature. As the temperature
rises even more, the heater appears bright red and is said to be red hot. When
the temperature reaches about 1500 K, the heater emits enough radiation in
the entire visible range of the spectrum to appear almost white to the eye, and
it is called white hot.
Although it cannot be sensed directly by the human eye, infrared radiation
can be detected by infrared cameras, which transmit the information to mi-
croprocessors to display visual images of objects at night. Rattlesnakes can
sense the infrared radiation or the "body heat" coming off warm-blooded ani-
mals, and thus they can see at night without using any instruments. Similarly,
honeybees are sensitive to ultraviolet radiation. A surface that reflects all of
the light appears white, while a surface that absorbs all of the light incident on
it appears black. (Then how do we see a black surface?)
It should be clear from this discussion that the color of an object is not due
to emission, which is primarily in the infrared region, unless the surface tem-
perature of the object exceeds about 1000 K. Instead, the color of a surface de-
pends on the absorption and reflection characteristics of the surface and is due
to selective absorption and reflection of the incident visible radiation coming
from a light source such as the sun or an incandescent lightbulb. A piece of
clothing containing a pigment that reflects red while absorbing the remaining
parts of the incident light appears "red" to the eye (Fig. 11-10). Leaves appear
"green" because their cells contain the pigment chlorophyll, which strongly
reflects green while absorbing other colors.
It is left as an exercise to show that integration of the spectral blackbody
emissive power E bK over the entire wavelength spectrum gives the total black-
body emissive power E b :
E b (T)
E bx (k, T)d\ = aT 4
(W/m 2 )
(11-6)
Thus, we obtained the Stefan-Boltzmann law (Eq. 11-3) by integrating
Planck's law (Eq. 11-4) over all wavelengths. Note that on an E bk -X chart, E bk
corresponds to any value on the curve, whereas E b corresponds to the area
under the entire curve for a specified temperature (Fig. 1 1-11). Also, the term
total means "integrated over all wavelengths."
EXAMPLE 11-1
Radiation Emission from a Black Ball
Consider a 20-cm-diameter spherical ball at 800 K suspended in air as shown
in Figure 11-12. Assuming the ball closely approximates a blackbody, deter-
mine (a) the total blackbody emissive power, (b) the total amount of radiation
emitted by the ball in 5 min, and (c) the spectral blackbody emissive power at
a wavelength of 3 u,m.
SOLUTION An isothermal sphere is suspended in air. The total blackbody
emissive power, the total radiation emitted in 5 minutes, and the spectral
blackbody emissive power at 3 mm are to be determined.
Assumptions The ball behaves as a blackbody.
cen58933_chll.qxd 9/9/2002 9:38 AM Page 569
Analysis (a) The total blackbody emissive power is determined from the
Stefan-Boltzmann law to be
E b = oT 4 = (5.67 X 1(T 8 W/m 2 • K 4 )(800 K) 4 = 23.2 X 10 3 W/m 2 = 23.2 kW/m 2
That is, the ball emits 23.2 kJ of energy in the form of electromagnetic radia-
tion per second per m 2 of the surface area of the ball.
(£>) The total amount of radiation energy emitted from the entire ball in 5 min
is determined by multiplying the blackbody emissive power obtained above by
the total surface area of the ball and the given time interval:
A s = ttD 2 = tt(0.2 m) 2 = 0.1257 m 2
60s
At = (5 min)
1 min
300 s
g rad = £A At = (23.2 kW/m 2 )(0.1257 m 2 )(300 s)l
= 876 kJ
lkJ
1000 W ■ s
That is, the ball loses 876 kJ of its internal energy in the form of electromag-
netic waves to the surroundings in 5 min, which is enough energy to raise the
temperature of 1 kg of water by 50 C C. Note that the surface temperature of the
ball cannot remain constant at 800 K unless there is an equal amount of energy
flow to the surface from the surroundings or from the interior regions of the ball
through some mechanisms such as chemical or nuclear reactions,
(c) The spectral blackbody emissive power at a wavelength of 3 fim is deter-
mined from Planck's distribution law to be
C,
3.743 X 10 8 W • |j,m 4 /m 2
exp
(3 |xm) 5
exp
1.4387 X 10 4 |xm ■ K
(3 |j,m)(800 K)
3848 W/m 2 • |xm
569
CHAPTER 11
800 K
FIGURE 11-12
The spherical ball considered in
Example 11-1.
The Stefan-Boltzmann law E b (T) = uT 4 gives the total radiation emitted
by a blackbody at all wavelengths from X = to X = °°. But we are often
interested in the amount of radiation emitted over some wavelength band.
For example, an incandescent lightbulb is judged on the basis of the radia-
tion it emits in the visible range rather than the radiation it emits at all
wavelengths.
The radiation energy emitted by a blackbody per unit area over a wave-
length band from X = to X. is determined from (Fig. 11-13)
E b ,^{T)
E bx {\, T) d\
(W/m 2 )
(11-7)
It looks like we can determine E bi _ x by substituting the E bx relation from
Eq. 11-4 and performing this integration. But it turns out that this integration
does not have a simple closed-form solution, and performing a numerical
integration each time we need a value of E b _ _ k is not practical. Therefore,
we define a dimensionless quantity / x called the blackbody radiation func-
tion as
E b ,o-x(T) = \E b} .(KT)dX
E bX (\ T)
FIGURE 11-13
On an E bk —\ chart, the area under
the curve to the left of the X = X : line
represents the radiation energy emitted
by a blackbody in the wavelength
range 0-X i for the given temperature.
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570
HEAT TRANSFER
TABLE 11-2
Blackbody radiation functions f x
XT,
XT,
|j,m • K
>x
|xm • K
fx
200
0.000000
6200
0.754140
400
0.000000
6400
0.769234
600
0.000000
6600
0.783199
800
0.000016
6800
0.796129
1000
0.000321
7000
0.808109
1200
0.002134
7200
0.819217
1400
0.007790
7400
0.829527
1600
0.019718
7600
0.839102
1800
0.039341
7800
0.848005
2000
0.066728
8000
0.856288
2200
0.100888
8500
0.874608
2400
0.140256
9000
0.890029
2600
0.183120
9500
0.903085
2800
0.227897
10,000
0.914199
3000
0.273232
10,500
0.923710
3200
0.318102
11,000
0.931890
3400
0.361735
11,500
0.939959
3600
0.403607
12,000
0.945098
3800
0.443382
13,000
0.955139
4000
0.480877
14,000
0.962898
4200
0.516014
15,000
0.969981
4400
0.548796
16,000
0.973814
4600
0.579280
18,000
0.980860
4800
0.607559
20,000
0.985602
5000
0.633747
25,000
0.992215
5200
0.658970
30,000
0.995340
5400
0.680360
40,000
0.997967
5600
0.701046
50,000
0.998953
5800
0.720158
75,000
0.999713
6000
0.737818
100,000
0.999905
ACO
E bk (X,T)dk
oT 4
(11-8)
The function f x represents the fraction of radiation emitted from a blackbody
at temperature T in the wavelength band from X = to X. The values of f x are
listed in Table 11-2 as a function of XT, where X is in |xm and T is in K.
The fraction of radiation energy emitted by a blackbody at tempera-
ture T over a finite wavelength band from X = X t to X = X 2 is determined
from (Fig. 11-14)
FIGURE 11-14
Graphical representation of the
fraction of radiation emitted in the
wavelength band from X { to \ 2 .
A^m=Am-AXT)
(11-9)
where / x (T) and/^(T) are blackbody radiation functions corresponding to
X^and X 2 T, respectively.
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EXAMPLE 11-2 Emission of Radiation from a Lightbulb
The temperature of the filament of an incandescent lightbulb is 2500 K. As-
suming the filament to be a blackbody, determine the fraction of the radiant en-
ergy emitted by the filament that falls in the visible range. Also, determine the
wavelength at which the emission of radiation from the filament peaks.
SOLUTION The temperature of the filament of an incandescent lightbulb is
given. The fraction of visible radiation emitted by the filament and the wave-
length at which the emission peaks are to be determined.
Assumptions The filament behaves as a blackbody.
Analysis The visible range of the electromagnetic spectrum extends from
X 1 = 0.4 (xm to X 2 = 0.76 |a,m. Noting that T= 2500 K, the blackbody radia-
tion functions corresponding to A^Tand \ 2 7" are determined from Table 11-2
to be
\ 2 T = (0.76 |xm)(2500 K)
1000 Lim • K
1900 |xm • K
0.000321
0.053035
That is, 0.03 percent of the radiation is emitted at wavelengths less than
0.4 |mm and 5.3 percent at wavelengths less than 0.76 |j,m. Then the fraction
of radiation emitted between these two wavelengths is (Fig. 11-15)
A,-\, A, A,
0.053035 - 0.000321 = 0.0527135
Therefore, only about 5 percent of the radiation emitted by the filament of the
lightbulb falls in the visible range. The remaining 95 percent of the radiation
appears in the infrared region in the form of radiant heat or "invisible light," as
it used to be called. This is certainly not a very efficient way of converting elec-
trical energy to light and explains why fluorescent tubes are a wiser choice for
lighting.
The wavelength at which the emission of radiation from the filament peaks is
easily determined from Wien's displacement law to be
(xr)„
2897.8 |xm • K
2897.8 (xm • K
2500 K
1.16 |xm
Discussion Note that the radiation emitted from the filament peaks in the in-
frared region.
571
CHAPTER 11
0.4 0.76 1.16 Ji,nm
FIGURE 11-15
Graphical representation of the
fraction of radiation emitted in the
visible range in Example 11-2.
11^ ■ RADIATION INTENSITY
Radiation is emitted by all parts of a plane surface in all directions into the
hemisphere above the surface, and the directional distribution of emitted (or
incident) radiation is usually not uniform. Therefore, we need a quantity that
describes the magnitude of radiation emitted (or incident) in a specified direc-
tion in space. This quantity is radiation intensity, denoted by /. Before we can
describe a directional quantity, we need to specify direction in space. The di-
rection of radiation passing through a point is best described in spherical co-
ordinates in terms of the zenith angle and the azimuth angle <]), as shown in
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572
HEAT TRANSFER
FIGURE 11-16
Radiation intensity is used to describe
the variation of radiation energy with
direction.
Plain angle,
Arc length
A slice of pizza of plain angle a
Solid angle,
A slice of watermelon of solid angle co
FIGURE 11-17
Describing the size of a slice of pizza
by a plain angle, and the size of a
watermelon slice by a solid angle.
Figure 11-16. Radiation intensity is used to describe how the emitted radia-
tion varies with the zenith and azimuth angles.
If all surfaces emitted radiation uniformly in all directions, the emissive
power would be sufficient to quantify radiation, and we would not need to
deal with intensity. The radiation emitted by a blackbody per unit normal area
is the same in all directions, and thus there is no directional dependence. But
this is not the case for real surfaces. Before we define intensity, we need to
quantify the size of an opening in space.
Solid Angle
Let us try to quantify the size of a slice of pizza. One way of doing that is to
specify the arc length of the outer edge of the slice, and to form the slice by
connecting the endpoints of the arc to the center. A more general approach is
to specify the angle of the slice at the center, as shown in Figure 11-17. An an-
gle of 90° (or tt/2 radians), for example, always represents a quarter pizza, no
matter what the radius is. For a circle of unit radius, the length of an arc is
equivalent in magnitude to the plane angle it subtends (both are 2ir for a com-
plete circle of radius r = 1).
Now consider a watermelon, and let us attempt to quantify the size of a
slice. Again we can do it by specifying the outer surface area of the slice (the
green part), or by working with angles for generality. Connecting all points at
the edges of the slice to the center in this case will form a three-dimensional
body (like a cone whose tip is at the center), and thus the angle at the center in
this case is properly called the solid angle. The solid angle is denoted by to,
and its unit is the steradian (sr). In analogy to plane angle, we can say that the
area of a surface on a sphere of unit radius is equivalent in magnitude to the
solid angle it subtends (both are 4tt for a sphere of radius r = 1).
This can be shown easily by considering a differential surface area on a
sphere dS = r 2 sin QdQdfy, as shown in Figure 11-18, and integrating it from
= to = it, and from cj) = to 4> = 2tt. We get
Je =
sin erfOcb = 2nr 2 sin QdQ = 4irr
(11-10)
dS ■
sphere
which is the formula for the area of a sphere. For r = 1 it reduces to S = 4tt,
and thus the solid angle associated with a sphere is w = 4ir sr. For a hemi-
sphere, which is more relevant to radiation emitted or received by a surface, it
is a) = 2ir sr.
The differential solid angle dco subtended by a differential area dS on a
sphere of radius r can be expressed as
da>
dS
sin QdQdfy
(11-11)
Note that the area dS is normal to the direction of viewing since dS is viewed
from the center of the sphere. In general, the differential solid angle du> sub-
tended by a differential surface area dA when viewed from a point at a dis-
tance r from dA is expressed as
dA„ dA cos a
du>
(11-12)
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CHAPTER 11
Solid angle:
du,= dS/r 2
Solid angle for a hemisphere
JP ti/2 p Ik
da>= sin 9 dQ d<b = 2iz
J 0=8 J 0=<I>
Hemisphere
dS = (r sin Q d<p)(r dB)
= r 2 sin6 d9 d<t>
rde
FIGURE 11-18
The emission of
radiation from
a differential
surface element
into the surrounding
hemispherical space
through a differential
solid angle.
where a is the angle between the normal of the surface and the direction of
viewing, and thus dA n = dA cos a is the normal (or projected) area to the di-
rection of viewing.
Small surfaces viewed from relatively large distances can approximately be
treated as differential areas in solid angle calculations. For example, the solid
angle subtended by a 5 cm 2 plane surface when viewed from a point O at a
distance of 80 cm along the normal of the surface is
r 2
5 cm 2
(80 cm) 2
7.81 X l(T 4 sr
If the surface is tilted so that the normal of the surface makes an angle of
a = 60° with the line connecting point O to the center of the surface, the pro-
jected area would be dA n = dA cos a = (5 cm 2 )cos 60° = 2.5 cm 2 , and the
solid angle in this case would be half of the value just determined.
Intensity of Emitted Radiation
Consider the emission of radiation by a differential area element dA of a sur-
face, as shown in Figure 11-18. Radiation is emitted in all directions into the
hemispherical space, and the radiation streaming though the surface area dS is
proportional to the solid angle dw subtended by dS. It is also proportional to
the radiating area dA as seen by an observer on dS, which varies from a max-
imum of dA when dS is at the top directly above dA (0 = 0°) to a minimum of
zero when dS is at the bottom (0 = 90°). Therefore, the effective area of dA for
emission in the direction of is the projection of dA on a plane normal to 0,
which is dA cos 0. Radiation intensity in a given direction is based on a unit
area normal to that direction to provide a common basis for the comparison of
radiation emitted in different directions.
The radiation intensity for emitted radiation / e (0, (j)) is defined as the rate
at which radiation energy dQ e is emitted in the (0, 4>) direction per unit area
normal to this direction and per unit solid angle about this direction. That is,
/,(6, <|>)
dQ e
dQ e
dA cos G • rfoj dA cos G sin QdQdfy
(W/m 2 • sr) (11-13)
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HEAT TRANSFER
The radiation flux for emitted radiation is the emissive power E (the rate at
which radiation energy is emitted per unit area of the emitting surface), which
can be expressed in differential form as
dE
(XL
dA
4(9, <|>) cos sin MQdfy
(11-14)
Projected area
A cos0
s
Solid angle, a
FIGURE 11-19
Radiation intensity is based on
projected area, and thus the
calculation of radiation emission
from a surface involves the projection
of the surface.
FIGURE 11-20
Radiation incident on a surface in the
direction (0, cf>).
Noting that the hemisphere above the surface will intercept all the radiation
rays emitted by the surface, the emissive power from the surface into the
hemisphere surrounding it can be determined by integration as
r r 2tt r -nil
E=\ dE= 4(0, <|>) cos sin MM§ (W/m 2 )
J. Jc|3=oJe=o
(11-15)
hemisphere
The intensity of radiation emitted by a surface, in general, varies with di-
rection (especially with the zenith angle 0). But many surfaces in practice can
be approximated as being diffuse. For a diffusely emitting surface, the inten-
sity of the emitted radiation is independent of direction and thus I e = constant.
rl-n rir/2
Noting that cos 9 sin QdQdfy = it, the emissive power relation in
Jc|j=oJe=o
Eq. 11-15 reduces in this case to
Diffusely emitting surface:
tt4
(W/m 2 )
(11-16)
Note that the factor in Eq. 11-16 is it. You might have expected it to be 2tt
since intensity is radiation energy per unit solid angle, and the solid angle as-
sociated with a hemisphere is 2tt. The reason for the factor being it is that the
emissive power is based on the actual surface area whereas the intensity is
based on the projected area (and thus the factor cos that accompanies it), as
shown in Figure 11-19.
For a blackbody, which is a diffuse emitter, Eq. 11-16 can be expressed as
Blackbody:
■nh
(11-17)
where E b = uT 4 is the blackbody emissive power. Therefore, the intensity of
the radiation emitted by a blackbody at absolute temperature T is
Blackbody:
h(T)
E b (T)
CT r 4
it
(W/m 2 • sr)
(11-18)
Incident Radiation
All surfaces emit radiation, but they also receive radiation emitted or reflected
by other surfaces. The intensity of incident radiation 4(0, <$>) is defined as the
rate at which radiation energy dG is incident from the (0, §) direction per unit
area of the receiving surface normal to this direction and per unit solid angle
about this direction (Fig. 11-20). Here is the angle between the direction of
incident radiation and the normal of the surface.
The radiation flux incident on a surface from all directions is called irradi-
ation G, and is expressed as
r rl-rr /-it/2
G = \dG= 7,(0, qV) cos sin QdQdfy (W/m 2 )
J Jcb=oJe=o
(11-19)
hemisphere
cen58933_chll.qxd 9/9/2002 9:38 AM Page 575
Therefore irradiation represents the rate at which radiation energy is incident
on a surface per unit area of the surface. When the incident radiation is diffuse
and thus /,- = constant, Eq. 11-19 reduces to
Diffusely incident radiation:
ttI,
(W/m 2 )
(11-20)
Again note that irradiation is based on the actual surface area (and thus the
factor cos 0), whereas the intensity of incident radiation is based on the pro-
jected area.
575
CHAPTER 11
Radiosity
Surfaces emit radiation as well as reflecting it, and thus the radiation leaving
a surface consists of emitted and reflected components, as shown in Figure
11-21. The calculation of radiation heat transfer between surfaces involves
the total radiation energy streaming away from a surface, with no regard for
its origin. Thus, we need to define a quantity that represents the rate at which
radiation energy leaves a unit area of a surface in all directions. This quantity
is called the radiosity /, and is expressed as
J4>=oJe=
I e+r (Q, d>) cos 6 sin QdQdfy (W/m 2 )
(11-21)
where I e+r is the sum of the emitted and reflected intensities. For a surface that
is both a diffuse emitter and a diffuse reflector, I e+r = constant, and the ra-
diosity relation reduces to
Diffuse emitter and reflector:
J = -nl.
(W/m 2 )
(11-22)
For a blackbody, radiosity J is equivalent to the emissive power E b since a
blackbody absorbs the entire radiation incident on it and there is no reflected
component in radiosity.
Spectral Quantities
So far we considered total radiation quantities (quantities integrated over all
wavelengths), and made no reference to wavelength dependence. This lumped
approach is adequate for many radiation problems encountered in practice.
But sometimes it is necessary to consider the variation of radiation with wave-
length as well as direction, and to express quantities at a certain wavelength X.
or per unit wavelength interval about X. Such quantities are referred to as
spectral quantities to draw attention to wavelength dependence. The modifier
"spectral" is used to indicate "at a given wavelength."
The spectral radiation intensity 7 X (X, 6, 4>), for example, is simply the total
radiation intensity 7(0, 4>) per unit wavelength interval about X. The spectral
intensity for emitted radiation I k e (X, 0, 4>) can be defined as the rate at which
radiation energy dQ e is emitted at the wavelength X in the (0, <§>) direction per
unit area normal to this direction, per unit solid angle about this direction,
and it can be expressed as
4,,(x,e,<|>)
dQ e
dA cos G • dtx> ■ dX
(W/m 2 • sr ■ (Am)
(11-23)
Radiosity, /
(Reflected
irradiation)
Irradiation, G
FIGURE 11-21
The three kinds of radiation flux
(in W/m 2 ): emissive power,
irradiation, and radiosity.
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HEAT TRANSFER
Area = / h, e dX=I e
dX
FIGURE 11-22
Integration of a "spectral" quantity for
all wavelengths gives the "total"
quantity.
Then the spectral emissive power becomes
-2-u r TT
4>=oJe=
I % e (\, 9, <|))cos sin QdQdfy (W/m 2 )
(11-24)
Similar relations can be obtained for spectral irradiation G x , and spectral ra-
diosity J x by replacing I k e in this equation by I x , and I x e+r , respectively.
When the variation of spectral radiation intensity / x with wavelength X is
known, the total radiation intensity / for emitted, incident, and emitted + re-
flected radiation can be determined by integration over the entire wavelength
spectrum as (Fig. 11-22)
/..
r
Jo
h e dX, I,
Jo
and
/,
Wo"
Jo
r dX
(11-25)
These intensities can then be used in Eqs. 11-15, 11-19, and 11-21 to deter-
mine the emissive power E, irradiation G, and radiosity J, respectively.
Similarly, when the variations of spectral radiation fluxes E x , G x , and J k
with wavelength X are known, the total radiation fluxes can be determined by
integration over the entire wavelength spectrum as
Jo
E x dX,
Jo
GidX, and
J
JidX
(11-26)
When the surfaces and the incident radiation are diffuse, the spectral radia-
tion fluxes are related to spectral intensities as
ifh
TT/,
and
Ji = tt/i
(11-27)
75 cm
A i = 3 cm 2
7, = 600 K
FIGURE 11-23
Schematic for Example 11-3.
Note that the relations for spectral and total radiation quantities are of the
same form.
The spectral intensity of radiation emitted by a blackbody at an absolute
temperature T at a wavelength X has been determined by Max Planck, and is
expressed as
hx(KT)
lhc\
X 5 [exp(hc Q /XkT) - 1]
(W/m 2 ■ sr • p,m)
(11-28)
where h = 6.6256 X 1(T 34 J • s is the Planck constant, k = 1.38065 X 1(T 23
J/K is the Boltzmann constant, and c = 2.9979 X 10 8 m/s is the speed of light
in a vacuum. Then the spectral blackbody emissive power is, from Eq. 11-27,
E bx (k,T) = irI bx (\,T) (11-29)
A simplified relation for E bx is given by Eq. 11-4.
EXAMPLE 11-3 Radiation Incident on a Small Surface
A small surface of area A x = 3 cm 2 emits radiation as a blackbody at T x = 600
K. Part of the radiation emitted by A x strikes another small surface of area A z =
5 cm 2 oriented as shown in Figure 1 1-23. Determine the solid angle subtended
by A 2 when viewed from A l7 and the rate at which radiation emitted by A 1 that
strikes A 2 .
cen58933_chll.qxd 9/9/2002 9:38 AM Page 577
SOLUTION A surface is subjected to radiation emitted by another surface. The
solid angle subtended and the rate at which emitted radiation is received are to
be determined.
Assumptions 1 Surface A 1 emits diffusely as a blackbody. 2 Both A 1 and A 2 can
be approximated as differential surfaces since both are very small compared to
the square of the distance between them.
Analysis Approximating both A 1 and A z as differential surfaces, the solid angle
subtended by A 2 when viewed from A 1 can be determined from Eq. 11-12 to be
A n ,2 _ A 2 cos 2 _ (5 cm 2 ) cos 40°
T 2 "" 7 2 ~ (75 cm) 2
6.81 X 10" 4 sr
since the normal of A 2 makes 40° with the direction of viewing. Note that solid
angle subtended by A 2 would be maximum if A 2 were positioned normal to the
direction of viewing. Also, the point of viewing on A 1 is taken to be a point in
the middle, but it can be any point since A 1 is assumed to be very small.
The radiation emitted by A x that strikes A 2 is equivalent to the radiation
emitted by A 1 through the solid angle a> 2 _ x . The intensity of the radiation
emitted by A x is
MTd
or?
(5.67 X 1(T 8 W/m 2 • K 4 )(600 K) 4
2339 W/m 2 • sr
This value of intensity is the same in all directions since a blackbody is a diffuse
emitter. Intensity represents the rate of radiation emission per unit area normal to
the direction of emission per unit solid angle. Therefore, the rate of radiation en-
ergy emitted by A x in the direction of : through the solid angle w 2 _j is deter-
mined by multiplying l x by the area of A x normal to : and the solid angle w 2 -i-
That is,
Qi-2 = h( A i cose,)<o 2 -i
= (2339 W/m 2 • sr)(3 X 1(T 4 cos 55° m 2 )(6.81 X 1(T 4 sr)
= 2.74 X 10" 4 W
Therefore, the radiation emitted from surface A x will strike surface A 2 at a rate
of 2.74 X 10- 4 W.
Discussion The total rate of radiation emission from surface A x is Q e = A x aT^ =
2.204 W. Therefore, the fraction of emitted radiation that strikes A 2 is 2.74 X
10~ 4 /2.204 = 0.00012 (or 0.012 percent). Noting that the solid angle associ-
ated with a hemisphere is 2ir, the fraction of the solid angle subtended by A 2 is
6.81 X 10- 4 /(2tt) = 0.000108 (or 0.0108 percent), which is 0.9 times the frac-
tion of emitted radiation. Therefore, the fraction of the solid angle a surface oc-
cupies does not represent the fraction of radiation energy the surface will receive
even when the intensity of emitted radiation is constant. This is because radiation
energy emitted by a surface in a given direction is proportional to the projected
area of the surface in that direction, and reduces from a maximum at = 0° (the
direction normal to surface) to zero at = 90° (the direction parallel to surface).
577
CHAPTER 11
11-5 ■ RADIATIVE PROPERTIES
Most materials encountered in practice, such as metals, wood, and bricks,
are opaque to thermal radiation, and radiation is considered to be a surface
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578
HEAT TRANSFER
phenomenon for such materials. That is, thermal radiation is emitted or ab-
sorbed within the first few microns of the surface, and thus we speak of radia-
tive properties of surfaces for opaque materials.
Some other materials, such as glass and water, allow visible radiation to
penetrate to considerable depths before any significant absorption takes place.
Radiation through such semitransparent materials obviously cannot be con-
sidered to be a surface phenomenon since the entire volume of the material in-
teracts with radiation. On the other hand, both glass and water are practically
opaque to infrared radiation. Therefore, materials can exhibit different behav-
ior at different wavelengths, and the dependence on wavelength is an impor-
tant consideration in the study of radiative properties such as emissivity,
absorptivity, reflectivity, and transmissivity of materials.
In the preceding section, we defined a blackbody as a perfect emitter and
absorber of radiation and said that no body can emit more radiation than a
blackbody at the same temperature. Therefore, a blackbody can serve as a
convenient reference in describing the emission and absorption characteristics
of real surfaces.
Emissivity
The emissivity of a surface represents the ratio of the radiation emitted by the
surface at a given temperature to the radiation emitted by a blackbody at the
same temperature. The emissivity of a surface is denoted by e, and it varies
between zero and one, ^ e < 1. Emissivity is a measure of how closely a
surface approximates a blackbody, for which e = 1 .
The emissivity of a real surface is not a constant. Rather, it varies with the
temperature of the surface as well as the wavelength and the direction of the
emitted radiation. Therefore, different emissivities can be defined for a sur-
face, depending on the effects considered. The most elemental emissivity of a
surface at a given temperature is the spectral directional emissivity, which is
defined as the ratio of the intensity of radiation emitted by the surface at a
specified wavelength in a specified direction to the intensity of radiation emit-
ted by a blackbody at the same temperature at the same wavelength. That is,
4 e (k, 9, <(>, T)
e K e (\, 6, 4>, T) = ' (1 1 -30)
where the subscripts \ and 9 are used to designate spectral and directional
quantities, respectively. Note that blackbody radiation intensity is independent
of direction, and thus it has no functional dependence on and c|).
The total directional emissivity is defined in a like manner by using total
intensities (intensities integrated over all wavelengths) as
/,(6, c|), T)
e9 (9,4>,7)= (11-31)
In practice, it is usually more convenient to work with radiation properties av-
eraged over all directions, called hemispherical properties. Noting that the in-
tegral of the rate of radiation energy emitted at a specified wavelength per unit
surface area over the entire hemisphere is spectral emissive power, the spec-
tral hemispherical emissivity can be expressed as
cen58933_chll.qxd 9/9/2002 9:38 AM Page 579
£x(X, T)
ErfX, T)
E bX (k, T)
(11-32)
Note that the emissivity of a surface at a given wavelength can be different at
different temperatures since the spectral distribution of emitted radiation (and
thus the amount of radiation emitted at a given wavelength) changes with
temperature.
Finally, the total hemispherical emissivity is defined in terms of the radia-
tion energy emitted over all wavelengths in all directions as
E(T)
e{T)
E b (T)
(11-33)
Therefore, the total hemispherical emissivity (or simply the "average emis-
sivity") of a surface at a given temperature represents the ratio of the total ra-
diation energy emitted by the surface to the radiation emitted by a blackbody
of the same surface area at the same temperature.
Noting from Eqs. 11-26 and 11-32 that E
E x d\ and E x (\, T)
e x (\, T)E bx (\, T), and the total hemispherical emissivity can also be ex-
pressed as
e(T)
E{T)
E b (T)
f
e x (X, T)E bX (X, T)dX
vT 4
(11-34)
since E b (T) = oT 4 . To perform this integration, we need to know the varia-
tion of spectral emissivity with wavelength at the specified temperature. The
integrand is usually a complicated function, and the integration has to be per-
formed numerically. However, the integration can be performed quite easily
by dividing the spectrum into a sufficient number of wavelength bands and as-
suming the emissivity to remain constant over each band; that is, by express-
ing the function e x (\, T) as a step function. This simplification offers great
convenience for little sacrifice of accuracy, since it allows us to transform the
integration into a summation in terms of blackbody emission functions.
As an example, consider the emissivity function plotted in Figure 11-24. It
seems like this function can be approximated reasonably well by a step func-
tion of the form
(11-35)
Then the average emissivity can be determined from Eq. 11-34 by breaking
the integral into three parts and utilizing the definition of the blackbody radi-
ation function as
6] = constant,
0<X<X,
e 2 = constant,
X, < X < X 2
e 3 = constant,
X 2 < X < °°
s(T)
E bX dX
Jo
E b idX e 3 1
h, A
+
E hi dX
579
CHAPTER 11
1 Actual
variation
E b E b E b FIGURE 11-24
= s ifo-k [ (T) + Sifkt-kiT) + s 3 / x ,_-^(7) (11-36) Approximating the actual variation
Radiation is a complex phenomenon as it is, and the consideration of the of emlssmt y wlth wavelength
wavelength and direction dependence of properties, assuming sufficient data " e P
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HEAT TRANSFER
Real surface:
e e
it constant
*
# constant
Diffuse surface:
e e
= constant
Gray surface:
*
= constant
Diffuse
gray surface:
8 =
= 6^=69 = constant
FIGURE 11-25
The effect of diffuse and gray
approximations on the emissivity
of a surface.
Nonconductor
FIGURE 11-26
Typical variations of emissivity with
direction for electrical conductors
and nonconductors.
exist, makes it even more complicated. Therefore, the gray and diffuse ap-
proximations are often utilized in radiation calculations. A surface is said to be
diffuse if its properties are independent of direction, and gray if its properties
are independent of wavelength. Therefore, the emissivity of a gray, diffuse
surface is simply the total hemispherical emissivity of that surface because of
independence of direction and wavelength (Fig. 11-25).
A few comments about the validity of the diffuse approximation are in order.
Although real surfaces do not emit radiation in a perfectly diffuse manner as a
blackbody does, they often come close. The variation of emissivity with direc-
tion for both electrical conductors and nonconductors is given in Figure 11-26.
Here is the angle measured from the normal of the surface, and thus = for
radiation emitted in a direction normal to the surface. Note that e e remains
nearly constant for about < 40° for conductors such as metals and for 9 < 70°
for nonconductors such as plastics. Therefore, the directional emissivity of a
surface in the normal direction is representative of the hemispherical emissivity
of the surface. In radiation analysis, it is common practice to assume the sur-
faces to be diffuse emitters with an emissivity equal to the value in the normal
(0 = 0) direction.
The effect of the gray approximation on emissivity and emissive power of a
real surface is illustrated in Figure 11-27. Note that the radiation emission from
a real surface, in general, differs from the Planck distribution, and the emission
curve may have several peaks and valleys. A gray surface should emit as much
radiation as the real surface it represents at the same temperature. Therefore, the
areas under the emission curves of the real and gray surfaces must be equal.
The emissivities of common materials are listed in Table A-18 in the appen-
dix, and the variation of emissivity with wavelength and temperature is illus-
trated in Figure 11-28. Typical ranges of emissivity of various materials are
given in Figure 11-29. Note that metals generally have low emissivities, as low
as 0.02 for polished surfaces, and nonmetals such as ceramics and organic ma-
terials have high ones. The emissivity of metals increases with temperature.
Also, oxidation causes significant increases in the emissivity of metals. Heav-
ily oxidized metals can have emissivities comparable to those of nonmetals.
FIGURE 11-27
Comparison of the emissivity (a) and
emissive power (b) of a real surface
with those of a gray surface and a
blackbody at the same temperature.
£>!
Blackbody, £ = 1
Gray surface, £ = const.
T = const.
, — Blackbody,
E„X
1 Gray surface, \
/ E x
= eE bl \
Real surface, V v ,
/,
/ s
I /
/
T
= const.
E X = £ \ E bX
\:
(a)
(b)
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CHAPTER 11
1.0
0.8
■3 0.6
o 0.4
0.2
Aluminum
^oxide,
1400 K
s Stainless steel,
1200 K
heavily oxidized
Stainless steel,
800 K
lightly oxidized
1.0
_ 0.8 -
Heavily oxidized
stainless steel
1 0.2 0.40.6 1
2 4 6 10
Wavelength, X, fim
(a)
20 40 60 100
1500 2000
Temperature, K
(in
3000 3500
FIGURE 11-28
The variation of normal emissivity with (a) wavelength and (b) temperature for various materials.
Care should be exercised in the use and interpretation of radiation property
data reported in the literature, since the properties strongly depend on the
surface conditions such as oxidation, roughness, type of finish, and cleanli-
ness. Consequently, there is considerable discrepancy and uncertainty in the
reported values. This uncertainty is largely due to the difficulty in character-
izing and describing the surface conditions precisely.
EXAMPLE 11-4 Emissivity of a Surface and Emissive Power
The spectral emissivity function of an opaque surface at 800 K is approximated
as (Fig. 11-30)
e, = 0.3,
< X < 3 jrni
e 2 = 0.8,
3 jjim < X < 7 |xm
e 3 = 0.1,
7 jjim < X < oo
Determine the average emissivity of the surface and its emissive power.
SOLUTION The variation of emissivity of a surface at a specified temperature
with wavelength is given. The average emissivity of the surface and its emissive
power are to be determined.
Analysis The variation of the emissivity of the surface with wavelength is given
as a step function. Therefore, the average emissivity of the surface can be de-
termined from Eq. 11-34 by breaking the integral into three parts,
e(T)
f
— ; 1 :
aT 4
ei/(WL,(r) + e 2 f h _ H (T) + e 3 A 2 -Xn
Bif h + S2(A 2 -A,) + R.(l -A 2 )
Vegetation, water, skin CZ1
Building materials, paints
Rocks, soil 1
Glasses, minerals I I
Carbon ■
Ceramics [
1
Oxidized
' metals
D Metals, unpolished
■ Polished metals
0.2
0.4
0.6 0.8 1.0
FIGURE 11-29
Typical ranges of emissivity for
various materials.
7 X, um
FIGURE 11-30
The spectral emissivity of the surface
considered in Example 11-4.
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HEAT TRANSFER
where f x and f x are blackbody radiation functions corresponding to X^Tand
\ Z T. These functions are determined from Table 11-2 to be
\ t T = (3 nm)(800 K) = 2400 |xm • K -> / x = 0.140256
\ 2 T = (7 p,m)(800 K) = 5600 p,m • K -> f k = 0.701046
Note that f _ h = f X} - f = f Xi , since f = 0, and f x , z _ rj , = f a - f x% = 1
since 4 = 1. Substituting,
L
s = 0.3 X 0.140256 + 0.8(0.701046 - 0.140256) + 0.1(1 - 0.701046)
= 0.521
That is, the surface will emit as much radiation energy at 800 K as a gray surface
having a constant emissivity of e = 0.521. The emissive power of the surface is
E = eoT 4 = 0.521(5.67 X lO" 8 W/m 2 • K 4 )(800 K) 4 = 12,100 W/m 2
Discussion Note that the surface emits 12.1 kJ of radiation energy per second
per m 2 area of the surface.
Reflected
FIGURE 11-31
The absorption, reflection, and
transmission of incident radiation by a
semitransparent material.
Absorptivity, Reflectivity, and Transmissivity
Everything around us constantly emits radiation, and the emissivity represents
the emission characteristics of those bodies. This means that every body, in-
cluding our own, is constantly bombarded by radiation coming from all direc-
tions over a range of wavelengths. Recall that radiation flux incident on a
surface is called irradiation and is denoted by G.
When radiation strikes a surface, part of it is absorbed, part of it is reflected,
and the remaining part, if any, is transmitted, as illustrated in Figure 11-31.
The fraction of irradiation absorbed by the surface is called the absorptivity
a, the fraction reflected by the surface is called the reflectivity p, and the frac-
tion transmitted is called the transmissivity t. That is,
Absorptivity:
Reflectivity:
Transmissivity:
Absorbed radiation _ "abs
Incident radiation G
Reflected radiation _ G rcf
Incident radiation G
G«
G
Transmitted radiation
Incident radiation
0<ct< 1
(11-37)
0< p< 1
(11-38)
0<T< 1
(11-39)
where G is the radiation energy incident on the surface, and G abs , G ref , and G tr
are the absorbed, reflected, and transmitted portions of it, respectively. The
first law of thermodynamics requires that the sum of the absorbed, reflected,
and transmitted radiation energy be equal to the incident radiation. That is,
Dividing each term of this relation by G yields
a + p + t = 1
For opaque surfaces, t = 0, and thus
(11-40)
(11-41)
cen58933_chll.qxd 9/9/2002 9:38 AM Page 583
a + p = 1
(11-42)
This is an important property relation since it enables us to determine both the
absorptivity and reflectivity of an opaque surface by measuring either of these
properties.
These definitions are for total hemispherical properties, since G represents
the radiation flux incident on the surface from all directions over the hemi-
spherical space and over all wavelengths. Thus, a, p, and t are the average
properties of a medium for all directions and all wavelengths. However, like
emissivity, these properties can also be defined for a specific wavelength
and/or direction. For example, the spectral directional absorptivity and
spectral directional reflectivity of a surface are defined, respectively, as the
absorbed and reflected fractions of the intensity of radiation incident at a spec-
ified wavelength in a specified direction as
ax,e(^9>4>)
/*..b.ft,e,4>)
4,,(^e><W
and p x ,e(^, 6, cf>)
/u(A..e,4»)
(11-43)
Likewise, the spectral hemispherical absorptivity and spectral hemispher-
ical reflectivity of a surface are defined as
a x (X)
G x (l)
and
Px(V>
GjA)
(11-44)
where G K is the spectral irradiation (in W/m 2 ■ |xm) incident on the surface,
and G K abs and G k ref are the reflected and absorbed portions of it, respectively.
Similar quantities can be defined for the transmissivity of semitransparent
materials. For example, the spectral hemispherical transmissivity of a
medium can be expressed as
txW
c x a)
(11-45)
The average absorptivity, reflectivity, and transmissivity of a surface can
also be defined in terms of their spectral counterparts as
G,d\
G,d\
P\G x dk
GxrfX
r,G x d\
G,d\
(11-46)
The reflectivity differs somewhat from the other properties in that it is bidi-
rectional in nature. That is, the value of the reflectivity of a surface depends not
only on the direction of the incident radiation but also the direction of reflec-
tion. Therefore, the reflected rays of a radiation beam incident on a real surface
in a specified direction will form an irregular shape, as shown in Figure 11-32.
Such detailed reflectivity data do not exist for most surfaces, and even if they
did, they would be of little value in radiation calculations since this would usu-
ally add more complication to the analysis than it is worth.
In practice, for simplicity, surfaces are assumed to reflect in a perfectly
specular or diffuse manner. In specular (or mirrorlike) reflection, the angle
of reflection equals the angle of incidence of the radiation beam. In diffuse
reflection, radiation is reflected equally in all directions, as shown in Figure
583
CHAPTER 11
Normal
Incident Reflected rays
(a)
Normal
(b)
Normal
FIGURE 11-32
Different types of reflection from
a surface: (a) actual or irregular,
(b) diffuse, and (c) specular
or mirrorlike.
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HEAT TRANSFER
1.0
0.8
I? 0.6
-2 0.4
<
0.2
300400 600 1000 2000 40006000
Source temperature, K
FIGURE 11-33
Variation of absorptivity with
the temperature of the source of
irradiation for various common
materials at room temperature.
7
9,
s^6
l.V
r'hite fireclay >
5 s
2. Asbestos
3. Cork
Ns^
_4. Wood
5. Porcelain
\X
6. Concrete
\4
7. Roof shingles
V 8>
9. C
raphit
r^-
11-32. Reflection from smooth and polished surfaces approximates specular
reflection, whereas reflection from rough surfaces approximates diffuse
reflection. In radiation analysis, smoothness is defined relative to wavelength.
A surface is said to be smooth if the height of the surface roughness is much
smaller than the wavelength of the incident radiation.
Unlike emissivity, the absorptivity of a material is practically independent
of surface temperature. However, the absorptivity depends strongly on the
temperature of the source at which the incident radiation is originating. This
is also evident from Figure 11-33, which shows the absorptivities of various
materials at room temperature as functions of the temperature of the radiation
source. For example, the absorptivity of the concrete roof of a house is about
0.6 for solar radiation (source temperature: 5780 K) and 0.9 for radiation orig-
inating from the surrounding trees and buildings (source temperature: 300 K),
as illustrated in Figure 11-34.
Notice that the absorptivity of aluminum increases with the source tempera-
ture, a characteristic for metals, and the absorptivity of electric nonconductors,
in general, decreases with temperature. This decrease is most pronounced for
surfaces that appear white to the eye. For example, the absorptivity of a white
painted surface is low for solar radiation, but it is rather high for infrared
radiation.
FIGURE 11-34
The absorptivity of a material may be
quite different for radiation originating
from sources at different temperatures.
Kirchhoffs Law
Consider a small body of surface area A s , emissivity e, and absorptivity a at
temperature T contained in a large isothermal enclosure at the same tempera-
ture, as shown in Figure 11-35. Recall that a large isothermal enclosure forms
a blackbody cavity regardless of the radiative properties of the enclosure sur-
face, and the body in the enclosure is too small to interfere with the blackbody
nature of the cavity. Therefore, the radiation incident on any part of the sur-
face of the small body is equal to the radiation emitted by a blackbody at tem-
perature T. That is, G = E b (T) = crT 4 , and the radiation absorbed by the small
body per unit of its surface area is
G abs = olG = aaT 4
The radiation emitted by the small body is
F = fctT 4
^emit fcu J
Considering that the small body is in thermal equilibrium with the enclosure,
the net rate of heat transfer to the body must be zero. Therefore, the radiation
emitted by the body must be equal to the radiation absorbed by it:
Thus, we conclude that
A r eaT 4 = A,aaT 4
e(T) = a(T)
(11-47)
That is, the total hemispherical emissivity of a surface at temperature T is
equal to its total hemispherical absorptivity for radiation coming from a black-
body at the same temperature. This relation, which greatly simplifies the radi-
ation analysis, was first developed by Gustav Kirchhoff in 1860 and is now
called Kirchhoffs law. Note that this relation is derived under the condition
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CHAPTER 11
that the surface temperature is equal to the temperature of the source of irradi-
ation, and the reader is cautioned against using it when considerable difference
(more than a few hundred degrees) exists between the surface temperature and
the temperature of the source of irradiation.
The derivation above can also be repeated for radiation at a specified wave-
length to obtain the spectral form of Kirchhoff 's law:
e x (T) = a x (D
(11-48)
This relation is valid when the irradiation or the emitted radiation is inde-
pendent of direction. The form of Kirchhoff 's law that involves no restrictions
is the spectral directional form expressed as e x B (T) = a x $(T). That is, the
emissivity of a surface at a specified wavelength, direction, and temperature
is always equal to its absorptivity at the same wavelength, direction, and
temperature.
It is very tempting to use Kirchhoff 's law in radiation analysis since the re-
lation e = a together with p = 1 — a enables us to determine all three
properties of an opaque surface from a knowledge of only one property.
Although Eq. 11-47 gives acceptable results in most cases, in practice, care
should be exercised when there is considerable difference between the surface
temperature and the temperature of the source of incident radiation.
The Greenhouse Effect
You have probably noticed that when you leave your car under direct sunlight
on a sunny day, the interior of the car gets much warmer than the air outside,
and you may have wondered why the car acts like a heat trap. The answer
lies in the spectral transmissivity curve of the glass, which resembles an in-
verted U, as shown in Figure 11-36. We observe from this figure that glass at
thicknesses encountered in practice transmits over 90 percent of radiation in
the visible range and is practically opaque (nontransparent) to radiation in the
longer-wavelength infrared regions of the electromagnetic spectrum (roughly
X > 3 |xm). Therefore, glass has a transparent window in the wavelength
range 0.3 |xm < \ < 3 |xm in which over 90 percent of solar radiation is
emitted. On the other hand, the entire radiation emitted by surfaces at room
temperature falls in the infrared region. Consequently, glass allows the solar
radiation to enter but does not allow the infrared radiation from the interior
surfaces to escape. This causes a rise in the interior temperature as a result of
the energy build-up in the car. This heating effect, which is due to the nongray
characteristic of glass (or clear plastics), is known as the greenhouse effect,
since it is utilized primarily in greenhouses (Fig. 11-37).
The greenhouse effect is also experienced on a larger scale on earth. The
surface of the earth, which warms up during the day as a result of the absorp-
tion of solar energy, cools down at night by radiating its energy into deep
space as infrared radiation. The combustion gases such as C0 2 and water
vapor in the atmosphere transmit the bulk of the solar radiation but absorb the
infrared radiation emitted by the surface of the earth. Thus, there is concern
that the energy trapped on earth will eventually cause global warming and
thus drastic changes in weather patterns.
In humid places such as coastal areas, there is not a large change between
the daytime and nighttime temperatures, because the humidity acts as a barrier
FIGURE 11-35
The small body contained in a large
isothermal enclosure used in the
development of Kirchhoff 's law.
0.25 0,
0.6 1.5 3.1 4.7 6.3 7.9
Wavelength X, [xm
FIGURE 11-36
The spectral transmissivity of low-iron
glass at room temperature for different
thicknesses.
FIGURE 11-37
A greenhouse traps energy by
allowing the solar radiation to come in
but not allowing the infrared radiation
to go out.
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HEAT TRANSFER
on the path of the infrared radiation coming from the earth, and thus slows
down the cooling process at night. In areas with clear skies such as deserts,
there is a large swing between the daytime and nighttime temperatures be-
cause of the absence of such barriers for infrared radiation.
G = G, cos
Earth's ^.--
surface ^^p
W/m 2
FIGURE 11-38
Solar radiation reaching the
earth's atmosphere and the total solar
irradiance.
-(4%L 2 )G S
FIGURE 11-39
The total solar energy passing through
concentric spheres remains constant,
but the energy falling per unit area
decreases with increasing radius.
11-6 - ATMOSPHERIC AND SOLAR RADIATION
The sun is our primary source of energy. The energy coming off the sun,
called solar energy, reaches us in the form of electromagnetic waves after
experiencing considerable interactions with the atmosphere. The radiation
energy emitted or reflected by the constituents of the atmosphere form the
atmospheric radiation. Below we give an overview of the solar and atmos-
pheric radiation because of their importance and relevance to daily life. Also,
our familiarity with solar energy makes it an effective tool in developing a
better understanding for some of the new concepts introduced earlier. Detailed
treatment of this exciting subject can be found in numerous books devoted to
this topic.
The sun is a nearly spherical body that has a diameter of D ~ 1.39 X 10 9 m
and a mass of m ~ 2 X 10 30 kg and is located at a mean distance of L = 1.50
X 10 11 m from the earth. It emits radiation energy continuously at a rate of
£ sun «3.8 X 10 26 W. Less than a billionth of this energy (about 1.7 X 10 17 W)
strikes the earth, which is sufficient to keep the earth warm and to maintain
life through the photosynthesis process. The energy of the sun is due to the
continuous/kMon reaction during which two hydrogen atoms fuse to form one
atom of helium. Therefore, the sun is essentially a nuclear reactor, with tem-
peratures as high as 40,000,000 K in its core region. The temperature drops to
about 5800 K in the outer region of the sun, called the convective zone, as a
result of the dissipation of this energy by radiation.
The solar energy reaching the earth's atmosphere is called the total solar
irradiance G s , whose value is
1373 W/m 2
(11-49)
The total solar irradiance (also called the solar constant) represents the rate
at which solar energy is incident on a surface normal to the sun's rays at the
outer edge of the atmosphere when the earth is at its mean distance from the
sun (Fig. 11-38).
The value of the total solar irradiance can be used to estimate the effective
surface temperature of the sun from the requirement that
(4ttL 2 )G s = (4-irr 2 ) o-Tl
(11-50)
where L is the mean distance between the sun's center and the earth and r is
the radius of the sun. The left-hand side of this equation represents the total
solar energy passing through a spherical surface whose radius is the mean
earth-sun distance, and the right-hand side represents the total energy that
leaves the sun's outer surface. The conservation of energy principle requires
that these two quantities be equal to each other, since the solar energy ex-
periences no attenuation (or enhancement) on its way through the vacuum
(Fig. 11-39). The effective surface temperature of the sun is determined
from Eq. 11-50 to be T st
5780 K. That is, the sun can be treated as a
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CHAPTER 11
blackbody at a temperature of 5780 K. This is also confirmed by the
measurements of the spectral distribution of the solar radiation just outside
the atmosphere plotted in Figure 11-40, which shows only small deviations
from the idealized blackbody behavior.
The spectral distribution of solar radiation on the ground plotted in Fig-
ure 11-40 shows that the solar radiation undergoes considerable attenua-
tion as it passes through the atmosphere as a result of absorption and scatter-
ing. About 99 percent of the atmosphere is contained within a distance of
30 km from the earth's surface. The several dips on the spectral distribution
of radiation on the earth's surface are due to absorption by the gases 2 ,
3 (ozone), H 2 0, and C0 2 . Absorption by oxygen occurs in a narrow band
about X = 0.76 |xm. The ozone absorbs ultraviolet radiation at wavelengths
below 0.3 (Jim almost completely, and radiation in the range 0.3-0.4 |xm
considerably. Thus, the ozone layer in the upper regions of the atmosphere
protects biological systems on earth from harmful ultraviolet radiation. In turn,
we must protect the ozone layer from the destructive chemicals commonly
used as refrigerants, cleaning agents, and propellants in aerosol cans. The use
of these chemicals is now banned in many countries. The ozone gas also ab-
sorbs some radiation in the visible range. Absorption in the infrared region is
dominated by water vapor and carbon dioxide. The dust particles and other
pollutants in the atmosphere also absorb radiation at various wavelengths.
As a result of these absorptions, the solar energy reaching the earth's sur-
face is weakened considerably, to about 950 W/m 2 on a clear day and much
less on cloudy or smoggy days. Also, practically all of the solar radiation
reaching the earth's surface falls in the wavelength band from 0.3 to 2.5 ujn.
Another mechanism that attenuates solar radiation as it passes through the
atmosphere is scattering or reflection by air molecules and the many other
kinds of particles such as dust, smog, and water droplets suspended in the at-
mosphere. Scattering is mainly governed by the size of the particle relative to
the wavelength of radiation. The oxygen and nitrogen molecules primarily
scatter radiation at very short wavelengths, comparable to the size of the
molecules themselves. Therefore, radiation at wavelengths corresponding to
violet and blue colors is scattered the most. This molecular scattering in all
directions is what gives the sky its bluish color. The same phenomenon is
responsible for red sunrises and sunsets. Early in the morning and late in the
afternoon, the sun's rays pass through a greater thickness of the atmosphere
than they do at midday, when the sun is at the top. Therefore, the violet and
blue colors of the light encounter a greater number of molecules by the time
they reach the earth's surface, and thus a greater fraction of them are scattered
(Fig. 11-41). Consequently, the light that reaches the earth's surface consists
primarily of colors corresponding to longer wavelengths such as red, orange,
and yellow. The clouds appear in reddish-orange color during sunrise and sun-
set because the light they reflect is reddish-orange at those times. For the same
reason, a red traffic light is visible from a longer distance than is a green light
under the same circumstances.
The solar energy incident on a surface on earth is considered to consist of
direct and diffuse parts. The part of solar radiation that reaches the earth's sur-
face without being scattered or absorbed by the atmosphere is called direct
solar radiation G D . The scattered radiation is assumed to reach the earth's
surface uniformly from all directions and is called diffuse solar radiation G d .
2500
1 1 1 1 —
- 5780 K blackbody
Solar irradiation
Extraterrestrial
Earth's surface
H,0
1.0 1.5 2.0 2.5 3.0
Wavelength, um
FIGURE 11-40
Spectral distribution of solar radiation
just outside the atmosphere, at the
surface of the earth on a typical day,
and comparison with blackbody
radiation at 5780 K.
Mostly
red i
FIGURE 11-41
Air molecules scatter blue light much
more than they do red light. At sunset,
the light travels through a thicker layer
of atmosphere, which removes much
of the blue from the natural light,
allowing the red to dominate.
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HEAT TRANSFER
Diffuse solar
radiation
W/m 2
FIGURE 11-42
The direct and diffuse radiation
incident on a horizontal surface at the
earth's surface.
Atmosphere
FIGURE 11-43
Radiation interactions of a surface
exposed to solar and atmospheric
radiation.
Then the total solar energy incident on the unit area of a horizontal surface on
the ground is (Fig. 11-42)
G D cos
(W/m 2 )
(11-51)
where is the angle of incidence of direct solar radiation (the angle that the
sun's rays make with the normal of the surface). The diffuse radiation varies
from about 10 percent of the total radiation on a clear day to nearly 100 per-
cent on a totally cloudy day.
The gas molecules and the suspended particles in the atmosphere emit
radiation as well as absorbing it. The atmospheric emission is primarily due
to the C0 2 and H 2 molecules and is concentrated in the regions from 5 to
8 (iin and above 13 jxm. Although this emission is far from resembling the
distribution of radiation from a blackbody, it is found convenient in radiation
calculations to treat the atmosphere as a blackbody at some lower fictitious
temperature that emits an equivalent amount of radiation energy. This ficti-
tious temperature is called the effective sky temperature r sky . Then the radi-
ation emission from the atmosphere to the earth's surface is expressed as
sky
C^sky
(W/m 2 )
(11-52)
The value of r sky depends on the atmospheric conditions. It ranges from about
230 K for cold, clear-sky conditions to about 285 K for warm, cloudy-sky
conditions.
Note that the effective sky temperature does not deviate much from the
room temperature. Thus, in the light of Kirchhoff's law, we can take the ab-
sorptivity of a surface to be equal to its emissivity at room temperature, a = e.
Then the sky radiation absorbed by a surface can be expressed as
J sky, absorbed
glG.
sky
aoT s 4 ky
eaTX y
(W/m 2 )
(11-53)
The net rate of radiation heat transfer to a surface exposed to solar and atmos-
pheric radiation is determined from an energy balance (Fig. 11-43):
•^ ^absorbed
77
solar, absorbed
a r G
^ ^emitted
+ E,
J solar
-'solar
e<T7;t y
sky, absorbed -^emitted
eoT 4
a, G„,,„ + ea(T* ky - T s 4 )
(W/m 2 )
(11-54)
where T s is the temperature of the surface in K and e is its emissivity at room
temperature. A positive result for q net rad indicates a radiation heat gain by the
surface and a negative result indicates a heat loss.
The absorption and emission of radiation by the elementary gases such as
H 2 , 2 , and N 2 at moderate temperatures are negligible, and a medium filled
with these gases can be treated as a vacuum in radiation analysis. The absorp-
tion and emission of gases with larger molecules such as H 2 and C0 2 , how-
ever, can be significant and may need to be considered when considerable
amounts of such gases are present in a medium. For example, a 1-m-thick
layer of water vapor at 1 aim pressure and 100°C emits more than 50 percent
of the energy that a blackbody would emit at the same temperature.
In solar energy applications, the spectral distribution of incident solar radi-
ation is very different than the spectral distribution of emitted radiation by
cen58933_chll.qxd 9/9/2002 9:39 AM Page 589
the surfaces, since the former is concentrated in the short-wavelength region
and the latter in the infrared region. Therefore, the radiation properties of sur-
faces will be quite different for the incident and emitted radiation, and the
surfaces cannot be assumed to be gray. Instead, the surfaces are assumed to
have two sets of properties: one for solar radiation and another for infrared
radiation at room temperature. Table 11-3 lists the emissivity e and the solar
absorptivity a s of the surfaces of some common materials. Surfaces that are
intended to collect solar energy, such as the absorber surfaces of solar col-
lectors, are desired to have high a s but low e values to maximize the absorp-
tion of solar radiation and to minimize the emission of radiation. Surfaces
that are intended to remain cool under the sun, such as the outer surfaces of
fuel tanks and refrigerator trucks, are desired to have just the opposite prop-
erties. Surfaces are often given the desired properties by coating them with
thin layers of selective materials. A surface can be kept cool, for example, by
simply painting it white.
We close this section by pointing out that what we call renewable energy is
usually nothing more than the manifestation of solar energy in different forms.
Such energy sources include wind energy, hydroelectric power, ocean thermal
energy, ocean wave energy, and wood. For example, no hydroelectric power
plant can generate electricity year after year unless the water evaporates by ab-
sorbing solar energy and comes back as a rainfall to replenish the water source
(Fig. 11^44). Although solar energy is sufficient to meet the entire energy needs
of the world, currently it is not economical to do so because of the low concen-
tration of solar energy on earth and the high capital cost of harnessing it.
EXAMPLE 11-5 Selective Absorber and Reflective Surfaces
Consider a surface exposed to solar radiation. At a given time, the direct and
diffuse components of solar radiation are G D = 400 and G d = 300 W/m 2 , and
the direct radiation makes a 20° angle with the normal of the surface. The sur-
face temperature is observed to be 320 K at that time. Assuming an effective
sky temperature of 260 K, determine the net rate of radiation heat transfer for
these cases (Fig. 11-45):
(a) a s = 0.9 and e = 0.9 (gray absorber surface)
(b) a s = 0.1 and e = 0.1 (gray reflector surface)
(c) a s = 0.9 and e = 0.1 (selective absorber surface)
{d) a s = 0.1 and e = 0.9 (selective reflector surface)
SOLUTION A surface is exposed to solar and sky radiation. The net rate of ra-
diation heat transfer is to be determined for four different combinations of
emissivities and solar absorptivities.
Analysis The total solar energy incident on the surface is
G D cos 6 + G d
(400 W/m 2 ) cos 20°
676 W/m 2
(300 W/m 2 )
Then the net rate of radiation heat transfer for each of the four cases is deter-
mined from:
<7net, rad = a s ^solar + e<T (-<sky ~~ * s )
589
CHAPTER 11
TABLE 11-3
Comparison of the solar absorptivity
a s of some surfaces with their
emissivity e at room temperature
Surface
a s
e
Aluminum
Polished
0.09
0.03
Anodized
0.14
0.84
Foil
0.15
0.05
Copper
Polished
0.18
0.03
Tarnished
0.65
0.75
Stainless steel
Polished
0.37
0.60
Dull
0.50
0.21
Plated metals
Black nickel oxide
0.92
0.08
Black chrome
0.87
0.09
Concrete
0.60
0.88
White marble
0.46
0.95
Red brick
0.63
0.93
Asphalt
0.90
0.90
Black paint
0.97
0.97
White paint
0.14
0.93
Snow
0.28
0.97
Human skin
(Caucasian)
0.62
0.97
Winds
Clouds
'•/,''i' ' ' > Rain
~ — -V Power lines Evaporation
"\
1PP
i
Solar
p^ '.. i energy
r/7///
vj-c Zl
FIGURE 11-44
The cycle that water undergoes in a
hydroelectric power plant.
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HEAT TRANSFER
0.9
(a)
E ♦
0.1
(c)
E ■■
0.9
0.1
3 |im
3 |im
X
(d)
FIGURE 11-45
Graphical representation of the spectral
emissivities of the four surfaces
considered in Example 11-5.
(a) ct s = 0.9 and e = 0.9 (gray absorber surface):
«„ct,rad = 0.9(676 W/m 2 ) + 0.9(5.67 X 10~ 8 W/m 2 ■ K 4 )[(260 K) 4 - (320 K) 4 ]
= 307 W/m 2
(b) a s = 0.1 and e = 0.1 (gray reflector surface):
■Wad = 0.1(676 W/m 2 ) + 0.1(5.67 X 10- 8 W/m 2 • K 4 )[(260 K) 4 - (320 K) 4 ]
= 34 W/m 2
(c) a s = 0.9 and e = 0.1 (selective absorber surface):
?„et.ra<i = 0.9(676 W/m 2 ) + 0.1(5.67 X 10- 8 W/m 2 ■ K 4 )[(260 K) 4 - (320 K) 4 ]
= 575 W/m 2
(d) a s = 0.1 and e = 0.9 (selective reflector surface):
■Wad = 0.1(676 W/m 2 ) + 0.9(5.67 X 10- 8 W/m 2 ■ K 4 )[(260 K) 4 - (320 K) 4 ]
= -234 W/m 2
Discussion Note that the surface of an ordinary gray material of high absorp-
tivity gains heat at a rate of 307 W/m 2 . The amount of heat gain increases to
575 W/m 2 when the surface is coated with a selective material that has the same
absorptivity for solar radiation but a low emissivity for infrared radiation. Also
note that the surface of an ordinary gray material of high reflectivity still gains
heat at a rate of 34 W/m 2 . When the surface is coated with a selective material
that has the same reflectivity for solar radiation but a high emissivity for infrared
radiation, the surface loses 234 W/m 2 instead. Therefore, the temperature of the
surface will decrease when a selective reflector surface is used.
TOPIC OF SPECIAL INTEREST
Solar Heat Gain Through Windows
The sun is the primary heat source of the earth, and the solar irradiance on
a surface normal to the sun's rays beyond the earth's atmosphere at the
mean earth-sun distance of 149.5 million km is called the total solar irra-
diance or solar constant. The accepted value of the solar constant is 1373
W/m 2 (435.4 Btu/h • ft 2 ), but its value changes by 3.5 percent from a max-
imum of 1418 W/m 2 on January 3 when the earth is closest to the sun, to a
minimum of 1325 W/m 2 on July 4 when the earth is farthest away from the
sun. The spectral distribution of solar radiation beyond the earth's atmos-
phere resembles the energy emitted by a blackbody at 5780°C, with about
9 percent of the energy contained in the ultraviolet region (at wavelengths
between 0.29 to 0.4 (Jim), 39 percent in the visible region (0.4 to 0.7 ujti),
and the remaining 52 percent in the near-infrared region (0.7 to 3.5 |juti).
The peak radiation occurs at a wavelength of about 0.48 |xm, which cor-
responds to the green color portion of the visible spectrum. Obviously a
"This section can be skipped without a loss in continuity.
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CHAPTER 11
glazing material that transmits the visible part of the spectrum while ab-
sorbing the infrared portion is ideally suited for an application that calls for
maximum daylight and minimum solar heat gain. Surprisingly, the ordinary
window glass approximates this behavior remarkably well (Fig. 11-46).
Part of the solar radiation entering the earth's atmosphere is scattered and
absorbed by air and water vapor molecules, dust particles, and water
droplets in the clouds, and thus the solar radiation incident on earth's sur-
face is less than the solar constant. The extent of the attenuation of solar
radiation depends on the length of the path of the rays through the atmos-
phere as well as the composition of the atmosphere (the clouds, dust, hu-
midity, and smog) along the path. Most ultraviolet radiation is absorbed by
the ozone in the upper atmosphere, and the scattering of short wavelength
radiation in the blue range by the air molecules is responsible for the blue
color of the clear skies. At a solar altitude of 41.8°, the total energy of di-
rect solar radiation incident at sea level on a clear day consists of about 3
percent ultraviolet, 38 percent visible, and 59 percent infrared radiation.
The part of solar radiation that reaches the earth's surface without being
scattered or absorbed is called direct radiation. Solar radiation that is scat-
tered or reemitted by the constituents of the atmosphere is called diffuse
radiation. Direct radiation comes directly from the sun following a straight
path, whereas diffuse radiation comes from all directions in the sky. The
entire radiation reaching the ground on an overcast day is diffuse radiation.
The radiation reaching a surface, in general, consists of three components:
direct radiation, diffuse radiation, and radiation reflected onto the surface
from surrounding surfaces (Fig. 11-47). Common surfaces such as grass,
trees, rocks, and concrete reflect about 20 percent of the radiation while ab-
sorbing the rest. Snow-covered surfaces, however, reflect 70 percent of the
incident radiation. Radiation incident on a surface that does not have a di-
rect view of the sun consists of diffuse and reflected radiation. Therefore,
at solar noon, solar radiations incident on the east, west, and north surfaces
of a south-facing house are identical since they all consist of diffuse and re-
flected components. The difference between the radiations incident on the
south and north walls in this case gives the magnitude of direct radiation in-
cident on the south wall.
When solar radiation strikes a glass surface, part of it (about 8 percent for
uncoated clear glass) is reflected back to outdoors, part of it (5 to 50 per-
cent, depending on composition and thickness) is absorbed within the
glass, and the remainder is transmitted indoors, as shown in Figure 11-48.
The conservation of energy principle requires that the sum of the transmit-
ted, reflected, and absorbed solar radiations be equal to the incident solar
radiation. That is,
g 1.00
| 0.80
| 0.60
& 0.40
§ 0.20
f~
~^s
\ /
\
A
h
4
^
A
1
/
1
s r
X
0.2 0.4 0.6 1 2
Wave length, um
3 4 5
1.3mm regular sheet
2. 6 mm gray heat-absorbing plate/float
3. 6 mm green heat-absorbing plate/float
FIGURE 11-46
The variation of the transmittance of
typical architectural glass with
wavelength (from ASHRAE
Handbook of Fundamentals,
Ref. 1, Chap. 27, Fig. 11).
FIGURE 11-47
Direct, diffuse, and reflected
components of solar radiation incident
on a window.
where t s is the transmissivity, p s is the reflectivity, and a s is the absorptiv-
ity of the glass for solar energy, which are the fractions of incident solar ra-
diation transmitted, reflected, and absorbed, respectively. The standard
3-mm- (|-in.) thick single -pane double-strength clear window glass trans-
mits 86 percent, reflects 8 percent, and absorbs 6 percent of the solar en-
ergy incident on it. The radiation properties of materials are usually given
for normal incidence, but can also be used for radiation incident at other
cen58933_chll.qxd 9/9/2002 9:39 AM Page 592
592
HEAT TRANSFER
4^"%
6-mm thick
clear glass
it Sun <r
/
%/fl^
/
Incident
solar ^^
radiation ^| h
100% ^
k. ^W
Transmitted
^^A. 80%
Reflected t^^^^
8% ^
Absorbed
12%
Outward transfer
Inward transfer
of absorbed ^«
~~ r of absorbed
radiation
radiation
8%
4%
FIGURE 11-48
Distribution of solar radiation incident
on a clear glass.
angles since the transmissivity, reflectivity, and absorptivity of the glazing
materials remain essentially constant for incidence angles up to about 60°
from the normal.
The hourly variation of solar radiation incident on the walls and windows
of a house is given in Table 11-4. Solar radiation that is transmitted indoors
is partially absorbed and partially reflected each time it strikes a surface,
but all of it is eventually absorbed as sensible heat by the furniture, walls,
people, and so forth. Therefore, the solar energy transmitted inside a build-
ing represents a heat gain for the building. Also, the solar radiation ab-
sorbed by the glass is subsequently transferred to the indoors and outdoors
by convection and radiation. The sum of the transmitted solar radiation and
the portion of the absorbed radiation that flows indoors constitutes the so-
lar heat gain of the building.
The fraction of incident solar radiation that enters through the glazing is
called the solar heat gain coefficient (SHGC) and is expressed as
SHGC
Solar heat gain through the window
Solar radiation incident on the window
H solar, gain
^ solar, incident
(11-55)
T s + /i«s
where a s is the solar absorptivity of the glass and/j is the inward flowing
fraction of the solar radiation absorbed by the glass. Therefore, the dimen-
sionless quantity SHGC is the sum of the fractions of the directly transmit-
ted (t s ) and the absorbed and reemitted (^a s ) portions of solar radiation
incident on the window. The value of SHGC ranges from to 1, with 1 cor-
responding to an opening in the wall (or the ceiling) with no glazing. When
the SHGC of a window is known, the total solar heat gain through that win-
dow is determined from
Q s
SHGC X A
glazing
x?
solar, incident
(W)
solar, incident
(11-56)
is the solar
where A glazing is the glazing area of the window and q
heat flux incident on the outer surface of the window, in W/m 2 .
Another way of characterizing the solar transmission characteristics of
different kinds of glazing and shading devices is to compare them to a well-
known glazing material that can serve as a base case. This is done by tak-
ing the standard 3-mm-(i-in.)-thick double-strength clear window glass
sheet whose SHGC is 0.87 as the reference glazing and defining a shading
coefficient SC as
SC
Solar heat gain of product
Solar heat gain of reference glazing
SHGC SHGC
(11-57)
SHGC„
0.87
1.15 X SHGC
Therefore, the shading coefficient of a single-pane clear glass window is
SC = 1.0. The shading coefficients of other commonly used fenestration
products are given in Table 1 1-5 for summer design conditions. The values
for winter design conditions may be slightly lower because of the higher
heat transfer coefficients on the outer surface due to high winds and thus
cen58933_chll.qxd 9/9/2002 9:39 AM Page 593
TABLE 11-4
Hourly variation of solar radiation incident on various surfaces and the daily totals throughout the year at 40° latitude
(from ASHRAE Handbook of Fundamentals, Ref. 1, Chap. 27, Table 15)
Solar Radiation Incident on
the Surface, *
W/m 2
Direction
Solar Time
of
12
Daily
Date Surface
5
6
7
8
9
10
11
noon
13
14
15
16
17
18
19
Total
Jan. N
20
43
66
68
71
68
66
43
20
446
NE
63
47
66
68
71
68
59
43
20
489
E
402
557
448
222
76
68
59
43
20
1863
SE
483
811
875
803
647
428
185
48
20
4266
S
271
579
771
884
922
884
771
579
271
5897
sw
20
48
185
428
647
803
875
811
483
4266
w
20
43
59
68
76
222
448
557
402
1863
NW
20
43
59
68
71
68
66
47
63
489
Horizontal
51
198
348
448
482
448
348
198
51
2568
Direct
446
753
865
912
926
912
865
753
446
—
Apr. N
41
57
79
97
110
120
122
120
110
97
79
57
41
1117
NE
262
508
462
291
134
123
122
120
110
97
77
52
17
2347
E
321
728
810
732
552
293
131
120
110
97
77
52
17
4006
SE
189
518
682
736
699
582
392
187
116
97
77
52
17
4323
S
18
59
149
333
437
528
559
528
437
333
149
59
18
3536
SW
17
52
77
97
116
187
392
582
699
736
682
518
189
4323
w
17
52
77
97
110
120
392
293
552
732
810
728
321
4006
NW
17
52
77
97
110
120
122
123
134
291
462
508
262
2347
Horizontal
39
222
447
640
786
880
911
880
786
640
447
222
39
6938
Direct
282
651
794
864
901
919
925
919
901
864
794
651
282
—
July N
3
133
109
103
117
126
134
138
134
126
117
103
109
133
3
1621
NE
8
454
590
540
383
203
144
138
134
126
114
95
71
39
3068
E
7
498
739
782
701
531
294
149
134
126
114
95
71
39
4313
SE
2
248
460
580
617
576
460
291
155
131
114
95
71
39
3849
S
39
76
108
190
292
369
395
369
292
190
108
76
39
2552
SW
39
71
95
114
131
155
291
460
576
617
580
460
248
2
3849
w
39
71
95
114
126
134
149
294
531
701
782
739
498
7
4313
NW
39
71
95
114
126
134
138
144
203
383
540
590
454
8
3068
Horizontal
1
115
320
528
702
838
922
949
922
838
702
528
320
115
1
3902
Direct
7
434
656
762
818
850
866
871
866
850
818
762
656
434
7
—
Oct. N
7
40
62
77
87
90
87
77
62
40
7
453
NE
74
178
84
80
87
90
87
87
62
40
7
869
E
163
626
652
505
256
97
87
87
62
40
7
2578
SE
152
680
853
864
770
599
364
137
66
40
7
4543
S
44
321
547
711
813
847
813
711
547
321
44
5731
SW
7
40
66
137
364
599
770
864
853
680
152
4543
w
7
40
62
87
87
97
256
505
652
626
163
2578
NW
7
40
62
87
87
90
87
80
84
178
74
869
Horizontal
14
156
351
509
608
640
608
509
351
156
14
3917
Direct
152
643
811
884
917
927
917
884
811
643
152
—
♦Multiply by 0.3171 to convert to Btu/h ■ ft 2 .
Values given are for the 21st of the month for average days with no clouds. The values can be up to 15 percent higher at high elevations under very
clear skies and up to 30 percent lower at very humid locations with very dusty industrial atmospheres. Daily totals are obtained using Simpson's rule for
integration with 10-min time intervals. Solar reflectance of the ground is assumed to be 0.2, which is valid for old concrete, crushed rock, and bright green
grass. For a specified location, use solar radiation data obtained for that location. The direction of a surface indicates the direction a vertical surface is
facing. For example, W represents the solar radiation incident on a west-facing wall per unit area of the wall.
Solar time may deviate from the local time. Solar noon at a location is the time when the sun is at the highest location (and thus when the shadows
are shortest). Solar radiation data are symmetric about the solar noon: the value on a west wall before the solar noon is equal to the value on an east wall two
hours after the solar noon.
593
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HEAT TRANSFER
TABLE 11-5
Shading coefficient SC and solar
transmissivity T S0 | ar for some
common glass types for summer
design conditions (from ASHRAE
Handbook of Fundamentals, Ref . 1 ,
Chap. 27, Table 11).
Nomina
Type of Thickness
Glazing mm
in.
T solar SC
(a) Single Glazing
Clear 3
1
0.86 1.0
6
1
0.78 0.95
10
3
8
0.72 0.92
13
1
?
0.67 0.88
Heat absorbing 3
1
8
0.64 0.85
6
1
0.46 0.73
10
3
8
0.33 0.64
13
1
2
0.24 0.58
(b) Double Glazing
Clear in, 3 a
1
a
0.71 b 0.88
clear out 5
1
0.61 0.82
Clear in, heat
absorbing out c 6
1
4
0.36 0.58
♦Multiply by 0.87 to obtain SHGC.
a The thickness of each pane of glass.
b Combined transmittance for assembled unit.
c Refers to gray-, bronze-, and green-tinted
heat-absorbing float glass.
higher rate of outward flow of solar heat absorbed by the glazing, but the
difference is small.
Note that the larger the shading coefficient, the smaller the shading ef-
fect, and thus the larger the amount of solar heat gain. A glazing material
with a large shading coefficient will allow a large fraction of solar radiation
to come in.
Shading devices are classified as internal shading and external shading,
depending on whether the shading device is placed inside or outside. Exter-
nal shading devices are more effective in reducing the solar heat gain since
they intercept the sun's rays before they reach the glazing. The solar heat
gain through a window can be reduced by as much as 80 percent by exterior
shading. Roof overhangs have long been used for exterior shading of win-
dows. The sun is high in the horizon in summer and low in winter. A prop-
erly sized roof overhang or a horizontal projection blocks off the sun's rays
completely in summer while letting in most of them in winter, as shown in
Figure 11^49. Such shading structures can reduce the solar heat gain on the
south, southeast, and southwest windows in the northern hemisphere con-
siderably. A window can also be shaded from outside by vertical or hori-
zontal architectural projections, insect or shading screens, and sun screens.
To be effective, air must be able to move freely around the exterior device
to carry away the heat absorbed by the shading and the glazing materials.
Some type of internal shading is used in most windows to provide pri-
vacy and aesthetic effects as well as some control over solar heat gain. In-
ternal shading devices reduce solar heat gain by reflecting transmitted solar
radiation back through the glazing before it can be absorbed and converted
into heat in the building.
Draperies reduce the annual heating and cooling loads of a building by
5 to 20 percent, depending on the type and the user habits. In summer,
they reduce heat gain primarily by reflecting back direct solar radiation
(Fig. 11-50). The semiclosed air space formed by the draperies serves as
an additional barrier against heat transfer, resulting in a lower [/-factor for
the window and thus a lower rate of heat transfer in summer and winter.
The solar optical properties of draperies can be measured accurately, or
they can be obtained directly from the manufacturers. The shading coeffi-
cient of draperies depends on the openness factor, which is the ratio of the
open area between the fibers that permits the sun's rays to pass freely, to
the total area of the fabric. Tightly woven fabrics allow little direct radia-
tion to pass through, and thus they have a small openness factor. The re-
flectance of the surface of the drapery facing the glazing has a major effect
on the amount of solar heat gain. Light-colored draperies made of closed
or tightly woven fabrics maximize the back reflection and thus minimize
the solar gain. Dark-colored draperies made of open or semi-open woven
fabrics, on the other hand, minimize the back reflection and thus maxi-
mize the solar gain.
The shading coefficients of drapes also depend on the way they are hung.
Usually, the width of drapery used is twice the width of the draped area to
allow folding of the drapes and to give them their characteristic "full" or
"wavy" appearance. A flat drape behaves like an ordinary window shade.
A flat drape has a higher reflectance and thus a lower shading coefficient
than a full drape.
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External shading devices such as overhangs and tinted glazings do not re-
quire operation, and provide reliable service over a long time without sig-
nificant degradation during their service life. Their operation does not
depend on a person or an automated system, and these passive shading de-
vices are considered fully effective when determining the peak cooling load
and the annual energy use. The effectiveness of manually operated shading
devices, on the other hand, varies greatly depending on the user habits, and
this variation should be considered when evaluating performance.
The primary function of an indoor shading device is to provide thermal
comfort for the occupants. An unshaded window glass allows most of the
incident solar radiation in, and also dissipates part of the solar energy it ab-
sorbs by emitting infrared radiation to the room. The emitted radiation and
the transmitted direct sunlight may bother the occupants near the window.
In winter, the temperature of the glass is lower than the room air tempera-
ture, causing excessive heat loss by radiation from the occupants. A shad-
ing device allows the control of direct solar and infrared radiation while
providing various degrees of privacy and outward vision. The shading de-
vice is also at a higher temperature than the glass in winter, and thus
reduces radiation loss from occupants. Glare from draperies can be
minimized by using off-white colors. Indoor shading devices, especially
draperies made of a closed-weave fabric, are effective in reducing sounds
that originate in the room, but they are not as effective against the sounds
coming from outside.
The type of climate in an area usually dictates the type of windows to be
used in buildings. In cold climates where the heating load is much larger
than the cooling load, the windows should have the highest transmissivity
for the entire solar spectrum, and a high reflectivity (or low emissivity) for
the far infrared radiation emitted by the walls and furnishings of the room.
Low-e windows are well suited for such heating-dominated buildings.
Properly designed and operated windows allow more heat into the build-
ing over a heating season than it loses, making them energy contributors
rather then energy losers. In warm climates where the cooling load is
much larger than the heating load, the windows should allow the visible
solar radiation (light) in, but should block off the infrared solar radiation.
Such windows can reduce the solar heat gain by 60 percent with no ap-
preciable loss in daylighting. This behavior is approximated by window
glazings that are coated with a heat-absorbing film outside and a low-e
film inside (Fig. 11-51). Properly selected windows can reduce the cool-
ing load by 15 to 30 percent compared to windows with clear glass.
Note that radiation heat transfer between a room and its windows is pro-
portional to the emissivity of the glass surface facing the room, e glass , and
can be expressed as
Q
rad, room-window
o A
°glass ^g
(t(T 4
i " V- 1 roorr
s )
(11-58)
Therefore, a low-e interior glass will reduce the heat loss by radiation in
winter (r glass < r room ) and heat gain by radiation in summer (T glass > T mom ).
Tinted glass and glass coated with reflective films reduce solar heat gain
in summer and heat loss in winter. The conductive heat gains or losses can
be minimized by using multiple-pane windows. Double-pane windows are
595
CHAPTER 11
Summer
"
FIGURE 11-49
A properly sized overhang blocks off
the sun's rays completely in summer
while letting them in in winter.
Reflected-'
by drapes
Window
FIGURE 11-50
Draperies reduce heat gain in summer
by reflecting back solar radiation, and
reduce heat loss in winter by forming
an air space before the window.
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Glass
(colder than room)
■ 2 rad ~ E
No
reflective
film
(a) Cold climates
Low-e film
(high infrared
reflectivity)
Glass
(warmer than room)
Infrared
Reflective film
Visible
Low-e film
(b) Warm climates
FIGURE 11-51
Radiation heat transfer between a
room and its windows is proportional
to the emissivity of the glass surface,
and low-e coatings on the inner
surface of the windows reduce heat
loss in winter and heat gain in
summer.
usually called for in climates where the winter design temperature is less
than 7°C (45°F). Double-pane windows with tinted or reflective films are
commonly used in buildings with large window areas. Clear glass is pre-
ferred for showrooms since it affords maximum visibility from outside, but
bronze-, gray-, and green-colored glass are preferred in office buildings
since they provide considerable privacy while reducing glare.
EXAMPLE 1 1-6 Installing Reflective Films on Windows
A manufacturing facility located at 40° N latitude has a glazing area of 40 m 2
that consists of double-pane windows made of clear glass (SHGC = 0.766). To
reduce the solar heat gain in summer, a reflective film that will reduce the
SHGC to 0.261 is considered. The cooling season consists of June, July, Au-
gust, and September, and the heating season October through April. The aver-
age daily solar heat fluxes incident on the west side at this latitude are 1.86,
2.66, 3.43, 4.00, 4.36, 5.13, 4.31, 3.93, 3.28, 2.80, 1.84, and 1.54
kWh/day • m 2 for January through December, respectively. Also, the unit cost of
the electricity and natural gas are $0.08/kWh and $0.50/therm, respectively. If
the coefficient of performance of the cooling system is 2.5 and efficiency of the
furnace is 0.8, determine the net annual cost savings due to installing reflec-
tive coating on the windows. Also, determine the simple payback period if the
installation cost of reflective film is $20/m 2 (Fig. 11-52).
SOLUTION The net annual cost savings due to installing reflective film on the
west windows of a building and the simple payback period are to be determined.
Assumptions 1 The calculations given below are for an average year. 2 The unit
costs of electricity and natural gas remain constant.
Analysis Using the daily averages for each month and noting the number of
days of each month, the total solar heat flux incident on the glazing during sum-
mer and winter months are determined to be
xZsolar, summer
k~so\m, winter
5.13 X 30 + 4.31 X 31 + 3.93 X 31 + 3.28 X 30 = 508 kWh/year
2.80 X 31 + 1.84 X 30 + 1.54 X 31 + 1.86 X 31
+ 2.66 X 28 + 3.43 X 31 + 4.00 X 30
548 kWh/year
Then the decrease in the annual cooling load and the increase in the annual
heating load due to the reflective film become
Cooling load decrease = 2 S0 , ar , summer A glazing (SHGC without fllm - SHGC wlth fllm )
= (508 kWh/year)(40 m 2 )(0.766 - 0.261)
= 10,262 kWh/year
Heating load increase = Q solal . wintcr A glazing (SHGC without fllm - SHGC withfflm )
= (548 kWh/year)(40 m 2 )(0.766 - 0.261)
= 11,070 kWh/year = 377.7 therms/year
since 1 therm = 29.31 kWh. The corresponding decrease in cooling costs and
the increase in heating costs are
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CHAPTER 11
Decrease in cooling costs = (Cooling load decrease)(Unit cost of electricity )/COP
= (10,262 kWh/year)($0.08/kWh)/2.5 = $328/year
Increase in heating costs = (Heating load increase)(Unit cost of fuel)/Efficiency
= (377.7 therms/year)($0.50/therm)/0.80 = $236/year
Then the net annual cost savings due to the reflective film become
Cost savings = Decrease in cooling costs — Increase in heating costs
= $328 - $236 = $92/year
The implementation cost of installing films is
Implementation cost = ($20/m 2 )(40 m 2 ) = $800
This gives a simple payback period of
Implementation cost $800
Simple payback period
Annual cost savings $92/year
8.7 years
Discussion The reflective film will pay for itself in this case in about nine years.
This may be unacceptable to most manufacturers since they are not usually
interested in any energy conservation measure that does not pay for itself within
three years. But the enhancement in thermal comfort and thus the resulting in-
crease in productivity often makes it worthwhile to install reflective film.
Glass
Reflected
Reflective film
Transmitted
FIGURE 11-52
Schematic for Example 11-6.
SUMMARY
Radiation propagates in the form of electromagnetic waves.
The frequency v and wavelength k of electromagnetic waves in
a medium are related by A. = civ, where c is the speed of prop-
agation in that medium. All matter whose temperature is above
absolute zero continuously emits thermal radiation as a result
of vibrational and rotational motions of molecules, atoms, and
electrons of a substance. Temperature is a measure of the
strength of these activities at the microscopic level.
A blackbody is defined as a perfect emitter and absorber of
radiation. At a specified temperature and wavelength, no sur-
face can emit more energy than a blackbody. A blackbody ab-
sorbs all incident radiation, regardless of wavelength and
direction. The radiation energy emitted by a blackbody per unit
time and per unit surface area is called the blackbody emissive
power E b and is expressed by the Stefan-Boltzmann law as
E b (T) = o-r 4
where cr = 5.670 X 10~ 8 W/m 2 ■ K 4 is the Stefan-Boltzmann
constant and T is the absolute temperature of the surface in K.
At any specified temperature, the spectral blackbody emissive
power E bx increases with wavelength, reaches a peak, and then
decreases with increasing wavelength. The wavelength at
which the peak occurs for a specified temperature is given by
Wien 's displacement law as
(^„
2897.8 |xm ■ K
The blackbody radiation function f x represents the fraction of
radiation emitted by a blackbody at temperature Tin the wave-
length band from \ = to \. The fraction of radiation energy
emitted by a blackbody at temperature T over a finite wave-
length band from k = kt to\ = k 2 is determined from
fx t -xXT)=f x XT)-f X[ (T)
where fx,(T) and /x/T) are the blackbody radiation functions
corresponding to X.Tand k 2 T, respectively.
The magnitude of a viewing angle in space is described by
solid angle expressed as dm = dAJr 2 . The radiation intensity
for emitted radiation I e (Q, 4>) is defined as the rate at which ra-
diation energy is emitted in the (G, 4>) direction per unit area
normal to this direction and per unit solid angle about this di-
rection. The radiation flux for emitted radiation is the emissive
power E, and is expressed as
r r 2ir r -nil
E= \dE=\ 7,(G, <)>) cos G sin MM§
J J<|)=oJe=o
hemisphere
For a diffusely emitting surface, intensity is independent of di-
rection and thus
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HEAT TRANSFER
E = nL
For a blackbody, we have
E b = tt/j and
E b (T) CT r 4
The radiation flux incident on a surface from all directions is
irradiation G, and for diffusely incident radiation of intensity /,
it is expressed as
tt/,
The rate at which radiation energy leaves a unit area of a sur-
face in all directions is radiosity J, and for a surface that is both
a diffuse emitter and a diffuse reflector it is expressed as
J = ivl e+r
where I e+r is the sum of the emitted and reflected intensities.
The spectral emitted quantities are related to total quantities as
h.dk
and
f
E x dX
The last relation reduces for a diffusely emitting surface and
for a blackbody to
TT/x
and E bX (\, T) = ir/ iX (\, T)
The emissivity of a surface represents the ratio of the radia-
tion emitted by the surface at a given temperature to the radia-
tion emitted by a blackbody at the same temperature. Different
emissivities are defined as
Spectral directional emissivity:
ex,e(M,<|>,7) =
Total directional emissivity:
B e (8, c|>, T) =
hx(X T)
W, 4>, T)
h(T)
Spectral hemispherical emissivity:
E X {K T)
e x (X, T) = — ^
E bX (l, T)
Total hemispherical emissivity
E(T)
e(T)
f e x (X, T)E bX (X, T)dX
E b {T)
oT 4
Emissivity can also be expressed as a step function by dividing
the spectrum into a sufficient number of wavelength bands of
constant emissivity as
b(T) = bJo-xST) + B 2 f x ,(T) + e 3 f x JT)
The total hemispherical emissivity e of a surface is the average
emissivity over all directions and wavelengths.
Radiation energy incident on a surface per unit surface area
per unit time is called irradiation G. When radiation strikes a
surface, part of it is absorbed, part of it is reflected, and the re-
maining part, if any, is transmitted. The fraction of incident ra-
diation (intensity /, or irradiation G) absorbed by the surface is
called the absorptivity, the fraction reflected by the surface is
called the reflectivity, and the fraction transmitted is called the
transmissivity. Various absorptivities, reflectivities, and trans-
missivities for a medium are expressed as
«x, e(^> 9' 40
,40
h
3bs (^ e, <w
,(M,«I>)
and
P\, o(^
e,4>)
i x ,i(KQ,4>)
G x ,
G
PxM =
ef(^)
(X) '
and
T X W
G X (X)
(
a = -
-'abs
G~' p =
G ref
G
and
T =
G tr
G
«xW
The consideration of wavelength and direction dependence of
properties makes radiation calculations very complicated.
Therefore, the gray and diffuse approximations are commonly
utilized in radiation calculations. A surface is said to be diffuse
if its properties are independent of direction and gray if its
properties are independent of wavelength.
The sum of the absorbed, reflected, and transmitted fractions
of radiation energy must be equal to unity,
a + p + t = 1
For opaque surfaces, t = 0, and thus
a + p = 1
Surfaces are usually assumed to reflect in a perfectly specular
or diffuse manner for simplicity. In specular (or mirrorlike) re-
flection, the angle of reflection equals the angle of incidence of
the radiation beam. In diffuse reflection, radiation is reflected
equally in all directions. Reflection from smooth and polished
surfaces approximates specular reflection, whereas reflection
from rough surfaces approximates diffuse reflection. Kirch-
hoff's law of radiation is expressed as
eCO
e (7), e k (T) = a x (T), and e{T) = a(T)
That is, the total hemispherical emissivity of a surface at tem-
perature Tis equal to its total hemispherical absorptivity for ra-
diation coming from a blackbody at the same temperature.
Gas molecules and the suspended particles in the atmosphere
emit radiation as well as absorbing it. The atmosphere can be
treated as a blackbody at some lower fictitious temperature,
called the effective sky temperature T sky that emits an equiva-
lent amount of radiation energy. Then the radiation emitted by
the atmosphere is expressed as
cen58933_chll.qxd 9/9/2002 9:39 AM Page 599
'sky
VT^
The net rate of radiation heat transfer to a surface exposed to
solar and atmospheric radiation is determined from an energy
balance expressed as
599
CHAPTER 11
#net,rad ~~ a ^solar + Sd(T
sky
71)
where T s is the surface temperature in K, and e is the surface
emissivity at room temperature.
REFERENCES AND SUGGESTED READING
1. American Society of Heating, Refrigeration, and Air
Conditioning Engineers, Handbook of Fundamentals,
Atlanta, ASHRAE, 1993.
2. A. G. H. Dietz. "Diathermanous Materials and Properties
of Surfaces." In Space Heating with Solar Energy, ed.
R. W. Hamilton. Cambridge, MA: MIT Press, 1954.
3. J. A. Duffy and W. A. Beckman. Solar Energy Thermal
Process. New York: John Wiley & Sons, 1974.
4. J. P. Holman. Heat Transfer. 9th ed. New York: McGraw-
Hill, 2002.
5. H. C. Hottel. "Radiant Heat Transmission," In Heat
Transmission, 3rd ed., ed. W. H. McAdams. New York:
McGraw-Hill, 1954.
6. F. P. Incropera and D. P. DeWitt. Introduction to Heat
Transfer. 4th ed. New York: John Wiley & Sons, 2002.
7. F. Kreith and M. S. Bonn. Principles of Heat Transfer. 6th
ed. Pacific Grove, CA: Brooks/Cole, 2001.
8. M. F. Modest, Radiative Heat Transfer. New York:
McGraw-Hill, 1993.
9. M. Planck. The Theory of Heat Radiation. New York:
Dover, 1959.
10. W. Sieber. Zeitschrift fur Technische Physics 22 (1941),
pp. 130-135.
11. R. Siegel and J. R. Howell. Thermal Radiation Heat
Transfer. 3rd ed. Washington, D.C.: Hemisphere, 1992.
12. N. V. Suryanarayana. Engineering Heat Transfer. St. Paul,
MN: West, 1995.
13. Y. S. Touloukain and D. P. DeWitt. "Nonmetallic Solids."
In Thermal Radiative Properties. Vol. 8. New York:
IFI/Plenum, 1970.
14. Y S. Touloukian and D. P. DeWitt. "Metallic Elements
and Alloys." In Thermal Radiative Properties, Vol. 7.
New York: IFI/Plenum, 1970.
PROBLEMS
Electromagnetic and Thermal Radiation
11-1C What is an electromagnetic wave? How does it differ
from a sound wave?
11-2C By what properties is an electromagnetic wave char-
acterized? How are these properties related to each other?
11-3C What is visible light? How does it differ from the
other forms of electromagnetic radiation?
11-4C How do ultraviolet and infrared radiation differ? Do
you think your body emits any radiation in the ultraviolet
range? Explain.
11-5C What is thermal radiation? How does it differ from
the other forms of electromagnetic radiation?
*Problems designated by a "C" are concept questions, and
students are encouraged to answer them all. Problems designated
by an "E" are in English units, and the SI users can ignore them.
Problems with an EES-CD icon ® are solved using EES, and
complete solutions together with parametric studies are included
on the enclosed CD. Problems with a computer-EES icon H are
comprehensive in nature, and are intended to be solved with a
computer, preferably using the EES software that accompanies
this text.
11-6C What is the cause of color? Why do some objects ap-
pear blue to the eye while others appear red? Is the color of a
surface at room temperature related to the radiation it emits?
11-7C Why is radiation usually treated as a surface
phenomenon?
11-8C Why do skiers get sunburned so easily?
11-9C How does microwave cooking differ from conven-
tional cooking?
11-10 Electricity is generated and transmitted in power lines
at a frequency of 60 Hz (1 Hz = 1 cycle per second). Deter-
mine the wavelength of the electromagnetic waves generated
by the passage of electricity in power lines.
11-11 A microwave oven is designed to operate at a fre-
quency of 2.8 X 10 9 Hz. Determine the wavelength of these
microwaves and the energy of each microwave.
11-12 A radio station is broadcasting radio waves at a wave-
length of 200 m. Determine the frequency of these waves.
Answer. 1.5 x 10 6 Hz
11-13 A cordless telephone is designed to operate at a fre-
quency of 8.5 X 10 8 Hz. Determine the wavelength of these
telephone waves.
cen58 933_chll.qxd 9/9/2002 9:39 AM Page 6C
600
HEAT TRANSFER
Blackbody Radiation
11-14C What is a blackbody? Does a blackbody actually
exist?
11-15C Define the total and spectral blackbody emissive pow-
ers. How are they related to each other? How do they differ?
11-16C Why did we define the blackbody radiation func-
tion? What does it represent? For what is it used?
11-17C Consider two identical bodies, one at 1000 K and the
other at 1500 K. Which body emits more radiation in the
shorter-wavelength region? Which body emits more radiation
at a wavelength of 20 |xm?
11-18 Consider a 20-cm X 20-cm X 20-cm cubical body at
1000 K suspended in the air. Assuming the body closely ap-
proximates a blackbody, determine (a) the rate at which the
cube emits radiation energy, in W, and (b) the spectral black-
body emissive power at a wavelength of 4 p,m.
11-19E The sun can be treated as a blackbody at an effective
surface temperature of 10,400 R. Determine the rate at which
infrared radiation energy (X = 0.76-100 jjum) is emitted by the
sun, in Btu/h • ft 2 .
11-20 fa'M The sun can be treated as a blackbody at
k^S 5780 K. Using EES (or other) software, calcu-
late and plot the spectral blackbody emissive power E bx of the
sun versus wavelength in the range of 0.01 p,m to 1000 p,m.
Discuss the results.
11-21 The temperature of the filament of an incandescent
lightbulb is 3200 K. Treating the filament as a blackbody, de-
termine the fraction of the radiant energy emitted by the fila-
ment that falls in the visible range. Also, determine the
wavelength at which the emission of radiation from the fila-
ment peaks.
11-22 Tu'M Reconsider Problem 11-21. Using EES (or
k^S other) software, investigate the effect of temper-
ature on the fraction of radiation emitted in the visible range.
Let the surface temperature vary from 1 000 K to 4000 K, and
plot fraction of radiation emitted in the visible range versus the
surface temperature.
11-23 An incandescent lightbulb is desired to emit at least
15 percent of its energy at wavelengths shorter than 1 |xm. De-
termine the minimum temperature to which the filament of the
lightbulb must be heated.
11-24 It is desired that the radiation energy emitted by a light
source reach a maximum in the blue range (k = 0.47 jjim). De-
termine the temperature of this light source and the fraction of
radiation it emits in the visible range (\ = 0.40-0.76 p,m).
11-25 A 3-mm-thick glass window transmits 90 percent of
the radiation between \ = 0.3 and 3.0 jjuti and is essentially
opaque for radiation at other wavelengths. Determine the rate
of radiation transmitted through a 2-m X 2-m glass window
from blackbody sources at (a) 5800 K and (b) 1000 K.
Answers: (a) 218,400 kW, (b) 55.8 kW
Radiation Intensity
11-26C What does a solid angle represent, and how does it
differ from a plane angle? What is the value of a solid angle as-
sociated with a sphere?
11-27C How is the intensity of emitted radiation defined?
For a diffusely emitting surface, how is the emissive power re-
lated to the intensity of emitted radiation?
11-28C For a surface, how is irradiation defined? For dif-
fusely incident radiation, how is irradiation on a surface related
to the intensity of incident radiation?
11-29C For a surface, how is radiosity defined? For diffusely
emitting and reflecting surfaces, how is radiosity related to the
intensities of emitted and reflected radiation?
11-30C When the variation of spectral radiation quantity
with wavelength is known, how is the corresponding total
quantity determined?
11-31 A small surface of area A \ = 4 cm 2 emits radiation as
a blackbody at T { = 800 K. Part of the radiation emitted by A \
strikes another small surface of area A 2 = 4 cm 2 oriented as
shown in the figure. Determine the solid angle subtended by A 2
when viewed from A u and the rate at which radiation emitted
by A] that strikes A 2 directly. What would your answer be if A 2
were directly above A t at a distance of 80 cm?
A 2 = 4 cm 2
800 K
FIGURE P1 1-31
11-32 A small circular surface of area A, = 2 cm 2 located at
the center of a 2-m-diameter sphere emits radiation as a black-
body at T t = 1000 K. Determine the rate at which radiation en-
ergy is streaming through a D 2 = 1 -cm-diameter hole located
(a) on top of the sphere directly above A , and (b) on the side of
sphere such that the line that connects the centers of A x and A 2
makes 45° with surface A x .
11-33 Repeat Problem 1 1-32 for a 4-m-diameter sphere.
11-34 A small surface of area A = 1 cm 2 emits radiation as a
blackbody at 1500 K. Determine the rate at which radiation en-
ergy is emitted through a band defined by < cf> < 2tt and 45
cen58933_chll.qxd 9/9/2002 9:39 AM Page 601
< < 60° where is the angle a radiation beam makes with
the normal of the surface and 4> is the azimuth angle.
11-35 A small surface of area A = 1 cm 2 is subjected to inci-
dent radiation of constant intensity /, = 2.2 X 10 4 W/m 2 • sr
over the entire hemisphere. Determine the rate at which radia-
tion energy is incident on the surface through (a) < G s 45°
and (b) 45 < < 90°, where is the angle a radiation beam
makes with the normal of the surface.
Radiation Properties
11-36C Define the properties emissivity and absorptivity.
When are these two properties equal to each other?
11-37C Define the properties reflectivity and transmissivity
and discuss the different forms of reflection.
11-38C What is a graybody? How does it differ from a
blackbody? What is a diffuse gray surface?
11-39C What is the greenhouse effect? Why is it a matter of
great concern among atmospheric scientists?
11-40C We can see the inside of a microwave oven during
operation through its glass door, which indicates that visible ra-
diation is escaping the oven. Do you think that the harmful mi-
crowave radiation might also be escaping?
11-41 The spectral emissivity function of an opaque surface
at 1000 K is approximated as
e, = 0.4,
e 2 = 0.7,
e, = 0.3,
< \ < 2 \xm
2 jjim < \ < 6 \im
6 jjun < \ < oo
Determine the average emissivity of the surface and the rate of
radiation emission from the surface, in W/m 2 .
Answers: 0.575, 32.6 kW/m 2
11-42 The reflectivity of aluminum coated with lead sulfate
is 0.35 for radiation at wavelengths less than 3 p,m and 0.95 for
radiation greater than 3 (xm. Determine the average reflectivity
of this surface for solar radiation (T ~ 5800 K) and radiation
coming from surfaces at room temperature (T ~ 300 K). Also,
determine the emissivity and absorptivity of this surface
at both temperatures. Do you think this material is suitable for
use in solar collectors?
11-43 A furnace that has a 25-cm X 25-cm glass window
can be considered to be a blackbody at 1200 K. If the trans-
missivity of the glass is 0.7 for radiation at wavelengths less
than 3 p,m and zero for radiation at wavelengths greater than 3
jjim, determine the fraction and the rate of radiation coming
from the furnace and transmitted through the window.
11-44 The emissivity of a tungsten filament can be approxi-
mated to be 0.5 for radiation at wavelengths less than 1 jjim and
0.15 for radiation at greater than 1 p,m. Determine the average
emissivity of the filament at (a) 2000 K and (b) 3000 K. Also,
601
CHAPTER 11
determine the absorptivity and reflectivity of the filament at
both temperatures.
11-45 The variations of the spectral emissivity of two sur-
faces are as given in Figure PI 1^45. Determine the average
emissivity of each surface at T = 3000 K. Also, determine the
average absorptivity and reflectivity of each surface for radia-
tion coming from a source at 3000 K. Which surface is more
suitable to serve as a solar absorber?
3
FIGURE P1 1-45
X, (im
11-46 The emissivity of a surface coated with aluminum ox-
ide can be approximated to be 0.2 for radiation at wavelengths
less than 5 |xm and 0.9 for radiation at wavelengths greater
than 5 p,m. Determine the average emissivity of this surface
at (a) 5800 K and (b) 300 K. What can you say about the ab-
sorptivity of this surface for radiation coming from sources at
5800 K and 300 K? Answers: (a) 0.203, (b) 0.89
11-47 The variation of the spectral absorptivity of a surface
is as given in Figure PI 1^-7. Determine the average absorptiv-
ity and reflectivity of the surface for radiation that originates
from a source at T = 2500 K. Also, determine the average
emissivity of this surface at 3000 K.
0.7
0.2
X, (ira
FIGURE P1 1^7
11-48E A 5-in. -diameter spherical ball is known to emit ra-
diation at a rate of 120 Btu/h when its surface temperature is
950 R. Determine the average emissivity of the ball at this
temperature.
11-49 The variation of the spectral transmissivity of a
0.6-cm-thick glass window is as given in Figure PI 1^-9. De-
termine the average transmissivity of this window for solar ra-
diation (T ~ 5800 K) and radiation coming from surfaces at
room temperature (7" ~ 300 K). Also, determine the amount of
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HEAT TRANSFER
solar radiation transmitted through the window for incident
solar radiation of 650 W/m 2 .
Answers: 0.848, 0.00015, 551.1 W/m 2
0.92
0.3
FIGURE P1 1-49
X, um
Atmospheric and Solar Radiation
11-50C What is the solar constant? How is it used to deter-
mine the effective surface temperature of the sun? How would
the value of the solar constant change if the distance between
the earth and the sun doubled?
11-51C What changes would you notice if the sun emitted ra-
diation at an effective temperature of 2000 K instead of 5762 K?
11-52C Explain why the sky is blue and the sunset is yellow-
orange.
11-53C When the earth is closest to the sun, we have winter
in the northern hemisphere. Explain why. Also explain why we
have summer in the northern hemisphere when the earth is far-
thest away from the sun.
11-54C What is the effective sky temperature?
11-55C You have probably noticed warning signs on the
highways stating that bridges may be icy even when the roads
are not. Explain how this can happen.
11-56C Unless you live in a warm southern state, you have
probably had to scrape ice from the windshield and windows of
your car many mornings. You may have noticed, with frustra-
tion, that the thickest layer of ice always forms on the wind-
shield instead of the side windows. Explain why this is the case.
11-57C Explain why surfaces usually have quite different
absorptivities for solar radiation and for radiation originating
from the surrounding bodies.
11-58 (~]b\ A surface has an absorptivity of a s = 0.85 for
xify solar radiation and an emissivity of e = 0.5 at
room temperature. The surface temperature is observed to be
350 K when the direct and the diffuse components of solar ra-
diation are G D = 350 and G d = 400 W/m 2 , respectively, and
the direct radiation makes a 30° angle with the normal of the
surface. Taking the effective sky temperature to be 280 K, de-
termine the net rate of radiation heat transfer to the surface at
that time.
11-59E Solar radiation is incident on the outer surface of a
spaceship at a rate of 400 Btu/h • ft 2 . The surface has an ab-
sorptivity of a s = 0.10 for solar radiation and an emissivity of
e = 0.8 at room temperature. The outer surface radiates heat
into space at R. If there is no net heat transfer into the space-
ship, determine the equilibrium temperature of the surface.
Answer: 413.3 R
11-60 The air temperature on a clear night is observed to re-
main at about 4°C. Yet water is reported to have frozen that
night due to radiation effect. Taking the convection heat trans-
fer coefficient to be 18 W/m 2 • °C, determine the value of the
maximum effective sky temperature that night.
11-61 The absorber surface of a solar collector is made of
aluminum coated with black chrome (a s = 0.87 and 8 = 0.09).
Solar radiation is incident on the surface at a rate of 600 W/m 2 .
The air and the effective sky temperatures are 25°C and 15°C,
respectively, and the convection heat transfer coefficient is
10 W/m 2 • °C. For an absorber surface temperature of 70°C,
determine the net rate of solar energy delivered by the absorber
plate to the water circulating behind it.
-=jl Sun it
600 W/m 2
T _ 1 Sop
J oc = 25°C sky-
T, = 70°C
Absorber plate
Water tubes
Insulation
FIGURE P1 1-61
11-62
Reconsider Problem 11-61. Using EES (or
other) software, plot the net rate of solar energy
transferred to water as a function of the absorptivity of the
absorber plate. Let the absorptivity vary from 0.5 to 1.0, and
discuss the results.
11-63 Determine the equilibrium temperature of the absorber
surface in Problem 11-61 if the back side of the absorber is
insulated.
Special Topic: Solar Heat Gain through Windows
11-64C What fraction of the solar energy is in the visible
range (a) outside the earth's atmosphere and (b) at sea level on
earth? Answer the same question for infrared radiation.
11-65C Describe the solar radiation properties of a window
that is ideally suited for minimizing the air-conditioning load.
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CHAPTER 11
11-66C Define the SHGC (solar heat gain coefficient), and
explain how it differs from the SC (shading coefficient). What
are the values of the SHGC and SC of a single -pane clear-glass
window?
11-67C What does the SC (shading coefficient) of a device
represent? How do the SCs of clear glass and heat-absorbing
glass compare?
11-68C What is a shading device? Is an internal or external
shading device more effective in reducing the solar heat gain
through a window? How does the color of the surface of a
shading device facing outside affect the solar heat gain?
11-69C What is the effect of a low-e coating on the inner
surface of a window glass on the (a) heat loss in winter and
(b) heat gain in summer through the window?
11-70C What is the effect of a reflective coating on the outer
surface of a window glass on the (a) heat loss in winter and
(b) heat gain in summer through the window?
11-71 A manufacturing facility located at 32° N latitude has
a glazing area of 60 m 2 facing west that consists of double-
pane windows made of clear glass (SHGC = 0.766). To re-
duce the solar heat gain in summer, a reflective film that will
reduce the SHGC to 0.35 is considered. The cooling season
consists of June, July, August, and September, and the heating
season, October through April. The average daily solar heat
fluxes incident on the west side at this latitude are 2.35, 3.03,
3.62, 4.00, 4.20, 4.24, 4.16, 3.93, 3.48, 2.94, 2.33, and 2.07
kWh/day • m 2 for January through December, respectively.
Also, the unit costs of electricity and natural gas are
$0.09/kWh and $0.45/therm, respectively. If the coefficient of
performance of the cooling system is 3.2 and the efficiency of
the furnace is 0.90, determine the net annual cost savings due
to installing reflective coating on the windows. Also, deter-
mine the simple payback period if the installation cost of re-
flective film is $20/m 2 . Answers: $53, 23 years
11-72 A house located in Boulder, Colorado (40° N latitude),
has ordinary double-pane windows with 6-mm-thick glasses
and the total window areas are 8, 6, 6, and 4 m 2 on the south,
west, east, and north walls, respectively. Determine the total
solar heat gain of the house at 9:00, 12:00, and 15:00 solar time
in July. Also, determine the total amount of solar heat gain per
day for an average day in January.
11-73 Repeat Problem 11-72 for double-pane windows that
are gray-tinted.
11-74 Consider a building in New York (40° N latitude) that
has 200 m 2 of window area on its south wall. The windows are
double-pane heat-absorbing type, and are equipped with light-
colored Venetian blinds with a shading coefficient of SC =
0.30. Determine the total solar heat gain of the building
through the south windows at solar noon in April. What would
your answer be if there were no blinds at the windows?
Venetian '
blinds
Light "
colored
Double-pane
window
Heat-absorbing
glass
FIGURE P1 1-74
11-75 A typical winter day in Reno, Nevada (39° N latitude),
is cold but sunny, and thus the solar heat gain through the win-
dows can be more than the heat loss through them during day-
time. Consider a house with double-door-type windows that
are double paned with 3-mm-thick glasses and 6.4 mm of air
space and have aluminum frames and spacers. The house is
maintained at 22°C at all times. Determine if the house is los-
ing more or less heat than it is gaining from the sun through an
east window on a typical day in January for a 24-h period if the
average outdoor temperature is 10°C. Answer: less
Double-pane
window
Solar
heat gain
22°C
Heat
loss
FIGURE P1 1-75
11-76 Repeat Problem 11-75 for a south window.
11-77E Determine the rate of net heat gain (or loss) through
a 9-ft-high, 15-ft-wide, fixed 4 -in. single-glass window with
aluminum frames on the west wall at 3 pm solar time during a
typical day in January at a location near 40° N latitude when
the indoor and outdoor temperatures are 70°F and 45°F,
respectively. Answer: 16,840 Btu/h gain
11-78 Consider a building located near 40° N latitude that
has equal window areas on all four sides. The building owner
is considering coating the south-facing windows with reflec-
tive film to reduce the solar heat gain and thus the cooling load.
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HEAT TRANSFER
But someone suggests that the owner will reduce the cooling
load even more if she coats the west-facing windows instead.
What do you think?
Review Problems
11-79 The spectral emissivity of an opaque surface at
1 200 K is approximated as
6]
for X < 2 jjim
0.85 for 2 < X < 6 |xm
for X > 6 jjim
Determine the total emissivity and the emissive flux of the
surface.
11-80 The spectral transmissivity of a 3-mm-thick regular
glass can be expressed as
t, = for A. < 0.35 |xm
t 2 = 0.85 for 0.35 < X < 2.5 |xm
t 3 = for X > 2.5 |xm
Determine the transmissivity of this glass for solar radiation.
What is the transmissivity of this glass for light?
11-81 A 1-m-diameter spherical cavity is maintained at a
uniform temperature of 600 K. Now a 5-mm-diameter hole is
drilled. Determine the maximum rate of radiation energy
streaming through the hole. What would your answer be if the
diameter of the cavity were 3 m?
11-82 The spectral absorptivity of an opaque surface is as
shown on the graph. Determine the absorptivity of the surface
for radiation emitted by a source at (a) 1000 K and (b) 3000 K.
0.3 1.2
FIGURE P1 1-82
X. nm
11-83 The surface in Problem 1 1-82 receives solar radiation
at a rate of 820 W/m 2 . Determine the solar absorptivity of the
surface and the rate of absorption of solar radiation.
11-84 The spectral transmissivity of a glass cover used in a
solar collector is given as
for X < 0.3 jjim
0.9 for 0.3 < X < 3 jjum
for X > 3 |xm
Solar radiation is incident at a rate of 950 W/m 2 , and the ab-
sorber plate, which can be considered to be black, is main-
tained at 340 K by the cooling water. Determine (a) the solar
flux incident on the absorber plate, (b) the transmissivity of the
glass cover for radiation emitted by the absorber plate, and
(c) the rate of heat transfer to the cooling water if the glass
cover temperature is also 340 K.
11-85 Consider a small black surface of area A = 2 cm 2
maintained at 600 K. Determine the rate at which radiation en-
ergy is emitted by the surface through a ring-shaped opening
defined by < 4> < 2ir and 40 < 9 < 50° where 4> is the az-
imuth angle and G is the angle a radiation beam makes with the
normal of the surface.
Design and Essay Problems
11-86 Write an essay on the radiation properties of selective
surfaces used on the absorber plates of solar collectors. Find
out about the various kinds of such surfaces, and discuss the
performance and cost of each type. Recommend a selective
surface that optimizes cost and performance.
11-87 According to an Atomic Energy Commission report,
a hydrogen bomb can be approximated as a large fireball at a
temperature of 7200 K. You are to assess the impact of such a
bomb exploded 5 km above a city. Assume the diameter of the
fireball to be 1 km, and the blast to last 15 s. Investigate the
level of radiation energy people, plants, and houses will be ex-
posed to, and how adversely they will be affected by the blast.
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RADIATION HEAT TRANSFER
CHAPTER
In Chapter 11, we considered the fundamental aspects of radiation and the
radiation properties of surfaces. We are now in a position to consider
radiation exchange between two or more surfaces, which is the primary
quantity of interest in most radiation problems.
We start this chapter with a discussion of view factors and the rules associ-
ated with them. View factor expressions and charts for some common config-
urations are given, and the crossed-strings method is presented. We then
discuss radiation heat transfer, first between black surfaces and then between
nonblack surfaces using the radiation network approach. We continue with ra-
diation shields and discuss the radiation effect on temperature measurements
and comfort. Finally, we consider gas radiation, and discuss the effective
emissivities and absorptivities of gas bodies of various shapes. We also dis-
cuss radiation exchange between the walls of combustion chambers and the
high-temperature emitting and absorbing combustion gases inside.
CONTENTS
12-1 The View Factor 606
12-2 View Factor Relations 609
12-3 Radiation Heat Transfer:
Black Surfaces 620
12-4 Radiation Heat Transfer:
Diffuse, Gray Surfaces 623
12-5 Radiation Shields and
the Radiation Effect 635
12-6 Radiation Exchange
with Emitting and
Absorbing Gases 639
Topic of Special Interest:
Heat Transfer from
the Human Body 649
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HEAT TRANSFER
Surface 2
Surface 1
Surface 3
Point
source
FIGURE 12-1
Radiation heat exchange between
surfaces depends on the orientation
of the surfaces relative to each other,
and this dependence on orientation is
accounted for by the view factor.
12-1 ■ THE VIEW FACTOR
Radiation heat transfer between surfaces depends on the orientation of the
surfaces relative to each other as well as their radiation properties and tem-
peratures, as illustrated in Figure 12—1. For example, a camper will make the
most use of a campfire on a cold night by standing as close to the fire as pos-
sible and by blocking as much of the radiation coming from the fire by turn-
ing her front to the fire instead of her side. Likewise, a person will maximize
the amount of solar radiation incident on him and take a sunbath by lying
down on his back instead of standing up on his feet.
To account for the effects of orientation on radiation heat transfer between
two surfaces, we define a new parameter called the view factor, which is a
purely geometric quantity and is independent of the surface properties and
temperature. It is also called the shape factor, configuration factor, and angle
factor. The view factor based on the assumption that the surfaces are diffuse
emitters and diffuse reflectors is called the diffuse view factor, and the view
factor based on the assumption that the surfaces are diffuse emitters but spec-
ular reflectors is called the specular view factor. In this book, we will consider
radiation exchange between diffuse surfaces only, and thus the term view fac-
tor will simply mean diffuse view factor.
The view factor from a surface i to a surface j is denoted by F { _^j or just F#,
and is defined as
the fraction of the radiation leaving surface i that strikes surface j directly
FIGURE 12-2
Geometry for the determination of the
view factor between two surfaces.
The notation -F,_>y is instructive for beginners, since it emphasizes that the
view factor is for radiation that travels from surface i to surface /'. However,
this notation becomes rather awkward when it has to be used many times in a
problem. In such cases, it is convenient to replace it by its shorthand ver-
sion Fjj.
Therefore, the view factor F u represents the fraction of radiation leaving
surface 1 that strikes surface 2 directly, and F 2I represents the fraction of the
radiation leaving surface 2 that strikes surface 1 directly. Note that the radia-
tion that strikes a surface does not need to be absorbed by that surface. Also,
radiation that strikes a surface after being reflected by other surfaces is not
considered in the evaluation of view factors.
To develop a general expression for the view factor, consider two differen-
tial surfaces dA x and dA 2 on two arbitrarily oriented surfaces A { and A 2 , re-
spectively, as shown in Figure 12-2. The distance between dA l and dA 2 is r,
and the angles between the normals of the surfaces and the line that connects
dA t and dA 2 are 6[ and 6 2 , respectively. Surface 1 emits and reflects radiation
diffusely in all directions with a constant intensity of I { , and the solid angle
subtended by dA 2 when viewed by dA l is da> 2l .
The rate at which radiation leaves dA x in the direction of 6 t is I x cos Q 1 dA l .
Noting that du> 2l = dA 2 cos Q 2 /r 2 , the portion of this radiation that strikes
dA 2 is
Q,i
I, cos 0,dA,d(x} 7 , = I, cos Q.dA,
dA-, cos 9
(12-1)
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The total rate at which radiation leaves dA x (via emission and reflection) in all
directions is the radiosity (which is J { = ir^) times the surface area,
QdA, = JldA x = TT/,rfA,
(12-2)
Then the differential view factor dF dA _^ dA , which is the fraction of radiation
leaving dA x that strikes dA 2 directly, becomes
dF,
QdAi-nJAi _ COS 0i COS 2
QdA, ™
-dA,
(12-3)
The differential view factor dF dA _^ dA can be determined from Eq. 12-3 by
interchanging the subscripts 1 and 2.
The view factor from a differential area dA x to a finite area A 2 can be
determined from the fact that the fraction of radiation leaving dA , that strikes
A 2 is the sum of the fractions of radiation striking the differential areas dA 2 .
Therefore, the view factor F dA _> A is determined by integrating dF dA _> rfAi
overA 2 ,
cos 0i cos 0,
dAj
(12-4)
607
CHAPTER 12
The total rate at which radiation leaves the entire A\ (via emission and re-
flection) in all directions is
Qa, = M\ = f^i
(12-5)
The portion of this radiation that strikes dA 2 is determined by considering the
radiation that leaves dA { and strikes dA 2 (given by Eq. 12-1), and integrating
it over A 1;
Qa, -^dA,~ QdA, -> dA, ~
JA, ' JA
I t cos 0, cos 2 <VA 2
dA,
(12-6)
Integration of this relation over A 2 gives the radiation that strikes the entire A 2 ,
r . r r I, cos 0, cos 0,
Qa, ->* = Q Al ^d Al = - 2 dA, dA 2 (12-7)
Dividing this by the total radiation leaving A, (from Eq. 12-5) gives the frac-
tion of radiation leaving A l that strikes A 2 , which is the view factor F A _> A (or
F l2 for short),
Qa,^a 2 l f f cos 0, cos Qi
Qa
Ja,ja,
■ dA , dAj
(12-8)
The view factor F A _> A is readily determined from Eq. 12-8 by interchanging
the subscripts 1 and 2,
F 2I = F
1 ~~ 1 A, -> A,
Qa 2 -
Qa
A 2 L J A
COS 0, COS 0t
■dA.dA^
(12-9)
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HEAT TRANSFER
(a) Plane surface
F,_„, =
(b) Convex surface
F 3 ^ 3 *0
(c) Concave surface
FIGURE 12-3
The view factor from a surface
to itself is zero for plane or
convex surfaces and nonzero
for concave surfaces.
Outer
sphere
FIGURE 12-4
In a geometry that consists of two
concentric spheres, the view factor
F] _, 2 = 1 since the entire radiation
leaving the surface of the smaller
sphere will be intercepted by the
larger sphere.
Note that /, is constant but r, 0,, and 2 are variables. Also, integrations can be
performed in any order since the integration limits are constants. These rela-
tions confirm that the view factor between two surfaces depends on their rel-
ative orientation and the distance between them.
Combining Eqs. 12-8 and 12-9 after multiplying the former by A { and the
latter by A 2 gives
A 2 F 2 ,
(12-10)
which is known as the reciprocity relation for view factors. It allows the cal-
culation of a view factor from a knowledge of the other.
The view factor relations developed above are applicable to any two sur-
faces i and; provided that the surfaces are diffuse emitters and diffuse reflec-
tors (so that the assumption of constant intensity is valid). For the special case
of j = i, we have
F ;-> j = the fraction of radiation leaving surface i that strikes itself directly
Noting that in the absence of strong electromagnetic fields radiation beams
travel in straight paths, the view factor from a surface to itself will be zero un-
less the surface "sees" itself. Therefore, F i ^ ti = for plane or convex surfaces
and F,-_> ,■ + for concave surfaces, as illustrated in Figure 12-3.
The value of the view factor ranges between zero and one. The limiting case
Fj _> 7 = indicates that the two surfaces do not have a direct view of each
other, and thus radiation leaving surface i cannot strike surface j directly. The
other limiting case F t ^ : = 1 indicates that surface j completely surrounds sur-
face i, so that the entire radiation leaving surface i is intercepted by surface j.
For example, in a geometry consisting of two concentric spheres, the entire
radiation leaving the surface of the smaller sphere (surface 1) will strike the
larger sphere (surface 2), and thus F { _> 2 = 1, as illustrated in Figure 12-4.
The view factor has proven to be very useful in radiation analysis because it
allows us to express the fraction of radiation leaving a surface that strikes an-
other surface in terms of the orientation of these two surfaces relative to each
other. The underlying assumption in this process is that the radiation a surface
receives from a source is directly proportional to the angle the surface sub-
tends when viewed from the source. This would be the case only if the
radiation coming off the source is uniform in all directions throughout its
surface and the medium between the surfaces does not absorb, emit, or scatter
radiation. That is, it will be the case when the surfaces are isothermal and
diffuse emitters and reflectors and the surfaces are separated by a non-
participating medium such as a vacuum or air.
The view factor F, _> 2 between two surfaces A l and A 2 can be determined in
a systematic manner first by expressing the view factor between two differen-
tial areas dA { and dA 2 in terms of the spatial variables and then by performing
the necessary integrations. However, this approach is not practical, since, even
for simple geometries, the resulting integrations are usually very complex and
difficult to perform.
View factors for hundreds of common geometries are evaluated and the re-
sults are given in analytical, graphical, and tabular form in several publica-
tions. View factors for selected geometries are given in Tables 12-1 and 12-2
in analytical form and in Figures 12-5 to 12-8 in graphical form. The view
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TABLE 12-1
View factor expressions for some common geometries of finite size (3D)
Geometry
Aligned parallel rectangles
j i N
i i
Coaxial parallel disks
r i L
Perpendicular rectangles
with a common edge
Relation
X = X/L and Y = Y/L
F.. = -^=lln
'^> kXY
{l+X 2 ){l + Y 2 )
1 + X 2 + Y 2
+ X(l + y 2 )" 2 tan-'-
+ y(l+X 2 ) 1/2 tan-
(1 + F 2 ) 1 ' 2
(1+X 2 )" 2
-Xtair'X-Ftan-' Y
R. = r./LandR. = r./L
• • 11
1 +fff
5=1 +
>->J 2
H = ZIX and W = Y/X
1 /,„._„-! 1
<->J
Wtan- 1 — +Htan'-
tzW\ W H
-(H 2 + W 2 ) xl2 lan
(H 2 + w 2 y
+ i ln l a + w 2 )a + H 2 )
1+W 2 + H 2
W 2 (l + W 2 + H 2 )
(1 + W 2 )(W 2 + H 2 )
H 2 (l+H 2 + W 2 )
(1 + H 2 )(H 2 + W 2 )
609
CHAPTER 12
factors in Table 12-1 are for three-dimensional geometries. The view factors
in Table 12-2, on the other hand, are for geometries that are infinitely long
in the direction perpendicular to the plane of the paper and are therefore
two-dimensional.
12-2 ■ VIEW FACTOR RELATIONS
Radiation analysis on an enclosure consisting of N surfaces requires the eval-
uation of N 2 view factors, and this evaluation process is probably the most
time-consuming part of a radiation analysis. However, it is neither practical
nor necessary to evaluate all of the view factors directly. Once a sufficient
number of view factors are available, the rest of them can be determined by
utilizing some fundamental relations for view factors, as discussed next.
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HEAT TRANSFER
TABLE 12-2
View factor expressions for some infinitely long (2D) geometries
Geometry
Relation
Parallel plates with midlines
connected by perpendicular line
TT
W. = w./L and W. = w. IL
ii j j
[(W t + Wp 2 + 4] " 2 - (Wj - W : ) 2 + 4] l/2
H.
Inclined plates of equal width
and with a common edge
t . .= 1 - sin —a
Perpendicular plates with a common edge
j
F. = ^1 + -
1 +
W:\2
Three-sided enclosure
2H\
Infinite plane and row of cylinders
© © © © © <z^ D
©D
F. =1
+ — tan =—
s D-
1 The Reciprocity Relation
The view factors F { ^j and Fj _> , are not equal to each other unless the areas of
the two surfaces are. That is,
Fj->i = Fj^,j when A t = Aj
Fj^i + Fj^j when A t + Aj
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CHAPTER 12
i-o
0.9 lA11
1
i::i:
0.8
/~ L 2~S
•-::- of
0.7 L
0.6 /
°- 5 :::: 1\
7\ ^/\
"s'^sSsfl
4~
\ ftlr:
= =S=H+
-lJ
0.4 J_ V
-<&
Inffc^*'
0.3
■ !
~"f;2^0.9^
'53=08 33^
0.2 -
: :::::: ?Sj^V
: — 1eHj _ 1 ^ — ny"
Sifc
tjitM ^r
^==0.6 «
.=-0.5-
1-4-
— ;-;0 4 _
— 5=======
4!
'/
yyyrLt
^[|f--
-;::=:0.3-
" _ -;=;0 25"
0.1 — y
0.09 //
%P''-.%
to
^
7
S —,;
i i
■-;;;:0.2-
- X-J Ul+A
' /
/
' ,:--0
18
16 =
14"
12"
™ 7 /VMf
t-t+Xfc 7 -
•'
,
- -0
4= ' ' 4+ X
.
,/
^ s* *
*- „
1 ■ u 4 IS,
\
S
' s\
L^*—
:::o
°' 0j AWj'Oft \M
' /
?^Sr
^^~
0.04 iMM
g^z z ^::^
*
-
^f'
.1
°-° 3 -Bill
¥
/
-^
"i+i 1 =:H =
l--P+ L [h-* ? '—
y
-+T h - M
' y / X /'/V/
' s
/
0.02 /j£f^-~
-- - -tSB-
'7 /fc'TOHTr
s
Z -/-JCL
/-.<■ \s
„„, , 7 /f J
s?S J X-
o.oi _? ^ii
^zyj Sk.
0.1 0.2 0.3 0.4 0.5 0.6 0.8 1 2
Ratio LJD
3 4 5 6
8 10
20
FIGURE 12-5
View factor between two
aligned parallel rectangles of
equal size.
0.2 0.3 0.4 0.5 0.6 0.8 1 2
Ratio LJW
FIGURE 12-6
View factor between two
perpendicular rectangles with
a common edge.
cen58933_chl2.qxd 9/9/2002 9:48 AM Page 612
612
HEAT TRANSFER
FIGURE 12-7
View factor between two
coaxial parallel disks.
1.0
0.9
0.8
0.7
0.6
2
0.5
0.4
0.3
0.2
0.1
r 2 IL = %y
ill
5 /
' 4
s- ?
( 1
^ 1
L
I
3
2
1.5'
III
1.25"""
1.0""""
1 1
0.8^
y y
0.5"
^0.4^^
■ r 2 /L =
= U.3
0.2 0.3 0.4 0.6
1.0
Llr,
2 3 4 5 6 8 10
F,
1.0
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
•?
%
^
^s
/
0.5
(
.25
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
FIGURE 12-8
View factors for two concentric cylinders of finite length: (a) outer cylinder to inner cylinder; (b) outer cylinder to itself.
cen58933_chl2.qxd 9/9/2002 9:48 AM Page 613
We have shown earlier the pair of view factors Fj_+j and F- _> , are related to
each other by
A i F i ^ j =A j F j ^ i
(12-11)
613
CHAPTER 12
This relation is referred to as the reciprocity relation or the reciprocity rule,
and it enables us to determine the counterpart of a view factor from a knowl-
edge of the view factor itself and the areas of the two surfaces. When deter-
mining the pair of view factors F { _^j and Ft _> ,, it makes sense to evaluate first
the easier one directly and then the more difficult one by applying the reci-
procity relation.
2 The Summation Rule
The radiation analysis of a surface normally requires the consideration of the
radiation coming in or going out in all directions. Therefore, most radiation
problems encountered in practice involve enclosed spaces. When formulating
a radiation problem, we usually form an enclosure consisting of the surfaces
interacting radiatively. Even openings are treated as imaginary surfaces with
radiation properties equivalent to those of the opening.
The conservation of energy principle requires that the entire radiation leav-
ing any surface i of an enclosure be intercepted by the surfaces of
the enclosure. Therefore, the sum of the view factors from surface i of an en-
closure to all surfaces of the enclosure, including to itself must equal unity.
This is known as the summation rule for an enclosure and is expressed as
(Fig. 12-9)
i
(12-12)
where N is the number of surfaces of the enclosure. For example, applying the
summation rule to surface 1 of a three-surface enclosure yields
2*
i
The summation rule can be applied to each surface of an enclosure by vary-
ing i from 1 to N. Therefore, the summation rule applied to each of the N sur-
faces of an enclosure gives N relations for the determination of the view
factors. Also, the reciprocity rule gives | N(N — 1) additional relations. Then
the total number of view factors that need to be evaluated directly for an
JV-surface enclosure becomes
N 2 - [N + \N(N - 1)] = iJV(JV - 1)
For example, for a six-surface enclosure, we need to determine only
2 X 6(6 — 1) = 15 of the 6 2 = 36 view factors directly. The remaining
21 view factors can be determined from the 21 equations that are obtained by
applying the reciprocity and the summation rules.
Surface i
FIGURE 12-9
Radiation leaving any surface i of
an enclosure must be intercepted
completely by the surfaces of the
enclosure. Therefore, the sum of
the view factors from surface i to
each one of the surfaces of the
enclosure must be unity.
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HEAT TRANSFER
FIGURE 12-10
The geometry considered
in Example 12-1.
EXAMPLE 12-1 View Factors Associated with Two Concentric
Spheres
Determine the view factors associated with an enclosure formed by two spheres,
shown in Figure 12-10.
SOLUTION The view factors associated with two concentric spheres are to be
determined.
Assumptions The surfaces are diffuse emitters and reflectors.
Analysis The outer surface of the smaller sphere (surface 1) and inner surface
of the larger sphere (surface 2) form a two-surface enclosure. Therefore, N = 2
and this enclosure involves N 2 = 2 2 = 4 view factors, which are F n , F 12 , F 21 ,
and F 22 . In this two-surface enclosure, we need to determine only
\N(N- 1) = i X 2(2 - 1) = 1
view factor directly. The remaining three view factors can be determined by the
application of the summation and reciprocity rules. But it turns out that we can
determine not only one but two view factors directly in this case by a simple
inspection:
F n = 0, since no radiation leaving surface 1 strikes itself
F| 2 = 1, since all radiation leaving surface 1 strikes surface 2
Actually it would be sufficient to determine only one of these view factors by
inspection, since we could always determine the other one from the summation
rule applied to surface 1 as F u + F 12 = 1.
The view factor F 21 is determined by applying the reciprocity relation to sur-
faces 1 and 2:
which yields
F 2 ,
A,F r
-F V2
A 2 F 2 ,
4irr, 2
X 1
Finally, the view factor F 22 is determined by applying the summation rule to sur-
face 2:
and thus
F 22 - 1 - F 21 - 1 - .^
Discussion Note that when the outer sphere is much larger than the inner
sphere (r 2 > r x ), F 22 approaches one. This is expected, since the fraction of
radiation leaving the outer sphere that is intercepted by the inner sphere will be
negligible in that case. Also note that the two spheres considered above do not
need to be concentric. However, the radiation analysis will be most accurate for
the case of concentric spheres, since the radiation is most likely to be uniform
on the surfaces in that case.
cen58933_chl2.qxd 9/9/2002 9:48 AM Page 615
3 The Superposition Rule
Sometimes the view factor associated with a given geometry is not available
in standard tables and charts. In such cases, it is desirable to express the given
geometry as the sum or difference of some geometries with known view fac-
tors, and then to apply the superposition rule, which can be expressed as the
view factor from a surface i to a surface j is equal to the sum of the view fac-
tors from surface i to the parts of surface j. Note that the reverse of this is not
true. That is, the view factor from a surface j to a surface i is not equal to the
sum of the view factors from the parts of surface j to surface ;'.
Consider the geometry in Figure 12-11, which is infinitely long in the
direction perpendicular to the plane of the paper. The radiation that leaves
surface 1 and strikes the combined surfaces 2 and 3 is equal to the sum of the
radiation that strikes surfaces 2 and 3. Therefore, the view factor from surface
1 to the combined surfaces of 2 and 3 is
> (2, 3)
+ F,
(12-13)
Suppose we need to find the view factor F 1 _> 3 . A quick check of the view fac-
tor expressions and charts in this section will reveal that such a view factor
cannot be evaluated directly. However, the view factor F { _> 3 can be deter-
mined from Eq. 12-13 after determining both F { _, 2 and F l _> {2 , 3) from the
chart in Figure 12-12. Therefore, it may be possible to determine some diffi-
cult view factors with relative ease by expressing one or both of the areas as
the sum or differences of areas and then applying the superposition rule.
To obtain a relation for the view factor F,
12-13 by A„
(2, 3)
we multiply Eq.
A,F,
> (2, 3)
and apply the reciprocity relation to each term to get
(A 2 + A 3 )Fp_ 3 ) _> j = A 2 F 2 _> i + A 3 F 3 _,. |
or
F{2, 3
(2, 3) -> 1
A 2 F 2 _> | + A 3 F 3
A 2 + A 3
(12-14)
Areas that are expressed as the sum of more than two parts can be handled in
a similar manner.
615
CHAPTER 12
^1 -^(2,3)~F 1 -^2 +F 1 ->3
FIGURE 12-11
The view factor from a surface to a
composite surface is equal to the sum
of the view factors from the surface to
the parts of the composite surface.
8 cm
FIGURE 12-12
The cylindrical enclosure
considered in Example 12-2.
EXAMPLE 12-2
Fraction of Radiation Leaving
through an Opening
Determine the fraction of the radiation leaving the base of the cylindrical en-
closure shown in Figure 12-12 that escapes through a coaxial ring opening
at its top surface. The radius and the length of the enclosure are r t = 10 cm
and L = 10 cm, while the inner and outer radii of the ring are r 2 = 5 cm and
r 3 = 8 cm, respectively.
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HEAT TRANSFER
SOLUTION The fraction of radiation leaving the base of a cylindrical enclosure
through a coaxial ring opening at its top surface is to be determined.
Assumptions The base surface is a diffuse emitter and reflector.
Analysis We are asked to determine the fraction of the radiation leaving the
base of the enclosure that escapes through an opening at the top surface.
Actually, what we are asked to determine is simply the view factor F, _^ ring from
the base of the enclosure to the ring-shaped surface at the top.
We do not have an analytical expression or chart for view factors between a
circular area and a coaxial ring, and so we cannot determine Fi^ ring directly.
However, we do have a chart for view factors between two coaxial parallel disks,
and we can always express a ring in terms of disks.
Let the base surface of radius r x = 10 cm be surface 1, the circular area of
r 2 = 5 cm at the top be surface 2, and the circular area of r 3 = 8 cm be sur-
face 3. Using the superposition rule, the view factor from surface 1 to surface 3
can be expressed as
F\ -> 3 = -M -> 2 + -r I -» ring
since surface 3 is the sum of surface 2 and the ring area. The view factors F,^ 2
and Fi^ 3 are determined from the chart in Figure 12-7.
L _ 10 cm
r \ ~ 10 cm
L = 10 cm
r i ~ 10 cm
Therefore,
and
and
r 2
L
L
5 cm
10 cm
8 cm
10 cm
0.5
0.8
(Fig. 12-7)
(Fig. 12-7)
0.11
0.28
F, -^nng =F^ 3 -Fi->2 = 0-28 - 0.11 = 0.17
which is the desired result. Note that F 1 ^ 2 an< ^ ^1^3 represent the fractions of
radiation leaving the base that strike the circular surfaces 2 and 3, respectively,
and their difference gives the fraction that strikes the ring area.
F 1 ->2 - ^l ->3
(Also, F 2 ^ 1 =F 3 ^ l )
FIGURE 12-13
Two surfaces that are symmetric about
a third surface will have the same
view factor from the third surface.
4 The Symmetry Rule
The determination of the view factors in a problem can be simplified further
if the geometry involved possesses some sort of symmetry. Therefore, it is
good practice to check for the presence of any symmetry in a problem before
attempting to determine the view factors directly. The presence of symmetry
can be determined by inspection, keeping the definition of the view factor in
mind. Identical surfaces that are oriented in an identical manner with respect
to another surface will intercept identical amounts of radiation leaving that
surface. Therefore, the symmetry rule can be expressed as two (or more) sur-
faces that possess symmetry about a third surface will have identical view fac-
tors from that surface (Fig. 12-13).
The symmetry rule can also be expressed as if the surfaces j and k are sym-
metric about the surface i then F,-.
can show that the relation F ; . , ■ = .
Fj _, k . Using the reciprocity rule, we
; is also true in this case.
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CHAPTER 12
EXAMPLE 12-3 View Factors Associated with a Tetragon
Determine the view factors from the base of the pyramid shown in Figure 12-14
to each of its four side surfaces. The base of the pyramid is a square, and its
side surfaces are isosceles triangles.
SOLUTION The view factors from the base of a pyramid to each of its four side
surfaces for the case of a square base are to be determined.
Assumptions The surfaces are diffuse emitters and reflectors.
Analysis The base of the pyramid (surface 1) and its four side surfaces (sur-
faces 2, 3, 4, and 5) form a five-surface enclosure. The first thing we notice
about this enclosure is its symmetry. The four side surfaces are symmetric
about the base surface. Then, from the symmetry rule, we have
Fll = ^13 = Fl* = ^15
Also, the summation rule applied to surface 1 yields
2 F v = F u + F l2 + F 13 + F 14 + Fis = 1
However, F n = 0, since the base is a flat surface. Then the two relations
above yield
12 - * 13 - ' 14 - ' 15
0.25
Discussion Note that each of the four side surfaces of the pyramid receive
one-fourth of the entire radiation leaving the base surface, as expected. Also
note that the presence of symmetry greatly simplified the determination of the
view factors.
FIGURE 12-14
The pyramid considered
in Example 12-3.
EXAMPLE 12-4 View Factors Associated with a Triangular Duct
Determine the view factor from any one side to any other side of the infinitely
long triangular duct whose cross section is given in Figure 12-15.
SOLUTION The view factors associated with an infinitely long triangular duct
are to be determined.
Assumptions The surfaces are diffuse emitters and reflectors.
Analysis The widths of the sides of the triangular cross section of the duct are
L\, L 2 , and L 3 , and the surface areas corresponding to them are A lt A 2 , and A 3 ,
respectively. Since the duct is infinitely long, the fraction of radiation leaving
any surface that escapes through the ends of the duct is negligible. Therefore,
the infinitely long duct can be considered to be a three-surface enclosure,
N= 3.
This enclosure involves N 2 = 3 2 = 9 view factors, and we need to determine
±N(N- 1) =iX3(3-l) = 3
FIGURE 12-15
The infinitely long triangular duct
considered in Example 12—4.
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HEAT TRANSFER
of these view factors directly. Fortunately, we can determine all three of them by
inspection to be
F 22
since all three surfaces are flat. The remaining six view factors can be deter-
mined by the application of the summation and reciprocity rules.
Applying the summation rule to each of the three surfaces gives
F u + F a + F a ~-
= 1
F 21 + F 22 + F 23 =
= 1
F 3] + F 32 + F 33 =
= 1
Noting that F n
and multiplying the first equation by A lt the
second by A 2 , and the third by A 3 gives
A x F l2 + A,F 13 = A,
I\ 2 r 91 ' ^^2 23 — 2
A 3 F 3] + A 3 t 32 = A 3
Finally, applying the three reciprocity relations A^F^
and A 2 F 23 = A 3 F 3Z gives
"Z'Zlr "1^13 — "3'31i
A,F I2
A,F I2
A 2 F 2i
A 2 r 23
A,
This is a set of three algebraic equations with three unknowns, which can be
solved to obtain
Fn
A] \ A9 ■'^3 1 2 ~S
2A,
2Li
/\i \ ^13 ^2 1 1 2
2A,
2L t
./it ~t" ^3 **-l 2 1 1
2A,
2L,
(12-15)
Discussion Note that we have replaced the areas of the side surfaces by their
corresponding widths for simplicity, since A = Ls and the length s can be fac-
tored out and canceled. We can generalize this result as the view factor from a
surface of a very long triangular duct to another surface is equal to the sum of
the widths of these two surfaces minus the width of the third surface, divided
by twice the width of the first surface.
View Factors between Infinitely Long Surfaces:
The Crossed-Strings Method
Many problems encountered in practice involve geometries of constant cross
section such as channels and ducts that are very long in one direction relative
cen58933_chl2.qxd 9/9/2002 9:48 AM Page 619
to the other directions. Such geometries can conveniently be considered to be
two-dimensional, since any radiation interaction through their end surfaces
will be negligible. These geometries can subsequently be modeled as being
infinitely long, and the view factor between their surfaces can be determined
by the amazingly simple crossed-strings method developed by H. C. Hottel in
the 1950s. The surfaces of the geometry do not need to be flat; they can be
convex, concave, or any irregular shape.
To demonstrate this method, consider the geometry shown in Figure 12-16,
and let us try to find the view factor F { _> 2 between surfaces 1 and 2. The first
thing we do is identify the endpoints of the surfaces (the points A, B, C, and D)
and connect them to each other with tightly stretched strings, which are
indicated by dashed lines. Hottel has shown that the view factor F 1 __ >2 can be
expressed in terms of the lengths of these stretched strings, which are straight
lines, as
(L 5 + L 6 ) - (L 3 + L 4 )
2L,
(12-16)
Note that L 5 + L 6 is the sum of the lengths of the crossed strings, and L 3 + L
is the sum of the lengths of the uncrossed strings attached to the endpoints
Therefore, Hottel's crossed-strings method can be expressed verbally as
2 (Crossed strings) — 2 (Uncrossed strings)
2 X (String on surface i)
619
CHAPTER 12
-IB
FIGURE 12-16
Determination of the view factor
^i -> 2 by the application of
the crossed-strings method.
(12-17)
The crossed-strings method is applicable even when the two surfaces consid-
ered share a common edge, as in a triangle. In such cases, the common edge
can be treated as an imaginary string of zero length. The method can also be
applied to surfaces that are partially blocked by other surfaces by allowing the
strings to bend around the blocking surfaces.
EXAMPLE 12-5 The Crossed-Strings Method for View Factors
Two infinitely long parallel plates of widths a = 12 cm and b = 5 cm are lo-
cated a distance c = 6 cm apart, as shown in Figure 12-17. (a) Determine the
view factor F 1 _> 2 from surface 1 to surface 2 by using the crossed-strings
method, (b) Derive the crossed-strings formula by forming triangles on the given
geometry and using Eq. 12-15 for view factors between the sides of triangles.
SOLUTION The view factors between two infinitely long parallel plates are to
be determined using the crossed-strings method, and the formula for the view
factor is to be derived.
Assumptions The surfaces are diffuse emitters and reflectors.
Analysis (a) First we label the endpoints of both surfaces and draw straight
dashed lines between the endpoints, as shown in Figure 12-17. Then we iden-
tify the crossed and uncrossed strings and apply the crossed-strings method
(Eq. 12-17) to determine the view factor F 1 ^ 2 '-
2 (Crossed strings) — 2 (Uncrossed strings) (L 5 + L 6 ) — (L 3 + L 4 )
2 X (String on surface 1)
2L,
b = L, = 5 cm
c = 6 cm
C FT
^2)
V /
X /
\
\
>
\
V
/
/
S N
\^4
As
L h
/
1 /
| /
N\
N\
UD
a = L , = 1 2 cm
FIGURE 12-17
The two infinitely long parallel
plates considered in Example 12-5.
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620
HEAT TRANSFER
where
= a
= b
= c
= 12 cm
= 5 cm
= 6 cm
u
u
= V7 2
+ 6 2 =
9.22 cm
= V5 2
+ 6 2 =
7.81 cm
= V12
2 + 6 2
= 13.42 cm
ubstituting,
Fl-,
2 =
[(7.81 +
13.42)
-(6 +
9.22)]
cm
2 X
12 cm
(£>) The geometry is infinitely long in the direction perpendicular to the plane of
the paper, and thus the two plates (surfaces 1 and 2) and the two openings
(imaginary surfaces 3 and 4) form a four-surface enclosure. Then applying the
summation rule to surface 1 yields
F» + F n + F 13 + F u = 1
But F n = since it is a flat surface. Therefore,
where the view factors F 13 and F 14 can be determined by considering the trian-
gles ABC and ABD, respectively, and applying Eq. 12-15 for view factors be-
tween the sides of triangles. We obtain
L, + L 3
L 4
2L,
2L,
Substituting,
L s
2L,
2Li
(L 5 + L 6 ) - (L3 + L 4 )
2L X
which is the desired result. This is also a miniproof of the crossed-strings
method for the case of two infinitely long plain parallel surfaces.
12-3 - RADIATION HEAT TRANSFER:
BLACK SURFACES
So far, we have considered the nature of radiation, the radiation properties of
materials, and the view factors, and we are now in a position to consider the
rate of heat transfer between surfaces by radiation. The analysis of radiation
exchange between surfaces, in general, is complicated because of reflection: a
radiation beam leaving a surface may be reflected several times, with partial
reflection occurring at each surface, before it is completely absorbed. The
analysis is simplified greatly when the surfaces involved can be approximated
cen58933_chl2.qxd 9/9/2002 9:48 AM Page 621
as blackbodies because of the absence of reflection. In this section, we con-
sider radiation exchange between black surfaces only; we will extend the
analysis to reflecting surfaces in the next section.
Consider two black surfaces of arbitrary shape maintained at uniform tem-
peratures T { and T 2 , as shown in Figure 12-18. Recognizing that radiation
leaves a black surface at a rate of E b = uT 4 per unit surface area and that the
view factor F { _> 2 represents the fraction of radiation leaving surface 1 that
strikes surface 2, the net rate of radiation heat transfer from surface 1 to sur-
face 2 can be expressed as
e,
I Radiation leaving \
the entire surface 1
I that strikes surface 2
I Radiation leaving '
the entire surface 2
I that strikes surface 1;
A, E h , F,
A 2 Em F 2 .
Applying the reciprocity relation A l F l
AnF,
fil
! ^(r, 4 - Ti)
(W)
yields
(W)
(12-18)
(12-19)
621
CHAPTER 12
FIGURE 12-18
Two general black surfaces maintained
at uniform temperatures T l and T 2 .
which is the desired relation. A negative value for Q , _> 2 indicates that net ra-
diation heat transfer is from surface 2 to surface 1.
Now consider an enclosure consisting of N black surfaces maintained at
specified temperatures. The net radiation heat transfer from any surface i of
this enclosure is determined by adding up the net radiation heat transfers from
surface i to each of the surfaces of the enclosure:
Qi
• 2 e.
-^A,F,_
F (T? - Tf)
(W)
(12-20)
Again a negative value for Q indicates that net radiation heat transfer is to
surface i (i.e., surface i gains radiation energy instead of losing). Also, the net
heat transfer from a surface to itself is zero, regardless of the shape of the
surface.
EXAMPLE 12-6
Radiation Heat Transfer in a Black Furnace
Consider the 5-m X 5-m X 5-m cubical furnace shown in Figure 12-19, whose
surfaces closely approximate black surfaces. The base, top, and side surfaces
of the furnace are maintained at uniform temperatures of 800 K, 1500 K, and
500 K, respectively. Determine (a) the net rate of radiation heat transfer be-
tween the base and the side surfaces, (b) the net rate of radiation heat transfer
between the base and the top surface, and (c) the net radiation heat transfer
from the base surface.
SOLUTION The surfaces of a cubical furnace are black and are maintained at
uniform temperatures. The net rate of radiation heat transfer between the base
and side surfaces, between the base and the top surface, from the base surface
are to be determined.
Assumptions The surfaces are black and isothermal.
>^JS>
r, = 500 K
T, = 800 K
FIGURE 12-19
The cubical furnace of black surfaces
considered in Example 12-6.
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HEAT TRANSFER
Analysis (a) Considering that the geometry involves six surfaces, we may be
tempted at first to treat the furnace as a six-surface enclosure. However, the
four side surfaces possess the same properties, and thus we can treat them as
a single side surface in radiation analysis. We consider the base surface to be
surface 1, the top surface to be surface 2, and the side surfaces to be surface
3. Then the problem reduces to determining Q 1 _^ 3 , Qi^ 2 > ar| d Qi-
The net rate of radiation heat transfer Qi^ 3 from surface 1 to surface 3 can
be determined from Eq. 12-19, since both surfaces involved are black, by re-
placing the subscript 2 by 3:
s,
mt? - n)
But first we need to evaluate the view factor Fi_> 3 . After checking the view fac-
tor charts and tables, we realize that we cannot determine this view factor di-
rectly. However, we can determine the view factor F x _, 2 directly from Figure
12-5 to be F 1 _, 2 = 0.2, and we know that F 1 _> 1 = since surface 1 is a
plane. Then applying the summation rule to surface 1 yields
F , . 1 + F, .-, + F, .■> = 1
Substituting,
Si _> 3 = (25 m 2 )(0.8)(5.67 X 1CT 8 W/m 2 • K 4 )[(800 K) 4 - (500 K) 4 ]
= 394 X 10 3 W = 394 kW
(£>) The net rate of radiation heat transfer Qi^ 2 f rom surface 1 to surface 2 is
determined in a similar manner from Eq. 12-19 to be
Si^2=A,F l ^ 2 a(r 1 4 - T£)
= (25 m 2 )(0.2)(5.67 X lO" 8 W/m 2 ■ K 4 )[(800 K) 4 - (1500 K) 4 ]
= - 1319 X 10 3 W = - 1319 kW
The negative sign indicates that net radiation heat transfer is from surface 2 to
surface 1.
(c) The net radiation heat transfer from the base surface Qi is determined from
Eq. 12-20 by replacing the subscript / by 1 and taking N = 3:
3
Si = 2Gi-j = Gi-n + ei-»2+Gi-»3
i = l
= + (-1319 kW) + (394 kW)
= -925 kW
Again the negative sign indicates that net radiation heat transfer is to surface 1.
That is, the base of the furnace is gaining net radiation at a rate of about
925 kW.
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623
CHAPTER 12
12^ - RADIATION HEAT TRANSFER:
DIFFUSE, GRAY SURFACES
The analysis of radiation transfer in enclosures consisting of black surfaces is
relatively easy, as we have seen above, but most enclosures encountered in
practice involve nonblack surfaces, which allow multiple reflections to occur.
Radiation analysis of such enclosures becomes very complicated unless some
simplifying assumptions are made.
To make a simple radiation analysis possible, it is common to assume the
surfaces of an enclosure to be opaque, diffuse, and gray. That is, the surfaces
are nontransparent, they are diffuse emitters and diffuse reflectors, and their
radiation properties are independent of wavelength. Also, each surface of the
enclosure is isothermal, and both the incoming and outgoing radiation are uni-
form over each surface. But first we review the concept of radiosity discussed
in Chap. 11.
Radiosity
Surfaces emit radiation as well as reflect it, and thus the radiation leaving a
surface consists of emitted and reflected parts. The calculation of radiation
heat transfer between surfaces involves the total radiation energy streaming
away from a surface, with no regard for its origin. The total radiation energy
leaving a surface per unit time and per unit area is the radiosity and is
denoted by / (Fig. 12-20).
For a surface i that is gray and opaque (e, = a, and a, + p, = 1), the
radiosity can be expressed as
, _ /Radiation emitted] , /Radiation reflected]
\ by surface i I \ by surface i I
= E,E bi + P;G;
= e,E bi + (1 - e,)G, (W/m 2 ) (12-21)
where E bj = uTf" is the blackbody emissive power of surface i and G, is
irradiation (i.e., the radiation energy incident on surface i per unit time per
unit area).
For a surface that can be approximated as a blackbody (e, = 1), the radios-
ity relation reduces to
Radiosity, /
Reflected Emitted
radiation radiation
"
Surface
FIGURE 12-20
Radiosity represents the sum of the
radiation energy emitted and
reflected by a surface.
Ji = E bi = ctT; 4
(blackbody)
(12-22)
That is, the radiosity of a blackbody is equal to its emissive power. This is
expected, since a blackbody does not reflect any radiation, and thus radiation
coming from a blackbody is due to emission only.
Net Radiation Heat Transfer to or from a Surface
During a radiation interaction, a surface loses energy by emitting radiation and
gains energy by absorbing radiation emitted by other surfaces. A surface ex-
periences a net gain or a net loss of energy, depending on which quantity is
larger. The net rate of radiation heat transfer from a surface i of surface area A,
is denoted by Q , and is expressed as
cen58933_chl2.qxd 9/9/2002 9:48 AM Page 624
624
HEAT TRANSFER
Q,
/Radiation leaving\ _ /Radiation incident\
\ entire surface i J \ on entire surface i)
A,(/ ; - G ; ) (W)
Solving for G, from Eq. 12-21 and substituting into Eq. 12-23 yields
Qi=A,[J,
1
A,e,
1
(E u -Jd
(W)
(12-23)
(12-24)
In an electrical analogy to Ohm's law, this equation can be rearranged as
Ei,i ~ Ji
Qi
R,
(W)
(12-25)
where
(12-26)
Surface
I — www — ♦ y,
A,e,
FIGURE 12-21
Electrical analogy of surface
resistance to radiation.
is the surface resistance to radiation. The quantity E bi — /, corresponds to a
potential difference and the net rate of radiation heat transfer corresponds to
current in the electrical analogy, as illustrated in Figure 12-21.
The direction of the net radiation heat transfer depends on the relative mag-
nitudes of J t (the radiosity) and E bi (the emissive power of a blackbody at the
temperature of the surface). It will be, from the surface if E bi > /, and to the
surface if /, > E bi . A negative value for 2, indicates that heat transfer is to
the surface. All of this radiation energy gained must be removed from the
other side of the surface through some mechanism if the surface temperature
is to remain constant.
The surface resistance to radiation for a blackbody is zero since e, = 1 and
/, = E bi . The net rate of radiation heat transfer in this case is determined
directly from Eq. 12-23.
Some surfaces encountered in numerous practical heat transfer applications
are modeled as being adiabatic since their back sides are well insulated and
the net heat transfer through them is zero. When the convection effects on the
front (heat transfer) side of such a surface is negligible and steady-state con-
ditions are reached, the surface must lose as much radiation energy as it gains,
and thus <2 , = 0. In such cases, the surface is said to reradiate all the radiation
energy it receives, and such a surface is called a reradiating surface. Setting
Q, = OinEq. 12-25 yields
Ji
UT; 4
(W/m 2 )
(12-27)
Therefore, the temperature of a reradiating surface under steady conditions
can easily be determined from the equation above once its radiosity is known.
Note that the temperature of a reradiating surface is independent of its emis-
sivity. In radiation analysis, the surface resistance of a reradiating surface is
disregarded since there is no net heat transfer through it. (This is like the fact
that there is no need to consider a resistance in an electrical network if no cur-
rent is flowing through it.)
cen58933_chl2.qxd 9/9/2002 9:48 AM Page 625
Net Radiation Heat Transfer
between Any Two Surfaces
Consider two diffuse, gray, and opaque surfaces of arbitrary shape maintained
at uniform temperatures, as shown in Figure 12-22. Recognizing that the radi-
osity J represents the rate of radiation leaving a surface per unit surface area
and that the view factor F t ^j represents the fraction of radiation leaving sur-
face ;' that strikes surface j, the net rate of radiation heat transfer from surface
i to surface j can be expressed as
Qi-
I Radiation leaving \
the entire surface i
\that strikes surface j
A J F - A J F
/ Radiation leaving \
the entire surface j
\that strikes surface i
(W)
(12-28)
Applying the reciprocity relation A, F^j = Aj f}_>,' yields
Q i ^j=A i F i ^ j (J i -Jj) (W)
Again in analogy to Ohm's law, this equation can be rearranged as
Qi-
J, - Jj
1 R;
(W)
(12-29)
(12-30)
625
CHAPTER 12
Surf ace j
'^t,i ^ Surface i
FIGURE 12-22
Electrical analogy of
space resistance to radiation.
where
<->J A. p. .
(12-31)
is the space resistance to radiation. Again the quantity J t — Jj corresponds to
a potential difference, and the net rate of heat transfer between two surfaces
corresponds to current in the electrical analogy, as illustrated in Figure 12-22.
The direction of the net radiation heat transfer between two surfaces de-
pends on the relative magnitudes of /, and J,. A positive value for Q,_> ; indi-
cates that net heat transfer is from surface i to surface j. A negative value
indicates the opposite.
In an TV-surface enclosure, the conservation of energy principle requires that
the net heat transfer from surface i be equal to the sum of the net heat transfers
from surface i to each of the N surfaces of the enclosure. That is,
Q, = 2 Qi->j = E A i p i^j ( J , ~ J j) = 2
] = i
; = i
; = i
Ri^j
(W) (12-32)
The network representation of net radiation heat transfer from surface i to the
remaining surfaces of an TV-surface enclosure is given in Figure 12-23. Note
that <2,-_> i (the net rate of heat transfer from a surface to itself) is zero regard-
less of the shape of the surface. Combining Eqs. 12-25 and 12-32 gives
I
R,
N J -
y —
■^ R;
J;
(W)
(12-33)
Surface i
FIGURE 12-23
Network representation of net
radiation heat transfer from surface i
to the remaining surfaces of an
TV-surface enclosure.
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HEAT TRANSFER
which has the electrical analogy interpretation that the net radiation flow from
a surface through its surface resistance is equal to the sum of the radiation
flows from that surface to all other surfaces through the corresponding space
resistances.
Methods of Solving Radiation Problems
In the radiation analysis of an enclosure, either the temperature or the net rate
of heat transfer must be given for each of the surfaces to obtain a unique solu-
tion for the unknown surface temperatures and heat transfer rates. There are
two methods commonly used to solve radiation problems. In the first method,
Eqs. 12-32 (for surfaces with specified heat transfer rates) and 12-33 (for sur-
faces with specified temperatures) are simplified and rearranged as
Surfaces with specified
net heat transfer rate Q t
Q i =A i ^F i ^ j (J i -J J ) (12-34)
; = i
Surfaces with specified I — e- N
temperature T, err, 4 = /, + e ' ^ F, _,/./,■ - Jj) (12-35)
j=i
Note that Q t = for insulated (or reradiating) surfaces, and oT, 4 = J, for
black surfaces since s, = 1 in that case. Also, the term corresponding to; = i
will drop out from either relation since /, — Jj = /, — J, ■ = in that case.
The equations above give N linear algebraic equations for the determination
of the N unknown radiosities for an TV-surface enclosure. Once the radiosities
J i, J 2 , ■ ■ ■ , Jn are available, the unknown heat transfer rates can be determined
from Eq. 12-34 while the unknown surface temperatures can be determined
from Eq. 12-35. The temperatures of insulated or reradiating surfaces can be
determined from uT? = Jj. A positive value for Q , indicates net radiation heat
transfer from surface i to other surfaces in the enclosure while a negative value
indicates net radiation heat transfer to the surface.
The systematic approach described above for solving radiation heat transfer
problems is very suitable for use with today's popular equation solvers such
as EES, Mathcad, and Matlab, especially when there are a large number of
surfaces, and is known as the direct method (formerly, the matrix method,
since it resulted in matrices and the solution required a knowledge of linear
algebra). The second method described below, called the network method, is
based on the electrical network analogy.
The network method was first introduced by A. K. Oppenheim in the 1950s
and found widespread acceptance because of its simplicity and emphasis on
the physics of the problem. The application of the method is straightforward:
draw a surface resistance associated with each surface of an enclosure and
connect them with space resistances. Then solve the radiation problem by
treating it as an electrical network problem where the radiation heat transfer
replaces the current and radiosity replaces the potential.
The network method is not practical for enclosures with more than three or
four surfaces, however, because of the increased complexity of the network.
Next we apply the method to solve radiation problems in two- and three-
surface enclosures.
cen58933_chl2.qxd 9/9/2002 9:48 AM Page 627
Radiation Heat Transfer in Two-Surface Enclosures
Consider an enclosure consisting of two opaque surfaces at specified temper-
atures T| and T 2 , as shown in Fig. 12-24, and try to determine the net rate of
radiation heat transfer between the two surfaces with the network method.
Surfaces 1 and 2 have emissivities e, and e 2 and surface areas A! and A 2 and
are maintained at uniform temperatures T x and T 2 , respectively. There are only
two surfaces in the enclosure, and thus we can write
Gl2=6l
Qi
That is, the net rate of radiation heat transfer from surface 1 to surface 2 must
equal the net rate of radiation heat transfer from surface 1 and the net rate of
radiation heat transfer to surface 2.
The radiation network of this two-surface enclosure consists of two surface
resistances and one space resistance, as shown in Figure 12-24. In an electri-
cal network, the electric current flowing through these resistances connected
in series would be determined by dividing the potential difference between
points A and B by the total resistance between the same two points. The net
rate of radiation transfer is determined in the same manner and is expressed as
627
CHAPTER 12
* VWVWA-
]-,
^www ♦
R
1
12 ■
R 2 -
1
2 e 2
FIGURE 12-24
Schematic of a two-surface
enclosure and the radiation
network associated with it.
e,
7vi + 7\p 4" R->
G,
G 2
or
6,
<j(Tf - r 2 4 )
1 -
" 6]
+
1
+
1 -
e 2
A,
<3|
A
Fu
A 2
e 2
(W)
(12-36)
This important result is applicable to any two gray, diffuse, opaque surfaces
that form an enclosure. The view factor F [2 depends on the geometry
and must be determined first. Simplified forms of Eq. 12-36 for some famil-
iar arrangements that form a two-surface enclosure are given in Table 12-3.
Note that F l2 = 1 for all of these special cases.
EXAMPLE 12-7 Radiation Heat Transfer between Parallel Plates
Two very large parallel plates are maintained at uniform temperatures 7"! =
800 K and T 2 = 500 K and have emissivities e 1 = 0.2 and e 2 = 0.7, respec-
tively, as shown in Figure 12-25. Determine the net rate of radiation heat trans-
fer between the two surfaces per unit surface area of the plates.
SOLUTION Two large parallel plates are maintained at uniform temperatures.
The net rate of radiation heat transfer between the plates is to be determined.
Assumptions Both surfaces are opaque, diffuse, and gray.
Analysis The net rate of radiation heat transfer between the two plates per unit
area is readily determined from Eq. 12-38 to be
= 0.2
e 2 = 0.7
T ? = 500 K
FIGURE 12-25
The two parallel plates
considered in Example 12-7.
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HEAT TRANSFER
TABLE 12-3
Small object in a large cavity
(A v T X S\
I <v \
^i
at-
!2 12 =A 1 ae 1 (7't-r 2 4 )
(12-37)
\
F 12 = l
^0 2 ,r 2 ,E 2
Infinitely large parallel plates
A„ T l .z l
F 12 =l
^ Aa(r 4 -r 4 )
12 " l l
— + — - 1
e, e 2
(12-38)
/i-), It, t t
Infinitely long concentric cylinders
'l
'P¥
-"2
A[ C[
'I -.J \
A 2 ~ r 2
F 12 = I
• A t o(T\-T*)
(12-39)
y ' 2 1 1-f r
e, e 2 I r 2 '
Concentric spheres
A 2 I r 2 '
F 12 = l
/ /Os 1 \
• A,o(r 4 -r 2 4 )
(12-40)
V_V/- 2
^12 . l_ F r 2
e, e 2 lr 2 j
2,2 «T(r, 4 -r 2 4 )
qn ~ A "i. + 1-i"
(5.67 X
10"
8 W/m 2 •
K 4 )[(800 K) 4 -
- (500 K) 4 ]
fr
■£-
= 3625 W/m 2
Discussion Note that heat at
a net rate of 3625 W
s transferred from plate 1
to plate 2 by radiation per uni
t surface
area
of either
plate.
cen58933_chl2.qxd 9/9/2002 9:49 AM Page 629
Radiation Heat Transfer
in Three-Surface Enclosures
We now consider an enclosure consisting of three opaque, diffuse, gray sur-
faces, as shown in Figure 12-26. Surfaces 1, 2, and 3 have surface areas
A h A 2 , and A 3 ; emissivities s b e 2 , and e 3 ; and uniform temperatures T u T 2 , and
r 3 , respectively. The radiation network of this geometry is constructed by fol-
lowing the standard procedure: draw a surface resistance associated with each
of the three surfaces and connect these surface resistances with space resis-
tances, as shown in the figure. Relations for the surface and space resistances
are given by Eqs. 12-26 and 12-31. The three endpoint potentials E bl , E b2 ,
and E b3 are considered known, since the surface temperatures are specified.
Then all we need to find are the radiosities J u J 2 , and J 3 . The three equations
for the determination of these three unknowns are obtained from the require-
ment that the algebraic sum of the currents (net radiation heat transfer) at
each node must equal zero. That is,
Em ~
- J
■ h~
•A
, h ~ Jx
R
1 R
i
' *!3
Ji-
h
Ebi ~
Ri
h
J 3 -J 2
+
^23
Jx-
h
^~
h
Ebl, ~ J 3
629
CHAPTER 12
R
R
R,
(12-41)
Once the radiosities J x , J 2 , and J 3 are available, the net rate of radiation heat
transfers at each surface can be determined from Eq. 12-32.
The set of equations above simplify further if one or more surfaces are "spe-
cial" in some way. For example, /, = E bi = oT,- 4 for a black ox reradiating sur-
face. Also, Qj = for a reradiating surface. Finally, when the net rate of
radiation heat transfer Q t is specified at surface i instead of the temperature,
the term (E bi — J,)/Rj should be replaced by the specified Q ,.
e,,A,,T
FIGURE 12-26
Schematic of a three-surface enclosure and the radiation network associated with it.
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HEAT TRANSFER
Black
FIGURE 12-27
The cylindrical furnace
considered in Example 12-
EXAMPLE 12-8 Radiation Heat Transfer in a Cylindrical Furnace
Consider a cylindrical furnace with r = H = 1 m, as shown in Figure 12-27.
The top (surface 1) and the base (surface 2) of the furnace has emissivities
8i = 0.8 and e 2 = 0.4, respectively, and are maintained at uniform tempera-
tures 7"! = 700 K and T 2 = 500 K. The side surface closely approximates a
blackbody and is maintained at a temperature of T 3 = 400 K. Determine the
net rate of radiation heat transfer at each surface during steady operation and
explain how these surfaces can be maintained at specified temperatures.
SOLUTION The surfaces of a cylindrical furnace are maintained at uniform
temperatures. The net rate of radiation heat transfer at each surface during
steady operation is to be determined.
Assumptions 1 Steady operating conditions exist. 2 The surfaces are opaque,
diffuse, and gray. 3 Convection heat transfer is not considered.
Analysis We will solve this problem systematically using the direct method to
demonstrate its use. The cylindrical furnace can be considered to be a three-
surface enclosure with surface areas of
A t =A 2 = Tir} = tt(1 m) 2 = 3.14 m 2
A 3 = 2-nrJJ = 2ir(l m)(l m) = 6.28 m 2
The view factor from the base to the top surface is, from Figure 12-7, F 12 =
0.38. Then the view factor from the base to the side surface is determined by
applying the summation rule to be
F 12 + F 13 = 1
1 - F„ - F„ = 1 - - 0.38 = 0.62
since the base surface is flat and thus F n = 0. Noting that the top and bottom
surfaces are symmetric about the side surface, F 21 = F 12 = 0.38 and F 23 =
F 13 = 0.62. The view factor F 31 is determined from the reciprocity relation,
AiF l3 =A 3 F 3
F 31 = F 13 (Ai/A 3 ) = (0.62)(0.314/0.628) = 0.31
Also, F 32 = F 31 = 0.31 because of symmetry. Now that all the view factors are
available, we apply Eq. 12-35 to each surface to determine the radiosities:
Top surface (i =1): oT 4 = J {
1 -e,
[F,^ 2 (7,-/ 2 ) + F,_ 3(7,-/3)]
1 - e 2
Bottom surface (i = 2): crF 2 = 7 2 H g — [F 2 _, , (/ 2 — 7|) + F 2 _, , (7 2 — 7 3 )]
Side surface (i = 3): ct/ 3 4 = 7 3 + (since surface 3 is black and thus e 3 = 1)
Substituting the known quantities,
(5.67 X 10- 8 W/m 2 • K 4 )(700 K) 4 = /, + 1 ~ °" 8 [0.38(7! - / 2 ) + 0.68(7, - / 3 )]
U.O
(5.67 X 10-s W/m 2 • K 4 )(500 K) 4 = 7 2 + 1 ~ °' 4 [0.28(7 2 - /,) + 0.68(/ 2 - / 3 )]
0.4
(5.67 X 10- 8 W/m 2 • K 4 )(400 K) 4 = / 3
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CHAPTER 12
Solving the equations above for J lt J 2 , and J 3 gives
7, = 1 1,418 W/m 2 , J 2 = 4562 W/m 2 , and J 3 = 1452 W/m 2
Then the net rates of radiation heat transfer at the three surfaces are deter-
mined from Eq. 12-34 to be
fii=A 1 [F 1 _ >2 (/ 1 -J 2 ) + F 1 _ >3 (J, -7 3 )]
= (3.14m 2 )[0.38(ll,418 -4562) + 0.62(11,418 - 1452)] W/m 2
= 27.6 x 10 3 W = 27.6 kW
Qt=A 2 [F 2 ^ l (J 2 -J l )
, (J 2 ~ /a)]
= (3.12 m 2 )[0.38(4562 - 11,418) + 0.62(4562 - 1452)] W/m 2
= -2.13 X 10 3 W = -2.13 kW
e 3 = a,[f 3 ^ i (/ 3 - Ji) + ^3^2 (J 3 - wi
= (6.28 m 2 )[0.31(1452 - 11,418) + 0.31(1452 - 4562)] W/m 2
= -25.5 X 10 3 W = -25.5 kW
Note that the direction of net radiation heat transfer is from the top surface to
the base and side surfaces, and the algebraic sum of these three quantities
must be equal to zero. That is,
Q1+Q2 + Q 3 = 27.6 + (-2.13) + (-25.5) =
Discussion To maintain the surfaces at the specified temperatures, we must
supply heat to the top surface continuously at a rate of 27.6 kW while removing
2.13 kW from the base and 25.5 kW from the side surfaces.
The direct method presented here is straightforward, and it does not require
the evaluation of radiation resistances. Also, it can be applied to enclosures
with any number of surfaces in the same manner.
EXAMPLE 12-9 Radiation Heat Transfer in a Triangular Furnace
A furnace is shaped like a long equilateral triangular duct, as shown in Figure
12-28. The width of each side is 1 m. The base surface has an emissivity of
0.7 and is maintained at a uniform temperature of 600 K. The heated left-side
surface closely approximates a blackbody at 1000 K. The right-side surface is
well insulated. Determine the rate at which heat must be supplied to the heated
side externally per unit length of the duct in order to maintain these operating
conditions.
► ♦ VWW\A
Insulated
e 3 :
FIGURE 12-28
The triangular furnace
considered in Example 12-9.
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HEAT TRANSFER
Solar
energy
V V V V V V Glass cover
e = 0.9
70°F
2 in
Aluminum tube
e = 0.95
Water
FIGURE 12-29
Schematic for Example 12-10.
SOLUTION Two of the surfaces of a long equilateral triangular furnace are
maintained at uniform temperatures while the third surface is insulated. The ex-
ternal rate of heat transfer to the heated side per unit length of the duct during
steady operation is to be determined.
Assumptions 1 Steady operating conditions exist. 2 The surfaces are opaque,
diffuse, and gray. 3 Convection heat transfer is not considered.
Analysis The furnace can be considered to be a three-surface enclosure with a
radiation network as shown in the figure, since the duct is very long and thus
the end effects are negligible. We observe that the view factor from any surface
to any other surface in the enclosure is 0.5 because of symmetry. Surface 3 is
a reradiating surface since the net rate of heat transfer at that surface is zero.
Then we must have Q 1 = -Q 2 > since the entire heat lost by surface 1 must
be gained by surface 2. The radiation network in this case is a simple series-
parallel connection, and we can determine Q 1 directly from
e.
i
where
A!
E b2
A 2 = A 3 = wL = 1 m X 1 m = 1 m 2
^*i3 = ^23 = 0.5 (symmetry)
oT? = (5.67 X 1(T 8 W/m 2 • K 4 )(600 K) 4 =
or, 4 = (5.67 X l(T 8 W/m 2 • K 4 )(1000 K) 4
1/A, F 13 + \IA 2 F 23
(per unit length of the duct)
7348 W/m 2
= 56,700 W/m 2
Substituting,
fit
(56,700 - 7348) W/m 2
1 - 0.7
+
0.7 X 1 m
28.0 X 10 3 = 28.0 kW
(0.5 X 1 m 2 ) +
1/(0.5 X 1 m 2 ) + 1/(0.5 X 1 m 2 )
Therefore, heat at a rate of 28 kW must be supplied to the heated surface per
unit length of the duct to maintain steady operation in the furnace.
EXAMPLE 12-10 Heat Transfer through a Tubular Solar Collector
A solar collector consists of a horizontal aluminum tube having an outer diam-
eter of 2 in. enclosed in a concentric thin glass tube of 4-in. diameter, as shown
in Figure 12-29. Water is heated as it flows through the tube, and the space
between the aluminum and the glass tubes is filled with air at 1 atm pressure.
The pump circulating the water fails during a clear day, and the water tempera-
ture in the tube starts rising. The aluminum tube absorbs solar radiation at a
rate of 30 Btu/h per foot length, and the temperature of the ambient air outside
is 70°F. The emissivities of the tube and the glass cover are 0.95 and 0.9,
respectively. Taking the effective sky temperature to be 50°F, determine the
cen58933_chl2.qxd 9/9/2002 9:49 AM Page 633
temperature of the aluminum tube when steady operating conditions are
established (i.e., when the rate of heat loss from the tube equals the amount of
solar energy gained by the tube).
633
CHAPTER 12
SOLUTION The circulating pump of a solar collector that consists of a hori-
zontal tube and its glass cover fails. The equilibrium temperature of the tube is
to be determined.
Assumptions 1 Steady operating conditions exist. 2 The tube and its cover are
isothermal. 3 Air is an ideal gas. 4 The surfaces are opaque, diffuse, and gray
for infrared radiation. 5 The glass cover is transparent to solar radiation.
Properties The properties of air should be evaluated at the average tempera-
ture. But we do not know the exit temperature of the air in the duct, and thus
we cannot determine the bulk fluid and glass cover temperatures at this point,
and thus we cannot evaluate the average temperatures. Therefore, we will as-
sume the glass temperature to be 110°F, and use properties at an anticipated
average temperature of (70 + 110)/2 = 90°F (Table A-15E),
k= 0.01505 Btu/h -ft- °F
v = 0.6310 ft 2 /h = 1.753 X 10~ 4 ft 2 /s
Pr = 0.7275
1 1
P
Analysis This problem was solved in Chapter 9 by disregarding radiation heat
transfer. Now we will repeat the solution by considering natural convection and
radiation occurring simultaneously.
We have a horizontal cylindrical enclosure filled with air at 1 atm pressure.
The problem involves heat transfer from the aluminum tube to the glass cover
and from the outer surface of the glass cover to the surrounding ambient air.
When steady operation is reached, these two heat transfer rates must equal the
rate of heat gain. That is,
e„
Q
glass-ambient
e s
30 Btu/h
The heat transfer surface area of the glass cover is
A = A glass = (ttD L) = 77(4/12 ft)(l ft) = 1.047 ft 2
(per foot of tube)
(per foot of tube)
To determine the Rayleigh number, we need to know the surface temperature of
the glass, which is not available. Therefore, it is clear that the solution will re-
quire a trial-and-error approach. Assuming the glass cover temperature to be
110°F, the Rayleigh number, the Nusselt number, the convection heat transfer
coefficient, and the rate of natural convection heat transfer from the glass cover
to the ambient air are determined to be
Ran
Nu
g&<T - rj Dl
Pr
(32.2 ft/s 2 )[l/(550 R)](l 10 - 70 R)(4/12 ft) 3
0.6
17.89
(1.753 X 10- 4 ft 2 /s) 2
0.387 Ra{f 12
[1 + (0.559/Pr) 9 " 6 ] 8
0.6 +
(0.7275) = 2.054 X 10 6
0.387(2.054 X 10 6 )" 6
[1 + (0.559/0.7275) 9 " 6 ] 8
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HEAT TRANSFER
/ k ™ 0.01505 Btu/h -fl-T ,,,^ nsm . m . , t2 oc
h„ = tt Nu = TTTTr^ (17.89) = 0.8075 Btu/h • ft 2 • F
° D 4/12 ft v '
fi D ,co„v = ^ A (r - r„) = (0.8075 Btu/h • ft 2 • °F)(1.047 ft 2 )(110 - 70)°F
= 33.8 Btu/h
Also,
Q
o, rad
-'sky)
e vA (T 4
(0.9)(0.1714 X 10- 8 Btu/h -ft 2 ■ R 4 )( 1.047 ft 2 )[(570 R) 4 - (510 R) 4 ]
61.2 Btu/h
Then the total rate of heat loss from the glass cover becomes
Qo, total = G,,co„v + Qe,. A = 33.8 + 61.2 = 95.0 Btu/h
which is much larger than 30 Btu/h. Therefore, the assumed temperature of
110°F for the glass cover is high. Repeating the calculations with lower tem-
peratures (including the evaluation of properties), the glass cover temperature
corresponding to 30 Btu/h is determined to be 78°F (it would be 106°F if radi-
ation were ignored).
The temperature of the aluminum tube is determined in a similar manner
using the natural convection and radiation relations for two horizontal concen-
tric cylinders. The characteristic length in this case is the distance between the
two cylinders, which is
L c = (D a - D;)/2 = (4 - 2)/2 = 1 in. = 1/12 ft
Also,
A = Aube = OA^) = ^(2 /12 ft)(l ft) = 0.5236 ft 2 (per foot of tube)
We start the calculations by assuming the tube temperature to be 122°F, and
thus an average temperature of (78 + 122)/2 = 100°F = 640 R. Using prop-
erties at 100°F,
gfW-TJLl
Ra L = Pr
_ (32.2 ft/s 2 )[l/(640 R)](122 - 78 R)(l/12 ft) 3
(1.809 X 10- 4 ft 2 /s) 2
The effective thermal conductivity is
[ln(ZVD,)] 4
(0.726) = 3.249 X 10 4
cyc L\ (A _3/5 + D~ m f
[ln(4/2)] 4
(1/12 ft) 3 [(2/12 ft)" 3 ' 5 + (4/12 ft)" 3 ' 5 ]
0.1466
& cff = 0.386£
Pr
0.861 + Pr
(F^RaJ" 4
0.386(0.01529 Btu/h • ft • °F)| . "^ „., 1(0.1466 X 3.249 X 10 4 )" 4
0.861 + 0.726
0.04032 Btu/h • ft • °F
cen58933_chl2.qxd 9/9/2002 9:49 AM Page 635
Then the rate of heat transfer between the cylinders by convection becomes
2TT£ cff
Also,
e,„
O = -^— IT - T)
_ 2ir(0.04032 Btu/h ■ ft °F)
~ ln(4/2)
ctA,. (Tf - 7?)
(122 - 78)°F = 16.1 Btu/h
1 1 ~ e P (D,
(0.1714 X 10- 8 Btu/h • ft 2 • R 4 )(0.5236 ft 2 )[(582 R) 4 - (538 R) 4 ]
1 1 - 0.9/2 in.
0.95 0.9 \4 in
25.1 Btu/h
Then the total rate of heat loss from the glass cover becomes
G, total = G,,co„v + e,. rad = 16.1 + 25.1 = 41.1 Btu/h
which is larger than 30 Btu/h. Therefore, the assumed temperature of 122°F for
the tube is high. By trying other values, the tube temperature corresponding
to 30 Btu/h is determined to be 112°F (it would be 180°F if radiation were
ignored). Therefore, the tube will reach an equilibrium temperature of 112°F
when the pump fails.
Discussion It is clear from the results obtained that radiation should always be
considered in systems that are heated or cooled by natural convection, unless
the surfaces involved are polished and thus have very low emissivities.
635
CHAPTER 12
12-5 - RADIATION SHIELDS
AND THE RADIATION EFFECT
Radiation heat transfer between two surfaces can be reduced greatly by in-
serting a thin, high-reflectivity (low-emissivity) sheet of material between the
two surfaces. Such highly reflective thin plates or shells are called radiation
shields. Multilayer radiation shields constructed of about 20 sheets per cm
thickness separated by evacuated space are commonly used in cryogenic and
space applications. Radiation shields are also used in temperature measure-
ments of fluids to reduce the error caused by the radiation effect when the
temperature sensor is exposed to surfaces that are much hotter or colder than
the fluid itself. The role of the radiation shield is to reduce the rate of radiation
heat transfer by placing additional resistances in the path of radiation heat
flow. The lower the emissivity of the shield, the higher the resistance.
Radiation heat transfer between two large parallel plates of emissivi-
ties G[ and e 2 maintained at uniform temperatures T { and T 2 is given by
Eq. 12-38:
cen58933_chl2.qxd 9/9/2002 9:49 AM Page 636
636
HEAT TRANSFER
FIGURE 12-30
The radiation shield placed between
two parallel plates and the radiation
network associated with it.
(l)
-e 12
o-^vwwv^-o-
A X F V
^VWWV^
Shield
£3 2 3
-OAAAAAA/^-CMWWW^-O-
A„F,
(2)
^WWW O — WWW — o
Q
Aa(Tf - r 2 4 )
12, no shield i
e7
1
Now consider a radiation shield placed between these two plates, as shown in
Figure 12-30. Let the emissivities of the shield facing plates 1 and 2 be e 3 j
and e 3 2 , respectively. Note that the emissivity of different surfaces of the
shield may be different. The radiation network of this geometry is constructed,
as usual, by drawing a surface resistance associated with each surface and
connecting these surface resistances with space resistances, as shown in the
figure. The resistances are connected in series, and thus the rate of radiation
heat transfer is
12, one shield i i „
1 - e, i 1 - e 3
H 1
A\ E x A, t n A 3 e 3 [
1 - e 3 2 1 1 — e.
^3 E 3, 2 ^3 ^32 "-2 e 2
ing that F j3 = F 2i = 1 and A, = A 2
= A 3 = A for infinite paralk
(12-42)
Eq. 12-42 simplifies to
*£■ 12, one shield
Av{T? - T*)
1 +
1
1
63 2
(12-43)
where the terms in the second set of parentheses in the denominator represent
the additional resistance to radiation introduced by the shield. The appearance
of the equation above suggests that parallel plates involving multiple radiation
shields can be handled by adding a group of terms like those in the second set
of parentheses to the denominator for each radiation shield. Then the radiation
heat transfer through large parallel plates separated by TV radiation shields
becomes
Q
A<T{Tf - r 2 4 )
12, AT shields
1 +
1 +
1
'-■N.l
(12-44)
cen58933_chl2.qxd 9/9/2002 9:49 AM Page 637
If the emissivities of all surfaces are equal, Eq. 12-44 reduces to
n) i
Gii
Aa(7, 4
■)li + i
N + 1
G,
(12-45)
637
CHAPTER 12
Therefore, when all emissivities are equal, 1 shield reduces the rate of radia-
tion heat transfer to one-half, 9 shields reduce it to one-tenth, and 19 shields
reduce it to one-twentieth (or 5 percent) of what it was when there were no
shields.
The equilibrium temperature of the radiation shield T 3 in Figure 12-30 can
be determined by expressing Eq. 12-43 for <2i3 or Q23 (which involves T 3 )
after evaluating Q 12 from Eq. 12^43 and noting that Q 12 = Q 13 = Q 2 i when
steady conditions are reached.
Radiation shields used to reduce the rate of radiation heat transfer between
concentric cylinders and spheres can be handled in a similar manner. In case
1 for both cases
of one shield, Eq. 12-42 can be used by taking F 13 =
and by replacing the A's by the proper area relations.
23
Radiation Effect on Temperature Measurements
A temperature measuring device indicates the temperature of its sensor, which
is supposed to be, but is not necessarily, the temperature of the medium that
the sensor is in. When a thermometer (or any other temperature measuring de-
vice such as a thermocouple) is placed in a medium, heat transfer takes place
between the sensor of the thermometer and the medium by convection until
the sensor reaches the temperature of the medium. But when the sensor is sur-
rounded by surfaces that are at a different temperature than the fluid, radiation
exchange will take place between the sensor and the surrounding surfaces.
When the heat transfers by convection and radiation balance each other, the
sensor will indicate a temperature that falls between the fluid and surface tem-
peratures. Below we develop a procedure to account for the radiation effect
and to determine the actual fluid temperature.
Consider a thermometer that is used to measure the temperature of a
fluid flowing through a large channel whose walls are at a lower temperature
than the fluid (Fig. 12-31). Equilibrium will be established and the reading of
the thermometer will stabilize when heat gain by convection, as measured
by the sensor, equals heat loss by radiation (or vice versa). That is, on a unit-
area basis,
1 conv, to sensor
H rad, fro
m sensor
h{T f - TJ = e lh a(T t i ~ T<)
or
e,h cr(r,h
Tt)
(K)
(12-46)
FIGURE 12-31
A thermometer used to measure the
temperature of a fluid in a channel.
cen58933_chl2.qxd 9/9/2002 9:49 AM Page 638
638
HEAT TRANSFER
where
Tf = actual temperature of the fluid, K
r ih = temperature value measured by the thermometer, K
T w = temperature of the surrounding surfaces, K
h = convection heat transfer coefficient, W/m 2 ■ K
e = emissivity of the sensor of the thermometer
The last term in Eq. 12-46 is due to the radiation effect and represents the
radiation correction. Note that the radiation correction term is most signifi-
cant when the convection heat transfer coefficient is small and the emissivity
of the surface of the sensor is large. Therefore, the sensor should be coated
with a material of high reflectivity (low emissivity) to reduce the radiation
effect.
Placing the sensor in a radiation shield without interfering with the fluid
flow also reduces the radiation effect. The sensors of temperature measure-
ment devices used outdoors must be protected from direct sunlight since the
radiation effect in that case is sure to reach unacceptable levels.
The radiation effect is also a significant factor in human comfort in heating
and air-conditioning applications. A person who feels fine in a room at a spec-
ified temperature may feel chilly in another room at the same temperature as
a result of the radiation effect if the walls of the second room are at a consid-
erably lower temperature. For example, most people will feel comfortable in
a room at 22°C if the walls of the room are also roughly at that temperature.
When the wall temperature drops to 5°C for some reason, the interior tem-
perature of the room must be raised to at least 27°C to maintain the same level
of comfort. Therefore, well-insulated buildings conserve energy not only by
reducing the heat loss or heat gain, but also by allowing the thermostats to be
set at a lower temperature in winter and at a higher temperature in summer
without compromising the comfort level.
©
©
= 0.2
= 0. 1
e 2 = 0.7
T 2 = 500 K
FIGURE 12-32
Schematic for Example 12-11.
EXAMPLE 12-11
Radiation Shields
A thin aluminum sheet with an emissivity of 0.1 on both sides is placed be-
tween two very large parallel plates that are maintained at uniform temperatures
Ti = 800 K and T 2 = 500 K and have emissivities e 1 = 0.2 and e 2 = 0.7, re-
spectively, as shown in Fig. 12-32. Determine the net rate of radiation heat
transfer between the two plates per unit surface area of the plates and compare
the result to that without the shield.
SOLUTION A thin aluminum sheet is placed between two large parallel plates
maintained at uniform temperatures. The net rates of radiation heat transfer be-
tween the two plates with and without the radiation shield are to be determined.
Assumptions The surfaces are opaque, diffuse, and gray.
Analysis The net rate of radiation heat transfer between these two plates with-
out the shield was determined in Example 12-7 to be 3625 W/m 2 . Heat trans-
fer in the presence of one shield is determined from Eq. 12-43 to be
cen58933_chl2.qxd 9/9/2002 9:49 AM Page 639
639
CHAPTER 12
1 12, one
*L. 12, one shield
shield A
<j(J? - r 2 4 )
(h
s 2 J y E 3, 1 e 3. 2 J
(5.67 X 10" f
W/m 2
K 4 )[(800 K) 4 - (500 K) 4 ]
u+
J--1
WCjl+jl-O
1,0.2
0.7
J \0.1 0.1 )
= 806 W/m 2
Discussion Note that the rate of radiation
heat transfer reduces to about
one-
fourth of what
it was as a result of
placing
a radiation shield between
the
two
parallel plates.
EXAMPLE 12-12 Radiation Effect on Temperature Measurements
A thermocouple used to measure the temperature of hot air flowing in a duct
whose walls are maintained at T w = 400 K shows a temperature reading of
T th = 650 K (Fig. 12-33). Assuming the emissivity of the thermocouple
junction to be e = 0.6 and the convection heat transfer coefficient to be h =
80 W/m 2 • °C, determine the actual temperature of the air.
SOLUTION The temperature of air in a duct is measured. The radiation effect
on the temperature measurement is to be quantified, and the actual air
temperature is to be determined.
Assumptions The surfaces are opaque, diffuse, and gray.
Analysis The walls of the duct are at a considerably lower temperature than
the air in it, and thus we expect the thermocouple to show a reading lower than
the actual air temperature as a result of the radiation effect. The actual air tem-
perature is determined from Eq. 12-46 to be
„ , e th «r(7a ~ T w)
(650 K) +
715 K
h
0.6 X (5.67 X 10" 8 W/m 2 • K 4 )[(650 K) 4 - (400 K) 4 ]
80 W/m 2 ■ °C
Note that the radiation effect causes a difference of 65°C (or 65 K since °C
for temperature differences) in temperature reading in this case.
T th = 650 K
L, • e = 0.6 „
T w = 400 K
FIGURE 12-33
Schematic for Example 12-12.
12-6 - RADIATION EXCHANGE WITH
EMITTING AND ABSORBING GASES
So far we considered radiation heat transfer between surfaces separated by a
medium that does not emit, absorb, or scatter radiation — a nonparticipating
medium that is completely transparent to thermal radiation. A vacuum satis-
fies this condition perfectly, and air at ordinary temperatures and pressures
cen58933_chl2.qxd 9/9/2002 9:49 AM Page 640
640
HEAT TRANSFER
comes very close. Gases that consist of monatomic molecules such as Ar and
He and symmetric diatomic molecules such as N 2 and 2 are essentially trans-
parent to radiation, except at extremely high temperatures at which ionization
occurs. Therefore, atmospheric air can be considered to be a nonparticipating
medium in radiation calculations.
Gases with asymmetric molecules such as H 2 0, C0 2 , CO, S0 2 , and hydro-
carbons H„C,„ may participate in the radiation process by absorption at mod-
erate temperatures, and by absorption and emission at high temperatures such
as those encountered in combustion chambers. Therefore, air or any other
medium that contains such gases with asymmetric molecules at sufficient con-
centrations must be treated as a participating medium in radiation calcula-
tions. Combustion gases in a furnace or a combustion chamber, for example,
contain sufficient amounts of H 2 and C0 2 , and thus the emission and ab-
sorption of gases in furnaces must be taken into consideration.
The presence of a participating medium complicates the radiation analysis
considerably for several reasons:
• A participating medium emits and absorbs radiation throughout its entire
volume. That is, gaseous radiation is a volumetric phenomena, and thus it
depends on the size and shape of the body. This is the case even if the
temperature is uniform throughout the medium.
• Gases emit and absorb radiation at a number of narrow wavelength bands.
This is in contrast to solids, which emit and absorb radiation over the
entire spectrum. Therefore, the gray assumption may not always be
appropriate for a gas even when the surrounding surfaces are gray.
• The emission and absorption characteristics of the constituents of a gas
mixture also depends on the temperature, pressure, and composition of
the gas mixture. Therefore, the presence of other participating gases
affects the radiation characteristics of a particular gas.
The propagation of radiation through a medium can be complicated further
by presence of aerosols such as dust, ice particles, liquid droplets, and soot
(unburned carbon) particles that scatter radiation. Scattering refers to the
change of direction of radiation due to reflection, refraction, and diffraction.
Scattering caused by gas molecules themselves is known as the Rayleigh scat-
tering, and it has negligible effect on heat transfer. Radiation transfer in scat-
tering media is considered in advanced books such as the ones by Modest
(1993, Ref. 12) and Siegel and Howell (1992, Ref. 14).
The participating medium can also be semitransparent liquids or solids such
as water, glass, and plastics. To keep complexities to a manageable level, we
will limit our consideration to gases that emit and absorb radiation. In partic-
ular, we will consider the emission and absorption of radiation by H 2 and
C0 2 only since they are the participating gases most commonly encountered
in practice (combustion products in furnaces and combustion chambers burn-
ing hydrocarbon fuels contain both gases at high concentrations), and they are
sufficient to demonstrate the basic principles involved.
Radiation Properties of a Participating Medium
Consider a participating medium of thickness L. A spectral radiation beam of
intensity I K is incident on the medium, which is attenuated as it propagates
cen58933_chl2.qxd 9/9/2002 9:49 AM Page 641
due to absorption. The decrease in the intensity of radiation as it passes
through a layer of thickness dx is proportional to the intensity itself and the
thickness dx. This is known as Beer's law, and is expressed as (Fig. 12-34)
dl k (x)
-K k I x (x)dx
(12-47)
where the constant of proportionality k x is the spectral absorption coeffi-
cient of the medium whose unit is m~ ' (from the requirement of dimensional
homogeneity). This is just like the amount of interest earned by a bank
account during a time interval being proportional to the amount of money in
the account and the time interval, with the interest rate being the constant of
proportionality.
Separating the variables and integrating from x = to x = L gives
4^
4.0
(12-48)
641
CHAPTER 12
I
FIGURE 12-34
The attenuation of a radiation
beam while passing through an
absorbing medium of thickness L.
where we have assumed the absorptivity of the medium to be independent
of x. Note that radiation intensity decays exponentially in accordance with
Beer's law.
The spectral transmissivity of a medium can be defined as the ratio of the
intensity of radiation leaving the medium to that entering the medium. That is,
»-K k i
(12-49)
J ,\.n
Note that t x = 1 when no radiation is absorbed and thus radiation intensity re-
mains constant. Also, the spectral transmissivity of a medium represents the
fraction of radiation transmitted by the medium at a given wavelength.
Radiation passing through a nonscattering (and thus nonreflecting) medium
is either absorbed or transmitted. Therefore a x + t x = 1, and the spectral ab-
sorptivity of a medium of thickness L is
1
1
(12-50)
From Kirchoff 's law, the spectral emissivity of the medium is
e x = a x = 1 — e~ KkL
(12-51)
Note that the spectral absorptivity, transmissivity, and emissivity of a medium
are dimensionless quantities, with values less than or equal to 1 . The spectral
absorption coefficient of a medium (and thus e x , a k , and i\), in general, vary
with wavelength, temperature, pressure, and composition.
For an optically thick medium (a medium with a large value of k x L), Eq.
12-51 gives e k ~ a x ~ 1. For k x L = 5, for example, e x = a k = 0.993. There-
fore, an optically thick medium emits like a blackbody at the given wave-
length. As a result, an optically thick absorbing-emitting medium with no
significant scattering at a given temperature T g can be viewed as a "black sur-
face" at T g since it will absorb essentially all the radiation passing through it,
and it will emit the maximum possible radiation that can be emitted by a sur-
face at T„ which is E bk (T„).
cen58933_chl2.qxd 9/9/2002 9:49 AM Page 642
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HEAT TRANSFER
Emissivity and Absorptivity
of Gases and Gas Mixtures
The spectral absorptivity of C0 2 is given in Figure 12-35 as a function of
wavelength. The various peaks and dips in the figure together with disconti-
nuities show clearly the band nature of absorption and the strong nongray
characteristics. The shape and the width of these absorption bands vary with
temperature and pressure, but the magnitude of absorptivity also varies with
the thickness of the gas layer. Therefore, absorptivity values without specified
thickness and pressure are meaningless.
The nongray nature of properties should be considered in radiation calcula-
tions for high accuracy. This can be done using a band model, and thus per-
forming calculations for each absorption band. However, satisfactory results
can be obtained by assuming the gas to be gray, and using an effective total
absorptivity and emissivity determined by some averaging process. Charts for
the total emissivities of gases are first presented by Hottel (Ref. 6), and they
have been widely used in radiation calculations with reasonable accuracy.
Alternative emissivity charts and calculation procedures have been developed
more recently by Edwards and Matavosian (Ref. 2). Here we present the
Hottel approach because of its simplicity.
Even with gray assumption, the total emissivity and absorptivity of a gas
depends on the geometry of the gas body as well as the temperature, pressure,
and composition. Gases that participate in radiation exchange such as C0 2 and
H 2 typically coexist with nonparticipating gases such as N 2 and 2 , and thus
radiation properties of an absorbing and emitting gas are usually reported for
a mixture of the gas with nonparticipating gases rather than the pure gas. The
emissivity and absorptivity of a gas component in a mixture depends pri-
marily on its density, which is a function of temperature and partial pressure
of the gas.
The emissivity of H 2 vapor in a mixture of nonparticipating gases is
plotted in Figure 12-36a for a total pressure of P = 1 atm as a function of gas
temperature T g for a range of values for P W L, where P w is the partial pressure
of water vapor and L is the mean distance traveled by the radiation beam.
FIGURE 12-35
Spectral absorptivity of
C0 2 at 830 K and 10 atm
for a path length of 38.8 cm
(from Siegel and Howell, 1992).
Band designation X, u.m
15 4.3
i.o r-y _/
10.4
0.!
S
&• 0.6
-S 0.4
0.2 -
4.8
VL
V
A I
20 10 8 6 5 4 3
Wavelength X, )im
2.5
1.67
cen58933_chl2.qxd 9/9/2002 9:49 AM Page 643
643
CHAPTER 12
0.8
0.6
0.4
0.3
0.2
J 0.1
|J 0.08
I 0.06
w
0.04
0.03
0.02
0.01
0.008
0.006
300
t
- 1 am
i
„L
*f°f
~10
5.
\
~~~>
v
1=,
-0.
4
6
^0.1
^0.06
\
\
v l
).04
\
s
\
s
\
o.c
)1
.02
if
s
■s
\
\
s o.
\
\
x
S
s c
.00
7
\
\
0.005 N
600 900 1200 1500
Gas temperature, 7 „(K)
(a) H 2
1800
2100
0.3
0.2
0.1
0.08
0.06
0.04
;0.03
0.02
0.01
0.008
0.006
0.004
0.003
0.002
0.001
p f
- 4 n ft . at,
c
_2.
I
-1.0
-0.85
-0.4^.
1
-0.
-0.
2-=
~0.06,
—0.04
""0.02,
-0.01
-0.006
~~0.004^
*» 1 T> —
0.003 ^
( 02
300 600 900 1200 1500
Gas temperature, 7 (K)
(b) C0 2
1800
2100
FIGURE 12-36
Emissivities of H 2 and C0 2 gases in a mixture of nonparticipating gases at a total pressure of 1 atm
for a mean beam length of L (1 m • atm = 3.28 ft • atm) (from Hottel, 1954, Ref. 6).
C„. 1.8
1.6
1.4
1.2
1.0
0.8
0.6
0.4
0.2
0.0
0.0
^
1^
0.2
0.25
0.50
1.00
2.50
5.00
10.0
0.4
0.6 0.8
(P w + P)/2 (atm)
(a) H 2
1.0
1.2
C c 2.0
1.5
1.0
0.8
0.6
0.5
0.4
0.3
s'
**
s
To-
0.5"
0.25
0.12
0.05
0-0
= 2.5t'
02
0.05 0.08 0.1
0.2 0.3 0.5 0.81.0
Total pressure, P (atm)
(b) C0 2
2.0 3.0
50
FIGURE 12-37
Correction factors for the emissivities of H 2 and C0 2 gases at pressures other than 1 atm for use in the relations
C u ,£
w, 1 atm
and e c = C c s c , atm (1 m • atm = 3.28 ft • atm) (from Hottel, 1954, Ref. 6).
Emissivity at a total pressure P other than P = 1 atm is determined by multi-
plying the emissivity value at 1 atm by a pressure correction factor C w ob-
tained from Figure 12-37a for water vapor. That is,
cen58933_chl2.qxd 9/9/2002 9:49 AM Page 644
644
HEAT TRANSFER
*— ii.' £ii
(12-52)
Note that C w = 1 for P = 1 aim and thus (P w + P)/2 = 0.5 (a very low con-
centration of water vapor is used in the preparation of the emissivity chart in
Fig. 12-36a and thus P H , is very low). Emissivity values are presented in a
similar manner for a mixture of C0 2 and nonparticipating gases in Fig.
12-366 and 12-376.
Now the question that comes to mind is what will happen if the C0 2 and
H 2 gases exist together in a mixture with nonparticipating gases. The emis-
sivity of each participating gas can still be determined as explained above us-
ing its partial pressure, but the effective emissivity of the mixture cannot be
determined by simply adding the emissivities of individual gases (although
this would be the case if different gases emitted at different wavelengths).
Instead, it should be determined from
s g = b c + b w — Ae
= C c b Ci | atra + C„, e„, , atm - Ae (1 2-53)
where Ae is the emissivity correction factor, which accounts for the overlap
of emission bands. For a gas mixture that contains both C0 2 and H 2 gases,
Ae is plotted in Figure 12-38.
The emissivity of a gas also depends on the mean length an emitted ra-
diation beam travels in the gas before reaching a bounding surface, and thus
the shape and the size of the gas body involved. During their experiments in
the 1930s, Hottel and his coworkers considered the emission of radiation from
a hemispherical gas body to a small surface element located at the center of
the base of the hemisphere. Therefore, the given charts represent emissivity
data for the emission of radiation from a hemispherical gas body of radius L
toward the center of the base of the hemisphere. It is certainly desirable to
extend the reported emissivity data to gas bodies of other geometries, and this
Ae
0.07
0.06
0.05
0.04
0.03
0.02
0.01
T=400K
T
= 800
K
P c L-\
- P W L = 5 ft •
atm
^,^-3.0
'0--4— 2.0.
1.0-^
^-0.75-^/ S
— 0.5-j x
ro2°^
T= 1200 K and above
P r L + P,X = 5 ft ■ atm
P +P ,
P + P ,
P +P ,
0.07
0.06
0.05
0.04
0.03
0.02
0.01
Ae
0.2 0.4 0.6 0.8 1.0 0.2 0.4 0.6 0.8 1.0 0.2 0.4 0.6 0.8 1.0
FIGURE 12-38
Emissivity correction Ae for use in e g = e„, + e r — Ae when both C0 2 and H 2 vapor are present in a gas mixture
(1 m ■ atm = 328 ft • atm) (from Hottel, 1954, Ref. 6).
cen58933_chl2.qxd 9/9/2002 9:49 AM Page 645
is done by introducing the concept of mean beam length L, which represents
the radius of an equivalent hemisphere. The mean beam lengths for various
gas geometries are listed in Table 12-4. More extensive lists are available in
the literature [such as Hottel (1954, Ref. 6), and Siegel and Howell, (1992,
Ref. 14)]. The emissivities associated with these geometries can be deter-
mined from Figures 12-36 through 12-38 by using the appropriate mean
beam length.
Following a procedure recommended by Hottel, the absorptivity of a gas
that contains C0 2 and H 2 gases for radiation emitted by a source at temper-
ature T s can be determined similarly from
a,, + a„
Aa
645
CHAPTER 12
(12-54)
where Aa = Ae and is determined from Figure 12-38 at the source tempera-
ture T s . The absorptivities of C0 2 and H 2 can be determined from the emis-
sivity charts (Figs. 12-36 and 12-37) as
CO,
Q X (TJT,) 0M X s c {T s , P c LT s IT e )
(12-55)
and
H 2 0:
a w = C w X (TJT S ) 0M X s w (T s , P K ,LT S /TJ
(12-56)
The notation indicates that the emissivities should be evaluated using T, in-
stead of T g (both in K or R), P C LT S /T g instead of P C L, and P W LT S /T g instead
of P W L. Note that the absorptivity of the gas depends on the source tempera-
ture T s as well as the gas temperature T g . Also, a = e when T s = T g , as ex-
pected. The pressure correction factors C c and C K are evaluated using P C L and
P w L, as in emissivity calculations.
When the total emissivity of a gas s g at temperature T is known, the emis-
sive power of the gas (radiation emitted by the gas per unit surface area) can
TABLE 12-4
Mean beam length L for various gas volume shapes
Gas Volume Geometry
Hemisphere of radius R radiating to the center of its base R
Sphere of diameter D radiating to its surface 0.65D
Infinite circular cylinder of diameter D radiating to curved surface 0.95D
Semi-infinite circular cylinder of diameter D radiating to its base 0.65D
Semi-infinite circular cylinder of diameter D radiating to center
of its base 0.90D
Infinite semicircular cylinder of radius R radiating to center
of its base 1.26/?
Circular cylinder of height equal to diameter D radiating to
entire surface 0.60D
Circular cylinder of height equal to diameter D radiating to center
of its base 0.7 ID
Infinite slab of thickness D radiating to either bounding plane 1.80D
Cube of side length L radiating to any face 0.66/.
Arbitrary shape of volume l/and surface area A s radiating to surface 3.6 VI A s
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646
HEAT TRANSFER
be expressed as E g = B g uT g . Then the rate of radiation energy emitted by a
gas to a bounding surface of area A s becomes
Qs
s„A,(tT*
(12-57)
If the bounding surface is black at temperature T s , the surface will emit ra-
diation to the gas at a rate of A s a T s 4 without reflecting any, and the gas will ab-
sorb this radiation at a rate of a. A s aT*, where a g is the absorptivity of the gas.
Then the net rate of radiation heat transfer between the gas and a black surface
surrounding it becomes
Black enclosure:
Q net = A sV( e g T g ~ a g T s 4 )
(12-58)
If the surface is not black, the analysis becomes more complicated because
of the radiation reflected by the surface. But for surfaces that are nearly
black with an emissivity s s > 0.7, Hottel (1954, Ref. 6), recommends this
modification,
t^net, £
1
Q
+ l
net, black
■W« ~ a s T ^)
(12-59)
The emissivity of wall surfaces of furnaces and combustion chambers are typ-
ically greater than 0.7, and thus the relation above provides great convenience
for preliminary radiation heat transfer calculations.
ff=5m
FIGURE 12-39
Schematic for Example 12-13.
EXAMPLE 12-13 Effective Emissivity of Combustion Gases
A cylindrical furnace whose height and diameter are 5 m contains combustion
gases at 1200 K and a total pressure of 2 atm. The composition of the com-
bustion gases is determined by volumetric analysis to be 80 percent N 2 , 8 per-
cent H 2 0, 7 percent 2 , and 5 percent C0 2 . Determine the effective emissivity
of the combustion gases (Fig. 12-39).
SOLUTION The temperature, pressure, and composition of a gas mixture is
given. The emissivity of the mixture is to be determined.
Assumptions 1 All the gases in the mixture are ideal gases. 2 The emissivity
determined is the mean emissivity for radiation emitted to all surfaces of the
cylindrical enclosure.
Analysis The volumetric analysis of a gas mixture gives the mole fractions y, of
the components, which are equivalent to pressure fractions for an ideal gas mix-
ture. Therefore, the partial pressures of C0 2 and H 2 are
Pc = yco,P = 0.05(2 atm) = 0.10 atm
Pw = y^oP = 0.08(2 atm) = 0.16 atm
The mean beam length for a cylinder of equal diameter and height for radiation
emitted to all surfaces is, from Table 12-4,
L = 0.60D = 0.60(5 m) = 3 m
cen58933_chl2.qxd 9/9/2002 9:49 AM Page 647
Then,
P C L = (0.10 atm)(3 m) = 0.30 m ■ atm = 0.98 ft • atm
P„.L = (0.16 atm)(3 m) = 0.48 m ■ atm = 1.57 ft • atm
The emissivities of C0 2 and H 2 corresponding to these values at the gas tem-
perature of T g = 1200 K and 1 atm are, from Figure 12-36,
0.16
and
0.23
These are the base emissivity values at 1 atm, and they need to be corrected for
the 2 atm total pressure. Noting that (P„ + P)I2 = (0.16 + 2)/2 = 1.08 atm,
the pressure correction factors are, from Figure 12-37,
C = 1.1
and
C v „
1.4
Both C0 2 and H 2 are present in the same mixture, and we need to correct for
the overlap of emission bands. The emissivity correction factor at T = T g =
1200 K is, from Figure 12-38,
P C L + P„L
P„
0.98 + 1.57
0.16
R, + P r 0.16 + 0.10
Ae = 0.048
Then the effective emissivity of the combustion gases becomes
s g = C c s c , atra + C >v ,e„, , atm - Ae = 1.1 X 0.16 + 1.4 X 0.23 - 0.048 = 0.45
Discussion This is the average emissivity for radiation emitted to all surfaces
of the cylindrical enclosure. For radiation emitted towards the center of the
base, the mean beam length is 0.710 instead of 0.600, and the emissivity
value would be different.
647
CHAPTER 12
EXAMPLE 12-14 Radiation Heat Transfer in a Cylindrical Furnace
Reconsider the cylindrical furnace discussed in Example 12-13. For a wall
temperature of 600 K, determine the absorptivity of the combustion gases and
the rate of radiation heat transfer from the combustion gases to the furnace
walls (Fig. 12-40).
SOLUTION The temperatures for the wall surfaces and the combustion gases
are given for a cylindrical furnace. The absorptivity of the gas mixture and the
rate of radiation heat transfer are to be determined.
Assumptions 1 All the gases in the mixture are ideal gases. 2 All interior
surfaces of furnace walls are black. 3 Scattering by soot and other particles is
negligible.
Analysis The average emissivity of the combustion gases at the gas tempera-
ture of T g = 1200 K was determined in the preceding example to be e g = 0.45.
fl = 5m
FIGURE 1 2-40
Schematic for Example 12-14.
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648
HEAT TRANSFER
For a source temperature of T s = 600 K, the absorptivity of the gas is again
determined using the emissivity charts as
P„L^ = (0.10 atm)(3 m) ^^% = 0.15 m • atm = 0.49 ft • atm
1200 K
L^ = (0.16 atm)(3 m) f,"^ = 0.24 m • atm = 0.79 ft • atm
1200 K.
T
T
The emissivities of C0 2 and H 2 corresponding to these values at a temperature
of T s = 600 K and 1 atm are, from Figure 12-36,
The pressure correction factors were determined in the preceding example to be
C c = l.landC w = 1.4, and they do not change with surface temperature. Then
the absorptivities of C0 2 and H 2 become
, T \0S5
a c = C C [ Y e c
d-i)(^s?#y' 65 (o.ii)=o.i9
a„. = C,
f \0.45
"' T,
(1.4)
600 K
1200 K
600 K
(0.25) = 0.48
Also Aa = As, but the emissivity correction factor is to be evaluated from
Figure 12-38 at T = T s = 600 K instead of T g = 1200 K. There is no chart
for 600 K in the figure, but we can read Ae values at 400 K and 800 K, and
take their average. At P W I(P W + P c ) = 0.615 and P C L + P W L = 2.55 we read
Ae = 0.027. Then the absorptivity of the combustion gases becomes
a g = a c + a w - Aa = 0.19 + 0.48 - 0.027 = 0.64
The surface area of the cylindrical surface is
ttD 2 tt(5 m) 2
A s = ttDH + 2 ^- = tt(5 m)(5 m) + 2 = 118 m 2
Then the net rate of radiation heat transfer from the combustion gases to the
walls of the furnace becomes
2 net = A sV( S g T g ~ a g T ?)
= (118 m 2 )(5.67 X lO" 8 W/m 2 • K 4 )[0.45(1200 K) 4 - 0.64(600 K) 4 ]
= 2.79 X 10 4 W
Discussion The heat transfer rate determined above is for the case of black
wall surfaces. If the surfaces are not black but the surface emissivity e s is
greater than 0.7, the heat transfer rate can be determined by multiplying the
rate of heat transfer already determined by (e s + l)/2.
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TOPIC OF SPECIAL INTEREST
649
CHAPTER 12
Heat Transfer from the Human Body
The metabolic heat generated in the body is dissipated to the environment
through the skin and the lungs by convection and radiation as sensible heat
and by evaporation as latent heat (Fig. 12-41). Latent heat represents the
heat of vaporization of water as it evaporates in the lungs and on the skin
by absorbing body heat, and latent heat is released as the moisture con-
denses on cold surfaces. The warming of the inhaled air represents sensible
heat transfer in the lungs and is proportional to the temperature rise of
inhaled air. The total rate of heat loss from the body can be expressed as
ti body, total \l skin ' \£ lungs
\x£ sensible x£ latent/skin \x£ sensible x£ latent/lungs
\x£ convection Eradiation x£ latent/skin \xi convection xi latent/lungs * ^"OUy
Therefore, the determination of heat transfer from the body by analysis
alone is difficult. Clothing further complicates the heat transfer from the
body, and thus we must rely on experimental data. Under steady condi-
tions, the total rate of heat transfer from the body is equal to the rate of
metabolic heat generation in the body, which varies from about 100 W for
light office work to roughly 1000 W during heavy physical work.
Sensible heat loss from the skin depends on the temperatures of the skin,
the environment, and the surrounding surfaces as well as the air motion.
The latent heat loss, on the other hand, depends on the skin wettedness and
the relative humidity of the environment as well. Clothing serves as insula-
tion and reduces both the sensible and latent forms of heat loss. The heat
transfer from the lungs through respiration obviously depends on the fre-
quency of breathing and the volume of the lungs as well as the environ-
mental factors that affect heat transfer from the skin.
Sensible heat from the clothed skin is first transferred to the clothing and
then from the clothing to the environment. The convection and radiation
heat losses from the outer surface of a clothed body can be expressed as
2c
h A (T
"cony -^clothing \-* clothing
,)
xi rad '^rad j ^clothing\-' clothing J surr/
(W)
(12-61)
(12-62)
Evaporation
30%
Conduction
3%
Floor
FIGURE 12-41
Mechanisms of heat loss
from the human body and relative
magnitudes for a resting person.
where
h com = convection heat transfer coefficient, as given in Table 12-5
h mi = radiation heat transfer coefficient, 4.7 W/m 2 ■ °C for typical indoor
conditions; the emissivity is assumed to be 0.95, which is typical
^clothing = outer surface area of a clothed person
Clothing = average temperature of exposed skin and clothing
^ambient = ambient air temperature
T surr = average temperature of the surrounding surfaces
*This section can be skipped without a loss in continuity.
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TABLE 12-5
Convection heat transfer coefficients
for a clothed body at 1 atm
(V is in m/s) (compiled
from various sources)
Activity
h *
"convi
W/m 2 • °C
Seated in air moving at
< V < 0.2 m/s
0.2 < Y < 4 m/s
Walking in still air at
0.5 < Y < 2 m/s
Walking on treadmill
in still air at
0.5 < y < 2 m/s
Standing in moving air at
< y < 0.15 m/s
0.15 < y < 1.5 m/s
3.1
g 3T 0.6
86 y0.53
5 5T 0.39
4.0
14. 8y - 59
*At pressures other than 1 atm, multiply by
P°- 55 , where Pis in atm.
T
SUIT
V
A J *-rad
U yj ^ conv
\\ \ T
\\ \ ambient
(a) Convection and radiation, separate
(ft) Convection and radiation, combined
FIGURE 1 2-42
Heat loss by convection and radiation
from the body can be combined
into a single term by defining an
equivalent operative temperature.
The convection heat transfer coefficients at 1 atm pressure are given in
Table 12-5. Convection coefficients at pressures P other than 1 atm are
obtained by multiplying the values at atmospheric pressure by p 055 where
P is in atm. Also, it is recognized that the temperatures of different surfaces
surrounding a person are probably different, and T saTT represents the mean
radiation temperature, which is the temperature of an imaginary iso-
thermal enclosure in which radiation heat exchange with the human body
equals the radiation heat exchange with the actual enclosure. Noting that
most clothing and building materials are essentially black, the mean radia-
tion temperature of an enclosure that consists of N surfaces at different
temperatures can be determined from
person- 1
T,
F J,+
* person-2 *■ 1
' r person-iV -* N
(12-63)
where T t is the temperature of the surface i and ^person-/ i s the view factor
between the person and surface i.
Total sensible heat loss can also be expressed conveniently by combin-
ing the convection and radiation heat losses as
2= h A
conv+rad combined clothing
V clothing
ve/
(Ac
^radXVlothing V-* clothing
=)
(W)
(12-64)
(12-65)
where the operative temperature ^operative i s the average of the mean radi-
ant and ambient temperatures weighed by their respective convection and
radiation heat transfer coefficients and is expressed as (Fig. 12-42)
h
conv J ambient
K
Ik
(12-66)
Note that the operative temperature will be the arithmetic average of the
ambient and surrounding surface temperatures when the convection and
radiation heat transfer coefficients are equal to each other. Another en-
vironmental index used in thermal comfort analysis is the effective tem-
perature, which combines the effects of temperature and humidity. Two
environments with the same effective temperature will evoke the same
thermal response in people even though they are at different temperatures
and humidities.
Heat transfer through the clothing can be expressed as
Qc
^clothing \ J skin J clothing/
D
' *flnth i n o
(12-67)
where R
clothing
is the unit thermal resistance of clothing in m 2 • °C/W,
which involves the combined effects of conduction, convection, and ra-
diation between the skin and the outer surface of clothing. The thermal
resistance of clothing is usually expressed in the unit clo where 1 clo =
0.155 m 2 • °C/W = 0.880 ft 2 • °F • h/Btu. The thermal resistance of trousers,
long-sleeve shirt, long-sleeve sweater, and T-shirt is 1.0 clo, or 0.155
m 2 • °C/W. Summer clothing such as light slacks and short-sleeved shirt has
an insulation value of 0.5 clo, whereas winter clothing such as heavy
slacks, long-sleeve shirt, and a sweater or jacket has an insulation value of
0.9 clo.
650
cen58933_chl2.qxd 9/9/2002 9:49 AM Page 651
Then the total sensible heat loss can be expressed in terms of the
skin temperature instead of the inconvenient clothing temperature as
(Fig. 12-43)
Gc
1 clothing V-* skin
(T*
1 operative
(12-68)
v clothing
At a state of thermal comfort, the average skin temperature of the body is
observed to be 33°C (91.5°F). No discomfort is experienced as the skin
temperature fluctuates by ±1.5°C (2.5°F). This is the case whether the
body is clothed or unclothed.
Evaporative or latent heat loss from the skin is proportional to the dif-
ference between the water vapor pressure at the skin and the ambient air,
and the skin wettedness, which is a measure of the amount of moisture on
the skin. It is due to the combined effects of the evaporation of sweat and
the diffusion of water through the skin, and can be expressed as
651
CHAPTER 12
FIGURE 1 2-43
Simplified thermal resistance
network for heat transfer
from a clothed person.
fii,
* vapor 'Vg
(12-69)
where
the rate of evaporation from the body, kg/s
h f = the enthalpy of vaporization of water = 2430 kJ/kg at 30°C
Heat loss by evaporation is maximum when the skin is completely wetted.
Also, clothing offers resistance to evaporation, and the rate of evaporation
in clothed bodies depends on the moisture permeability of the clothes. The
maximum evaporation rate for an average man is about 1 L/h (0.3 g/s),
which represents an upper limit of 730 W for the evaporative cooling rate.
A person can lose as much as 2 kg of water per hour during a workout on a
hot day, but any excess sweat slides off the skin surface without evaporat-
ing (Fig. 12-44).
During respiration, the inhaled air enters at ambient conditions and ex-
haled air leaves nearly saturated at a temperature close to the deep body
temperature (Fig. 12-45). Therefore, the body loses both sensible heat by
convection and latent heat by evaporation from the lungs, and these can be
expressed as
2c
Q
latent, lungs
'air, lungs p, airV-* exhale J ambit
'vapor, lungs fg '"air, lungs \"e;
,t)
e*t)hfi
(12-70)
(12-71)
where
air, lungs
c,
rate of air intake to the lungs, kg/s
p. air
specific heat of air = 1 .0 kJ/kg • °C
Exhale = temperature of exhaled air
w = humidity ratio (the mass of moisture per unit mass of dry air)
m = 0.3 g/s
vapor, max °
2|atent, max ~~ m latent, max "fg <8> 30°C
= (0.3 g/s)(2430 kJ/kg)
= 730W
FIGURE 1 2-44
An average person can lose heat at
a rate of up to 730 W by evaporation.
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HEAT TRANSFER
Warm and moist
exhaled air
35°C
37°C
FIGURE 1 2-45
Part of the metabolic heat generated in
the body is rejected to the air from
the lungs during respiration.
FIGURE 1 2-46
Schematic for Example 12-15.
The rate of air intake to the lungs is directly proportional to the metabolic
rate <2 met . The rate of total heat loss from the lungs through respiration can
be expressed approximately as
2c
0.0014g met (34 - r arablent ) + 0.0173e mct (5.87 - P v
mbient/
(12-72)
where P v ambieilt is the vapor pressure of ambient air in kPa.
The fraction of sensible heat varies from about 40 percent in the case of
heavy work to about 70 percent during light work. The rest of the energy is
rejected from the body by perspiration in the form of latent heat.
EXAMPLE 12-15 Effect of Clothing on Thermal Comfort
It is well established that a clothed or unclothed person feels comfortable when
the skin temperature is about 33°C. Consider an average man wearing summer
clothes whose thermal resistance is 0.6 do. The man feels very comfortable
while standing in a room maintained at 22°C. The air motion in the room is neg-
ligible, and the interior surface temperature of the room is about the same as
the air temperature. If this man were to stand in that room unclothed, deter-
mine the temperature at which the room must be maintained for him to feel
thermally comfortable.
SOLUTION A man wearing summer clothes feels comfortable in a room at
22 C C. The room temperature at which this man would feel thermally comfort-
able when unclothed is to be determined.
Assumptions 1 Steady conditions exist. 2 The latent heat loss from the person
remains the same. 3 The heat transfer coefficients remain the same.
Analysis The body loses heat in sensible and latent forms, and the sensible
heat consists of convection and radiation heat transfer. At low air velocities, the
convection heat transfer coefficient for a standing man is given in Table 12-5
to be 4.0 W/m 2 • °C. The radiation heat transfer coefficient at typical indoor
conditions is 4.7 W/m 2 • °C. Therefore, the surface heat transfer coefficient for
a standing person for combined convection and radiation is
K,
4.0 + 4.7 = 8.7 W/m 2 ■ °C
The thermal resistance of the clothing is given to be
^ clothing = 0.6 clo = 0.6 X 0.155 m 2 • °C/W = 0.093 m 2 • °C/W
Noting that the surface area of an average man is 1.8 m 2 , the sensible heat loss
from this person when clothed is determined to be (Fig. 12-46)
a— sensible, clothed
A S (T,
skin -* ambient
(1.8m 2 )(33 - 22)°C
R.
clothing ' fa
95.2 W
0.093 m 2 • °C/W
1
,1 W/m 2 • °C
From a heat transfer point of view, taking the clothes off is equivalent to re-
moving the clothing insulation or setting ff C | 0th = 0. The heat transfer in this
case can be expressed as
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653
CHAPTER 12
k£ sensible, uncloti
^jUskin -<ambient) (1.8m)(33 r ambicnt )C
1
1
K,
1.7 W/m 2 ■ °C
To maintain thermal comfort after taking the clothes off, the skin temperature
of the person and the rate of heat transfer from him must remain the same.
Then setting the equation above equal to 95.2 W gives
-'ambient = 26.9 L
Therefore, the air temperature needs to be raised from 22 to 26.9 C C to ensure
that the person will feel comfortable in the room after he takes his clothes off
(Fig. 12-47). Note that the effect of clothing on latent heat is assumed to be
negligible in the solution above. We also assumed the surface area of the
clothed and unclothed person to be the same for simplicity, and these two
effects should counteract each other.
/ Unclothed
/ person
FIGURE 1 2-47
Clothing serves as insulation,
and the room temperature needs to be
raised when a person is unclothed to
maintain the same comfort level.
SUMMARY
Radiaton heat transfer between surfaces depends on the ori-
entation of the surfaces relative to each other, the effects of
orientation are accounted for by the geometric parameter view
factor. The view factor from a surface i to a surface / is denoted
by F t _> j or F«, and is defined as the fraction of the radiation
leaving surface i that strikes surface / directly. The view factors
between differential and finite surfaces are expressed as differ-
ential view factor dF M] _> dA ^ is expressed as
dF,
Qa
£?,,
dA 2
F n = F A
L
cos 6 1 cos 0t
dAi
Q,
cos G, cos 0-,
■dA, dA-,
unity. This is known as the summation rule for an enclosure.
The superposition rule is expressed as the view factor from a
surface i to a surface j is equal to the sum of the view factors
from surface i to the parts of surface / The symmetry rule is
expressed as if the surfaces j and k are symmetric about the
surface ;' then F, _> j = F t _> k .
The rate of net radiation heat transfer between two black sur-
faces is determined from
G,
A,F^Mn
r 2 4 )
(W)
The net radiation heat transfer from any surface i of a black
enclosure is determined by adding up the net radiation heat
transfers from surface i to each of the surfaces of the enclosure:
e^EeMrS^a^
Tf)
(W)
;=i
;=i
where r is the distance between dA { and dA 2 , and 0! and G 2 are
the angles between the normals of the surfaces and the line that
connects dA t and dA 2 .
The view factor F, ^ , represents the fraction of the radiation
leaving surface i that strikes itself directly; F ; _, ,■ = for plane
or convex surfaces and F i _> ; # for concave surfaces. For
view factors, the reciprocity rule is expressed as
At F>
AjFj-
The total radiation energy leaving a surface per unit time and
per unit area is called the radiosity and is denoted by J. The
net rate of radiation heat transfer from a surface i of surface
area A, is expressed as
Qi
where
J,
(W)
The sum of the view factors from surface i of an enclosure to
all surfaces of the enclosure, including to itself, must equal
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654
HEAT TRANSFER
is the surface resistance to radiation. The net rate of radiation
heat transfer from surface i to surface j can be expressed as
The radiation effect in temperature measurements can be prop-
erly accounted for by the relation
Qi-
(W)
h
(K)
where
A,F,
is the space resistance to radiation. The network method is
applied to radiation enclosure problems by drawing a surface
resistance associated with each surface of an enclosure and
connecting them with space resistances. Then the problem is
solved by treating it as an electrical network problem where the
radiation heat transfer replaces the current and the radiosity
replaces the potential. The direct method is based on the
following two equations:
Surfaces with specified
net heat transfer rate g,
Surfaces with specified
temperature T t
Qi=A i ^F i ^ J (J i -J J )
1 " e - ^
The first group (for surfaces with specified heat transfer rates)
and the second group (for surfaces with specified tempera-
tures) of equations give N linear algebraic equations for the de-
termination of the N unknown radiosities for an JV-surface
enclosure. Once the radiosities J,, J 2 , . . . , J N are available, the
unknown surface temperatures and heat transfer rates can be
determined from the equations just shown.
The net rate of radiation transfer between any two gray, dif-
fuse, opaque surfaces that form an enclosure is given by
Q^
v(T? - T$)
1 -
" El
+
A
1
F 12
+
1 -
A 2
e 2
A,
Si
e 2
(W)
Radiation heat transfer between two surfaces can be reduced
greatly by inserting between the two surfaces thin, high-
reflectivity (low-emissivity) sheets of material called ra-
diation shields. Radiation heat transfer between two large
parallel plates separated by N radiation shields is
Q
Aa(Tt ~ T 2 4 )
12, A< shields
h + l
1 +
^ + ^
where T f is the actual temperature of the fluid, T th is the tem-
perature value measured by the thermometer, and T w is the
temperature of the surrounding walls, all in K.
Gases with asymmetric molecules such as H 2 0, C0 2 CO,
S0 2 , and hydrocarbons H„C„, participate in the radiation pro-
cess by absorption and emission. The spectral transmissivity,
absorptivity, and emissivity of a medium are expressed as
OLs = 1
1
1
and
where k x is the spectral absorption coefficient of the medium.
The emissivities of H 2 and C0 2 gases are given in Figure
12-36 for a total pressure of P = 1 atm. Emissivities at other
pressures are determined from
and
S c *--c E c, 1 atm
where C„. and C c are the pressure correction factors. For gas
mixtures that contain both of Ff 2 and C0 2 , the emissivity is
determined from
e g = s c + 8„. — As = C c e^ , alm + C w e w [ alm — Ae
where Ae is the emissivity correction factor, which accounts
for the overlap of emission bands. The gas absorptivities for ra-
diation emitted by a source at temperature T s axe determined
similarly from
a„ = a,. + a„
Aa
where Aa = Ae at the source temperature T s and
CO.:
a c = C c X (T g /T s ) ' 65 X e c (T s , P c LT s IT g )
a„, = C w X (T g /T s f M X e„,(7; P w LT s IT g )
The rate of radiation heat transfer between a gas and a sur-
rounding surface is
Black enclosure:
Gray enclosure,
with e s > 0.7: g net .
2 „et =A s cr(G g T* - a g T s 4 )
s, + 1
■A^eJ. 4 -aJ, 4 )
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REFERENCES AND SUGGESTED READING
655
CHAPTER 12
1. D. K. Edwards. Radiation Heat Transfer Notes.
Washington, D.C.: Hemisphere, 1981.
2. D. K. Edwards and R. Matavosian. "Scaling Rules for
Total Absorptivity and Emissivity of Gases." Journal
of Heat Transfer 106 (1984), pp. 684-689.
3. D. K. Edwards and R. Matavosian. "Emissivity Data for
Gases." Section 5.5.5, in Hemisphere Handbook of
Heat Exchanger Design, ed. G. F. Hewitt. New York:
Hemisphere, 1990.
4. D. C. Hamilton and W. R. Morgan. "Radiation
Interchange Configuration Factors." National Advisory
Committee for Aeronautics, Technical Note 2836, 1952.
5. J. P. Holman. Heat Transfer. 9th ed. New York:
McGraw-Hill, 2002.
6. H. C. Hottel. "Radiant Heat Transmission." In Heat
Transmission, ed. W. H. McAdams. 3rd ed. New York:
McGraw-Hill, 1954.
7. H. C. Hottel. "Heat Transmission by Radiation from
Non-luminous Gases," Transaction of the AIChE (1927),
pp. 173-205.
8. H. C. Hottel and R. B. Egbert. "Radiant Heat
Transmission from Water Vapor." Transactions of
the AIChE 38 (1942), pp. 531-565.
9. J. R. Howell. A Catalog of Radiation Configuration
Factors. New York: McGraw-Hill, 1982.
10. F. P. Incropera and D. P. DeWitt. Introduction to Heat
Transfer. 4th ed. New York: John Wiley & Sons, 2002.
11. F. Kreith and M. S. Bohn. Principles of Heat Transfer.
6th ed. Pacific Grove, CA: Brooks/Cole, 2001.
12. M. F Modest. Radiative Heat Transfer. New York:
McGraw-Hill, 1993.
13. A. K. Oppenheim. "Radiation Analysis by the Network
Method." Transactions oftheASME18 (1956),
pp. 725-735.
14. R. Siegel and J. R. Howell. The rmal Radiation Heat
Transfer. 3rd ed. Washington, D.C.: Hemisphere, 1992.
15. N. V. Suryanaraya. Engineering Heat Transfer. St. Paul,
Minn.: West, 1995.
PROBLEMS
The View Factor
12-1C What does the view factor represent? When is the
view factor from a surface to itself not zero?
12-2C How can you determine the view factor F n when the
view factor F 2I and the surface areas are available?
12-3C What are the summation rule and the superposition
rule for view factors?
12— 4C What is the crossed-strings method? For what kind of
geometries is the crossed-strings method applicable?
12-5 Consider an enclosure consisting of six surfaces. How
many view factors does this geometry involve? How many
of these view factors can be determined by the application of
the reciprocity and the summation rules?
*Problems designated by a "C" are concept questions, and
students are encouraged to answer them all. Problems designated
by an "E" are in English units, and the SI users can ignore them.
Problems with an EES-CD icon ® are solved using EES, and
complete solutions together with parametric studies are included
on the enclosed CD. Problems with a computer-EES icon H are
comprehensive in nature, and are intended to be solved with a
computer, preferably using the EES software that accompanies
this text.
12-6 Consider an enclosure consisting of five surfaces. How
many view factors does this geometry involve? How many
of these view factors can be determined by the application of
the reciprocity and summation rules?
12-7 Consider an enclosure consisting of 12 surfaces. How
many view factors does this geometry involve? How many of
these view factors can be determined by the application of the
reciprocity and the summation rules? Answers: 144, 78
12-8 Determine the view factors F, 3 and F 23 between the
rectangular surfaces shown in Figure P12-8.
« 2 m ••
A 2
f
lm
1
^1
\
f
lm
\,l
\ *
\lm
\ \
FIGURE P1 2-8
12-9 Consider a cylindrical enclosure whose height is twice
the diameter of its base. Determine the view factor from the
side surface of this cylindrical enclosure to its base surface.
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HEAT TRANSFER
FIGURE P1 2-11
(a) Semicylindrical duct.
(b) Triangular duct,
(c) Rectangular duct, (a)
(b)
(c)
(a)
FIGURE P1 2-1 2
(a) Semicylindrical groove
(b) Triangular groove.
(c) Rectangular groove.
12-10 Consider a hemispherical furnace with a flat circular
base of diameter D. Determine the view factor from the dome
of this furnace to its base. Answer: 0.5
12-11 Determine the view factors F l2 and F 2l for the very
long ducts shown in Figure P12-11 without using any view
factor tables or charts. Neglect end effects.
12-12 Determine the view factors from the very long
grooves shown in Figure PI 2-1 2 to the surroundings without
using any view factor tables or charts. Neglect end effects.
12-13 Determine the view factors from the base of a cube to
each of the other five surfaces.
12-14 Consider a conical enclosure of height h and base di-
ameter D. Determine the view factor from the conical side sur-
face to a hole of diameter d located at the center of the base.
12-15 Determine the four view factors associated with an en-
closure formed by two very long concentric cylinders of radii
r x and r 2 . Neglect the end effects.
12-16 Determine the view factor F l2 between the rectangular
surfaces shown in Figure P12-16.
12-17 Two infinitely long parallel cylinders of diameter D
are located a distance 5 apart from each other. Determine the
view factor F n between these two cylinders.
12-18 Three infinitely long parallel cylinders of diameter D
are located a distance 5 apart from each other. Determine the
view factor between the cylinder in the middle and the sur-
roundings.
FIGURE P1 2-14
FIGURE P1 2-1 8
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657
CHAPTER 12
2 in
FIGURE P1 2-1 6
Radiation Heat Transfer between Surfaces
12-19C Why is the radiation analysis of enclosures that con-
sist of black surfaces relatively easy? How is the rate of radia-
tion heat transfer between two surfaces expressed in this case?
12-20C How does radiosity for a surface differ from the
emitted energy? For what kind of surfaces are these two quan-
tities identical?
12-21C What are the radiation surface and space resis-
tances? How are they expressed? For what kind of surfaces is
the radiation surface resistance zero?
12-22C What are the two methods used in radiation analy-
sis? How do these two methods differ?
12-23C What is a reradiating surface? What simplifications
does a reradiating surface offer in the radiation analysis?
12-24E Consider a 10-ft X 10-ft X 10-ft cubical furnace
whose top and side surfaces closely approximate black sur-
faces and whose base surface has an emissivity e = 0.7. The
base, top, and side surfaces of the furnace are maintained at
uniform temperatures of 800 R, 1600 R, and 2400 R, respec-
tively. Determine the net rate of radiation heat transfer between
(a) the base and the side surfaces and (b) the base and the top
surfaces. Also, determine the net rate of radiation heat transfer
to the base surface.
12-25E JEa| Reconsider Problem 12-24E. Using EES (or
k^S other) software, investigate the effect of base
surface emissivity on the net rates of radiation heat transfer be-
tween the base and the side surfaces, between the base and top
surfaces, and to the base surface. Let the emissivity vary from
0.1 to 0.9. Plot the rates of heat transfer as a function of emis-
sivity, and discuss the results.
12-26 Two very large parallel plates are maintained at uni-
form temperatures of T t = 600 K and T 2 = 400 K and have
emissivities e, = 0.5 and e 2 = 0.9, respectively. Determine the
net rate of radiation heat transfer between the two surfaces per
unit area of the plates.
12-27 rra| Reconsider Problem 12-26. Using EES (or
|^i£ other) software, investigate the effects of the
temperature and the emissivity of the hot plate on the net rate
of radiation heat transfer between the plates. Let the tempera-
ture vary from 500 K to 1000 K and the emissivity from 0.1 to
0.9. Plot the net rate of radiation heat transfer as functions of
temperature and emissivity, and discuss the results.
12-28 A furnace is of cylindrical shape with R = H = 2 m.
The base, top, and side surfaces of the furnace are all black and
are maintained at uniform temperatures of 500, 700, and 1200
K, respectively. Determine the net rate of radiation heat trans-
fer to or from the top surface during steady operation.
FIGURE P1 2-28
12-29 Consider a hemispherical furnace of diameter D = 5
m with a flat base. The dome of the furnace is black, and the
base has an emissivity of 0.7. The base and the dome of the fur-
nace are maintained at uniform temperatures of 400 and 1000
K, respectively. Determine the net rate of radiation heat trans-
fer from the dome to the base surface during steady operation.
Answer: 759 kW
Black
■ 5 m-
FIGURE P1 2-29
12-30 Two very long concentric cylinders of diameters D| =
0.2 m and D 2 = 0.5 m are maintained at uniform temperatures
of T\ = 950 K and T 2 = 500 K and have emissivities e t = 1
and e 2 = 0.7, respectively. Determine the net rate of radiation
heat transfer between the two cylinders per unit length of the
cylinders.
12-31 This experiment is conducted to determine the emis-
sivity of a certain material. A long cylindrical rod of diameter
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D| = 0.01 m is coated with this new material and is placed in
an evacuated long cylindrical enclosure of diameter D 2 =
0.1 m and emissivity e 2 = 0.95, which is cooled externally and
maintained at a temperature of 200 K at all times. The rod is
heated by passing electric current through it. When steady op-
erating conditions are reached, it is observed that the rod is dis-
sipating electric power at a rate of 8 W per unit of its length
and its surface temperature is 500 K. Based on these measure-
ments, determine the emissivity of the coating on the rod.
12-32E A furnace is shaped like a long semicylindrical duct
of diameter D = 15 ft. The base and the dome of the furnace
have emissivities of 0.5 and 0.9 and are maintained at uniform
temperatures of 550 and 1 800 R, respectively. Determine the
net rate of radiation heat transfer from the dome to the base
surface per unit length during steady operation.
\~D= 15 ft -|
FIGURE P12-32E
12-33 Two parallel disks of diameter D = 0.6 m separated
by L = 0.4 m are located directly on top of each other. Both
disks are black and are maintained at a temperature of 700 K.
The back sides of the disks are insulated, and the environment
that the disks are in can be considered to be a blackbody at
T„ = 300 K. Determine the net rate of radiation heat transfer
from the disks to the environment. Answer: 5505 W
12-34 A furnace is shaped like a long equilateral-triangular
duct where the width of each side is 2 m. Heat is supplied from
the base surface, whose emissivity is s t = 0.8, at a rate of
800 W/m 2 while the side surfaces, whose emissivities are 0.5,
are maintained at 500 K. Neglecting the end effects, determine
the temperature of the base surface. Can you treat this geome-
try as a two-surface enclosure?
12-35 Tu'M Reconsider Problem 12-34. Using EES (or
b^ti other) software, investigate the effects of the
rate of the heat transfer at the base surface and the temperature
of the side surfaces on the temperature of the base surface. Let
the rate of heat transfer vary from 500 W/m 2 to 1000 W/m 2 and
the temperature from 300 K to 700 K. Plot the temperature of
the base surface as functions of the rate of heat transfer and the
temperature of the side surfaces, and discuss the results.
12-36 Consider a 4-m X 4-m X 4-m cubical furnace whose
floor and ceiling are black and whose side surfaces are reradi-
ating. The floor and the ceiling of the furnace are maintained at
temperatures of 550 K and 1100 K, respectively. Determine
the net rate of radiation heat transfer between the floor and the
ceiling of the furnace.
12-37 Two concentric spheres of diameters D t = 0.3 m
and D 2 = 0.8 m are maintained at uniform temperatures
T\ = 700 K and T 2 = 400 K and have emissivities s t = 0.5 and
e 2 = 0.7, respectively. Determine the net rate of radiation heat
transfer between the two spheres. Also, determine the convec-
tion heat transfer coefficient at the outer surface if both the sur-
rounding medium and the surrounding surfaces are at 30°C.
Assume the emissivity of the outer surface is 0.35.
12-38 A spherical tank of diameter D = 2 m that is filled
with liquid nitrogen at 100 K is kept in an evacuated cubic en-
closure whose sides are 3 m long. The emissivities of the
spherical tank and the enclosure are s l =0.1 and e 2
().:
respectively. If the temperature of the cubic enclosure is mea-
sured to be 240 K, determine the net rate of radiation heat
transfer to the liquid nitrogen. Answer: 228 W
FIGURE P1 2-38
12-39 Repeat Problem 12-38 by replacing the cubic enclo-
sure by a spherical enclosure whose diameter is 3 m.
12-40 r^| Reconsider Problem 12-38. Using EES (or
kS other) software, investigate the effects of the
side length and the emissivity of the cubic enclosure, and the
emissivity of the spherical tank on the net rate of radiation heat
transfer. Let the side length vary from 2.5 m to 5.0 m and both
emissivities from 0.1 to 0.9. Plot the net rate of radiation heat
transfer as functions of side length and emissivities, and dis-
cuss the results.
12-41 Consider a circular grill whose diameter is 0.3 m. The
bottom of the grill is covered with hot coal bricks at 1100 K,
while the wire mesh on top of the grill is covered with steaks
initially at 5°C. The distance between the coal bricks and the
steaks is 0.20 m. Treating both the steaks and the coal bricks as
blackbodies, determine the initial rate of radiation heat transfer
from the coal bricks to the steaks. Also, determine the initial
rate of radiation heat transfer to the steaks if the side opening
of the grill is covered by aluminum foil, which can be approx-
imated as a reradiating surface. Answers: 1674 W, 3757 W
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CHAPTER 12
Steaks
0.20 m
FIGURE P1 2-41
12-42E A 19-ft-high room with a base area of 12 ft X 12 ft
is to be heated by electric resistance heaters placed on the ceil-
ing, which is maintained at a uniform temperature of 90°F at
all times. The floor of the room is at 65°F and has an emissiv-
ity of 0.8. The side surfaces are well insulated. Treating the
ceiling as a blackbody, determine the rate of heat loss from the
room through the floor.
12-43 Consider two rectangular surfaces perpendicular to
each other with a common edge which is 1 .6 m long. The hor-
izontal surface is 0.8 m wide and the vertical surface is 1.2 m
high. The horizontal surface has an emissivity of 0.75 and is
maintained at 400 K. The vertical surface is black and is main-
tained at 550 K. The back sides of the surfaces are insulated.
The surrounding surfaces are at 290 K, and can be considered
to have an emissivity of 0.85. Determine the net rate of radia-
tion heat transfers between the two surfaces, and between the
horizontal surface and the surroundings.
T 2 = 550 K
82 = 1
W=l.6m
L 7 = 1.2 m
A 2 ©
©
r 3 = 290 K
e 3 = 0.85
r, = 400 K
e, =0.75
300 K
FIGURE P1 2^4
12-45 Consider a long semicylindrical duct of diameter
1 .0 m. Heat is supplied from the base surface, which is black,
at a rate of 1200 W/m 2 , while the side surface with an emissiv-
ity of 0.4 are is maintained at 650 K. Neglecting the end ef-
fects, determine the temperature of the base surface.
12-46 Consider a 20-cm-diameter hemispherical enclosure.
The dome is maintained at 600 K and heat is supplied from the
dome at a rate of 50 W while the base surface with an emissiv-
ity is 0.55 is maintained at 400 K. Determine the emissivity of
the dome.
Radiation Shields and the Radiation Effect
12-47C What is a radiation shield? Why is it used?
12-48C What is the radiation effect? How does it influence
the temperature measurements?
12-49C Give examples of radiation effects that affect human
comfort.
12-50 Consider a person whose exposed surface area is
1.7 m 2 , emissivity is 0.85, and surface temperature is 30°C.
Determine the rate of heat loss from that person by radiation in
a large room whose walls are at a temperature of (a) 300 K and
(b) 280 K.
12-51 A thin aluminum sheet with an emissivity of 0.15 on
both sides is placed between two very large parallel plates,
which are maintained at uniform temperatures T t = 900 K and
650 K and have emissivities e,
0.5 and e,
().:
FIGURE P12-43
respectively. Determine the net rate of radiation heat transfer
between the two plates per unit surface area of the plates and
compare the result with that without the shield.
12-44 Two long parallel 16-cm -diameter cylinders are lo-
cated 50 cm apart from each other. Both cylinders are black,
and are maintained at temperatures 425 K and 275 K. The
surroundings can be treated as a blackbody at 300 K. For a
1-m-long section of the cylinders, determine the rates of radia-
tion heat transfer between the cylinders and between the hot
cylinder and the surroundings.
T, = 900 K
e ;=o. 5
= 0.15
© '2
&, = 0.8
650 K
FIGURE P1 2-51
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HEAT TRANSFER
12-52 [?(,■>! Reconsider Problem 12-51. Using EES (or
|^£^ other) software, plot the net rate of radiation
heat transfer between the two plates as a function of the emis-
sivity of the aluminum sheet as the emissivity varies from 0.05
to 0.25, and discuss the results.
12-53 Two very large parallel plates are maintained at uni-
form temperatures of T { = 1000 K and T 2 = 800 K and have
emissivities of s t = e 2 = 0.2, respectively. It is desired to
reduce the net rate of radiation heat transfer between the two
plates to one-fifth by placing thin aluminum sheets with an
emissivity of 0.15 on both sides between the plates. Determine
the number of sheets that need to be inserted.
12-54 Five identical thin aluminum sheets with emissivities
of 0. 1 on both sides are placed between two very large parallel
plates, which are maintained at uniform temperatures of T { =
800 K and T 2 = 450 K and have emissivities of s t = e 2 = 0.1,
respectively. Determine the net rate of radiation heat transfer
between the two plates per unit surface area of the plates and
compare the result to that without the shield.
12-55 Tu'M Reconsider Problem 12-54. Using EES (or
k^S other) software, investigate the effects of the
number of the aluminum sheets and the emissivities of the
plates on the net rate of radiation heat transfer between the two
plates. Let the number of sheets vary from 1 to 10 and the
emissivities of the plates from 0.1 to 0.9. Plot the rate of ra-
diation heat transfer as functions of the number of sheets and
the emissivities of the plates, and discuss the results.
12-56E Two parallel disks of diameter D = 3 f t separated by
L = 2 f t are located directly on top of each other. The disks are
separated by a radiation shield whose emissivity is 0.15. Both
disks are black and are maintained at temperatures of 1200 R
and 700 R, respectively. The environment that the disks are in
can be considered to be a blackbody at 540 R. Determine the
net rate of radiation heat transfer through the shield under
steady conditions. Answer: 866 Btu/h
FIGURE P12-56E
12-57 A radiation shield that has the same emissivity e 3 on
both sides is placed between two large parallel plates, which
are maintained at uniform temperatures of T t = 650 K and
T 2 = 400 K and have emissivities of s { = 0.6 and s 2 = 0.9,
respectively. Determine the emissivity of the radiation shield if
the radiation heat transfer between the plates is to be reduced to
15 percent of that without the radiation shield.
12-58 rfi£M Reconsider Problem 12-57. Using EES (or
)SZ2 other) software, investigate the effect of the per-
cent reduction in the net rate of radiation heat transfer between
the plates on the emissivity of the radiation shields. Let the per-
cent reduction vary from 40 to 95 percent. Plot the emissivity
versus the percent reduction in heat transfer, and discuss the
results.
12-59 Two coaxial cylinders of diameters D l = 0.10 m and
D 2 = 0.30 m and emissivities ej = 0.7 and e 2 = 0.4 are main-
tained at uniform temperatures of T { = 750 K and T 2 = 500 K,
respectively. Now a coaxial radiation shield of diameter D 3 =
0.20 m and emissivity s 3 = 0.2 is placed between the two
cylinders. Determine the net rate of radiation heat transfer be-
tween the two cylinders per unit length of the cylinders and
compare the result with that without the shield.
12-60 r^| Reconsider Problem 12-59. Using EES (or
kS other) software, investigate the effects of the di-
ameter of the outer cylinder and the emissivity of the radiation
shield on the net rate of radiation heat transfer between the two
cylinders. Let the diameter vary from 0.25 m to 0.50 m and the
emissivity from 0.05 to 0.35. Plot the rate of radiation heat
transfer as functions of the diameter and the emissivity, and
discuss the results.
Radiation Exchange with Absorbing and Emitting Gases
12-61 C How does radiation transfer through a participating
medium differ from that through a nonparticipating medium?
12-62C Define spectral transmissivity of a medium of thick-
ness L in terms of (a) spectral intensities and (b) the spectral
absorption coefficient.
12-63C Define spectral emissivity of a medium of thickness
L in terms of the spectral absorption coefficient.
12-64C How does the wavelength distribution of radiation
emitted by a gas differ from that of a surface at the same tem-
perature?
12-65 Consider an equimolar mixture of C0 2 and 2 gases at
500 K and a total pressure of 0.5 atm. For a path length of
1 .2 m, determine the emissivity of the gas.
12-66 A cubic furnace whose side length is 6 m contains
combustion gases at 1000 K and a total pressure of 1 atm. The
composition of the combustion gases is 75 percent N 2 , 9 per-
cent H 2 0, 6 percent O,, and 10 percent C0 2 . Determine the
effective emissivity of the combustion gases.
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12-67 A cylindrical container whose height and diameter are
8 m is filled with a mixture of C0 2 and N 2 gases at 600 K and
1 atm. The partial pressure of C0 2 in the mixture is 0.15 atm.
If the walls are black at a temperature of 450 K, determine
the rate of radiation heat transfer between the gas and the con-
tainer walls.
12-68 Repeat Problem 12-67 by replacing C0 2 by the
H 2 gas.
12-69 A 2-m-diameter spherical furnace contains a mixture
of C0 2 and N 2 gases at 1200 K and 1 atm. The mole fraction of
C0 2 in the mixture is 0.15. If the furnace wall is black and its
temperature is to be maintained at 600 K, determine the net
rate of radiation heat transfer between the gas mixture and the
furnace walls.
12-70 A flow-through combustion chamber consists of
15-cm diameter long tubes immersed in water. Compressed air
is routed to the tube, and fuel is sprayed into the compressed
air. The combustion gases consist of 70 percent N 2 , 9 percent
H 2 0, 15 percent 2 , and 6 percent C0 2 , and are maintained at
1 atm and 1500 K. The tube surfaces are near black, with an
emissivity of 0.9. If the tubes are to be maintained at a temper-
ature of 600 K, determine the rate of heat transfer from com-
bustion gases to tube wall by radiation per m length of tube.
12-71 Repeat Problem 12-70 for a total pressure of 3 atm.
12-72 In a cogeneration plant, combustion gases at 1 atm
and 800 K are used to preheat water by passing them through
6-m-long 10-cm-diameter tubes. The inner surface of the tube
is black, and the partial pressures of C0 2 and H 2 in combus-
tion gases are 0.12 atm and 0.18 atm, respectively. If the tube
temperature is 500 K, determine the rate of radiation heat trans-
fer from the gases to the tube.
12-73 A gas at 1200 K and 1 atm consists of 10 percent CO z ,
10 percent H 2 0, 10 percent N 2 , and 70 percent N, by volume.
The gas flows between two large parallel black plates main-
tained at 600 K. If the plates are 20 cm apart, determine the rate
of heat transfer from the gas to each plate per unit surface area.
Special Topic: Heat Transfer from the Human Body
12-74C Consider a person who is resting or doing light
work. Is it fair to say that roughly one-third of the metabolic
heat generated in the body is dissipated to the environment by
convection, one-third by evaporation, and the remaining one-
third by radiation?
12-75C What is sensible heat? How is the sensible heat loss
from a human body affected by (a) skin temperature, (b) envi-
ronment temperature, and (c) air motion?
12-76C What is latent heat? How is the latent heat loss from
the human body affected by (a) skin wettedness and (b) rela-
tive humidity of the environment? How is the rate of evapora-
tion from the body related to the rate of latent heat loss?
661
CHAPTER 12
12-77C How is the insulating effect of clothing expressed?
How does clothing affect heat loss from the body by convec-
tion, radiation, and evaporation? How does clothing affect heat
gain from the sun?
12-78C Explain all the different mechanisms of heat transfer
from the human body (a) through the skin and (b) through
the lungs.
12-79C What is operative temperature? How is it related to
the mean ambient and radiant temperatures? How does it differ
from effective temperature?
12-80 The convection heat transfer coefficient for a clothed
person while walking in still air at a velocity of 0.5 to 2 m/s is
given by h = 8.6T 053 , where T is in m/s and h is in W/m 2 • °C.
Plot the convection coefficient against the walking velocity,
and compare the convection coefficients in that range to the
average radiation coefficient of about 5 W/m 2 ■ °C.
12-81 A clothed or unclothed person feels comfortable when
the skin temperature is about 33°C. Consider an average man
wearing summer clothes whose thermal resistance is 0.7 clo.
The man feels very comfortable while standing in a room
maintained at 20°C. If this man were to stand in that room un-
clothed, determine the temperature at which the room must be
maintained for him to feel thermally comfortable. Assume the
latent heat loss from the person to remain the same.
Answer-. 26.4°C
12-82E An average person produces 0.50 lbm of moisture
while taking a shower and 0.12 lbm while bathing in a tub.
Consider a family of four who shower once a day in a bath-
room that is not ventilated. Taking the heat of vaporization
of water to be 1050 Btu/lbm, determine the contribution of
showers to the latent heat load of the air conditioner in summer
per day.
12-83 An average (1.82 kg or 4.0 lbm) chicken has a basal
metabolic rate of 5.47 W and an average metabolic rate of
10.2 W (3.78 W sensible and 6.42 W latent) during normal ac-
tivity. If there are 100 chickens in a breeding room, determine
the rate of total heat generation and the rate of moisture pro-
duction in the room. Take the heat of vaporization of water to
be 2430 kJ/kg.
12-84 Consider a large classroom with 150 students on a hot
summer day. All the lights with 4.0 kW of rated power are kept
on. The room has no external walls, and thus heat gain through
the walls and the roof is negligible. Chilled air is available at
15°C, and the temperature of the return air is not to exceed
25°C. Determine the required flow rate of air, in kg/s, that
needs to be supplied to the room. Answer-. 1.45 kg/s
12-85 A person feels very comfortable in his house in light
clothing when the thermostat is set at 22°C and the mean radi-
ation temperature (the average temperature of the surrounding
surfaces) is also 22°C. During a cold day, the average mean
radiation temperature drops to 18°C. To what level must the
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HEAT TRANSFER
FIGURE P1 2-85
indoor air temperature be raised to maintain the same level of
comfort in the same clothing?
12-86 Repeat Problem 12-85 for a mean radiation tempera-
ture of 12°C.
12-87 A car mechanic is working in a shop whose interior
space is not heated. Comfort for the mechanic is provided by
two radiant heaters that radiate heat at a total rate of 10 kJ/s.
About 5 percent of this heat strikes the mechanic directly. The
shop and its surfaces can be assumed to be at the ambient tem-
perature, and the emissivity and absorptivity of the mechanic
can be taken to be 0.95 and the surface area to be 1.8 m 2 . The
mechanic is generating heat at a rate of 350 W, half of which is
latent, and is wearing medium clothing with a thermal resis-
tance of 0.7 clo. Determine the lowest ambient temperature in
which the mechanic can work comfortably.
FIGURE P1 2-87
Review Problems
12-88 A thermocouple used to measure the temperature
of hot air flowing in a duct whose walls are maintained at
T w = 500 K shows a temperature reading of T th = 850 K.
Assuming the emissivity of the thermocouple junction to be
; Thermocouple
! r th = 850 K
:
Air « e = 0.6
rJ w = 500 K
FIGURE P1 2-88
e = 0.6 and the convection heat transfer coefficient to be
h = 60 W/m 2 ■ °C, determine the actual temperature of air.
Answer: 1111 K
12-89 A thermocouple shielded by aluminum foil of emis-
sivity 0.15 is used to measure the temperature of hot gases
flowing in a duct whose walls are maintained at T w = 380 K.
The thermometer shows a temperature reading of T th = 530 K.
Assuming the emissivity of the thermocouple junction to
be e = 0.7 and the convection heat transfer coefficient to be
h = 120 W/m 2 • °C, determine the actual temperature of the
gas. What would the thermometer reading be if no radiation
shield was used?
12-90E Consider a sealed 8-in.-high electronic box whose
base dimensions are 12 in. X 12 in. placed in a vacuum cham-
ber. The emissivity of the outer surface of the box is 0.95. If the
electronic components in the box dissipate a total of 100 W of
power and the outer surface temperature of the box is not to ex-
ceed 130°F, determine the highest temperature at which the
surrounding surfaces must be kept if this box is to be cooled by
radiation alone. Assume the heat transfer from the bottom sur-
face of the box to the stand to be negligible. Answer: 43°F
kP
FIGURE P12-90E
12-91 A 2-m-internal-diameter double-walled spherical tank
is used to store iced water at 0°C. Each wall is 0.5 cm thick,
and the 1.5 -cm-thick air space between the two walls of the
tank is evacuated in order to minimize heat transfer. The sur-
faces surrounding the evacuated space are polished so that each
surface has an emissivity of 0.15. The temperature of the outer
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:; C
■20°C
E = 0.15
Vacuum
0.5 cm
FIGURE P1 2-91
0.5 cm
wall of the tank is measured to be 20°C. Assuming the inner
wall of the steel tank to be at 0°C, determine (a) the rate of heat
transfer to the iced water in the tank and (b) the amount of ice
at 0°C that melts during a 24-h period.
12-92 Two concentric spheres of diameters D t = 15 cm and
D 2 = 25 cm are separated by air at 1 atm pressure. The surface
temperatures of the two spheres enclosing the air are T l =
350 K and T 2 = 275 K, respectively, and their emissivities
are 0.5. Determine the rate of heat transfer from the inner
sphere to the outer sphere by (a) natural convection and
(b) radiation.
12-93 Consider a 1.5-m-high and 3 -m- wide solar collector
that is tilted at an angle 20° from the horizontal. The distance
between the glass cover and the absorber plate is 3 cm, and the
back side of the absorber is heavily insulated. The absorber
plate and the glass cover are maintained at temperatures of
80°C and 32°C, respectively. The emissivity of the glass sur-
face is 0.9 and that of the absorber plate is 0.8. Determine the
rate of heat loss from the absorber plate by natural convection
and radiation. Answers: 750 W, 1289 W
Glass
( cover
Absorber
plate
Air space
Insulation
\8 = 20
FIGURE P1 2-93
12-94E
►, A solar collector consists of a horizontal
xsgy aluminum tube having an outer diameter of
2.5 in. enclosed in a concentric thin glass tube of diameter 5 in.
663
CHAPTER 12
Water is heated as it flows through the tube, and the annular
space between the aluminum and the glass tube is filled with
air at 0.5 atm pressure. The pump circulating the water fails
during a clear day, and the water temperature in the tube starts
rising. The aluminum tube absorbs solar radiation at a rate of
30 Btu/h per foot length, and the temperature of the ambient air
outside is 75 °F. The emissivities of the tube and the glass cover
are 0.9. Taking the effective sky temperature to be 60°F, deter-
mine the temperature of the aluminum tube when thermal equi-
librium is established (i.e., when the rate of heat loss from the
tube equals the amount of solar energy gained by the tube).
12-95 A vertical 2-m-high and 3-m-wide double-pane win-
dow consists of two sheets of glass separated by a 5-cm-thick
air gap. In order to reduce heat transfer through the window,
the air space between the two glasses is partially evacuated to
0.3 atm pressure. The emissivities of the glass surfaces are 0.9.
Taking the glass surface temperatures across the air gap to be
15°C and 5°C, determine the rate of heat transfer through the
window by natural convection and radiation.
2 m
Frame
FIGURE P1 2-95
12-96 f~fc\ A simple solar collector is built by placing a
xj«£/ 6-cm-diameter clear plastic tube around a gar-
den hose whose outer diameter is 2 cm. The hose is painted
black to maximize solar absorption, and some plastic rings
are used to keep the spacing between the hose and the clear
plastic cover constant. The emissivities of the hose surface and
the glass cover are 0.9, and the effective sky temperature is
Solar T, = 15°C
sky
radiation
\ V \ \ 25°C
11*11 - Clear plastic tube
/ 40°C
Water /H^~ "^
-T>
w'J) ;
V y » v y
K y v y
Garden hose
FIGURE P 12-96
T
Spacer
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HEAT TRANSFER
estimated to be 15°C. The temperature of the plastic tube is
measured to be 40°C, while the ambient air temperature is
25°C. Determine the rate of heat loss from the water in the
hose by natural convection and radiation per meter of its length
under steady conditions.
Answers: 5.2 W, 26.2 W
12-97 A solar collector consists of a horizontal copper tube
of outer diameter 5 cm enclosed in a concentric thin glass tube
of diameter 9 cm. Water is heated as it flows through the tube,
and the annular space between the copper and the glass tubes is
filled with air at 1 atm pressure. The emissivities of the tube
surface and the glass cover are 0.85 and 0.9, respectively. Dur-
ing a clear day, the temperatures of the tube surface and the
glass cover are measured to be 60°C and 40°C, respectively.
Determine the rate of heat loss from the collector by natural
convection and radiation per meter length of the tube.
12-98 A furnace is of cylindrical shape with a diameter of
1 .2 m and a length of 1 .2 m. The top surface has an emissivity
of 0.70 and is maintained at 500 K. The bottom surface has an
emissivity of 0.50 and is maintained at 650 K. The side surface
has an emissivity of 0.40. Heat is supplied from the base
surface at a net rate of 1400 W. Determine the temperature of
the side surface and the net rates of heat transfer between the
top and the bottom surfaces, and between the bottom and side
surfaces.
1.2 m
FIGURE P 12-98
12-99 Consider a cubical furnace with a side length of 3 m.
The top surface is maintained at 700 K. The base surface has
an emissivity of 0.90 and is maintained at 950 K. The side sur-
face is black and is maintained at 450 K. Heat is supplied from
the base surface at a rate of 340 kW. Determine the emissivity
of the top surface and the net rates of heat transfer between the
top and the bottom surfaces, and between the bottom and side
surfaces.
12-100 A thin aluminum sheet with an emissivity of 0.12 on
both sides is placed between two very large parallel plates
maintained at uniform temperatures of T { = 750 K and T 2 =
550 K. The emissivities of the plates are e, = 0.8 and s 2 = 0.9.
Determine the net rate of radiation heat transfer between the
two plates per unit surface area of the plates, and the tempera-
ture of the radiation shield in steady operation.
12-101 Two thin radiation shields with emissivities of s 3 =
0.10 and e 4 = 0.15 on both sides are placed between two very
large parallel plates, which are maintained at uniform tem-
peratures 7| = 600 K and T 2 = 300 K and have emissivities
g| = 0.6 and e 2 = 0.7, respectively. Determine the net rates of
radiation heat transfer between the two plates with and without
the shields per unit surface area of the plates, and the tempera-
tures of the radiation shields in steady operation.
^- 7\ = 600 K
e, =0.6
^e 3 = 0.10
T 2 = 300 K
^~e 2 ~=0.7
^e 4 = 0.15
FIGURE P12-101
12-102 In a natural-gas fired boiler, combustion gases pass
through 6-m-long 15-cm-diameter tubes immersed in water at
1 atm pressure. The tube temperature is measured to be 105°C,
and the emissivity of the inner surfaces of the tubes is esti-
mated to be 0.9. Combustion gases enter the tube at 1 atm and
1200 K at a mean velocity of 3 m/s. The mole fractions of C0 2
and H 2 in combustion gases are 8 percent and 16 percent, re-
spectively. Assuming fully developed flow and using proper-
ties of air for combustion gases, determine (a) the rates of heat
transfer by convection and by radiation from the combustion
gases to the tube wall and (b) the rate of evaporation of water.
12-103 Repeat Problem 12-102 for a total pressure of 3 atm
for the combustion gases.
Computer, Design, and Essay Problems
12-104 Consider an enclosure consisting of N diffuse and
gray surfaces. The emissivity and temperature of each surface
as well as all the view factors between the surfaces are speci-
fied. Write a program to determine the net rate of radiation heat
transfer for each surface.
12-105 Radiation shields are commonly used in the design
of superinsulations for use in space and cryogenic applications.
Write an essay on superinsulations and how they are used in
different applications.
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12-106 Thermal comfort in a house is strongly affected by
the so-called radiation effect, which is due to radiation heat
transfer between the person and surrounding surfaces. A person
feels much colder in the morning, for example, because of the
665
CHAPTER 12
lower surface temperature of the walls at that time, although
the thermostat setting of the house is fixed. Write an essay on
the radiation effect, how it affects human comfort, and how it
is accounted for in heating and air-conditioning applications.
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HEAT EXCHANGERS
CHAPTER
Heat exchangers are devices that facilitate the exchange of heat between
two fluids that are at different temperatures while keeping them from
mixing with each other. Heat exchangers are commonly used in prac-
tice in a wide range of applications, from heating and air-conditioning systems
in a household, to chemical processing and power production in large plants.
Heat exchangers differ from mixing chambers in that they do not allow the
two fluids involved to mix. In a car radiator, for example, heat is transferred
from the hot water flowing through the radiator tubes to the air flowing
through the closely spaced thin plates outside attached to the tubes.
Heat transfer in a heat exchanger usually involves convection in each fluid
and conduction through the wall separating the two fluids. In the analysis of
heat exchangers, it is convenient to work with an overall heat transfer coeffi-
cient U that accounts for the contribution of all these effects on heat transfer.
The rate of heat transfer between the two fluids at a location in a heat ex-
changer depends on the magnitude of the temperature difference at that
location, which varies along the heat exchanger. In the analysis of heat ex-
changers, it is usually convenient to work with the logarithmic mean temper-
ature difference LMTD, which is an equivalent mean temperature difference
between the two fluids for the entire heat exchanger.
Heat exchangers are manufactured in a variety of types, and thus we start
this chapter with the classification of heat exchangers. We then discuss the de-
termination of the overall heat transfer coefficient in heat exchangers, and the
LMTD for some configurations. We then introduce the correction factor F to
account for the deviation of the mean temperature difference from the LMTD
in complex configurations. Next we discuss the effectiveness-NTU method,
which enables us to analyze heat exchangers when the outlet temperatures of
the fluids are not known. Finally, we discuss the selection of heat exchangers.
CONTENTS
13-1 Types of Heat
Exchangers 668
13-2 The Overall Heat Transfer
Coefficient 671
13-3 Analysis of Heat
Exchangers 678
13-4 The Log Mean Temperature
Difference Method 680
13-5 The Effectiveness-NTU
Method 690
13-6 Selection of Heat
Exchangers 700
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HEAT TRANSFER
13-1 ■ TYPES OF HEAT EXCHANGERS
Different heat transfer applications require different types of hardware and
different configurations of heat transfer equipment. The attempt to match the
heat transfer hardware to the heat transfer requirements within the specified
constraints has resulted in numerous types of innovative heat exchanger
designs.
The simplest type of heat exchanger consists of two concentric pipes of dif-
ferent diameters, as shown in Figure 13—1, called the double-pipe heat
exchanger. One fluid in a double-pipe heat exchanger flows through the
smaller pipe while the other fluid flows through the annular space between
the two pipes. Two types of flow arrangement are possible in a double-pipe
heat exchanger: in parallel flow, both the hot and cold fluids enter the heat
exchanger at the same end and move in the same direction. In counter flow,
on the other hand, the hot and cold fluids enter the heat exchanger at opposite
ends and flow in opposite directions.
Another type of heat exchanger, which is specifically designed to realize a
large heat transfer surface area per unit volume, is the compact heat ex-
changer. The ratio of the heat transfer surface area of a heat exchanger to its
volume is called the area density 0. A heat exchanger with > 700 m 2 /m 3
(or 200 ft 2 /ft 3 ) is classified as being compact. Examples of compact heat
exchangers are car radiators (0 ~ 1000 m 2 /m 3 ), glass ceramic gas turbine
heat exchangers (0 ~ 6000 m 2 /m 3 ), the regenerator of a Stirling engine
(0 « 15,000 m 2 /m 3 ), and the human lung (0 « 20,000 m 2 /m 3 ). Compact heat
exchangers enable us to achieve high heat transfer rates between two fluids in
Cold
out
Cold
in
Hot
in
FIGURE 13-1
Different flow regimes and
associated temperature profiles in
a double-pipe heat exchanger.
Hot
out
Hot
in
Hot
t
Cold
in
(a) Parallel flow
I
Cold
out
(b) Counter flow
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CHAPTER 13
a small volume, and they are commonly used in applications with strict limi-
tations on the weight and volume of heat exchangers (Fig. 13-2).
The large surface area in compact heat exchangers is obtained by attaching
closely spaced thin plate or corrugated fins to the walls separating the two flu-
ids. Compact heat exchangers are commonly used in gas-to-gas and gas-to-
liquid (or liquid-to-gas) heat exchangers to counteract the low heat transfer
coefficient associated with gas flow with increased surface area. In a car radi-
ator, which is a water-to-air compact heat exchanger, for example, it is no sur-
prise that fins are attached to the air side of the tube surface.
In compact heat exchangers, the two fluids usually move perpendicular to
each other, and such flow configuration is called cross-flow. The cross-flow
is further classified as unmixed and mixed flow, depending on the flow con-
figuration, as shown in Figure 13-3. In (a) the cross-flow is said to be un-
mixed since the plate fins force the fluid to flow through a particular interfin
spacing and prevent it from moving in the transverse direction (i.e., parallel to
the tubes). The cross-flow in (b) is said to be mixed since the fluid now is free
to move in the transverse direction. Both fluids are unmixed in a car radiator.
The presence of mixing in the fluid can have a significant effect on the heat
transfer characteristics of the heat exchanger.
Cross-flow
(unmixed)
(a) Both fluids unmixed
Tube flow Tube flow
(unmixed) (unmixed)
(b) One fluid mixed, one fluid unmixed
FIGURE 13-2
A gas-to-liquid compact heat
exchanger for a residential air-
conditioning system.
FIGURE 13-3
Different flow configurations in
cross-flow heat exchangers.
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HEAT TRANSFER
FIGURE 13-4
The schematic of
a shell-and-tube
heat exchanger
(one-shell pass
and one-tube
pass).
Tube
outlet
Baffles
Shell-side fluid
In
(c
Tube-side
fluid
-D-^Out
ni-«- In
{Out
(a) One-shell pass and two-tube passes
Shell-side fluid
In
A
Out
Tube-
side
fluid
Out
(ft) Two-shell passes and four-tube passes
FIGURE 13-5
Multipass flow arrangements in shell-
and-tube heat exchangers.
Perhaps the most common type of heat exchanger in industrial applications
is the shell-and-tube heat exchanger, shown in Figure 13-4. Shell-and-tube
heat exchangers contain a large number of tubes (sometimes several hundred)
packed in a shell with their axes parallel to that of the shell. Heat transfer takes
place as one fluid flows inside the tubes while the other fluid flows outside the
tubes through the shell. Baffles are commonly placed in the shell to force the
shell-side fluid to flow across the shell to enhance heat transfer and to main-
tain uniform spacing between the tubes. Despite their widespread use, shell-
and-tube heat exchangers are not suitable for use in automotive and aircraft
applications because of their relatively large size and weight. Note that the
tubes in a shell-and-tube heat exchanger open to some large flow areas called
headers at both ends of the shell, where the tube-side fluid accumulates before
entering the tubes and after leaving them.
Shell-and-tube heat exchangers are further classified according to the num-
ber of shell and tube passes involved. Heat exchangers in which all the tubes
make one U-turn in the shell, for example, are called one-shell-pass and two-
tube-passes heat exchangers. Likewise, a heat exchanger that involves two
passes in the shell and four passes in the tubes is called a two-shell-passes and
four-tube-passes heat exchanger (Fig. 13-5).
An innovative type of heat exchanger that has found widespread use is the
plate and frame (or just plate) heat exchanger, which consists of a series of
plates with corrugated flat flow passages (Fig. 13-6). The hot and cold fluids
flow in alternate passages, and thus each cold fluid stream is surrounded by
two hot fluid streams, resulting in very effective heat transfer. Also, plate heat
exchangers can grow with increasing demand for heat transfer by simply
mounting more plates. They are well suited for liquid-to-liquid heat exchange
applications, provided that the hot and cold fluid streams are at about the same
pressure.
Another type of heat exchanger that involves the alternate passage of the hot
and cold fluid streams through the same flow area is the regenerative heat ex-
changer. The static-type regenerative heat exchanger is basically a porous
mass that has a large heat storage capacity, such as a ceramic wire mesh. Hot
and cold fluids flow through this porous mass alternatively. Heat is transferred
from the hot fluid to the matrix of the regenerator during the flow of the hot
fluid, and from the matrix to the cold fluid during the flow of the cold fluid.
Thus, the matrix serves as a temporary heat storage medium.
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CHAPTER 13
Special gaskets on end plates
prevent fluids from contacting-
the frames.
A gasket mounted on each
plate seals the channel
between it and the next
plate
FIGURE 13-6
A plate-and-frame
, J2BSSSSS*. liquid-to-liquid heat
prevent™, lateral movement exchanger (courtesy of
Trante Corp.).
The lower
assures absolute
The dynamic -type regenerator involves a rotating drum and continuous flow
of the hot and cold fluid through different portions of the drum so that any
portion of the drum passes periodically through the hot stream, storing heat,
and then through the cold stream, rejecting this stored heat. Again the drum
serves as the medium to transport the heat from the hot to the cold fluid
stream.
Heat exchangers are often given specific names to reflect the specific appli-
cation for which they are used. For example, a condenser is a heat exchanger
in which one of the fluids is cooled and condenses as it flows through the heat
exchanger. A boiler is another heat exchanger in which one of the fluids ab-
sorbs heat and vaporizes. A space radiator is a heat exchanger that transfers
heat from the hot fluid to the surrounding space by radiation.
13-2 - THE OVERALL HEAT TRANSFER COEFFICIENT
A heat exchanger typically involves two flowing fluids separated by a solid
wall. Heat is first transferred from the hot fluid to the wall by convection,
through the wall by conduction, and from the wall to the cold fluid again by
convection. Any radiation effects are usually included in the convection heat
transfer coefficients.
The thermal resistance network associated with this heat transfer process
involves two convection and one conduction resistances, as shown in Figure
13-7. Here the subscripts i and o represent the inner and outer surfaces of the
Hot
fluid
Heat
\^*
transfer
r '*^l
| Cold
ot fluid J
1 fluid
A ._y Wall
V /
K
t »—miH-^-wmh-i^-wm~»
«wall R = j^j
n , A , h o A o
FIGURE 13-7
Thermal resistance network
associated with heat transfer
in a double-pipe heat exchanger.
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HEAT TRANSFER
inner tube. For a double -pipe heat exchanger, we have A, = ttD,L and A c
TrD L, and the thermal resistance of the tube wall in this case is
R„
In (DJDj)
lukh
(13-1)
Heat 1
transfer \
x Outer tube
Outer
fluid
Inner \
Inner tube
fluid \
-~A =kD L
V A;
=%D-L
FIGURE 13-8
The two heat transfer surface areas
associated with a double-pipe heat
exchanger (for thin tubes, Z), ~ D„
and thus A, ~ A ).
where k is the thermal conductivity of the wall material and L is the length of
the tube. Then the total thermal resistance becomes
R = R„
R, + R,
R„
1
ll;A,
In(ZVA)
2irkL
+
1
h„A„
(13-2)
The A, is the area of the inner surface of the wall that separates the two fluids,
and A is the area of the outer surface of the wall. In other words, A t and A B are
surface areas of the separating wall wetted by the inner and the outer fluids,
respectively. When one fluid flows inside a circular tube and the other outside
of it, we have A { = ttDjL and A = ttD L (Fig. 13-8).
In the analysis of heat exchangers, it is convenient to combine all the ther-
mal resistances in the path of heat flow from the hot fluid to the cold one into
a single resistance R, and to express the rate of heat transfer between the two
fluids as
Q
AT
R
UA AT = U;A, AT = U„A„ AT
(13-3)
where U is the overall heat transfer coefficient, whose unit is W/m 2 • °C,
which is identical to the unit of the ordinary convection coefficient h. Cancel-
ing AT, Eq. 13-3 reduces to
1
UA,
1
U,A:
1
U n A n
R
1
h,A
+ R.
1
h n A„
(13-4)
Perhaps you are wondering why we have two overall heat transfer coefficients
Uj and U for a heat exchanger. The reason is that every heat exchanger has
two heat transfer surface areas A, and A , which, in general, are not equal to
each other.
Note that £/,A, = U A , but [/,■ =£ U unless A, = A . Therefore, the overall
heat transfer coefficient U of a heat exchanger is meaningless unless the area
on which it is based is specified. This is especially the case when one side of
the tube wall is finned and the other side is not, since the surface area of the
finned side is several times that of the unfinned side.
When the wall thickness of the tube is small and the thermal conductivity of
the tube material is high, as is usually the case, the thermal resistance of the
tube is negligible (R mii ~ 0) and the inner and outer surfaces of the tube are
almost identical (A, ~ A ~ A s ). Then Eq. 13-4 for the overall heat transfer co-
efficient simplifies to
1 = 1 + 1
U h, h n
(13-5)
where U ~ Uj ~ U . The individual convection heat transfer coefficients
inside and outside the tube, h t and h , are determined using the convection
relations discussed in earlier chapters.
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CHAPTER 13
The overall heat transfer coefficient U in Eq. 13-5 is dominated by the
smaller convection coefficient, since the inverse of a large number is small.
When one of the convection coefficients is much smaller than the other (say,
hj < h ), we have I//2, > l//z a , and thus U ~ h t . Therefore, the smaller heat
transfer coefficient creates a bottleneck on the path of heat flow and seriously
impedes heat transfer. This situation arises frequently when one of the fluids
is a gas and the other is a liquid. In such cases, fins are commonly used on the
gas side to enhance the product UA S and thus the heat transfer on that side.
Representative values of the overall heat transfer coefficient U are given in
Table 13-1. Note that the overall heat transfer coefficient ranges from about
10 W/m 2 • °C for gas-to-gas heat exchangers to about 10,000 W/m 2 • °C for
heat exchangers that involve phase changes. This is not surprising, since gases
have very low thermal conductivities, and phase-change processes involve
very high heat transfer coefficients.
When the tube is finned on one side to enhance heat transfer, the total heat
transfer surface area on the finned side becomes
A f ,
(13-6)
where A fin is the surface area of the fins and A unfmned is the area of the unfinned
portion of the tube surface. For short fins of high thermal conductivity, we can
use this total area in the convection resistance relation ^ conv = \lhA s since the
fins in this case will be very nearly isothermal. Otherwise, we should deter-
mine the effective surface area A from
^unfinned + Tlfin^fin
(13-7)
Representative values of the overall heat tra
nsfer coeff
cients in
heat exchangers
Type of heat exchanger
U, W/m 2 • °C*
Water-to-water
850-1700
Water-to-oil
100-350
Water-to-gasoline or kerosene
300-1000
Feedwater heaters
1000-8500
Steam-to- light fuel oil
200-400
Steam-to-heavy fuel oil
50-200
Steam condenser
1000-6000
Freon condenser (water cooled)
300-1000
Ammonia condenser (water cooled)
800-1400
Alcohol condensers (water cooled)
250-700
Gas-to-gas
10-40
Water-to-air in finned tubes (water in t
ubes)
30-60*
400-850*
Steam-to-air in finned tubes (steam in
tubes)
30-300*
400-4000*
'Multiply the listed values by 0.176 to convert them to Btu/h • ft 2 ■ °F.
'Based on air-side surface area.
•Based on water- or steam-side surface area.
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HEAT TRANSFER
FIGURE 13-9
Precipitation fouling of
ash particles on superheater tubes
(from Steam, Its Generation, and Use,
Babcock and Wilcox Co., 1978).
where T| fin is the fin efficiency. This way, the temperature drop along the fins
is accounted for. Note that % in = 1 for isothermal fins, and thus Eq. 13-7
reduces to Eq. 13-6 in that case.
Fouling Factor
The performance of heat exchangers usually deteriorates with time as a result
of accumulation of deposits on heat transfer surfaces. The layer of deposits
represents additional resistance to heat transfer and causes the rate of heat
transfer in a heat exchanger to decrease. The net effect of these accumulations
on heat transfer is represented by a fouling factor R f , which is a measure of
the thermal resistance introduced by fouling.
The most common type of fouling is the precipitation of solid deposits in a
fluid on the heat transfer surfaces. You can observe this type of fouling even
in your house. If you check the inner surfaces of your teapot after prolonged
use, you will probably notice a layer of calcium- based deposits on the surfaces
at which boiling occurs. This is especially the case in areas where the water is
hard. The scales of such deposits come off by scratching, and the surfaces can
be cleaned of such deposits by chemical treatment. Now imagine those min-
eral deposits forming on the inner surfaces of fine tubes in a heat exchanger
(Fig. 13-9) and the detrimental effect it may have on the flow passage area
and the heat transfer. To avoid this potential problem, water in power and
process plants is extensively treated and its solid contents are removed before
it is allowed to circulate through the system. The solid ash particles in the flue
gases accumulating on the surfaces of air preheaters create similar problems.
Another form of fouling, which is common in the chemical process indus-
try, is corrosion and other chemical fouling. In this case, the surfaces are
fouled by the accumulation of the products of chemical reactions on the sur-
faces. This form of fouling can be avoided by coating metal pipes with glass
or using plastic pipes instead of metal ones. Heat exchangers may also be
fouled by the growth of algae in warm fluids. This type of fouling is called
biological fouling and can be prevented by chemical treatment.
In applications where it is likely to occur, fouling should be considered in
the design and selection of heat exchangers. In such applications, it may be
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CHAPTER 13
necessary to select a larger and thus more expensive heat exchanger to ensure
that it meets the design heat transfer requirements even after fouling occurs.
The periodic cleaning of heat exchangers and the resulting down time are ad-
ditional penalties associated with fouling.
The fouling factor is obviously zero for a new heat exchanger and increases
with time as the solid deposits build up on the heat exchanger surface. The
fouling factor depends on the operating temperature and the velocity of the
fluids, as well as the length of service. Fouling increases with increasing tem-
perature and decreasing velocity.
The overall heat transfer coefficient relation given above is valid for clean
surfaces and needs to be modified to account for the effects of fouling on both
the inner and the outer surfaces of the tube. For an unfinned shell-and-tube
heat exchanger, it can be expressed as
1
UA,
1
U;A;
1
U„A„
R
h,A, A
v , In (A/A) , Rf,o
1 1
1
2-nkL
h„A„
(13-8)
where A, = ir£),L and A = irD L are the areas of inner and outer surfaces,
and Rf t and R f are the fouling factors at those surfaces.
Representative values of fouling factors are given in Table 13-2. More com-
prehensive tables of fouling factors are available in handbooks. As you would
expect, considerable uncertainty exists in these values, and they should be
used as a guide in the selection and evaluation of heat exchangers to account
for the effects of anticipated fouling on heat transfer. Note that most fouling
factors in the table are of the order of 10~ 4 m 2 • °C/W, which is equivalent to
the thermal resistance of a 0.2-mm-thick limestone layer (k = 2.9 W/m • °C)
per unit surface area. Therefore, in the absence of specific data, we can as-
sume the surfaces to be coated with 0.2 mm of limestone as a starting point to
account for the effects of fouling.
EXAMPLE 13-1 Overall Heat Transfer Coefficient of a
Heat Exchanger
Hot oil is to be cooled in a double-tube counter-flow heat exchanger. The copper
inner tubes have a diameter of 2 cm and negligible thickness. The inner diame-
ter of the outer tube (the shell) is 3 cm. Water flows through the tube at a rate of
0.5 kg/s, and the oil through the shell at a rate of 0.8 kg/s. Taking the average
temperatures of the water and the oil to be 45 C C and 80 C C, respectively, deter-
mine the overall heat transfer coefficient of this heat exchanger.
SOLUTION Hot oil is cooled by water in a double-tube counter-flow heat
exchanger. The overall heat transfer coefficient is to be determined.
Assumptions 1 The thermal resistance of the inner tube is negligible since
the tube material is highly conductive and its thickness is negligible. 2 Both
the oil and water flow are fully developed. 3 Properties of the oil and water are
constant.
Properties The properties of water at 45°C are (Table A-9)
p = 990kg/m 3 Pr = 3.91
k = 0.637 W/m • °C v = uVp = 0.602 X lO" 6 m 2 /s
TABLE 13-2
Representative fouling
factors (thermal resistance due
to fouling for a unit surface area)
(Source: Tubular Exchange
Manufacturers
Association.)
Fluid
R t , m 2 ■ °C/W
Distilled water, sea
water, river water,
boiler feedwater:
Below 50°C
0.0001
Above 50°C
0.0002
Fuel oil
0.0009
Steam (oil-free)
0.0001
Refrigerants (liquid)
0.0002
Refrigerants (vapor)
0.0004
Alcohol vapors
0.0001
Air
0.0004
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HEAT TRANSFER
Hot oil
0.8kg/s I
Cold
water
- 0~ 12cm 3 (
0.5 kg/s
ti
B—
FIGURE 13-10
Schematic for Example 13-1.
TABLE 13-3
Nusselt number for fully developed
laminar flow in a circular annulus
with one surface insulated and the
other isothermal (Kays and Perkins,
Ref. 8.)
D,/D
Nu,
Nu
0.00
—
3.66
0.05
17.46
4.06
0.10
11.56
4.11
0.25
7.37
4.23
0.50
5.74
4.43
1.00
4.86
4.86
The properties of oil at 80°C are (Table A-16).
p = 852 kg/m 3 Pr = 490
k = 0.138 W/m ■ °C v = 37.5 X 10- 6 m 2 /s
Analysis The schematic of the heat exchanger is given in Figure 13-10. The
overall heat transfer coefficient U can be determined from Eq. 13-5:
U
i+1
h, K
where ft, and /? are the convection heat transfer coefficients inside and outside
the tube, respectively, which are to be determined using the forced convection
relations.
The hydraulic diameter for a circular tube is the diameter of the tube itself,
D h = D = 0.02 m. The mean velocity of water in the tube and the Reynolds
number are
m
0.5 kg/s
pQ-rrZ) 2 ) (990 kg/m 3 )[±ir (0.02 m) 2 ]
1.61 m/s
and
Re
r,„D h _ (1.61 m/s)(0.02m)
v ~ 0.602 X 10- 6 m 2 /s
53,490
which is greater than 4000. Therefore, the flow of water is turbulent. Assuming
the flow to be fully developed, the Nusselt number can be determined from
hD h
Nu
0.023 Re°- 8 Pr°
0.023(53,490)°- 8 (3.91)°
240.6
Then,
h
■Nu
0.637 W/m
: (240.6) = 7663 W/m 2 ■ °C
D h 0.02 m
Now we repeat the analysis above for oil. The properties of oil at 80°C are
m 2 /s
p = 852 kg/m 3
jk = 0.138 W/m- °C
v
Pr
37.5 X 10- 6
490
The hydraulic diameter for the annular space is
D,, = D„ - D, = 0.03 - 0.02 = 0.01 m
The mean velocity and the Reynolds number in this case are
m m 0-8 kg/s
'" pA c p [Itt(/) 2 -D 2 )] (852kg/m 3 )[iTr(0.03 2 -0.0:
and
Y,„D h (2.39 m/s)(0.01 m)
2.39 m/s
Re
37.5 X 10- 6 m 2 /s
637
which is less than 4000. Therefore, the flow of oil is laminar. Assuming fully
developed flow, the Nusselt number on the tube side of the annular space
Nu, corresponding to D,ID = 0.02/0.03 = 0.667 can be determined from
Table 13-3 by interpolation to be
Nu = 5.45
cen58933_chl3.qxd 9/9/2002 9:57 AM Page 677
and
■Nu
0.138 W/m • °C
(5.45) = 75.2 W/m 2 °C
° D h 0.01m
Then the overall heat transfer coefficient for this heat exchanger becomes
mA e 1X7/ 2 , °(~*
u
i+i
hi h„
1
+
1
74.5 W/m 2
7663 W/m 2 • °C 75.2 W/m 2 ■ °C
Discussion Note that L/ ~ fr in this case, since h, > h . This confirms our ear-
lier statement that the overall heat transfer coefficient in a heat exchanger is
dominated by the smaller heat transfer coefficient when the difference between
the two values is large.
To improve the overall heat transfer coefficient and thus the heat transfer in
this heat exchanger, we must use some enhancement techniques on the oil
side, such as a finned surface.
677
CHAPTER 13
EXAMPLE 13-2 Effect of Fouling on the Overall Heat
Transfer Coefficient
A double-pipe (shell-and-tube) heat exchanger is constructed of a stainless steel
(k = 15.1 W/m • °C) inner tube of inner diameter D, = 1.5 cm and outer diam-
eter D = 1.9 cm and an outer shell of inner diameter 3.2 cm. The convection
heat transfer coefficient is given to be h, = 800 W/m 2 • °C on the inner surface
of the tube and h = 1200 W/m 2 • °C on the outer surface. For a fouling factor
of R f: , = 0.0004 m 2 • °C/W on the tube side and R t = 0.0001 m 2 ■ °C/W on
the shell side, determine (a) the thermal resistance of the heat exchanger per
unit length and (b) the overall heat transfer coefficients, U, and U based on the
inner and outer surface areas of the tube, respectively.
SOLUTION The heat transfer coefficients and the fouling factors on the tube
and shell sides of a heat exchanger are given. The thermal resistance and the
overall heat transfer coefficients based on the inner and outer areas are to be
determined.
Assumptions The heat transfer coefficients and the fouling factors are constant
and uniform.
Analysis (a) The schematic of the heat exchanger is given in Figure 13-11.
The thermal resistance for an unfinned shell-and-tube heat exchanger with foul-
ing on both heat transfer surfaces is given by Eq. 13-8 as
R
1
UA,
1
U;A;
1
where
A n
U n A n
-nDfL
-nDL
1 | «/., | ln(ZVA) | Rf.o
h : A t A, 2-nkL A„
1
h n A„
ir(0.015 m)(l m) = 0.0471 m 2
ir(0.019 m)(l m) = 0.0597 m 2
Substituting, the total thermal resistance is determined to be
Cold fluid
Outer layer of fouling
Tube wall
Inner layer of fouling
Cold fluid
\
a
Do
K
R f.o
D j = 1.5 cm
ft. = 800 W/m 2 -°C
R fi = 0.0004 m 2 -°C/W
: 1.9 cm
= 1200W/m 2 -°C
= 0.0001 m 2 -°C/W
FIGURE 13-11
Schematic for Example 13-2.
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HEAT TRANSFER
1
(800 W/m 2 • °C)(0.0471 m 2 )
In (0.019/0.015)
+ 2tt(15.1 W/m ■ °C)(1 m)
+
0.0001 m 2 ■ °C/W
0.0004 m 2 ■ °C/W
0.0471 m 2
1
0.0597 m 2 (1200 W/m 2 • °C)(0.0597 m 2 )
= (0.02654 + 0.00849 + 0.0025 + 0.00168 + 0.01396)°C/W
= 0.0532°C/W
Note that about 19 percent of the total thermal resistance in this case is due to
fouling and about 5 percent of it is due to the steel tube separating the two flu-
ids. The rest (76 percent) is due to the convection resistances on the two sides
of the inner tube.
(b) Knowing the total thermal resistance and the heat transfer surface areas,
the overall heat transfer coefficient based on the inner and outer surfaces of the
tube are determined again from Eq. 13-8 to be
and
U ;
U n
1
1
RA) (0.0532 °C/W)(0.0471 m 2 )
1
1
(0.0532 °C/W)(0.0597 m 2 )
399 W/m 2 • °C
315 W/m 2 • °C
Discussion Note that the two overall heat transfer coefficients differ signifi-
cantly (by 27 percent) in this case because of the considerable difference be-
tween the heat transfer surface areas on the inner and the outer sides of the
tube. For tubes of negligible thickness, the difference between the two overall
heat transfer coefficients would be negligible.
13-3 - ANALYSIS OF HEAT EXCHANGERS
Heat exchangers are commonly used in practice, and an engineer often finds
himself or herself in a position to select a heat exchanger that will achieve a
specified temperature change in a fluid stream of known mass flow rate, or to
predict the outlet temperatures of the hot and cold fluid streams in a specified
heat exchanger.
In upcoming sections, we will discuss the two methods used in the analysis
of heat exchangers. Of these, the log mean temperature difference (or LMTD)
method is best suited for the first task and the effectiveness— NTU method for
the second task as just stated. But first we present some general considerations.
Heat exchangers usually operate for long periods of time with no change in
their operating conditions. Therefore, they can be modeled as steady-flow de-
vices. As such, the mass flow rate of each fluid remains constant, and the fluid
properties such as temperature and velocity at any inlet or outlet remain the
same. Also, the fluid streams experience little or no change in their velocities
and elevations, and thus the kinetic and potential energy changes are negligi-
ble. The specific heat of a fluid, in general, changes with temperature. But, in
cen58933_chl3.qxd 9/9/2002 9:57 AM Page 679
a specified temperature range, it can be treated as a constant at some average
value with little loss in accuracy. Axial heat conduction along the tube is usu-
ally insignificant and can be considered negligible. Finally, the outer surface
of the heat exchanger is assumed to be perfectly insulated, so that there is no
heat loss to the surrounding medium, and any heat transfer occurs between the
two fluids only.
The idealizations stated above are closely approximated in practice, and
they greatly simplify the analysis of a heat exchanger with little sacrifice of
accuracy. Therefore, they are commonly used. Under these assumptions, the
first law of thermodynamics requires that the rate of heat transfer from the hot
fluid be equal to the rate of heat transfer to the cold one. That is,
Q
AC P c(T c ,
T )
x c, in/
and
Q ~ m hCph\Th,m 7/i, out)
(13-9)
(13-10)
where the subscripts c and h stand for cold and hot fluids, respectively, and
m c , m h = mass flow rates
= specific heats
outlet temperatures
inlet temperatures
c c
^pc-> *~ph
T T
c, out' h. out
Note that the heat transfer rate Q is taken to be a positive quantity, and its di-
rection is understood to be from the hot fluid to the cold one in accordance
with the second law of thermodynamics.
In heat exchanger analysis, it is often convenient to combine the product of
the mass flow rate and the specific heat of a fluid into a single quantity. This
quantity is called the heat capacity rate and is defined for the hot and cold
fluid streams as
C h = rii,,C,
h^-ph
and
C, = m,.C„,
(13-11)
The heat capacity rate of a fluid stream represents the rate of heat transfer
needed to change the temperature of the fluid stream by 1°C as it flows
through a heat exchanger. Note that in a heat exchanger, the fluid with a large
heat capacity rate will experience a small temperature change, and the fluid
with a small heat capacity rate will experience a large temperature change.
Therefore, doubling the mass flow rate of a fluid while leaving everything else
unchanged will halve the temperature change of that fluid.
With the definition of the heat capacity rate above, Eqs. 13-9 and 13-10 can
also be expressed as
and
Q = C c (T c , mt -T c/m )
Q = C„(T„.
(13-12)
(13-13)
That is, the heat transfer rate in a heat exchanger is equal to the heat capacity
rate of either fluid multiplied by the temperature change of that fluid. Note
that the only time the temperature rise of a cold fluid is equal to the tempera-
ture drop of the hot fluid is when the heat capacity rates of the two fluids are
equal to each other (Fig. 13-12).
679
CHAPTER 13
Inlet Outlet
FIGURE 13-12
Two fluids that have the same mass
flow rate and the same specific heat
experience the same temperature
change in a well-insulated heat
exchanger.
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680
HEAT TRANSFER
Inlet
Outlet
(b) Boiler (C c -> t»)
FIGURE 13-13
Variation of fluid temperatures in a
heat exchanger when one of the fluids
condenses or boils.
Two special types of heat exchangers commonly used in practice are con-
densers and boilers. One of the fluids in a condenser or a boiler undergoes a
phase-change process, and the rate of heat transfer is expressed as
Q = mh f .
(13-14)
where m is the rate of evaporation or condensation of the fluid and hf„ is the
enthalpy of vaporization of the fluid at the specified temperature or pressure.
An ordinary fluid absorbs or releases a large amount of heat essentially
at constant temperature during a phase-change process, as shown in Figure
13-13. The heat capacity rate of a fluid during a phase-change process must
approach infinity since the temperature change is practically zero. That is,
C = rhCp — > °° when AT — > 0, so that the heat transfer rate Q = mC p AT is a
finite quantity. Therefore, in heat exchanger analysis, a condensing or boiling
fluid is conveniently modeled as a fluid whose heat capacity rate is infinity.
The rate of heat transfer in a heat exchanger can also be expressed in an
analogous manner to Newton's law of cooling as
Q = UA S AT„,
(13-15)
where U is the overall heat transfer coefficient, A s is the heat transfer area, and
AT m is an appropriate average temperature difference between the two fluids.
Here the surface area A s can be determined precisely using the dimensions of
the heat exchanger. However, the overall heat transfer coefficient U and the
temperature difference AT between the hot and cold fluids, in general, are not
constant and vary along the heat exchanger.
The average value of the overall heat transfer coefficient can be determined
as described in the preceding section by using the average convection coeffi-
cients for each fluid. It turns out that the appropriate form of the mean tem-
perature difference between the two fluids is logarithmic in nature, and its
determination is presented in Section 13-4.
13^ - THE LOG MEAN TEMPERATURE
DIFFERENCE METHOD
Earlier, we mentioned that the temperature difference between the hot and
cold fluids varies along the heat exchanger, and it is convenient to have a
mean temperature difference AT m for use in the relation Q = UA S AT m .
In order to develop a relation for the equivalent average temperature differ-
ence between the two fluids, consider the parallel-flow double-pipe heat ex-
changer shown in Figure 13-14. Note that the temperature difference AT
between the hot and cold fluids is large at the inlet of the heat exchanger but
decreases exponentially toward the outlet. As you would expect, the tempera-
ture of the hot fluid decreases and the temperature of the cold fluid increases
along the heat exchanger, but the temperature of the cold fluid can never
exceed that of the hot fluid no matter how long the heat exchanger is.
Assuming the outer surface of the heat exchanger to be well insulated so
that any heat transfer occurs between the two fluids, and disregarding any
cen58933_chl3.qxd 9/9/2002 9:57 AM Page 681
changes in kinetic and potential energy, an energy balance on each fluid in a
differential section of the heat exchanger can be expressed as
so
-m h C ph dT h
and
&G = rh c C dT c
(13-16)
(13-17)
That is, the rate of heat loss from the hot fluid at any section of a heat ex-
changer is equal to the rate of heat gain by the cold fluid in that section. The
temperature change of the hot fluid is a negative quantity, and so a negative
sign is added to Eq. 13-16 to make the heat transfer rate Q a positive quantity.
Solving the equations above for clT h and dT c gives
dT„
and
dT r
Sg
m h C ph
_§e_
m c C pc
Taking their difference, we get
dT h - dT c = d(T h - T c )
-86
m h C ph
m,.C„
(13-18)
(13-19)
(13-20)
The rate of heat transfer in the differential section of the heat exchanger can
also be expressed as
8g = U(T„ - T c ) dA s
Substituting this equation into Eq. 13-20 and rearranging gives
d(T„-T c ) „ / 1 1
(13-21)
-UdA
m h C ph
m c C pc
(13-22)
Integrating from the inlet of the heat exchanger to its outlet, we obtain
In-
-UA
*■ h, in -* r, in \J^h ^ph ^c ^pc
1
(13-23)
Finally, solving Eqs. 13-9 and 13-10 for m c C pc and m h C ph and substituting into
Eq. 13-23 gives, after some rearrangement,
Q = UA S Ar ln
(13-24)
681
CHAPTER 13
SQ=U(T h -T c )dA s
Hot
fluid
-31
-dA s 2 A,
T k
c.out T
r d \
h, in
h, out
t
Cold fluid
T ■
c, in
FIGURE 13-14
Variation of the fluid temperatures in a
parallel-flow double-pipe heat
exchanger.
where
Ar,„
Ar, - Ar 2
In (A7,/Ar 2 )
(13-25)
is the log mean temperature difference, which is the suitable form of the
average temperature difference for use in the analysis of heat exchangers.
Here AT^ and AT 2 represent the temperature difference between the two fluids
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682
HEAT TRANSFER
4 -*f,out
Hot
fluid
vA7;
AT
Cold fluid
(a) Parallel-flow heat exchangers
Cold
fluid
Hot
fluid
.AT,
¥
T
c, out
A7 'l= 7 /,,i„- :r C ,o l ,t
^T 2 = T hoat -T c - m
1 h,oui
at:
(b) Counter-flow heat exchangers
FIGURE 13-15
The AT, and AT 2 expressions in
parallel-flow and counter-flow heat
exchangers.
at the two ends (inlet and outlet) of the heat exchanger. It makes no difference
which end of the heat exchanger is designated as the inlet or the outlet
(Fig. 13-15).
The temperature difference between the two fluids decreases from AT l at
the inlet to AT 2 at the outlet. Thus, it is tempting to use the arithmetic mean
temperature Ar am = \(AT { + AT 2 ) as the average temperature difference. The
logarithmic mean temperature difference Ar lm is obtained by tracing the ac-
tual temperature profile of the fluids along the heat exchanger and is an exact
representation of the average temperature difference between the hot and
cold fluids. It truly reflects the exponential decay of the local temperature
difference.
Note that AT lm is always less than AT am . Therefore, using Ar am in calcula-
tions instead of Ar lm will overestimate the rate of heat transfer in a heat ex-
changer between the two fluids. When ATi differs from AT 2 by no more than
40 percent, the error in using the arithmetic mean temperature difference is
less than 1 percent. But the error increases to undesirable levels when AT,
differs from AT 2 by greater amounts. Therefore, we should always use the
logarithmic mean temperature difference when determining the rate of heat
transfer in a heat exchanger.
Counter-Flow Heat Exchangers
The variation of temperatures of hot and cold fluids in a counter-flow heat ex-
changer is given in Figure 13-16. Note that the hot and cold fluids enter the
heat exchanger from opposite ends, and the outlet temperature of the cold
fluid in this case may exceed the outlet temperature of the hot fluid. In the lim-
iting case, the cold fluid will be heated to the inlet temperature of the hot fluid.
However, the outlet temperature of the cold fluid can never exceed the inlet
temperature of the hot fluid, since this would be a violation of the second law
of thermodynamics.
The relation above for the log mean temperature difference is developed us-
ing a parallel-flow heat exchanger, but we can show by repeating the analysis
above for a counter-flow heat exchanger that is also applicable to counter-
flow heat exchangers. But this time, AT X and AT 2 are expressed as shown in
Figure 13-15.
For specified inlet and outlet temperatures, the log mean temperature
difference for a counter-flow heat exchanger is always greater than that for a
parallel-flow heat exchanger. That is, AT im ^ CF > AT [mPF , and thus a smaller
surface area (and thus a smaller heat exchanger) is needed to achieve a speci-
fied heat transfer rate in a counter-flow heat exchanger. Therefore, it is com-
mon practice to use counter-flow arrangements in heat exchangers.
In a counter-flow heat exchanger, the temperature difference between the
hot and the cold fluids will remain constant along the heat exchanger when
the heat capacity rates of the two fluids are equal (that is, AT = constant
when C h = C c or m h C ph = m c C pc ). Then we have A 7^ = AT 2 , and the last log
mean temperature difference relation gives AT lm = jj, which is indeterminate.
It can be shown by the application of l'Hopital's rule that in this case we have
Ar ta = ATi = &T 2 , as expected.
A condenser or a boiler can be considered to be either a parallel- or counter-
flow heat exchanger since both approaches give the same result.
cen58933_chl3.qxd 9/9/2002 9:57 AM Page 683
Multipass and Cross-Flow Heat Exchangers:
Use of a Correction Factor
The log mean temperature difference AT lm relation developed earlier is limited
to parallel-flow and counter- flow heat exchangers only. Similar relations are
also developed for cross-flow and multipass shell-and-tube heat exchangers,
but the resulting expressions are too complicated because of the complex flow
conditions.
In such cases, it is convenient to relate the equivalent temperature dif-
ference to the log mean temperature difference relation for the counter-flow
case as
683
CHAPTER 13
Ar,„
f Ar,„
(13-26)
where F is the correction factor, which depends on the geometry of the heat
exchanger and the inlet and outlet temperatures of the hot and cold fluid
streams. The Ar lm CF is the log mean temperature difference for the case of
a counter-flow heat exchanger with the same inlet and outlet temperatures
and is determined from Eq. 13-25 by taking AT X = T h in — T c out and AT 2 =
7/,, out" ^.mCFig. 13-17).
The correction factor is less than unity for a cross-flow and multipass shell-
and-tube heat exchanger. That is, F < 1. The limiting value of F = 1 corre-
sponds to the counter-flow heat exchanger. Thus, the correction factor F for a
heat exchanger is a measure of deviation of the AT Xm from the corresponding
values for the counter-flow case.
The correction factor F for common cross-flow and shell-and-tube heat ex-
changer configurations is given in Figure 13-18 versus two temperature ratios
P and R defined as
(13-27)
and
R
J 1 i 2 ^ P -'tube side
(mC p ) s]
(13-28)
where the subscripts 1 and 2 represent the inlet and outlet, respectively. Note
that for a shell-and-tube heat exchanger, T and t represent the shell- and
tube-side temperatures, respectively, as shown in the correction factor
charts. It makes no difference whether the hot or the cold fluid flows
through the shell or the tube. The determination of the correction factor F
requires the availability of the inlet and the outlet temperatures for both the
cold and hot fluids.
Note that the value of P ranges from to 1. The value of R, on the other
hand, ranges from to infinity, with R = corresponding to the phase-change
(condensation or boiling) on the shell-side and R — > °° to phase-change on the
tube side. The correction factor is F = 1 for both of these limiting cases.
Therefore, the correction factor for a condenser or boiler is F = 1, regardless
of the configuration of the heat exchanger.
Cold
} fluid
Hot
fluid
IB-
FIGURE 13-16
The variation of the fluid temperatures
in a counter-flow double-pipe heat
exchanger.
Cold | t c
fluid T
Hot
fluid
-QZ
AT*,
Cross-flow or multipass
shell-and-tube heat exchanger
Heat transfer rate:
Q = UA s FAT lmCF
where
^
,CF~
AT r -AT 2
ln(AT { /AT 2 )
Ar,
= T H,
-T
n c,out
A7,
- T -T
ft, out c,in
and F=... (Fig. 13-18)
FIGURE 13-17
The determination of the heat transfer
rate for cross-flow and multipass
shell-and-tube heat exchangers
using the correction factor.
cen58933_chl3.qxd 9/9/2002 9:57 AM Page 684
684
HEAT TRANSFER
^
S^,
^S-S;
$:
=r
^
— —
.
—-.
Vv
X
>
>
\
^
\
AT
\
^
\
>>
\
\
ii_
\
\
v
\
*
\
\
h
4.0 3.0 2.0
[If \~ L
.5
\
1.0 0.8
\ \
0.6
0.4
0.2
\
\
\
7
T-
\
\
\
\
R = .
\
\
\
I
'1 "
1
\
\
\
1
1.0
k, 0.9
| 0.8
=
O
% 0.7
(3 0.6
0.5
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
(a) One-shell pass and 2, 4, 6, etc. (any multiple of 2), tube passes
1.0
k. 0.9
Q
| 0.8
=
o
tj 0.7
<u
(3 0.6
0.5
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
(b) Two-shell passes and 4, 8, 12, etc. (any multiple of 4), tube passes
1.0
k, 0.9
p
| 0.8
=
o
tj 0.7
8 0.6
0.5
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
(c) Single-pass cross-flow with both fluids unmixed
I I
I I
^
\
s
.
\
\
\
R = 4.0 3.0
2.C
1.5
1.0 0.8 0.6 0.4 0.2
\|
3
\
\
\
\
\
\
\
7
T
1
\
\
]
K 1
• 1
\
I I
I I
I I
I I
>
^
^
-«,
■ — .
\
\
\
o
\
"S
s
\
s
\
\
\
\
\
R
= 4
n 3t
)
">(
\
) 1 5
\
n
n
8
16
0.4 0.2
1
I
\
L
\
\
N
\
\l\
\
\
s
V
\
\\
7
T
\
\
V
\
\
\
\
\
}
\
\
\
\
V
K 1
2 ~
'{
\
s
\
\
\
A
V
FIGURE 13-18
Correction factor F charts
for common shell-and-tube and
cross-flow heat exchangers (from
k, 0.9
3
I 0-8
=
o
t5 0.7
Q 0.6
0.5
\
h
4
\
3f
2.0
1
5
1
in
8
tfi
4
n
1
\
\|
\
\
\
\
\
\
\
\
\
)
\
\
7
T
|
\
\
\
\
-r=:
I
"
]
I
\
"
' 1
A
V
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 P --
Bowman, Mueller, and Nagle, Ref. 2). (A) Single-pass cross-flow with one fluid mixed and the other
unmixed
cen58933_chl3.qxd 9/9/2002 9:57 AM Page 685
685
CHAPTER 13
EXAMPLE 13-3 The Condensation of Steam in a Condenser
Steam in the condenser of a power plant is to be condensed at a temperature of
30°C with cooling water from a nearby lake, which enters the tubes of the con-
denser at 14 C C and leaves at 22°C. The surface area of the tubes is 45 m 2 , and
the overall heat transfer coefficient is 2100 W/m 2 • °C. Determine the mass flow
rate of the cooling water needed and the rate of condensation of the steam in
the condenser.
SOLUTION Steam is condensed by cooling water in the condenser of a power
plant. The mass flow rate of the cooling water and the rate of condensation are
to be determined.
Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well
insulated so that heat loss to the surroundings is negligible and thus heat trans-
fer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes
in the kinetic and potential energies of fluid streams are negligible. 4 There is
no fouling. 5 Fluid properties are constant.
Properties The heat of vaporization of water at 30°C is h fg = 2431 kJ/kg
and the specific heat of cold water at the average temperature of 18°C is
C p = 4184 J/kg • °C (Table A-9).
Analysis The schematic of the condenser is given in Figure 13-19. The con-
denser can be treated as a counter-flow heat exchanger since the temperature
of one of the fluids (the steam) remains constant.
The temperature difference between the steam and the cooling water at the
two ends of the condenser is
i — T h>in T c oul — (30
- 22)°C = 8°C
2 = **, out ~~ T c - m = (30 -
- 14)°C = 16°C
That is, the temperature difference between the two fluids varies from 8°C at
one end to 16°C at the other. The proper average temperature difference be-
tween the two fluids is the logarithmic mean temperature difference (not the
arithmetic), which is determined from
AT,.
AT, - AT,
8-16
In (Ar,/Ar 2 ) In (8/16)
11.5°C
This is a little less than the arithmetic mean temperature difference of
i(8 + 16) = 12°C. Then the heat transfer rate in the condenser is determined
from
Q = UA S AT lm = (2100 W/m 2 • °C)(45 m 2 )(11.5°C) = 1.087 X 10 6 W = 1087 kW
Therefore, the steam will lose heat at a rate of 1,087 kW as it flows through the
condenser, and the cooling water will gain practically all of it, since the con-
denser is well insulated.
The mass flow rate of the cooling water and the rate of the condensation of the
steam are determined from Q = [mC p (7" 0Ut - 7; n )] C00 | ing water = {mh fg ) steam to be
Q
cooling water (~t /y T \
'-'n \* out 1 m)
1,087 kJ/s
(4.184 kJ/kg ■ °C)(22 - 14)°C
32.5 kg/s
30°C
FIGURE 13-19
Schematic for Example 13-3.
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686
HEAT TRANSFER
and
Q_
1,087 kJ/s
2431 kJ/ka
0.45 kg/s
Therefore, we need to circulate about 72 kg of cooling water for each 1 kg of
steam condensing to remove the heat released during the condensation process.
EXAMPLE 13-4 Heating Water in a Counter-Flow Heat Exchanger
A counter-flow double-pipe heat exchanger is to heat water from 20°C to 80°C
at a rate of 1.2 kg/s. The heating is to be accomplished by geothermal water
available at 160°C at a mass flow rate of 2 kg/s. The inner tube is thin-walled
and has a diameter of 1.5 cm. If the overall heat transfer coefficient of the heat
exchanger is 640 W/m 2 • °C, determine the length of the heat exchanger re-
quired to achieve the desired heating.
Cold
water
^o:
20°C
1.2 kg/s
FIGURE 13
Schematic
Hot
geothermal
water 1
2 kg/s f<
160°C
80°C
V D= 1.5 cm
-20
for Example 13-4.
SOLUTION Water is heated in a counter-flow double-pipe heat exchanger by
geothermal water. The required length of the heat exchanger is to be determined.
Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well
insulated so that heat loss to the surroundings is negligible and thus heat trans-
fer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes
in the kinetic and potential energies of fluid streams are negligible. 4 There is
no fouling. 5 Fluid properties are constant.
Properties We take the specific heats of water and geothermal fluid to be 4.18
and 4.31 kJ/kg ■ C C, respectively.
Analysis The schematic of the heat exchanger is given in Figure 13-20. The
rate of heat transfer in the heat exchanger can be determined from
Q = [mC p (T 0M - r,„)] watcr = (1.2 kg/s)(4.18 kJ/kg • °C)(80 - 20)°C = 301 kW
Noting that all of this heat is supplied by the geothermal water, the outlet
temperature of the geothermal water is determined to be
Q = [mCJT m - T J]
geothermal
Q
mC„
160°C
125°C
301 kW
(2kg/s)(4.31kJ/kg -°C)
Knowing the inlet and outlet temperatures of both fluids, the logarithmic mean
temperature difference for this counter-flow heat exchanger becomes
and
AT,
Ar,
Ar,„
AT, - AT,
(160 - 80)°C
(125 - 20)°C
80 - 105
80°C
105°C
92.0°C
In (AT,/Ar 2 ) In (80/105)
Then the surface area of the heat exchanger is determined to be
Q 301,000 W
Q = UA s AT ]m > A s
UAT lm (640 W/m 2 • °C)(92.0°C)
5.11 m 2
cen58933_chl3.qxd 9/9/2002 9:57 AM Page 687
To provide this much heat transfer surface area, the length of the tube must be
ttDL
A^ = 5.11m 2
itD ir(0.015 m)
108 m
Discussion The inner tube of this counter-flow heat exchanger (and thus the
heat exchanger itself) needs to be over 100 m long to achieve the desired heat
transfer, which is impractical. In cases like this, we need to use a plate heat
exchanger or a multipass shell-and-tube heat exchanger with multiple passes of
tube bundles.
687
CHAPTER 13
EXAMPLE 13-5
Heating of Glycerin in a
Multipass Heat Exchanger
A 2-shell passes and 4-tube passes heat exchanger is used to heat glycerin from
20°C to 50°C by hot water, which enters the thin-walled 2-cm-diameter tubes
at 80°C and leaves at 40°C (Fig. 13-21). The total length of the tubes in the
heat exchanger is 60 m. The convection heat transfer coefficient is 25 W/m 2 •
°C on the glycerin (shell) side and 160 W/m 2 ■ °C on the water (tube) side. De-
termine the rate of heat transfer in the heat exchanger (a) before any fouling oc-
curs and (£>) after fouling with a fouling factor of 0.0006 m 2 • °C/W occurs on
the outer surfaces of the tubes.
SOLUTION Glycerin is heated in a 2-shell passes and 4-tube passes heat
exchanger by hot water. The rate of heat transfer for the cases of fouling and no
fouling are to be determined.
Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well
insulated so that heat loss to the surroundings is negligible and thus heat trans-
fer from the hot fluid is equal to heat transfer to the cold fluid. 3 Changes in the
kinetic and potential energies of fluid streams are negligible. 4 Heat transfer co-
efficients and fouling factors are constant and uniform. 5 The thermal resis-
tance of the inner tube is negligible since the tube is thin-walled and highly
conductive.
Analysis The tubes are said to be thin-walled, and thus it is reasonable to
assume the inner and outer surface areas of the tubes to be equal. Then the
heat transfer surface area becomes
A s = nDL = tt(0.02 m)(60 m) = 3.77 m 2
The rate of heat transfer in this heat exchanger can be determined from
Q = UA s FAT lm , CF
where Fis the correction factor and A7^ m CF is the log mean temperature differ-
ence for the counter-flow arrangement. These two quantities are determined
from
Aio — If, nilt
T c , ont = (80 - 50)°C = 30°C
T c m = (40 -20)°C = 20°C
Ar,„
Ar, - Ar,
30 - 20
In (Ar,/A7 2 ) In (30/20)
24.7°C
Cold
glycerin
20°C
,Jl_
-+ —
— ft
40 C ^
Hot /(
i;
—
=*=»
rc\
50°C
FIGURE 13-21
Schematic for Example 13-5.
cen58933_chl3.qxd 9/9/2002 9:57 AM Page 688
688
HEAT TRANSFER
and
R =
l 2 H
T, - T 2
40-
- 80
20-
- 80
20-
-50
40- 80
0.67
= 0.75
}F= 0.91
(Fig. 13-1 1
(a) In the case of no fouling, the overall heat transfer coefficient U is deter-
mined from
U
1
1
1 , 1
1
1
21.6 W/m 2 -°C
hi h„ 160W/m 2 - °C 25 W/m 2 ■ °C
Then the rate of heat transfer becomes
Q = UA s FAT lmCF = (21.6 W/m 2 ■ °C)(3.77m 2 )(0.91)(24.7°C) = 1830 W
(b) When there is fouling on one of the surfaces, the overall heat transfer coef-
ficient U is
U
1
1
i + 1
h, h n
R
1
l /
+
1
160W/m 2 -°C 25W/m 2 -°C
+ 0.0006 m 2 ■ °C/W
= 21.3 W/m 2 -°C
The rate of heat transfer in this case becomes
Q = UA s FAT lmCF = (21.3 W/m 2 ■ °C)(3.77 m 2 )(0.91)(24.7°C) = 1805 W
Discussion Note that the rate of heat transfer decreases as a result of fouling,
as expected. The decrease is not dramatic, however, because of the relatively
low convection heat transfer coefficients involved.
EXAMPLE 13-6 Cooling of an Automotive Radiator
A test is conducted to determine the overall heat transfer coefficient in an au-
tomotive radiator that is a compact cross-flow water-to-air heat exchanger with
both fluids (air and water) unmixed (Fig. 13-22). The radiator has 40 tubes of
internal diameter 0.5 cm and length 65 cm in a closely spaced plate-finned
matrix. Hot water enters the tubes at 90°C at a rate of 0.6 kg/s and leaves at
65°C. Air flows across the radiator through the interfin spaces and is heated
from 20°C to 40°C. Determine the overall heat transfer coefficient L/,-of this ra-
diator based on the inner surface area of the tubes.
SOLUTION During an experiment involving an automotive radiator, the inlet
and exit temperatures of water and air and the mass flow rate of water are mea-
sured. The overall heat transfer coefficient based on the inner surface area is to
be determined.
Assumptions 1 Steady operating conditions exist. 2 Changes in the kinetic
and potential energies of fluid streams are negligible. 3 Fluid properties are
constant.
cen58933_chl3.qxd 9/9/2002 9:57 AM Page 689
Air flow
(unmixed)
20°C
90°C
XX
-40°C
65°C
Water flow
(unmixed)
689
CHAPTER 13
FIGURE 13-22
Schematic for Example 13-6.
Properties The specific heat of water at the average temperature of (90 + 65)/
2 = 77.5°C is 4.195 kJ/kg ■ °C.
Analysis The rate of heat transfer in this radiator from the hot water to the air
is determined from an energy balance on water flow,
Q = [mC p (T m - r out )] watcr = (0.6 kg/s)(4.195 kJ/kg • °C)(90 - 65)°C = 62.93 kW
The tube-side heat transfer area is the total surface area of the tubes, and is
determined from
A, = wttD,L = (40)tt(0.005 m)(0.65 m) = 0.408 m 2
Knowing the rate of heat transfer and the surface area, the overall heat transfer
coefficient can be determined from
Q = U i A i FAT ln
U;
Q
A;FAT ln
where Fis the correction factor and A7^ m CF is the log mean temperature differ-
ence for the counter-flow arrangement. These two quantities are found to be
A7\ =
AT 2 =
AT l[Rt CF
h, out
AT,
T c ,ou t = (90 - 40)°C = 50°C
T c<in = (65 -20)°C = 45°C
- Ar 2 50 - 45
In (AT { IAT 2 ) In (50/45)
47.6°C
and
P
R =
h
- h
T,
-h
r,
- T 2
65 -
-90
20 -
-90
20 -
-40
65 - 90
0.36
0.80
>F = 0.97
(Fig. 13-1 8c)
Substituting, the overall heat transfer coefficient U, is determined to be
Q 62,930 W
U,
A t F AT lmCF (0.408 m 2 )(0.97)(47.6°C)
3341 W/m 2
Note that the overall heat transfer coefficient on the air side will be much lower
because of the large surface area involved on that side.
cen58933_chl3.qxd 9/9/2002 9:57 AM Page 69C
690
HEAT TRANSFER
13-5 ■ THE EFFECTIVENESS-NTU METHOD
The log mean temperature difference (LMTD) method discussed in Section
13-4 is easy to use in heat exchanger analysis when the inlet and the outlet
temperatures of the hot and cold fluids are known or can be determined from
an energy balance. Once AT lm , the mass flow rates, and the overall heat trans-
fer coefficient are available, the heat transfer surface area of the heat ex-
changer can be determined from
Q = UA S Ar lra
Therefore, the LMTD method is very suitable for determining the size
of a heat exchanger to realize prescribed outlet temperatures when the mass
flow rates and the inlet and outlet temperatures of the hot and cold fluids are
specified.
With the LMTD method, the task is to select a heat exchanger that will meet
the prescribed heat transfer requirements. The procedure to be followed by the
selection process is:
1. Select the type of heat exchanger suitable for the application.
2. Determine any unknown inlet or outlet temperature and the heat transfer
rate using an energy balance.
3. Calculate the log mean temperature difference Ar lm and the correction
factor F, if necessary.
4. Obtain (select or calculate) the value of the overall heat transfer co-
efficient U.
5. Calculate the heat transfer surface area A s .
The task is completed by selecting a heat exchanger that has a heat transfer
surface area equal to or larger than A s .
A second kind of problem encountered in heat exchanger analysis is the de-
termination of the heat transfer rate and the outlet temperatures of the hot and
cold fluids for prescribed fluid mass flow rates and inlet temperatures when
the type and size of the heat exchanger are specified. The heat transfer surface
area A of the heat exchanger in this case is known, but the outlet temperatures
are not. Here the task is to determine the heat transfer performance of a spec-
ified heat exchanger or to determine if a heat exchanger available in storage
will do the job.
The LMTD method could still be used for this alternative problem, but the
procedure would require tedious iterations, and thus it is not practical. In an
attempt to eliminate the iterations from the solution of such problems, Kays
and London came up with a method in 1955 called the effectiveness-NTU
method, which greatly simplified heat exchanger analysis.
This method is based on a dimensionless parameter called the heat trans-
fer effectiveness e, defined as
Q Actual heat transfer rate
<2n, ax Maximum possible heat transfer rate
The actual heat transfer rate in a heat exchanger can be determined from an
energy balance on the hot or cold fluids and can be expressed as
Q = C C (T C , out - r Ci J = C h (T K in - T K „J (1 3-30)
cen58933_chl3.qxd 9/9/2002 9:57 AM Page 691
where C c = rh c C pc and C h = th c C ph are the heat capacity rates of the cold and
the hot fluids, respectively.
To determine the maximum possible heat transfer rate in a heat exchanger,
we first recognize that the maximum temperature difference in a heat ex-
changer is the difference between the inlet temperatures of the hot and cold
fluids. That is,
AT
(13-31)
The heat transfer in a heat exchanger will reach its maximum value when
(1) the cold fluid is heated to the inlet temperature of the hot fluid or (2) the
hot fluid is cooled to the inlet temperature of the cold fluid. These two limit-
ing conditions will not be reached simultaneously unless the heat capacity
rates of the hot and cold fluids are identical (i.e., C c = C h ). When C c + C h ,
which is usually the case, the fluid with the smaller heat capacity rate will ex-
perience a larger temperature change, and thus it will be the first to experience
the maximum temperature, at which point the heat transfer will come to a halt.
Therefore, the maximum possible heat transfer rate in a heat exchanger is
(Fig. 13-23)
where C min is the smaller of C h =
clarified by the following example.
!cmax ^'minv-' /;. in -* c, iny
m h C ph and C c
(13-32)
rh c C pc . This is further
EXAMPLE 13-7 Upper Limit for Heat Transfer in a
Heat Exchanger
Cold water enters a counter-flow heat exchanger at 10°C at a rate of 8 kg/s,
where it is heated by a hot water stream that enters the heat exchanger at 70°C
at a rate of 2 kg/s. Assuming the specific heat of water to remain constant at
C p = 4.18 kJ/kg • °C, determine the maximum heat transfer rate and the outlet
temperatures of the cold and the hot water streams for this limiting case.
SOLUTION Cold and hot water streams enter a heat exchanger at specified
temperatures and flow rates. The maximum rate of heat transfer in the heat ex-
changer is to be determined.
Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well
insulated so that heat loss to the surroundings is negligible and thus heat trans-
fer from the hot fluid is equal to heat transfer to the cold fluid. 3 Changes in the
kinetic and potential energies of fluid streams are negligible. 4 Heat transfer co-
efficients and fouling factors are constant and uniform. 5 The thermal resis-
tance of the inner tube is negligible since the tube is thin-walled and highly
conductive.
Properties The specific heat of water is given to be C p = 4.18 kJ/kg • °C.
Analysis A schematic of the heat exchanger is given in Figure 13-24. The heat
capacity rates of the hot and cold fluids are determined from
and
C h = m h C ph = (2 kg/s)(4.18 kJ/kg ■ °C) = 8.36 kW/°C
C c = m c C pc = (8 kg/s)(4.18 kJ/kg ■ °C) = 33.4 kW/°C
691
CHAPTER 13
20°C Cold
25 kg/s \ water
Hot
oil
130°C
40 kg/s
J>
C c = m c C pc = 104.5 kW/°C
C„ = m c C pl =92kWrC
-C AT
10,120 kW
FIGURE 13-23
The determination of the maximum
rate of heat transfer in a heat
exchanger.
in°c Cold
8 kg/s \ water
Hot
water
70°C
2 kg/s
D
V
FIGURE 13-24
Schematic for Example 13-7.
cen58933_chl3.qxd 9/9/2002 9:57 AM Page 692
692
HEAT TRANSFER
m r ,C
Cold
P C L fluid
Hot
fluid
X
m h- c P h m
D
e =
m h C P h
^h
=
m c C pc
Ar,
if
m c C pc
= m h
C ri:
then
AT,=
AT,
FIGURE 13-25
The temperature rise of the cold fluid
in a heat exchanger will be equal to
the temperature drop of the hot fluid
when the mass flow rates and the
specific heats of the hot and cold
fluids are identical.
Therefore
C m
C = 8.36 kW/°C
which is the smaller of the two heat capacity rates. Then the maximum heat
transfer rate is determined from Eq. 13-32 to be
xi max min\ J h, in c, in/
= (8.36 kW/°C)(70 - 10)°C
= 502 kW
That is, the maximum possible heat transfer rate in this heat exchanger is
502 kW. This value would be approached in a counter-flow heat exchanger with
a very large heat transfer surface area.
The maximum temperature difference in this heat exchanger is A7" max =
' h in ' r
(70 - 10)°C = 60°C. Therefore, the hot water cannot be cooled
by more than 60 C C (to 10°C) in this heat exchanger, and the cold water cannot
be heated by more than 60 C C (to 70°C), no matter what we do. The outlet tem-
peratures of the cold and the hot streams in this limiting case are determined
to be
\J ^ ( \L C out L c in J
2 = C h {T h - tn — T/ lout )
_> T =T + 6 = 10 oc + 502kW
^,out J r ,,n c iu ^ 33.4kW/°C
25°C
_> T = T
h, out k, in f
Q 70°C-^^ = 10 o C
8.38 kW/°C
Discussion Note that the hot water is cooled to the limit of 10°C (the inlet
temperature of the cold water stream), but the cold water is heated to 25°C only
when maximum heat transfer occurs in the heat exchanger. This is not surpris-
ing, since the mass flow rate of the hot water is only one-fourth that of the cold
water, and, as a result, the temperature of the cold water increases by 0.25°C
for each 1°C drop in the temperature of the hot water.
You may be tempted to think that the cold water should be heated to 70°C in
the limiting case of maximum heat transfer. But this will require the tempera-
ture of the hot water to drop to -170°C (below 10°C), which is impossible.
Therefore, heat transfer in a heat exchanger reaches its maximum value when
the fluid with the smaller heat capacity rate (or the smaller mass flow rate when
both fluids have the same specific heat value) experiences the maximum tem-
perature change. This example explains why we use C min in the evaluation of
Qmax instead of C max .
We can show that the hot water will leave at the inlet temperature of the cold
water and vice versa in the limiting case of maximum heat transfer when the mass
flow rates of the hot and cold water streams are identical (Fig. 13-25). We can
also show that the outlet temperature of the cold water will reach the 70°C limit
when the mass flow rate of the hot water is greater than that of the cold water.
The determination of g max requires the availability of the inlet temperature
of the hot and cold fluids and their mass flow rates, which are usually speci-
fied. Then, once the effectiveness of the heat exchanger is known, the actual
heat transfer rate Q can be determined from
G = eg n
e *-'minA-'/7,
T ■ )
-* c, in/
(13-33)
cen58933_chl3.qxd 9/9/2002 9:57 AM Page 693
Therefore, the effectiveness of a heat exchanger enables us to determine the
heat transfer rate without knowing the outlet temperatures of the fluids.
The effectiveness of a heat exchanger depends on the geometry of the heat
exchanger as well as the flow arrangement. Therefore, different types of heat
exchangers have different effectiveness relations. Below we illustrate the de-
velopment of the effectiveness e relation for the double-pipe parallel-flow
heat exchanger.
Equation 13-23 developed in Section 13-4 for a parallel-flow heat ex-
changer can be rearranged as
693
CHAPTER 13
T — T
-* h, out ■* c, out
-* h, in J c, in
UA S I C (
, solving Eq. 13-30 for T h out gives
c„
T = T (T — T \
A, out h, in /-• \ c, out c, in/
(13-34)
(13-35)
Substituting this relation into Eq. 13-34 after adding and subtracting
T c , hl gives
C
Th, in T c ; n
In-
T — T -(T — T )
c, in c, out r^ V c , out c,\nJ
H
T — T
/;, in c, in
c c c h
which simplifies to
In
C \ T — T
^h I h, in c , in
UAs
C r
1 +
(13-36)
We now manipulate the definition of effectiveness to obtain
Q ^cVV,out * c, in/
l[/ maN min\ A,in c, in/
T r
c, in '-rnin
C
Substituting this result into Eq. 13-36 and solving for e gives the following
relation for the effectiveness of a parallel-flow heat exchanger:
1 — exp
UAs
C,
^parallel flow
c \ c
(13-37)
Taking either C c or C h to be C min (both approaches give the same result), the
relation above can be expressed more conveniently as
1 — exp
UK
c
r
v -'min
c
^parallel flow
1 +
(13-38)
Again C min is the smaller heat capacity ratio and C max is the larger one, and it
makes no difference whether C min belongs to the hot or cold fluid.
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HEAT TRANSFER
Effectiveness relations of the heat exchangers typically involve the dimen-
sionless group UA S IC m[n . This quantity is called the number of transfer units
NTU and is expressed as
UA 1= _UA ±
NTU
(13-39)
(mC p ) miB
where U is the overall heat transfer coefficient and A s is the heat transfer surface
area of the heat exchanger. Note that NTU is proportional to A s . Therefore, for
specified values of U and C min , the value of NTU is a measure of the heat trans-
fer surface area A s . Thus, the larger the NTU, the larger the heat exchanger.
In heat exchanger analysis, it is also convenient to define another dimen-
sionless quantity called the capacity ratio c as
(13-40)
It can be shown that the effectiveness of a heat exchanger is a function of the
number of transfer units NTU and the capacity ratio c. That is,
e = function (UA S /C ml „, C mw /C mn% ) = function (NTU, c)
Effectiveness relations have been developed for a large number of heat ex-
changers, and the results are given in Table 13-4. The effectivenesses of some
common types of heat exchangers are also plotted in Figure 13-26. More
TABLE 13-4
Effectiveness relations for heat exchangers: NTU = UA S /C m] „ and
c = C min /C max = (mC p ) min /(mC p ) max (Kays and London, Ref. 5.)
Heat exchanger
type
Effectiveness relation
1 Double pipe:
Parallel-flow
Counter-flow
2 Shell and tube:
One-shell pass
2, 4, . . . tube
passes
3 Cross- flow
{single-pass)
Both fluids
unmixed
4 All heat
exchangers
with c =
1 - exp[-NTU(l + c)]
1 + c
1 -exp [-NTU(1 - c)]
1 - cexp[-NTU(l - c)]
1 + exp [-NTUV1 + c 2 ]]" 1
e = 2 1 + c + Vl + c 2
1 - exp[-NTUVl + c 2 ]
e = 1 - exp
f NTU c
[exp(-c NTU 078 )- 1]
e = i(l - exp [l-c[l - exp (-NTU)]})
e = 1 - exp] -|[1 - exp (-c NTU)]
e = 1 - exp(-NTU)
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CHAPTER 13
100
60
40
Eh
a
20
i ^
4>&f
c x
' 0^-
"O^O.
.- — 0?75
■ — Too'
Tube
t
r
-
flui
d 1 — 1
-0 — ■:
— »-
—
-
u
f Shell fluid
12 3 4 5
Number of transfer units NTU = A.U/C-
(a) Parallel-flow
^0
80
jr;
0^
G^
.0
*
*-"0~T5
"' 60
-"Too"
>
% 40
'
Shell fluid
-
**d
1
'f 1 -
20
N *=
1
Tube fluid
'l 1
n
J
1
12 3 4 5
Number of transfer units NTU = AU/C ^
(c) One-shell pass and 2, 4, 6, ... tube passes
100
12 3 4 5
Number of transfer units NTU = AVIC„-
(e) Cross-flow with both fluids unmixed
1 00
60
40
20
y
O
i^fe c
Shell fluid
Tube p
flu
d 1 — '1
— »-
I
-m —
r
111
12 3 4 5
Number of transfer units NTU =A,UIC.
(b) Counter-flow
100
12 3 4 5
Number of transfer units NTU = A s U/C mln
(d) Two-shell passes and 4, 8, 12, ... tube passes
100
12 3 4 5
Number of transfer units NTU = A S U /C ■
(f) Cross-flow with one fluid mixed and the
other unmixed
FIGURE 13-26
Effectiveness for heat exchangers (from Kays and London, Ref. 5).
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696
HEAT TRANSFER
Counter-flow
FIGURE 13-27
For a specified NTU and capacity
ratio c, the counter-flow heat
exchanger has the highest
effectiveness and the parallel-flow the
lowest.
0.5
/^— e = 1 - e
-NTU
/ (All heat exchangers
/ with c =
= 0)
1 1
2
NTU:
3
UAJC„
FIGURE 13-28
The effectiveness relation reduces to
e = e max = 1 — exp(— NTU) for all
heat exchangers when the capacity
ratio c = 0.
extensive effectiveness charts and relations are available in the literature. The
dashed lines in Figure 13-26/ are for the case of C min unmixed and C max mixed
and the solid lines are for the opposite case. The analytic relations for the ef-
fectiveness give more accurate results than the charts, since reading errors in
charts are unavoidable, and the relations are very suitable for computerized
analysis of heat exchangers.
We make these following observations from the effectiveness relations and
charts already given:
1
The value of the effectiveness ranges from to 1. It increases rapidly
with NTU for small values (up to about NTU = 1 .5) but rather slowly
for larger values. Therefore, the use of a heat exchanger with a large
NTU (usually larger than 3) and thus a large size cannot be justified
economically, since a large increase in NTU in this case corresponds to
a small increase in effectiveness. Thus, a heat exchanger with a very
high effectiveness may be highly desirable from a heat transfer point of
view but rather undesirable from an economical point of view.
For a given NTU and capacity ratio c = C min /C max , the counter-flow
heat exchanger has the highest effectiveness, followed closely by the
cross-flow heat exchangers with both fluids unmixed. As you might
expect, the lowest effectiveness values are encountered in parallel-flow
heat exchangers (Fig. 13-27).
The effectiveness of a heat exchanger is independent of the capacity
ratio c for NTU values of less than about 0.3.
The value of the capacity ratio c ranges between and 1 . For a given
NTU, the effectiveness becomes a maximum for c = and a minimum
for c = 1. The case c = C min /C max — > corresponds to C max — > °°, which
is realized during a phase-change process in a condenser or boiler. All
effectiveness relations in this case reduce to
exp(-NTU)
(13-41)
regardless of the type of heat exchanger (Fig. 13-28). Note that the
temperature of the condensing or boiling fluid remains constant in
this case. The effectiveness is the lowest in the other limiting case of
c = C min /C max = 1 , which is realized when the heat capacity rates of
the two fluids are equal.
Once the quantities c = C min /C max and NTU = UA S /C min have been evalu-
ated, the effectiveness s can be determined from either the charts or (prefer-
ably) the effectiveness relation for the specified type of heat exchanger. Then
the rate of heat transfer Q and the outlet temperatures T h out and T c out can be
determined from Eqs. 13-33 and 13-30, respectively. Note that the analysis of
heat exchangers with unknown outlet temperatures is a straightforward matter
with the effectiveness-NTU method but requires rather tedious iterations with
the LMTD method.
We mentioned earlier that when all the inlet and outlet temperatures
are specified, the size of the heat exchanger can easily be determined
using the LMTD method. Alternatively, it can also be determined from the
effectiveness-NTU method by first evaluating the effectiveness e from its
definition (Eq. 13-29) and then the NTU from the appropriate NTU relation in
Table 13-5.
cen58933_chl3.qxd 9/9/2002 9:57 AM Page 697
TABLE 13-5
(mC p ) mjn /(mC p ) max (Kays and London, Ref. 5.)
Heat exchanger type
NTU relation
1 Double-pipe:
Parallel-flow
Counter-flow
2 Shell and tube:
One-shell pass
2, 4, . . . tube passes
3 Cross- flow {single-pass)
C max mixed,
C min unmixed
C min mixed,
C max unmixed
4 All heat exchangers
with c =
NTU =
NTU =
NTU =
In [1 - e(l + c)]
1
In
1
1 + c
e- 1
EC
In
/2/e - 1 - c - VYT
NTU = -In
Vl + c 2 \2/e- 1
In (1 - ec)
+ VYT
1 +
NTU =
In [cln (1 - e)+ 1]
NTU = -ln(l - E )
Note that the relations in Table 13-5 are equivalent to those in Table 13-4.
Both sets of relations are given for convenience. The relations in Table 13-4
give the effectiveness directly when NTU is known, and the relations in
Table 13-5 give the NTU directly when the effectiveness e is known.
697
CHAPTER 13
EXAMPLE 13-8 Using the Effectiveness-NTU Method
Repeat Example 13-4, which was solved with the LMTD method, using the
effectiveness-NTU method.
SOLUTION The schematic of the heat exchanger is redrawn in Figure 13-29,
and the same assumptions are utilized.
Analysis In the effectiveness-NTU method, we first determine the heat capac-
ity rates of the hot and cold fluids and identify the smaller one:
m h C ph
■ AC P c
Therefore,
and
(2 kg/s)(4.31 kJ/kg • °C) = 8.62 kW/°C
(1.2 kg/s)(4.18 kJ/kg • °C) = 5.02 kW/°C
C mln = C= 5.02 kW/°C
c = C min /C max = 5.02/8.62 = 0.583
Then the maximum heat transfer rate is determined from Eq. 13-32 to be
ximax minv /z, in c, in/
= (5.02 kW/°C)(160 - 20)°C
= 702.8 kW
Cold
water
Hot
geothermal . 160°C
brine |
2kg/s
-KE
20°C
1.2kg/s
3>
V
80°C
~D= 1.5 cm
FIGURE 13-29
Schematic for Example 13-8.
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HEAT TRANSFER
That is, the maximum possible heat transfer rate in this heat exchanger is
702.8 kW. The actual rate of heat transfer in the heat exchanger is
Q = [mC p (T 0Ut - 7/ in )] water = (1.2 kg/s)(4.18 kJ/kg • °C)(80 - 20)°C= 301.0 kW
Thus, the effectiveness of the heat exchanger is
Q 301.0 kW
Qr
702.8 kW
0.428
Knowing the effectiveness, the NTU of this counter-flow heat exchanger can be
determined from Figure 13-26£> or the appropriate relation from Table 13-5.
We choose the latter approach for greater accuracy:
NTU
1
■In
1
I
■In
0.428 - 1
c - 1 \ec - \j 0.583 - 1 \ 0.428 X 0.583
Then the heat transfer surface area becomes
NTU
r
^min
A,
NTU C ir
(0.651)(5020 W/°C)
0.651
5.11m 2
U 640 W/m 2 • °C
To provide this much heat transfer surface area, the length of the tube must be
A s 5.11m 2
A.
ttDL
ttD TT(0.015m)
108 m
Discussion Note that we obtained the same result with the effectiveness-NTU
method in a systematic and straightforward manner.
150 ° C I aLg/s
H
C
C
c
J
J
c
3>
20°C
Y
Water
0.2 kg/s
FIGURE 13-30
Schematic for Example 13-9.
EXAMPLE 13-9 Cooling Hot Oil by Water in a Multipass
Heat Exchanger
Hot oil is to be cooled by water in a 1-shell-pass and 8-tube-passes heat
exchanger. The tubes are thin-walled and are made of copper with an internal
diameter of 1.4 cm. The length of each tube pass in the heat exchanger is 5 m,
and the overall heat transfer coefficient is 310 W/m 2 • °C. Water flows through
the tubes at a rate of 0.2 kg/s, and the oil through the shell at a rate of 0.3 kg/s.
The water and the oil enter at temperatures of 20°C and 150°C, respectively.
Determine the rate of heat transfer in the heat exchanger and the outlet tem-
peratures of the water and the oil.
SOLUTION Hot oil is to be cooled by water in a heat exchanger. The mass flow
rates and the inlet temperatures are given. The rate of heat transfer and the out-
let temperatures are to be determined.
Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well
insulated so that heat loss to the surroundings is negligible and thus heat trans-
fer from the hot fluid is equal to the heat transfer to the cold fluid. 3 The thick-
ness of the tube is negligible since it is thin-walled. 4 Changes in the kinetic
and potential energies of fluid streams are negligible. 5 The overall heat trans-
fer coefficient is constant and uniform.
Analysis The schematic of the heat exchanger is given in Figure 13-30. The
outlet temperatures are not specified, and they cannot be determined from an
cen58933_chl3.qxd 9/9/2002 9:57 AM Page 699
energy balance. The use of the LMTD method in this case will involve tedious
iterations, and thus the e-NTU method is indicated. The first step in the e-NTU
method is to determine the heat capacity rates of the hot and cold fluids and
identify the smaller one:
C h = m h C ph = (0.3 kg/s)(2.13 kJ/kg • °C) = 0.639 kW/°C
Therefore,
and
C c = m c C = (0.2 kg/s)(4.18 kJ/kg • °C) = 0.836 kW/°C
C m
0.639
0.836
0.764
Then the maximum heat transfer rate is determined from Eq. 13-32 to be
tiraax ^minv /y, in -* c, in/
= (0.639 kW/°C)(150 - 20)°C = 83.1 kW
That is, the maximum possible heat transfer rate in this heat exchanger is 83.1
kW. The heat transfer surface area is
A s = ii(ttDL) = 8tt(0.014 m)(5 m) = 1.76 m 2
Then the NTU of this heat exchanger becomes
UA S (310 W/m 2 • °C)(1.76 m 2 )
NTU = C"f = 639WC = °- 853
The effectiveness of this heat exchanger corresponding to c = 0.764 and
NTU = 0.853 is determined from Figure 13-26c to be
s = 0.47
We could also determine the effectiveness from the third relation in Table 13-4
more accurately but with more labor. Then the actual rate of heat transfer
becomes
e = e6 n
(0.47)(83.1 kW) = 39.1 kW
Finally, the outlet temperatures of the cold and the hot fluid streams are deter-
mined to be
G = c c (t; .-r )
Q ~ C h (T h - m T u om )
-> T„
-> T h
20°C +
150°C
e
c c
39.1 kW
0.836 kW/°C
Q
c„
39.1 kW
0.639 kW/°C
66.8°C
88.8°C
Therefore, the temperature of the cooling water will rise from 20°C to 66.8°C as
it cools the hot oil from 150°C to 88.8°C in this heat exchanger.
699
CHAPTER 13
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HEAT TRANSFER
13-6 - SELECTION OF HEAT EXCHANGERS
Heat exchangers are complicated devices, and the results obtained with the
simplified approaches presented above should be used with care. For example,
we assumed that the overall heat transfer coefficient U is constant throughout
the heat exchanger and that the convection heat transfer coefficients can be
predicted using the convection correlations. However, it should be kept in
mind that the uncertainty in the predicted value of U can even exceed 30 per-
cent. Thus, it is natural to tend to overdesign the heat exchangers in order to
avoid unpleasant surprises.
Heat transfer enhancement in heat exchangers is usually accompanied by
increased pressure drop, and thus higher pumping power. Therefore, any gain
from the enhancement in heat transfer should be weighed against the cost of
the accompanying pressure drop. Also, some thought should be given to
which fluid should pass through the tube side and which through the shell
side. Usually, the more viscous fluid is more suitable for the shell side (larger
passage area and thus lower pressure drop) and the fluid with the higher pres-
sure for the tube side.
Engineers in industry often find themselves in a position to select heat
exchangers to accomplish certain heat transfer tasks. Usually, the goal is to
heat or cool a certain fluid at a known mass flow rate and temperature to a
desired temperature. Thus, the rate of heat transfer in the prospective heat
exchanger is
t^max Wt-'pyl in ~ * out)
which gives the heat transfer requirement of the heat exchanger before having
any idea about the heat exchanger itself.
An engineer going through catalogs of heat exchanger manufacturers will
be overwhelmed by the type and number of readily available off-the-shelf heat
exchangers. The proper selection depends on several factors.
Heat Transfer Rate
This is the most important quantity in the selection of a heat exchanger. A heat
exchanger should be capable of transferring heat at the specified rate in order
to achieve the desired temperature change of the fluid at the specified mass
flow rate.
Cost
Budgetary limitations usually play an important role in the selection of heat
exchangers, except for some specialized cases where "money is no object."
An off-the-shelf heat exchanger has a definite cost advantage over those made
to order. However, in some cases, none of the existing heat exchangers will
do, and it may be necessary to undertake the expensive and time-consuming
task of designing and manufacturing a heat exchanger from scratch to suit the
needs. This is often the case when the heat exchanger is an integral part of the
overall device to be manufactured.
The operation and maintenance costs of the heat exchanger are also impor-
tant considerations in assessing the overall cost.
cen58933_chl3.qxd 9/9/2002 9:57 AM Page 701
Pumping Power
In a heat exchanger, both fluids are usually forced to flow by pumps or fans
that consume electrical power. The annual cost of electricity associated with
the operation of the pumps and fans can be determined from
Operating cost = (Pumping power, kW) X (Hours of operation, h)
X (Price of electricity, $/kWh)
where the pumping power is the total electrical power consumed by the
motors of the pumps and fans. For example, a heat exchanger that involves a
1-hp pump and a |-hp fan (1 hp = 0.746 kW) operating 8 h a day and 5 days
a week will consume 2017 kWh of electricity per year, which will cost $161.4
at an electricity cost of 8 cents/kWh.
Minimizing the pressure drop and the mass flow rate of the fluids will min-
imize the operating cost of the heat exchanger, but it will maximize the size of
the heat exchanger and thus the initial cost. As a rule of thumb, doubling the
mass flow rate will reduce the initial cost by half but will increase the pump-
ing power requirements by a factor of roughly eight.
Typically, fluid velocities encountered in heat exchangers range between 0.7
and 7 m/s for liquids and between 3 and 30 m/s for gases. Low veloci-
ties are helpful in avoiding erosion, tube vibrations, and noise as well as pres-
sure drop.
Size and Weight
Normally, the smaller and the lighter the heat exchanger, the better it is. This
is especially the case in the automotive and aerospace industries, where size
and weight requirements are most stringent. Also, a larger heat exchanger nor-
mally carries a higher price tag. The space available for the heat exchanger in
some cases limits the length of the tubes that can be used.
Type
The type of heat exchanger to be selected depends primarily on the type of
fluids involved, the size and weight limitations, and the presence of any phase-
change processes. For example, a heat exchanger is suitable to cool a liquid
by a gas if the surface area on the gas side is many times that on the liquid
side. On the other hand, a plate or shell-and-tube heat exchanger is very suit-
able for cooling a liquid by another liquid.
Materials
The materials used in the construction of the heat exchanger may be an im-
portant consideration in the selection of heat exchangers. For example, the
thermal and structural stress effects need not be considered at pressures below
15 atm or temperatures below 150°C. But these effects are major considera-
tions above 70 atm or 550°C and seriously limit the acceptable materials of
the heat exchanger.
A temperature difference of 50°C or more between the tubes and the shell
will probably pose differential thermal expansion problems and needs to be
considered. In the case of corrosive fluids, we may have to select expensive
701
CHAPTER 13
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HEAT TRANSFER
corrosion-resistant materials such as stainless steel or even titanium if we are
not willing to replace low-cost heat exchangers frequently.
Other Considerations
There are other considerations in the selection of heat exchangers that may
or may not be important, depending on the application. For example, being
leak-tight is an important consideration when toxic or expensive fluids are in-
volved. Ease of servicing, low maintenance cost, and safety and reliability are
some other important considerations in the selection process. Quietness is one
of the primary considerations in the selection of liquid-to-air heat exchangers
used in heating and air-conditioning applications.
Hot
water
(c
Cold
-_-— water
;^_15°C
I I I
FIGURE 13-31
Schematic for Example 13-10.
EXAMPLE 13-10
Installing a Heat Exchanger to Save Energy
and Money
In a dairy plant, milk is pasteurized by hot water supplied by a natural gas fur-
nace. The hot water is then discharged to an open floor drain at 80°C at a rate
of 15 kg/min. The plant operates 24 h a day and 365 days a year. The furnace
has an efficiency of 80 percent, and the cost of the natural gas is $0.40 per
therm (1 therm = 105,500 kJ). The average temperature of the cold water en-
tering the furnace throughout the year is 15°C. The drained hot water cannot be
returned to the furnace and recirculated, because it is contaminated during the
process.
In order to save energy, installation of a water-to-water heat exchanger to pre-
heat the incoming cold water by the drained hot water is proposed. Assuming
that the heat exchanger will recover 75 percent of the available heat in the hot
water, determine the heat transfer rating of the heat exchanger that needs to be
purchased and suggest a suitable type. Also, determine the amount of money
this heat exchanger will save the company per year from natural gas savings.
SOLUTION A water-to-water heat exchanger is to be installed to transfer energy
from drained hot water to the incoming cold water to preheat it. The rate of heat
transfer in the heat exchanger and the amount of energy and money saved per
year are to be determined.
Assumptions 1 Steady operating conditions exist. 2 The effectiveness of the
heat exchanger remains constant.
Properties We use the specific heat of water at room temperature, C p = 4.18 kJ7
kg ■ °C (Table A-9), and treat it as a constant.
Analysis A schematic of the prospective heat exchanger is given in Figure
13-31. The heat recovery from the hot water will be a maximum when it leaves
the heat exchanger at the inlet temperature of the cold water. Therefore,
Q
^h^py* h, in
15
60
67.9 kJ/s
T )
*■ c, in/
kg/s 1(4.18 kJ/kg • °C)(80 - 15)°C
That is, the existing hot water stream has the potential to supply heat at a rate
of 67.9 kJ/s to the incoming cold water. This value would be approached in a
counter-flow heat exchanger with a very large heat transfer surface area. A heat
exchanger of reasonable size and cost can capture 75 percent of this heat
cen58933_chl3.qxd 9/9/2002 9:57 AM Page 703
transfer potential. Thus, the heat transfer rating of the prospective heat ex-
changer must be
Q = sQ n
(0.75)(67.9 kJ/s) = 50.9 kj/s
That is, the heat exchanger should be able to deliver heat at a rate of 50.9 kJ/s
from the hot to the cold water. An ordinary plate or shell-and-tube heat exchanger
should be adequate for this purpose, since both sides of the heat exchanger in-
volve the same fluid at comparable flow rates and thus comparable heat transfer
coefficients. (Note that if we were heating air with hot water, we would have to
specify a heat exchanger that has a large surface area on the air side.)
The heat exchanger will operate 24 h a day and 365 days a year. Therefore,
the annual operating hours are
Operating hours = (24 h/day)(365 days/year) = 8760 h/year
Noting that this heat exchanger saves 50.9 kJ of energy per second, the energy
saved during an entire year will be
Energy saved = (Heat transfer rate)(Operation time)
= (50.9 kJ/s)(8760 h/year)(3600 s/h)
= 1.605 X 10"kJ/year
The furnace is said to be 80 percent efficient. That is, for each 80 units of heat
supplied by the furnace, natural gas with an energy content of 100 units must
be supplied to the furnace. Therefore, the energy savings determined above re-
sult in fuel savings in the amount of
Fuel saved
Energy saved 1.605 X 10 9 kJ/year / \ therm
O80 [ 105,500 kJ
Furnace efficiency
19,020 therms/year
Noting that the price of natural gas is $0.40 per therm, the amount of money
saved becomes
Money saved = (Fuel saved) X (Price of fuel)
= (19,020 therms/year)($0.40/therm)
= $7607/ year
Therefore, the installation of the proposed heat exchanger will save the com-
pany $7607 a year, and the installation cost of the heat exchanger will proba-
bly be paid from the fuel savings in a short time.
703
CHAPTER 13
SUMMARY
Heat exchangers are devices that allow the exchange of heat
between two fluids without allowing them to mix with each
other. Heat exchangers are manufactured in a variety of
types, the simplest being the double-pipe heat exchanger. In a
parallel-flow type, both the hot and cold fluids enter the heat
exchanger at the same end and move in the same direction,
whereas in a counter-flow type, the hot and cold fluids enter
the heat exchanger at opposite ends and flow in opposite
directions. In compact heat exchangers, the two fluids move
perpendicular to each other, and such a flow configuration is
called cross-flow. Other common types of heat exchangers in
industrial applications are the plate and the shell-and-tube heat
exchangers.
Heat transfer in a heat exchanger usually involves convection
in each fluid and conduction through the wall separating the two
fluids. In the analysis of heat exchangers, it is convenient to
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704
HEAT TRANSFER
work with an overall heat transfer coefficient U or a total ther-
mal resistance R, expressed as
1
UA,
1
U,A;
1
R
1
h;A;
R„
+
1
where the subscripts ?' and o stand for the inner and outer sur-
faces of the wall that separates the two fluids, respectively.
When the wall thickness of the tube is small and the thermal
conductivity of the tube material is high, the last relation sim-
plifies to
U
where U ~ U i ~ U . The effects of fouling on both the inner
and the outer surfaces of the tubes of a heat exchanger can be
accounted for by
= R
In (£>„/£>,-) | R f . | i
2-nkL A„ h n A„
1
UA
1
; u t A
1
U A
1
hi A;
+ A" +
ttD: L and A„ = ti
where A, = ttZ), L and A a = itD L are the areas of the inner and
outer surfaces and R f , and R f B are the fouling factors at those
surfaces.
In a well-insulated heat exchanger, the rate of heat transfer
from the hot fluid is equal to the rate of heat transfer to the cold
one. That is,
%J t?l c L- pc (l collt ^ c , in/ ^cV-'c.out *c.'m)
and
Q ~ m hC p h(T htin
t) — C h {T h i„ T h out)
where the subscripts c and h stand for the cold and hot fluids,
respectively, and the product of the mass flow rate and the spe-
cific heat of a fluid mC p is called the heat capacity rate.
Of the two methods used in the analysis of heat exchangers,
the log mean temperature difference (or LMTD) method is
best suited for determining the size of a heat exchanger
when all the inlet and the outlet temperatures are known. The
effectiveness-NTU method is best suited to predict the outlet
temperatures of the hot and cold fluid streams in a specified
heat exchanger. In the LMTD method, the rate of heat transfer
is determined from
Q = UA S A7 ln
where
AT,
AT, - AT 2
In (ATJAT 2 )
is the log mean temperature difference, which is the suitable
form of the average temperature difference for use in the analy-
sis of heat exchangers. Here AT { and AT 2 represent the temper-
ature differences between the two fluids at the two ends (inlet
and outlet) of the heat exchanger. For cross-flow and multipass
shell-and-tube heat exchangers, the logarithmic mean temper-
ature difference is related to the counter-flow one A71 m C f as
AT,
FAT„
where F is the correction factor, which depends on the geome-
try of the heat exchanger and the inlet and outlet temperatures
of the hot and cold fluid streams.
The effectiveness of a heat exchanger is defined as
Q
Qu
where
Actual heat transfer rate
Maximum possible heat transfer rate
s^max minv li. in c, in/
m h C ph and C c
m c C pc . The ef-
fectiveness of heat exchangers can be determined from effec-
tiveness relations or charts.
The selection or design of a heat exchanger depends on
several factors such as the heat transfer rate, cost, pressure
drop, size, weight, construction type, materials, and operating
environment.
REFERENCES AND SUGGESTED READING
1. N. Afgan and E. U. Schlunder. Heat Exchanger: Design
and Theory Sourcebook. Washington D.C.: McGraw-
Hill/Scripta, 1974.
2. R. A. Bowman, A. C. Mueller, and W. M. Nagle. "Mean
Temperature Difference in Design."' Transactions of the
ASME 62 (1940), p. 283.
3. A. P. Fraas. Heat Exchanger Design. 2d ed. New York:
John Wiley & Sons, 1989.
4. K. A. Gardner. "Variable Heat Transfer Rate Correction in
Multipass Exchangers, Shell Side Film Controlling."
Transactions of the ASME 67 (1945), pp. 31-38.
5. W. M. Kays and A. L. London. Compact Heat
Exchangers. 3rd ed. New York: McGraw-Hill, 1984.
6. W. M. Kays and H. C. Perkins. In Handbook of Heat
Transfer, ed. W. M. Rohsenow and J. P. Hartnett. New
York: McGraw-Hill, 1972, Chap. 7.
cen58933_chl3.qxd 9/9/2002 9:57 AM Page 705
7. A. C. Mueller. "Heat Exchangers." In Handbook of Heat
Transfer, ed. W. M. Rohsenow and J. P. Hartnett. New
York: McGraw-Hill, 1972, Chap. 18.
8. M. N. Ozisik. Heat Transfer — A Basic Approach. New
York: McGraw-Hill, 1985.
9. E. U. Schlunder. Heat Exchanger Design Handbook.
Washington, D.C.: Hemisphere, 1982.
10. Standards of Tubular Exchanger Manufacturers
Association. New York: Tubular Exchanger
Manufacturers Association, latest ed.
705
CHAPTER 13
11. R. A. Stevens, J. Fernandes, and J. R. Woolf. "Mean
Temperature Difference in One, Two, and Three Pass
Crossflow Heat Exchangers." Transactions of the ASME
79 (1957), pp. 287-297.
12. J. Taborek, G. F. Hewitt, and N. Afgan. Heat Exchangers:
Theory and Practice. New York: Hemisphere, 1983.
13. G. Walker. Industrial Heat Exchangers. Washington,
D.C.: Hemisphere, 1982.
PROBLEMS
Types of Heat Exchangers
13-1C Classify heat exchangers according to flow type and
explain the characteristics of each type.
13-2C Classify heat exchangers according to construction
type and explain the characteristics of each type.
13-3C When is a heat exchanger classified as being com-
pact? Do you think a double-pipe heat exchanger can be classi-
fied as a compact heat exchanger?
13-4C How does a cross-flow heat exchanger differ from a
counter-flow one? What is the difference between mixed and
unmixed fluids in cross-flow?
13-5C What is the role of the baffles in a shell-and-tube heat
exchanger? How does the presence of baffles affect the heat
transfer and the pumping power requirements? Explain.
13-6C Draw a 1 -shell-pass and 6-tube-passes shell-and-tube
heat exchanger. What are the advantages and disadvantages of
using 6 tube passes instead of just 2 of the same diameter?
13-7C Draw a 2-shell-passes and 8-tube-passes shell-and-
tube heat exchanger. What is the primary reason for using so
many tube passes?
13-8C What is a regenerative heat exchanger? How does a
static type of regenerative heat exchanger differ from a dy-
namic type?
The Overall Heat Transfer Coefficient
13-9C What are the heat transfer mechanisms involved dur-
ing heat transfer from the hot to the cold fluid?
*Problems designated by a "C" are concept questions, and
students are encouraged to answer them all. Problems designated
by an "E" are in English units, and the SI users can ignore them.
Problems with an EES-CD icon ® are solved using EES, and
complete solutions together with parametric studies are included
on the enclosed CD. Problems with a computer-EES icon H are
comprehensive in nature, and are intended to be solved with a
computer, preferably using the EES software that accompanies
this text.
13-10C Under what conditions is the thermal resistance of
the tube in a heat exchanger negligible?
13-11C Consider a double-pipe parallel-flow heat exchanger
of length L. The inner and outer diameters of the inner tube are
Dj and D 2 , respectively, and the inner diameter of the outer
tube is D 3 . Explain how you would determine the two heat
transfer surface areas A ;
assume A,- ~ A„~ AJl
and A„. When is it reasonable to
13-12C Is the approximation h : ~ h ~ h for the convection
heat transfer coefficient in a heat exchanger a reasonable one
when the thickness of the tube wall is negligible?
13-13C Under what conditions can the overall heat transfer
coefficient of a heat exchanger be determined from U = {\lh {
+ 1/AJ- 1 ?
13-14C What are the restrictions on the relation UA S = l/ ; A,
= U A for a heat exchanger? Here A s is the heat transfer sur-
face area and U is the overall heat transfer coefficient.
13-1 5C In a thin-walled double-pipe heat exchanger, when
is the approximation U = h : a reasonable one? Here U is the
overall heat transfer coefficient and h t is the convection heat
transfer coefficient inside the tube.
13-16C What are the common causes of fouling in a heat
exchanger? How does fouling affect heat transfer and pres-
sure drop?
13-17C How is the thermal resistance due to fouling in a
heat exchanger accounted for? How do the fluid velocity and
temperature affect fouling?
13-18 A double-pipe heat exchanger is constructed of a cop-
per (k = 380 W/m • °C) inner tube of internal diameter Z), =
1 .2 cm and external diameter D = 1 .6 cm and an outer tube of
diameter 3.0 cm. The convection heat transfer coefficient is re-
ported to be h t = 700 W/m 2 ■ °C on the inner surface of the
tube and h = 1400 W/m 2 • °C on its outer surface. For a foul-
ing factor R f , = 0.0005 m 2 ■ °C/W on the tube side and R f =
0.0002 m 2 ■ °C/W on the shell side, determine (a) the thermal
resistance of the heat exchanger per unit length and (b) the
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706
HEAT TRANSFER
overall heat transfer coefficients £/,- and U a based on the inner
and outer surface areas of the tube, respectively.
13-19 fitt'M Reconsider Problem 13-18. Using EES (or
k^S other) software, investigate the effects of pipe
conductivity and heat transfer coefficients on the thermal resis-
tance of the heat exchanger. Let the thermal conductivity vary
from 10 W/m • °C to 400 W/m ■ °C, the convection heat trans-
fer coefficient from 500 W/m 2 ■ °C to 1500 W/m 2 • °C on the in-
ner surface, and from 1000 W/m 2 • °C to 2000 W/m 2 • °C on the
outer surface. Plot the thermal resistance of the heat exchanger
as functions of thermal conductivity and heat transfer coeffi-
cients, and discuss the results.
13-20 Water at an average temperature of 107°C and an av-
erage velocity of 3.5 m/s flows through a 5-m-long stainless
steel tube (k = 14.2 W/m ■ °C) in a boiler. The inner and outer
diameters of the tube are D, = 1.0 cm and D a = 1 .4 cm, re-
spectively. If the convection heat transfer coefficient at the
outer surface of the tube where boiling is taking place is h =
8400 W/m 2 • °C, determine the overall heat transfer coefficient
Uj of this boiler based on the inner surface area of the tube.
13-21 Repeat Problem 13-20, assuming a fouling factor
R ft , = 0.0005 m 2 • °C/W on the inner surface of the tube.
13-22 fu'M Reconsider Problem 13-21. Using EES (or
1^13 other) software, plot the overall heat transfer
coefficient based on the inner surface as a function of fouling
factor F; as it varies from 0.0001 m 2 • °C/W to 0.0008 m 2 ■ °C/W,
and discuss the results.
13-23 A long thin-walled double -pipe heat exchanger with
tube and shell diameters of 1.0 cm and 2.5 cm, respectively, is
used to condense refrigerant 134a by water at 20°C. The re-
frigerant flows through the tube, with a convection heat trans-
fer coefficient of /z ; = 5000 W/m 2 • °C. Water flows through the
shell at a rate of 0.3 kg/s. Determine the overall heat transfer
coefficient of this heat exchanger. Answer: 2020 W/m 2 • °C
13-24 Repeat Problem 13-23 by assuming a 2-mm-thick
layer of limestone (k = 1.3 W/m ■ °C) forms on the outer sur-
face of the inner tube.
13-25 Tu'M Reconsider Problem 13-24. Using EES (or
b^ti other) software, plot the overall heat transfer
coefficient as a function of the limestone thickness as it varies
from 1 mm to 3 mm, and discuss the results.
13-26E Water at an average temperature of 140°F and an
average velocity of 8 ft/s flows through a thin-walled |-in.-
diameter tube. The water is cooled by air that flows across the
tube with a velocity of T„ =12 ft/s at an average temperature
of 80°F. Determine the overall heat transfer coefficient.
Analysis of Heat Exchangers
13-27C What are the common approximations made in the
analysis of heat exchangers?
13-28C Under what conditions is the heat transfer relation
Q = m c C (T
valid for a heat exchanger?
m hCph (Xh.
13-29C What is the heat capacity rate? What can you say
about the temperature changes of the hot and cold fluids in a
heat exchanger if both fluids have the same capacity rate?
What does a heat capacity of infinity for a fluid in a heat ex-
changer mean?
13-30C Consider a condenser in which steam at a specified
temperature is condensed by rejecting heat to the cooling
water. If the heat transfer rate in the condenser and the tem-
perature rise of the cooling water is known, explain how the
rate of condensation of the steam and the mass flow rate of
the cooling water can be determined. Also, explain how the
total thermal resistance R of this condenser can be evaluated
in this case.
13-31C Under what conditions will the temperature rise of
the cold fluid in a heat exchanger be equal to the temperature
drop of the hot fluid?
The Log Mean Temperature Difference Method
13-32C In the heat transfer relation Q = UA S AT [m for a heat
exchanger, what is Ar )m called? How is it calculated for a
parallel-flow and counter-flow heat exchanger?
13-33C How does the log mean temperature difference for a
heat exchanger differ from the arithmetic mean temperature
difference (AMTD)? For specified inlet and outlet tempera-
tures, which one of these two quantities is larger?
13-34C The temperature difference between the hot and cold
fluids in a heat exchanger is given to be AT t at one end and A7 2
at the other end. Can the logarithmic temperature difference
Ar )m of this heat exchanger be greater than both A 7^ and AT^?
Explain.
13-35C Can the logarithmic mean temperature difference
Ar lm of a heat exchanger be a negative quantity? Explain.
13-36C Can the outlet temperature of the cold fluid in a heat
exchanger be higher than the outlet temperature of the hot fluid
in a parallel-flow heat exchanger? How about in a counter-flow
heat exchanger? Explain.
13-37C For specified inlet and outlet temperatures, for what
kind of heat exchanger will the AT [m be greatest: double-pipe
parallel-flow, double -pipe counter-flow, cross-flow, or multi-
pass shell-and-tube heat exchanger?
13-38C In the heat transfer relation Q = UA S F AT [m for a
heat exchanger, what is the quantity F called? What does it rep-
resent? Can F be greater than one?
13-39C When the outlet temperatures of the fluids in a heat
exchanger are not known, is it still practical to use the LMTD
method? Explain.
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13-40C Explain how the LMTD method can be used to de-
termine the heat transfer surface area of a multipass shell-and-
tube heat exchanger when all the necessary information,
including the outlet temperatures, is given.
13-41 Steam in the condenser of a steam power plant is to be
condensed at a temperature of 50°C (h fg = 2305 kJ/kg) with
cooling water (C p = 4180 J/kg • °C) from a nearby lake, which
enters the tubes of the condenser at 18°C and leaves at 27°C.
The surface area of the tubes is 58 m 2 , and the overall heat
transfer coefficient is 2400 W/m 2 • °C. Determine the mass flow
rate of the cooling water needed and the rate of condensation of
the steam in the condenser. Answers: 101 kg/s, 1.65 kg/s
Steam
1 50°C
n
c
c
c
27°C
18°C
50°C *
Water
FIGURE P1 3-41
13—42 A double -pipe parallel-flow heat exchanger is to heat
water (C p = 4180 J/kg ■ °C) from 25°C to 60°C at a rate of 0.2
kg/s. The heating is to be accomplished by geothermal water
(C p = 4310 J/kg • °C) available at 140°C at a mass flow rate of
0.3 kg/s. The inner tube is thin-walled and has a diameter of 0.8
cm. If the overall heat transfer coefficient of the heat exchanger
is 550 W/m 2 ■ °C, determine the length of the heat exchanger
required to achieve the desired heating.
13-43 [ttSS Reconsider Problem \~h-\2. Using EES (or
1^2 other) software, investigate the effects of tem-
perature and mass flow rate of geothermal water on the length
of the heat exchanger. Let the temperature vary from 100°C to
200°C, and the mass flow rate from 0.1 kg/s to 0.5 kg/s. Plot
the length of the heat exchanger as functions of temperature
and mass flow rate, and discuss the results.
13-44E A 1 -shell-pass and 8-tube-passes heat exchanger is
used to heat glycerin (C p = 0.60 Btu/lbm • °F) from 65°F to
140°F by hot water (C p = 1 .0 Btu/lbm ■ °F) that enters the thin-
walled 0.5-in. -diameter tubes at 175°F and leaves at 120°F.
The total length of the tubes in the heat exchanger is 500 ft.
The convection heat transfer coefficient is 4 Btu/h • ft 2 • °F
on the glycerin (shell) side and 50 Btu/h • ft 2 ■ °F on the water
(tube) side. Determine the rate of heat transfer in the heat ex-
changer (a) before any fouling occurs and {b) after fouling with
a fouling factor of 0.002 h • ft 2 • °F/Btu occurs on the outer sur-
faces of the tubes.
707
CHAPTER 13
13-45 A test is conducted to determine the overall heat trans-
fer coefficient in a shell-and-tube oil-to-water heat exchanger
that has 24 tubes of internal diameter 1.2 cm and length 2 m
in a single shell. Cold water (C p =4180 J/kg ■ °C) enters the
tubes at 20°C at a rate of 5 kg/s and leaves at 55°C. Oil
(C p = 2150 J/kg ■ °C) flows through the shell and is cooled
from 120°C to 45°C. Determine the overall heat transfer coef-
ficient Uj of this heat exchanger based on the inner surface area
of the tubes. Answer: 13.9 kW/m 2 • °C
13-46 A double -pipe counter-flow heat exchanger is to cool
ethylene glycol (C p = 2560 J/kg • °C) flowing at a rate of
3.5 kg/s from 80°C to 40°C by water (C p = 4180 J/kg ■ °C) that
enters at 20°C and leaves at 55°C. The overall heat transfer co-
efficient based on the inner surface area of the tube is
250 W/m 2 • °C. Determine (a) the rate of heat transfer, (b) the
mass flow rate of water, and (c) the heat transfer surface area
on the inner side of the tube.
Cold water
20°C
Hot glycol
80°C
3.5 kg/s
FIGURE P1 3-46
D
40°C
13-47 Water {C p = 4180 J/kg ■ °C) enters the 2.5-cm-
internal-diameter tube of a double -pipe counter-flow heat ex-
changer at 17°C at a rate of 3 kg/s. It is heated by steam
condensing at 120°C (h fg = 2203 kJ/kg) in the shell. If the
overall heat transfer coefficient of the heat exchanger is 1500
W/m 2 • °C, determine the length of the tube required in order to
heat the water to 80°C.
13-48 A thin-walled double-pipe counter-flow heat ex-
changer is to be used to cool oil (C p = 2200 J/kg ■ °C) from
150°C to 40°C at a rate of 2 kg/s by water (C p = 4180 J/kg ■
°C) that enters at 22°C at a rate of 1 .5 kg/s. The diameter of the
tube is 2.5 cm, and its length is 6 m. Determine the overall heat
transfer coefficient of this heat exchanger.
13-49 PS^I Reconsider Problem 13-48. Using EES (or
1^2 other) software, investigate the effects of oil
exit temperature and water inlet temperature on the overall heat
transfer coefficient of the heat exchanger. Let the oil exit tem-
perature vary from 30°C to 70°C and the water inlet tempera-
ture from 5°C to 25°C. Plot the overall heat transfer coefficient
as functions of the two temperatures, and discuss the results.
13-50 Consider a water-to-water double -pipe heat exchanger
whose flow arrangement is not known. The temperature mea-
surements indicate that the cold water enters at 20°C and
leaves at 50°C, while the hot water enters at 80°C and leaves at
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708
HEAT TRANSFER
45°C. Do you think this is a parallel-flow or counter-flow heat
exchanger? Explain.
13-51 Cold water (C p = 4180 J/kg • °C) leading to a shower
enters a thin-walled double-pipe counter-flow heat exchanger
at 15°C at a rate of 0.25 kg/s and is heated to 45°C by hot water
(C p = 4190 J/kg ■ °C) that enters at 100°C at a rate of 3 kg/s. If
the overall heat transfer coefficient is 1210 W/m 2 • °C, deter-
mine the rate of heat transfer and the heat transfer surface area
of the heat exchanger.
13-52 Engine oil (C p = 2100 J/kg • °C) is to be heated from
20°C to 60°C at a rate of 0.3 kg/s in a 2-cm-diameter thin-
walled copper tube by condensing steam outside at a temper-
ature of 130°C (h fg = 2174 kJ/kg). For an overall heat
transfer coefficient of 650 W/m 2 • °C, determine the rate of
heat transfer and the length of the tube required to achieve it.
Answers: 25.2 kW, 7.0 m
Steam
130°C
Oil ,
(L
20°C | ' 60°C
0.3 kg/s [J
\ 55°C
FIGURE P1 3-52
13-53E Geothermal water (C p = 1.03 Btu/lbm • °F) is to be
used as the heat source to supply heat to the hydronic heating
system of a house at a rate of 30 Btu/s in a double-pipe
counter-flow heat exchanger. Water [C p = 1.0 Btu/lbm ■ °F) is
heated from 140°F to 200°F in the heat exchanger as the geo-
thermal water is cooled from 310°F to 180°F. Determine the
mass flow rate of each fluid and the total thermal resistance of
this heat exchanger.
13-54 Glycerin (C p = 2400 J/kg • °C) at 20°C and 0.3 kg/s is
to be heated by ethylene glycol (C p = 2500 J/kg • °C) at 60°C
in a thin-walled double-pipe parallel-flow heat exchanger. The
temperature difference between the two fluids is 15°C at
the outlet of the heat exchanger. If the overall heat transfer co-
efficient is 240 W/m 2 • °C and the heat transfer surface area is
3.2 m 2 , determine (a) the rate of heat transfer, (b) the outlet
temperature of the glycerin, and (c) the mass flow rate of the
ethylene glycol.
13-55 Air (C,, = 1 005 J/kg • °C) is to be preheated by hot ex-
haust gases in a cross-flow heat exchanger before it enters the
furnace. Air enters the heat exchanger at 95 kPa and 20°C at a
rate of 0.8 m 3 /s. The combustion gases (C p = 1 100 J/kg ■ °C)
enter at 180°C at a rate of 1.1 kg/s and leave at 95°C. The prod-
uct of the overall heat transfer coefficient and the heat transfer
surface area is AU = 1200 W/°C. Assuming both fluids to be
unmixed, determine the rate of heat transfer and the outlet tem-
perature of the air.
Air -
95kPa
20°C
0.8 m 3 /s
Exhaust gases
1.1 kg/s
95°C
FIGURE P1 3-55
13-56 A shell-and-tube heat exchanger with 2-shell passes
and 12-tube passes is used to heat water (C. = 4180 J/kg • °C)
in the tubes from 20°C to 70°C at a rate of 4.5 kg/s. Heat is
supplied by hot oil (C p = 2300 J/kg • °C) that enters the shell
side at 170°C at a rate of 10 kg/s. For a tube-side overall heat
transfer coefficient of 600 W/m 2 • °C, determine the heat trans-
fer surface area on the tube side. Answer. 15 m 2
13-57 Repeat Problem 1 3-56 for a mass flow rate of 2 kg/s
for water.
13-58 A shell-and-tube heat exchanger with 2-shell passes and
8-tube passes is used to heat ethyl alcohol (C p = 2670 J/kg • °C)
in the tubes from 25°C to 70°C at a rate of 2.1 kg/s. The heating
is to be done by water (C p = 4190 J/kg • °C) that enters the shell
side at 95°C and leaves at 45°C. If the overall heat transfer coef-
ficient is 950 W/m 2 • °C, determine the heat transfer surface area
of the heat exchanger.
Water
95°C
70°C
Ethyl
alcohol
25°C -»
2.1 kg/s III
45°C
FIGURE P 13-58
(8-tube passes)
13-59 A shell-and-tube heat exchanger with 2-shell passes and
12-tube passes is used to heat water (C p = 4180 J/kg • °C) with
ethylene glycol (C p = 2680 J/kg • °C). Water enters the tubes at
22°C at a rate of 0.8 kg/s and leaves at 70°C. Ethylene glycol en-
ters the shell at 110°C and leaves at 60°C. If the overall heat
transfer coefficient based on the tube side is 280 W/m 2 • °C,
determine the rate of heat transfer and the heat transfer surface
area on the tube side.
cen58933_chl3.qxd 9/9/2002 9:57 AM Page 709
13-60 fifiFM Reconsider Problem 13-59. Using EES (or other)
k^S software, investigate the effect of the mass flow
rate of water on the rate of heat transfer and the tube-side surface
area. Let the mass flow rate vary from 0.4 kg/s to 2.2 kg/s. Plot
the rate of heat transfer and the surface area as a function of the
mass flow rate, and discuss the results.
13-61E Steam is to be condensed on the shell side of a
1 -shell-pass and 8-tube-passes condenser, with 50 tubes in
each pass at 90°F (h fg = 1043 Btu/lbm). Cooling water (C p =
1.0 Btu/lbm • °F) enters the tubes at 60°F and leaves at 73°R
The tubes are thin-walled and have a diameter of 3/4 in. and
length of 5 ft per pass. If the overall heat transfer coefficient
is 600 Btu/h • ft 2 • °F, determine (a) the rate of heat transfer,
(b) the rate of condensation of steam, and (c) the mass flow
rate of cold water.
Steam
90°F
1 201bm/s
n
c
c
c
3>
73°F
60°F
}f
Water
FIGURE P13-61E
13-62E [c?^l Reconsider Problem 13-61E. Using EES (or
k^S other) software, investigate the effect of the
condensing steam temperature on the rate of heat transfer, the
rate of condensation of steam, and the mass flow rate of cold
water. Let the steam temperature vary from 80°F to 1 20°F. Plot
the rate of heat transfer, the condensation rate of steam, and the
mass flow rate of cold water as a function of steam tempera-
ture, and discuss the results.
13-63 A shell-and-tube heat exchanger with 1 -shell pass and
20-tube passes is used to heat glycerin {C p = 2480 J/kg • °C)
in the shell, with hot water in the tubes. The tubes are thin-
walled and have a diameter of 1 .5 cm and length of 2 m per
pass. The water enters the tubes at 100°C at a rate of 5 kg/s and
leaves at 55°C. The glycerin enters the shell at 15°C and leaves
at 55°C. Determine the mass flow rate of the glycerin and the
overall heat transfer coefficient of the heat exchanger.
13-64 In a binary geofhermal power plant, the working fluid
isobutane is to be condensed by air in a condenser at 75°C
(h fg = 255.7 kJ/kg) at a rate of 2.7 kg/s. Air enters the con-
denser at 21°C and leaves at 28°C. The heat transfer surface
709
CHAPTER 13
Air
28°C
Isobutane
75°C
2.7 kg/s
Air
21°C
FIGURE P1 3-64
area based on the isobutane side is 24 m 2 . Determine the mass
flow rate of air and the overall heat transfer coefficient.
13-65 Hot exhaust gases of a stationary diesel engine are to
be used to generate steam in an evaporator. Exhaust gases
(C p = 1051 J/kg • °C) enter the heat exchanger at 550°C at a
rate of 0.25 kg/s while water enters as saturated liquid and
evaporates at 200°C (h fg = 1941 kJ/kg). The heat transfer sur-
face area of the heat exchanger based on water side is 0.5 m 2
and overall heat transfer coefficient is 1780 W/m 2 ■ °C. Deter-
mine the rate of heat transfer, the exit temperature of exhaust
gases, and the rate of evaporation of water.
13-66 Fct^S R econs ider Problem 13-65. Using EES (or
kS other) software, investigate the effect of the ex-
haust gas inlet temperature on the rate of heat transfer, the exit
temperature of exhaust gases, and the rate of evaporation of
water. Let the temperature of exhaust gases vary from 300°C to
600°C. Plot the rate of heat transfer, the exit temperature of ex-
haust gases, and the rate of evaporation of water as a function
of the temperature of the exhaust gases, and discuss the results.
13-67 In a textile manufacturing plant, the waste dyeing wa-
ter (C p = 4295 J/g • °C) at 75°C is to be used to preheat fresh
water (C p = 4180 J/kg • °C) at 15°C at the same flow rate in a
double-pipe counter-flow heat exchanger. The heat transfer
surface area of the heat exchanger is 1.65 m 2 and the overall
heat transfer coefficient is 625 W/m 2 • °C. If the rate of heat
transfer in the heat exchanger is 35 kW, determine the outlet
temperature and the mass flow rate of each fluid stream.
Fresh
water
15°C
Dyeing
water
d
75°C
I T
FIGURE P 13-67
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HEAT TRANSFER
The Effectiveness-NTU Method
13-68C Under what conditions is the effectiveness-NTU
method definitely preferred over the LMTD method in heat ex-
changer analysis?
13-69C What does the effectiveness of a heat exchanger rep-
resent? Can effectiveness be greater than one? On what factors
does the effectiveness of a heat exchanger depend?
13-70C For a specified fluid pair, inlet temperatures, and
mass flow rates, what kind of heat exchanger will have the
highest effectiveness: double-pipe parallel-flow, double-pipe
counter-flow, cross-flow, or multipass shell-and-tube heat
exchanger?
13-71C Explain how you can evaluate the outlet tempera-
tures of the cold and hot fluids in a heat exchanger after its ef-
fectiveness is determined.
13-72C Can the temperature of the hot fluid drop below the
inlet temperature of the cold fluid at any location in a heat ex-
changer? Explain.
13-73C Can the temperature of the cold fluid rise above the
inlet temperature of the hot fluid at any location in a heat ex-
changer? Explain.
13-74C Consider a heat exchanger in which both fluids have
the same specific heats but different mass flow rates. Which
fluid will experience a larger temperature change: the one with
the lower or higher mass flow rate?
13-75C Explain how the maximum possible heat transfer
rate Q max in a heat exchanger can be determined when the mass
flow rates, specific heats, and the inlet temperatures of the two
fluids are specified. Does the value of <2 max depend on the type
of the heat exchanger?
13-76C Consider two double-pipe counter-flow heat ex-
changers that are identical except that one is twice as long as
the other one. Which heat exchanger is more likely to have a
higher effectiveness?
13-77C Consider a double-pipe counter-flow heat ex-
changer. In order to enhance heat transfer, the length of the heat
exchanger is now doubled. Do you think its effectiveness will
also double?
13-78C Consider a shell-and-tube water-to-water heat ex-
changer with identical mass flow rates for both the hot and cold
water streams. Now the mass flow rate of the cold water is re-
duced by half. Will the effectiveness of this heat exchanger in-
crease, decrease, or remain the same as a result of this
modification? Explain. Assume the overall heat transfer coeffi-
cient and the inlet temperatures remain the same.
13-79C Under what conditions can a counter-flow heat ex-
changer have an effectiveness of one? What would your an-
swer be for a parallel-flow heat exchanger?
13-80C How is the NTU of a heat exchanger defined? What
does it represent? Is a heat exchanger with a very large NTU
(say, 10) necessarily a good one to buy?
13-81C Consider a heat exchanger that has an NTU of 4.
Someone proposes to double the size of the heat exchanger and
thus double the NTU to 8 in order to increase the effectiveness
of the heat exchanger and thus save energy. Would you support
this proposal?
13-82C Consider a heat exchanger that has an NTU of 0.1.
Someone proposes to triple the size of the heat exchanger and
thus triple the NTU to 0.3 in order to increase the effectiveness
of the heat exchanger and thus save energy. Would you support
this proposal?
13-83 Air (C p = 1005 J/kg • °C) enters a cross-flow heat ex-
changer at 10°C at a rate of 3 kg/s, where it is heated by a hot
water stream (C p = 4190 J/kg • °C) that enters the heat ex-
changer at 95°C at a rate of 1 kg/s. Determine the maximum
heat transfer rate and the outlet temperatures of the cold and
the hot water streams for that case.
13-84 Hot oil (C p = 2200 J/kg ■ °C) is to be cooled by water
(C p = 4180 J/kg • °C) in a 2-shell-pass and 12-tube-pass heat
exchanger. The tubes are thin-walled and are made of copper
with a diameter of 1 .8 cm. The length of each tube pass in the
heat exchanger is 3 m, and the overall heat transfer coefficient
is 340 W/m 2 • °C. Water flows through the tubes at a total rate
of 0.1 kg/s, and the oil through the shell at a rate of 0.2 kg/s.
The water and the oil enter at temperatures 18°C and 160°C,
respectively. Determine the rate of heat transfer in the heat ex-
changer and the outlet temperatures of the water and the oil.
Answers: 36.2 kW, 104. 6°C, 77.7°C
Oil
160°C
0.2 kg/s
ill
Water
18°C -►
o.i kg/s hi—
FIGURE P 13-84
(12-tube passes)
13-85 Consider an oil-to-oil double-pipe heat exchanger
whose flow arrangement is not known. The temperature mea-
surements indicate that the cold oil enters at 20°C and leaves at
55°C, while the hot oil enters at 80°C and leaves at 45°C. Do
you think this is a parallel-flow or counter-flow heat ex-
changer? Why? Assuming the mass flow rates of both fluids to
be identical, determine the effectiveness of this heat exchanger.
13-86E Hot water enters a double -pipe counter-flow water-
to-oil heat exchanger at 220°F and leaves at 100°F. Oil enters
at 70°F and leaves at 150°F. Determine which fluid has the
smaller heat capacity rate and calculate the effectiveness of this
heat exchanger.
13-87 A thin-walled double-pipe parallel-flow heat ex-
changer is used to heat a chemical whose specific heat is 1800
cen58933_chl3.qxd 9/9/2002 9:57 AM Page 711
J/kg ■ °C with hot water (C p = 4180 J/kg • °C). The chemical
enters at 20°C at a rate of 3 kg/s, while the water enters at
110°C at a rate of 2 kg/s. The heat transfer surface area
of the heat exchanger is 7 m 2 and the overall heat transfer co-
efficient is 1200 W/m 2 • °C. Determine the outlet temperatures
of the chemical and the water.
H
Chemical
^C
20°C
3 kg/s
° l --a Hot water
* 110°C
' 2 kg/s
FIGURE P1 3-87
13-88 tu'M Reconsider Problem 13-87. Using EES (or
k^S other) software, investigate the effects of the in-
let temperatures of the chemical and the water on their outlet
temperatures. Let the inlet temperature vary from 10°C to 50°C
for the chemical and from 80°C to 150°C for water. Plot the
outlet temperature of each fluid as a function of the inlet tem-
perature of that fluid, and discuss the results.
13-89 A cross-flow air-to-water heat exchanger with an ef-
fectiveness of 0.65 is used to heat water (C p = 4180 J/kg • °C)
with hot air (C p = 1010 J/kg • °C). Water enters the heat ex-
changer at 20°C at a rate of 4 kg/s, while air enters at 100°C at
a rate of 9 kg/s. If the overall heat transfer coefficient based on
the water side is 260 W/m 2 • °C, determine the heat transfer
surface area of the heat exchanger on the water side. Assume
both fluids are unmixed. Answer: 52.4 m 2
13-90 Water (C p = 4180 J/kg ■ °C) enters the 2.5-cm-
internal-diameter tube of a double-pipe counter-flow heat ex-
changer at 17°C at a rate of 3 kg/s. Water is heated by steam
condensing at 120°C (h fg = 2203 kJ/kg) in the shell. If the
overall heat transfer coefficient of the heat exchanger is 900
W/m 2 • °C, determine the length of the tube required in order to
heat the water to 80°C using (a) the LMTD method and (b) the
e-NTU method.
13-91 Ethanol is vaporized at 78°C (h fg = 846 kJ/kg) in a
double-pipe parallel-flow heat exchanger at a rate of 0.03 kg/s
Oil
120°C '
Ethanol
78°C
0.03 kg/s
711
CHAPTER 13
by hot oil (C p = 2200 J/kg • °C) that enters at 120°C. If the heat
transfer surface area and the overall heat transfer coefficients
are 6.2 m 2 and 320 W/m 2 • °C, respectively, determine the out-
let temperature and the mass flow rate of oil using (a) the
LMTD method and (b) the e-NTU method.
13-92 Water (C p = 4180 J/kg • °C) is to be heated by solar-
heated hot air (C p = 1010 J/kg • °C) in a double -pipe counter-
flow heat exchanger. Air enters the heat exchanger at 90°C at a
rate of 0.3 kg/s, while water enters at 22°C at a rate of 0. 1 kg/s.
The overall heat transfer coefficient based on the inner side of
the tube is given to be 80 W/m 2 ■ °C. The length of the tube is
12 m and the internal diameter of the tube is 1 .2 cm. Determine
the outlet temperatures of the water and the air.
13-93 Tu'M Reconsider Problem 13-92. Using EES (or
l^tl other) software, investigate the effects of the
mass flow rate of water and the tube length on the outlet tem-
peratures of water and air. Let the mass flow rate vary from
0.05 kg/s to 1 .0 kg/s and the tube length from 5 m to 25 m. Plot
the outlet temperatures of the water and the air as the func-
tions of the mass flow rate and the tube length, and discuss the
results.
13-94E A thin-walled double-pipe heat exchanger is to be
used to cool oil (C p = 0.525 Btu/lbm • °F) from 300°F to 105°F
at a rate of 5 lbm/s by water (C p = 1 .0 Btu/lbm • °F) that enters
at 70°F at a rate of 3 lbm/s. The diameter of the tube is 1 in. and
its length is 20 ft. Determine the overall heat transfer coeffi-
cient of this heat exchanger using (a) the LMTD method and
(b) the e-NTU method.
13-95 Cold water (C p = 4180 J/kg • °C) leading to a
shower enters a thin-walled double-pipe counter-flow heat
exchanger at 15°C at a rate of 0.25 kg/s and is heated to 45°C
by hot water (C p = 4190 J/kg • °C) that enters at 100°C at a
rate of 3 kg/s. If the overall heat transfer coefficient is 950
W/m 2 • °C, determine the rate of heat transfer and the heat
transfer surface area of the heat exchanger using the e-NTU
method. Answers: 31.35 kW, 0.482 m 2
Cold water
15°C I
0.25 kg/s
Hot water
100°C
3 kg/s
\ 45°C
FIGURE P1 3-95
13-96
Reconsider Problem 13-95. Using EES (or
FIGURE P1 3-91
other) software, investigate the effects of the
inlet temperature of hot water and the heat transfer coeffi-
cient on the rate of heat transfer and surface area. Let the inlet
temperature vary from 60°C to 120°C and the overall heat
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HEAT TRANSFER
transfer coefficient from 750 W/m 2 ■ °C to 1250 W/m 2 ■ °C.
Plot the rate of heat transfer and surface area as functions of
inlet temperature and the heat transfer coefficient, and discuss
the results.
13-97 Glycerin (C p = 2400 J/kg • °C) at 20°C and 0.3 kg/s
is to be heated by ethylene glycol (C p = 2500 J/kg • °C) at
60°C and the same mass flow rate in a thin-walled double-
pipe parallel-flow heat exchanger. If the overall heat transfer
coefficient is 380 W/m 2 ■ °C and the heat transfer surface area
is 5.3 m 2 , determine (a) the rate of heat transfer and (b) the
outlet temperatures of the glycerin and the glycol.
13-98 A cross-flow heat exchanger consists of 40 thin-
walled tubes of 1 -cm diameter located in a duct oil m X 1 m
cross-section. There are no fins attached to the tubes. Cold
water (C p = 4180 J/kg • °C) enters the tubes at 18°C with an
average velocity of 3 m/s, while hot air (C p = 1010 J/kg ■ °C)
enters the channel at 130°C and 105 kPa at an average veloc-
ity of 12 m/s. If the overall heat transfer coefficient is 130
W/m 2 ■ °C, determine the outlet temperatures of both fluids
and the rate of heat transfer.
FIGURE P1 3-98
13-99 /^fe> A shell-and-tube heat exchanger with 2-shell
KsHy passes and 8-tube passes is used to heat ethyl
alcohol (C p = 2670 J/kg • °C) in the tubes from 25°C to 70°C
at a rate of 2.1 kg/s. The heating is to be done by water (C p =
4190 J/kg ■ °C) that enters the shell at 95°C and leaves at
60°C. If the overall heat transfer coefficient is 800 W/m 2 • °C,
determine the heat transfer surface area of the heat exchanger
using (a) the LMTD method and (b) the e-NTU method.
Answer (a): 11.4 m 2
13-100 Steam is to be condensed on the shell side of a
1 -shell-pass and 8-tube-passes condenser, with 50 tubes in
each pass, at 30°C {h fg = 2430 kJ/kg). Cooling water (C p =
4180 J/kg ■ °C) enters the tubes at 15 °C at a rate of 1800 kg/h.
The tubes are thin-walled, and have a diameter of 1 .5 cm and
length of 2 m per pass. If the overall heat transfer coefficient
is 3000 W/m 2 • °C, determine (a) the rate of heat transfer and
(b) the rate of condensation of steam.
Steam
1 30°C
r
(c
»
((
>J
l<
»
(c
15°C
Water
T
30°C i
1800 kg/h
FIGURE P1 3-1 00
13-101 r^| Reconsider Problem 13-100. Using EES (or
I^H other) software, investigate the effects of the
condensing steam temperature and the tube diameters on the
rate of heat transfer and the rate of condensation of steam. Let
the steam temperature vary from 20°C to 70°C and the tube
diameter from 1 .0 cm to 2.0 cm. Plot the rate of heat transfer
and the rate of condensation as functions of steam temperature
and tube diameter, and discuss the results.
13-102 Cold water (C p = 4180 J/kg • °C) enters the tubes of
a heat exchanger with 2-shell-passes and 13-tube-passes at
20°C at a rate of 3 kg/s, while hot oil (C p = 2200 J/kg • °C) en-
ters the shell at 1 30°C at the same mass flow rate. The overall
heat transfer coefficient based on the outer surface of the tube
is 300 W/m 2 • °C and the heat transfer surface area on that side
is 20 m 2 . Determine the rate of heat transfer using (a) the
LMTD method and (b) the e-NTU method.
Selection of Heat Exchangers
13-1 03C A heat exchanger is to be selected to cool a hot liq-
uid chemical at a specified rate to a specified temperature. Ex-
plain the steps involved in the selection process.
13-104C There are two heat exchangers that can meet the
heat transfer requirements of a facility. One is smaller and
cheaper but requires a larger pump, while the other is larger
and more expensive but has a smaller pressure drop and thus
requires a smaller pump. Both heat exchangers have the same
life expectancy and meet all other requirements. Explain which
heat exchanger you would choose under what conditions.
13-105C There are two heat exchangers that can meet the
heat transfer requirements of a facility. Both have the same
pumping power requirements, the same useful life, and the
same price tag. But one is heavier and larger in size. Under
what conditions would you choose the smaller one?
13-106 A heat exchanger is to cool oil (C p = 2200 J/kg • °C)
at a rate of 13 kg/s from 120°C to 50°C by air. Determine the
heat transfer rating of the heat exchanger and propose a suit-
able type.
cen58933_chl3.qxd 9/9/2002 9:57 AM Page 713
13-107 A shell-and-tube process heater is to be selected to
heat water (C p = 4190 J/kg • °C) from 20°C to 90°C by steam
flowing on the shell side. The heat transfer load of the heater is
600 kW. If the inner diameter of the tubes is 1 cm and the ve-
locity of water is not to exceed 3 m/s, determine how many
tubes need to be used in the heat exchanger.
Steam
A
(c
))
(c
>)
(c
>)
«
20°C
Water
FIGURE P1 3-1 07
13-108
Reconsider Problem 13-107. Using EES (or
other) software, plot the number of tube passes
as a function of water velocity as it varies from 1 m/s to 8 m/s,
and discuss the results.
13-109 The condenser of a large power plant is to remove
500 MW of heat from steam condensing at 30°C (h fg = 2430
kJ/kg). The cooling is to be accomplished by cooling water
(C p = 4180 J/kg ■ °C) from a nearby river, which enters the
tubes at 18°C and leaves at 26°C. The tubes of the heat ex-
changer have an internal diameter of 2 cm, and the overall heat
transfer coefficient is 3500 W/m 2 • °C. Determine the total
length of the tubes required in the condenser. What type of heat
exchanger is suitable for this task? Answer: 312.3 km
13-110 Repeat Problem 13-109 for a heat transfer load of
300 MW.
Review Problems
13-111 Hot oil is to be cooled in a multipass shell-and-tube
heat exchanger by water. The oil flows through the shell, with
a heat transfer coefficient of h a = 35 W/m 2 • °C, and the water
flows through the tube with an average velocity of 3 m/s. The
tube is made of brass (k = 110 W/m • °C) with internal and ex-
ternal diameters of 1 .3 cm and 1 .5 cm, respectively. Using wa-
ter properties at 25°C, determine the overall heat transfer
coefficient of this heat exchanger based on the inner surface.
13-112 Repeat Problem 1 3-1 1 1 by assuming a fouling factor
R
f.o
0.0004 m 2 ■ °C/W on the outer surface of the tube.
13-113 Cold water (C p = 4180 J/kg • °C) enters the tubes of
a heat exchanger with 2-shell passes and 20-tube passes at
20°C at a rate of 3 kg/s, while hot oil (C p = 2200 J/kg ■ °C) en-
ters the shell at 130°C at the same mass flow rate and leaves at
713
CHAPTER 13
60°C. If the overall heat transfer coefficient based on the outer
surface of the tube is 300 W/m 2 • °C, determine (a) the rate of
heat transfer and (b) the heat transfer surface area on the outer
side of the tube. Answers: (a) 462 kW, (b) 29.2 m 2
Hot oil
130°C
3 kg/s
Cold water
20°C -^
3 kg/s
2)
r
60°C
(20-tube passes)
FIGURE P1 3-1 13
13-114E Water (C„ = 1.0 Btu/lbm • °F) is to be heated by
solar-heated hot air (C p = 0.24 Btu/lbm • °F) in a double-pipe
counter-flow heat exchanger. Air enters the heat exchanger at
190°F at a rate of 0.7 lbm/s and leaves at 135°F Water enters at
70°F at a rate of 0.35 lbm/s. The overall heat transfer coeffi-
cient based on the inner side of the tube is given to be 20 Btu/h
• ft 2 • °F Determine the length of the tube required for a tube in-
ternal diameter of 0.5 in.
13-115 By taking the limit as A7 2 — > AT|, show that when
A7\ = ATn for a heat exchanger, the &T [m relation reduces to
Ar lm = AT, = A7 2 .
13-116 The condenser of a room air conditioner is designed
to reject heat at a rate of 15,000 kJ/h from Refrigerant- 134a
as the refrigerant is condensed at a temperature of 40°C. Air
(C p = 1005 J/kg ■ °C) flows across the finned condenser
coils, entering at 25°C and leaving at 35°C. If the overall heat
transfer coefficient based on the refrigerant side is 150 W/m 2
■ °C, determine the heat transfer area on the refrigerant side.
Answer: 3.05 m 2
35°C
40°C
FIGURE P1 3-1 16
13-117 Air (C p = 1005 J/kg • °C) is to be preheated by hot
exhaust gases in a cross-flow heat exchanger before it enters
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HEAT TRANSFER
the furnace. Air enters the heat exchanger at 95 kPa and 20°C
at a rate of 0.8 m 3 /s. The combustion gases {C p = 1 100 J/kg ■
°C) enter at 180°C at a rate of 1 .1 kg/s and leave at 95°C. The
product of the overall heat transfer coefficient and the heat
transfer surface area is UA S = 1620 W/°C. Assuming both flu-
ids to be unmixed, determine the rate of heat transfer.
13-118 In a chemical plant, a certain chemical is heated by
hot water supplied by a natural gas furnace. The hot water
(C p = 4180 J/kg ■ °C) is then discharged at 60°C at a rate of
8 kg/min. The plant operates 8 h a day, 5 days a week, 52
weeks a year. The furnace has an efficiency of 78 percent, and
the cost of the natural gas is $0.54 per therm (1 therm =
100,000 Btu = 105,500 kJ). The average temperature of the
cold water entering the furnace throughout the year is 14°C. In
order to save energy, it is proposed to install a water-to-water
heat exchanger to preheat the incoming cold water by the
drained hot water. Assuming that the heat exchanger will re-
cover 72 percent of the available heat in the hot water, deter-
mine the heat transfer rating of the heat exchanger that needs to
be purchased and suggest a suitable type. Also, determine the
amount of money this heat exchanger will save the company
per year from natural gas savings.
13-119 A shell-and-tube heat exchanger with 1 -shell pass
and 14-tube passes is used to heat water in the tubes with geo-
fhermal steam condensing at 120°C (h fg = 2203 kJ/kg) on the
shell side. The tubes are thin-walled and have a diameter of 2.4
cm and length of 3.2 m per pass. Water (C p = 4180 J/kg ■ °C)
enters the tubes at 22°C at a rate of 3.9 kg/s. If the temperature
difference between the two fluids at the exit is 46°C, determine
(a) the rate of heat transfer, (b) the rate of condensation of
steam, and (c) the overall heat transfer coefficient.
Steam
, 120°C
n.
c
c
c
3>
14 tubes
120°C
FIGURE P1 3-1 19
22°C
Y
Water
3.9 kg/s
13-120 Geothermal water (C p = 4250 J/kg • °C) at 95°C is to
be used to heat fresh water (C p = 4180 J/kg ■ °C) at 12°C at a
rate of 1 .2 kg/s in a double -pipe counter-flow heat exchanger.
The heat transfer surface area is 25 m 2 , the overall heat transfer
coefficient is 480 W/m 2 • °C, and the mass flow rate of geo-
thermal water is larger than that of fresh water. If the effective-
ness of the heat exchanger is desired to be 0.823, determine the
mass flow rate of geothermal water and the outlet temperatures
of both fluids.
13-121 Air at 18°C (C p = 1006 J/kg • °C) is to be heated to
70°C by hot oil at 80°C (C p = 2150 J/kg • °C) in a cross-flow
heat exchanger with air mixed and oil unmixed. The product of
heat transfer surface area and the overall heat transfer coeffi-
cient is 750 W/m 2 ■ °C and the mass flow rate of air is twice
that of oil. Determine (a) the effectiveness of the heat ex-
changer, (b) the mass flow rate of air, and (c) the rate of heat
transfer.
13-122 Consider a water-to-water counter-flow heat ex-
changer with these specifications. Hot water enters at 95°C
while cold water enters at 20°C. The exit temperature of hot
water is 15°C greater than that of cold water, and the mass flow
rate of hot water is 50 percent greater than that of cold water.
The product of heat transfer surface area and the overall heat
transfer coefficient is 1400 W/m 2 • °C. Taking the specific heat
of both cold and hot water to be C p = 41 80 J/kg ■ °C, determine
(a) the outlet temperature of the cold water, (b) the effective-
ness of the heat exchanger, (c) the mass flow rate of the cold
water, and (d) the heat transfer rate.
Cold water
20°C I
Hot water
95°C
FIGURE P1 3-1 22
Computer, Design, and Essay Problems
13-123 Write an interactive computer program that will give
the effectiveness of a heat exchanger and the outlet tempera-
tures of both the hot and cold fluids when the type of fluids, the
inlet temperatures, the mass flow rates, the heat transfer sur-
face area, the overall heat transfer coefficient, and the type of
heat exchanger are specified. The program should allow the
user to select from the fluids water, engine oil, glycerin, ethyl
alcohol, and ammonia. Assume constant specific heats at about
room temperature.
13-124 Water flows through a shower head steadily at a rate
of 8 kg/min. The water is heated in an electric water heater
from 15°C to 45°C. In an attempt to conserve energy, it is pro-
posed to pass the drained warm water at a temperature of 38°C
through a heat exchanger to preheat the incoming cold water.
Design a heat exchanger that is suitable for this task, and dis-
cuss the potential savings in energy and money for your area.
13-125 Open the engine compartment of your car and search
for heat exchangers. How many do you have? What type are
they? Why do you think those specific types are selected? If
cen58933_chl3.qxd 9/9/2002 9:58 AM Page 715
you were redesigning the car, would you use different kinds?
Explain.
13-126 Write an essay on the static and dynamic types of re-
generative heat exchangers and compile information about the
manufacturers of such heat exchangers. Choose a few models
by different manufacturers and compare their costs and perfor-
mance.
13-127 Design a hydrocooling unit that can cool fruits and
vegetables from 30°C to 5°C at a rate of 20,000 kg/h under the
following conditions:
The unit will be of flood type that will cool the products as
they are conveyed into the channel filled with water. The prod-
ucts will be dropped into the channel filled with water at one end
and picked up at the other end. The channel can be as wide as 3
m and as high as 90 cm. The water is to be circulated and cooled
by the evaporator section of a refrigeration system. The refriger-
ant temperature inside the coils is to be -2°C, and the water tem-
perature is not to drop below 1 °C and not to exceed 6°C.
Assuming reasonable values for the average product density,
specific heat, and porosity (the fraction of air volume in a box),
recommend reasonable values for the quantities related to the
thermal aspects of the hydrocooler, including (a) how long the
fruits and vegetables need to remain in the channel, (b) the
length of the channel, (c) the water velocity through the chan-
nel, (d) the velocity of the conveyor and thus the fruits and
vegetables through the channel, (e) the refrigeration capacity of
the refrigeration system, and (f) the type of heat exchanger for
the evaporator and the surface area on the water side.
13-128 Design a scalding unit for slaughtered chicken to
loosen their feathers before they are routed to feather-picking
machines with a capacity of 1200 chickens per hour under the
following conditions:
The unit will be of immersion type filled with hot water at
an average temperature of 53°C at all times. Chickens with an
average mass of 2.2 kg and an average temperature of 36°C
will be dipped into the tank, held in the water for 1 .5 min, and
taken out by a slow-moving conveyor. Each chicken is ex-
pected to leave the tank 15 percent heavier as a result of the
water that sticks to its surface. The center-to-center distance
between chickens in any direction will be at least 30 cm. The
tank can be as wide as 3 m and as high as 60 cm. The water is
715
CHAPTER 13
to be circulated through and heated by a natural gas furnace,
but the temperature rise of water will not exceed 5°C as it
passes through the furnace. The water loss is to be made up by
the city water at an average temperature of 16°C. The ambient
air temperature can be taken to be 20°C. The walls and the
floor of the tank are to be insulated with a 2.5-cm-thick ure-
thane layer. The unit operates 24 h a day and 6 days a week.
Assuming reasonable values for the average properties, rec-
ommend reasonable values for the quantities related to the
thermal aspects of the scalding tank, including (a) the mass
flow rate of the make-up water that must be supplied to the
tank; (b) the length of the tank; (c) the rate of heat transfer from
the water to the chicken, in kW; (d) the velocity of the con-
veyor and thus the chickens through the tank; (e) the rate of
heat loss from the exposed surfaces of the tank and if it is sig-
nificant; (/) the size of the heating system in kJ/h; (g) the type
of heat exchanger for heating the water with flue gases of the
furnace and the surface area on the water side; and (h) the op-
erating cost of the scalding unit per month for a unit cost of
$0.56 therm of natural gas (1 therm = 105,000 kJ).
13-129 A company owns a refrigeration system whose re-
frigeration capacity is 200 tons (1 ton of refrigeration = 211
kJ/min), and you are to design a forced-air cooling system for
fruits whose diameters do not exceed 7 cm under the following
conditions:
The fruits are to be cooled from 28°C to an average temper-
ature of 8°C. The air temperature is to remain above -2°C and
below 10°C at all times, and the velocity of air approaching the
fruits must remain under 2 m/s. The cooling section can be as
wide as 3.5 m and as high as 2 m.
Assuming reasonable values for the average fruit density,
specific heat, and porosity (the fraction of air volume in a box),
recommend reasonable values for the quantities related to the
thermal aspects of the forced-air cooling, including (a) how
long the fruits need to remain in the cooling section; (b) the
length of the cooling section; (c) the air velocity approaching
the cooling section; (d) the product cooling capacity of the sys-
tem, in kg ■ fruit/h; (e) the volume flow rate of air; and (/*) the
type of heat exchanger for the evaporator and the surface area
on the air side.
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CHAPTER
MASS TRANSFER
To this point we have restricted our attention to heat transfer problems
that did not involve any mass transfer. However, many significant heat
transfer problems encountered in practice involve mass transfer. For ex-
ample, about one-third of the heat loss from a resting person is due to evapo-
ration. It turns out that mass transfer is analogous to heat transfer in many
respects, and there is close resemblance between heat and mass transfer rela-
tions. In this chapter we discuss the mass transfer mechanisms and develop re-
lations for the mass transfer rate for some situations commonly encountered
in practice.
Distinction should be made between mass transfer and the bulk fluid motion
(or fluid flow) that occurs on a macroscopic level as a fluid is transported from
one location to another. Mass transfer requires the presence of two regions at
different chemical compositions, and mass transfer refers to the movement of
a chemical species from a high concentration region toward a lower concen-
tration one relative to the other chemical species present in the medium. The
primary driving force for fluid flow is the pressure difference, whereas for
mass transfer it is the concentration difference. Therefore, we do not speak of
mass transfer in a homogeneous medium.
We begin this chapter by pointing out numerous analogies between heat and
mass transfer and draw several parallels between them. We then discuss
boundary conditions associated with mass transfer and one-dimensional
steady and transient mass diffusion. Following is a discussion of mass trans-
fer in a moving medium. Finally, we consider convection mass transfer and
simultaneous heat and mass transfer.
CONTENTS
14-1 Introduction 718
14-2 Analogy between Heat
and Mass Transfer 719
14-3 Mass Diffusion 721
14-4 Boundary Conditions 727
14-5 Steady Mass Diffusion
through a Wall 732
14-6 Water Vapor Migration
in Buildings 736
14-7 Transient Mass Diffusion 740
14-8 Diffusion in a
Moving Medium 743
14-9 Mass Convection 754
14-10 Simultaneous Heat
and Mass Transfer 763
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718
HEAT TRANSFER
Water
JXLLL
Salt
Salty
water
(a) Before (b) After
FIGURE 14-1
Whenever there is concentration
difference of a physical quantity in
a medium, nature tends to equalize
things by forcing a flow from the
high to the low concentration region.
14-1 - INTRODUCTION
It is a common observation that whenever there is an imbalance of a
commodity in a medium, nature tends to redistribute it until a "balance" or
"equality" is established. This tendency is often referred to as the driving
force, which is the mechanism behind many naturally occurring transport
phenomena.
If we define the amount of a commodity per unit volume as the concentra-
tion of that commodity, we can say that the flow of a commodity is always in
the direction of decreasing concentration; that is, from the region of high con-
centration to the region of low concentration (Fig. 14-1). The commodity sim-
ply creeps away during redistribution, and thus the flow is a diffusion process.
The rate of flow of the commodity is proportional to the concentration gradi-
ent dCldx, which is the change in the concentration C per unit length in the
flow direction x, and the area A normal to flow direction and is expressed as
Flow rate «■ (Normal area)(Concentration gradient)
or
Q
"/r H ;ff A
ciC
dx
(14-1)
Initial N 2 J"
concentration I
O A ft L O
Initial 2
' concentration
0.79
0.21
"ool
<. o o o
O O O <>
« »o «<>ol
> A» c ,
,oo .«»<>Air% o»o°
*0»0 2 O ° #'
oN,
o,
FIGURE 14-2
A tank that contains N 2 and air in its
two compartments, and the diffusion
of N 2 into the air when the partition
is removed.
Here the proportionality constant k dm is the diffusion coefficient of the
medium, which is a measure of how fast a commodity diffuses in the medium,
and the negative sign is to make the flow in the positive direction a positive
quantity (note that dCldx is a negative quantity since concentration decreases
in the flow direction). You may recall that Fourier's law of heat conduction,
Ohm's law of electrical conduction, and Newton's law of viscosity are all in the
form of Equation 14—1.
To understand the diffusion process better, consider a tank that is divided
into two equal parts by a partition. Initially, the left half of the tank contains
nitrogen N 2 gas while the right half contains air (about 21 percent 2 and
79 percent N 2 ) at the same temperature and pressure. The 2 and N 2 mole-
cules are indicated by dark and light circles, respectively. When the partition
is removed, we know that the N 2 molecules will start diffusing into the air
while the 2 molecules diffuse into the N 2 , as shown in Figure 14-2. If we
wait long enough, we will have a homogeneous mixture of N 2 and 2 in the
tank. This mass diffusion process can be explained by considering an imagi-
nary plane indicated by the dashed line in the figure as: Gas molecules move
randomly, and thus the probability of a molecule moving to the right or to the
left is the same. Consequently, half of the molecules on one side of the dashed
line at any given moment will move to the other side. Since the concentration
of N 2 is greater on the left side than it is on the right side, more N 2 molecules
will move toward the right than toward the left, resulting in a net flow of N 2
toward the right. As a result, N 2 is said to be transferred to the right. A similar
argument can be given for 2 being transferred to the left. The process con-
tinues until uniform concentrations of N 2 and 2 are established throughout
the tank so that the number of N 2 (or 2 ) molecules moving to the right equals
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719
CHAPTER 14
the number moving to the left, resulting in zero net transfer of N 2 or 2 across
an imaginary plane.
The molecules in a gas mixture continually collide with each other, and the
diffusion process is strongly influenced by this collision process. The collision
of like molecules is of little consequence since both molecules are identical
and it makes no difference which molecule crosses a certain plane. The colli-
sions of unlike molecules, however, influence the rate of diffusion since un-
like molecules may have different masses and thus different momentums, and
thus the diffusion process will be dominated by the heavier molecules. The
diffusion coefficients and thus diffusion rates of gases depend strongly on
temperature since the temperature is a measure of the average velocity of gas
molecules. Therefore, the diffusion rates will be higher at higher temperatures.
Mass transfer can also occur in liquids and solids as well as in gases. For ex-
ample, a cup of water left in a room will eventually evaporate as a result of
water molecules diffusing into the air (liquid-to-gas mass transfer). Apiece of
solid C0 2 (dry ice) will also get smaller and smaller in time as the C0 2 mole-
cules diffuse into the air (solid-to-gas mass transfer). A spoon of sugar in a
cup of coffee will eventually move up and sweeten the coffee although the
sugar molecules are much heavier than the water molecules, and the mole-
cules of a colored pencil inserted into a glass of water will diffuse into the
water as evidenced by the gradual spread of color in the water (solid-to-liquid
mass transfer). Of course, mass transfer can also occur from a gas to a liquid
or solid if the concentration of the species is higher in the gas phase. For ex-
ample, a small fraction of 2 in the air diffuses into the water and meets the
oxygen needs of marine animals. The diffusion of carbon into iron during
case-hardening, doping of semiconductors for transistors, and the migration of
doped molecules in semiconductors at high temperature are examples of solid-
to-solid diffusion processes (Fig. 14-3).
Another factor that influences the diffusion process is the molecular
spacing. The larger the spacing, in general, the higher the diffusion rate.
Therefore, the diffusion rates are typically much higher in gases than they are
in liquids and much higher in liquids than in solids. Diffusion coefficients in
gas mixtures are a few orders of magnitude larger than these of liquid or solid
solutions.
14-2 ■ ANALOGY BETWEEN
HEAT AND MASS TRANSFER
We have spent a considerable amount of time studying heat transfer, and we
could spend just as much time (perhaps more) studying mass transfer. How-
ever, the mechanisms of heat and mass transfer are analogous to each other,
and thus we can develop an understanding of mass transfer in a short time
with little effort by simply drawing parallels between heat and mass transfer.
Establishing those "bridges" between the two seemingly unrelated areas will
make it possible to use our heat transfer knowledge to solve mass transfer
problems. Alternately, gaining a working knowledge of mass transfer will help
us to better understand the heat transfer processes by thinking of heat as
a massless substance as they did in the nineteenth century. The short-lived
caloric theory of heat is the origin of most heat transfer terminology used
Ail-
Water vapor
inn
Liquid
water
(a) Liquid to gas (b) Solid to gas
(c) Solid to liquid (d) Solid to solid
FIGURE 14-3
Some examples of mass transfer that
involve a liquid and/or a solid.
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720
HEAT TRANSFER
70°C
10°C
- Heat concentration
- Mass concentration
FIGURE 14-4
Analogy between
heat and mass transfer.
today and served its purpose well until it was replaced by the kinetic theory.
Mass is, in essence, energy since mass and energy can be converted to each
other according to Einstein's formula E = mc 2 , where c is the speed of light.
Therefore, we can look at mass and heat as two different forms of energy and
exploit this to advantage without going overboard.
Temperature
The driving force for heat transfer is the temperature difference. In contrast,
the driving force for mass transfer is the concentration difference. We can
view temperature as a measure of "heat concentration," and thus a high tem-
perature region as one that has a high heat concentration (Fig. 14-4). There-
fore, both heat and mass are transferred from the more concentrated regions to
the less concentrated ones. If there is no temperature difference between two
regions, then there is no heat transfer. Likewise, if there is no difference be-
tween the concentrations of a species at different parts of a medium, there will
be no mass transfer.
No mass
radiation
FIGURE 14-5
Unlike heat radiation, there is
no such thing as mass radiation.
Concentration profile
of species A
FIGURE 14-6
Analogy between heat conduction
and mass diffusion.
Conduction
You will recall that heat is transferred by conduction, convection, and radia-
tion. Mass, however, is transferred by conduction (called diffusion) and con-
vection only, and there is no such thing as "mass radiation" (unless there is
something Scotty knows that we don't when he "beams" people to anywhere
in space at the speed of light) (Fig. 14-5). The rate of heat conduction in a di-
rection x is proportional to the temperature gradient dTld.x in that direction and
is expressed by Fourier's law of heat conduction as
6 c
-kA
cJT
dx
(14-2)
where k is the thermal conductivity of the medium and A is the area normal to
the direction of heat transfer. Likewise, the rate of mass diffusion m diff of a
chemical species A in a stationary medium in the direction x is proportional to
the concentration gradient dCldx in that direction and is expressed by Fick's
law of diffusion by (Fig. 14-6)
dC A
-D AB A — A
AB dx
(14-3)
where D AB is the diffusion coefficient (or mass diffusivity) of the species in
the mixture and C A is the concentration of the species in the mixture at that
location.
It can be shown that the differential equations for both heat conduction and
mass diffusion are of the same form. Therefore, the solutions of mass dif-
fusion equations can be obtained from the solutions of corresponding heat
conduction equations for the same type of boundary conditions by simply
switching the corresponding coefficients and variables.
Heat Generation
Heat generation refers to the conversion of some form of energy such as elec-
trical, chemical, or nuclear energy into sensible heat energy in the medium.
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CHAPTER 14
Heat generation occurs throughout the medium and exhibits itself as a rise in
temperature. Similarly, some mass transfer problems involve chemical reac-
tions that occur within the medium and result in the generation of a species
throughout. Therefore, species generation is a volumetric phenomenon, and
the rate of generation may vary from point to point in the medium. Such reac-
tions that occur within the medium are called homogeneous reactions and are
analogous to internal heat generation. In contrast, some chemical reactions
result in the generation of a species at the surface as a result of chemical re-
actions occurring at the surface due to contact between the medium and the
surroundings. This is a surface phenomenon, and as such it needs to be treated
as a boundary condition. In mass transfer studies, such reactions are called
heterogeneous reactions and are analogous to specified surface heat flux.
Convection
You will recall that heat convection is the heat transfer mechanism that in-
volves both heat conduction (molecular diffusion) and bulk fluid motion.
Fluid motion enhances heat transfer considerably by removing the heated
fluid near the surface and replacing it by the cooler fluid further away. In the
limiting case of no bulk fluid motion, convection reduces to conduction. Like-
wise, mass convection (or convective mass transfer) is the mass transfer
mechanism between a surface and a moving fluid that involves both mass dif-
fusion and bulk fluid motion. Fluid motion also enhances mass transfer con-
siderably by removing the high concentration fluid near the surface and
replacing it by the lower concentration fluid further away. In mass convection,
we define a concentration boundary layer in an analogous manner to the ther-
mal boundary layer and define new dimensionless numbers that are counter-
parts of the Nusselt and Prandtl numbers.
The rate of heat convection for external flow was expressed conveniently by
Newton's law of cooling as
fie
"conv^A^s *•»)
(14-4)
where h conv is the heat transfer coefficient, A s is the surface area, and T s — T„
is the temperature difference across the thermal boundary layer. Likewise, the
rate of mass convection can be expressed as (Fig. 14-7)
"mass ^A ^s *-W
(14-5)
where /z mass is the mass transfer coefficient, A s is the surface area, and C s — C ro
is a suitable concentration difference across the concentration boundary layer.
Various aspects of the analogy between heat and mass convection are ex-
plored in Section 14-9. The analogy is valid for low mass transfer rate cases
in which the flow rate of species undergoing mass flow is low (under 10 per-
cent) relative to the total flow rate of the liquid or gas mixture.
14-3 - MASS DIFFUSION
Fick's law of diffusion, proposed in 1855, states that the rate of diffusion of a
chemical species at a location in a gas mixture (or liquid or solid solution) is
proportional to the concentration gradient of that species at that location.
Mass
transfer
Concentration
coefficient
difference
\
Mass
\
convection:
m conv = h
^ S ,A,(C S - C„)
Heat
convection:
Gconv = h<
/
Heat
transfer
Temperature
coefficient
difference
FIGURE 14-7
Analogy between convection heat
transfer and convection mass transfer.
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722
HEAT TRANSFER
A + B mixture
O O Of ft > O O/s ° o
' # ^o o ooo o oo 4<
OOO O o o °
;.*;>'
♦o I v ; v <>
P = Pa + Pa
Mass basis:
P.4 :
Mole basis:
'".4
I''
V' y A :
Pa
zA.
C
Relation between them:
Ca ~m a ' w *-^m
FIGURE 14-8
Different ways of expressing the
concentration of species A of
a binary mixture A and B.
Although a higher concentration for a species means more molecules of that
species per unit volume, the concentration of a species can be expressed in
several ways. Next we describe two common ways.
1 Mass Basis
On a mass basis, concentration is expressed in terms of density (or mass con-
centration), which is mass per unit volume. Considering a small volume V at
a location within the mixture, the densities of a species (subscript i) and of the
mixture (no subscript) at that location are given by (Fig. 14-8)
Partial density of species i: p,
Total density of mixture: p
m,IV
m I V = ^ m l I V = ^ p,
(kg/m 3 )
Therefore, the density of a mixture at a location is equal to the sum of the
densities of its constituents at that location. Mass concentration can also be
expressed in dimensionless form in terms of mass fraction w as
Mass fraction of species i:
in l
Hi
m/V
Pi
P
(14-6)
Note that the mass fraction of a species ranges between and 1, and the con-
servation of mass requires that the sum of the mass fractions of the con-
stituents of a mixture be equal to 1. That is, 2w ; = 1. Also note that the density
and mass fraction of a constituent in a mixture, in general, vary with location
unless the concentration gradients are zero.
2 Mole Basis
On a mole basis, concentration is expressed in terms of molar concentration
(or molar density), which is the amount of matter in kmol per unit volume.
Again considering a small volume V at a location within the mixture, the mo-
lar concentrations of a species (subscript i) and of the mixture (no subscript)
at that location are given by
Partial molar concentration of species i: C, = NjTV
Total molar concentration of mixture: C = N/V = ^ NJV = ^ C ;
(kmol/m 3 )
Therefore, the molar concentration of a mixture at a location is equal to the
sum of the molar concentrations of its constituents at that location. Molar
concentration can also be expressed in dimensionless form in terms of mole
fraction y as
Mole fraction of species i:
y,
N, _N t IV _ C,
N~~NJV~C
(14-7)
Again the mole fraction of a species ranges between and 1, and the sum of
the mole fractions of the constituents of a mixture is unity, Sy, = 1.
The mass m and mole number N of a substance are related to each other by
m = NM (or, for a unit volume, p = CM) where M is the molar mass (also
called the molecular weight) of the substance. This is expected since the mass
of 1 kmol of the substance is M kg, and thus the mass of N kmol is NM kg.
Therefore, the mass and molar concentrations are related to each other by
cen58 93 3_chl4.qxd 9/9/2002 10:05 AM Page 723
P, P
C, = 77 (for species i) and C = -rz (for the mixture)
(14-8)
723
CHAPTER 14
where M is the molar mass of the mixture which can be determined from
M
N
5>,m, N
N
(14-9)
The mass and mole fractions of species i of a mixture are related to each
other by
W > P CM Ji M
(14-10)
Two different approaches are presented above for the description of con-
centration at a location, and you may be wondering which approach is better
to use. Well, the answer depends on the situation on hand. Both approaches
are equivalent, and the better approach for a given problem is the one that
yields the desired solution more easily.
Special Case: Ideal Gas Mixtures
At low pressures, a gas or gas mixture can conveniently be approximated as
an ideal gas with negligible error. For example, a mixture of dry air and water
vapor at atmospheric conditions can be treated as an ideal gas with an error
much less than 1 percent. The total pressure of a gas mixture P is equal to the
sum of the partial pressures P t of the individual gases in the mixture and is ex-
pressed as P = HP j. Here P t is called the partial pressure of species i, which
is the pressure species i would exert if it existed alone at the mixture temper-
ature and volume. This is known as Dalton's law of additive pressures. Then
using the ideal gas relation PV = NR U T where R u is the universal gas constant
for both the species i and the mixture, the pressure fraction of species i can
be expressed as (Fig. 14-9)
P,
P
NR„T/V N
y>
(14-11)
Therefore, the pressure fraction of species (' of an ideal gas mixture is equiva-
lent to the mole fraction of that species and can be used in place of it in mass
transfer analysis.
Fick's Law of Diffusion:
Stationary Medium Consisting of Two Species
We mentioned earlier that the rate of mass diffusion of a chemical species in
a stagnant medium in a specified direction is proportional to the local concen-
tration gradient in that direction. This linear relationship between the rate of
diffusion and the concentration gradient proposed by Fick in 1855 is known
as Fick's law of diffusion and can be expressed as
2molA
6molB
P = 120 kPa
A mixture of two ideal gases A and B
Pa
N A
= 0.25
N 2 + 6
y A P = 0.25 X 120 = 30 kPa
FIGURE 14-9
For ideal gas mixtures, pressure
fraction of a gas is equal to
its mole fraction.
Mass flux = Constant of proportionality X Concentration gradient
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HEAT TRANSFER
Higher
Lower
concentration
concentration
of species A
of species A
Area
A
C A (x)
Concentration
profile of species A
Mass basis:
- P AD A
-pAD A
dw A
' dx
d(p A /p)
dx
Mole basis:
dpA
-AD AB —2- (if p = constant)
4fi dx
d(C A /C)
-CAD
'AB'
dx
-AD,
dC^
'~dx
(if C = constant)
FIGURE 14-10
Various expressions of Fick's law of
diffusion for a binary mixture.
Mass Concentration
diffusivity gradient
\ i
. \, dp A
m A = -D AB A —
Mass diffusion:
Heat conduction
cy<
/ t
Thermal Temperature
conductivity gradient
FIGURE 14-11
Analogy between Fourier's law of
heat conduction and Fick's
law of mass diffusion.
But the concentration of a species in a gas mixture or liquid or solid solution
can be defined in several ways such as density, mass fraction, molar concen-
tration, and mole fraction, as already discussed, and thus Fick's law can be ex-
pressed mathematically in many ways. It turns out that it is best to express the
concentration gradient in terms of the mass or mole fraction, and the most ap-
propriate formulation of Fick's law for the diffusion of a species A in a sta-
tionary binary mixture of species A and B in a specified direction x is given by
(Fig. 14-10)
Mass basis: j ii{
Mole basis: j iiSi/
1 diff, A
N.
diff, A
-pD.
CD AB
d{p A lp)
AB dx "
d(C A IC)
dx
-pD.
dw A
[B dx
dy A
dx
(kg/s ■ m 2 )
(mol/s-m 2 ) (14-12)
Here j diff A is the (diffusive) mass flux of species A (mass transfer by diffusion
per unit time and per unit area normal to the direction of mass transfer, in
kg/s • m 2 ) and y diff A is the (diffusive) molar flux (in kmol/s • m 2 ). The mass
flux of a species at a location is proportional to the density of the mixture at
that location. Note that p = p A + p B is the density and C = C A + C B is the
molar concentration of the binary mixture, and in general, they may vary
throughout the mixture. Therefore, pd(p A /p) =fc dp A or Cd(C A /C) i= dC A . But
in the special case of constant mixture density p or constant molar concentra-
tion C, the relations above simplify to
Mass basis (p = constant): 7diff,A = ~~ D AB
dp A
dx
dC A
Mole basis (C = constant): /diff, a = ~^ab~T~
(kg/s ■ m 2 )
(kmol/s -m 2 ) (14-13)
The constant density or constant molar concentration assumption is usually
appropriate for solid and dilute liquid solutions, but often this is not the case
for gas mixtures or concentrated liquid solutions. Therefore, Eq. 14-12 should
be used in the latter case. In this introductory treatment we will limit our con-
sideration to one-dimensional mass diffusion. For two- or three-dimensional
cases, Fick's law can conveniently be expressed in vector form by simply re-
placing the derivatives in the above relations by the corresponding gradients
(such as j A = -pD AB V w A ).
Remember that the constant of proportionality in Fourier's law was defined
as the transport property thermal conductivity. Similarly, the constant of pro-
portionality in Fick's law is defined as another transport property called the
binary diffusion coefficient or mass diffusivity, D AB . The unit of mass diffu-
sivity is m 2 /s, which is the same as the units of thermal diffusivity or momen-
tum diffusivity (also called kinematic viscosity) (Fig. 14-11).
Because of the complex nature of mass diffusion, the diffusion coefficients
are usually determined experimentally. The kinetic theory of gases indicates
that the diffusion coefficient for dilute gases at ordinary pressures is essen-
tially independent of mixture composition and tends to increase with temper-
ature while decreasing with pressure as
D A
J 3/2
~P~
Di
Pi IT,
(14-14)
cen58 93 3_chl4.qxd 9/9/2002 10:05 AM Page 725
This relation is useful in determining the diffusion coefficient for gases at dif-
ferent temperatures and pressures from a knowledge of the diffusion coeffi-
cient at a specified temperature and pressure. More general but complicated
relations that account for the effects of molecular collisions are also available.
The diffusion coefficients of some gases in air at 1 aim pressure are given in
Table 14-1 at various temperatures.
The diffusion coefficients of solids and liquids also tend to increase with
temperature while exhibiting a strong dependence on the composition. The
diffusion process in solids and liquids is a great deal more complicated than
that in gases, and the diffusion coefficients in this case are almost exclusively
determined experimentally.
The binary diffusion coefficient for several binary gas mixtures and solid
and liquid solutions are given in Tables 14-2 and 14-3. We make these two
observations from these tables:
1. The diffusion coefficients, in general, are highest in gases and lowest
in solids. The diffusion coefficients of gases are several orders of
magnitude greater than those of liquids.
2. Diffusion coefficients increase with temperature. The diffusion
coefficient (and thus the mass diffusion rate) of carbon through iron
during a hardening process, for example, increases by 6000 times as
the temperature is raised from 500°C to 1000°C.
Due to its practical importance, the diffusion of water vapor in air has been
the topic of several studies, and some empirical formulas have been developed
for the diffusion coefficient D H o-air- Marrero and Mason proposed this popu-
lar formula (Table 14-4):
725
CHAPTER 14
TABLE 14-1
Binary diffusion coefficients of
some gases in air at 1 atm pressure
(from Mills, Ref. 13, Table A. 17a,
p. 869)
Binary Diffusion Coefficient, *
m 2 /s
x 10 s
T, K
2
C0 2
H 2
NO
200
0.95
0.74
3.75
0.88
300
1.88
1.57
7.77
1.80
400
5.25
2.63
12.5
3.03
500
4.75
3.85
17.1
4.43
600
6.46
5.37
24.4
6.03
700
8.38
6.84
31.7
7.82
800
10.5
8.57
39.3
9.78
900
12.6
10.5
47.7
11.8
1000
15.2
12.4
56.9
14.1
1200
20.6
16.9
77.7
19.2
1400
26.6
21.7
99.0
24.5
1600
33.2
27.5
125
30.4
1800 40.3
32.8
152
37.0
2000 48.0
39.4
180
44.8
*Mu Iti ply by 10.76 to convert to ft 2 /s.
TABLE 14-2
Binary diffusion coefficients of dilute gas mixtures at 1 atm
(from Barrer, Ref. 2; Geankoplis, Ref. 5; Perry, Ref. 14; and Reid et al.
Ref. 15)
Substance
Substance
r,
D AB or
D m ,
Substance
r,
D AB or D BA ,
A
B
K
m 2 /s
Substance A
B
K
m 2 /s
Air
Acetone
273
1.1 X
10- 5
Argon, Ar
Nitrogen, N 2
293
1.9 X 10- 5
Air
Ammonia, NH 3
298
2.6 x
10- 5
Carbon dioxide
C0 2
Benzene
318
0.72 X 10- 5
Air
Benzene
298
0.88 x
10- 5
Carbon dioxide
C0 2
Hydrogen, H 2
273
5.5 X 10- 5
Air
Carbon dioxide
298
1.6 x
10- 5
Carbon dioxide
C0 2
Nitrogen, N 2
293
1.6 X 10- 5
Air
Chlorine
273
1.2 x
10- 5
Carbon dioxide
C0 2
Oxygen, 2
273
1.4 x 10- 5
Air
Ethyl alcohol
298
1.2 x
10- 5
Carbon dioxide
C0 2
Water vapor
298
1.6 x 10- 5
Air
Ethyl ether
298
0.93 x
10- 5
Hydrogen, H 2
Nitrogen, N 2
273
6.8 x 10- 5
Air
Helium, He
298
7.2 x
10- 5
Hydrogen, H 2
Oxygen, 2
273
7.0 x 10- 5
Air
Hydrogen, H 2
298
7.2 x
10- 5
Oxygen, 2
Ammonia
293
2.5 x 10- 5
Air
Iodine, l 2
298
0.83 x
10- 5
Oxygen, 2
Benzene
296
0.39 x 10- 5
Air
Methanol
298
1.6 x
10- 5
Oxygen, 2
Nitrogen, N 2
273
1.8 x 10- 5
Air
Mercury
614
4.7 x
10- 5
Oxygen, 2
Water vapor
298
2.5 x 10- 5
Air
Napthalene
300
0.62 x
10- 5
Water vapor
Argon, Ar
298
2.4 x 10- 5
Air
Oxygen, 2
298
2.1 x
10- 5
Water vapor
Helium, He
298
9.2 x 10- 5
Air
Water vapor
298
2.5 x
10- 5
Water vapor
Nitrogen, N 2
298
2.5 x 10- 5
Note: The effect of pressure and temperature on D AB can be accounted for through D AB ~ P I2 IP. Also, multiply D AB values by 10.76 to convert them to ft 2 /s.
cen58 93 3_chl4.qxd 9/9/2002 10:05 AM Page 726
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HEAT TRANSFER
TABLE 14-3
Binary diffusion coefficients of dilute liquid solutions and solid solutions at 1 atm
(from Barrer, Ref. 2; Reid et al., Ref. 15; Thomas, Ref. 19; and van Black, Ref. 20)
(a) Diffusion through Liquids
(b) Diffusion through Solids
Substance
Substance
T,
Substance
Substance
B
r,
A (Solute)
B (Solvent)
K
D AB , m 2 /s
A (Solute)
(Solvent)
K
D AB ,
m 2 /s
Ammonia
Water
285
1.6 X
10- 9
Carbon dioxide
Natural ru
)ber
298
1.1 X
io- 10
Benzene
Water
293
1.0 X
10-9
Nitrogen
Natural ru
)ber
298
1.5 x
io- 10
Carbon dioxic
e Water
298
2.0 X
10-9
Oxygen
Natural ru
)ber
298
2.1 x
io- 10
Chlorine
Water
285
1.4 X
10-9
Helium
Pyrex
773
2.0 x
io- 12
Ethanol
Water
283
0.84 x
10-9
Helium
Pyrex
293
4.5 x
io- 15
Ethanol
Water
288
1.0 x
10-9
Helium
Silicon dioxide
298
4.0 x
io- 14
Ethanol
Water
298
1.2 x
10-9
Hydrogen
Iron
298
2.6 x
io- 13
Glucose
Water
298
0.69 x
10-9
Hydrogen
Nickel
358
1.2 x
io- 12
Hydrogen
Water
298
6.3 x
10-9
Hydrogen
Nickel
438
1.0 x
io- 11
Methane
Water
275
0.85 x
10-9
Cadmium
Copper
293
2.7 x
io- 19
Methane
Water
293
1.5 x
10-9
Zinc
Copper
773
4.0 x
io- 18
Methane
Water
333
3.6 x
10-9
Zinc
Copper
1273
5.0 x
io- 13
Methanol
Water
288
1.3 x
10-9
Antimony
Silver
293
3.5 x
io- 25
Nitrogen
Water
298
2.6 x
10-9
Bismuth
Lead
293
1.1 X
io- 20
Oxygen
Water
298
2.4 x
10-9
Mercury
Lead
293
2.5 x
io- 19
Water
Ethanol
298
1.2 x
10-9
Copper
Aluminum
773
4.0 x
io- 14
Water
Ethylene glycol
298
0.18 x
10-9
Copper
Aluminum
1273
1.0 x
io- 10
Water
Methanol
298
1.8 x
10-9
Carbon
Iron (fee)
773
5.0 x
io- 15
Chloroform
Methanol
288
2.1 x
10-9
Carbon
Iron (fee)
1273
3.0 x
io- 11
V
J
^H,0-Air
= 1.87 X 10-'°-
D (m 2 /s),
280 K < T < 450 K
(14-15)
TABLE 14-4
In a binary ideal gas mixture of
species A and 6, the diffusion
coefficient of A in B is equal to
the diffusion coefficient of B in A,
and both increase with temperature
"|H 2 0-Air 0r "Air-H 2
at 1 atm, in m 2 /s
7", °C
(from Eq. 14-15)
2.09 X IO- 5
5
2.17 X IO" 5
10
2.25 X IO- 5
15
2.33 X IO- 5
20
2.42 X IO- 5
25
2.50 X IO- 5
30
2.59 X IO- 5
35
2.68 X IO- 5
40
2.77 X IO- 5
50
2.96 X IO- 5
100
3.99 X IO- 5
150
5.18 X IO- 5
where P is total pressure in atm and T is the temperature in K.
The primary driving mechanism of mass diffusion is the concentration
gradient, and mass diffusion due to a concentration gradient is known as the
ordinary diffusion. However, diffusion may also be caused by other effects.
Temperature gradients in a medium can cause thermal diffusion (also called
the soret effect), and pressure gradients may result in pressure diffusion.
Both of these effects are usually negligible, however, unless the gradients are
very large. In centrifuges, the pressure gradient generated by the centrifugal
effect is used to separate liquid solutions and gaseous isotopes. An external
force field such as an electric or magnetic field applied on a mixture or solu-
tion can be used successfully to separate electrically charged or magnetized
molecules (as in an electrolyte or ionized gas) from the mixture. This is called
forced diffusion. Also, when the pores of a porous solid such as silica-gel are
smaller than the mean free path of the gas molecules, the molecular collisions
may be negligible and a free molecule flow may be initiated. This is known as
Knudsen diffusion. When the size of the gas molecules is comparable to the
pore size, adsorbed molecules move along the pore walls. This is known as
surface diffusion. Finally, particles whose diameter is under 0.1 (xm such as
mist and soot particles act like large molecules, and the diffusion process of
such particles due to the concentration gradient is called Brownian motion.
Large particles (those whose diameter is greater than 1 |xm) are not affected
cen58 93 3_chl4.qxd 9/9/2002 10:05 AM Page 727
727
CHAPTER 14
by diffusion as the motion of such particles is governed by Newton's laws. In
our elementary treatment of mass diffusion, we will assume these additional
effects to be nonexistent or negligible, as is usually the case, and refer the in-
terested reader to advanced books on these topics.
EXAMPLE 14-1 Determining Mass Fractions from Mole Fractions
The composition of dry standard atmosphere is given on a molar basis to be
78.1 percent N 2 , 20.9 percent 2 , and 1.0 percent Ar and other constituents
(Fig. 14-12). Treating other constituents as Ar, determine the mass fractions of
the constituents of air.
SOLUTION The molar fractions of the constituents of air are given. The mass
fractions are to be determined.
Assumptions The small amounts of other gases in air are treated as argon.
Properties The molar masses of N 2 , 2 , and Ar are 28.0, 32.0, and 39.9
kg/kmol, respectively (Table A-l).
Analysis The molar mass of air is determined to be
M = 2 y,M, = 0.781 X 28.0 + 0.209 X 32.0 + 0.01 X 39.9 = 29.0 kg/kmol
Then the mass fractions of constituent gases are determined from Eq. 14-10 to
be
0.754
N 2 :
w Nj
-y», M
= (0.781) §1
2 :
w .
M ,
-*>, M
32
= (0.209) —
Ar:
w At
~ y * M
39.9
= (0.01) ^^ =
v ; 29.0
0.231
0.014
Therefore, the mass fractions of N 2 , 2 , and Ar in dry standard atmosphere are
75.4 percent, 23.1 percent, and 1.4 percent, respectively.
14^ - BOUNDARY CONDITIONS
We mentioned earlier that the mass diffusion equation is analogous to the heat
diffusion (conduction) equation, and thus we need comparable boundary con-
ditions to determine the species concentration distribution in a medium. Two
common types of boundary conditions are the (1) specified species concen-
tration, which corresponds to specified temperature, and (2) specified species
flux, which corresponds to specified heat flux.
Despite their apparent similarity, an important difference exists between
temperature and concentration: temperature is necessarily a continuous func-
tion, but concentration, in general, is not. The wall and air temperatures at a
wall surface, for example, are always the same. The concentrations of air on
the two sides of a water-air interface, however, are obviously very different
(in fact, the concentration of air in water is close to zero). Likewise, the
concentrations of water on the two sides of a water-air interface are also dif-
ferent even when air is saturated (Fig. 14-13). Therefore, when specifying a
AIR
78.1% N,
20.9% O,
1.0% Ar
FIGURE 14-12
Schematic for Example 14—1.
X
i
\ Air
\ S - V H,0, gas side (")
1 —
Jump in
concentration
Water
Concentration
profile
V
-'H-,0, liquid side
...
FIGURE 14-13
Unlike temperature, the concentration
of species on the two sides of a
liquid-gas (or solid-gas or
solid-liquid) interface are
usually not the same.
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728
HEAT TRANSFER
boundary condition, specifying the location is not enough. We also need to
specify the side of the boundary. To do this, we consider two imaginary sur-
faces on the two sides of the interface that are infinitesimally close to the
interface. Whenever there is a doubt, we indicate the desired side of the inter-
face by specifying its phase as a subscript. For example, the water (liquid
or vapor) concentration at the liquid and gas sides of a water-air interface at
x = can be expressed on a molar basis is
^HiO, liquid side (0) ~~ "9\
and
J'HjO, gas side (0) — ^2
(14-16)
Impermeable
surface
FIGURE 14-14
An impermeable surface in
mass transfer is analogous to an
insulated surface in heat transfer.
Using Fick's law, the constant species flux boundary condition for a diffus-
ing species A at a boundary at x = is expressed, in the absence of any blow-
ing or suction, as
CD,
dy A
1 dx
z Ja.o
PD A
dw A
' dx
~~ Ja, o
(14-17)
where j A and j A are the specified mole and mass fluxes of species A at the
boundary, respectively. The special case of zero mass flux (] A = j A = 0)
corresponds to an impermeable surface for which dy A (0)/dx = dw A (0)/
dx = (Fig. 14-14).
To apply the specified concentration boundary condition, we must know the
concentration of a species at the boundary. This information is usually ob-
tained from the requirement that thermodynamic equilibrium must exist at the
interface of two phases of a species. In the case of air-water interface, the con-
centration values of water vapor in the air are easily determined from satura-
tion data, as shown in Example 14-2.
92 kPa, 15°C
Saturated
air
/■ ^H 2 0, air side =
0.0185
- V H,0, liquid
Lake
side^ 1 -
15°C
FIGURE 14-15
Schematic for Example 14-
-2.
EXAMPLE 14-2 Mole Fraction of Water Vapor
at the Surface of a Lake
Determine the mole fraction of the water vapor at the surface of a lake whose
temperature is 15°C and compare it to the mole fraction of water in the lake
(Fig. 14-15). Take the atmospheric pressure at lake level to be 92 kPa.
SOLUTION The mole fraction of the water vapor at the surface of a lake and
the mole fraction of water in the lake are to be determined and compared.
Assumptions 1 Both the air and water vapor are ideal gases. 2 The mole frac-
tion of dissolved air in water is negligible.
Properties The saturation pressure of water at 15°C is 1.705 kPa (Table A-9).
Analysis The air at the water surface will be saturated. Therefore, the partial
pressure of water vapor in the air at the lake surface will simply be the satura-
tion pressure of water at 15°C,
P = P
1 vapor J si
1.705 kPa
Assuming both the air and vapor to be ideal gases, the mole fraction of water
vapor in the air at the surface of the lake is determined from Eq. 14-11 to be
1.705 kPa
92kPa
= 0.0185 (or 1.85 percent)
cen58 93 3_chl4.qxd 9/9/2002 10:05 AM Page 729
Water contains some dissolved air, but the amount is negligible. Therefore, we
can assume the entire lake to be liquid water. Then its mole fraction becomes
^water, liquid side = 1-0 (or 1 00 percent)
Discussion Note that the concentration of water on a molar basis is 100 per-
cent just beneath the air-water interface and 1.85 percent just above it, even
though the air is assumed to be saturated (so this is the highest value at 15°C).
Therefore, huge discontinuities can occur in the concentrations of a species
across phase boundaries.
729
CHAPTER 14
The situation is similar at solid-liquid interfaces. Again, at a given temper-
ature, only a certain amount of solid can be dissolved in a liquid, and the sol-
ubility of the solid in the liquid is determined from the requirement that
thermodynamic equilibrium exists between the solid and the solution at the in-
terface. The solubility represents the maximum amount of solid that can be
dissolved in a liquid at a specified temperature and is widely available in
chemistry handbooks. In Table 14-5 we present sample solubility data for
sodium chloride (NaCl) and calcium bicarbonate [Ca(HC0 3 ) 2 ] at various
temperatures. For example, the solubility of salt (NaCl) in water at 310 K is
36.5 kg per 100 kg of water. Therefore, the mass fraction of salt in the brine at
the interface is simply
salt, liquid side
/W salt
m
36.5 kg
(100 + 36.5) kg
0.267 (or 26.7 percent)
whereas the mass fraction of salt in the pure solid salt is w = 1.0. Note that
water becomes saturated with salt when 36.5 kg of salt are dissolved in
100kgofwaterat310K.
Many processes involve the absorption of a gas into a liquid. Most gases are
weakly soluble in liquids (such as air in water), and for such dilute solutions
the mole fractions of a species i in the gas and liquid phases at the interface
are observed to be proportional to each other. That is, v,
gas side "(', liquid side
or
Pi, gas s,de 0C P Ji, liquid side since y u gas side = Pj gas side IP for ideal gas mixtures. This
is known as Henry's law and is expressed as
)',. i
quid side
i, gas side
(at interface)
(14-18)
where H is Henry's constant, which is the product of the total pressure of the
gas mixture and the proportionality constant. For a given species, it is a func-
tion of temperature only and is practically independent of pressure for pres-
sures under about 5 arm. Values of Henry's constant for a number of aqueous
solutions are given in Table 14-6 for various temperatures. From this table
and the equation above we make the following observations:
1. The concentration of a gas dissolved in a liquid is inversely proportional
to Henry's constant. Therefore, the larger Henry's constant, the smaller
the concentration of dissolved gases in the liquid.
TABLE
Solubility of two inorganic
compounds in water at various
temperatures, in kg, in 100 kg of
water [from Handbook of Chemistry
(New York: McGraw-Hill, 1961)]
Solute
Calcium
Tempera-
Salt,
Bicarbonate,
ture, K
NaCl
Ca(HC0 3 ) 2
273.15
35.7
16.15
280
35.8
16.30
290
35.9
16.53
300
36.2
16.75
310
36.5
16.98
320
36.9
17.20
330
37.2
17.43
340
37.6
17.65
350
38.2
17.88
360
38.8
18.10
370
39.5
18.33
373.15
39.8
18.40
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HEAT TRANSFER
Gas A
\\\\y
-A, gas side
Gas: A
Liquid: B
~ -M, liquid side
- A, gas side -'A, liquid side
p
A, gas side
-A, liquid side
A, gas side -A, liquid side
FIGURE 14-16
Dissolved gases in a liquid can
be driven off by heating the liquid.
Air
TABLE 14-6
Henry's constant H (in bars) for selected gases in water at low to moderate
pressures (for gas /, H = P /igas side /y,, wate r side)
(from Mills, Ref. 13, Table A. 21, p. 874)
Solute
290 K
300 K
310 K
320 K
330 K
340 K
H 2 S
440
560
700
830
980
1140
C0 2
1280
1710
2170
2720
3220
—
2
38,000
45,000
52,000
57,000
61,000
65,000
H 2
67,000
72,000
75,000
76,000
77,000
76,000
CO
51,000
60,000
67,000
74,000
80,000
84,000
Air
62,000
74,000
84,000
92,000
99,000
104,000
N 2
76,000
89,000
101,000
110,000
118,000
124,000
2. Henry's constant increases (and thus the fraction of a dissolved gas
in the liquid decreases) with increasing temperature. Therefore, the
dissolved gases in a liquid can be driven off by heating the liquid
(Fig. 14-16).
3. The concentration of a gas dissolved in a liquid is proportional to the
partial pressure of the gas. Therefore, the amount of gas dissolved in a
liquid can be increased by increasing the pressure of the gas. This can
be used to advantage in the carbonation of soft drinks with C0 2 gas.
Strictly speaking, the result obtained from Eq. 14-18 for the mole fraction of
dissolved gas is valid for the liquid layer just beneath the interface and not
necessarily the entire liquid. The latter will be the case only when thermo-
dynamic phase equilibrium is established throughout the entire liquid body.
Saturated air>
p
/ dry air, gas side
Lake
17°C
-dry air, liquid side
FIGURE 14-17
Schematic for
example 14-3.
EXAMPLE 14-3 Mole Fraction of Dissolved Air in Water
Determine the mole fraction of air dissolved in water at the surface of a lake
whose temperature is 17 C C (Fig. 14-17). Take the atmospheric pressure at lake
level to be 92 kPa.
SOLUTION The mole fraction of air dissolved in water at the surface of a lake
is to be determined.
Assumptions 1 Both the air and water vapor are ideal gases. 2 Air is weakly sol-
uble in water so that Henry's law is applicable.
Properties The saturation pressure of water at 17°C is 1.92 kPa (Table A-9).
Henry's constant for air dissolved in water at 290 K is H = 62,000 bar (Table
14-6).
Analysis This example is similar to the previous example. Again the air at the
water surface will be saturated, and thus the partial pressure of water vapor in
the air at the lake surface will be the saturation pressure of water at 17°C,
°vapor = "sat <8> 17°C = 1-92 kPa
Assuming both the air and vapor to be ideal gases, the partial pressure of dry air
is determined to be
dry air
92 - 1.92 = 90.08 kPa = 0.9008 bar
cen58 93 3_chl4.qxd 9/9/2002 10:05 AM Page 731
Note that with little loss in accuracy (an error of about 2 percent), we could
have ignored the vapor pressure since the amount of vapor in air is so small.
Then the mole fraction of air in the water becomes
.Miry air, liquid state
dry air, gas side 0.9008 bar
62,000 bar
H
1.45 x 10" s
which is very small, as expected. Therefore, the concentration of air in water
just below the air-water interface is 1.45 moles per 100,0000 moles. But ob-
viously this is enough oxygen for fish and other creatures in the lake. Note that
the amount of air dissolved in water will decrease with increasing depth.
731
CHAPTER 14
We mentioned earlier that the use of Henry's law is limited to dilute gas-
liquid solutions; that is, a liquid with a small amount of gas dissolved in it.
Then the question that arises naturally is, what do we do when the gas is
highly soluble in the liquid (or solid), such as ammonia in water? In this case
the linear relationship of Henry's law does not apply, and the mole fraction of
a gas dissolved in the liquid (or solid) is usually expressed as a function of the
partial pressure of the gas in the gas phase and the temperature. An approxi-
mate relation in this case for the mole fractions of a species on the liquid and
gas sides of the interface is given by Raoult's law as
a = V ;
i, gas side J ;, gas side
}).
p
liquid side i, sat
(n
(14-19)
where P, smCO i s the saturation pressure of the species i at the interface tem-
perature and P is the total pressure on the gas phase side. Tabular data are
available in chemical handbooks for common solutions such as the ammonia-
water solution that is widely used in absorption-refrigeration systems.
Gases may also dissolve in solids, but the diffusion process in this case can
be very complicated. The dissolution of a gas may be independent of the
structure of the solid, or it may depend strongly on its porosity. Some dissolu-
tion processes (such as the dissolution of hydrogen in titanium, similar to the
dissolution of C0 2 in water) are reversible, and thus maintaining the gas con-
tent in the solid requires constant contact of the solid with a reservoir of that
gas. Some other dissolution processes are irreversible. For example, oxygen
gas dissolving in titanium forms Ti0 2 on the surface, and the process does not
reverse itself.
The concentration of the gas species i in the solid at the interface C, so i idside
is proportional to the partial pressure of the species i in the gas P t gas side on the
gas side of the interface and is expressed as
C,
;', solid side
Vxp,
i. gas side
(kmol/m 3 )
(14-20)
where SP is the solubility. Expressing the pressure in bars and noting that the
unit of molar concentration is kmol of species i per m 3 , the unit of solubility is
kmol/m 3 • bar. Solubility data for selected gas-solid combinations are given in
Table 14-7. The product of the solubility of a gas and the diffusion coefficient
of the gas in a solid is referred to as the permeability 9\ which is a measure
of the ability of the gas to penetrate a solid. That is, 2P = ifD AB where D AB is
TABLE 14-7
Solubility of selected
gases and solids
(for gas /, if = C, ; solid M JP i: gasside )
(from Barrer, Ref. 2)
Gas
Solid
7", K
y kmol/m 3 • bar
?
Rubber
298
0.00312
No
Rubber
298
0.00156
C0 ?
Rubber
298
0.04015
He
Si0 2
293
0.00045
H 2
Ni
358
0.00901
cen58 93 3_chl4.qxd 9/9/2002 10:05 AM Page 732
732
HEAT TRANSFER
the diffusivity of the gas in the solid. Permeability is inversely proportional to
thickness and has the unit kmol/s • bar.
Finally, if a process involves the sublimation of a pure solid (such as ice or
solid C0 2 ) or the evaporation of a pure liquid (such as water) in a different
medium such as air, the mole (or mass) fraction of the substance in the liquid
or solid phase is simply taken to be 1.0, and the partial pressure and thus the
mole fraction of the substance in the gas phase can readily be determined from
the saturation data of the substance at the specified temperature. Also, the as-
sumption of thermodynamic equilibrium at the interface is very reasonable for
pure solids, pure liquids, and solutions, except when chemical reactions are
occurring at the interface.
Nickel
plate
358 K
300 kPa ■
Air
L x
FIGURE 14-18
Schematic for Example 14-4.
TABLE 14-8
Analogy between heat conduction
and mass diffusion in a
stationary medium
Mass Diffusion
Heat
Mass
Molar
Conduction
Basis
Basis
T
k
pD AB
y,
CD AB
q
a
L
j,
Dab
L
S,
D AB
L
EXAMPLE 14-4 Diffusion of Hydrogen Gas into a Nickel Plate
Consider a nickel plate that is in contact with hydrogen gas at 358 K and
300 kPa. Determine the molar and mass density of hydrogen in the nickel at
the interface (Fig. 14-18).
SOLUTION A nickel plate is exposed to hydrogen. The molar and mass density
of hydrogen in the nickel at the interface is to be determined.
Assumptions Nickel and hydrogen are in thermodynamic equilibrium at the
interface.
Properties The molar mass of hydrogen is M = 2 kg/kmol (Table A-l). The sol-
ubility of hydrogen in nickel at 358 K is 0.00901 kmol/m 3 ■ bar (Table 14-7).
Analysis Noting that 300 kPa = 3 bar, the molar density of hydrogen in the
nickel at the interface is determined from Eq. 14-20 to be
*-"H,. solid side J ^ -* H,, gas side
= (0.00901 kmol/m 3 • bar)(3 bar) = 0.027 kmol/m 3
It corresponds to a mass density of
PH,. solid side = Ql,, solid sidc H i
= (0.027 kmol/m 3 )(2) = 0.054 kg/m 3
That is, there will be 0.027 kmol (or 0.054 kg) of H 2 gas in each m 3 volume of
nickel adjacent to the interface.
14-5 ■ STEADY MASS DIFFUSION
THROUGH A WALL
Many practical mass transfer problems involve the diffusion of a species
through a plane-parallel medium that does not involve any homogeneous
chemical reactions under one-dimensional steady conditions. Such mass
transfer problems are analogous to the steady one-dimensional heat conduc-
tion problems in a plane wall with no heat generation and can be analyzed
similarly. In fact, many of the relations developed in Chapter 3 can be used
for mass transfer by replacing temperature by mass (or molar) fraction, ther-
mal conductivity by pD AB (or CD AB ), and heat flux by mass (or molar) flux
(Table 14-8).
cen58 93 3_chl4.qxd 9/9/2002 10:05 AM Page 733
Consider a solid plane wall (medium B) of area A, thickness L, and
density p. The wall is subjected on both sides to different concentrations of a
species A to which it is permeable. The boundary surfaces at x = and x = L
are located within the solid adjacent to the interfaces, and the mass fractions
of A at those surfaces are maintained at w At [ and w A> 2 , respectively, at all times
(Fig. 14-19). The mass fraction of species A in the wall will vary in the
x-direction only and can be expressed as w A (x). Therefore, mass transfer
through the wall in this case can be modeled as steady and one-dimensional.
Here we determine the rate of mass diffusion of species A through the wall
using a similar approach to that used in Chapter 3 for heat conduction.
The concentration of species A at any point will not change with time since
operation is steady, and there will be no production or destruction of species
A since no chemical reactions are occurring in the medium. Then the conser-
vation of mass principle for species A can be expressed as the mass flow rate
of species A through the wall at any cross section is the same. That is
"'cliff, a = j/A = constant (kg/s)
Then Fick's law of diffusion becomes
Ja
A
PD A
dw A
' dx
constant
Separating the variables in this equation and integrating across the wall from
x = 0, where w(0) = w A u to x = L, where w(L) = w A 2 > we get
iiff,A c L (""'a.
— \ dx = - P D A
A Jo Jw,,
,dw.
(14-21)
733
CHAPTER 14
L x
FIGURE 14-19
Schematic for steady one-dimensional
mass diffusion of species A
through a plane wall.
where the mass transfer rate m dm A and the wall area A are taken out of the
integral sign since both are constants. If the density p and the mass diffusion
coefficient D AB vary little along the wall, they can be assumed to be constant.
The integration can be performed in that case to yield
~ w a,2 „ . Pa,i - Pa, 2
PD AB A
This relation can be rearranged as
L/pD AB A
(kg/s) (14-22)
(14-23)
where
R,
PD AB A
is the diffusion resistance of the wall, in s/kg, which is analogous to the elec-
trical or conduction resistance of a plane wall of thickness L and area A (Fig.
14-20). Thus, we conclude that the rate of mass diffusion through a plane
wall is proportional to the average density, the wall area, and the concentra-
tion difference across the wall, but is inversely proportional to the wall thick-
ness. Also, once the rate of mass diffusion is determined, the mass fraction
w A (x) at any location x can be determined by replacing w At 2 in Eq. 14-22 by
w A (x) and L by x.
T,*-
■ T, - T,
Q = — — -
R
(a) Heat flow
T, ♦-
Tj-T 2
R e
R.
(b) Current flow
-+T 2
-»T,
W A , \- m A,2
«„
Miff, A ■
' Y mass
^VWVW - " W A,2
D
(c) Mass flow
FIGURE 14-20
Analogy between
thermal, electrical, and mass
diffusion resistance concepts.
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734
HEAT TRANSFER
FIGURE 14-21
One-dimensional mass diffusion
through a cylindrical or
spherical shell.
The preceding analysis can be repeated on a molar basis with this result,
N.
diff, A, wall
CD AB A
?A,1 -VA.2
D AB A
C A ,
C
A, 2 yA, 1 yA, 2
R
(14-24)
diff, wall
where i? diff wal] = LICD AB A is the molar diffusion resistance of the wall in
s/kmol. Note that mole fractions are accompanied by molar concentrations
and mass fractions are accompanied by density. Either relation can be used to
determine the diffusion rate of species A across the wall, depending on
whether the mass or molar fractions of species A are known at the boundaries.
Also, the concentration gradients on both sides of an interface are different,
and thus diffusion resistance networks cannot be constructed in an analogous
manner to thermal resistance networks.
In developing these relations, we assumed the density and the diffusion co-
efficient of the wall to be nearly constant. This assumption is reasonable when
a small amount of species A diffuses through the wall and thus the concentra-
tion of A is small. The species A can be a gas, a liquid, or a solid. Also, the
wall can be a plane layer of a liquid or gas provided that it is stationary.
The analogy between heat and mass transfer also applies to cylindrical and
spherical geometries. Repeating the approach outlined in Chapter 3 for heat
conduction, we obtain the following analogous relations for steady one-
dimensional mass transfer through nonreacting cylindrical and spherical
layers (Fig. 14-21)
'diff, A, cyl
2irLpD AB ■
ln(r 2 /r,)
2ttLD a
Pa, i ~ Pa. 2
' \n(r 2 /r { )
A.sph =4irr I r 2 pD 4
"A, 2
r,
4 TIT, 1'nD,
Pa, i ~ Pa, 2
(14-25)
(14-26)
P AA
Solid wall
\ P A,2
\
V
Gas
Gas
A
j . P A,l- P A,2
I TV
I diff, A
\L x
FIGURE 14-22
The diffusion rate of a gas species
through a solid can be determined
from a knowledge of the partial
pressures of the gas on both sides
and the permeability of the solid
to that gas.
or, on a molar basis,
Va 1 ~~ ^a •>
#diff,A,cyi = 2ttLCD ab ^ '" = 2ttLD a
Ca, i Ca, 2
ln(r 2 /r,)
Ndiff.A.sph = 4Trr,r 2 CD
?a,i ~yA,2
ai> r, — r.
4irr, nD A
Ca, i Ca, 2
r 2 ~ r,
(14-27)
(14-28)
Here, L is the length of the cylinder, r { is the inner radius, and r 2 is the outer
radius for the cylinder or the sphere. Again, the boundary surfaces at r = r t
and r = r 2 are located within the solid adjacent to the interfaces, and the mass
fractions of A at those surfaces are maintained at w A t and w A> 2 , respectively,
at all times. (We could make similar statements for the density, molar concen-
tration, and mole fraction of species A at the boundaries.)
We mentioned earlier that the concentration of the gas species in a solid at
the interface is proportional to the partial pressure of the adjacent gas and was
expressed as C A
solid side
J arP
AB r A, gas side
where iP AB is the solubility (in kmol/m 3
bar) of the gas A in the solid B. We also mentioned that the product of solu-
bility and the diffusion coefficient is called the permeability, 2P A6 = *if AB D AB
(in kmol/m • s • bar). Then the molar flow rate of a gas through a solid under
steady one-dimensional conditions can be expressed in terms of the partial
pressures of the adjacent gas on the two sides of the solid by replacing C A in
these relations by if AB P A or ^abPaIDab- in me case of & plane wall, for ex-
ample, it gives (Fig. 14-22)
cen58 93 3_chl4.qxd 9/9/2002 10:05 AM Page 735
iY,i
Dar^arA '
A, 2
Jifi"
(kmol/s)
(14-29)
where P A j and P A 2 are the partial pressures of gas A on the two sides of the
wall. Similar relations can be obtained for cylindrical and spherical walls by
following the same procedure. Also, if the permeability is given on a mass
basis (in kg/m • s • bar), then Eq. 14-29 will give the diffusion mass flow rate.
Noting that 1 kmol of an ideal gas at the standard conditions of 0°C and
1 atm occupies a volume of 22.414 m 3 , the volume flow rate of the gas
through the wall by diffusion can be determined from
22.414%:
(standard m 3 /s, at 0°C and 1 atm)
The volume flow rate at other conditions can be determined from the ideal gas
relation P A V = N A R„T.
735
CHAPTER 14
EXAMPLE 14-5 Diffusion of Hydrogen through a
Spherical Container
Pressurized hydrogen gas is stored at 358 K in a 4.8-m-outer-diameter spheri-
cal container made of nickel (Fig. 14-23). The shell of the container is 6 cm
thick. The molar concentration of hydrogen in the nickel at the inner surface is
determined to be 0.087 kmol/m 3 . The concentration of hydrogen in the nickel
at the outer surface is negligible. Determine the mass flow rate of hydrogen by
diffusion through the nickel container.
SOLUTION Pressurized hydrogen gas is stored in a spherical container. The
diffusion rate of hydrogen through the container is to be determined.
Assumptions 1 Mass diffusion is steady and one-dimensional since the hydro-
gen concentration in the tank and thus at the inner surface of the container is
practically constant, and the hydrogen concentration in the atmosphere and
thus at the outer surface is practically zero. Also, there is thermal symmetry
about the center. 2 There are no chemical reactions in the nickel shell that re-
sult in the generation or depletion of hydrogen.
Properties The binary diffusion coefficient for hydrogen in the nickel at the
specified temperature is 1.2 X 10~ 12 m 2 /s (Table 14-3b).
Analysis We can consider the total molar concentration to be constant (C =
C A + C B = C B = constant), and the container to be a stationary medium since
there is no diffusion of nickel molecules (A/ B = 0) and the concentration of the
hydrogen in the container is extremely low {C A <^ D.Then the molar flow rate of
hydrogen through this spherical shell by diffusion can readily be determined
from Eq. 14-28 to be
C A
Ca.2
4-ir(2.34m)(2.40m)(1.2 X 10-' 2 m 2 /s)
1.228 X 10" 10 kmol/s
(0.087 - 0) kmol/m 3
2.40 - 2.34
0.087
kmol
FIGURE 14-23
Schematic for Example 14-5.
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736
HEAT TRANSFER
The mass flow rate is determined by multiplying the molar flow rate by the
molar mass of hydrogen, which is M = 2 kg/kmol,
MN„
(2 kg/kmol)( 1.228 X lO" 10 kmol/s) = 2.46 X 10" 10 kg/s
Therefore, hydrogen will leak out through the shell of the container by diffusion
at a rate of 2.46 X 10~ 10 kg/s or 7.8 g/year. Note that the concentration of
hydrogen in the nickel at the inner surface depends on the temperature and
pressure of the hydrogen in the tank and can be determined as explained in
Example 14-4. Also, the assumption of zero hydrogen concentration in nickel
at the outer surface is reasonable since there is only a trace amount of hydro-
gen in the atmosphere (0.5 part per million by mole numbers).
Dry
insulation
Wet
insulation
1.25g
]
0% 5%
moisture moisture
FIGURE 14-24
A 5 percent moisture content can
increase heat transfer through wall
insulation by 25 percent.
14-6 - WATER VAPOR MIGRATION IN BUILDINGS
Moisture greatly influences the performance and durability of building mate-
rials, and thus moisture transmission is an important consideration in the con-
struction and maintenance of buildings.
The dimensions of wood and other hygroscopic substances change with
moisture content. For example, a variation of 4.5 percent in moisture content
causes the volume of white oak wood to change by 2.5 percent. Such cyclic
changes of dimensions weaken the joints and can jeopardize the structural in-
tegrity of building components, causing "squeaking" at the minimum. Excess
moisture can also cause changes in the appearance and physical properties of
materials: corrosion and rusting in metals, rotting in woods, and peeling of
paint on the interior and exterior wall surfaces. Soaked wood with a water
content of 24 to 31 percent is observed to decay rapidly at temperatures 10 to
38°C. Also, molds grow on wood surfaces at relative humidities above 85 per-
cent. The expansion of water during freezing may damage the cell structure of
porous materials.
Moisture content also affects the effective conductivity of porous mediums
such as soils, building materials, and insulations, and thus heat transfer
through them. Several studies have indicated that heat transfer increases al-
most linearly with moisture content, at a rate of 3 to 5 percent for each percent
increase in moisture content by volume. Insulation with 5 percent moisture
content by volume, for example, increases heat transfer by 15 to 25 percent
relative to dry insulation (ASHRAE Handbook of Fundamentals, Ref. 1,
Chap. 20) (Fig. 14-24). Moisture migration may also serve as a transfer mech-
anism for latent heat by alternate evaporation and condensation. During a hot
and humid day, for example, water vapor may migrate through a wall and con-
dense on the inner side, releasing the heat of vaporization, with the process re-
versing during a cool night. Moisture content also affects the specific heat and
thus the heat storage characteristics of building materials.
Moisture migration in the walls, floors, or ceilings of buildings and in other
applications is controlled by either vapor barriers or vapor retarders. Vapor
barriers are materials that are impermeable to moisture, such as sheet metals,
cen58 93 3_chl4.qxd 9/9/2002 10:05 AM Page 737
737
CHAPTER 14
heavy metal foils, and thick plastic layers, and they effectively bar the vapor
from migrating. Vapor retarders, on the other hand, retard or slow down the
flow of moisture through the structures but do not totally eliminate it. Vapor
retarders are available as solid, flexible, or coating materials, but they usually
consist of a thin sheet or coating. Common forms of vapor retarders are rein-
forced plastics or metals, thin foils, plastic films, treated papers, coated felts,
and polymeric or asphaltic paint coatings. In applications such as the building
of walls where vapor penetration is unavoidable because of numerous open-
ings such as electrical boxes, telephone lines, and plumbing passages, vapor
retarders are used instead of vapor barriers to allow the vapor that somehow
leaks in to exit to the outside instead of trapping it in. Vapor retarders with
a permeance of 57.4 X 10~ 9 kg/s • m 2 are commonly used in residential
buildings.
The insulation on chilled water lines and other impermeable surfaces that
are always cold must be wrapped with a vapor barrier jacket, or such cold
surfaces must be insulated with a material that is impermeable to moisture.
This is because moisture that migrates through the insulation to the cold
surface will condense and remain there indefinitely with no possibility of
vaporizing and moving back to the outside. The accumulation of moisture in
such cases may render the insulation useless, resulting in excessive energy
consumption.
Atmospheric air can be viewed as a mixture of dry air and water vapor, and
the atmospheric pressure is the sum of the pressure of dry air and the pressure
of water vapor, which is called the vapor pressure P v . Air can hold a certain
amount of moisture only, and the ratio of the actual amount of moisture in the
air at a given temperature to the maximum amount air can hold at that tem-
perature is called the relative humidity <j>. The relative humidity ranges from
for dry air to 100 percent for saturated air (air that cannot hold any more
moisture). The partial pressure of water vapor in saturated air is called the
saturation pressure P sat . Table 14-9 lists the saturation pressure at various
temperatures.
The amount of moisture in the air is completely specified by the tempera-
ture and the relative humidity, and the vapor pressure is related to relative hu-
midity c|) by
4^*
(14-30)
where F sat is the saturation (or boiling) pressure of water at the specified tem-
perature. Then the mass flow rate of moisture through a plain layer of thick-
ness L and normal area A can be expressed as
SPA-
SPA
$1^
foP,
sat, 2
(kg/s)
(14-31)
where 9P is the vapor permeability of the material, which is usually expressed
on a mass basis in the unit ng/s • m • Pa, where ng = 10~ 12 kg and 1 Pa = 10~ 5
bar. Note that vapor migrates or diffuses from a region of higher vapor pres-
sure toward a region of lower vapor pressure.
TABLE 14-9
Saturation pressure of water at
various temperatures
Saturation
Temperature, °C Pressure, Pa
-40
13
-36
20
-32
31
-28
47
-24
70
-20
104
-16
151
-12
218
-8
310
-4
438
611
5
872
10
1,228
15
1,705
20
2,339
25
3,169
30
4,246
35
5,628
40
7,384
50
12,349
100
101,330
200
1.55 X 10 6
300
8.58 X 10 6
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738
HEAT TRANSFER
TABLE 14-10
Typical vapor permeance of common
building materials (from ASHRAE,
Ref. 1, Chap. 22, Table 9)*
Materials and Permeance
Its Thickness ng/s
• m 2 • Pa
Concrete (1:2:4 mix, 1 m)
4.7
Brick, masonry, 100 mm
46
Plaster on metal lath,
19 mm
860
Plaster on wood lath,
19mm
630
Gypsum wall board,
9.5 mm
2860
Plywood, 6.4 mm 40-109
Still air, 1 m
174
Mineral wool insulation
(unprotected), 1 m
245
Expanded polyurethane
insulation board, 1 m 0.58-2.3
Aluminum foil, 0.025 mm
0.0
Aluminum foil, 0.009 mm
2.9
Polyethylene, 0.051 mm
9.1
Polyethylene, 0.2 mm
2.3
Polyester, 0.19 mm
4.6
Vapor retarder latex paint,
0.070 mm
26
Exterior acrylic house and
trim paint, 0.040 mm
313
Building paper, unit mass
of 0.16-0. 68 kg/m 2 0.1-2400
*Data vary greatly. Consult manufacturer for more
accurate data. Multiply by 1.41 X 10~ 6 to convert
to Ibm/s ■ ft 2 • psi. Also, 1 ng = 1C-- 12 kg.
The permeability of most construction materials is usually expressed for a
given thickness instead of per unit thickness. It is called the permeance M,
which is the ratio of the permeability of the material to its thickness. That is,
Permeance
M
Permeability
Thickness
(kg/s • m 2 • Pa)
(14-32)
The reciprocal of permeance is called (unit) vapor resistance and is ex-
pressed as
Vapor resistance
1
Permeance
L = k
(s • m 2 ■ Pa/kg)
(14-33)
Note that vapor resistance represents the resistance of a material to water
vapor transmission.
It should be pointed out that the amount of moisture that enters or leaves a
building by diffusion is usually negligible compared to the amount that enters
with infiltrating air or leaves with exfiltrating air. The primary cause of inter-
est in the moisture diffusion is its impact on the performance and longevity of
building materials.
The overall vapor resistance of a composite building structure that consists
of several layers in series is the sum of the resistances of the individual layers
and is expressed as
«.
Rv. i + R v . 2
r = "V R
v, n / > v, ;
(14-34)
Then the rate of vapor transmission through a composite structure can be de-
termined in an analogous manner to heat transfer from
AP„
(kg/s)
(14-35)
Vapor permeance of common building materials is given in Table 14-10.
EXAMPLE 14-6 Condensation and Freezing of Moisture in Walls
The condensation and even freezing of moisture in walls without effective vapor
retarders is a real concern in cold climates, and it undermines the effectiveness
of insulations. Consider a wood frame wall that is built around 38 mm X
90 mm (2x4 nominal) wood studs. The 90-mm-wide cavity between the studs
is filled with glass fiber insulation. The inside is finished with 13-mm gypsum
wallboard and the outside with 13-mm wood fiberboard and 13-mm X 200-mm
wood bevel lapped siding. Using manufacturer's data, the thermal and vapor re-
sistances of various components for a unit wall area are determined to be as:
cen58 93 3_chl4.qxd 9/9/2002 10:05 AM Page 739
Construction
fl-Value,
°C/W
fl„-Value,
nr
s • rrr
Pa/ng
1. Outside surface, 24 km/h wind
2. Painted wood bevel lapped siding
3. Wood fiberboard sheeting, 13 mm
4. Glass fiber insulation, 90 mm
5. Painted gypsum wallboard, 13 mm
6. Inside surface, still air
TOTAL
0.030
—
0.14
0.019
0.23
0.0138
2.45
0.0004
0.079
0.012
0.12
—
3.05
0.0452
The indoor conditions are 20°C and 60 percent relative humidity while the
outside conditions are — 16°C and 70 percent relative humidity. Determine if
condensation or freezing of moisture will occur in the insulation.
SOLUTION The thermal and vapor resistances of different layers of a wall are
given. The possibility of condensation or freezing of moisture in the wall is to be
investigated.
Assumptions 1 Steady operating conditions exist. 2 Heat transfer through the
wall is one-dimensional. 3 Thermal and vapor resistances of different layers of
the wall and the heat transfer coefficients are constant.
Properties The thermal and vapor resistances are as given in the problem
statement. The saturation pressures of water at 20 C C and -16°C are 2339 Pa
and 151 Pa, respectively (Table 14-9).
Analysis The schematic of the wall as well as the different elements used in its
construction are shown in Figure 14-25. Condensation is most likely to occur
at the coldest part of insulation, which is the part adjacent to the exterior
sheathing. Noting that the total thermal resistance of the wall is 3.05
m 2 ■ °C/W, the rate of heat transfer through a unit area A = 1 m 2 of the wall is
Gv
(lm 2 )
[20 - (-16)°C]
3.05 m 2 • °C/W
11.8W
The thermal resistance of the exterior part of the wall beyond the insulation is
0.03 + 0.14 + 0.23 = 0.40 m z • °C/W. Then the temperature of the insula-
tion-outer sheathing interface is
Ti — T + <2 wa ii« c .
-16°C + (11.8 W)(0.40°C/W)
■11.3°C
The saturation pressure of water at -11.3°C is 234 Pa, as shown in Table
14-9, and if there is condensation or freezing, the vapor pressure at the insu-
lation-outer sheathing interface will have to be this value. The vapor pressure at
the indoors and the outdoors is
P v , i = biP**, i = 0.60 X (2340 Pa) = 1404 Pa
Pv,2 = <Wsat,2 = °- 70 x U 51 Pa ) = 106 Pa
Then the rate of moisture flow through the interior and exterior parts of the wall
becomes
739
CHAPTER 14
FIGURE 14-25
Schematic for Example 14-6.
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740
HEAT TRANSFER
Carbonaceous
material
FIGURE 14-26
The surface hardening of a mild steel
component by the diffusion of carbon
molecules is a transient mass
diffusion process.
v, interior
Al
AP
R
(lm 2 )
°v, 1 °v, /
v / interior v, interior
(1404 - 234) Pa
(0.012 + 0.0004) Pa • m 2 • s/ng
Al
A/A
A
"v, 1 "v, /
(lm 2 )
/? / A?
v /exterior v, exterior
(234 - 106) Pa
(0.019 + 0.0138) Pa • m 2 ■ s/ng
94,355 ng/s = 94.4 |jug/s
3902 ng/s = 3.9 p,g/s
That is, moisture is flowing toward the interface at a rate of 94.4 |xg/s but flow-
ing from the interface to the outdoors at a rate of only 3.9 fig/s. Noting that the
interface pressure cannot exceed 234 Pa, these results indicate that moisture
is freezing in the insulation at a rate of
V, freezing
94.4 - 3.9 = 90.5 |xg/s
This corresponds to 7.82 g during a 24-h period, which can be absorbed by the
insulation or sheathing, and then flows out when the conditions improve. How-
ever, excessive condensation (or frosting at temperatures below C C) of moisture
in the walls during long cold spells can cause serious problems. This problem
can be avoided or minimized by installing vapor barriers on the interior side of
the wall, which will limit the moisture flow rate to 3.9 |xg/s. Note that if there
were no condensation or freezing, the flow rate of moisture through aim 2 sec-
tion of the wall would be 28.7 |xg/s (can you verify this?).
14-7 - TRANSIENT MASS DIFFUSION
The steady analysis discussed earlier is useful when determining the leakage
rate of a species through a stationary layer. But sometimes we are interested in
the diffusion of a species into a body during a limited time before steady op-
erating conditions are established. Such problems are studied using transient
analysis. For example, the surface of a mild steel component is commonly
hardened by packing the component in a carbonaceous material in a furnace
at high temperature. During the short time period in the furnace, the carbon
molecules diffuse through the surface of the steel component, but they pene-
trate to a depth of only a few millimeters. The carbon concentration decreases
exponentially from the surface to the inner parts, and the result is a steel com-
ponent with a very hard surface and a relatively soft core region (Fig. 14-26).
The same process is used in the gem industry to color clear stones. For ex-
ample, a clear sapphire is given a brilliant blue color by packing it in titanium
and iron oxide powders and baking it in an oven at about 2000°C for about a
month. The titanium and iron molecules penetrate less than 0.5 mm in the sap-
phire during this process. Diffusion in solids is usually done at high tempera-
tures to take advantage of the high diffusion coefficients at high temperatures
and thus to keep the diffusion time at a reasonable level. Such diffusion or
"doping" is also commonly practiced in the production of n- or p-type semi-
conductor materials used in the manufacture of electronic components. Dry-
ing processes such as the drying of coal, timber, food, and textiles constitute
another major application area of transient mass diffusion.
cen58 93 3_chl4.qxd 9/9/2002 10:05 AM Page 741
Transient mass diffusion in a stationary medium is analogous to transient
heat transfer provided that the solution is dilute and thus the density of the
medium p is constant. In Chapter 4 we presented analytical and graphical so-
lutions for one-dimensional transient heat conduction problems in solids with
constant properties, no heat generation, and uniform initial temperature. The
analogous one-dimensional transient mass diffusion problems satisfy these
requirements:
1. The diffusion coefficient is constant. This is valid for an isothermal
medium since D AB varies with temperature (corresponds to constant
thermal diffusivity).
2. There are no homogeneous reactions in the medium that generate or
deplete the diffusing species A (corresponds to no heat generation).
3. Initially (t = 0) the concentration of species A is constant throughout
the medium (corresponds to uniform initial temperature).
Then the solution of a mass diffusion problem can be obtained directly from
the analytical or graphical solution of the corresponding heat conduction prob-
lem given in Chapter 4. The analogous quantities between heat and mass
transfer are summarized in Table 14-11 for easy reference. For the case of a
semi-infinite medium with constant surface concentration, for example, the
solution can be expressed in an analogous manner to Eq. 4-24 as
C A (x, t) - C A
Ca.s ~ C A j
erfc
(14-36)
and C A , is the
where C A , is the initial concentration of species A at time t
concentration at the inner side of the exposed surface of the medium. By using
the definitions of molar fraction, mass fraction, and density, it can be shown
that for dilute solutions,
C A (x, t) - C Ai { p A (x, t) - p A] , w A (x, t) - w A< i y A (x, i) - y A ,
C A
C\
Pa,
Pa,;
yA, s -y A ,i
(14-37)
since the total density or total molar concentration of dilute solutions is usu-
ally constant (p = constant or C = constant). Therefore, other measures of
concentration can be used in Eq. 14-36.
A quantity of interest in mass diffusion processes is the depth of diffusion at
a given time. This is usually characterized by the penetration depth defined
as the location x where the tangent to the concentration profile at the surface
(x = 0) intercepts the C A = C A>i line, as shown in Figure 14-27. Obtaining the
concentration gradient at x = by differentiating Eq. 14-36, the penetration
depth is determined to be
C\
C\
c\
c t
-(dC A /dx) x=0 (C Atl - Q, ,)/V^Sa7
V^D AB J
(14-38)
Therefore, the penetration depth is proportional to the square root of both the
diffusion coefficient and time. The diffusion coefficient of zinc in copper at
1000°C, for example, is 5.0 X 10~ 12 m 2 /s. Then the penetration depth of zinc
in copper in 10 h is
741
CHAPTER 14
TABLE 14-11
Analogy between the quantities
that appear in the formulation
and solution of transient heat
conduction and transient mass
diffusion in a stationary medium
Heat
Mass
Conduction
Diffusion
T
C, y, p or w
a
Dab
TCx, t)
6 = T,
T r _
T,_
"mass
w A (x, t) - w A ^
T(x, t)
-
T s
w A {x, t) - w A
T,-
L
w A , i - w A
x
Smass
X
^ 2yat
2VDaJ
n . h mm L
^'mass
''mass *-
Cl k
Dab
at
T
D AB t
L 2
■Miff Tangent line to
concentration
gradient at x =
Semi-infinite
medium
Slope of
tangent line
l/.V
C,
= "diff
FIGURE 14-27
The concentration profile of species A
in a semi-infinite medium during
transient mass diffusion and the
penetration depth.
cen58 93 3_chl4.qxd 9/9/2002 10:05 AM Page 742
742
HEAT TRANSFER
V^D AB t
Vit(5.0 X l(r 12 m 2 /s)(10 X 3600 s)
0.00038 m = 0.38 mm
That is, zinc will penetrate to a depth of about 0.38 mm in an appreciable
amount in 10 h, and there will hardly be any zinc in the copper block beyond
a depth of 0.38 mm.
The diffusion coefficients in solids are typically very low (on the order of
10~ 9 to 10~ 15 m 2 /s), and thus the diffusion process usually affects a thin layer
at the surface. A solid can conveniently be treated as a semi-infinite medium
during transient mass diffusion regardless of its size and shape when the pen-
etration depth is small relative to the thickness of the solid. When this is not
the case, solutions for one-dimensional transient mass diffusion through a
plane wall, cylinder, and sphere can be obtained from the solutions of analo-
gous heat conduction problems using the Heisler charts or one-term solutions
presented in Chapter 4.
Furnace
Carbonaceous
material
Carbon ■
-H^=1.2%
■^w A (x,t) = V
i
0.5 mm W A, i
Steel
component
w. . = 0.15% x
f
FIGURE 14-28
Schematic for Example 14-7.
EXAMPLE 14-7 Hardening of Steel by the Diffusion of Carbon
The surface of a mild steel component is commonly hardened by packing the
component in a carbonaceous material in a furnace at a high temperature for a
predetermined time. Consider such a component with a uniform initial carbon
concentration of 0.15 percent by mass. The component is now packed in a
carbonaceous material and is placed in a high-temperature furnace. The diffu-
sion coefficient of carbon in steel at the furnace temperature is given to be
4.8 X 10~ 10 m 2 /s, and the equilibrium concentration of carbon in the iron
at the interface is determined from equilibrium data to be 1.2 percent by
mass. Determine how long the component should be kept in the furnace for the
mass concentration of carbon 0.5 mm below the surface to reach 1 percent
(Fig. 14-28).
SOLUTION A steel component is to be surface hardened by packing it in a
carbonaceous material in a furnace. The length of time the component should
be kept in the furnace is to be determined.
Assumptions Carbon penetrates into a very thin layer beneath the surface of
the component, and thus the component can be modeled as a semi-infinite
medium regardless of its thickness or shape.
Properties The relevant properties are given in the problem statement.
Analysis This problem is analogous to the one-dimensional transient heat con-
duction problem in a semi-infinite medium with specified surface temperature,
and thus can be solved accordingly. Using mass fraction for concentration since
the data are given in that form, the solution can be expressed as
w A (x, t) - w A> ,
erfc
Substituting the specified quantities gives
0.01 - 0.0015
0.012 - 0.0015
0.81 = erfcl
cen58 93 3_chl4.qxd 9/9/2002 10:05 AM Page 743
The argument whose complementary error function is 0.81 is determined from
Table 4-3 to be 0.17. That is,
^= = 0.17
Then solving for the time t gives
x 2 (0.0005 m) 2
4D AB (0.n) 2 4 X (4.8 X 10-'°m 2 /s)(0.17) 2
4505 s = 1 h 15 min
Discussion The steel component in this case must be held in the furnace for
1 h and 15 min to achieve the desired level of hardening. The diffusion coeffi-
cient of carbon in steel increases exponentially with temperature, and thus this
process is commonly done at high temperatures to keep the diffusion time at a
reasonable level.
743
CHAPTER 14
14-8 - DIFFUSION IN A MOVING MEDIUM
To this point we have limited our consideration to mass diffusion in a station-
ary medium, and thus the only motion involved was the creeping motion of
molecules in the direction of decreasing concentration, and there was no mo-
tion of the mixture as a whole. Many practical problems, such as the evapora-
tion of water from a lake under the influence of the wind or the mixing of two
fluids as they flow in a pipe, involve diffusion in a moving medium where the
bulk motion is caused by an external force. Mass diffusion in such cases is
complicated by the fact that chemical species are transported both by diffusion
and by the bulk motion of the medium (i.e., convection). The velocities and
mass flow rates of species in a moving medium consist of two components:
one due to molecular diffusion and one due to convection (Fig. 14-29).
Diffusion in a moving medium, in general, is difficult to analyze since var-
ious species can move at different velocities in different directions. Turbu-
lence will complicate the things even more. To gain a firm understanding of
the physical mechanism while keeping the mathematical complexities to a
minimum, we limit our consideration to systems that involve only two com-
ponents (species A and B) in one-dimensional flow (velocity and other prop-
erties change in one direction only, say the x-direction). We also assume the
total density (or molar concentration) of the medium remains constant. That
is, p = p A + p B = constant (or C = C A + C B = constant) but the densities of
species A and B may vary in the x-direction.
Several possibilities are summarized in Figure 14-30. In the trivial case
(case a) of a stationary homogeneous mixture, there will be no mass transfer
by molecular diffusion or convection since there is no concentration gradient
or bulk motion. The next case (case b) corresponds to the flow of a well-mixed
fluid mixture through a pipe. Note that there is no concentration gradients and
thus molecular diffusion in this case, and all species move at the bulk flow ve-
locity of T. The mixture in the third case (case c) is stationary (T = 0) and
thus it corresponds to ordinary molecular diffusion in stationary mediums,
which we discussed before. Note that the velocity of a species at a location in
Convection
FIGURE 14-29
In a moving medium, mass transfer is
due to both diffusion and convection.
cen58 93 3_chl4.qxd 9/9/2002 10:05 AM Page 744
744
HEAT TRANSFER
• A OB
Species
Density
Velocity
Mass flow rate
(a) Homogeneous mixture •0»0»0«0»0
without bulk motion 0»0«0»0»0«
(no concentration gradients •0»0*0»0»0„ / ,_
j ,u Ate • i 0»0»0»0»0»
and thus no diffusion)
•o»o»o«o»o
o»o»o»o»o»
Species A
Species B
Mixture of
AandB
p A = constant
p B = constant
P = Pa + Pb
= constant
T A =
T s =
y = o
m A =
m B =
m =
(6) Homogeneous mixture •0»0»0»0»0
with bulk motion 0«0«0«0«0«
(no concentration gradients •0»0»0»0»0 — „
j ,u j« x o»o»o»o»o»
and thus no diffusion) „ „
' •o«o»o«o«o
o»o»o«o»o»
Species A
Species B
Mixture of
AandB
p A = constant
p B = constant
P = Pa + Ps
= constant
y A =y
y s = y
y =y
™a = Pa t a a
' il s = Ps° |/ s A
m = pyA
= ">A + " ! fl
(c) Nonhomogeneous mixture ••0«00«000
without bulk motion •••••00»00
(stationary medium with •••00»00»0 j / ,_
concentration gradients) _ _ _ „ „
5 ' •••ooo»ooo
•OtttlOOOO
'iiB.A "* "" Xliff.S
Species A
Species B
Mixture of
A and B
p A # constant
p B ^ constant
P = Pa + Ps
= constant
T4=^diff,A
'B = ' diff, S
y =
» 1 A = PA T diff,A A
™S=Pfl T diff,S A
m = p TA =
(thus m A = -fn B )
(d) Nonhomogeneous mixture ••0«00«000
with bulk motion •••••00»00
(moving medium with •••00000*0 — ^ y
concentration gradients)
6 •••ooo«ooo
•o»»»»oooo
^diff.A *" "" 'iiS.B
Species A
Species B
Mixture of
AandB
p A # constant
p B # constant
P = Pa + Ps
= constant
^A=^+^diff,A
^S = ^+W
t = y
™A = PA r am,A A
" , S=Ps T diff,S A
m = p°VA
= ™A + > h B
FIGURE 14-30
Various quantities associated with a mixture of two species A and B at a location .r under one-dimensional flow or no-flow
conditions. (The density of the mixture p = p A + p B is assumed to remain constant.)
this case is simply the diffusion velocity, which is the average velocity of a
group of molecules at that location moving under the influence of concentra-
tion gradient. Finally, the last case (case d) involves both molecular diffusion
and convection, and the velocity of a species in this case is equal to the sum of
the bulk flow velocity and the diffusion velocity. Note that the flow and the
diffusion velocities can be in the same or opposite directions, depending on
the direction of the concentration gradient. The diffusion velocity of a species
is negative when the bulk flow is in the positive .r-direction and the concen-
tration gradient is positive (i.e., the concentration of the species increases in
the jc-direction).
Noting that the mass flow rate at any flow section is expressed as m = pYA
where p is the density, T is the velocity, and A is the cross sectional area, the
conservation of mass relation for the flow of a mixture that involves two
species A and B can be expressed as
or
pTA = p/V A A + p B Y B A
Canceling A and solving for T gives
cen58 93 3_chl4.qxd 9/9/2002 10:05 AM Page 745
M , PaVa + Pb°Vb Pa „.
T = p = ~^ r A
Pb
r R
w A °V A + w B T B
(14-39)
745
CHAPTER 14
where T is called the mass-average velocity of the flow, which is the veloc-
ity that would be measured by a velocity sensor such as a pitot tube, a turbine
device, or a hot wire anemometer inserted into the flow.
The special case T = corresponds to a stationary medium, which can
now be defined more precisely as a medium whose mass-average velocity is
zero. Therefore, mass transport in a stationary medium is by diffusion only,
and zero mass-average velocity indicates that there is no bulk fluid motion.
When there is no concentration gradient (and thus no molecular mass
diffusion) in the fluid, the velocity of all species will be equal to the mass-
average velocity of the flow. That is, T = T A = °V B . But when there is a con-
centration gradient, there will also be a simultaneous flow of species in the
direction of decreasing concentration at a diffusion velocity of T diff . Then the
average velocity of the species A and B can be determined by superimposing
the average flow velocity and the diffusion velocity as (Fig. 14-31)
(14-40)
Similarly, we apply the superposition principle to the species mass flow rates
to get
m A = p A °V A A = p A (°V+ °Vhb,a)A = p A TA + p A °V iiB _ A A = m conv , A + m diffA
m B = p B °V B A = p B (¥ + T diff , B )A = p B TA + p B T iiB _ B A = m conv , B + m diff>B (14-41)
Using Fick's law of diffusion, the total mass fluxes j = ml A can be ex-
pressed as
Pa dw A dw A
Ja = Pa°V + P^diff.A = y P°V ~ pD AB — = w A (j A + j B ) - pD AB —
Pb dw B dw B
Jb = Pb'V' + PB°Vm, B = -p-p°V- PD BA -fa = WbQa + Jb) ~ P d ba ~^ (14-42)
Note that the diffusion velocity of a species is negative when the molecular
diffusion occurs in the negative x-direction (opposite to flow direction). The
mass diffusion rates of the species A and B at a specified location x can be ex-
pressed as
■a = PaV^aA = PA (T A - Y)A
,b = Pfl^diff.a^ = p B (¥ B - T)A
(14-43)
By substituting the Trelation from Eq. 14-39 into Eq. 1 1^43, it can be shown
that at any cross section
' cliff, A ' "' diff. B
-pAD A
dw A
pAD B
dw R
(14-44)
^d,ff,A =
T A = T
• O • O •
o
o
o
o
o
l 1
o • o • o
u
o
o
o
o
— -T,
• o|» o- 4 *
o
o
o
o
o
|_o •_]o • o
e
o
o
o
• T
• o • o •
o
o
o
o
O Flow
o • o • o
o
o
o
o
^velocity
(a) No concentration gradient
\=^^Mf,A
• • o • o o
• •"]• • • o
• o> o^o o
o
o
o
o
o
o
o
o
o
o
o
o
diff ' A |_» •Jo • • • o • o o y
• • •OOO0OOO Flow
• •••••ooo o yelocit y
(b) Mass concentration gradient and thus
mass diffusion
FIGURE 14-31
The velocity of a species at a point is
equal to the sum of the bulk
flow velocity and the diffusion
velocity of that species at that point.
which indicates that the rates of diffusion of species A and B must be equal
in magnitude but opposite in sign. This is a consequence of the assumption
cen58 93 3_chl4.qxd 9/9/2002 10:05 AM Page 746
746
HEAT TRANSFER
p = p A + p B = constant, and it indicates that anytime the species A diffuses in
one direction, an equal amount of species B must diffuse in the opposite di-
rection to maintain the density (or the molar concentration) constant. This
behavior is closely approximated by dilute gas mixtures and dilute liquid
or solid solutions. For example, when a small amount of gas diffuses into a
liquid, it is reasonable to assume the density of the liquid to remain constant.
Note that for a binary mixture, w A + w B = 1 at any location x. Taking the
derivative with respect to x gives
dw A
dx
dw B
dx
(14-45)
• •o«ooooooo
• • • • • 0,;,» • O • O
"Miff, A
•o«oo«oo«oo
• • o • • o A » #000
-« m diff, B
• •••0*00000
v^
o • o o o o o
£
dw A
dx
dWg
dx
Thus we conclude from Eq. 14-44 that (Fig. 14-32)
(14-46)
That is, in the case of constant total concentration, the diffusion coefficient of
species A into B is equal to the diffusion coefficient of species B into A.
We now repeat the analysis presented above with molar concentration C and
the molar flow rate N. The conservation of matter in this case is expressed as
or
P TA = p A Y A A + p B Y B A
Canceling A and solving for T gives
c a y a + c b y b c
c R
c r A + -£r B = y A r A + y B r B
(14-47)
(14-48)
FIGURE 14-32
In a binary mixture of species A and B
with p = p A + p B = constant, the rates
of mass diffusion of species A and B
are equal magnitude and opposite
in direction.
where T is called the molar-average velocity of the flow. Note that T + T
unless the mass and molar fractions are the same. The molar flow rates of
species are determined similarly to be
N A = C A T A A = C A (¥ + T diff , A )A = C A TA + C A % m , A A = W conv , A + iV diflU
N B = C B Y B A = C B (V + T diff , B )A = C B YA + C B Y dm , B A = W conv>B + N dlff , B (14-49)
Using Fick's law of diffusion, the total molar fluxes j = NIA and diffusion
molar flow rates N m can be expressed as
C\
dy A
j A = C A T + C A T diff , A = -g CV - CD AB -£ = y A ( j A + j B ) - CD AB ±
] B = C B T + C fi T dlff , s = § CT - CD B
dy B
dx
y B ( ja + jb) - cd b
dy A
dx
dy B
dx
(14-50)
and
N W , A = C A Y^ A A = C A (Y A - T)A
N*
C B Y^ B A = C B (Y B -T)A
(14-51)
cen58 93 3_chl4.qxd 9/9/2002 10:05 AM Page 747
By substituting the T relation from Eq. 14^-8 into these two equations, it can
be shown that
N dlff , A + N 6lff , B = -> N imA = -N imB (14-52)
which again indicates that the rates of diffusion of species A and B must be
equal in magnitude but opposite in sign.
It is important to note that when working with molar units, a medium is said
to be stationary when the molar-average velocity is zero. The average veloc-
ity of the molecules will be zero in this case, but the apparent velocity of the
mixture as measured by a velocimeter placed in the flow will not necessarily
be zero because of the different masses of different molecules. In a mass-
based stationary medium, for each unit mass of species A moving in one
direction, a unit mass of species B moves in the opposite direction. In a mole-
based stationary medium, however, for each mole of species A moving in one
direction, one mole of species B moves in the opposite direction. But this may
result in a net mass flow rate in one direction that can be measured by a
velocimeter since the masses of different molecules are different.
You may be wondering whether to use the mass analysis or molar analysis
in a problem. The two approaches are equivalent, and either approach can be
used in mass transfer analysis. But sometimes it may be easier to use one of
the approaches, depending on what is given. When mass-average velocity is
known or can easily be obtained, obviously it is more convenient to use the
mass-based formulation. When the total pressure and temperature of a mix-
ture are constant, however, it is more convenient to use the molar formulation,
as explained next.
Special Case: Gas Mixtures
at Constant Pressure and Temperature
Consider a gas mixture whose total pressure and temperature are constant
throughout. When the mixture is homogeneous, the mass density p, the molar
density (or concentration) C, the gas constant R, and the molar mass M of the
mixture are the same throughout the mixture. But when the concentration of
one or more gases in the mixture is not constant, setting the stage for mass dif-
fusion, then the mole fractions y t of the species will vary throughout the mix-
ture. As a result, the gas constant R, the molar mass M, and the mass density p
of the mixture will also vary since, assuming ideal gas behavior,
M = ^ J y i M i , R = Jj> and P = Rf
where R u = 8.314 kJ/kmol • K is the universal gas constant. Therefore, the as-
sumption of constant mixture density (p = constant) in such cases will not be
accurate unless the gas or gases with variable concentrations constitute a very
small fraction of the mixture. However, the molar density C of a mixture re-
mains constant when the mixture pressure P and temperature T are constant
since
R„
P = pRT= PirjT= CRJ (14-53)
747
CHAPTER 14
cen58 93 3_chl4.qxd 9/9/2002 10:05 AM Page 74E
748
HEAT TRANSFER
Gas
mixture
T
= constant
P
= constant
Independent
of composition
of mixture
C =
R„T
RT (RJM)T
Depends on
composition
of mixture
FIGURE 14-33
When the total pressure P and
temperature T of a binary mixture
of ideal gases is held constant, then
the molar concentration C of the
mixture remains constant.
Gas mixture
A+B
ca
-
Liquid A
FIGURE 14-34
Diffusion of a vapor A
through a stagnant gas B.
The condition C = constant offers considerable simplification in mass
transfer analysis, and thus it is more convenient to use the molar formulation
when dealing with gas mixtures at constant total pressure and temperature
(Fig. 14-33).
Diffusion of Vapor through a Stationary Gas:
Stefan Flow
Many engineering applications such as heat pipes, cooling ponds, and the fa-
miliar perspiration involve condensation, evaporation, and transpiration in the
presence of a noncondensable gas, and thus the diffusion of a vapor through a
stationary (or stagnant) gas. To understand and analyze such processes, con-
sider a liquid layer of species A in a tank surrounded by a gas of species B,
such as a layer of liquid water in a tank open to the atmospheric air (Fig.
14-34), at constant pressure P and temperature T. Equilibrium exists between
the liquid and vapor phases at the interface (x = 0), and thus the vapor pres-
sure at the interface must equal the saturation pressure of species A at the
specified temperature. We assume the gas to be insoluble in the liquid, and
both the gas and the vapor to behave as ideal gases.
If the surrounding gas at the top of the tank (x = L) is not saturated, the
vapor pressure at the interface will be greater than the vapor pressure at the
top of the tank (P A > P A L and thus y Aj > y A L since y A = P A /P), and this
pressure (or concentration) difference will drive the vapor upward from the
air-water interface into the stagnant gas. The upward flow of vapor will be
sustained by the evaporation of water at the interface. Under steady condi-
tions, the molar (or mass) flow rate of vapor throughout the stagnant gas col-
umn remains constant. That is,
Ja
(orj A
constant)
The pressure and temperature of the gas-vapor mixture are said to be con-
stant, and thus the molar density of the mixture must be constant throughout
the mixture, as shown earlier. That is, C = C A + C B = constant, and it is more
convenient to work with mole fractions or molar concentrations in this case
instead of mass fractions or densities since p =£ constant.
Noting that y A + y B = 1 and that y A > Va, u we must have y B < y B L . That
is, the mole fraction of the gas must be decreasing downward by the same
amount that the mole fraction of the vapor is increasing. Therefore, gas must
be diffusing from the top of the column toward the liquid interface. However,
the gas is said to be insoluble in the liquid, and thus there can be no net mass
flow of the gas downward. Then under steady conditions, there must be an
upward bulk fluid motion with an average velocity T that is just large enough
to balance the diffusion of air downward so that the net molar (or mass) flow
rate of the gas at any point is zero. In other words, the upward bulk motion
offsets the downward diffusion, and for each air molecule that moves down-
ward, there is another air molecule that moves upward. As a result, the air ap-
pears to be stagnant (it does not move). That is,
j B = N B /A =
(or j B = m B /A = 0)
The diffusion medium is no longer stationary because of the bulk motion. The
implication of the bulk motion of the gas is that it transports vapor as well as
cen58 93 3_chl4.qxd 9/9/2002 10:05 AM Page 749
the gas upward with a velocity of T, which results in additional mass flow of
vapor upward. Therefore, the molar flux of the vapor can be expressed as
Ja = N A IA = y A>conv + L im = y A ( j A + j B ) - CD AB
djA
dx
(14-54)
749
CHAPTER 14
Noting that j B = 0, it simplifies to
j A = y A j A - cd a
dy A
1 dx
(14-55)
Solving for j A gives
Ja
CD A b dy A
\-y A dx
_J dy^ = j A
1 — y A dx CD A
constant
(14-56)
since j A = constant, C = constant, and D AB = constant. Separating the vari-
ables and integrating from x = 0, where y A (0) = y A , to x = L, where
?a(£) =y/L,L give*
C >•'■•■ dy A
Ja.o 1 ~y A
f
Jo
Ja
CD A
dx
(14-57)
Performing the integrations,
In
i -y A ,i
Ja
1 -3>A,0 CD A
(14-58)
Then the molar flux of vapor A, which is the evaporation rate of species A per
unit interface area, becomes
Ja = N A IA
CD,
-In
i -y A ,i
i ~y A ,i
(kmol/s • m 2 )
(14-59)
This relation is known as Stefan's law, and the induced convective flow de-
scribed that enhances mass diffusion is called the Stefan flow.
An expression for the variation of the mole fraction of A with x can be de-
termined by performing the integration in Eq. 14-57 to the upper limit of x
where y A (x) = y A (instead of to L where y A (L) = y A ^ L ). It yields
In
i -y A ,o
Ja
CD A
(14-60)
Substituting the j A expression from Eq. 14-59 into this relation and rearrang-
ing gives
i-y A /1-Xa.
1 -yx.0 v 1 -yA
and
y B Pb,l
ys. o lyfl, o
(14-61)
The second relation for the variation of the mole fraction of the stationary
gas B is obtained from the first one by substituting 1 — y A = y B since
yA + yB = i-
cen58 93 3_chl4.qxd 9/9/2002 10:05 AM Page 75C
750
HEAT TRANSFER
To maintain isothermal conditions in the tank during evaporation, heat must
be supplied to the tank at a rate of
Q = m A h^ A =JAA s hf g , A = (JAM A )A s h fg:A
(kJ/s)
(14-62)
where A s is the surface area of the liquid-vapor interface, h fg A is the latent
heat of vaporization, and M A is the molar mass of species A.
Gas
mixture
A+B
y A > >'h
oB
Gas
mixture
A+B
y» < vj
• •<>••« O » o °o °o • o o»O n°o°.
o_» o •! l„»o°. , <
2,0 o •
,o°
..V
o « u
T, P
)°0
Oo°
oo
O'OO
o oo
T, P
Equimolar Counterdiffusion
Consider two large reservoirs connected by a channel of length L, as shown in
Figure 14-35. The entire system contains a binary mixture of gases A and B at
a uniform temperature T and pressure P throughout. The concentrations of
species are maintained constant in each of the reservoirs such that y A > y A , L
and y B < y B L . The resulting concentration gradients will cause the species A
to diffuse in the positive x-direction and the species B in the opposite direc-
tion. Assuming the gases to behave as ideal gases and thus P = CR U T, the total
molar concentration of the mixture C will remain constant throughout the
mixture since P and T are constant. That is,
FIGURE 14-35
Equimolar isothermal counterdiffusion
of two gases A and B.
C
constant
(kmol/m 3 )
This requires that for each molecule of A that moves to the right, a molecule
of B moves to the left, and thus the molar flow rates of species A and B must
be equal in magnitude and opposite in sign. That is,
Ni
-N„
(kmol/s)
This process is called equimolar counterdiffusion for obvious reasons. The
net molar flow rate of the mixture for such a process, and thus the molar-
average velocity, is zero since
N = N A + N B = -> CAT = -> T =
Therefore, the mixture is stationary on a molar basis and thus mass transfer is
by diffusion only (there is no mass transfer by convection) so that
Ja
-CD.
dy A
: dx
and
j B = N B /A = -CD E
dy B
dx
(14-63)
Under steady conditions, the molar flow rates of species A and B can be
determined directly from Eq. 14-24 developed earlier for one-dimensional
steady diffusion in a stationary medium, noting that P = CR lt T and thus
C = PIR U T for each constituent gas and the mixture. For one-dimensional
flow through a channel of uniform cross sectional area A with no homoge-
neous chemical reactions, they are expressed as
AW = CD AB A
yA, i y~A, 2 „ Ca, i C A 2 D AB Pa, o "a, l
N dlff , B = CD BA A
D A bA ■
yB, i y^ 2
R„T
DbaA
Cfl, i C s 2 D BA P B , o Pb. l
— = - 5 ^=A (14-64)
cen58 93 3_chl4.qxd 9/9/2002 10:05 AM Page 751
These relations imply that the mole fraction, molar concentration, and the par-
tial pressure of either gas vary linearly during equimolar counterdiffusion.
It is interesting to note that the mixture is stationary on a molar basis, but
it is not stationary on a mass basis unless the molar masses of A and B are
equal. Although the net molar flow rate through the channel is zero, the net
mass flow rate of the mixture through the channel is not zero and can be
determined from
m A + m B = N A M A + N B M B = N A (M A - M B )
(14-65)
751
CHAPTER 14
since N B = ~N A . Note that the direction of net mass flow rate is the flow
direction of the gas with the larger molar mass. A velocity measurement
device such as an anemometer placed in the channel will indicate a velocity
of T = rh/pA where p is the total density of the mixture at the site of
measurement.
EXAMPLE 14-8 Venting of Helium into the Atmosphere
by Diffusion
The pressure in a pipeline that transports helium gas at a rate of 2 kg/s is main-
tained at 1 atm by venting helium to the atmosphere through a 5-mm-internal-
diameter tube that extends 15 m into the air, as shown in Figure 14-36.
Assuming both the helium and the atmospheric air to be at 25°C, determine
(a) the mass flow rate of helium lost to the atmosphere through an individual
tube, (b) the mass flow rate of air that infiltrates into the pipeline, and (c) the
flow velocity at the bottom of the tube where it is attached to the pipeline that
will be measured by an anemometer in steady operation.
SOLUTION The pressure in a helium pipeline is maintained constant by vent-
ing to the atmosphere through a long tube. The mass flow rates of helium and
air through the tube and the net flow velocity at the bottom are to be deter-
mined.
Assumptions 1 Steady operating conditions exist. 2 Helium and atmospheric
air are ideal gases. 3 No chemical reactions occur in the tube. 4 Air concentra-
tion in the pipeline and helium concentration in the atmosphere are negligible
so that the mole fraction of the helium is 1 in the pipeline and in the atmo-
sphere (we will check this assumption later).
Properties The diffusion coefficient of helium in air (or air in helium) at normal
atmospheric conditions is D AB = 7.20 X 10~ 5 m 2 /s (Table 14-2). The molar
masses of air and helium are 29 and 4 kg/kmol, respectively (Table A-l).
Analysis This is a typical equimolar counterdiffusion process since the prob-
lem involves two large reservoirs of ideal gas mixtures connected to each other
by a channel, and the concentrations of species in each reservoir (the pipeline
and the atmosphere) remain constant.
(a) The flow area, which is the cross sectional area of the tube, is
A = ttD 2 /4 = tt(0.005 m) 2 /4 = 1.963 X 1(T 5 m 2
Noting that the pressure of helium is 1 atm at the bottom of the tube (x = 0)
and at the top (x = L), its molar flow rate is determined from Eq. 14-64 to be
He
Air
1 atm
25°C
Air
5 mm
He
Helium
(A)
\i
L=15m
Air
2 kg/s
1 atm
25°C
f
_
FIGURE 14-36
Schematic for Example 14-8
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752
HEAT TRANSFER
M
N A ;
D AB A P.
A,
A,L
helium diff, A r> r r j
(7.20 X l(r 5 m 2 /s)( 1.963 X 10" 5 m 2 ) l\ a tm - 0\/l01.3 kPa
(8.314 kPa • m 3 /kmol ■ K)(298 K)
3.85 X 1(T 12 kmol/s
15 m
1 atm
Therefore,
(NM) b
(3.85 X 10- 12 kmol/s)(4 kg/kmol) = 1.54 X 10 11 kg/s
which corresponds to about 0.5 g per year.
(b) Noting that N B = -N A during an equimolar counterdiffusion process, the
molar flow rate of air into the helium pipeline is equal to the molar flow rate of
helium. The mass flow rate of air into the pipeline is
m alr = (NM) ml = (-3.85 X lO" 12 kmol/s)(29 kg/kmol) = -112 X 10 12 kg/s
The mass fraction of air in the helium pipeline is
|m air | 112 X 10-' 2 kg/s
(2 + 112 X 10-
1.54 X 10- u )kg/s
5.6 X 10"
which validates our original assumption of negligible air in the pipeline,
(c) The net mass flow rate through the tube is
'"net = ^helium + *air = 1-54 X 10-" ~ 112 X lO" 12 = ~9.66 X 10-" kg/S
The mass fraction of air at the bottom of the tube is very small, as shown above,
and thus the density of the mixture at x = can simply be taken to be the den-
sity of helium, which is
Phclu
101.325 kPa
RT (2.0769 kPa ■ m 3 /kg • K)(298 K)
0.1637 kg/m 3
Then the average flow velocity at the bottom part of the tube becomes
-9.66 X 10-' 1 kg/s
Y
pA (0.01637 kg/m 3 )(l. 963 X 10- 5 m 2 )
-3.02 X 10- 5 m/s
which is difficult to measure by even the most sensitive velocity measurement
devices. The negative sign indicates flow in the negative x-direction (toward the
pipeline).
EXAMPLE 14-9 Measuring Diffusion Coefficient by the Stefan
Tube
A 3-cm-diameter Stefan tube is used to measure the binary diffusion coefficient
of water vapor in air at 20°C at an elevation of 1600 m where the atmospheric
cen58 93 3_chl4.qxd 9/9/2002 10:05 AM Page 753
753
CHAPTER 14
pressure is 83.5 kPa. The tube is partially filled with water, and the distance
from the water surface to the open end of the tube is 40 cm (Fig. 14-37). Dry
air is blown over the open end of the tube so that water vapor rising to the top
is removed immediately and the concentration of vapor at the top of the tube is
zero. In 15 days of continuous operation at constant pressure and temperature,
the amount of water that has evaporated is measured to be 1.23 g. Determine
the diffusion coefficient of water vapor in air at 20°C and 83.5 kPa.
SOLUTION The amount of water that evaporates from a Stefan tube at a spec-
ified temperature and pressure over a specified time period is measured. The
diffusion coefficient of water vapor in air is to be determined.
Assumptions 1 Water vapor and atmospheric air are ideal gases. 2 The amount
of air dissolved in liquid water is negligible. 3 Heat is transferred to the water
from the surroundings to make up for the latent heat of vaporization so that the
temperature of water remains constant at 20°C.
Properties The saturation pressure of water at 20°C is 2.34 kPa (Table A-9).
Analysis The vapor pressure at the air-water interface is the saturation pres-
sure of water at 20°C, and the mole fraction of water vapor (species A) at the
interface is determined from
y
vapor,
^,0
vapor, o _ 2.34 kPa
P ~ 83.5 kPa
0.0280
Dry air is blown on top of the tube and, thus, y vapori i = y^ i = 0. Also, the total
molar density throughout the tube remains constant because of the constant
temperature and pressure conditions and is determined to be
C
83.5 kPa
0.0343 kmol/m 3
R„T (8.314 kPa ■ m 3 /kmol • K)(293 K)
The cross-sectional area of the tube is
A = ttD 2 /4 = ir(0.03 m) 2 /4 = 7.069 X 10~ 4 m 2
The evaporation rate is given to be 1.23 g per 15 days. Then the molar flow rate
of vapor is determined to be
1.23 X 10- 3 kg
A N ™ p01 M (15 X 24 X 3600s)(18kg/kmol)
5.27 X 10-" kmol/s
Finally, substituting the information above into Eq. 14-59 we get
5.27 X 10-" kmol/s (0.0343 kmol/m 3 )D / , s i-o
— In ■
7.069 X 10- 4 m 2
0.4 m
1 - 0.028
which gives
D AR = 3.06 X 10" 5 m 2 /s
for the binary diffusion coefficient of water vapor in air at 20°C and 83.5 kPa.
Air, B
a S
rt
=
O O "*
o
"53 G "O
3 -J 9
s
Q m '3
a
\
I
Water, A
3k, !
FIGURE 14-37
Schematic for Example 14-9.
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754
HEAT TRANSFER
Pa, =
i
Concentration
boundary
layer
Concentration
profile
Pa,- y s?*&
Species A
FIGURE 14-38
The development of a concentration
boundary layer for species A during
external flow on a flat surface.
Concentration
entry length
Fully developed
region
Species A
Concentration boundary layer
Thermal boundary layer
Velocity boundary layer
FIGURE 14-39
The development of the velocity,
thermal, and concentration boundary
layers in internal flow.
14-9 - MASS CONVECTION
So far we have considered mass diffusion, which is the transfer of mass due to
a concentration gradient. Now we consider mass convection (or convective
mass transfer), which is the transfer of mass between a surface and a moving
fluid due to both mass diffusion and bulk fluid motion. We mentioned ear-
lier that fluid motion enhances heat transfer considerably by removing the
heated fluid near the surface and replacing it by the cooler fluid further away.
Likewise, fluid motion enhances mass transfer considerably by removing
the high-concentration fluid near the surface and replacing it by the lower-
concentration fluid further away. In the limiting case of no bulk fluid motion,
mass convection reduces to mass diffusion, just as convection reduces to con-
duction. The analogy between heat and mass convection holds for both forced
and natural convection, laminar and turbulent flow, and internal and ex-
ternal flow.
Like heat convection, mass convection is also complicated because of the
complications associated with fluid flow such as the surface geometry, flow
regime, flow velocity, and the variation of the fluid properties and composi-
tion. Therefore, we will have to rely on experimental relations to determine
mass transfer. Also, mass convection is usually analyzed on a mass basis
rather than on a molar basis. Therefore, we will present formulations in terms
of mass concentration (density p or mass fraction w) instead of molar concen-
tration (molar density C or mole fraction y). But the formulations on a molar
basis can be obtained using the relation C = p/M where M is the molar mass.
Also, for simplicity, we will restrict our attention to convection in fluids that
are (or can be treated as) binary mixtures.
Consider the flow of air over the free surface of a water body such as a lake
under isothermal conditions. If the air is not saturated, the concentration of
water vapor will vary from a maximum at the water surface where the air is al-
ways saturated to the free steam value far from the surface. In heat convection,
we defined the region in which temperature gradients exist as the thermal
boundary layer. Similarly, in mass convection, we define the region of the
fluid in which concentration gradients exist as the concentration boundary
layer, as shown in Figure 14-38. In external flow, the thickness of the con-
centration boundary layer 8 C for a species A at a specified location on the sur-
face is defined as the normal distance y from the surface at which
Pa,
Pa
Pa, s ~ Pa,
0.99
where p A s and p A „ are the densities of species A at the surface (on the fluid
side) and the free stream, respectively.
In internal flow, we have a concentration entrance region where the
concentration profile develops, in addition to the hydrodynamic and thermal
entry regions (Fig. 14-39). The concentration boundary layer continues to
develop in the flow direction until its thickness reaches the tube center and
the boundary layers merge. The distance from the tube inlet to the location
where this merging occurs is called the concentration entry length L c , and
the region beyond that point is called the fully developed region, which is
characterized by
cen58 93 3_chl4.qxd 9/9/2002 10:05 AM Page 755
dx\P A ,
Pa,
Pa
PA.b
where p A b is the bulk mean density of species A defined as
(14-66)
(14-67)
Therefore, the nondimensionalized concentration difference profile as well as
the mass transfer coefficient remain constant in the fully developed region.
This is analogous to the friction and heat transfer coefficients remaining con-
stant in the fully developed region.
In heat convection, the relative magnitudes of momentum and heat diffusion
in the velocity and thermal boundary layers are expressed by the dimension-
less Prandtl number, defined as (Fig. 14-40)
Prandtl number:
Pr
v
a
Momentum diffusivity
Thermal diffusivity
(14-68)
The corresponding quantity in mass convection is the dimensionless Schmidt
number, defined as
Schmidt number:
Sc
v Momentum diffusivity
D A
Mass diffusivity
(14-69)
which represents the relative magnitudes of molecular momentum and mass
diffusion in the velocity and concentration boundary layers, respectively.
The relative growth of the velocity and thermal boundary layers in laminar
flow is governed by the Prandtl number, whereas the relative growth of the
velocity and concentration boundary layers is governed by the Schmidt num-
ber. A Prandtl number of near unity (Pr ~ 1) indicates that momentum and
heat transfer by diffusion are comparable, and velocity and thermal boundary
layers almost coincide with each other. A Schmidt number of near unity
(Sc ~ 1) indicates that momentum and mass transfer by diffusion are com-
parable, and velocity and concentration boundary layers almost coincide with
each other.
It seems like we need one more dimensionless number to represent the rel-
ative magnitudes of heat and mass diffusion in the thermal and concentration
boundary layers. That is the Lewis number, defined as (Fig. 14-41)
Lewis number:
Le
Sc a Thermal diffusivity
Pr D AB Mass diffusivity
(14-70)
The relative thicknesses of velocity, thermal, and concentration boundary
layers in laminar flow are expressed as
J velocity
Pr",
J velocity
"concentration
Sc", and
"concentration
Le"
(14-71)
where n = | for most applications in all three relations. These relations, in
general, are not applicable to turbulent boundary layers since turbulent mix-
ing in this case may dominate the diffusion processes.
755
CHAPTER 14
Heat transfer: Pr = ■
>
Mass transfer: Sc =
FIGURE 14-40
In mass transfer, the Schmidt number
plays the role of the Prandtl
number in heat transfer.
Le
Sc = _a
Pr ~D.
Thermal diffusivity
~Mass diffusivity
FIGURE 14-41
Lewis number is a measure of heat
diffusion relative to mass diffusion.
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756
HEAT TRANSFER
Concentration
profile
^ rr
'"t
* *
"V-c
y_ r
C d\
*■
*■
».
Mass diff
,
1 i J
Species A
1 By
,( W A
J
y=0
FIGURE 14-42
Mass transfer at a surface occurs by
diffusion because of the no-slip
boundary condition, just like heat
transfer occurring by conduction.
Note that species transfer at the surface (y = 0) is by diffusion only because
of the no-slip boundary condition, and mass flux of species A at the surface
can be expressed by Fick's law as (Fig. 14-42)
dw A
-PD AB ^
(kg/s • m 2 )
(14-72)
This is analogous to heat transfer at the surface being by conduction only and
expressing it by Fourier's law.
The rate of heat convection for external flow was expressed conveniently by
Newton's law of cooling as
e c
hoo m A(T s - T„)
where /z conv is the average heat transfer coefficient, A is the surface area, and
T s — T„ is the temperature difference across the thermal boundary layer.
Likewise, the rate of mass convection can be expressed as
"mass^HP/i
PA, ») = ^ma Ss pA(W A , , - W A _ «,)
(kg/s)
(14-73)
where h mdSS is the average mass transfer coefficient, in m/s; A is the surface
area; p A , — p A „ is the mass concentration difference of species A across the
concentration boundary layer; and p is the average density of the fluid in the
boundary layer. The product h mass p, whose unit is kg/m 2 • s, is called the mass
transfer conductance. If the local mass transfer coefficient varies in the flow
direction, the average mass transfer coefficient can be determined from
h = — \ h rIA
mass, ave A \ mass
(14-74)
In heat convection analysis, it is often convenient to express the heat trans-
fer coefficient in a nondimensionalized form in terms of the dimensionless
Nusselt number, defined as
Nusselt number:
Nu
(14-75)
Heat transfer: Nu = ■
Mass transfer: Sh :
k
FIGURE 14-43
In mass transfer, the Sherwood
number plays the role the Nusselt
number plays in heat transfer.
where L is the characteristic length of k is the thermal conductivity of the
fluid. The corresponding quantity in mass convection is the dimensionless
Sherwood number, defined as (Fig. 14-43)
Sherwood number:
Sh
D t
(14-76)
where /z mass is the mass transfer coefficient and D AB is the mass diffusivity. The
Nusselt and Sherwood numbers represent the effectiveness of heat and mass
convection at the surface, respectively.
Sometimes it is more convenient to express the heat and mass transfer co-
efficients in terms of the dimensionless Stanton number as
Heat transfer Stanton number:
St
K
pre,
Nu
1
RePr
(14-77)
cen58 93 3_chl4.qxd 9/9/2002 10:05 AM Page 757
and
757
CHAPTER 14
Mass transfer Stanton number:
_ ''mass _, 1
mass ~ Y ' Re Sc
(14-78)
where T is the free steam velocity in external flow and the bulk mean fluid
velocity in internal flow.
For a given geometry, the average Nusselt number in forced convection
depends on the Reynolds and Prandtl numbers, whereas the average Sher-
wood number depends on the Reynolds and Schmidt numbers. That is,
Nusselt number:
Sherwood number:
Nu = /(Re, Pr)
Sh =/(Re, Sc)
where the functional form of/is the same for both the Nusselt and Sherwood
numbers in a given geometry, provided that the thermal and concentration
boundary conditions are of the same type. Therefore, the Sherwood number
can be obtained from the Nusselt number expression by simply replacing the
Prandtl number by the Schmidt number. This shows what a powerful tool
analogy can be in the study of natural phenomena (Table 14-12).
In natural convection mass transfer, the analogy between the Nusselt and
Sherwood numbers still holds, and thus Sh = /(Gr, Sc). But the Grashof num-
ber in this case should be determined directly from
Gr
£(P«
g(A P /p) Ll
pv-
(14-79)
which is applicable to both temperature- and/or concentration-driven natural
convection flows. Note that in homogeneous fluids (i.e., fluids with no con-
centration gradients), density differences are due to temperature differences
only, and thus we can replace Ap/p by (BArfor convenience, as we did in nat-
ural convection heat transfer. However, in nonhomogeneous fluids, density
differences are due to the combined effects of temperature and concentration
differences, and Ap/p cannot be replaced by (3ATin such cases even when all
we care about is heat transfer and we have no interest in mass transfer. For ex-
ample, hot water at the bottom of a pond rises to the top. But when salt is
placed at the bottom, as it is done in solar ponds, the salty water (brine) at the
bottom will not rise because it is now heavier than the fresh water at the top
(Fig. 14-44).
Concentration-driven natural convection flows are based on the densities of
different species in a mixture being different. Therefore, at isothermal condi-
tions, there will be no natural convection in a gas mixture that is composed of
gases with identical molar masses. Also, the case of a hot surface facing up
corresponds to diffusing fluid having a lower density than the mixture (and
thus rising under the influence of buoyancy), and the case of a hot surface fac-
ing down corresponds to the diffusing fluid having a higher density. For ex-
ample, the evaporation of water into air corresponds to a hot surface facing up
since water vapor is lighter than the air and it will tend to rise. But this will not
be the case for gasoline unless the temperature of the gasoline-air mixture at
the gasoline surface is so high that thermal expansion overwhelms the density
differential due to higher gasoline concentration near the surface.
TABLE 14-12
Analogy between the quantities
that appear in the formulation
and solution of heat convection
and mass convection
Heat
Convection
Mass
Convection
Re =
Gr =
Pr =
St =
T
C, y, p, or w
''conv
''mass
^thermal
^concentration
YL C
V
Re
YL C
~ V
V 2
7-J Ll
Gr
g(p„ - p s ) L 3 C
pV-
V
a
Sc
V
P Y C p
.. ''conv *- c
NU = ;
k
Nu = f(Re, Pr)
Nu = f(Gr, Pr)
St„
T
Sh =
Dab
Sh = f(Re, Sc)
Sh = f(Gr, Sc)
20°C
Fresh water
No convection
SOLAR
currents
POND
70°C
rbrine r W ater
Brine
Salt
FIGURE 14-44
A hot fluid at the bottom will rise and
initiate natural convection currents
only if its density is lower.
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758
HEAT TRANSFER
Velocity,
temperature, or
concentration
profile
Tangent line
at v =
FIGURE 14-45
The friction, heat, and mass transfer
coefficients for flow over a surface are
proportional to the slope of the tangent
line of the velocity, temperature, and
concentration profiles, respectively,
at the surface.
Analogy between Friction, Heat Transfer,
and Mass Transfer Coefficients
Consider the flow of a fluid over a flat plate of length L with free steam con-
ditions of Too, To,, and w At m (Fig. 14^15). Noting that convection at the surface
(y = 0) is equal to diffusion because of the no-slip condition, the friction, heat
transfer, and mass transfer conditions at the surface can be expressed as
Wall friction:
Heat transfer:
Mass transfer:
J A.
-D A
dT
k dy
dw A
■ dy
f
pT-i
2
"heatUs ~~ *»)
h m *J.w A , s - *>A, »)
(14-80)
(14-81)
(14-82)
These relations can be rewritten for internal flow by using bulk mean proper-
ties instead of free stream properties. After some simple mathematical manip-
ulations, the three relations above can be rearranged as
Wall friction:
Heat transfer:
Mass transfer:
d(T/T„)
diylL)
d[(T - T S )I(T X - T s )]
/pV.L /
;y =
d(y/L)
d[(w A - w AiS )/(w Ai „ - w AtS )]
d(ylL)
2 V
"heat *-*
D A
2
Nu
Sh
Re
(14-83)
(14-84)
(14-85)
The left sides of these three relations are the slopes of the normalized veloc-
ity, temperature, and concentration profiles at the surface, and the right sides
are the dimensionless numbers discussed earlier.
Normalized
velocity, Velocity,
temperature, or temperature, or
concentration concentration
profile boundary layer
_v
Reynolds analogy
v = a = D AB
(or Pr = Sc = Le)
FIGURE 14-46
When the molecular diffusivities of
momentum, heat, and mass are equal
to each other, the velocity,
temperature, and concentration
boundary layers coincide.
Special Case: Pr ~ Sc ~ 1 (Reynolds Analogy)
Now consider the hypothetical case in which the molecular diffusivities of
momentum, heat, and mass are identical. That is, v = a = D AB and thus Pr =
Sc = Le = 1. In this case the normalized velocity, temperature, and concen-
tration profiles will coincide, and thus the slope of these three curves at the
surface (the left sides of Eqs. 14-83 through 14-85) will be identical (Fig.
14-46). Then we can set the right sides of those three equations equal to each
other and obtain
/
Re = Nu = Sh
2 v
K
K
D,
Noting that Pr = Sc = 1, we can also write this equation as
Nu Sh
2
Re Pr Re Sc
/
2
St = SL
(14-86)
(14-87)
This relation is known as the Reynolds analogy, and it enables us to deter-
mine the seemingly unrelated friction, heat transfer, and mass transfer coeffi-
cients when only one of them is known or measured. (Actually the original
cen58 93 3_chl4.qxd 9/9/2002 10:05 AM Page 759
Reynolds analogy proposed by O. Reynolds in 1874 is St = f/2, which is then
extended to include mass transfer.) However, it should always be remembered
that the analogy is restricted to situations for which Pr ~ Sc ~ 1. Of course
the first part of the analogy between friction and heat transfer coefficients can
always be used for gases since their Prandtl number is close to unity.
759
CHAPTER 14
General Case: Pr ^ Sc ^ 1 (Chilton-Colburn Analogy)
The Reynolds analogy is a very useful relation, and it is certainly desirable to
extend it to a wider range of Pr and Sc numbers. Several attempts have been
made in this regard, but the simplest and the best known is the one suggested
by Chilton and Colburn in 1934 as
2 '
St Pr 2 '
(14-88)
for 0.6 < Pr < 60 and 0.6 < Sc < 3000. This equation is known as the
Chilton-Colburn analogy. Using the definition of heat and mass Stanton
numbers, the analogy between heat and mass transfer can be expressed more
conveniently as (Fig. 14-47)
St = /Sc\ 2/3
St mass \Prj
K
h
"heat /Sc\
«
x2/3
yD
'abI
= pC,Le 2 «
(14-89)
For air-water vapor mixtures at 298 K, the mass and thermal diffusivities are
D AB = 2.5 X 10~ 5 m 2 /s and a = 2.18 X 10~ 5 m 2 /s and thus the Lewis number
is Le = a/D AB = 0.872. (We simply use the a value of dry air instead of the
moist air since the fraction of vapor in the air at atmospheric conditions is
low.) Then (a/D AB ) m = 0.872 2 ' 3 = 0.913, which is close to unity. Also, the
Lewis number is relatively insensitive to variations in temperature. Therefore,
for air-water vapor mixtures, the relation between heat and mass transfer co-
efficients can be expressed with a good accuracy as
K
pCphn
(air-water vapor mixtures)
(14-90)
where p and C p are the density and specific heat of air at mean conditions (or
pC ; , is the specific heat of air per unit volume). Equation 14-90 is known as
the Lewis relation and is commonly used in air-conditioning applications.
Another important consequence of Le = 1 is that the wet-bulb and adiabatic
saturation temperatures of moist air are nearly identical. In turbulent flow, the
Lewis relation can be used even when the Lewis number is not 1 since eddy
mixing in turbulent flow overwhelms any molecular diffusion, and heat and
mass are transported at the same rate.
The Chilton-Colburn analogy has been observed to hold quite well in lam-
inar or turbulent flow over plane surfaces. But this is not always the case for
internal flow and flow over irregular geometries, and in such cases specific re-
lations developed should be used. When dealing with flow over blunt bodies,
it is important to note that/in these relations is the skin friction coefficient, not
the total drag coefficient, which also includes the pressure drag.
Chilton-Colburn Analogy
General
=H^f
Special
case: v = a = D AB
^hcal 1
FIGURE 14-47
When the friction or heat transfer
coefficient is known, the mass transfer
coefficient can be determined directly
from the Chilton-Colburn analogy.
cen58 93 3_chl4.qxd 9/9/2002 10:05 AM Page 76C
760
HEAT TRANSFER
Ail-
Saturated
air
_L_L
Evaporation
t t t
Lake
FIGURE 14-48
Evaporation from the free surface of
water into air.
Limitation on the Heat-Mass Convection Analogy
Caution should be exercised when using the analogy in Eq. 14-88 since there
are a few factors that put some shadow on the accuracy of that relation. For
one thing, the Nusselt numbers are usually evaluated for smooth surfaces, but
many mass transfer problems involve wavy or roughened surfaces. Also,
many Nusselt relations are obtained for constant surface temperature situa-
tions, but the concentration may not be constant over the entire surface be-
cause of the possible surface dryout. The blowing or suction at the surface
during mass transfer may also cause some deviation, especially during high
speed blowing or suction.
Finally, the heat-mass convection analogy is valid for low mass flux cases
in which the flow rate of species undergoing mass flow is low relative to the
total flow rate of the liquid or gas mixture so that the mass transfer between
the fluid and the surface does not affect the flow velocity. (Note that convec-
tion relations are based on zero fluid velocity at the surface, which is true only
when there is no net mass transfer at the surface.) Therefore, the heat-mass
convection analogy is not applicable when the rate of mass transfer of a
species is high relative to the flow rate of that species.
Consider, for example, the evaporation and transfer of water vapor into air
in an air washer, an evaporative cooler, a wet cooling tower, or just at the free
surface of a lake or river (Fig. 14-48). Even at a temperature of 40°C, the
vapor pressure at the water surface is the saturation pressure of 7.4 kPa, which
corresponds to a mole fraction of 0.074 or a mass fraction of w Ai s = 0.047 for
the vapor. Then the mass fraction difference across the boundary layer will be,
at most, Aw = w AtS — w^ „ = 0.047 — = 0.047. For the evaporation of
water into air, the error involved in the low mass flux approximation is
roughly Aw/2, which is 2.5 percent in the worst case considered above. There-
fore, in processes that involve the evaporation of water into air, we can use the
heat-mass convection analogy with confidence. However, the mass fraction
of vapor approaches 1 as the water temperature approaches the saturation tem-
perature, and thus the low mass flux approximation is not applicable to mass
transfer in boilers, condensers, and the evaporation of fuel droplets in com-
bustion chambers. In this chapter we limit our consideration to low mass flux
applications.
Mass Convection Relations
Under low mass flux conditions, the mass convection coefficients can be de-
termined by either (1) determining the friction or heat transfer coefficient and
then using the Chilton-Colburn analogy or (2) picking the appropriate Nusselt
number relation for the given geometry and analogous boundary conditions,
replacing the Nusselt number by the Sherwood number and the Prandtl num-
ber by the Schmidt number, as shown in Table 14-13 for some representative
cases. The first approach is obviously more convenient when the friction or
heat transfer coefficient is already known. Otherwise, the second approach
should be preferred since it is generally more accurate, and the Chilton-
Colburn analogy offers no significant advantage in this case. Relations for
convection mass transfer in other geometries can be written similarly using
the corresponding heat transfer relation in Chapters 6 and 7.
cen58 93 3_chl4.qxd 9/9/2002 10:05 AM Page 761
761
CHAPTER 14
TABLE 14-13
Sherwood number relations in mass convection for specified concentration at the surface corresponding to the Nusselt
number relations in heat convection for specified surface temperature
Convective Heat Transfer
Convective Mass Transfer
1 . Forced Convection over a Flat Plate
(a) Laminar flow (Re < 5 X 10 5 )
Nu = 0.664 Re? 5 Pr 1/3 , Pr > 0.6
(b) Turbulent flow (5 X 10 5 < Re < 10 7 )
Nu = 0.037 Re° L 8 Pr 1/3 , Pr > 0.6
2. Fully Developed Flow in Smooth Circular Pipes
(a) Laminar flow (Re < 2300)
Nu = 3.66
{b) Turbulent flow (Re > 10,000)
Nu = 0.023 Re 08 Pr 04 , 0.7 < Pr < 160
3. Natural Convection over Surfaces
(a) Vertical plate
Nu = 0.59(Gr Pr) 1 ' 4 , 10 5 < Gr Pr < 10 9
Nu = O.KGr Pr) 1 ' 3 , 10 9 < Gr Pr < 10 13
(£>) Upper surface of a horizontal plate
Surface is hot (T s > 7"J
Nu = 0.54(Gr Pr) 1 ' 4 , 10 4 < Gr Pr < 10 7
Nu = 0.15(Gr Pr) 1 ' 3 , 10 7 < Gr Pr < 10 11
(c) Lower surface of a horizontal plate
Surface is hot (T s > TJ
Nu = 0.27(Gr Pr) 1 ' 4 , 10 5 < Gr Pr < 10 11
Sh = 0.664 RePSc 1 ' 3 , So 0.5
Sh = 0.037 Re?- 8 Sc 1/3 , So 0.5
Sh = 3.66
Sh = 0.023 Re°- 8 Sc°
Sh = 0.59(Gr Sc) 1 ' 4 ,
Sh = O.KGr Sc) 1 ' 3 ,
0.7 < Sc 160
10 5 < GrSc < 10 9
10 9 < GrSc < 10 13
Fluid near the surface is light (p s < pj
Sh = 0.54(Gr Sc) 1 ' 4 , 10 4 < Gr Sc < 10 7
Sh = 0.15(GrSc) 1/3 , 10 7 < Gr Sc < 10 11
Fluid near the surface is light (p s < p rx )
Sh = 0.27(Gr Sc) 1 ' 4 , 10 5 < Gr Sc < 10 11
EXAMPLE 14-10 Mass Convection inside a Circular Pipe
Consider a circular pipe of inner diameter D = 0.015 m whose inner surface is
covered with a layer of liquid water as a result of condensation (Fig. 14-49). In
order to dry the pipe, air at 300 K and 1 atm is forced to flow through it with an
average velocity of 1.2 m/s. Using the analogy between heat and mass transfer,
determine the mass transfer coefficient inside the pipe for fully developed flow.
SOLUTION The liquid layer on the inner surface of a circular pipe is dried by
blowing air through it. The mass transfer coefficient is to be determined.
Assumptions 1 The low mass flux model and thus the analogy between heat
and mass transfer is applicable since the mass fraction of vapor in the air is low
(about 2 percent for saturated air at 300 K). 2 The flow is fully developed.
Properties Because of low mass flux conditions, we can use dry air properties
for the mixture at the specified temperature of 300 K and 1 atm, for which v =
1.58 X 10~ 5 m 2 /s (Table A-15). The mass diffusivity of water vapor in the air at
300 K is determined from Eq. 14-15 to be
D,
£V
1.87 X 10-
1.87 X 10"
, 300 2C
2.54 X 10- 5 m 2 /s
Wet pipe
Air
300 K\
1 atm
-y = 1.2 m/s
FIGURE 14-49
Schematic for Example 14-10.
cen58 93 3_chl4.qxd 9/9/2002 10:05 AM Page 762
762
HEAT TRANSFER
Analysis The Reynolds number for this internal flow is
YD (1.2m/s)(0.015m)
Re = -^ = , ' = 1139
v 1.58 X 10- 5 m 2 /s
which is less than 2300 and thus the flow is laminar. Therefore, based on the
analogy between heat and mass transfer, the Nusselt and the Sherwood num-
bers in this case are Nu = Sh = 3.66. Using the definition of Sherwood num-
ber, the mass transfer coefficient is determined to be
ShD AB (3.66)(2.54 X lO" 5 m 2 /s)
D
0.015 m
0.00620 m/s
The mass transfer rate (or the evaporation rate) in this case can be determined
by defining the logarithmic mean concentration difference in an analogous
manner to the logarithmic mean temperature difference.
Naphthalene
vapor
0.3 m 2
Air
1 atm
T o0 = 25°C ;
y„ = 2 m/s ■
FIGURE 14-50
Schematic for Example 14-11.
EXAMPLE 14-11 Analogy between Heat and Mass Transfer
Heat transfer coefficients in complex geometries with complicated boundary
conditions can be determined by mass transfer measurements on similar
geometries under similar flow conditions using volatile solids such as naphtha-
lene and dichlorobenzene and utilizing the Chilton-Colburn analogy between
heat and mass transfer at low mass flux conditions. The amount of mass trans-
fer during a specified time period is determined by weighing the model or mea-
suring the surface recession.
During a certain experiment involving the flow of dry air at 25°C and 1 atm at
a free stream velocity of 2 m/s over a body covered with a layer of naphthalene,
it is observed that 12 g of naphthalene has sublimated in 15 min (Fig. 14-50).
The surface area of the body is 0.3 m 2 . Both the body and the air were kept at
25°C during the study. The vapor pressure of naphthalene at 25°C is 11 Pa and
the mass diffusivity of naphthalene in air at 25°C is D AB = 0.61 X 10~ 5 m 2 /s.
Determine the heat transfer coefficient under the same flow conditions over the
same geometry.
SOLUTION Air is blown over a body covered with a layer of naphthalene, and
the rate of sublimation is measured. The heat transfer coefficient under the
same flow conditions over the same geometry is to be determined.
Assumptions 1 The low mass flux conditions exist so that the Chilton-Colburn
analogy between heat and mass transfer is applicable (will be verified). 2 Both
air and naphthalene vapor are ideal gases.
Properties The molar mass of naphthalene is 128.2 kg/kmol. Because of low
mass flux conditions, we can use dry air properties for the mixture at the spec-
ified temperature of 25°C and 1 atm, at which p = 1.184 kg/m 3 , C P = 1007
J/kg ■ K, and a = 2.141 X lO" 5 m 2 /s (Table A-15).
Analysis The incoming air is free of naphthalene, and thus the mass fraction
of naphthalene at free stream conditions is zero, w A „ = 0. Noting that the
vapor pressure of naphthalene at the surface is 11 Pa, its mass fraction at the
surface is determined to be
cen58 93 3_chl4.qxd 9/9/2002 10:05 AM Page 763
M,
11 Pa /128.2 kg/kmol\
- =4.8 X 1(T 4
P \M U J 101,325 Pa \ 29 kg/kmol
which confirms that the low mass flux approximation is valid. The rate of evap-
oration of naphthalene in this case is
m 0.012 kg
1.33 X 10- 5 kg/s
cvap A? (15 X 60s)
Then the mass convection coefficient becomes
,n 1.33 X 10- 5 kg/s
pA s (wa,s ~ w a,J (1.184 kg/m 3 )(0.3 m 2 )(4.8 X 10"
0)
0.0780 m/s
Using the analogy between heat and mass transfer, the average heat transfer
coefficient is determined from Eq. 14-89 to be
pC p h n
(-T
■ \D AB )
/? 141 X 1 f) — 5 m 2 Ai
(1.184 kg/m 3 )(1007 J/kg • °C)(0.0776 m/s)(
\ 0.61 X 10- 5 m 2 /s
215 W/m 2
Discussion Because of the convenience it offers, naphthalene has been
used in numerous heat transfer studies to determine convection heat transfer
coefficients.
14-10 - SIMULTANEOUS HEAT AND
MASS TRANSFER
Many mass transfer processes encountered in practice occur isothermally, and
thus they do not involve any heat transfer. But some engineering applications
involve the vaporization of a liquid and the diffusion of this vapor into the sur-
rounding gas. Such processes require the transfer of the latent heat of vapor-
ization hj g to the liquid in order to vaporize it, and thus such problems involve
simultaneous heat and mass transfer. To generalize, any mass transfer problem
involving phase change (evaporation, sublimation, condensation, melting,
etc.) must also involve heat transfer, and the solution of such problems needs
to be analyzed by considering simultaneous heat and mass transfer. Some ex-
amples of simultaneous heat and mass problems are drying, evaporative cool-
ing, transpiration (or sweat) cooling, cooling by dry ice, combustion of fuel
droplets, and ablation cooling of space vehicles during reentry, and even ordi-
nary events like rain, snow, and hail. In warmer locations, for example, the
snow melts and the rain evaporates before reaching the ground (Fig. 14-51).
To understand the mechanism of simultaneous heat and mass transfer, con-
sider the evaporation of water from a swimming pool into air. Let us assume
that the water and the air are initially at the same temperature. If the air is
saturated (a relative humidity of cj) = 100 percent), there will be no heat or
mass transfer as long as the isothermal conditions remain. But if the air is not
763
CHAPTER 14
Plastic
or glass
Evaporation
r^Heat
S
(a) Ablation
Evaporation
(b) Evaporation of
rain droplet
jy Heat
^swrejection
Condensation
Vapor
Liquid
yw Evaporation
Heat
^Heat
absorption
(c) Drying of clothes (d) Heat pipes
FIGURE 14-51
Many problems
encountered in practice involve
simultaneous heat and mass transfer.
cen58 93 3_chl4.qxd 9/9/2002 10:05 AM Page 764
764
HEAT TRANSFER
Surroundings
20°C
FIGURE 14-52
Various mechanisms of heat transfer
involved during the evaporation of
water from the surface of a lake.
saturated (d) < 100 percent), there will be a difference between the concen-
tration of water vapor at the water-air interface (which is always saturated)
and some distance above the interface (the concentration boundary layer).
Concentration difference is the driving force for mass transfer, and thus this
concentration difference will drive the water into the air. But the water must
vaporize first, and it must absorb the latent heat of vaporization in order to va-
porize. Initially, the entire heat of vaporization will come from the water near
the interface since there is no temperature difference between the water and
the surroundings and thus there cannot be any heat transfer. The temperature
of water near the surface must drop as a result of the sensible heat loss, which
also drops the saturation pressure and thus vapor concentration at the inter-
face.
This temperature drop creates temperature differences within the water at
the top as well as between the water and the surrounding air. These tempera-
ture differences drive heat transfer toward the water surface from both the air
and the deeper parts of the water, as shown in Figure 14-52. If the evaporation
rate is high and thus the demand for the heat of vaporization is higher than the
amount of heat that can be supplied from the lower parts of the water body
and the surroundings, the deficit will be made up from the sensible heat of the
water at the surface, and thus the temperature of water at the surface will drop
further. The process will continue until the latent heat of vaporization is equal
to the rate of heat transfer to the water at the surface. Once the steady opera-
tion conditions are reached and the interface temperature stabilizes, the energy
balance on a thin layer of liquid at the surface can be expressed as
Q
sensible, transferred
*£ latent, a
or Q = m v h f
(14-91)
where m „ is the rate of evaporation and hj g is the latent heat of vaporization of
water at the surface temperature. Various expressions for m v under various ap-
proximations are given in Table 14-14. The mixture properties such as the
specific heat C p and molar mass M should normally be evaluated at the mean
film composition and mean film temperature. However, when dealing with
air-water vapor mixtures at atmospheric conditions or other low mass flux sit-
uations, we can simply use the properties of the gas with reasonable accuracy.
TABLE 14-14
Various expressions for evaporation rate of a liquid into a gas through an
interface area A s under various approximations (subscript v stands for vapor,
s for liquid-gas interface, and qq away from surface)
Assumption
Evaporation Rate
General
Assuming vapor to be an ideal gas,
Pv = p v RJ
Using Chilton-Colburn analogy,
"heat = pCp"massLe
111 T s + T a
Using — - — ~ — , where T = — - —
and P= P RT= p(R u /M)T
n\, = h mass A s {p V:S - p Vi J
n\
m, =
<A,=
"mass "s
K Ts
s
p,
T
•)
R,
I
"mass "
• I
S
Pv,»
pC p Le 2 ' 3
R v \
r.
h A
"mass ^s
M v
p,
S
p^
P C p Le 2/3 M
cen58 93 3_chl4.qxd 9/9/2002 10:05 AM Page 765
The Q in Eq. 14-91 represents all forms of heat from all sources transferred
to the surface, including convection and radiation from the surroundings and
conduction from the deeper parts of the water due to the sensible energy of the
water itself or due to heating the water body by a resistance heater, heating
coil, or even chemical reactions in the water. If heat transfer from the water
body to the surface as well as radiation from the surroundings is negligible,
which is often the case, then the heat loss by evaporation must equal heat gain
by convection. That is,
Q
v fg
or h com A s (T x , - T s )
h mm A s h h M v P v ,
C p Lc™
M
765
CHAPTER 14
Canceling h com A s from both sides of the second equation gives
T =T kfs M * P "-'- p ^
" C„Le 2 ' 3 M P
(14-92)
which is a relation for the temperature difference between the liquid and the
surrounding gas under steady conditions.
EXAMPLE 14-12 Evaporative Cooling of a Canned Drink
During a hot summer day, a canned drink is to be cooled by wrapping it in a
cloth that is kept wet continually, and blowing air to it with a fan (Fig. 14-53).
If the environment conditions are 1 atm, 30°C, and 40 percent relative humid-
ity, determine the temperature of the drink when steady conditions are reached.
SOLUTION Air is blown over a canned drink wrapped in a wet cloth to cool it
by simultaneous heat and mass transfer. The temperature of the drink when
steady conditions are reached is to be determined.
Assumptions 1 The low mass flux conditions exist so that the Chilton-Colburn
analogy between heat and mass transfer is applicable since the mass fraction of
vapor in the air is low (about 2 percent for saturated air at 300 K). 2 Both air
and water vapor at specified conditions are ideal gases (the error involved in this
assumption is less than 1 percent). 3 Radiation effects are negligible.
Properties Because of low mass flux conditions, we can use dry air properties
for the mixture at the average temperature of (7" x + 7~ s )/2 which cannot be de-
termined at this point because of the unknown surface temperature 7" s . We know
that T s < T„ and, for the purpose of property evaluation, we take 7" s to be 20°C.
Then the properties of water at 20°C and the properties of dry air at the average
temperature of 25°C and 1 atm are (Tables A-9 and A-15)
Water: h fg = 2454 kJ/kg, P v = 2.34 kPa; also, P v = 4.25 kPa at 30°C
Dry air: C P = 1.007 kJ/kg • °C, a = 2.141 X lO" 5 m 2 /s
The molar masses of water and air are 18 and 29 kg/kmol, respectively (Table
A-l). Also, the mass diffusivity of water vapor in air at 25°C is D Hz0 _ air = 2.50 X
10- 5 m 2 /s (Table 14-4).
Analysis Utilizing the Chilton-Colburn analogy, the surface temperature of the
drink can be determined from Eq. 14-92,
1 atm
30°C
40% RH
FIGURE 14-53
Schematic for Example 14-12.
cen58 93 3_chl4.qxd 9/9/2002 10:05 AM Page 766
766
HEAT TRANSFER
'78
M V P V , S ~P V ,
C p Le 2B M
where the Lewis number is
Le
D.
2.141 X lQ- 5 m 2 /s
2.5 X 10- 5 m 2 /s
0.856
Note that we could take the Lewis number to be 1 for simplicity, but we chose
to incorporate it for better accuracy.
The air at the surface is saturated, and thus the vapor pressure at the sur-
face is simply the saturation pressure of water at the surface temperature
(2.34 kPa). The vapor pressure of air far from the surface is determined from
W
sat ® T,_ '
(0.40)P SI
,. c = (0.40)(4.25 kPa) = 1.70 kPa
Noting that the atmospheric pressure is 1 atm = 101.3 kPa, substituting gives
2454 kJ/kg 18 kg/kmol (2.34 - 1.70) kPa
30°C
(1.007 kJ/kg • °C)(0.872) 2 ' 3 29 kg/kmol 101.3 kPa
= 19.4°C
Therefore, the temperature of the drink can be lowered to 19.4°C by this
process.
Surrounding
surfaces
20°C
Water
bath
50°C
A
F\
AAA
, Aerosol
j=k can
■» Heat supplied
Resistance heater
FIGURE 14-54
Schematic for Example 14-13.
EXAMPLE 14-13
Heat Loss from Uncovered Hot Water Baths
Hot water baths with open tops are commonly used in manufacturing facilities
for various reasons. In a plant that manufactures spray paints, the pressurized
paint cans are temperature tested by submerging them in hot water at 50°C in
a 40-cm-deep rectangular bath and keeping them there until the cans are
heated to 50°C to ensure that the cans can withstand temperatures up to 50°C
during transportation and storage (Fig. 14-54). The water bath is 1 m wide and
3.5 m long, and its top surface is open to ambient air to facilitate easy obser-
vation for the workers. If the average conditions in the plant are 92 kPa, 25°C,
and 52 percent relative humidity, determine the rate of heat loss from the top
surface of the water bath by (a) radiation, {b) natural convection, and (c) evap-
oration. Assume the water is well agitated and maintained at a uniform temper-
ature of 50°C at all times by a heater, and take the average temperature of the
surrounding surfaces to be 20 C C.
SOLUTION Spray paint cans are temperature tested by submerging them in an
uncovered hot water bath. The rates of heat loss from the top surface of the
bath by radiation, natural convection, and evaporation are to be determined.
Assumptions 1 The low mass flux conditions exist so that the Chilton-Colburn
analogy between heat and mass transfer is applicable since the mass fraction of
vapor in the air is low (about 2 percent for saturated air at 300 K). 2 Both air
and water vapor at specified conditions are ideal gases (the error involved in this
assumption is less than 1 percent). 3 Water is maintained at a uniform temper-
ature of 50°C.
cen58 93 3_chl4.qxd 9/9/2002 10:05 AM Page 767
Properties Relevant properties for each mode of heat transfer are determined
below in respective sections.
Analysis (a) The emissivity of liquid water is given in Table A-18 to be
0.95. Then the radiation heat loss from the water to the surrounding surfaces
becomes
Grad = ^M's — 'suit)
= (0.95)(3.5 m 2 )(5.67 X 1(T 8 W/m 2 X K 4 )[(323 K) 4 - (293K) 4 ]
= 663W
(£>) The air-water vapor mixture is dilute and thus we can use dry air properties
for the mixture at the average temperature of {T a + T S )I2 = (25 + 50)/2 =
37.5°C. Noting that the total atmospheric pressure is 92/101.3 = 0.9080 atm,
the properties of dry air at 37.5°C and 0.908 atm are
k = 0.02644 W/m • °C, Pr = 0.7261 (independent of pressure)
a = (2.311 X 10- 5 m 2 /s)/0.9080 = 2.545 X 10- 5 m 2 /s
v = (1.678 X 10- 5 m 2 /s)/0.9080 = 1.848 X 10- 5 m 2 /s
The properties of water at 50°C are
h fg = 2383 kJ/kg and P v = 12.35 kPa
The air at the surface is saturated, and thus the vapor pressure at the surface is
simply the saturation pressure of water at the surface temperature (12.35 kPa).
The vapor pressure of air far from the water surface is determined from
Pv.- = *P««r„= (0.52)P sat(825 . c = (0.52)(3.17 kPa) = 1.65 kPa
Treating the water vapor and the air as ideal gases and noting that the total at-
mospheric pressure is the sum of the vapor and dry air pressures, the densities
of the water vapor, dry air, and their mixture at the water-air interface and far
from the surface are determined to be
At the
surface:
Pv,.
12.35 kPa
Ps = Pv,,
(0.4615 kPa ■ m 3 /kg ■ K)(323 K)
(92 - 12.35) kPa
" (0.287 kPa ■ nrVkg ■ K)(323 K) ~
p as = 0.0828 + 0.8592 = 0.9420 kg/m 3
= 0.0828 kg/m 3
0.8592 kg/m 3
and
Away from
the surface:
Pv,*
P«
1.65 kPa
Pv,»
(0.4615 kPa ■ m 3 /kg ■ K)(298 K)
(92-1.65)kPa
" (0.287 kPa ■ m 3 /kg ■ K)(298 K) ~
p n , K = 0.0120 + 1.0564 = 1.0684 kg/m 3
: 0.0120 kg/m 3
1.0564 kg/m 3
The area of the top surface of the water bath is A s = (3.5 m)(l m) = 3.5 m 2 and
its perimeter is p = 2(3.5 + 1) = 9 m. Therefore, the characteristic length is
767
CHAPTER 14
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768
HEAT TRANSFER
4 = "7T = £ 7T ilL = 0.3889 m
P 9 m
Then using densities (instead of temperatures) since the mixture is not homo-
geneous, the Grashof number is determined to be
Gr
g(P°° ~ Ps)Ll
pv 2
(9.81 m/s 2 )(1.0684 - 0.9420 kg/m 3 )(0.3889 m) 3
~ [(0.9420 + 1.0684)/2kg/m 3 ]( 1.848 X 10- 5 m 2 /s) 2
= 2.125 X 10 8
Recognizing that this is a natural convection problem with hot horizontal sur-
face facing up, the Nusselt number and the convection heat transfer coeffi-
cients are determined to be
Nu = 0.15(GrPr) 1/3 = 0.15(2.125 X 10 8 X 0.7261) 1
80.45
and
Nu/fc _ (80.45)(0.02644 W/m ■ °C)
~T~ 0.3889 m
5.47 W/m 2 -°C
Then the natural convection heat transfer rate becomes
= (5.47 W/m 2 ■ °C)(3.5 m 2 )(50 - 25)°C = 479 W
Note that the magnitude of natural convection heat transfer is comparable to
that of radiation, as expected.
(c) Utilizing the analogy between heat and mass convection, the mass transfer
coefficient is determined the same way by replacing Pr by Sc. The mass diffu-
sivity of water vapor in air at the average temperature of 310.5 K is determined
from Eq. 14-15 to be
7-2.072
= 3.00 X 10- 5 m 2 /s
1.87 X 10
310.5 207
0.908
The Schmidt number is determined from its definition to be
1.848 X 10- 5 m 2 /s
Sc
D A
3.00 X 10- 5 m 2 /s
0.616
The Sherwood number and the mass transfer coefficients are determined to be
Sh = 0.15(GrSc) 1/3 = 0.15(2.125 X 10 8 X 0.616) 1 ' 3 = 76.2
and
ShD AB _ (76.2)(3.00 X 10" 5 m 2 /s)
L. 0.3889 m
0.00588 m/s
cen58 93 3_chl4.qxd 9/9/2002 10:05 AM Page 769
Then the evaporation rate and the rate of heat transfer by evaporation become
™« = "mass^APv, s ~ Pv, »)
= (0.00588 m/s)(3.5 m 2 )(0.0828 - 0.0120)kg/m 3
= 0.00146 kg/s = 5.25 kg/h
and
fie
,h f = (0.00146 kg/s)(2383 kJ/kg) = 3.479 kW = 3479 W
which is more than seven times the rate of heat transfer by natural convection.
Finally, noting that the direction of heat transfer is always from high to low
temperature, all forms of heat transfer determined above are in the same direc-
tion, and the total rate of heat loss from the water to the surrounding air and
surfaces is
fit
firad + Gc
fie
663 + 479 + 3479 = 4621 W
Discussion Note that if the water bath is heated electrically, a 4.6 kW resis-
tance heater will be needed just to make up for the heat loss from the top sur-
face. The total heater size will have to be larger to account for the heat losses
from the side and bottom surfaces of the bath as well as the heat absorbed by
the spray paint cans as they are heated to 50°C. Also note that water needs to
be supplied to the bath at a rate of 5.25 kg/h to make up for the water loss by
evaporation. Also, in reality, the surface temperature will probably be a little
lower than the bulk water temperature, and thus the heat transfer rates will be
somewhat lower than indicated here.
769
CHAPTER 14
SUMMARY
Mass transfer is the movement of a chemical species from a
high concentration region toward a lower concentration one
relative to the other chemical species present in the medium.
Heat and mass transfer are analogous to each other, and several
parallels can be drawn between them. The driving forces are
the temperature difference in heat transfer and the concentra-
tion difference in mass transfer. Fick's law of mass diffusion is
of the same form as Fourier's law of heat conduction. The
species generation in a medium due to homogeneous reactions
is analogous to heat generation. Also, mass convection due to
bulk fluid motion is analogous to heat convection. Constant
surface temperature corresponds to constant concentration at
the surface, and an adiabatic wall corresponds to an imperme-
able wall. However, concentration is usually not a continuous
function at a phase interface.
The concentration of a species A can be expressed in
terms of density p A or molar concentration C A . It can also be
expressed in dimensionless form in terms of mass or molar
fraction as
Mass fraction of species A:
Mole fraction of species A:
>'a
»<A
m A /V
Pa
m
N A
m/V
N A /V
P
c A
N
N/V
c
In the case of an ideal gas mixture, the mole fraction of a gas is
equal to its pressure fraction. Fick's law for the diffusion of a
species A in a stationary binary mixture of species A and B in a
specified direction x is expressed as
basis:
m im,A d(p A /p)
Job. a A P^ab dx
dw A
basis:
t N as,A d(C A /c)
Jdiff.A" A LV AB ^
dyA
= ~CD AB ^
AU dx
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770
HEAT TRANSFER
where D AB is the diffusion coefficient (or mass diffusivity) of
the species in the mixture, _/ diff A is the diffusive mass flux of
species A, andj diff A is the molar flux.
The mole fractions of a species i in the gas and liquid phases
at the interface of a dilute mixture are proportional to each
other and are expressed by Henry 's law as
ji, liquid side
i, gas side
~~H
where //is Henry's constant. When the mixture is not dilute, an
approximate relation for the mole fractions of a species on the
liquid and gas sides of the interface are expressed approxi-
mately by Raoult's law as
i, gas side
J\. gas side J i, liquid side /, satv /
where P, sat (I) is the saturation pressure of the species i at the
interface temperature and P is the total pressure on the gas
phase side.
The concentration of the gas species i in the solid at the in-
terface C, sond side is proportional to the partial pressure of the
species i in the gas P L gas sidc on the gas side of the interface and
is expressed as
C,
JfXR
i. gas side
where if is the solubility. The product of the solubility of a gas
and the diffusion coefficient of the gas in a solid is referred to
as the permeability 2P, which is a measure of the ability of the
gas to penetrate a solid.
In the absence of any chemical reactions, the mass transfer
rates m diff A through a plane wall of area A and thickness L and
cylindrical and spherical shells of inner and outer radii r x and r 2
under one-dimensional steady conditions are expressed as
pD AB A
P.4.
Pa,]
w A , — w A 2 Pa, l — Pa, 2
m A ,„ A „„, = 2TrLpD AB —r~, — rr^- = 2ttLD a
'diff,A,cyl
'Wdiff.A.sph = 4Trr,r 2 pD
4Tiri r 2 D
\n{r 2 lr{)
wa.i - w A , 2
ab r 2 - r l
Pa,i ~ Pa. 2
ln(r 2 /V,)
l'2^AS ,-, - r .
where P A { and P A 2 are the partial pressures of gas A on the
two sides of the wall.
During mass transfer in a moving medium, chemical species
are transported both by molecular diffusion and by the bulk
fluid motion, and the velocities of the species are expressed as
T A = T + T dlff , A
T fi = T + T diff , 5
where T is the mass-average velocity of the flow. It is the
velocity that would be measured by a velocity sensor and is
expressed as
T = WaYa + Wb y b
The special case T = corresponds to a stationary medium.
Using Fick's law of diffusion, the total mass fluxes j = m/A in
a moving medium are expressed as
h = Pa°V + PA^diff.A = Wa(Ja + Jb) ~ PD AB
dw A
dx
dw R
Jb = 9b°V + Ps^diff.s = w B (j A + j B ) - pD BA —
The rate of mass convection of species A in a binary mixture
is expressed in an analogous manner to Newton's law of cool-
ing as
>Weo„v = ^mass^ S (PA, s ~ Pa, J = ''mass P^jO^ s ~ W A „)
where h m!iss is the average mass transfer coefficient, in m/s.
The counterparts of the Prandtl and Nusselt numbers in mass
convection are the Schmidt number Sc and the Sherwood num-
ber Sh, defined as
Sc
v Momentum diffusivity
D AB Mass diffusivity
and Sh
h J
mass
D iR
The relative magnitudes of heat and mass diffusion in the ther-
mal and concentration boundary layers are represented by the
Lewis number, defined as
Le
Sc a Thermal diffusivity
Pr D AB Mass diffusivity
The mass flow rate of a gas through a solid plane wall under
steady one-dimensional conditions can also be expressed in
terms of the partial pressures of the adjacent gas on the two
sides of the solid as
^diff.A.wall — D AB if AB A
Pa.2
Heat and mass transfer coefficients are sometimes expressed in
terms of the dimensionless Stanton number, defined as
St,
h
conv
— - = Nu and St mass = -^— = Sh -
C„ Re Pr mass T Re Sc
where T is the free-stream velocity in external flow and the
bulk mean fluid velocity in internal flow. For a given geometry
and boundary conditions, the Sherwood number in natural or
cen58 93 3_chl4.qxd 9/9/2002 10:06 AM Page 771
forced convection can be determined from the corresponding
Nusselt number expression by simply replacing the Prandtl
number by the Schmidt number. But in natural convection, the
Grashof number should be expressed in terms of density dif-
ference instead of temperature difference.
When the molecular diffusivities of momentum, heat,
and mass are identical, we have v = a = D AB , and thus Pr =
Sc = Le = 1 . The similarity between momentum, heat, and
mass transfer in this case is given by the Reynolds analogy,
expressed as
/
Re = Nu = Sh
2 v
D,
St = SL
771
CHAPTER 14
which is known as the Chllton-Colburn analogy. The analogy
between heat and mass transfer is expressed more conve-
niently as
h
pQLe 2 ' 3 K
9 CJa/D AB f a h n
For air-water vapor mixtures, Le = 1, and thus this relation
simplifies further. The heat-mass convection analogy is limited
to low mass flux cases in which the flow rate of species under-
going mass flow is low relative to the total flow rate of the
liquid or gas mixture. The mass transfer problems that involve
phase change (evaporation, sublimation, condensation, melt-
ing, etc.) also involve heat transfer, and such problems are an-
alyzed by considering heat and mass transfer simultaneously.
For the general case of Pr + Sc # 1, it is modified as
{ = StPr 2 « = St raass Sc^
REFERENCES AND SUGGESTED READING
1. American Society of Heating, Refrigeration, and Air
Conditioning Engineers. Handbook of Fundamentals.
Atlanta: ASHRAE, 1993.
2. R. M. Barrer. Diffusion in and through Solids. New York:
Macmillan, 1941.
3. R. B. Bird. "Theory of Diffusion." Advances in Chemical
Engineering 1 (1956), p. 170.
4. R. B. Bird, W. E. Stewart, and E. N. Lightfoot. Transport
Phenomena. New York: John Wiley & Sons, 1960.
5. C. J. Geankoplis. Mass Transport Phenomena. New York:
Holt, Rinehart, and Winston, 1972.
6. Handbook of Chemistry and Physics 56th ed. Cleveland,
OH: Chemical Rubber Publishing Co., 1976.
7. J. O. Hirshfelder, F. Curtis, and R. B. Bird. Molecular
Theory of Gases and Liquids. New York: John Wiley &
Sons, 1954.
8. J. P. Holman. Heat Transfer. 7th ed. New York:
McGraw-Hill, 1990.
9. F. P. Incropera and D. P. De Witt. Fundamentals of Heat
and Mass Transfer. 2nd ed. New York: John Wiley &
Sons, 1985.
10. International Critical Tables. Vol. 3. New York:
McGraw-Hill, 1928.
11. W. M. Kays and M. E. Crawford. Convective Heat and
Mass Transfer. 2nd ed. New York: McGraw-Hill, 1980.
12. T R. Marrero and E. A. Mason. "Gaseous Diffusion
Coefficients." Journal ofPhys. Chem. Ref Data 1 (1972),
pp. 3-118.
13. A. F Mills. Basic Heat and Mass Transfer. Burr Ridge,
IL: Richard D. Irwin, 1995.
14. J. H. Perry, ed. Chemical Engineer's Handbook. 4th ed.
New York: McGraw-Hill, 1963.
15. R. D. Reid, J. M. Prausnitz, and T. K. Sherwood. The
Properties of Gases and Liquids. 3rd ed. New York:
McGraw-Hill, 1977.
16. A. H. P. Skelland. Diffusional Mass Transfer. New York:
John Wiley & Sons, 1974.
17. D. B. Spalding. Convective Mass Transfer. New York:
McGraw-Hill, 1963.
18. W. F Stoecker and J. W. Jones. Refrigeration and Air
Conditioning. New York: McGraw-Hill, 1982.
19. L. C. Thomas. Mass Transfer Supplement — Heat Transfer.
Englewood Cliffs, NJ: Prentice Hall, 1991.
20. L. Van Black. Elements of Material Science and
Engineering. Reading, MA: Addison-Wesley, 1980.
cen58 93 3_chl4.qxd 9/9/2002 10:06 AM Page 772
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HEAT TRANSFER
PROBLEMS
Analogy between Heat and Mass Transfer
14-1C How does mass transfer differ from bulk fluid flow?
Can mass transfer occur in a homogeneous medium?
14-2C How is the concentration of a commodity defined?
How is the concentration gradient defined? How is the diffu-
sion rate of a commodity related to the concentration gradient?
14-3C Give examples for (a) liquid-to-gas, (b) solid-to-
liquid, (c) solid-to-gas, and (d) gas-to-liquid mass transfer.
14-4C Someone suggests that thermal (or heat) radiation can
also be viewed as mass radiation since, according to Einstein's
formula, an energy transfer in the amount of E corresponds to
a mass transfer in the amount of m = Elc 1 . What do you think?
14-5C What is the driving force for (a) heat transfer,
(b) electric current flow, (c) fluid flow, and (d) mass transfer?
14-6C What do (a) homogeneous reactions and (b) hetero-
geneous reactions represent in mass transfer? To what do they
correspond in heat transfer?
Mass Diffusion
14-7C Both Fourier's law of heat conduction and Fick's law
of mass diffusion can be expressed as Q = —kA{dTldx). What
do the quantities Q, k, A, and T represent in (a) heat conduction
and (b) mass diffusion?
14-8C Mark these statements as being True or False for a
binary mixture of substances A and B.
(a) The density of a mixture is always equal to the sum
of the densities of its constituents.
(b) The ratio of the density of component A to the den-
sity of component B is equal to the mass fraction of
component A.
(c) If the mass fraction of component A is greater than
0.5, then at least half of the moles of the mixture are
component A.
(d) If the molar masses of A and B are equal to each
other, then the mass fraction of A will be equal to the
mole fraction of A.
(e) If the mass fractions of A and B are both 0.5, then the
molar mass of the mixture is simply the arithmetic
average of the molar masses of A and B.
*Problems designated by a "C" are concept questions, and
students are encouraged to answer them all. Problems designated
by an "E" are in English units, and the SI users can ignore them.
Problems with an EES-CD icon ® are solved using EES, and
complete solutions together with parametric studies are included
on the enclosed CD. Problems with a computer-EES icon H are
comprehensive in nature, and are intended to be solved with a
computer, preferably using the EES software that accompanies
this text.
14-9C Mark these statements as being True or False for a
binary mixture of substances A and B.
(a) The molar concentration of a mixture is always
equal to the sum of the molar concentrations of its
constituents.
(b) The ratio of the molar concentration of A to the molar
concentration of B is equal to the mole fraction of
component A.
(c) If the mole fraction of component A is greater than
0.5, then at least half of the mass of the mixture is
component A.
(cl) If both A and B are ideal gases, then the pressure frac-
tion of A is equal to its mole fraction.
(e) If the mole fractions of A and B are both 0.5, then the
molar mass of the mixture is simply the arithmetic
average of the molar masses of A and B.
14-10C Fick's law of diffusion is expressed on the mass
and mole basis as rh im A = —pAD AB (dw A /dx) and N im A =
— CAD AB (dy A /dx), respectively. Are the diffusion coefficients
D AB in the two relations the same or different?
14-11C How does the mass diffusivity of a gas mixture
change with (a) temperature and (b) pressure?
14-12C At a given temperature and pressure, do you think
the mass diffusivity of air in water vapor will be equal to the
mass diffusivity of water vapor in air? Explain.
14-13C At a given temperature and pressure, do you think
the mass diffusivity of copper in aluminum will be equal to the
mass diffusivity of aluminum in copper? Explain.
14-14C In a mass production facility, steel components are
to be hardened by carbon diffusion. Would you carry out the
hardening process at room temperature or in a furnace at a high
temperature, say 900°C? Why?
14-15C Someone claims that the mass and the mole frac-
tions for a mixture of C0 2 and N 2 gases are identical. Do you
agree? Explain.
14-16 The composition of moist air is given on a molar basis
to be 78 percent N 2 , 20 percent 2 , and 2 percent water vapor.
Determine the mass fractions of the constituents of air.
Answers: 76.4 percent N 2 , 22.4 percent 2 , 1.2 percent H 2
14-1 7E A gas mixture consists of 5 lbm of 2 , 8 lbm of N 2 ,
and 10 lbm of C0 2 . Determine (a) the mass fraction of each
component, (b) the mole fraction of each component, and
(c) the average molar mass of the mixture.
14-18 A gas mixture consists of 8 kmol of H 2 and 2 kmol of
N 2 . Determine the mass of each gas and the apparent gas con-
stant of the mixture.
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14-19 The molar analysis of a gas mixture at 290 K and
250 kPa is 65 percent N 2 , 20 percent 2 , and 15 percent C0 2 .
Determine the mass fraction and partial pressure of each gas.
14-20 Determine the binary diffusion coefficient of C0 2
in air at (a) 200 K and 1 atm, (b) 400 K and 0.8 atm, and
(c) 600 K and 3 atm.
1 4-21 Repeat Problem 1 4-20 for 2 in N 2 .
14-22E The relative humidity of air at 80°F and 14.7 psia is
increased from 30 percent to 90 percent during a humidifica-
tion process at constant temperature and pressure. Determine
the percent error involved in assuming the density of air to
have remained constant. Answer: 2.1 percent
FIGURE P14-22E
14-23 The diffusion coefficient of hydrogen in steel is given
as a function of temperature as
D AB = 1.65 X 10- 6 expHt630/r) (m 2 /s)
where T is in K. Determine the diffusion coefficients from
200 K to 1200 K in 200 K increments and plot the results.
14-24 rra| Reconsider Problem 14-23. Using EES (or
k^S other) software, plot the diffusion coefficient as
a function of the temperature in the range of 200 K to 1200 K.
Boundary Conditions
14-25C Write three boundary conditions for mass transfer
(on a mass basis) for species A at x = that correspond to
specified temperature, specified heat flux, and convection
boundary conditions in heat transfer.
14-26C What is an impermeable surface in mass transfer?
How is it expressed mathematically (on a mass basis)? To what
does it correspond in heat transfer?
14-27C Consider the free surface of a lake exposed to the
atmosphere. If the air at the lake surface is saturated, will the
mole fraction of water vapor in air at the lake surface be
773
CHAPTER 14
the same as the mole fraction of water in the lake (which is
nearly 1)?
14-28C When prescribing a boundary condition for mass
transfer at a solid-gas interface, why do we need to specify the
side of the surface (whether the solid or the gas side)? Why did
we not do it in heat transfer?
14-29C Using properties of saturated water, explain how
you would determine the mole fraction of water vapor at the
surface of a lake when the temperature of the lake surface and
the atmospheric pressure are specified.
14-30C Using solubility data of a solid in a specified liquid,
explain how you would determine the mass fraction of the
solid in the liquid at the interface at a specified temperature.
14-31C Using solubility data of a gas in a solid, explain how
you would determine the molar concentration of the gas in the
solid at the solid-gas interface at a specified temperature.
14-32C Using Henry"s constant data for a gas dissolved in a
liquid, explain how you would determine the mole fraction of
the gas dissolved in the liquid at the interface at a specified
temperature.
14-33C What is permeability? How is the permeability of a
gas in a solid related to the solubility of the gas in that solid?
14-34E Determine the mole fraction of the water vapor at
the surface of a lake whose temperature at the surface is 60°F,
and compare it to the mole fraction of water in the lake. Take
the atmospheric pressure at lake level to be 13.8 psia.
14-35 Determine the mole fraction of dry air at the surface of
a lake whose temperature is 15°C. Take the atmospheric pres-
sure at lake level to be 100 kPa. Answer: 98.3 percent
14—36 TS^M Reconsider Problem 14-35. Using EES (or
)SZ2 other) software, plot the mole fraction of dry air
at the surface of the lake as a function of the lake temperature
as the temperatue varies from 5°C to 25°C, and discuss the
results.
14-37 Consider a rubber plate that is in contact with nitrogen
gas at 298 K and 250 kPa. Determine the molar and mass den-
sities of nitrogen in the rubber at the interface.
Answers: 0.0039 kmol/m 3 , 0.1092 kg/m 3
Rubber
plate
Pn 2 = ?-
IGURI P 14-37
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HEAT TRANSFER
14-38 A wall made of natural rubber separates 2 and N 2
gases at 25°C and 500 kPa. Determine the molar concentra-
tions of 2 and N 2 in the wall.
14-39 Consider a glass of water in a room at 20°C and
97 kPa. If the relative humidity in the room is 100 percent and
the water and the air are in thermal and phase equilibrium, de-
termine (a) the mole fraction of the water vapor in the air and
(b) the mole fraction of air in the water.
14-40E Water is sprayed into air at 80°F and 14.3 psia, and
the falling water droplets are collected in a container on the
floor. Determine the mass and mole fractions of air dissolved
in the water.
14-41 Consider a carbonated drink in a bottle at 27°C and
130 kPa. Assuming the gas space above the liquid consists of a
saturated mixture of C0 2 and water vapor and treating the
drink as water, determine (a) the mole fraction of the water
vapor in the C0 2 gas and (b) the mass of dissolved C0 2 in a
200-ml drink. Answers: (a) 2.77 percent, {b) 0.36 g
(d) Other things being equal, doubling the mass fraction
of the diffusing species at the high concentration side
will double the rate of mass transfer.
14-44C Consider one-dimensional mass diffusion of species
A through a plane wall of thickness L. Under what conditions
will the concentration profile of species A in the wall be a
straight line?
14-45C Consider one-dimensional mass diffusion of species
A through a plane wall. Does the species A content of the wall
change during steady mass diffusion? How about during tran-
sient mass diffusion?
14-46 Helium gas is stored at 293 K in a 3-m-outer-diameter
spherical container made of 5-cm-thick Pyrex. The molar con-
centration of helium in the Pyrex is 0.00073 kmol/m 3 at the
inner surface and negligible at the outer surface. Determine
the mass flow rate of helium by diffusion through the Pyrex
container. Answer: 7.2 x 10~ 15 kg/s
5 cm
co 2
11,0
FIGURE P 14-41
Steady Mass Diffusion through a Wall
14-42C Write down the relations for steady one -dimensional
heat conduction and mass diffusion through a plane wall, and
identify the quantities in the two equations that correspond to
each other.
14-43C Consider steady one-dimensional mass diffusion
through a wall. Mark these statements as being True or False.
(a) Other things being equal, the higher the density of the
wall, the higher the rate of mass transfer.
(b) Other things being equal, doubling the thickness of
the wall will double the rate of mass transfer.
(c) Other things being equal, the higher the temperature,
the higher the rate of mass transfer.
Air
FIGURE P 14-46
14-47 A thin plastic membrane separates hydrogen from air.
The molar concentrations of hydrogen in the membrane at the
inner and outer surfaces are determined to be 0.065 and 0.003
kmol/m 3 , respectively. The binary diffusion coefficient of hy-
drogen in plastic at the operation temperature is 5.3 X 10~ 10
m 2 /s. Determine the mass flow rate of hydrogen by diffusion
through the membrane under steady conditions if the thickness
of the membrane is (a) 2 mm and (b) 0.5 mm.
14-48 The solubility of hydrogen gas in steel in terms of its
mass fraction is given as w ft = 2.09 X 10~ 4 exp(-3950/7)Pj| 5
where P Wi is the partial pressure of hydrogen in bars and T is
the temperature in K. If natural gas is transported in a 1 -cm-
thick, 3-m-internal-diameter steel pipe at 500 kPa pressure and
the mole fraction of hydrogen in the natural gas is 8 percent,
determine the highest rate of hydrogen loss through a 100-m-
long section of the pipe at steady conditions at a temperature
of 293 K if the pipe is exposed to air. Take the diffusivity of hy-
drogen in steel to be 2.9 X 10~ 13 m 2 /s.
Answer: 3.98 x 10" 14 kg/s
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14-49 figM Reconsider Problem 14-48. Using EES (or
si3 other) software, plot the highest rate of hydro-
gen loss as a function of the mole fraction of hydrogen in nat-
ural gas as the mole fraction varies from 5 to 15 percent, and
discuss the results.
14-50 Helium gas is stored at 293 K and 500 kPa in a 1 -cm-
thick, 2-m-inner-diameter spherical tank made of fused silica
(Si0 2 ). The area where the container is located is well venti-
lated. Determine (a) the mass flow rate of helium by diffusion
through the tank and (b) the pressure drop in the tank in one
week as a result of the loss of helium gas.
14-51 You probably have noticed that balloons inflated with
helium gas rise in the air the first day during a party but they
fall down the next day and act like ordinary balloons filled with
air. This is because the helium in the balloon slowly leaks out
through the wall while air leaks in by diffusion.
Consider a balloon that is made of 0. 1 -mm-thick soft rubber
and has a diameter of 15 cm when inflated. The pressure and
temperature inside the balloon are initially 110 kPa and 25°C.
The permeability of rubber to helium, oxygen, and nitrogen
at 25°C are 9.4 X lO" 13 , 7.05 X lO" 13 , and 2.6 X lO" 13
kmol/m • s ■ bar, respectively. Determine the initial rates of dif-
fusion of helium, oxygen, and nitrogen through the balloon
wall and the mass fraction of helium that escapes the balloon
during the first 5 h assuming the helium pressure inside the bal-
loon remains nearly constant. Assume air to be 21 percent oxy-
gen and 79 percent nitrogen by mole numbers and take the
room conditions to be 100 kPa and 25°C.
FIGURE P14-51
14-52 Reconsider the balloon discussed in Problem 14-51.
Assuming the volume to remain constant and disregarding
the diffusion of air into the balloon, obtain a relation for the
variation of pressure in the balloon with time. Using the results
obtained and the numerical values given in the problem,
775
CHAPTER 14
determine how long it will take for the pressure inside the
balloon to drop to 100 kPa.
14-53 Pure N 2 gas at 1 atm and 25°C is flowing through a
10-m-long, 3-cm-inner diameter pipe made of 1-mm-thick
rubber. Determine the rate at which N 2 leaks out of the pipe if
the medium surrounding the pipe is (a) a vacuum and (b) at-
mospheric air at 1 atm and 25°C with 21 percent O z and
79 percent N 2 .
Answers: (a) 4.48 x ICT 10 kmol/s, (£>) 9.4 x lO" 11 kmol/s
Vacuum
N 2
-c
N, gas
1 atm
25°C
Rubber pipe
FIGURE P1 4-53
Water Vapor Migration in Buildings
14-54C Consider a tank that contains moist air at 3 atm and
whose walls are permeable to water vapor. The surrounding air
at 1 atm pressure also contains some moisture. Is it possible for
the water vapor to flow into the tank from surroundings?
Explain.
14-55C Express the mass flow rate of water vapor through a
wall of thickness L in terms of the partial pressure of water
vapor on both sides of the wall and the permeability of the wall
to the water vapor.
14-56C How does the condensation or freezing of water
vapor in the wall affect the effectiveness of the insulation in the
wall? How does the moisture content affect the effective ther-
mal conductivity of soil?
14-57C Moisture migration in the walls, floors, and ceilings
of buildings is controlled by vapor barriers or vapor retarders.
Explain the difference between the two, and discuss which is
more suitable for use in the walls of residential buildings.
14-58C What are the adverse effects of excess moisture
on the wood and metal components of a house and the paint on
the walls?
14-59C Why are the insulations on the chilled water lines
always wrapped with vapor barrier jackets?
14-60C Explain how vapor pressure of the ambient air is
determined when the temperature, total pressure, and relative
humidity of the air are given.
14-61 The diffusion of water vapor through plaster boards
and its condensation in the wall insulation in cold weather are
of concern since they reduce the effectiveness of insulation.
Consider a house that is maintained at 20°C and 60 percent
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HEAT TRANSFER
.5 mm
Outdoors
Vapor
diffusion
FIGURE P14-61
relative humidity at a location where the atmospheric pressure
is 97 kPa. The inside of the walls is finished with 9. 5 -mm -thick
gypsum wallboard. Taking the vapor pressure at the outer side
of the wallboard to be zero, determine the maximum amount of
water vapor that will diffuse through a 3-m X 8-m section of a
wall during a 24-h period. The permeance of the 9.5-mm-thick
gypsum wallboard to water vapor is 2.86 X 10~ 9 kg/s ■ m 2 ■ Pa.
14-62 Reconsider Problem 14-61. In order to reduce the
migration of water vapor through the wall, it is proposed to use
a 0.2-mm-thick polyethylene film with a permeance of 2.3 X
10~ 12 kg/s • m 2 • Pa. Determine the amount of water vapor that
will diffuse through the wall in this case during a 24-h period.
Answer: 6.7 g
14-63 The roof of a house is 15 m X 8 m and is made of a
20-cm-thick concrete layer. The interior of the house is main-
tained at 25°C and 50 percent relative humidity and the local
atmospheric pressure is 100 kPa. Determine the amount of
water vapor that will migrate through the roof in 24 h if the av-
erage outside conditions during that period are 3°C and 30 per-
cent relative humidity. The permeability of concrete to water
vapor is 24.7 X 10~ 12 kg/s • m ■ Pa.
14-64 Tu'M Reconsider Problem 14-63. Using EES (or
k^S other) software, investigate the effects of tem-
perature and relative humidity of air inside the house on the
amount of water vapor that will migrate through the roof. Let
the temperature vary from 15°C to 30°C and the relative hu-
midity from 30 to 70 percent. Plot the amount of water vapor
that will migrate as functions of the temperature and the rela-
tive humidity, and discuss the results.
14-65 Reconsider Problem 14-63. In order to reduce the mi-
gration of water vapor, the inner surface of the wall is painted
with vapor retarder latex paint whose permeance is 26 X 10~ 12
kg/s ■ m 2 • Pa. Determine the amount of water vapor that will
diffuse through the roof in this case during a 24-h period.
14-66 A glass of milk left on top of a counter in the kitchen
at 25°C, 88 kPa, and 50 percent relative humidity is tightly
sealed by a sheet of 0.009-mm-thick aluminum foil whose
25°C
88kPa
50% RH Moisture
y \ «■ migration
FIGURE P 14-66
permeance is 2.9 X 10~ 12 kg/s ■ m 2 • Pa. The inner diameter of
the glass is 12 cm. Assuming the air in the glass to be saturated
at all times, determine how much the level of the milk in the
glass will recede in 12 h. Answer: 0.00079 mm
Transient Mass Diffusion
14-67C In transient mass diffusion analysis, can we treat the
diffusion of a solid into another solid of finite thickness (such
as the diffusion of carbon into an ordinary steel component) as
a diffusion process in a semi-infinite medium? Explain.
14-68C Define the penetration depth for mass transfer, and
explain how it can be determined at a specified time when the
diffusion coefficient is known.
14-69C When the density of a species A in a semi-infinite
medium is known at the beginning and at the surface, explain
how you would determine the concentration of the species A at
a specified location and time.
14-70 A steel part whose initial carbon content is 0.12 per-
cent by mass is to be case-hardened in a furnace at 1 1 50 K by
exposing it to a carburizing gas. The diffusion coefficient of
carbon in steel is strongly temperature dependent, and at the
furnace temperature it is given to be D AB = 7.2 X 10~ 12 m 2 /s.
1150K
. Carbon
FIGURE P1 4-70
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CHAPTER 14
Also, the mass fraction of carbon at the exposed surface of the
steel part is maintained at 0.011 by the carbon-rich environ-
ment in the furnace. If the hardening process is to continue
until the mass fraction of carbon at a depth of 0.7 mm is raised
to 0.32 percent, determine how long the part should be held in
the furnace. Answer-. 9 h
14-71 Repeat Problem 14-70 for a furnace temperature of
500 K at which the diffusion coefficient of carbon in steel is
D AB = 2.1 X 10- 20 m 2 /s.
14-72 A pond with an initial oxygen content of zero is to be
oxygenated by forming a tent over the water surface and filling
the tent with oxygen gas at 25°C and 130 kPa. Determine the
mole fraction of oxygen at a depth of 2 cm from the surface
after 12 h.
Tent
25°C
130 kPa
* I 1
Or, diffusion
FIGURE P 14-72
14-73 A long nickel bar with a diameter of 5 cm has been
stored in a hydrogen-rich environment at 358 K and 300 kPa
for a long time, and thus it contains hydrogen gas throughout
uniformly. Now the bar is taken into a well-ventilated area so
that the hydrogen concentration at the outer surface remains at
almost zero at all times. Determine how long it will take for the
hydrogen concentration at the center of the bar to drop by half.
The diffusion coefficient of hydrogen in the nickel bar at the
room temperature of 298 K can be taken to be D AB = 1 .2 X
10~ 12 m 2 /s. Answer: 3.3 years
Diffusion in a Moving Medium
14-74C Define the following terms: mass-average velocity,
diffusion velocity, stationary medium, and moving medium.
14-75C What is diffusion velocity? How does it affect the
mass-average velocity? Can the velocity of a species in a mov-
ing medium relative to a fixed reference point be zero in a
moving medium? Explain.
14-76C What is the difference between mass-average veloc-
ity and mole-average velocity during mass transfer in a moving
medium? If one of these velocities is zero, will the other also
necessarily be zero? Under what conditions will these two ve-
locities be the same for a binary mixture?
14-77C Consider one-dimensional mass transfer in a moving
medium that consists of species A and B with p = p A + p s =
constant. Mark these statements as being True or False.
-(a)
-(b)
-(c)
The rates of mass diffusion of species A and B are
equal in magnitude and opposite in direction.
D,
D„
-(d)
During equimolar counterdiffusion through a tube,
equal numbers of moles of A and B move in opposite
directions, and thus a velocity measurement device
placed in the tube will read zero.
The lid of a tank containing propane gas (which is
heavier than air) is left open. If the surrounding air
and the propane in the tank are at the same tempera-
ture and pressure, no propane will escape the tank
and no air will enter.
14-78C What is Stefan flow? Write the expression for
Stefan's law and indicate what each variable represents.
14-79E The pressure in a pipeline that transports helium gas
at a rate of 5 lbm/s is maintained at 14.5 psia by venting helium
to the atmosphere through a i-in. internal diameter tube that
extends 30 ft into the air. Assuming both the helium and the
atmospheric air to be at 80°F, determine (a) the mass flow rate
of helium lost to the atmosphere through an individual tube,
(b) the mass flow rate of air that infiltrates into the pipeline,
and (c) the flow velocity at the bottom of the tube where it is
attached to the pipeline that will be measured by an anemome-
ter in steady operation.
Air
80°F
He
V
Aii-
He
0.25 in.
30 ft
Helium
Air
lm
14.5 psia
80°F
5 lbm/s
FIGURE P14-79E
14-80E Repeat Problem 14-79E for a pipeline that trans-
ports carbon dioxide instead of helium.
14-81 A tank with a 2-cm thick shell contains hydrogen gas
at the atmospheric conditions of 25°C and 90 kPa. The charg-
ing valve of the tank has an internal diameter of 3 cm and ex-
tends 8 cm above the tank. If the lid of the tank is left open so
that hydrogen and air can undergo equimolar counterdiffusion
through the 10-cm-long passageway, determine the mass flow
rate of hydrogen lost to the atmosphere through the valve at the
initial stages of the process. Answer: 4.20 x 10~ 8 kg/s
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HEAT TRANSFER
14-82 KO| Reconsider Problem 14-81. Using EES (or
1^2 other) software, plot the mass flow rate of
hydrogen lost as a function of the diameter of the charging
valve as the diameter varies from 1 cm to 5 cm, and discuss the
results.
14-83E A 1 -in. -diameter Stefan tube is used to measure the
binary diffusion coefficient of water vapor in air at 70°F and
13.8 psia. The tube is partially filled with water with a distance
from the water surface to the open end of the tube of 10 in. Dry
air is blown over the open end of the tube so that water vapor
rising to the top is removed immediately and the concentration
of vapor at the top of the tube is zero. During 10 days of con-
tinuous operation at constant pressure and temperature, the
amount of water that has evaporated is measured to be 0.0015
lbm. Determine the diffusion coefficient of water vapor in air
at 70°F and 13.8 psia.
14-84 An 8-cm-internal-diameter, 30-cm-high pitcher half
filled with water is left in a dry room at 15°C and 87 kPa with
its top open. If the water is maintained at 15°C at all times also,
determine how long it will take for the water to evaporate
completely. Answer: 1125 days
FIGURE P 14-84
14-85 A large tank containing ammonia at 1 arm and 25°C is
vented to the atmosphere through a 3-m-long tube whose inter-
nal diameter is 1 cm. Determine the rate of loss of ammonia
and the rate of infiltration of air into the tank.
Mass Convection
14-86C Heat convection is expressed by Newton's law of
cooling as Q = hA(T s — TJ). Express mass convection in an
analogous manner on a mass basis, and identify all the quanti-
ties in the expression and state their units.
14-87C What is a concentration boundary layer? How is it
defined for flow over a plate?
14-88C What is the physical significance of the Schmidt
number? How is it defined? To what dimensionless number
does it correspond in heat transfer? What does a Schmidt num-
ber of 1 indicate?
14-89C What is the physical significance of the Sherwood
number? How is it defined? To what dimensionless number
does it correspond in heat transfer? What does a Sherwood
number of 1 indicate for a plain fluid layer?
14-90C What is the physical significance of the Lewis
number? How is it defined? What does a Lewis number of 1
indicate?
14-91C In natural convection mass transfer, the Grashof
number is evaluated using density difference instead of tem-
perature difference. Can the Grashof number evaluated this
way be used in heat transfer calculations also?
14-92C Using the analogy between heat and mass transfer,
explain how the mass transfer coefficient can be determined
from the relations for the heat transfer coefficient.
14-93C It is well known that warm air in a cooler environ-
ment rises. Now consider a warm mixture of air and gasoline
(C g H Ig ) on top of an open gasoline can. Do you think this gas
mixture will rise in a cooler environment?
14-94C Consider two identical cups of coffee, one with no
sugar and the other with plenty of sugar at the bottom. Initially,
both cups are at the same temperature. If left unattended, which
cup of coffee will cool faster?
14-95C Under what conditions will the normalized velocity,
thermal, and concentration boundary layers coincide during
flow over a flat plate?
14-96C What is the relation ( //2) Re = Nu = Sh known as?
Under what conditions is it valid? What is the practical im-
portance of it?
14-97C What is the name of the relation //2 = St Pr 2/3 =
St mass Sc 2/3 and what are the names of the variables in it? Under
what conditions is it valid? What is the importance of it in
engineering?
14-98C What is the relation /! heat = pC p h mnss known as? For
what kind of mixtures is it valid? What is the practical im-
portance of it?
14-99C What is the low mass flux approximation in mass
transfer analysis? Can the evaporation of water from a lake be
treated as a low mass flux process?
14-100E Consider a circular pipe of inner diameter D =
0.5 in. whose inner surface is covered with a thin layer of
liquid water as a result of condensation. In order to dry the
pipe, air at 540 R and 1 atm is forced to flow through it with an
average velocity of 4 ft/s. Using the analogy between heat and
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mass transfer, determine the mass transfer coefficient inside the
pipe for fully developed flow. Answer-. 0.024 ft/s
14-101 The average heat transfer coefficient for air flow
over an odd-shaped body is to be determined by mass transfer
measurements and using the Chilton-Colburn analogy between
heat and mass transfer. The experiment is conducted by blow-
ing dry air at 1 atm at a free stream velocity of 2 m/s over a
body covered with a layer of naphthalene. The surface area of
the body is 0.75 m 2 , and it is observed that 100 g of naph-
thalene has sublimated in 45 min. During the experiment, both
the body and the air were kept at 25°C, at which the vapor
pressure and mass diffusivity of naphthalene are 1 1 Pa and
D AB = 0.61 X 10~ 5 m 2 /s, respectively. Determine the heat
transfer coefficient under the same flow conditions over the
same geometry.
Air
0.75 m-
1 atm
2 m/s
25°C
FIGURE P14-101
Naphthalene
vapor
14-102 Consider a 15-cm-internal-diameter, 10-m-long cir-
cular duct whose interior surface is wet. The duct is to be dried
by forcing dry air at 1 atm and 15°C through it at an average
velocity of 3 m/s. The duct passes through a chilled room, and
it remains at an average temperature of 15°C at all times. De-
termine the mass transfer coefficient in the duct.
14-103 Ka| Reconsider Problem 14-102. Using EES (or
|^i£ other) software, plot the mass transfer coeffi-
cient as a function of the air velocity as the velocity varies from
1 m/s to 8 m/s, and discuss the results.
14-104 Dry air at 15°C and 92 kPa flows over a 2-m-long
wet surface with a free stream velocity of 4 m/s. Determine the
average mass transfer coefficient. Answer: 0.00514 m/s
Dry air
15°C, 92kPa
4 m/s
Evaporation
Wet
i
779
CHAPTER 14
14-105 Consider a 5-m X 5-m wet concrete patio with an
average water film thickness of 0.3 mm. Now wind at 50 km/h
is blowing over the surface. If the air is at 1 atm, 15°C, and
35 percent relative humidity, determine how long it will take
for the patio to dry completely. Answer: 18.6 min
14-106E A 2 -in. -diameter spherical naphthalene ball is sus-
pended in a room at 1 atm and 80°F. Determine the average
mass transfer coefficient between the naphthalene and the air
if air is forced to flow over naphthalene with a free stream ve-
locity of 15 ft/s. The Schmidt number of naphthalene in air at
room temperature is 2.35. Answer: 0.0525 ft/s
14-107 Consider a 3-mm-diameter raindrop that is falling
freely in atmospheric air at 25°C. Taking the temperature of the
raindrop to be 9°C, determine the terminal velocity of the rain-
drop at which the drag force equals the weight of the drop and
the average mass transfer coefficient at that time.
14-108 In a manufacturing facility, wet brass plates coming
out of a water bath are to be dried by passing them through a
section where dry air at 1 atm and 25°C is blown parallel to
their surfaces. If the plates are at 20°C and there are no dry
spots, determine the rate of evaporation from both sides of
a plate.
40 cm
FIGURE P1 4-1 04
Brass plate
20°C
FIGURE P14-108
14-109E Air at 80°F, 1 atm, and 30 percent relative humid-
ity is blown over the surface of a 15 -in. X 15 -in. square pan
filled with water at a free stream velocity of 10 ft/s. If the water
is maintained at a uniform temperature of 80°F, determine the
rate of evaporation of water and the amount of heat that needs
to be supplied to the water to maintain its temperature constant.
14-110E Repeat Problem 14-109E for temperature of 60°F
for both the air and water.
Simultaneous Heat and Mass Transfer
14-111C Does a mass transfer process have to involve heat
transfer? Describe a process that involves both heat and mass
transfer.
14-112C Consider a shallow body of water. Is it possible for
this water to freeze during a cold and dry night even when the
ambient air and surrounding surface temperatures never drop
to 0°C? Explain.
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HEAT TRANSFER
14-113C During evaporation from a water body to air, under
what conditions will the latent heat of vaporization be equal to
convection heat transfer from the air?
14-114 Jugs made of porous clay were commonly used to
cool water in the past. A small amount of water that leaks out
keeps the outer surface of the jug wet at all times, and hot and
relatively dry air flowing over the jug causes this water to
evaporate. Part of the latent heat of evaporation comes from the
water in the jug, and the water is cooled as a result. If the envi-
ronment conditions are 1 atm, 30°C, and 35 percent relative
humidity, determine the temperature of the water when steady
conditions are reached.
Hot, dry air
30°C
35% RH
FIGURE P14-1 14
Water that
leaks out
14-115 P?ja| Reconsider Problem 14-114. Using EES (or
1^13 other) software, plot the water temperature as a
function of the relative humidity of air as the relative humidity
varies from 10 to 100 percent, and discuss the results.
14-116E During a hot summer day, a 2-L bottle drink is to be
cooled by wrapping it in a cloth kept wet continually and blow-
ing air to it with a fan. If the environment conditions are 1 atm,
80°F, and 30 percent relative humidity, determine the tempera-
ture of the drink when steady conditions are reached.
14-117 (~fe\ A glass bottle washing facility uses a well-
xHJy agitated hot water bath at 55°C with an open
top that is placed on the ground. The bathtub is 1 m high, 2 m
wide, and 4 m long and is made of sheet metal so that the outer
side surfaces are also at about 55°C. The bottles enter at a rate
of 800 per minute at ambient temperature and leave at the wa-
ter temperature. Each bottle has a mass of 150 g and removes
0.6 g of water as it leaves the bath wet. Makeup water is sup-
plied at 15°C. If the average conditions in the plant are 1 atm,
25°C, and 50 percent relative humidity, and the average tem-
perature of the surrounding surfaces is 15°C, determine (a) the
amount of heat and water removed by the bottles themselves
per second; (b) the rate of heat loss from the top surface of the
water bath by radiation, natural convection, and evaporation;
(c) the rate of heat loss from the side surfaces by natural con-
vection and radiation; and (d) the rate at which heat and water
must be supplied to maintain steady operating conditions. Dis-
regard heat loss through the bottom surface of the bath and take
the emissivities of sheet metal and water to be 0.61 and 0.95,
respectively.
Answers: (a) 61,337 W, (b) 1480 W, (c) 3773 W, (of) 79,960 W,
44.9 kg/h
14-118
of50°C.
Repeat Problem 14-117 for a water bath temperature
14-119 One way of increasing heat transfer from the head on
a hot summer day is to wet it. This is especially effective in
windy weather, as you may have noticed. Approximating the
head as a 30-cm-diameter sphere at 30°C with an emissivity of
0.95, determine the total rate of heat loss from the head at am-
bient air conditions of 1 atm, 25°C, 40 percent relative humid-
ity, and 25 km/h winds if the head is (a) dry and (b) wet. Take
the surrounding temperature to be 25°C.
Answers: (a) 40.6 W, (b) 352 W
Evaporation
1 atm
25°C
40% RH
25 km/h
FIGURE P14-1 19
14-120 A 2-m-deep 20-m X 20-m heated swimming pool is
maintained at a constant temperature of 30°C at a location
where the atmospheric pressure is 1 atm. If the ambient air is at
20 C C and 60 percent relative humidity and the effective sky
temperature is 0°C, determine the rate of heat loss from the top
surface of the pool by (a) radiation, (b) natural convection, and
(c) evaporation, (d) Assuming the heat losses to the ground to
be negligible, determine the size of the heater.
14-121 Repeat Problem 14-120 for a pool temperature
of25°C.
Review Problems
14-122C Mark these statements as being True or False.
(a) The units of mass diffusivity, heat diffusivity, and
momentum diffusivity are all the same.
(b) If the molar concentration (or molar density) C of a
mixture is constant, then its density p must also be
constant.
(c) If the mass-average velocity of a binary mixture is
zero, then the mole-average velocity of the mixture
must also be zero.
(d) If the mole fractions of A and B of a mixture are both
0.5, then the molar mass of the mixture is simply the
arithmetic average of the molar masses of A and B.
14-123 Using Henry's law, show that the dissolved gases in
a liquid can be driven off by heating the liquid.
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CHAPTER 14
14-124 Show that for an ideal gas mixture maintained at a
constant temperature and pressure, the molar concentration C
of the mixture remains constant but this is not necessarily the
case for the density p of the mixture.
14-125E A gas mixture in a tank at 600 R and 20 psia con-
sists of 1 lbm of C0 2 and 3 lbm of CH 4 . Determine the volume
of the tank and the partial pressure of each gas.
14-126 Dry air whose molar analysis is 78.1 percent N 2 , 20.9
percent 2 , and 1 percent Ar flows over a water body until it is
saturated. If the pressure and temperature of air remain con-
stant at 1 atm and 25°C during the process, determine (a) the
molar analysis of the saturated air and (b) the density of air
before and after the process. What do you conclude from your
results?
14-127 Consider a glass of water in a room at 25°C and
100 kPa. If the relative humidity in the room is 70 percent and
the water and the air are at the same temperature, determine
(a) the mole fraction of the water vapor in the room air, (b) the
mole fraction of the water vapor in the air adjacent to the water
surface, and (c) the mole fraction of air in the water near the
surface.
Answers: (a) 2.22 percent, (b) 3.17 percent, (c) 1.34 x lO" 5
percent
FIGURE P14-1 27
14-128 The diffusion coefficient of carbon in steel is
given as
D AB = 2.67 X 10" 5 exp(-l 7,400/7) m 2 /s
where T is in K. Determine the diffusion coefficient from
300 K to 1500 K in 100 K increments and plot the results.
14-129 A carbonated drink is fully charged with C0 2 gas at
17°C and 600 kPa such that the entire bulk of the drink is in
thermodynamic equilibrium with the C0 2 -water vapor mix-
ture. Now consider a 2-L soda bottle. If the C0 2 gas in that bot-
tle were to be released and stored in a container at 25 °C and
1 00 kPa, determine the volume of the container.
Answer: 12.7 L
FIGURE P14-1 29
14-130 Consider a brick house that is maintained at 20°C
and 60 percent relative humidity at a location where the atmos-
pheric pressure is 85 kPa. The walls of the house are made of
20-cm thick brick whose permeance is 23 X 10~ 9 kg/s ■ m 2 ■
Pa. Taking the vapor pressure at the outer side of the wallboard
to be zero, determine the maximum amount of water vapor that
will diffuse through a 4-m X 7-m section of a wall during a
24-h period.
14-131E Consider a masonry cavity wall that is built around
6-in. -thick concrete blocks. The outside is finished with 4-in.
face brick with y-in. cement mortar between the bricks and
concrete blocks. The inside finish consists of ^-in. gypsum
wallboard separated from the concrete block by |-in. -thick
air space. The thermal and vapor resistances of various compo-
nents for a unit wall area are as follows:
FIGURE P1 4-1 31 E
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HEAT TRANSFER
ff-Value,
fl„-Value,
Construction
h • ft 2 • °F/Btu
s ■ ft 2 ■ psi/lbm
1. Outside surface,
15 mph wind
0.17
—
2. Face brick, 4 in.
0.43
15,000
3. Cement mortar,
0.5 in.
0.10
1930
4. Concrete block,
6-in.
5. Air space,
|-in.
4.20
1.02
23,000
77.6
6. Gypsum wallboard,
0.5 in.
0.45
332
7. Inside surface,
still air
0.68
—
The indoor conditions are 70°F and 65 percent relative humid-
ity while the outdoor conditions are 32°F and 40 percent rela-
tive humidity. Determine the rates of heat and water vapor
transfer through a 9 -ft X 25-ft section of the wall.
Answers: 1436 Btu/h, 4.03 Ibm/h
14-132 The oxygen needs of fish in aquariums are usually
met by forcing air to the bottom of the aquarium by a compres-
sor. The air bubbles provide a large contact area between the
water and the air, and as the bubbles rise, oxygen and nitrogen
gases in the air dissolve in water while some water evaporates
into the bubbles. Consider an aquarium that is maintained at
room temperature of 25°C at all times. The air bubbles are ob-
served to rise to the free surface of water in 2 s. If the air enter-
ing the aquarium is completely dry and the diameter of the air
bubbles is 4 mm, determine the mole fraction of water vapor at
the center of the bubble when it leaves the aquarium. Assume
no fluid motion in the bubble so that water vapor propagates in
the bubble by diffusion only.
Answer: 3.13 percent
1 atm
25°C
- Air bubbles
Aquarium
FIGURE P1 4-1 32
14-133 Oxygen gas is forced into an aquarium at 1 atm and
25°C, and the oxygen bubbles are observed to rise to the free
surface in 2 s. Determine the penetration depth of oxygen into
water from a bubble during this time period.
14-134 Consider a 30-cm-diameter pan filled with water at
15°C in a room at 20°C, 1 atm, and 30 percent relative humid-
ity. Determine (a) the rate of heat transfer by convection, (b)
the rate of evaporation of water, and (c) the rate of heat trans-
fer to the water needed to maintain its temperature at 15°C.
Disregard any radiation effects.
14-135 Repeat Problem 14-134 assuming a fan blows air
over the water surface at a velocity of 3 m/s. Take the radius of
the pan to be the characteristic length.
14-136 Naphthalene is commonly used as a repellent against
moths to protect clothing during storage. Consider a 1 -cm-
diameter spherical naphthalene ball hanging in a closet at 25 °C
and 1 atm. Considering the variation of diameter with time, de-
termine how long it will take for the naphthalene to sublimate
completely. The density and vapor pressure of naphthalene at
25°C are 0.11 Pa and 1100 kg/m 3 and 11 Pa, respectively, and
the mass diffusivity of naphthalene in air at 25°C is D AB = 0.61
X 10~ 5 m 2 /s.
Answer: 45.7 days
Closet
25°C
Sublimation
1 atm
Naphthalene \
\ 25°C /
FIGURE P1 4-1 36
14-137E A swimmer extends his wet arms into the windy air
outside at 1 atm, 40°F, 50 percent relative humidity, and 20
mph. If the average skin temperature is 80°F, determine the
rate at which water evaporates from both arms and the corre-
sponding rate of heat transfer by evaporation. The arm can be
modeled as a 2-ft-long and 3-in. -diameter cylinder with adia-
batic ends.
14-138 A thick part made of nickel is put into a room filled
with hydrogen at 3 atm and 85 °C. Determine the hydrogen
concentration at a depth of 2 -mm from the surface after 24 h.
Answer: 4.1 x 10~ 7 kmol/m 3
14-139 A membrane made of 0.1-mm-thick soft rubber sep-
arates pure 2 at 1 atm and 25°C from air at 1.2 atm pressure.
Determine the mass flow rate of 2 through the membrane per
unit area and the direction of flow.
14-140E The top section of an 8-ft-deep 100-ft X 100-ft
heated solar pond is maintained at a constant temperature of
80°F at a location where the atmospheric pressure is 1 atm. If
the ambient air is at 70°F and 100 percent relative humidity
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and wind is blowing at an average velocity of 40 mph, deter-
mine the rate of heat loss from the top surface of the pond by
(a) forced convection, (b) radiation, and (c) evaporation. Take
the average temperature of the surrounding surfaces to be 60°F.
14-141E Repeat Problem 14-140E for a solar pond surface
temperature of 90°F.
Answers: (a) 299,400 Btu/h, (b) 1,057,000 Btu/h, (c) 3,396,000
Btu/h
Computer, Design, and Essay Problems
14-142 Write an essay on diffusion caused by effects other
than the concentration gradient such as thermal diffusion, pres-
sure diffusion, forced diffusion, knodsen diffusion, and surface
diffusion.
14-143 Write a computer program that will convert the mole
fractions of a gas mixture to mass fractions when the molar
masses of the components of the mixture are specified.
14-144 One way of generating electricity from solar energy
involves the collection and storage of solar energy in large ar-
tificial lakes of a few meters deep, called solar ponds. Solar en-
ergy is stored at the bottom part of the pond at temperatures
close to boiling, and the rise of hot water to the top is prevented
by planting salt to the bottom of the pond. Write an essay on
the operation of solar pond power plants, and find out how
much salt is used per year per m 2 . If the cost is not a factor, can
sugar be used instead of salt to maintain the concentration gra-
dient? Explain.
14-145 The condensation and even freezing of moisture in
building walls without effective vapor retarders is a real con-
783
CHAPTER 14
cern in cold climates as it undermines the effectiveness of the
insulation. Investigate how the builders in your area are coping
with this problem, whether they are using vapor retarders or
vapor barriers in the walls, and where they are located in the
walls. Prepare a report on your findings and explain the rea-
soning for the current practice.
14-146 You are asked to design a heating system for a swim-
ming pool that is 2 m deep, 25 m long, and 25 m wide. Your
client desires that the heating system be large enough to raise
the water temperature from 20°C to 30°C in 3 h. The heater
must also be able to maintain the pool at 30°C at the outdoor
design conditions of 15°C, 1 atm, 35 percent relative humidity,
40 mph winds, and effective sky temperature of 10°C. Heat
losses to the ground are expected to be small and can be disre-
garded. The heater considered is a natural gas furnace whose
efficiency is 80 percent. What heater size (in Btu/h input)
would you recommend that your client buy?
15°C
1 atm
Heat
Evaporation
35% RH
loss
\
t
t t
tit
> m
30°C
Heating
fluid
Pool
— »-|_
FIGURE P1 4-1 46
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COOLING OF ELECTRONIC
EOUIPMENT
CHAPTER
Electronic equipment has made its way into practically every aspect of
modern life, from toys and appliances to high-power computers. The re-
liability of the electronics of a system is a major factor in the overall re-
liability of the system. Electronic components depend on the passage of
electric current to perform their duties, and they become potential sites for ex-
cessive heating, since the current flow through a resistance is accompanied by
heat generation. Continued miniaturization of electronic systems has resulted
in a dramatic increase in the amount of heat generated per unit volume, com-
parable in magnitude to those encountered at nuclear reactors and the surface
of the sun. Unless properly designed and controlled, high rates of heat gener-
ation result in high operating temperatures for electronic equipment, which
jeopardizes its safety and reliability.
The failure rate of electronic equipment increases exponentially with
temperature. Also, the high thermal stresses in the solder joints of electronic
components mounted on circuit boards resulting from temperature variations
are major causes of failure. Therefore, thermal control has become increas-
ingly important in the design and operation of electronic equipment.
In this chapter, we discuss several cooling techniques commonly used
in electronic equipment such as conduction cooling, natural convection and
radiation cooling, forced-air cooling, liquid cooling, and immersion cooling.
This chapter is intended to familiarize the reader with these techniques and put
them into perspective. The reader interested in an in-depth coverage of any of
these topics can consult numerous other sources available, such as those listed
in the references.
CONTENTS
15-1 Introduction and
History 786
15-2 Manufacturing of Electronic
Equipment 787
15-3 Cooling Load of Electronic
Equipment 793
15-4 Thermal Environment 794
15-5 Electronics Cooling in Different
Applications 795
15-6 Conduction Cooling 797
15-7 Air Cooling: Natural Convection
and Radiation 812
15-8 Air Cooling: Forced
Convection 820
15-9 Liquid Cooling 833
15-10 Immersion Cooling 836
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786
HEAT TRANSFER
FIGURE 15-1
The increase in the number of
components packed on a chip over
the years.
15-1 - INTRODUCTION AND HISTORY
The field of electronics deals with the construction and utilization of devices
that involve current flow through a vacuum, a gas, or a semiconductor. This
exciting field of science and engineering dates back to 1883, when Thomas
Edison invented the vacuum diode. The vacuum tube served as the foun-
dation of the electronics industry until the 1950s, and played a central role in
the development of radio, TV, radar, and the digital computer. Of the several
computers developed in this era, the largest and best known is the ENIAC
(Electronic Numerical Integrator and Computer), which was built at the Uni-
versity of Pennsylvania in 1946. It had over 18,000 vacuum tubes and occu-
pied a room 7 m X 14 m in size. It consumed a large amount of power, and its
reliability was poor because of the high failure rate of the vacuum tubes.
The invention of the bipolar transistor in 1948 marked the beginning of a
new era in the electronics industry and the obsolescence of vacuum tube tech-
nology. Transistor circuits performed the functions of the vacuum tubes with
greater reliability, while occupying negligible space and consuming negligible
power compared with vacuum tubes. The first transistors were made from ger-
manium, which could not function properly at temperatures above 100°C.
Soon they were replaced by silicon transistors, which could operate at much
higher temperatures.
The next turning point in electronics occurred in 1959 with the introduction
of the integrated circuits (IC), where several components such as diodes,
transistors, resistors, and capacitors are placed in a single chip. The number of
components packed in a single chip has been increasing steadily since then at
an amazing rate, as shown in Figure 15—1. The continued miniaturization of
electronic components has resulted in medium- scale integration (MSI) in the
1960s with 50-1000 components per chip, large-scale integration (LSI) in
the 1970s with 1000-100,000 components per chip, and very large-scale inte-
gration (VLSI) in the 1980s with 100,000-10,000,000 components per chip.
Today it is not unusual to have a chip 3 cm X 3 cm in size with several mil-
lion components on it.
The development of the microprocessor in the early 1970s by the Intel
Corporation marked yet another beginning in the electronics industry. The ac-
companying rapid development of large-capacity memory chips in this decade
made it possible to introduce capable personal computers for use at work or at
home at an affordable price. Electronics has made its way into practically
everything from watches to household appliances to automobiles. Today it is
difficult to imagine a new product that does not involve any electronic parts.
The current flow through a resistance is always accompanied by heat
generation in the amount of I 2 R, where / is the electric current and R is the
resistance. When the transistor was first introduced, it was touted in the news-
papers as a device that "produces no heat." This certainly was a fair statement,
considering the huge amount of heat generated by vacuum tubes. Obviously,
the little heat generated in the transistor was no match to that generated in its
predecessor. But when thousands or even millions of such components are
packed in a small volume, the heat generated increases to such high levels that
its removal becomes a formidable task and a major concern for the safety and
reliability of the electronic devices. The heat fluxes encountered in electronic
devices range from less than 1 W/cm 2 to more than 100 W/cm 2 .
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CHAPTER 15
Heat is generated in a resistive element for as long as current continues to
flow through it. This creates a heat build-up and a subsequent temperature
rise at and around the component. The temperature of the component will
continue rising until the component is destroyed unless heat is transferred
away from it. The temperature of the component will remain constant when
the rate of heat removal from it equals the rate of heat generation.
Individual electronic components have no moving parts, and thus nothing to
wear out with time. Therefore, they are inherently reliable, and it seems as if
they can operate safely for many years. Indeed, this would be the case if com-
ponents operated at room temperature. But electronic components are ob-
served to fail under prolonged use at high temperatures. Possible causes of
failure are diffusion in semiconductor materials, chemical reactions, and creep
in the bonding materials, among other things. The failure rate of electronic de-
vices increases almost exponentially with the operating temperature, as shown
in Figure 15-2. The cooler the electronic device operates, the more reliable it
is. A rule of thumb is that the failure rate of electronic components is halved
for each 10°C reduction in their junction temperature.
15-2 - MANUFACTURING OF ELECTRONIC
EQUIPMENT
The narrow band where two different regions of a semiconductor (such as the
p-type and n-type regions) come in contact is called a junction. A transistor,
for example, involves two such junctions, and a diode, which is the simplest
semiconductor device, is based on a single p-n junction. In heat transfer analy-
sis, the circuitry of an electronic component through which electrons flow
and thus heat is generated is also referred to as the junction. That is, junctions
are the sites of heat generation and thus the hottest spots in a component. In
silicon-based semiconductor devices, the junction temperature is limited to
125°C for safe operation. However, lower junction temperatures are desirable
for extended life and lower maintenance costs. In a typical application, nu-
merous electronic components, some smaller than 1 |xm in size, are formed
from a silicon wafer into a chip.
20 40
The
60 /80 100 120 140
75
Temperature, °C
FIGURE 15-2
increase in the failure rate of
bipolar digital devices with
temperature (from Ref. 15).
The Chip Carrier
The chip is housed in a chip carrier or substrate made of ceramic, plastic, or
glass in order to protect its delicate circuitry from the detrimental effects of
the environment. The chip carrier provides a ragged housing for the safe han-
dling of the chip during the manufacturing process, as well as the connectors
between the chip and the circuit board. The various components of the chip
carrier are shown in Figure 15-3. The chip is secured in the carrier by bond-
ing it to the bottom surface. The thermal expansion coefficient of the plastic
is about 20 times that of silicon. Therefore, bonding the silicon chip directly
to the plastic case would result in such large thermal stresses that the reliabil-
ity would be seriously jeopardized. To avoid this problem, a lead frame made
of a copper alloy with a thermal expansion coefficient close to that of silicon
is used as the bonding surface.
The design of the chip carrier is the first level in the thermal control of elec-
tronic devices, since the transfer of heat from the chip to the chip carrier is the
Lid Bond wires
Case
I Chip |
/
Leads
Lead frame Bond Pins
FIGURE 15-3
The components of a chip carrier.
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HEAT TRANSFER
FIGURE 15-4
The chip carrier for a high-power
transistor attached to a flange for
enhanced heat transfer.
first step in the dissipation of the heat generated on the chip. The heat gener-
ated on the chip is transferred to the case of the chip carrier by a combination
of conduction, convection, and radiation. However, it is obvious from the fig-
ure that the common chip carrier is designed with the electrical aspects in
mind, and little consideration is given to the thermal aspects. First of all, the
cavity of the chip carrier is filled with a gas, which is a poor heat conductor,
and the case is often made of materials that are also poor conductors of heat.
This results in a relatively large thermal resistance between the chip and the
case, called the junction-to-case resistance, and thus a large temperature dif-
ference. As a result, the temperature of the chip will be much higher than that
of the case for a specified heat dissipation rate. The junction-to-case thermal
resistance depends on the geometry and the size of the chip and the chip car-
rier as well as the material properties of the bonding and the case. It varies
considerably from one device to another and ranges from about 10°C/W to
more than 100°C/W.
Moisture in the cavity of the chip carrier is highly undesirable, since it
causes corrosion on the wiring. Therefore, chip carriers are made of materials
that prevent the entry of moisture by diffusion and are hermetically sealed in
order to prevent the direct entry of moisture through cracks. Materials that
outgas are also not permitted in the chip cavity, because such gases can also
cause corrosion. In products with strict hermeticity requirements, the more ex-
pensive ceramic cases are used instead of the plastic ones.
A common type of chip carrier for high-power transistors is shown in Fig-
ure 15-4. The transistor is formed on a small silicon chip housed in the disk-
shaped cavity, and the I/O pins come out from the bottom. The case of the
transistor carrier is usually attached directly to a flange, which provides a
large surface area for heat dissipation and reduces the junction-to-case thermal
resistance.
It is often desirable to house more than one chip in a single chip carrier. The
result is a hybrid or multichip package. Hybrid packages house several chips,
individual electronic components, and ordinary circuit elements connected to
each other in a single chip carrier. The result is improved performance due to
the shortening of the wiring lengths, and enhanced reliability. Lower cost
would be an added benefit of multichip packages if they are produced in suf-
ficiently large quantity.
Junction
\
J3
junction-case
/
v^ir- '
• — Case
III
III 1
FIGURE 15-5
Schematic for Example 15-1.
EXAMPLE 15-1 Predicting the Junction Temperature of a
Transistor
The temperature of the case of a power transistor that is dissipating 3 W is mea-
sured to be 50°C. If the junction-to-case resistance of this transistor is speci-
fied by the manufacturer to be 15°C/W, determine the temperature at the
junction of the transistor.
SOLUTION The case temperature of a power transistor and the junction-to-
case resistance are given. The junction temperature is to be determined.
Assumptions Steady operating conditions exist.
Analysis The schematic of the transistor is given in Figure 15-5. The rate of
heat transfer between the junction and the case in steady operation can be
expressed as
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CHAPTER 15
Q
-ffl,
inction-case
T -
*■ junction
T
*■ case
D
^junction
case
Then the
unci
ion
temperature becomes
T
*■ junction
= T +
* case
«- junction
case
= 50°C -+
(3W)(15°C/W)
= 95°C
Therefore,
the
terr
perature of the
transistor
junction
will be 95°C when
its
case
is at 50°C
EXAMPLE 15-2 Determining the Junction-to-Case Thermal
Resistance
This experiment is conducted to determine the junction-to-case thermal resis-
tance of an electronic component. Power is supplied to the component from a
15-V source, and the variations in the electric current and in the junction and
the case temperatures with time are observed. When things are stabilized, the
current is observed to be 0.1 A and the temperatures to be 80°C and 55°C at
the junction and the case, respectively. Calculate the junction-to-case resis-
tance of this component.
SOLUTION The power dissipated by an electronic component as well as the
junction and case temperatures are measured. The junction-to-case resistance
is to be determined.
Assumptions Steady operating conditions exist.
Analysis The schematic of the component is given in Figure 15-6. The elec-
tric power consumed by this electronic component is
W e = VI = (15 V)(0.1 A) = 1.5 W
In steady operation, this is equivalent to the heat dissipated by the component.
That is,
Q
R j.
/junc
1 junction
R„
1.5W
iction-casc ''junction-case
Then the junction-to-case resistance is determined to be
R,
1 junction
r caS e (80 - 55)°C
e
1.5 W
16.7°C/W
Discussion Note that a temperature difference of 16.7°C will occur between
the electronic circuitry and the case of the chip carrier for each W of power con-
sumed by the component.
"QI
>\ j^80°C
-55°C
ITT
w
FIGURE 15-6
Schematic for Example 15-2.
Printed Circuit Boards
A printed circuit board (PCB) is a properly wired plane board made of poly-
mers and glass-epoxy materials on which various electronic components such
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HEAT TRANSFER
as the ICs, diodes, transistors, resistors, and capacitors are mounted to perform
a certain task, as shown in Figure 15-7. The PCBs are commonly called cards,
and they can be replaced easily during a repair. The PCBs are plane boards,
usually 10 cm wide and 15 cm long and only a few millimeters thick, and they
are not suitable for heavy components such as transformers. Usually a copper
cladding is added on one or both sides of the board. The cladding on one side
is subjected to an etching process to form wiring strips and attachment pads
for the components. The power dissipated by a PCB usually ranges from 5 W
to about 30 W.
A typical electronic system involves several layers of PCBs. The PCBs are
usually cooled by direct contact with a fluid such as air flowing between the
boards. But when the boards are placed in a hermetically sealed enclosure,
they must be cooled by a cold plate (a heat exchanger) in contact with the
edge of the boards. The device-to-board edge thermal resistance of a PCB is
usually high (about 20 to 60°C/W) because of the small thickness of the board
and the low thermal conductivity of the board material. In such cases, even a
thin layer of copper cladding on one side of the board can decrease the device-
to-board edge thermal resistance in the plane of the board and enhance heat
transfer in that direction drastically.
In the thermal design of a PCB, it is important to pay particular attention to
the components that are not tolerant of high temperatures, such as certain
high-performance capacitors, and to ensure their safe operation. Often when
one component on a PCB fails, the whole board fails and must be replaced.
Printed circuit boards come in three types: single- sided, double-sided, and
multilayer boards. Each type has its own strengths and weaknesses. Single-
sided PCBs have circuitry lines on one side of the board only and are suitable
for low-density electronic devices (10 to 20 components). Double-sided PCBs
FIGURE 15-7
A printed circuit board (PCB)
with a variety of components on it
(courtesy of Litton Systems, Inc.).
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CHAPTER 15
have circuits on both sides and are best suited for intermediate-density
devices. Multilayer PCBs contain several layers of circuitry and are suitable
for high-density devices. They are equivalent to several PCBs sandwiched
together.
The single-sided PCB has the lowest cost, as expected, and it is easy to
maintain, but it occupies a lot of space. The multilayer PCB, on the other
hand, allows the placement of a large number of components in a three-
dimensional configuration, but it has the highest initial cost and is difficult to
repair. Also, temperatures are most likely to be the highest in multilayer PCBs.
In critical applications, the electronic components are placed on boards at-
tached to a conductive metal, called the heat frame, that serves as a conduc-
tion path to the edge of the circuit board and thus to the cold plate for the heat
generated in the components. Such boards are said to be conduction-cooled.
The temperature of the components in this case will depend on the location of
the components on the boards: it will be highest for the components in the
middle and lowest for those near the edge, as shown in Figure 15-8.
Materials used in the fabrication of circuit boards should be (1) effective
electrical insulators to prevent electrical breakdown and (2) good heat con-
ductors to conduct away the heat generated. They should also have (3) high
material strength to withstand forces and to maintain dimensional stability;
(4) thermal expansion coefficients that closely match that of copper, to pre-
vent cracking in the copper cladding during thermal cycling; (5) resistance to
moisture absorption, since moisture can affect both mechanical and electrical
properties and degrade performance; (6) stability in properties at temperature
levels encountered in electronic applications; (7) ready availability and manu-
facturability; and, of course, (8) low cost. As you might have already guessed,
no existing material has all of these desirable characteristics.
Glass-epoxy laminates made of an epoxy or polymide matrix reinforced by
several layers of woven glass cloth are commonly used in the production of
circuit boards. Polymide matrices are more expensive than epoxy but can
withstand much higher temperatures. Polymer or polymide films are also used
without reinforcement for flexible circuits.
The Enclosure
An electronic system is not complete without a rugged enclosure (a case or a
cabinet) that will house the circuit boards and the necessary peripheral equip-
ment and connectors, protect them from the detrimental effects of the envi-
ronment, and provide a cooling mechanism (Fig. 15-9). In a small electronic
system such as a personal computer, the enclosure can simply be an inexpen-
sive box made of sheet metal with proper connectors and a small fan. But for
a large system with several hundred PCBs, the design and construction of the
enclosure are challenges for both electronic and thermal designers. An en-
closure must provide easy access for service personnel so that they can iden-
tify and replace any defective parts easily and quickly in order to minimize
down time, which can be very costly. But, at the same time, the enclosure
must prevent any easy access by unauthorized people in order to protect the
sensitive electronics from them as well as the people from possible electrical
hazards. Electronic circuits are powered by low voltages (usually under
±15 V), but the currents involved may be very high (sometimes a few hun-
dred amperes).
Maximum
temperature
Electronic component
Circuit board
Heat frame
^- Cold plate
FIGURE 15-8
The path of heat flow in a conduction-
cooled PCB and the temperature
distribution.
FIGURE 15-9
A cabinet-style enclosure.
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HEAT TRANSFER
FIGURE 15-10
Different stages involved in the
production of an electronic system
(from Dally, Ref. 3).
Plug-in-type circuit boards make it very easy to replace a defective board
and are commonly used in low-power electronic equipment. High-power cir-
cuit boards in large systems, however, are tightly attached to the racks of the
cabinet with special brackets. A well-designed enclosure also includes
switches, indicator lights, a screen to display messages and present informa-
tion about the operation, and a key pad for user interface.
The printed circuit boards in a large system are plugged into a back panel
through their edge connectors. The back panel supplies power to the PCBs
and interconnects them to facilitate the passage of current from one board
to another. The PCBs are assembled in an orderly manner in card racks or
chassis. One or more such assemblies are housed in a cabinet, as shown in
Figure 15-10.
Electronic enclosures come in a wide variety of sizes and shapes. Sheet
metals such as thin-gauge aluminum or steel sheets are commonly used in the
production of enclosures. The thickness of the enclosure walls depends on the
shock and vibration requirements. Enclosures made of thick metal sheets or
by casting can meet these requirements, but at the expense of increased weight
and cost.
Electronic boxes are sometimes sealed to prevent the fluid inside (usually
air) from leaking out and the water vapor outside from leaking in. Sealing
against moisture migration is very difficult because of the small size of the
water molecule and the large vapor pressure outside the box relative to that
within the box. Sealing adds to the size, weight, and cost of an electronic box,
especially in space or high-altitude operation, since the box in this case must
withstand the larger forces due to the higher pressure differential between in-
side and outside the box.
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15-3 - COOLING LOAD OF ELECTRONIC EQUIPMENT
The first step in the selection and design of a cooling system is the determina-
tion of the heat dissipation, which constitutes the cooling load. The easiest
way to determine the power dissipation of electronic equipment is to measure
the voltage applied V and the electric current / at the entrance of the electronic
device under full-load conditions and to substitute them into the relation
w = vi = r-R
(W)
(15-1)
where W e is the electric power consumption of the electronic device, which
constitutes the energy input to the device.
The first law of thermodynamics requires that in steady operation the en-
ergy input into a system be equal to the energy output from the system. Con-
sidering that the only form of energy leaving the electronic device is heat
generated as the current flows through resistive elements, we conclude that
the heat dissipation or cooling load of an electronic device is equal to its
power consumption. That is, Q = W e , as shown in Figure 15—11. The excep-
tion to this rule is equipment that outputs other forms of energy as well, such
as the emitter tubes of a radar, radio, or TV installation emitting radiofre-
quency (RF) electromagnetic radiation. In such cases, the cooling load will be
equal to the difference between the power consumption and the RF power
emission. An equivalent but cumbersome way of determining the cooling load
of an electronic device is to determine the heat dissipated by each component
in the device and then to add them up.
The discovery of superconductor materials that can operate at room tempera-
ture will cause drastic changes in the design of electronic devices and cooling
techniques, since such devices will generate hardly any heat. As a result, more
components can be packed into a smaller volume, resulting in enhanced speed
and reliability without having to resort to exotic cooling techniques.
Once the cooling load has been determined, it is common practice to inflate
this number to leave some safety margin, or a "cushion," and to make some al-
lowance for future growth. It is not uncommon to add another card to an ex-
isting system (such as adding a fax/modem card to a PC) to perform an
additional task. But we should not go overboard in being conservative, since
an oversized cooling system will cost more, occupy more space, be heavier,
and consume more power. For example, there is no need to install a large and
noisy fan in an electronic system just to be "safe" when a smaller one will do.
For the same reason, there is no need to use an expensive and failure-prone
liquid cooling system when air cooling is adequate. We should always keep in
mind that the most desirable form of cooling is natural convection cooling,
since it does not require any moving parts, and thus it is inherently reliable,
quiet, and, best of all, free.
The cooling system of an electronic device must be designed considering
the actual field operating conditions. In critical applications such as those in
the military, the electronic device must undergo extensive testing to satisfy
stringent requirements for safety and reliability. Several such codes exist to
specify the minimum standards to be met in some applications.
The duty cycle is another important consideration in the design and selec-
tion of a cooling technique. The actual power dissipated by a device can be
FIGURE 15-11
In the absence of other energy
interactions, the heat output of an
electronic device in steady operation is
equal to the power input to the device.
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HEAT TRANSFER
Time, t
FIGURE 15-12
The temperature change of an
electronic component with time as it
reaches steady operating temperature
after it is turned on.
much less than the rated power, depending on its duty cycle (the fraction of
time it is on). A 5-W power transistor, for example, will dissipate an average
of 2 W of power if it is active only 40 percent of the time. If the chip of this
transistor is 1.5 mm wide, 1.5 mm high, and 0.1 mm thick, then the heat flux
on the chip will be (2 W)/(0.15 cm) 2 = 89 W/cm 2 .
An electronic device that is not running is in thermal equilibrium with its sur-
roundings, and thus is at the temperature of the surrounding medium. When the
device is turned on, the temperature of the components and thus the device starts
rising as a result of absorbing the heat generated. The temperature of the device
stabilizes at some point when the heat generated equals the heat removed by the
cooling mechanism. At this point, the device is said to have reached steady op-
erating conditions. The warming-up period during which the component tem-
perature rises is called the transient operation stage (Fig. 15-12).
Another thermal factor that undermines the reliability of electronic devices
is the thermal stresses caused by temperature cycling. In an experimental
study (see Hilbert and Kube, Ref. 10), the failure rate of electronic devices
subjected to deliberate temperature cycling of more than 20°C is observed to
increase eightfold. Shock and vibration are other common causes of failure for
electronic devices and should be considered in the design and manufacturing
process for increased reliability.
Most electronic devices operate for long periods of time, and thus their
cooling mechanism is designed for steady operation. But electronic devices in
some applications never run long enough to reach steady operation. In such
cases, it may be sufficient to use a limited cooling technique, such as thermal
storage for a short period, or not to use one at all. Transient operation can
also be caused by large swings in the environmental conditions. A common
cooling technique for transient operation is to use a double-wall construction
for the enclosure of the electronic equipment, with the space between the
walls filled with a wax with a suitable melting temperature. As the wax melts,
it absorbs a large amount of heat and thus delays overheating of the electronic
components considerably. During off periods, the wax solidifies by rejecting
heat to the environment.
Ventilation
slits
Cool
/ ,r)
FIGURE 15-13
Strategically located ventilation holes
are adequate to cool low-power
electronics such as a TV or VCR.
15^ - THERMAL ENVIRONMENT
An important consideration in the selection of a cooling technique is the envi-
ronment in which the electronic equipment is to operate. Simple ventilation
holes on the case may be all we need for the cooling of low-power-density elec-
tronics such as a TV or a VCR in a room, and a fan may be adequate for the safe
operation of a home computer (Fig. 15-13). But the thermal control of the elec-
tronics of an aircraft will challenge thermal designers, since the environmental
conditions in this case will swing from one extreme to another in a matter of
minutes. The expected duration of operation in a hostile environment is also an
important consideration in the design process. The thermal design of the elec-
tronics for an aircraft that cruises for hours each time it takes off will be quite
different than that of a missile that has an operation time of a few minutes.
The thermal environment in marine applications is relatively stable, since
the ultimate heat sink in this case is water with a temperature range of 0°C to
30°C. For ground applications, however, the ultimate heat sink is the atmos-
pheric air, whose temperature varies from — 50°C at polar regions to +50°C
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CHAPTER 15
in desert climates, and whose pressure ranges from about 70 kPa (0.7 atm) at
3000 m elevation to 107 kPa (1.08 atm) at 500 m below sea level. The com-
bined convection and radiation heat transfer coefficient can range from
10 W/m 2 • °C in calm weather to 80 W/m 2 • °C in 100 km/h (62 mph) winds.
Also, the surfaces of the devices facing the sun directly can be subjected to
solar radiation heat flux of 1000 W/m 2 on a clear day.
In airborne applications, the thermal environment can change from 1 atm
and 35°C on the ground to 19 kPa (0.2 atm) and — 60°C at a typical cruising
altitude of 12,000 m in minutes (Fig. 15-14). At supersonic velocities, the sur-
face temperature of some part of the aircraft may rise 200°C above the envi-
ronment temperature.
Electronic devices are rarely exposed to uncontrolled environmental condi-
tions directly because of the wide variations in the environmental variables.
Instead, a conditioned fluid such as air, water, or a dielectric fluid is used to
serve as a local heat sink and as an intermediary between the electronic equip-
ment and the environment, just like the air-conditioned air in a building pro-
viding thermal comfort to the human body. Conditioned air is the preferred
cooling medium, since it is benign, readily available, and not prone to leakage.
But its use is limited to equipment with low power densities, because of the
low thermal conductivity of air. The thermal design of electronic equipment in
military applications must comply with strict military standards in order to
satisfy the utmost reliability requirements.
FIGURE 15-14
The thermal environment of a
spacecraft changes drastically in a
short time, and this complicates the
thermal control of the electronics.
15-5 - ELECTRONICS COOLING IN DIFFERENT
APPLICATIONS
The cooling techniques used in the cooling of electronic equipment vary
widely, depending on the particular application. Electronic equipment
designed for airborne applications such as airplanes, satellites, space vehicles,
and missiles offers challenges to designers because it must fit into odd-shaped
spaces because of the curved shape of the bodies, yet be able to provide ade-
quate paths for the flow of fluid and heat. Most such electronic equipment are
cooled by forced convection using pressurized air bled off a compressor. This
compressed air is usually at a high temperature, and thus it is cooled first by
expanding it through a turbine. The moisture in the air is also removed before
the air is routed to the electronic boxes. But the removal process may not be
adequate under rainy conditions. Therefore, electronics in some cases are
placed in sealed finned boxes that are externally cooled to eliminate any direct
contact with electronic components.
The electronics of short-range missiles do not need any cooling because of
their short cruising times (Fig. 15-15). The missiles reach their destinations
before the electronics reach unsafe temperatures. Long-range missiles such as
cruise missiles, however, may have a flight time of several hours. Therefore,
they must utilize some form of cooling mechanism. The first thing that comes
to mind is to use forced convection with the air that rams the missile by uti-
lizing its large dynamic pressure. However, the dynamic temperature of air,
which is the rise in the temperature of the air as a result of the ramming effect,
may be more than 50°C at speeds close to the speed of sound (Fig. 15-16).
For example, at a speed of 320 m/s, the dynamic temperature of air is
FIGURE 15-15
The electronics of short-range missiles
may not need any cooling because of
the short flight time involved.
Temperature
rise during
stagnation
J 305 K
300 K
FIGURE 15-16
The temperature of a gas having a
specific heat C p flowing at a velocity
of V rises by T 2 /2C p when it is
brought to a complete stop.
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HEAT TRANSFER
Y 2 (320 m/s) 2 / 1 J/kg
^mic 2 C p 2(1005 J/kg • °C) [ 1 m 2 /s 2 J 51 ° C (15 " 2)
Therefore, the temperature of air at a velocity of 320 m/s and a temperature of
30°C will rise to 81 °C as a result of the conversion of kinetic energy to inter-
nal energy. Air at such high temperatures is not suitable for use as a cooling
medium. Instead, cruise missiles are often cooled by taking advantage of the
cooling capacity of the large quantities of liquid fuel they carry. The electron-
ics in this case are cooled by passing the fuel through the cold plate of the
electronic enclosure as it flows toward the combustion chamber.
Electronic equipment in space vehicles is usually cooled by a liquid circu-
lated through the components, where heat is picked up, and then through a
space radiator, where the waste heat is radiated into deep space at about K.
Note that radiation is the only heat transfer mechanism for rejecting heat to the
vacuum environment of space, and radiation exchange depends strongly on
surface properties. Desirable radiation properties on surfaces can be obtained
by special coatings and surface treatments. When electronics in sealed boxes
are cooled by a liquid flowing through the outer surface of the electronics box,
it is important to run a fan in the box to circulate the air, since there are no nat-
ural convection currents in space because of the absence of a gravity field.
Electronic equipment in ships and submarines is usually housed in rugged
cabinets to protect it from vibrations and shock during stormy weather. Be-
cause of easy access to water, water-cooled heat exchangers are commonly
used to cool shipboard electronics. This is usually done by cooling air in a
closed- or open-loop air-to-water heat exchanger and forcing the cool air to
the electronic cabinet by a fan. When forced-air cooling is used, it is important
to establish a flow path for air such that no trapped hot-air pockets will be
formed in the cabinets.
Communication systems located at remote locations offer challenges to ther-
mal designers because of the extreme conditions under which they operate.
These electronic systems operate for long periods of time under adverse con-
ditions such as rain, snow, high winds, solar radiation, high altitude, high
humidity, and extremely high or low temperatures. Large communication
systems are housed in specially built shelters. Sometimes it is necessary to air-
condition these shelters to safely dissipate the large quantities of heat dissi-
pated by the electronics of communication systems.
Electronic components used in high-power microwave equipment such as
radars generate enormous amounts of heat because of the low conversion
efficiency of electrical energy to microwave energy. Klystron tubes of high-
power radar systems where radio-frequency energy is generated can yield lo-
cal heat fluxes as high as 2000 W/cm 2 , which is close to one-third of the heat
flux on the sun's surface. The safe and reliable dissipation of these high heat
fluxes usually requires the immersion of such equipment in a suitable dielec-
tric fluid that can remove large quantities of heat by boiling.
The manufacturers of electronic devices usually specify the rate of heat dis-
sipation and the maximum allowable component temperature for reliable
operation. These two numbers help us determine the cooling techniques that
are suitable for the device under consideration.
The heat fluxes attainable at specified temperature differences are plotted in
Figure 15-17 for some common heat transfer mechanisms. When the power
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CHAPTER 15
0.01 0.02 0.04
0.1 0.2 0.4 1
Surface heat flux, W/cm 2
rating of a device or component is given, the heat flux is determined by divid-
ing the power rating by the exposed surface area of the device or component.
Then suitable heat transfer mechanisms can be determined from Figure 15-17
from the requirement that the temperature difference between the surface of
the device and the surrounding medium not exceed the allowable maximum
value. For example, a heat flux of 0.5 W/cm 2 for an electronic component
would result in a temperature difference of about 500°C between the com-
ponent surface and the surrounding air if natural convection in air is used.
Considering that the maximum allowable temperature difference is typically
under 80°C, the natural convection cooling of this component in air is out of
the question. But forced convection with air is a viable option if using a fan is
acceptable. Note that at heat fluxes greater than 1 W/cm 2 , even forced con-
vection with air will be inadequate, and we must use a sufficiently large heat
sink or switch to a different cooling fluid such as water. Forced convection
with water can be used effectively for cooling electronic components with
high heat fluxes. Also note that dielectric liquids such as fluorochemicals can
remove high heat fluxes by immersing the component directly in them.
FIGURE 15-17
Heat fluxes that can be attained at
specified temperature differences with
various heat transfer mechanisms and
fluids (from Kraus and Bar-Cohen,
Ref. 14, p. 22; reproduced
with permission).
15-6 - CONDUCTION COOLING
Heat is generated in electronic components whenever electric current flows
through them. The generated heat causes the temperature of the components to
rise, and the resulting temperature difference drives the heat away from the
components through a path of least thermal resistance. The temperature of the
components stabilizes when the heat dissipated equals the heat generated. In or-
der to minimize the temperature rise of the components, effective heat transfer
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HEAT TRANSFER
FIGURE 15-18
The thermal resistance of a medium
is proportional to its length in the
direction of heat transfer, and
inversely proportional to its heat
transfer surface area and
thermal conductivity.
paths must be established between the components and the ultimate heat sink,
which is usually the atmospheric air.
The selection of a cooling mechanism for electronic equipment depends on
the magnitude of the heat generated, reliability requirements, environmental
conditions, and cost. For low-cost electronic equipment, inexpensive cooling
mechanisms such as natural or forced convection with air as the cooling
medium are commonly used. For high-cost, high-performance electronic
equipment, however, it is often necessary to resort to expensive and compli-
cated cooling techniques.
Conduction cooling is based on the diffusion of heat through a solid, liquid,
or gas as a result of molecular interactions in the absence of any bulk fluid
motion. Steady one-dimensional heat conduction through a plane medium of
thickness L, heat transfer surface area A, and thermal conductivity k is given
by (Fig. 15-18).
A . .AT AT _..,
Q = kA — = — (W)
where
R
L_
kA
Length
Thermal conductivity X Heat transfer area
(15-3)
(15-4)
is the thermal resistance of the medium and AT is the temperature difference
across the medium. Note that this is analogous to the electric current being
equal to the potential difference divided by the electrical resistance.
The thermal resistance concept enables us to solve heat transfer problems in
an analogous manner to electric circuit problems using the thermal resistance
network, as discussed in Chapter 3. When the rate of heat conduction Q is
known, the temperature drop along a medium whose thermal resistance is R is
simply determined from
AT= QR
(°C)
(15-5)
Therefore, the greatest temperature drops along the path of heat conduction will
occur across portions of the heat flow path with the largest thermal resistances.
Conduction in Chip Carriers
The conduction analysis of an electronic device starts with the circuitry or
junction of a chip, which is the site of heat generation. In order to understand
the heat transfer mechanisms at the chip level, consider the DIP (dual in-line
package) type chip carrier shown in Figure 15-19.
The heat generated at the junction spreads throughout the chip and is con-
ducted across the thickness of the chip. The spread of heat from the junction
to the body of the chip is three-dimensional in nature, but can be approxi-
mated as one-dimensional by adding a constriction thermal resistance to the
thermal resistance network. For a small heat generation area of diameter d on
a considerably larger body, the constriction resistance is given by
R
1
constriction
ndk
(°C/W)
(15-6)
where k is the thermal conductivity of the larger body.
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CHAPTER 15
FIGURE 15-19
The schematic for the internal
geometry and the cross-sectional view
of a DIP (dual in-line package) type
electronic device with 14 leads.
The chip is attached to the lead frame with a highly conductive bonding ma-
terial that provides a low-resistance path for heat flow from the chip to the lead
frame. There is no metal connection between the lead frame and the leads,
since this would short-circuit the entire chip. Therefore, heat flow from the lead
frame to the leads is through the dielectric case material such as plastic or ce-
ramic. Heat is then transported outside the electronic device through the leads.
When solving a heat transfer problem, it is often necessary to make some
simplifying assumptions regarding the primary heat flow path and the magni-
tudes of heat transfer in other directions (Fig. 15-20). In the chip carrier dis-
cussed above, for example, heat transfer through the top is disregarded since
it is very small because of the large thermal resistance of the stagnant air space
between the chip and the lid. Heat transfer from the base of the electronic de-
vice is also considered to be negligible because of the low thermal conductiv-
ity of the case material and the lack of effective convection on the base
surface.
EXAMPLE 15-3 Analysis of Heat Conduction in a Chip
A chip is dissipating 0.6 W of power in a DIP with 12 pin leads. The materials
and the dimensions of various sections of this electronic device are as given in
the table below. If the temperature of the leads is 40°C, estimate the tempera-
ture at the junction of the chip.
Thermal
Section and
Conductivity,
Thickness,
Heat Transfer
Material
W/m • °C
mm
Surface Area
Junction constriction
—
—
diameter 0.4 mm
Silicon chip
120+
0.4
3 mm X 3 mm
Eutectic bond
296
0.03
3 mm X 3 mm
Copper lead frame
386
0.25
3 mm X 3 mm
Plastic separator
1
0.2
12
X 1 mm X 0.25 mm
Copper leads
386
5
12
X 1 mm X 0.25 mm
The thermal conductivity of silicon varies greatly with temperature from 153.5 W/m • °C at 27°C to
113.7 W/m • °C at 100°C, and the value 120 W/m • °C reflects the anticipation that the temperature
I of the silicon chip will be close to 100°C.
Air gap
Junction
/ Bond wires
Case
Lead frame
FIGURE 15-20
Heat generated at the junction of an
electronic device flows through the
path of least resistance.
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HEAT TRANSFER
Junction
constriction
chip
lead
frame
R,
plastic
FIGURE 15-21
Thermal resistance network for the
electronic device considered
in Example 15-3.
SOLUTION The dimensions and power dissipation of a chip are given. The
junction temperature of the chip is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Heat transfer through var-
ious components is one-dimensional. 3 Heat transfer through the air gap and
the lid on top of the chip is negligible because of the very large thermal resis-
tance involved along this path.
Analysis The geometry of the device is as shown in Figure 15-20. We take the
primary heat flow path to be the chip, the eutectic bond, the lead frame, the
plastic insulator, and the 12 leads. When the constriction resistance between
the junction and the chip is considered, the thermal resistance network for this
problem becomes as shown in Figure 15-21.
The various thermal resistances on the path of primary heat flow are deter-
mined as follows.
1
1
v constnction
2vW& Vtt(0.4 X l(r 3 m)(120W/m- °C)
_ (_L_\ = 0.4 X lQ- 3 m
chip \ M /chi P (120 W/m • °C)(9 X 10- 6 m 2 )
L\ 0.03 X 10-3 m
5.88°C/W
R
-"bond
R\r
^Vbond (296 W/m • °C)(9 X lO" 6 m 2 )
L_\ 0.25 X 10- 3 m
*i
kVicd tame (386 W/m ■ °C)(9 X lO" 6 m 2 )
_L_\ = 0.2 X !Q- 3 m
' M /pi asl ic (1 W/m • °C)(12 X 0.25 X lO" 6 m 2 )
L\ 5 X 10~ 3 m
M/ Ieads (386 W/m ■ °C)(12 X 0.25 X lQ- 6 m 2 )
= 0.37°C/W
0.01 °c/w
0.07°C/W
= 66.67°C/W
4.32°C/W
Note that for heat transfer purposes, all 12 leads can be considered as a single
lead whose cross-sectional area is 12 times as large. The alternative is to find
the resistance of a single lead and to calculate the equivalent resistance for 12
such resistances connected in parallel. Both approaches give the same result.
All the resistances determined here are in series. Thus the total thermal
resistance between the junction and the leads is determined by simply adding
them up:
total junction-lead
R
constriction
"chip + "bond
R] r
R,
plastic
R,,
(5.88 + 0.37 + 0.01 + 0.07 + 66.67 + 4.32)°C/W
77.32°C/W
Heat transfer through the chip can be expressed as
'AT\
Q
T — T
± junction x leads
R
/junc
R
junction-leads
Solving for 7i Unction and substituting the given values, the junction temperature
is determined to be
function = T |cads + e^„„ct,o„-leads = 40°C + (0.6 W)(77.32°C/ W) = 86.4°C
Note that the plastic layer between the lead frame and the leads accounts
for 66.67/77.32 = 86 percent of the total thermal resistance and thus the 86
cen58 93 3_chl5.qxd 9/9/2002 10:20 AM Page 801
percent of the temperature drop (0.6 X 66.67 = 40 C C) between the junction
and the leads. In other words, the temperature of the junction would be just
86.5 - 40 = 46.5°C if the thermal resistance of the plastic was eliminated.
Discussion The simplified analysis given here points out that any attempt to
reduce the thermal resistance in the chip carrier and thus improve the heat
flow path should start with the plastic layer. We also notice from the magni-
tudes of individual resistances that some sections, such as the eutectic bond
and the lead frame, have negligible thermal resistances, and any attempt to
improve them further will have practically no effect on the junction tempera-
ture of the chip.
801
CHAPTER 15
The analytical determination of the junction-to-case thermal resistance of
an electronic device can be rather complicated and can involve considerable
uncertainty, as already shown. Therefore, the manufacturers of electronic de-
vices usually determine this value experimentally and list it as part of their
product description. When the thermal resistance is known, the temperature
difference between the junction and the outer surface of the device can be de-
termined from
AT„,
!e^*\junction-(
(°C)
(15-7)
where Q is the power consumed by the device.
The determination of the actual junction temperature depends on the
ambient temperature T. imbient as well as the thermal resistance ^ case - a mbient be-
tween the case and the ambient (Fig. 15-22). The magnitude of this resis-
tance depends on the type of ambient (such as air or water) and the fluid
velocity. The two thermal resistances discussed above are in series, and the
total resistance between the junction and the ambient is determined by sim-
ply adding them up:
R„
R
junction-ambient
R„
+ R r
(°C/W)
(15-8)
Many manufacturers of electronic devices go the extra step and list the total
resistance between the junction and the ambient for various chip configura-
tions and ambient conditions likely to be encountered. Once the total thermal
resistance is available, the junction temperature corresponding to the specified
power consumption (or heat dissipation rate) of Q is determined from
T = T
junction ambient
QR t u
junction-ambient
(°C)
(15-9)
A typical chart for the total junction-to-ambient thermal resistance for a
single DIP-type electronic device mounted on a circuit board is given in
Figure 15-23 for various air velocities and lead numbers. The values at the
intersections of the curves and the vertical axis represent the thermal resis-
tances corresponding to natural convection conditions (zero air velocity).
Note that the thermal resistance and thus the junction temperature decrease
with increasing air velocity and the number of leads extending from the elec-
tronic device, as expected.
Junction
D D ■ p
total — junction-case case-ambient
junction — ambient *^ total
FIGURE 15-22
The junction temperature of a chip
depends on the external case-to-
ambient thermal resistance as well as
the internal junction-to-case
resistance.
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HEAT TRANSFER
FIGURE 15-23
Total thermal resistance between the
junction of a plastic DIP device
mounted on a circuit board and the
ambient air as a function of the air
velocity and the number of leads
(courtesy of Motorola Semiconductor
Products, Inc.).
200
*
c3
1 6U
o
'-i
140
=
1 Ml
1 Ul)
T-,
a
SO
a
<u
on
ci
o
40
<
20
z— axis (transverse)
"jfiflfi.
Airflow direction
8-lead
-24-lead
50 100 150 200
Airflow velocity, m/min
250 300
Air
30°C
1.2 W
KPt
FIGURE 15-24
Schematic for Example 15-4.
EXAMPLE 15-4 Predicting the Junction Temperature of a Device
A fan blows air at 30°C and a velocity of 200 m/min over a 1.2-W plastic DIP
with 16 leads mounted on a PCB, as shown in Figure 15-24. Using data from
Figure 15-23, determine the junction temperature of the electronic device.
What would the junction temperature be if the fan were to fail?
SOLUTION A plastic DIP with 16 leads is cooled by forced air. Using data sup-
plied by the manufacturer, the junction temperature is to be determined.
Assumptions Steady operating conditions exist.
Analysis The junction-to-ambient thermal resistance of the device with 16
leads corresponding to an air velocity of 200 m/min is determined from Figure
15-23 to be
R,
junction-ambient
55°C/W
Then the junction temperature can be determined from Eq. 15-9 to be
nt t^ junction-!
30°C + (1.2 W)(55°C/W) = 96°C
When the fan fails, the airflow velocity over the device will be zero. The total
thermal resistance in this case is determined from the same chart by reading
the value at the intersection of the curve and the vertical axis to be
R,
junction-ambient
70°C/W
which gives
junction ambient x£ junction-ambient
30°C + (1.2 W)(70°C/W) = 114°C
Discussion Note that the temperature of the junction will rise by 18°C when
the fan fails. Of course, this analysis assumes the temperature of the sur-
rounding air still to be 30°C, which may no longer be the case. Any increase in
the ambient temperature as a result of inadequate airflow will reflect on the
junction temperature, which will seriously jeopardize the safety of the elec-
tronic device.
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CHAPTER 15
Conduction in Printed Circuit Boards
Heat-generating electronic devices are commonly mounted on thin rectangu-
lar boards, usually 10 cm X 15 cm in size, made of electrically insulating ma-
terials such as glass-epoxy laminates, which are also poor conductors of heat.
The resulting printed circuit boards are usually cooled by blowing air or pass-
ing a dielectric liquid through them. In such cases, the components on the
PCBs are cooled directly, and we are not concerned about heat conduction
along the PCBs. But in some critical applications such as those encountered in
the military, the PCBs are contained in sealed enclosures, and the boards pro-
vide the only effective heat path between the components and the heat sink at-
tached to the sealed enclosure. In such cases, heat transfer from the side faces
of the PCBs is negligible, and the heat generated in the components must be
conducted along the PCB toward its edges, which are clamped to cold plates
for removing the heat externally.
Heat transfer along a PCB is complicated in nature because of the multidi-
mensional effects and nonuniform heat generation on the surfaces. We can
still obtain sufficiently accurate results by using the thermal resistance net-
work in one or more dimensions.
Copper or aluminum cladding, heat frames, or cores are commonly used to
enhance heat conduction along the PCBs. The thickness of the copper
cladding on the PCB is usually expressed in terms of ounces of copper, which
is the thickness of 1-ft 2 copper sheet made of one ounce of copper. An ounce
of copper is equivalent to 0.03556-mm (1.4-mil) thickness of a copper layer.
When analyzing heat conduction along a PCB with copper (or aluminum)
cladding on one or both sides, often the question arises whether heat transfer
along the epoxy laminate can be ignored relative to that along the copper
layer, since the thermal conductivity of copper is about 1500 times that of
epoxy. The answer depends on the relative cross-sectional areas of each layer,
since heat conduction is proportional to the cross-sectional area as well as the
thermal conductivity.
Consider a copper-cladded PCB of width w and length L, across which the
temperature difference is AT, as shown in Figure 15-25. Assuming heat con-
duction is along the length L only and heat conduction in other dimensions is
negligible, the rate of heat conduction along this PCB is the sum of the heat
conduction along the epoxy board and the copper layer and is expressed as
i^PCB s^epoxy i^copper
= [(M) cpoxy + (M)
= [(^)epoxy + (M
«¥) +(«x)
^ I epoxy \ / co PP er
i —
copperJ j
(15-10)
,wM
copperJ j
PCB epoxy copper
FIGURE 15-25
Schematic of a copper-cladded epoxy
board and heat conduction along it.
where t denotes the thickness. Therefore, the relative magnitudes of heat con-
duction along the two layers depend on the relative magnitudes of the thermal
conductivity-thickness product kt of the layer. Therefore, if the kt product of
the copper is 100 times that of epoxy, then neglecting heat conduction along
the epoxy board will involve an error of just 1 percent, which is negligible.
We can also define an effective thermal conductivity for metal-cladded
PCBs as
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HEAT TRANSFER
(kt\
(kt) a
* epoxy ^copper
(W/m ■ °C)
so that the rate of heat conduction along the PCB can be expressed as
wAT , , AT
GpCB ~~ ^effUepoxy + ^copper) r ~~ *eff ^PCB j
(W)
(15-11)
(15-12)
where A PCB = w(/ epoxy + t copper ) is the area normal to the direction of heat
transfer. When there are holes or discontinuities along the copper cladding, the
above analysis needs to be modified to account for their effect.
Epoxy
Copper
.0.16 mm
Q 0.04 mm
FIGURE 15-26
Schematic for Example 15-5.
EXAMPLE 15-5 Heat Conduction along a PCB with
Copper Cladding
Heat is to be conducted along a PCB with copper cladding on one side. The
PCB is 10 cm long and 10 cm wide, and the thickness of the copper and epoxy
layers are 0.04 mm and 0.16 mm, respectively, as shown in Figure 15-26. Dis-
regarding heat transfer from side surfaces, determine the percentages of heat
conduction along the copper (k = 386 W/m • °C) and epoxy (k = 0.26 W/m ■
°C) layers. Also, determine the effective thermal conductivity of the PCB.
SOLUTION A PCB with copper cladding is given. The percentages of heat
conduction along the copper and epoxy layers as well as the effective thermal
conductivity of the PCB are to be determined.
Assumptions 1 Heat conduction along the PCB is one-dimensional since heat
transfer from side surfaces is negligible. 2 The thermal properties of epoxy and
copper layers are constant.
Analysis The length and width of both layers are the same, and so is the tem-
perature difference across each layer. Heat conduction along a layer is propor-
tional to the thermal conductivity-thickness product kt, which is determined for
each layer and the entire PCB to be
(fe)cop PCT = (386 W/m ■ °C)(0.04 X 10- 3 m) = 15.44 X 10" 3 W/°C
Wepoxy = (0.26 W/m • °C)(0.16 X 10- 3 m) = 0.04 X lO" 3 W/°C
(A:Opcb = (toLpper + (« )e P ox y = (1544 + 0.04) X 10 3 m = 15.48 X 10- 3 W/°C
Therefore, heat conduction along the epoxy board will constitute
(^koxy 0.04 X 10- 3 W/°C
/ =
(*Op
15.48 X 10- 3 W/°C
0.0026
or 0.26 percent of the thermal conduction along the PCB, which is negligible.
Therefore, heat conduction along the epoxy layer in this case can be disre-
garded without any reservations.
The effective thermal conductivity of the board is determined from Eq. 15-11
to be
\ /epoxy \ /copper
(15.44 + 0.04) X 10- 3 W/°C
(0.16 + 0.04) X 10- 3 m
77.4 W/m • °C
That is, the entire PCB can be treated as a 0.20-mm-thick single homogeneous
layer whose thermal conductivity is 77.4 W/m • °C for heat transfer along its
length.
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CHAPTER 15
Discussion Note that a very thin layer of copper cladding on a PCB improves
heat conduction along the PCB drastically, and thus it is commonly used in
conduction-cooled electronic devices.
Heat Frames
In applications where direct cooling of circuit boards by passing air or a di-
electric liquid over the electronic components is not allowed, and the junction
temperatures are to be maintained relatively low to meet strict safety require-
ments, a thick heat frame is used instead of a thin layer of copper cladding.
This is especially the case for multilayer PCBs that are packed with high-
power output chips.
The schematic of a PCB that is conduction-cooled via a heat frame is shown
in Figure 15-27. Heat generated in the chips is conducted through the circuit
board, through the epoxy adhesive, to the center of the heat frame, along the
heat frame, and to a heat sink or cold plate, where heat is externally removed.
The heat frame provides a low-resistance path for the flow of heat from the
circuit board to the heat sink. The thicker the heat frame, the lower the ther-
mal resistance, and thus the smaller the temperature difference between the
center and the ends of the heat frame. When the heat load is evenly distrib-
uted on the PCB, there will be thermal symmetry about the centerline, and
the temperature distribution along the heat frame and the PCB will be para-
bolic in nature, with the chips in the middle of the PCB (farthest away from
the edges) operating at the highest temperatures and the chips near the edges
operating at the lowest temperatures. Also, when the PCB is cooled from two
edges, heat generated in the left half of the PCB will flow toward the left
edge and heat generated in the right half will flow toward the right edge of
the heat frame. But when the PCB is cooled from all four edges, the heat
transfer along the heat frame as well as the resistance network will be two-
dimensional.
When a heat frame is used, heat conduction in the epoxy layer of the PCB
is through its thickness instead of along its length. The epoxy layer in this case
offers a much smaller resistance to heat flow because of the short distance in-
volved. This resistance can be made even smaller by drilling holes in the
epoxy and filling them with copper, as shown in Figure 15-28. These copper
fillings are usually 1 mm in diameter and their centers are a few millimeters
apart. Such highly conductive fillings provide easy passageways for heat from
one side of the PCB to the other and result in considerable reduction in the
thermal resistance of the board along its thickness, as shown in Examples
15-6, 15-7, and 15-8.
EXAMPLE 15-6 Thermal Resistance of an Epoxy Glass Board
Consider a 10-cm X 15-cm glass-epoxy laminate (k = 0.26 W/m ■ °C) whose
thickness is 0.8 mm, as shown in Figure 15-29. Determine the thermal resis-
tance of this epoxy layer for heat flow (a) along the 15-cm-long side and
(b) across its thickness.
Temperature
distribution
Electronic
components
PCB
Qm
luvicr i.'j-\.
Cold
plate
Heat
frame
Epoxy
adhesive
FIGURE 15-27
Conduction cooling of a printed
circuit board with a heat frame, and
the typical temperature distribution
along the frame.
Epoxy board
Copper
fillings
FIGURE 15-28
Planting the epoxy board with copper
fillings decreases the thermal
resistance across its thickness
considerably.
0.8 mm
FIGURE 15-29
Schematic for Example 15-6.
cen58 93 3_chl5.qxd 9/9/2002 10:21 AM Page 806
806
HEAT TRANSFER
SOLUTION The dimensions of an epoxy-glass laminate are given. The thermal
resistances for heat flow along the layers and across the thickness are to be
determined.
Assumptions 1 Heat conduction in the laminate is one-dimensional in either
case. 2 Thermal properties of the laminate are constant.
Analysis The thermal resistance of a plane parallel medium in the direction of
heat conduction is given by
R
L_
kA
where L is the length in the direction of heat flow, k is the thermal conductiv-
ity, and A is the area normal to the direction of heat conduction. Substituting
the given values, the thermal resistances of the board for both cases are deter-
mined to be
(a)
(b)
R
along length
^ /along length
0.15 m
(0.26 W/m • °C)(0.1 m)(0.8 X 10- 3 m)
7212°C/W
across thickness I ka I
V^Vacross th
L\
.ckness
0.8 X 10~ 3 m
(0.26 W/m • °C)(0.1 m)(0.15 m)
0.21 °C/W
Discussion Note that heat conduction at a rate of 1 W along this PCB would
cause a temperature difference of 7212°C across a length of 15 cm. But the
same rate of heat conduction would cause a temperature difference of only
0.21 C C across the thickness of the epoxy board.
1 mm
2.5 mm
Copper Epoxy board
filling
FIGURE 15-30
Schematic for Example 15-7.
EXAMPLE 15-7 Planting Cylindrical Copper Fillings in an
Epoxy Board
Reconsider the 10-cm X 15-cm glass-epoxy laminate (k = 0.26 W/m • °C) of
thickness 0.8 mm discussed in Example 15-6. In order to reduce the thermal
resistance across its thickness from the current value of 0.21 C C/W, cylindrical
copper fillings (k = 386 W/m • C C) of 1-mm diameter are to be planted through-
out the board with a center-to-center distance of 2.5 mm, as shown in Figure
15-30. Determine the new value of the thermal resistance of the epoxy board
for heat conduction across its thickness as a result of this modification.
SOLUTION Cylindrical copper fillings are planted throughout an epoxy glass
board. The thermal resistance of the board across its thickness is to be
determined.
Assumptions 1 Heat conduction along the board is one-dimensional. 2 Thermal
properties of the board are constant.
Analysis Heat flow through the thickness of the board in this case will take
place partly through the copper fillings and partly through the epoxy in parallel
paths. The thickness of both materials is the same and is given to be 0.8 mm.
cen58 93 3_chl5.qxd 9/9/2002 10:21 AM Page 807
But we also need to know the surface area of each material before we can de-
termine the thermal resistances.
It is stated that the distance between the centers of the copper fillings
is 2.5 mm. That is, there is only one 1-mm-diameter copper filling in every
2.5-mm X 2.5-mm square section of the board. The number of such squares
and thus the number of copper fillings on the board are
Area of the board ( 1 00 mm)( 1 50 mm)
Area of one square (2.5 mm)(2.5 mm)
2400
Then the surface areas of the copper fillings and the remaining epoxy layer
become
—
-^copper "
(2400)
ir(l X 10- 3 m) 2
0.001885 m 2
1 total
(Length)(Width) = (0.1 m)(0.15 m) = 0.015 m 2
Aep„x y = A touU - A coppcr = (0.015 - 0.001885) m 2 = 0.013115 m 2
The thermal resistance of each material is
0.8 X 10- 3 m
R,,
R?
^/copper
fcA/epoxy (0- 26 w/m • °C)(0.0131 15 m 2 )
(386 W/m • °C)(0.001885m 2 )
0.8 X 10 3 m
0.001 1°C/W
0.2346°C/W
Noting that these two resistances are in parallel, the equivalent thermal resis-
tance of the entire board is determined from
1
1
0.001 1°C/W 0.2346°C/W
which gives
Discussion Note that the thermal resistance of the epoxy board has dropped
from 0.21°C/W by a factor of almost 200 to just 0.00109°C/W as a result of
implanting 1-mm-diameter copper fillings into it. Therefore, implanting copper
pins into the epoxy laminate has virtually eliminated the thermal resistance of
the epoxy across its thickness.
807
CHAPTER 15
EXAMPLE 15-8 Conduction Cooling of PCBs by a Heat Frame
A 10-cm X 12-cm circuit board dissipating 24 W of heat is to be conduction-
cooled by a 1.2-mm-thick copper heat frame [k = 386 W/m • C C) 10 cm X 14
cm in size. The epoxy laminate (k = 0.26 W/m • °C) has a thickness of 0.8 mm
and is attached to the heat frame with conductive epoxy adhesive (k = 1 .8 W/m
• °C) of 0.13-mm thickness, as shown in Figure 15-31. The PCB is attached to
a heat sink by clamping a 5-mm-wide portion of the edge to the heat sink from
both ends. The temperature of the heat frame at this point is 20°C. Heat is uni-
formly generated on the PCB at a rate of 2 W per 1-cm X 10-cm strip. Consid-
ering only one-half of the PCB board because of symmetry, determine the
cen58 93 3_chl5.qxd 9/9/2002 10:21 AM Page 808
Epoxy
board
808
HEAT TRANSFER
' 1 cm / 1 cm / 1 cm /\ cm / 1 cm / 1 cm / *
►4 *4 *4 — *4 *4 ^ ^
Symmetry .
plane / /
copper II
FIGURE 15-31
The schematic and thermal resistance network for Example 15-8.
Clamp
area
maximum temperature on the PCB and the temperature distribution along the
heat frame.
SOLUTION A circuit board with uniform heat generation is to be conduction-
cooled by a copper heat frame. Temperature distribution along the heat frame
and the maximum temperature in the PCB are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Thermal properties are
constant. 3 There is no direct heat dissipation from the surface of the PCB, and
thus all the heat generated is conducted by the heat frame to the heat sink.
Analysis The PCB under consideration possesses thermal symmetry about the
centerline. Therefore, the heat generated on the left half of the PCB is con-
ducted to the left heat sink, and the heat generated on the right half is con-
ducted to the right heat sink. Thus we need to consider only half of the board in
the analysis.
The maximum temperature will occur at a location furthest away from the
heat sinks, which is the symmetry line. Therefore, the temperature of the elec-
tronic components located at the center of the PCB will be the highest, and
their reliability will be the lowest.
Heat generated in the components on each strip is conducted through the
epoxy layer underneath. Heat is then conducted across the epoxy adhesive and
to the middle of the copper heat frame. Finally, heat is conducted along the
heat frame to the heat sink.
The thermal resistance network associated with heat flow in the right half of
the PCB is also shown in Figure 15-31. Note that all vertical resistances are
identical and are equal to the sum of the three resistances in series. Also note
that heat conduction toward the heat sink is assumed to be predominantly
along the heat frame, and conduction along the epoxy adhesive is considered
to be negligible. This assumption is quite reasonable, since the conductivity-
thickness product of the heat frame is much larger than those of the other two
layers.
The properties and dimensions of various sections of the PCB are sum-
marized in this table.
cen58 93 3_chl5.qxd 9/9/2002 10:21 AM Page 809
Thermal
Section and
Conductivity,
Thickness,
Heat Transfer
Material
W/m • °C
mm
Surface area
Epoxy board
0.26
0.8
10 mm X 100 mm
Epoxy adhesive
1.8
0.13
10 mm X 100 mm
Copper heat frame,
1
(normal to frame
)
386
0.6
10 mm X 100 mm
Copper heat frame,
II
(along the frame]
386
10
1.2 mm X 100 mm
Using the values in the table, the various thermal resistances are determined
to be
R„
R.
copper, _L
kA
^/adhesive
kA
0.8 X 10-
(0.26 W/m
■ °C)(0.01
rn X 0.1
m)
0.13 X 10~
3 m
(1.8 W/m
• °C)(0.01
m X 0.1
m)
0.6 X 10-
3 m
R„
(386 W/m • °C)(0.01 m X 0.1 m)
L_\ _ 0.01 m
M /copper, ||
3.077°C/W
0.072°C/W
0.002°C/W
(386 W/m • °C)(0.0012 m X 0.1 m)
0.216°C/W
The combined resistance between the electronic components on each strip
and the heat frame can be determined, by adding the three resistances in se-
ries, to be
«.
R.
R,
R„
epoxy " "-adhesive ' "copper, _L
= (3.077 + 0.072 + 0.002)°C/W
= 3.151°C/W
The various temperatures along the heat frame can be determined from the
relation
AT-
high
QR
where R is the thermal resistance between two specified points, Q is the heat
transfer rate through that resistance, and A T is the temperature difference
across that resistance.
The temperature at the location where the heat frame is clamped to the heat
sink is given as T = 20 C C. Noting that the entire 12 W of heat generated on
the right half of the PCB must pass through the last thermal resistance adjacent
to the heat sink, the temperature Tj can be determined from
r, = T + Q,- Ri-o = 20°C + (12W)(0.216°C/W) = 22.59°C
Following the same line of reasoning, the temperatures at specified locations
along the heat frame are determined to be
T 2 = T\ + G2-1 #2-1 = 22.59°C + (10 W)(0.216°C/W) = 24.75°C
Ti = T 2 + Q 3Z R 3~2 = 24.75°C + (8 W)(0.216°C/W) = 26.48°C
T 4 = T 3 + e 4 _3^4_ 3 = 26.48°C + (6 W)(0.216°C/W) = 27.78°C
T 5 = T 4 + Q 5 ^R 5 _ 4 = 27.78°C + (4 W)(0.216°C/W) = 28.64°C
T 6 = T 5 + 6 6 5^6-s = 28.64°C + (2 W)(0.216°C/W) = 29.07°C
809
CHAPTER 15
cen58 93 3_chl5.qxd 9/9/2002 10:21 AM Page 81C
810
HEAT TRANSFER
Finally, T 7 , which is the maximum temperature on the PCB, is determined from
T 7 = T 6 + g vcrtical fl vcrtical = 29.07°C + (2 W)(3.151°C/W) = 35.37°C
Discussion The maximum temperature difference between the PCB and the
heat sink is only 15.37 C C, which is very impressive considering that the PCB
has no direct contact with the cooling medium. The junction temperatures in
this case can be determined by calculating the temperature difference between
the junction and the leads of the chip carrier at the point of contact to the PCB
and adding 35.37°C to it. The maximum temperature rise of 15.37 C C can be
reduced, if necessary, by using a thicker heat frame.
Metal core
rA
I
Epoxy
lamina
Electronic
— J/"^ components
J
Heat
sink
FIGURE 15-32
A two-sided printed circuit board with
a metal core for conduction cooling.
Conduction cooling can also be used when electronic components are
mounted on both sides of the PCB by using a copper or aluminum core plate
in the middle of the PCB, as shown in Figure 15-32. The heat load in this case
will be twice that of a PCB that has components on one side only. Again, heat
generated in the components will be conducted through the thickness of the
epoxy layer to the metal core, which serves as a channel for effective heat
removal. The thickness of the core is selected such that the maximum com-
ponent temperatures remain below specified values to meet a prescribed reli-
ability criterion.
The thermal expansion coefficients of aluminum and copper are about twice
as large as that of the glass-epoxy. This large difference in thermal expansion
coefficients can cause warping on the PCBs if the epoxy and the metal are not
bonded properly. One way of avoiding warping is to use PCBs with compo-
nents on both sides, as discussed. Extreme care should be exercised during the
bonding and curing process when components are mounted on only one side
of the PCB.
The Thermal Conduction Module (TCM)
The heat flux for logic chips has been increasing steadily as a result of the
increasing circuit density in the chips. For example, the peak flux at the chip
level has increased from 2 W/cm 2 on IBM System 370 to 20 W/cm 2 on IBM
System 3081, which was introduced in the early 1980s. The conventional
forced-air cooling technique used in earlier machines was inadequate for re-
moving such high heat fluxes, and it was necessary to develop a new and
more effective cooling technique. The result was the thermal conduction
module, shown in Figure 15-33. The TCM was different from previous chip
packaging designs in that it incorporated both electrical and thermal consider-
ations in early stages of chip design. Previously, a chip would be designed pri-
marily by electrical designers, and the thermal designer would be told to come
up with a cooling scheme for the chip. That approach resulted in unnecessar-
ily high junction temperatures, and reduced reliability, since the thermal de-
signer had no direct access to the chip. The TCM reflects a new philosophy in
electronic packaging in that the thermal and electrical aspects are given equal
treatment in the design process, and a successful thermal design starts at the
chip level.
cen58 93 3_chl5.qxd 9/9/2002 10:21 AM Page 811
811
CHAPTER 15
Cold plate
Spring
Helium
reservoir
C-ring seal
Piston
Ceramic
substrate
Spherical
radius —
Chip
Cold plate
Interposer
«... ,,
Helium
(or gas mixture)
>R-
p-h
>*r*
'gp
>
K
-p
>R;.
FIGURE 15-33
Cutaway view of the thermal conduction module (TCM), and the thermal resistance network between a single chip and the
cooling fluid (courtesy of IBM Corporation).
In the TCM, one side of the chip is reserved for electrical connections and
the other side for heat rejection. The chip is cooled by direct contact to the
cooling system to minimize the junction-to-case thermal resistance.
The TCM houses 100 to 118 logic chips, which are bonded to a multilayer
ceramic substrate 90 mm X 90 mm in size with solder balls, which also pro-
vide the electrical connections between the chips and the substrate. Each chip
dissipates about 4 W of power. The heat flow path from the chip to the metal
casing is provided by a piston, which is pressed against the back surface of the
chip by a spring. The tip of the piston is slightly curved to ensure good ther-
mal contact even when the chip is tilted or misaligned.
Heat conduction between the chip and the piston occurs primarily through
the gas space between the chip and the piston because of the limited contact
area between them. To maximize heat conduction through the gas, the air in
the TCM cavity is evacuated and is replaced by helium gas, whose thermal
conductivity is about six times that of air. Heat is then conducted through the
piston, across the surrounding helium gas layer, through the module housing,
and finally to the cooling water circulating through the cold plate attached to
the top surface of the TCM.
The total internal thermal resistance R- mt of the TCM is about 8°C/ W, which
is rather impressive. This means that the temperature difference between the
chip surface and the outer surface of the housing of the module will be only
24°C for a 3-W chip. The external thermal resistance R ext between the hous-
ing of the module and the cooling fluid is usually comparable in magnitude to
R int . Also, the thermal resistance between the junction and the surface of the
chip can be taken to be 1°C/W.
cen58 93 3_chl5.qxd 9/9/2002 10:21 AM Page 812
812
HEAT TRANSFER
The compact design of the TCM significantly reduces the distance between
the chips, and thus the signal transmission time between the chips. This, in
turn, increases the operating speed of the electronic device.
Substrate
FIGURE 15-34
Thermal resistance network for
Example 15-9.
EXAMPLE 15-9 Cooling of Chips by the Thermal Conduction
Module
Consider a thermal conduction module with 100 chips, each dissipating 3 W of
power. The module is cooled by water at 25°C flowing through the cold plate on
top of the module. The thermal resistances in the path of heat flow are ff chip =
1°C/W between the junction and the surface of the chip, R mi = 8°C/W between
the surface of the chip and the outer surface of the thermal conduction module,
and ff ext = 6°C/W between the outer surface of the module and the cooling wa-
ter. Determine the junction temperature of the chip.
SOLUTION A thermal conduction module TCM with 100 chips is cooled by wa-
ter. The junction temperature of the chip is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Heat transfer through var-
ious components is one-dimensional.
Analysis Because of symmetry, we will consider only one of the chips in our
analysis. The thermal resistance network for heat flow is given in Figure 15-34.
Noting that all resistances are in series, the total thermal resistance between
the junction and the cooling water is
**W = ^noion-w^- = R Mp + *W + #ex t = (1 + 8 + 6)°C/W = 15°C/W
Noting that the total power dissipated by the chip is 3 W and the water temper-
ature is 25°C, the junction temperature of the chip in steady operation can be
determined from
Q
AT\
R . ,
/junction-'
T — T
*■ junction water
v junction-water
Solving for 7" junction and substituting the specified values gives
T = T
*■ junction *■ water
t^^junction-'
25°C + (3 W)(15°C/W) = 70°C
Therefore, the circuits of the chip will operate at about 70°C, which is consid-
ered to be a safe operating temperature for silicon chips.
Cold plates are usually made of metal plates with fluid channels running
through them, or copper tubes attached to them by brazing. Heat transferred to
the cold plate is conducted to the tubes, and from the tubes to the fluid flow-
ing through them. The heat carried away by the fluid is finally dissipated to
the ambient in a heat exchanger.
15-7 - AIR COOLING: NATURAL CONVECTION
AND RADIATION
Low-power electronic systems are conveniently cooled by natural convection
and radiation. Natural convection cooling is very desirable, since it does not
involve any fans that may break down.
cen58 93 3_chl5.qxd 9/9/2002 10:21 AM Page 813
813
CHAPTER 15
Natural convection is based on the fluid motion caused by the density dif-
ferences in a fluid due to a temperature difference. A fluid expands when
heated and becomes less dense. In a gravitational field, this lighter fluid rises
and initiates a motion in the fluid called natural convection currents (Fig.
15-35). Natural convection cooling is most effective when the path of the
fluid is relatively free of obstacles, which tend to slow down the fluid, and is
least effective when the fluid has to pass through narrow flow passages and
over many obstacles.
The magnitude of the natural convection heat transfer between a surface and
a fluid is directly related to the flow rate of the fluid. The higher the flow rate,
the higher the heat transfer rate. In natural convection, no blowers are used
and therefore the flow rate cannot be controlled externally. The flow rate in
this case is established by the dynamic balance of buoyancy and friction. The
larger the temperature difference between the fluid adjacent to a hot surface
and the fluid away from it, the larger the buoyancy force, and the stronger the
natural convection currents, and thus the higher the heat transfer rate. Also,
whenever two bodies in contact move relative to each other, a friction force
develops at the contact surface in the direction opposite to that of the motion.
This opposing force slows down the fluid, and thus reduces the flow rate of
the fluid. Under steady conditions, the airflow rate driven by buoyancy is es-
tablished at the point where these two effects balance each other. The friction
force increases as more and more solid surfaces are introduced, seriously dis-
rupting the fluid flow and heat transfer.
Electronic components or PCBs placed in enclosures such as a TV or VCR
are cooled by natural convection by providing a sufficient number of vents on
the case to enable the cool air to enter and the heated air to leave the case
freely, as shown in Figure 15-36. From the heat transfer point of view,
the vents should be as large as possible to minimize the flow resistance
and should be located at the bottom of the case for air entering and at the top
for air leaving. But equipment and human safety requirements dictate that
the vents should be quite narrow to discourage unintended entry into the box.
Also, concern about human habits such as putting a cup of coffee on the
closest flat surface make it very risky to place vents on the top surface. The
narrow clearance allowed under the case also offers resistance to airflow.
Therefore, vents on the enclosures of natural convection-cooled electronic
equipment are usually placed at the lower section of the side or back surfaces
for air inlet and at the upper section of those surfaces for air exit.
The heat transfer from a surface at temperature T s to a fluid at temperature
Tfiuid by convection is expressed as
Heated air
(light)
Warm air
rising
\\ Velocity
profile
^vmw
Unheated air
(dense)
Friction
at the
surface
FIGURE 15-35
Natural convection currents
around a hot object in air.
FIGURE 15-36
Natural convection cooling of
electronic components in an enclosure
with air vents.
Geo
h A AT ■
conv **s
Tfluid)
(W)
(15-13)
where h com is the convection heat transfer coefficient and A s is the heat trans-
fer surface area. The value of h conv depends on the geometry of the surface and
the type of fluid flow, among other things.
Natural convection currents start out as laminar (smooth and orderly) and
turn turbulent when the dimension of the body and the temperature difference
between the hot surface and the fluid are large. For air, the flow remains lam-
inar when the temperature differences involved are less than 100°C and the
characteristic length of the body is less than 0.5 m, which is almost always the
cen58 93 3_chl5.qxd 9/9/2002 10:21 AM Page 814
814
HEAT TRANSFER
Cool surface
FIGURE 15-37
Simultaneous natural convection heat
transfer to air and radiation heat
transfer to the surrounding surfaces
from a hot electronic component
mounted on the wall of an enclosure.
FIGURE 15-38
A chassis with an array of
vertically oriented PCBs
cooled by natural convection.
case in electronic equipment. Therefore, the airflow in the analysis of elec-
tronic equipment can be assumed to be laminar.
The natural convection heat transfer coefficient for laminar flow of air at at-
mospheric pressure is given by a simplified relation of the form
K
K(fT 5
(W/m 2 • °C)
(15-14)
where AT = T s — 7* fluid is the temperature difference between the surface and
the fluid, L is the characteristic length (the length of the body along the heat
flow path), and K is a constant whose value depends on the geometry and ori-
entation of the body.
The heat transfer coefficient relations are given in Table 15-1 for some
common geometries encountered in electronic equipment in both SI and Eng-
lish unit systems. Once h com has been determined from one of these relations,
the rate of heat transfer can be determined from Eq. 15-13. The relations in
Table 15-1 can also be used at pressures other than 1 atm by multiplying them
by \/p, where P is the air pressure in atm (1 atm = 101.325 kPa = 14.696
psia). That is,
/)„
K
,Vp
(W/m 2 • °C)
(15-15)
When hot surfaces are surrounded by cooler surfaces such as the walls and
ceilings of a room or just the sky, the surfaces are also cooled by radiation, as
shown in Figure 15-37. The magnitude of radiation heat transfer, in general,
is comparable to the magnitude of natural convection heat transfer. This is es-
pecially the case for surfaces whose emissivity is close to unity, such as plas-
tics and painted surfaces (regardless of color). Radiation heat transfer is
negligible for polished metals because of their very low emissivity and for
bodies surrounded by surfaces at about the same temperature.
Radiation heat transfer between a surface at temperature T s completely sur-
rounded by a much larger surface at temperature T sim can be expressed as
Grad ~~ S A S (J(T S T saII )
(W)
(15-16)
where e is the emissivity of the surface, A s is the heat transfer surface area,
and ct is the Stefan-Boltzmann constant, whose value is ct = 5.67 X 10~ 8
W/m 2 • K 4 = 0.1714 X 10~ 8 Btu/h • ft 2 • R 4 . Here, both temperatures must be
expressed in K or R. Also, if the hot surface analyzed has only a partial view
of the surrounding cooler surface at r surr , the result obtained from Eq. 15-16
must be multiplied by a view factor, which is the fraction of the view of the hot
surface blocked by the cooler surface. The value of the view factor ranges
from (the hot surface has no direct view of the cooler surface) to 1 (the hot
surface is completely surrounded by the cooler surface). In preliminary analy-
sis, the surface is usually assumed to be completely surrounded by a single hy-
pothetical surface whose temperature is the equivalent average temperature of
the surrounding surfaces.
Arrays of low-power PCBs are often cooled by natural convection by
mounting them within a chassis with adequate openings at the top and at the
bottom to facilitate airflow, as shown in Figure 15-38. The air between the
PCBs rises when heated by the electronic components and is replaced by
the cooler air entering from below. This initiates the natural convection flow
cen58 93 3_chl5.qxd 9/9/2002 10:21 AM Page 815
TABLE 15-1
Simplified relations for natural convection heat transfer coefficients for various
geometries in air at atmospheric pressure for laminar flow conditions
(From Refs. 4 and 5.)
Natural convection heat transfer coefficient
Geometry
W/m 2 • °C
(A 7" in °C, Lor Din m)
Btu/h • ft 2 ■ °F
(A7"in°F, Lor Din ft)
Vertical
Pi
ate c
>r cylinder
T
L
1
h am = 1.42(f) 025
, conv =0.29(f)°- 25
Horizontal cylinder
"-=1.32(f) 025
^=0.27(f)°- 25
(
n
)
1
Horizontal plate (L =
AA/p, where A is surface
area and p is perimeter
s Hot surface
h am = 1.32 (f ) 025
/,_= 0.59(f) 025
. con ^0.27(f) - 25
. con ^0.12(f)°- 25
/ A
,1
a) Hot surface fa
(1 "• — ' "
/ i
/Hi ,
cing up
d
/ Hot
'•" -' surface
b) Hot surface fa
cing down
Components
on a circuit
board
1
3
3
ft =2 4 4 /AIV- 25
"conv *-.-r-ri . 1
^ m =0.50(f)°- 25
Small compon
or wires in
free air
en
1
T
ts
L
, CO /A7"\0.25
h conv = 3.531 — 1
^^ 0.72(f) 025
Sphere / x
h = ! 92 (^> 25
n com = 0.39(f) 025
815
CHAPTER 15
cen58 93 3_chl5.qxd 9/9/2002 10:21 AM Page 816
816
HEAT TRANSFER
Warm air
out
FIGURE 15-39
The PCBs in a chassis must be
oriented vertically and spaced
adequately to maximize heat transfer
by natural convection.
through the parallel flow passages formed by the PCBs. The PCBs must be
placed vertically to take advantage of natural convection currents and to min-
imize trapped air pockets (Fig. 15-39). Placing the PCBs too far from each
other wastes valuable cabinet space, and placing them too close tends to
"choke" the flow because of the increased resistance. Therefore, there should
be an optimum spacing between the PCBs. It turns out that a distance of about
2 cm between the PCBs provides adequate air flow for effective natural con-
vection cooling.
In the heat transfer analysis of PCBs, radiation heat transfer is disregarded,
since the view of the components is largely blocked by other heat-generating
components. As a result, hot components face other hot surfaces instead of a
cooler surface. The exceptions are the two PCBs at the ends of the chassis that
view the cooler side surfaces. Therefore, it is wise to mount any high-power
components on the PCBs facing the walls of the chassis to take advantage of
the additional cooling provided by radiation.
Circuit boards that dissipate up to about 5 W of power (or that have a power
density of about 0.02 W/cm 2 ) can be cooled effectively by natural convection.
Heat transfer from PCBs can be analyzed by treating them as rectangular
plates with uniformly distributed heat sources on one side, and insulated on
the other side, since heat transfer from the back surfaces of the PCBs is usu-
ally small. For PCBs with electronic components mounted on both sides, the
rate of heat transfer and the heat transfer surface area will be twice as large.
It should be remembered that natural convection currents occur only in
gravitational fields. Therefore, there can be no heat transfer in space by nat-
ural convection. This will also be the case when the air passageways are
blocked and hot air cannot rise. In such cases, there will be no air motion, and
heat transfer through the air will be by convection.
The heat transfer from hot surfaces by natural convection and radiation can
be enhanced by attaching fins to the surfaces. The heat transfer in this case can
best be determined by using the data supplied by the manufacturers, as dis-
cussed in Chapter 3, especially for complex geometries.
Electronic box
FIGURE 15-40
Schematic for Example 15-10.
EXAMPLE 15-10 Cooling of a Sealed Electronic Box
Consider a sealed electronic box whose dimensions are 15 cm X 30 cm X 40
cm placed on top of a stand in a room at 35°C, as shown in Figure 15-40. The
box is painted, and the emissivity of its outer surface is 0.85. If the electronic
components in the box dissipate 75 W of power and the outer surface tempera-
ture of the box is not to exceed 65°C, determine if this box can be cooled by
natural convection and radiation alone. Assume the heat transfer from the bot-
tom surface of the box to the stand to be negligible.
SOLUTION The surface temperature of a sealed electronic box placed on top
of a stand is not to exceed 65°C. It is to be determined if this box can be cooled
by natural convection and radiation alone.
Assumptions 1 The box is located at sea level so that the local atmospheric
pressure is 1 atm. 2 The temperature of the surrounding surfaces is the same
as the air temperature in the room.
Analysis The sealed electronic box will lose heat from the top and the side
surfaces by natural convection and radiation. All four side surfaces of the box
cen58 93 3_chl5.qxd 9/9/2002 10:21 AM Page 817
can be treated as 0.15-m-high vertical surfaces. Then the natural convection
heat transfer from these surfaces is determined to be
L = 0.15 m
(2 X 0.4 m + 2 X 0.3 m)(0.15 m) = 0.21 m 2
K
AT
L
1.42i
65 - 35
0.15
5.34 W/m 2 -°C
t^conv, side "conv, side -^sidev-* 5 -* fluid/
= (5.34 W/m 2 • °C)(0.21 m 2 )(65 - 35)°C
= 33.6 W
Similarly, heat transfer from the horizontal top surface by natural convection is
determined to be
4A top 4(0.3 m)(0.4 m)
L = — =2(03 + 0.4) m = °- 34m
A top = (0.3 m)(0.4 m) = 0.12 m 2
h
= 1 32 :
conv. top A . — i-i j i
m -a
65- 35
0.34
4.05 W/m 2 ■ °C
s^conv, top "conv, top -^topV-* s J fluid/
= (4.05 W/m 2 • °C)(0.12 m 2 )(65 - 35)°C
= 14.6 W
Therefore, the natural convection heat transfer from the entire box is
t^conv x^conv, side s^c
33.6 + 14.6 = 48.2W
The box is completely surrounded by the surfaces of the room, and it is stated
that the temperature of the surfaces facing the box is equal to the air
temperature in the room. Then the rate of heat transfer from the box by radia-
tion can be determined from
Q md = sA s a(7? - T*J
= 0.85[(0.21 + 0.12)m 2 ](5.67 X 10- 8 W/m 2 X K 4 )
X [(65 + 273) 4 - (35 + 273) 4 ]K 4
= 64.5 W
Note that we must use absolute temperatures in radiation calculations. Then
the total heat transfer from the box is simply
titotal i^-c
Q,
48.2 + 64.5 = 112.7 W
which is greater than 75 W. Therefore, this box can be cooled by combined
natural convection and radiation, and there is no need to install any fans. There
is even some safety margin left for occasions when the air temperature rises
above 35°C.
EXAMPLE 15-11 Cooling of a Component by Natural Convection
A 0.2-W small cylindrical resistor mounted on a PCB is 1 cm long and has a
diameter of 0.3 cm, as shown in Figure 15-41. The view of the resistor is
817
CHAPTER 15
FIGURE 15-41
Schematic for Example 15-11.
cen58 93 3_chl5.qxd 9/9/2002 10:21 AM Page 818
818
HEAT TRANSFER
largely blocked by the PCB facing it, and the heat transfer from the connecting
wires is negligible. The air is free to flow through the parallel flow passages be-
tween the PCBs. If the air temperature at the vicinity of the resistor is 50°C, de-
termine the surface temperature of the resistor.
20 cm
oooooooo
□
-c
ooo oooo
15 cm
-7W
PCB
FIGURE 15-42
Schematic for Example 15-12.
SOLUTION A small cylindrical resistor mounted on a PCB is being cooled by
natural convection and radiation. The surface temperature of the resistor is to
be determined.
Assumptions 1 Steady operating conditions exist. 2 The device is located at
sea level so that the local atmospheric pressure is 1 atm. 3 Radiation is negli-
gible in this case since the resistor is surrounded by surfaces that are at about
the same temperature, and the net radiation heat transfer between two surfaces
at the same temperature is zero. This leaves natural convection as the only
mechanism of heat transfer from the resistor.
Analysis Using the relation for components on a circuit board from Table
15-1, the natural convection heat transfer coefficient for this cylindrical com-
ponent can be determined from
0.25
K
2.44
D
where the diameter D = 0.003 m, which is the length in the heat flow path, is
the characteristic length. We cannot determine h com yet since we do not know
the surface temperature of the component and thus AT. But we can substitute
this relation into the heat transfer relation to get
0.25
/i s -'fluicA
ficonv = "conv 4Uj ~~ •'fluid) = 2.44 — A(T „ ~ i fl uid )
D
2 AAA
(X s 7fi uid )
£,0.25
The heat transfer surface area of the component is
A, = 2X IttD 2 + ttDL = 2 X i-ir(0.3 cm) 2 + tt(0.3 cm)(l cm) = 1.084 cm 2
Substituting this and other known quantities in proper units (W for Q, °C for T,
m 2 for A, and m for D) into this equation and solving for T s yields
0.2 = 2.44(1.084 X lO" 4 )
(T s - 50) 12
0.003 - 25
-> T = 113°C
Therefore, the surface temperature of the resistor on the PCB will be 113°C,
which is considered to be a safe operating temperature for the resistors. Note
that blowing air to the circuit board will lower this temperature considerably as a
result of increasing the convection heat transfer coefficient and decreasing the
air temperature at the vicinity of the components due to the larger flow rate of air.
EXAMPLE 15-12 Cooling of a PCB in a Box by Natural Convection
A 15-cm X 20-cm PCB has electronic components on one side, dissipating a
total of 7 W, as shown in Figure 15-42. The PCB is mounted in a rack vertically
together with other PCBs. If the surface temperature of the components is not
■ to exceed 100 C C, determine the maximum temperature of the environment in
cen58 93 3_chl5.qxd 9/9/2002 10:21 AM Page 819
which this PCB can operate safely at sea level. What would your answer be if
this rack is located at a location at 4000 m altitude where the atmospheric
pressure is 61.66 kPa?
SOLUTION The surface temperature of a PCB is not to exceed 100°C. The
maximum environment temperatures for safe operaton at sea level and at 4000
m altitude are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Radiation heat transfer
is negligible since the PCB is surrounded by other PCBs at about the same tem-
perature. 3 Heat transfer from the back surface of the PCB will be very small
and thus negligible.
Analysis The entire heat load of the PCB is dissipated to the ambient air by nat-
ural convection from its front surface, which can be treated as a vertical flat plate.
Using the simplified relation for a vertical surface from Table 15-1, the nat-
ural convection heat transfer coefficient for this PCB can be determined from
K,
^f)
1.42
The characteristic length in this case is the height (L = 0.15 m) of the PCB,
which is the length in the path of heat flow. We cannot determine h com yet,
since we do not know the ambient temperature and thus A 7". But we can sub-
stitute this relation into the heat transfer relation to get
0.25
e c
"conv A S (T S
(T.
1.42A.
-'fluid) —
1 fluid'
1.42
A(T S 7fluid)
L 0.25
The heat transfer surface area of the PCB is
A s = (Width)(Height) = (0.2 m)(0.15 m) = 0.03 m 2
Substituting this and other known quantities in proper units (W for Q, °C for T,
m 2 for A s , and m for L) into this equation and solving for 7" fluid yields
(100 - T mii ) 125
7 = 1.42(0.03) ^j^ > 7/ fluid = 59.5°C
Therefore, the PCB will operate safely in environments with temperatures up to
59.4°C by relying solely on natural convection.
At an altitude of 4000 m, the atmospheric pressure is 61.66 kPa, which is
equivalent to
P = (61.66 kPa)
1 atm
101.325 kPa
0.609 atm
The heat transfer coefficient in this case is obtained by multiplying the value at
sea level by \fp, where P is in atm. Substituting
(100 - T^y) 1 - 25 ,
7 = 1.42(0.03)- " " — V0.609
50.6°C
which is about 10°C lower than the value obtained at 1 atm pressure. There-
fore, the effect of altitude on convection should be considered in high-altitude
applications.
819
CHAPTER 15
cen58 93 3_chl5.qxd 9/9/2002 10:21 AM Page 82C
820
HEAT TRANSFER
Electronic
components
in v Air inlet
Power consumed:
IV,,
■-™C p (T ml
TJ
-^-absorbed
FIGURE 15-43
In steady operation, the heat absorbed
by air per unit time as it flows through
an electronic box is equal to the power
consumed by the electronic
components in the box.
15-8 - AIR COOLING: FORCED CONVECTION
We mentioned earlier that convection heat transfer between a solid surface
and a fluid is proportional to the velocity of the fluid. The higher the velocity,
the larger the flow rate and the higher the heat transfer rate. The fluid veloci-
ties associated with natural convection currents are naturally low, and thus
natural convection cooling is limited to low-power electronic systems.
When natural convection cooling is not adequate, we simply add a fan and
blow air through the enclosure that houses the electronic components. In other
words, we resort to forced convection in order to enhance the velocity and thus
the flow rate of the fluid as well as the heat transfer. By doing so, we can in-
crease the heat transfer coefficient by a factor of up to about 10, depending on
the size of the fan. This means we can remove heat at much higher rates for a
specified temperature difference between the components and the air, or we
can reduce the surface temperature of the components considerably for a spec-
ified power dissipation.
The radiation heat transfer in forced-convection-cooled electronic systems
is usually disregarded for two reasons. First, forced convection heat transfer
is usually much larger than that due to radiation, and the consideration of ra-
diation causes no significant change in the results. Second, the electronic
components and circuit boards in convection-cooled systems are mounted so
close to each other that a component is almost entirely surrounded by other
components at about the same high temperature. That is, the components have
hardly any direct view of a cooler surface. This results in little or no radiation
heat transfer from the components. The components near the edges of circuit
boards with a large view of a cooler surface may benefit somewhat from the
additional cooling by radiation, and it is a good design practice to reserve
those spots for high-power components to have a thermally balanced system.
When heat transfer from the outer surface of the enclosure of the electronic
equipment is negligible, the amount of heat absorbed by the air becomes equal
to the amount of heat rejected (or power dissipated) by the electronic compo-
nents in the enclosure, and can be expressed as (Fig. 15-43)
Q = mC p (T m
(W)
(15-17)
where Q is the rate of heat transfer to the air; C p is the specific heat of air;
T in and T oat are the average temperatures of air at the inlet and exit of the
enclosure, respectively; and rh is the mass flow rate of air.
Note that for a specified mass flow rate and power dissipation, the temper-
ature rise of air, r out — T in , remains constant as it flows through the enclosure.
Therefore, the higher the inlet temperature of the air, the higher the exit tem-
perature, and thus the higher the surface temperature of the components. It is
considered a good design practice to limit the temperature rise of air to 10°C
and the maximum exit temperature of air to 70°C. In a properly designed
forced-air-cooled system, this results in a maximum component surface tem-
perature of under 100°C.
The mass flow rate of air required for cooling an electronic box depends on
the temperature of air available for cooling. In cool environments, such as an
air-conditioned room, a smaller flow rate will be adequate. However, in hot
environments, we may need to use a larger flow rate to avoid overheating the
components and the potential problems associated with it.
cen58 93 3_chl5.qxd 9/9/2002 10:21 AM Page 821
Forced convection is covered in detail in a separate chapter. For those who
skipped that chapter because of time limitations, here we present a brief
review of basic concepts and relations.
The fluid flow over a body such as a transistor is called external flow, and
flow through a confined space such as inside a tube or through the parallel pas-
sage area between two circuit boards in an enclosure is called internal flow
(Fig. 15-44). Both types of flow are encountered in a typical electronic system.
Fluid flow is also categorized as being laminar (smooth and streamlined)
or turbulent (intense eddy currents and random motion of chunks of fluid).
Turbulent flow is desirable in heat transfer applications since it results in a
much larger heat transfer coefficient. But it also comes with a much larger
friction coefficient, which requires a much larger fan (or pump for liquids).
Numerous experimental studies have shown that turbulence tends to occur
at larger velocities, during flow over larger bodies or flow through larger
channels, and with fluids having smaller viscosities. These effects are com-
bined into the dimensionless Reynolds number, defined as
Re
YD
v
(15-18)
where
T = velocity of the fluid (free-stream velocity for external flow and average
velocity for internal flow), m/s
D = characteristic length of the geometry (the length the fluid flows over in
external flow, and the equivalent diameter in internal flow), m
v = |x/p = kinematic viscosity of the fluid, m 2 /s.
The Reynolds number at which the flow changes from laminar to turbulent is
called the critical Reynolds number, whose value is 2300 for internal flow,
500,000 for flow over a flat plate, and 200,000 for flow over a cylinder or
sphere.
The equivalent (or hydraulic) diameter for internal flow is defined as
D„
AA C
(m)
(15-19)
where A c is the cross-sectional area of the flow passage and p is the perimeter.
Note that for a circular pipe, the hydraulic diameter is equivalent to the ordi-
nary diameter.
The convection heat transfer is expressed by Newton's law of cooling as
e c
hA s (T s
(W)
(15-20)
where
h
average convection heat transfer coefficient, W/m 2 • °C
A s = heat transfer surface area, m 2
T s = temperature of the surface, °C
Tfluid = temperature of the fluid sufficiently far from the surface for external
flow, and average temperature of the fluid at a specified location in
internal flow, °C
When the heat load is distributed uniformly on the surfaces with a constant
heat flux q, the total rate of heat transfer can also be expressed as Q = qA s .
821
CHAPTER 15
FIGURE 15-44
Internal flow through a circular tube
and external flow over it.
cen58 93 3_chl5.qxd 9/9/2002 10:21 AM Page 822
822
HEAT TRANSFER
^
r
Entry **-
-»- Fully developed ^>^
"I
1
region
region ^ — '
1
1
1 2 out
1 i fluid
1
1
1
T
m
n
AT- T -T =2s.
al ~ 's J fluid /,
1
1
1
1
1
-I— r
q = constant
I I I I I I I I I I I
TTTTTTTTTTT
FIGURE 15-45
Under constant heat flux conditions,
the surface and fluid temperatures
increase linearly, but their difference
remains constant in the fully
developed region.
In fully developed flow through a pipe or duct (i.e., when the entrance ef-
fects are negligible) subjected to constant heat flux on the surfaces, the con-
vection heat transfer coefficient h remains constant. In this case, both the
surface temperature T s and the fluid temperature T fiuid increase linearly, as
shown in Figure 15-45, but the difference between them, T s — T md , remains
constant. Then the temperature rise of the surface above the fluid temperature
can be determined from Eq. 15-20 to be
rp m XL- COflV
Ar
hA.
(°C)
(15-21)
Note that the temperature rise of the surface is inversely proportional to the
convection heat transfer coefficient. Therefore, the greater the convection
coefficient, the lower the surface temperature of the electronic components.
When the exit temperature of the fluid, T mt , is known, the highest surface
temperature that will occur at the end of the flow channel can be determined
from Eq. 15-21 to be
Q Q
T = T H — — = T H — —
s, max fluid, max u\ out fa a
(°C)
(15-22)
If this temperature is within the safe range, then we don't need to worry about
temperatures at other locations. But if it is not, it may be necessary to use a
larger fan to increase the flow rate of the fluid.
In convection analysis, the convection heat transfer coefficient h is usually
expressed in terms of the dimensionless Nusselt number Nu as
D
Nu
(W/m 2 • °C)
(15-23)
where k is the thermal conductivity of the fluid and D is the characteristic
length of the geometry. Relations for the average Nusselt number based on
experimental data are given in Table 15-2 for external flow and in Table 15-3
for laminar (Re < 2300) internal flow under a uniform heat flux condition,
which is closely approximated by electronic equipment. For turbulent flow
(Re > 2300) through smooth tubes and channels, the Nusselt number can be
determined from the Dittus-Boelter correlation,
Nu = 0.023 Re 08 Pr°
(15-24)
for any geometry. Here Pr is the dimensionless Prandtl number, and its value
is about 0.7 for air at room temperature.
The fluid properties in the above relations are to be evaluated at the
bulk mean fluid temperature T a ,
~(T in + T out ) for internal flow, which is
the arithmetic average of the mean fluid temperatures at the inlet and the exit
of the tube, and at the film temperature T n[m = UT S + T t[uid ) for external flow,
which is the arithmetic average of the surface temperature and free-stream
temperature of the fluid.
The relations in Table 15-3 for internal flow assume fully developed flow
over the entire flow section, and disregard the heat transfer enhancement
effects of the development region at the entrance. Therefore, the results
obtained from these relations are on the conservative side. We don't mind this
much, however, since it is common practice in engineering design to have
some safety margin to fall back to "just in case," as long as it does not result
cen58 93 3_chl5.qxd 9/9/2002 10:21 AM Page 823
TABLE 15-2
Empirical correlations for the average Nusselt number for forced convection
over a flat plate and circular and noncircular cylinders in cross flow
(From Jakob, Ref. 12, and Zukauskas, Ref. 17.)
Cross-section of the cylinder
Circle
Square
Square (tilted 45°)
Flat plate
Vertical plate
Fluid
Gas or
liquid
Gas
Gas
Gas or
liquid
Gas
Range of Re
0.4-4
4-40
40-4000
4000-40,000
40,000-400,000
5000-100,000
5000-100,000
0-5 X 10 5
5 X 10 5 -10 7
4000-15,000
Nusselt number
Nu = 0.989 Re°- 330 Pr 1/3
Nu = 0.911 Re°- 385 Pr 1/3
Nu = 0.683 Re°- 466 Pr 1/3
Nu = 0.193 Re°- 618 Pr 1/3
Nu = 0.027 Re°- 805 Pr 1/3
Nu = 0.102 Re°- 675 Pr L
Nu = 0.246 Re
8p r L
Nu = 0.664 Re 1/2 Pr 1/3
Nu = (0.037 Re 4 ' 5 - 871)Pr L
Nu = 0.228 Re°- 731 Pr 1/3
in a grossly overdesigned system. Also, it may sometimes be necessary to do
some local analysis for critical components with small surface areas to assure
reliability and to incorporate solutions to local problems such as attaching heat
sinks to high power components.
Fan Selection
Air can be supplied to electronic equipment by one or several fans. Although
the air is free and abundant, the fans are not. Therefore, a few words about the
fan selection are in order.
A fan at a fixed speed (or fixed rpm) will deliver a fixed volume of air re-
gardless of the altitude and pressure. But the mass flow rate of air will be less at
high altitude as a result of the lower density of air. For example, the atmospheric
823
CHAPTER 15
TABLE 15
-3
Nusselt number of fully developed
laminar flow in circular tubes and
rectangular channels
Cross-section
Aspect
Nusselt
of the tube
ratio
number
Circle
4.36
Square
—
3.61
Rectangle
alb
S^ ^^^
—
s'
1
3.61
f ,
2
4.12
b ^L^
3
4.79
4
5.33
a
6
6.05
8
6.49
00
8.24
cen58 93 3_chl5.qxd 9/9/2002 10:21 AM Page 824
824
HEAT TRANSFER
Crack
FIGURE 15-46
A fan placed at the exit of an
electronic box draws in air as well as
contaminants in the air through cracks.
pressure of air drops by more than 50 percent at an altitude of 6000 m from its
value at sea level. This means that the fan will deliver half as much air mass at
this altimde at the same rpm and temperature, and thus the temperature rise of
air cooling will double. This may create serious reliability problems and cata-
strophic failures of electronic equipment if proper precautions are not taken.
Variable-speed fans that automatically increase speed when the air density de-
creases are available to avoid such problems. Expensive electronic systems are
usually equipped with thermal cutoff switches to prevent overheating due to in-
adequate airflow rate or the failure of the cooling fan.
Fans draw in not only cooling air but also all kinds of contaminants that are
present in the air, such as lint, dust, moisture, and even oil. If unattended,
these contaminants can pile up on the components and plug up narrow pas-
sageways, causing overheating. It should be remembered that the dust that set-
tles on the electronic components acts as an insulation layer that makes it very
difficult for the heat generated in the component to escape. To minimize the
contamination problem, air filters are commonly used. It is good practice
to use the largest air filter practical to minimize the pressure drop of air and to
maximize the dust capacity.
Often the question arises about whether to place the fan at the inlet or the
exit of an electronic box. The generally preferred location is the inlet. A fan
placed at the inlet draws air in and pressurizes the electronic box and prevents
air infiltration into the box from cracks or other openings. Having only one lo-
cation for air inlet makes it practical to install a filter at the inlet to clean the
air from all the dust and dirt before they enter the box. This allows the elec-
tronic system to operate in a clean environment. Also, a fan placed at the inlet
handles cooler and thus denser air, which results in a higher mass flow rate for
the same volume flow rate or rpm. Since the fan is always subjected to cool
air, this has the added benefit that it increased the reliability and extends the
life of the fan. The major disadvantage associated with having a fan mounted
at the inlet is that the heat generated by the fan and its motor is picked up by
air on its way into the box, which adds to the heat load of the system.
When the fan is placed at the exit, the heat generated by the fan and its mo-
tor is immediately discarded to the atmosphere without getting blown first into
the electronic box. However, a fan at the exit creates a vacuum inside the box,
which draws air into the box through inlet vents as well as any cracks and
openings (Fig. 15^16). Therefore, the air is difficult to filter, and the dirt and
dust that collect on the components undermine the reliability of the system.
There are several types of fans available on the market for cooling elec-
tronic equipment, and the right choice depends on the situation on hand. There
are two primary considerations in the selection of the fan: the static pressure
head of the system, which is the total resistance an electronic system offers to
air as it passes through, and the volume flow rate of the air. Axial fans are sim-
ple, small, light, and inexpensive, and they can deliver a large flow rate. How-
ever, they are suitable for systems with relatively small pressure heads. Also,
axial fans usually run at very high speeds, and thus they are noisy. The radial
or centrifugal fans, on the other hand, can deliver moderate flow rates to sys-
tems with high-static pressure heads at relatively low speeds. But they are
larger, heavier, more complex, and more expensive than axial fans.
The performance of a fan is represented by a set of curves called the char-
acteristic curves, which are provided by fan manufacturers to help engineers
cen58 93 3_chl5.qxd 9/9/2002 10:21 AM Page 825
825
CHAPTER 15
with the selection of fans. A typical static pressure head curve for a fan is
given in Figure 15-47 together with a typical system flow resistance curve
plotted against the flow rate of air. Note that a fan creates the highest pressure
head at zero flow rate. This corresponds to the limiting case of blocked exit
vents of the enclosure. The flow rate increases with decreasing static head and
reaches its maximum value when the fan meets no flow resistance.
Any electronic enclosure will offer some resistance to flow. The system re-
sistance curve is parabolic in shape, and the pressure or head loss due to this
resistance is nearly proportional to the square of the flow rate. The fan must
overcome this resistance to maintain flow through the enclosure. The design
of a forced convection cooling system requires the determination of the total
system resistance characteristic curve. This curve can be generated accurately
by measuring the static pressure drop at different flow rates. It can also be de-
termined approximately by evaluating the pressure drops.
A fan will operate at the point where the fan static head curve and the sys-
tem resistance curve intersect. Note that a fan will deliver a higher flow rate
to a system with a low flow resistance. The required airflow rate for a system
can be determined from heat transfer requirements alone, using the design
heat load of the system and the allowable temperature rise of air. Then the
flow resistance of the system at this flow rate can be determined analytically
or experimentally. Knowing the flow rate and the needed pressure head, it is
easy to select a fan from manufacturers' catalogs that will meet both of these
requirements.
Below we present some general guidelines associated with the forced-air
cooling of electronic systems.
Fan static
pressure curve
System resistance
curve (head loss)
Airflow rate
FIGURE 15-47
The airflow rate a fan delivers into an
electronic enclosure depends on the
flow resistance of that system as well
as the variation of the static head of
the fan with flow rate.
1. Before deciding on forced-air cooling, check to see if natural convection
cooling is adequate. If it is, which may be the case for low-power
systems, incorporate it and avoid all the problems associated with fans
such as cost, power consumption, noise, complexity, maintenance, and
possible failure.
2. Select a fan that is neither too small nor too large. An undersized fan
may cause the electronic system to overheat and fail. An oversized
fan will definitely provide adequate cooling, but it will needlessly
be larger and more expensive and will consume more power.
3. If the temperature rise of air due to the power consumed by the motor of
the fan is acceptable, mount the fan at the inlet of the box to pressurize
the box and filter the air to keep dirt and dust out (Fig. 15-48).
4. Position and size the air exit vents so that there is adequate airflow
throughout the entire box. More air can be directed to a certain
area by enlarging the size of the vent at that area. The total exit
areas should be at least as large as the inlet flow area to avoid the
choking of the airflow, which may result in a reduced airflow rate.
5. Place the most critical electronic components near the entrance, where
the air is coolest. Place the non-critical components that consume a lot
of power near the exit (Fig. 15-49).
6. Arrange the circuit boards and the electronic components in the box
such that the resistance of the box to airflow is minimized and thus
the flow rate of air through the box is maximized for the same fan speed.
Make sure that no hot air pockets are formed during operation.
Filter
Electronic
"\
^
I -
I -
Electronic
components __J^*
^__^r_ \
►Air
^outlet
Heat
Fan
-Motor
FIGURE 15-48
Installing the fan at the inlet keeps the
dirt and dust out, but the heat
generated by the fan motor in.
cen58 93 3_chl5.qxd 9/9/2002 10:21 AM Page 826
826
HEAT TRANSFER
Air outlet
urn
iC
,c
ic
iTr— 1
lc
18
oo
High-power
component
Critical
component
ttttt
Air inlet
FIGURE 15-49
Sensitive components should be
located near the inlet and high-power
components near the exit.
FIGURE 15-50
A desktop personal computer with
monitor and keyboard.
7. Consider the effect of altitude in high-altitude applications.
8. Try to avoid any flow sections that increase the flow resistance of
the systems, such as unnecessary corners, sharp turns, sudden
expansions and contractions, and very high velocities (greater than
7 m/s), since the flow resistance is nearly proportional to the flow
rate. Also, avoid very low velocities since these result in a poor
heat transfer performance and allow the dirt and the dust in the air
to settle on the components.
9. Arrange the system such that natural convection helps forced
convection instead of hurting it. For example, mount the PCBs
vertically, and blow the air from the bottom toward the top instead
of the other way around.
10. When the design calls for the use of two or more fans, a decision
needs to be made about mounting the fans in parallel or in series.
Fans mounted in series will boost the pressure head available and
are best suited for systems with a high flow resistance. Fans
connected in parallel will increase the flow rate of air and are
best suited for systems with small flow resistance.
Cooling Personal Computers
The introduction of the 4004 chip, the first general-purpose microprocessor,
by the Intel Corporation in the early 1970s marked the beginning of the elec-
tronics era in consumer goods, from calculators and washing machines to per-
sonal computers. The microprocessor, which is the "brain" of the personal
computer, is basically a DIP-type LSE package that incorporates a central pro-
cessing unit (CPU), memory, and some input/output capabilities.
A typical desktop personal computer consists of a few circuit boards
plugged into a mother board, which houses the microprocessor and the mem-
ory chips, as well as the network of interconnections enclosed in a formed
sheet metal chassis, which also houses the disk and CD-ROM drives. Con-
nected to this "magic" box are the monitor, a keyboard, a printer, and other
auxiliary equipment (Fig. 15-50). The PCBs are normally mounted vertically
on a mother board, since this facilitates better cooling.
A small and quiet fan is usually mounted to the rear or side of the chassis to
cool the electronic components. There are also louvers and openings on the
side surfaces to facilitate air circulation. Such openings are not placed on the
top surface, since many users would block them by putting books or other
things there, which will jeopardize safety, and a coffee or soda spill can cause
major damage to the system.
EXAMPLE 15-13 Forced-Air Cooling of a Hollow-Core PCB
Some strict specifications of electronic equipment require that the cooling air
not come into direct contact with the electronic components, in order to protect
them from exposure to the contaminants in the air. In such cases, heat gener-
ated in the components on a PCB must be conducted a long way to the walls of
the enclosure through a metal core strip or a heat frame attached to the PCB.
An alternative solution is the hollow-core PCB, which is basically a narrow duct
cen58 93 3_chl5.qxd 9/9/2002 10:21 AM Page 827
" of rectangular cross section made of thin glass-epoxy board with electronic
components mounted on both sides, as shown in Figure 15-51. Heat generated
in the components is conducted to the hollow core through a thin layer of epoxy
board and is then removed by the cooling air flowing through the core. Effective
* sealing is provided to prevent air leakage into the component chamber.
Consider a hollow-core PCB 12 cm high and 18 cm long, dissipating a total
of 40 W. The width of the air gap between the two sides of the PCB is 0.3 cm.
The cooling air enters the core at 20°C at a rate of 0.72 L/s. Assuming the heat
generated to be uniformly distributed over the two side surfaces of the PCB, de-
■ termine (a) the temperature at which the air leaves the hollow core and {b) the
highest temperature on the inner surface of the core.
SOLUTION A hollow-core PCB is cooled by forced air. The outlet temperature
of air and the highest surface temperature are to be determined.
Assumptions 1 Steady operating conditions exist. 2 The inner surfaces of
the duct are smooth. 3 Air is an ideal gas. 4 Operation is at sea level and thus
the atmospheric pressure is 1 atm. 5 The entire heat generated in electronic
components is removed by the air flowing through the hollow core.
Properties The temperature of air varies as it flows through the core, and so do
its properties. We will perform the calculations using property values at 25°C
from Table A-15 since the air enters at 20°C and its temperature will increase.
C„
1.184 kg/m 3
1007 J/ka • °C
Pr = 0.7296
k = 0.02551 W/m-°C
v = 1.562 X 10- 5 m 2 /s
After we calculate the exit temperature of air, we can repeat the calculations, if
necessary, using properties at the average temperature.
Analysis The cross-sectional area of the channel and its hydraulic diameter are
A c = (Height)(Width) = (0.12 m)(0.003 m) = 3.6 X 10- 4 m 2
4A C _ 4 X (3.6 X 10- 4 m 3 )
7"2X (0.12 + 0.003) m
D,
0.005854 m
The average velocity and the mass flow rate of air are
0.72 X 10- 3 m 2 /s
V_
3.6 X 10~ 4 m 2
2.0 m/s
m = pV = (1.184 kg/m 3 )(0.72 X 10" 3 m 3 /s) = 0.8525 X lO" 3 kg/s
(a) The temperature of air at the exit of the hollow core can be determined from
Q = mC p (T 0Ut - TJ
Solving for 7" out and substituting the given values, we obtain
Q
mC„
20°C +
40J/s
(0.8525 X 10- 3 kg/s)(1007J/kg • °C)
66.6°C
(b) The surface temperature of the channel at any location can be deter-
mined from
Gconv = hA s {T s — 7fl u id)
where the heat transfer surface area is
827
CHAPTER 15
Cooling
Electronic
component
FIGURE 15-51
The hollow-core PCB discussed in
Example 15-13.
cen58 93 3_chl5.qxd 9/9/2002 10:21 AM Page 828
828
HEAT TRANSFER
A s = 2A side = 2(Height)(Length) = 2(0.12 m)(0.18 m) = 0.0432 m 2
To determine the convection heat transfer coefficient, we first need to calculate
the Reynolds number:
Re
YD,, _ (2 m/s)(0.005854 m)
~ lr ~ 1.562 X 10- 5 m 2 /s
750 < 2300
Therefore, the flow is laminar, and, assuming fully developed flow, the Nusselt
number for the airflow in this rectangular cross section corresponding to the
aspect ratio alb = (12 cm)/(0.3 cm) = 40 ~ <» is determined from Table 15-3
to be
Nu = 8.24
and thus
h = -£- Nu
0.02551 W/m
; (8.24) = 35.9 W/m 2 • °C
0.005854 m
Then the surface temperature of the hollow core near the exit becomes
Q ^™„ , 40 w
T = T +
s, max out ^ a
66.6°C +
(35.9 W/m 2 • °C)(0.0432 m 2 )
92.4°C
Discussion Note that the temperature difference between the surface and the
air at the exit of the hollow core is 25.8°C. This temperature difference between
the air and the surface remains at that value throughout the core, since the heat
generated on the side surfaces is uniform and the convection heat transfer co-
efficient is constant. Therefore, the surface temperature of the core at the inlet
will be 20°C + 25.8°C = 45.8°C. In reality, however, this temperature will be
somewhat lower because of the entrance effects, which affect heat transfer fa-
vorably. The fully developed flow assumption gives somewhat conservative re-
sults but is commonly used in practice because it provides considerable
simplification in calculations.
fi-
PCB
TO 71
transistor
0.44 cm
0.53 cm
Itttttt
Air flow
T = 90 m/min
65°C
FIGURE 15-52
Schematic for Example 15-14.
EXAMPLE 15-14
Forced-Air Cooling of a Transistor Mounted
on a PCB
A TO 71 transistor with a height of 0.53 cm and a diameter of 0.44 cm is
mounted on a circuit board, as shown in Figure 15-52. The transistor is cooled
by air flowing over it at a velocity of 90 m/min. If the air temperature is 65°C
and the transistor case temperature is not to exceed 95°C, determine the
amount of power this transistor can dissipate safely.
SOLUTION A transistor mounted on a circuit board is cooled by air flowing
over it. The power dissipated when its case temperature is 90°C is to be deter-
mined.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas. 3 Oper-
ation is at sea level and thus the atmospheric pressure is 1 atm.
Properties The properties of air at 1 atm pressure and the film temperature of
T,^(T S + r fluid )/2 = (95 + 65)/2 = 80°C are
cen58 93 3_chl5.qxd 9/9/2002 10:21 AM Page 829
p = 0.9994 kg/m 3
C„ = 1008 J/kg • °C
0.7154
p
Pr
k = 0.02953 W/m • °C
v = 2.097 X 10- 5 m 2 /s
Analysis The transistor is cooled by forced convection through its cylindrical
surface as well as its flat top and bottom surfaces. The characteristic length for
flow over a cylinder is the diameter D = 0.0044 m. Then the Reynolds number
becomes
TD (fm/s)(0.0044m)
Re = — — = ; — — = 315
v 2.097 X 10- 5 m 2 /s
which falls into the range 40 to 4000. Using the corresponding relation from
Table 15-2 for the Nusselt number, we obtain
Nu = 0.683 Re 0466 Pr" 3 = 0.683(315) 0466 (0.7154)'
;.9i
and
Also,
D
Nu
0.02953 W/m ■ °C
0.0044 m
(8.91) = 59.8 W/m 2 • °C
A cyl = irDL = tt(0.0044 m)(0.0053 m) = 0.7326 X 10~ 4 m 2
Then the rate of heat transfer from the cylindrical surface becomes
Gcyl = "Acy](7j ~~ Tfluid)
= (59.8 W/m 2 • °C)(0.7326 X 10- 4 m 2 )(95 - 65)°C = 0.131 W
We now repeat the calculations for the top and bottom surfaces of the transis-
tor, which can be treated as flat plates of length L = 0.0044 m in the flow di-
rection (which is the diameter), and, using the proper relation from Table 15-2,
and
Also,
^flat
6 flat
Re =
Nu =
= k_
D
v
(|m/s)(0.0044m)
315
2.097 X 10- 5 m 2 /s
0.664 Re" 2 Pr" 3 = 0.664(315)" 2 (0.7154) 1/3
10.5
Nu
0.02953 W/m ■ °C
0.0044 m
(10.5) = 70.7 W/m 2 °C
A
top ' bottom
hAftJj s — ifl uid )
2Xi-irD 2
2 X iir(0.0044 m) 2 = 0.3041 X 10~ 4 m 2
(70.7 W/m 2 ■ °C)(0.3041 X 10- 4 m 2 )(95 - 65)°C
0.065 W
Therefore, the total rate of heat that can be dissipated from all surfaces of the
transistor is
G,<
Gey, + Gfl
(0.131 + 0.065) W = 0.196 W
which seems to be low. This value can be increased considerably by attaching
a heat sink to the transistor to enhance the heat transfer surface area and thus
heat transfer, or by increasing the air velocity, which will increase the heat
transfer coefficient.
829
CHAPTER 15
cen58 93 3_chl5.qxd 9/9/2002 10:21 AM Page 83C
830
HEAT TRANSFER
Exhaust
inlet
FIGURE 15-53
Schematic for Example 15-15.
EXAMPLE 15-15 Choosing a Fan to Cool a Computer
The desktop computer shown in Figure 15-53 is to be cooled by a fan. The
electronics of the computer consume 75 W of power under full-load conditions.
The computer is to operate in environments at temperatures up to 40°C and at
elevations up to 2000 m, where the atmospheric pressure is 79.50 kPa. The
exit temperature of air is not to exceed 70°C to meet reliability requirements.
Also, the average velocity of air is not to exceed 75 m/min at the exit of the
computer case, where the fan is installed, to keep the noise level down. Deter-
mine the flow rate of the fan that needs to be installed and the diameter of the
casing of the fan.
SOLUTION A desktop computer is to be cooled by a fan safely in hot environ-
ments and high elevations. The airflow rate of the fan and the diameter of the
casing are to be determined.
Assumptions 1 Steady operation under worst conditions is considered. 2 Air is
an ideal gas.
Properties The specific heat of air at the average temperature of (40 + 70)/2
= 55°C is 1007 J/kg ■ °C (Table A-15).
Analysis We need to determine the flow rate of air for the worst-case scenario.
Therefore, we assume the inlet temperature of air to be 40°C and the atmos-
pheric pressure to be 79.50 kPa and disregard any heat transfer from the outer
surfaces of the computer case. Note that any direct heat loss from the computer
case will provide a safety margin in the design.
Noting that all the heat dissipated by the electronic components is absorbed
by air, the required mass flow rate of air to absorb heat at a rate of 75 W can be
determined from
Q = mC p (T mt - TJ
Solving for m and substituting the given values, we obtain
fi 75 J/s
C P (T 0M - Tj (1007 J/kg ■ °C)(70 - 40)°C
= 0.00249 kg/s = 0.149 kg/min
In the worst case, the exhaust fan will handle air at 70°C. Then the density of
air entering the fan and the volume flow rate become
V
P_
RT
m
P~~
79.50 kPa
(0.287 kPa ■ mVkg • K)(70 + 275) K
0.149 kg/min
0.8076 ks/m 3
0.8076 kg/m 3
0.184 m 3 /min
Therefore, the fan must be able to provide a flow rate of 0.184 m 3 /min or 6.5
cfm (cubic feet per minute). Note that if the fan were installed at the inlet in-
stead of the exit, then we would need to determine the flow rate using the den-
sity of air at the inlet temperature of 40°C, and we would need to add the power
consumed by the motor of the fan to the heat load of 75 W. The result may be
a slightly smaller or larger fan, depending on which effect dominates.
For an average velocity of 75 m/min, the diameter of the duct in which the
fan is installed can be determined from
V = AT
ttD 2 T
cen58 93 3_chl5.qxd 9/9/2002 10:21 AM Page 831
Solving for D and substituting the known values, we obtain
\A X 0.184 m 3 /min)
D
4V
V"
tt(75 m/min)
0.056 m = 5.6 cm
Therefore, a fan with a casing diameter of 5.6 cm and a flow rate of 0.184
m 3 /min will meet the design requirements.
831
CHAPTER 15
EXAMPLE 15-16 Cooling of a Computer by a Fan
A computer cooled by a fan contains six PCBs, each dissipating 15 W of power,
as shown in Figure 15-54. The height of the PCBs is 15 cm and the length is
20 cm. The clearance between the tips of the components on the PCB and the
back surface of the adjacent PCB is 0.4 cm. The cooling air is supplied by a
20-W fan mounted at the inlet. If the temperature rise of air as it flows through
the case of the computer is not to exceed 10°C, determine (a) the flow rate of
the air the fan needs to deliver, (b) the fraction of the temperature rise of air
due to the heat generated by the fan and its motor, and (c) the highest allow-
able inlet air temperature if the surface temperature of the components is not
to exceed 90°C anywhere in the system.
SOLUTION A computer cooled by a fan, and the temperature rise of air is lim-
ited to 10°C. The flow rate of the air, the fraction of the temperature rise of air
caused by the fan and its motor, and maximum allowable air inlet temperature
are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas. 3 The
computer is located at sea level so that the local atmospheric pressure is 1 atm.
4 The entire heat generated by the electronic components is removed by air
flowing through the opening between the PCBs. 5 The entire power consumed
by the fan motor is transferred as heat to the cooling air. This is a conservative
approach since the fan and its motor are usually mounted to the chassis of the
electronic system, and some of the heat generated in the motor may be con-
ducted to the chassis through the mounting brackets.
Properties We use properties of air at 30 C C since the air enters at room tem-
perature, and the temperature rise of air is limited to just 10°C:
p = 1.164 kg/m 3
C p = 1005 J/kg • °C
Pr = 0.7282
k = 0.02588 W/m • °C
v = 1.608 X 10- 5 m 2 /s
Analysis Because of symmetry, we consider the flow area between the two
adjacent PCBs only. We assume the flow rate of air through all six channels to
be identical, and to be equal to one-sixth of the total flow rate,
(a) Noting that the temperature rise of air is limited to 10°C and that the power
consumed by the fan is also absorbed by the air, the total mass flow rate of air
through the computer can be determined from
Q = mC p (T aat - TJ
Solving for rh and substituting the given values, we obtain
Q (6 X 15 + 20)J/s
C P (T 0M - TJ (1007 J/kg • °C)(10°C)
0.01092 kg/s
20 cm
PCB, 15 W
FIGURE 15-54
Schematic of the computer discussed
in Example 15-16.
cen58 93 3_chl5.qxd 9/9/2002 10:21 AM Page 832
832
HEAT TRANSFER
Then the volume flow rate of air and the air velocity become
m 0.01092 kg/s
V
T
V_
A,.
1.164 kg/m 3
0.009381 m 3 /s
6 X (6 X 10- 4 m 2 )
0.009381 m 3 /s = 0.563 nrVmin
2.606 m/s
Therefore, the fan needs to supply air at a rate of 0.563 m 3 /min, or about
20 cfm.
(b) The temperature rise of air due to the power consumed by the fan motor can
be determined by assuming the entire 20 W of power consumed by the motor to
be transferred to air as heat:
AT
20J/s
Q fan _
~ihC v ~ (0.01092 kg/s)(1007 J/kg • °C)
1.8°C
Therefore, 18 percent of the temperature rise of air is due to the heat generated
by the fan motor. Note that the fraction of the power consumed by the fan is
also 18 percent of the total, as expected.
(c) The surface temperature of the channel at any location can be deter-
mined from
<7conv = Gconv'-^s = "\Ta ~ 'fluid)
where the heat transfer surface area is
A s = A sidc = (Height)(Length) = (0.15 m)(0.20 m) = 0.03 m 2
To determine the convection heat transfer coefficient, we first need to calculate
the Reynolds number. The cross-sectional area of the channel and its hydraulic
diameter are
A c = (Height)(Width) = (0.15 m)(0.004 m)
_ AA C 4 X (6 X 10~ 4 m 2 )
D ''~~P~~ 2 X (0.15 + 0.004) m
Then the Reynolds number becomes
6 X 10- 4 m 2
0.007792 m
Re
VD h _ (2.606 m/s)(0.007792 m)
~ ir ~ 1.606 X 10- 5 m 2 /s
1263 < 2300
Therefore, the flow is laminar, and, assuming fully developed flow, the Nusselt
number for the airflow in this rectangular cross-section corresponding to the as-
pect ratio alb = (15 cm)/(0.4 cm) = 37.5 ~ <* is determined from Table 15-3
to be
Nu = 8.24
and thus
D,
Nu
0.02588 W/m ■
0.007792 m
; (8.24) = 27.4 W/m 2 • °C
Disregarding the entrance effects, the temperature difference between the
surface of the PCB and the air anywhere along the channel is determined to be
cen58 93 3_chl5.qxd 9/9/2002 10:21 AM Page 833
(15W)/(0.03m 2 )
27.4 W/m 2 • °C
18.3°C
That is, the surface temperature of the components on the PCB will be 18.3°C
higher than the temperature of air passing by.
The highest air and component temperatures will occur at the exit. Therefore,
in the limiting case, the component surface temperature at the exit will be
90°C. The air temperature at the exit in this case will be
AT
90°C - 18.3°C = 71.7°C
Noting that the air experiences a temperature rise of 10°C between the inlet
and the exit, the inlet temperature of air is
Tkma* = ^u,, raa x - 10°C = (71.7 - 10)°C = 61.7°C
This is the highest allowable air inlet temperature if the surface temperature of
the components is not to exceed 90°C anywhere in the system.
It should be noted that the analysis presented above is approximate since we
have made some simplifying assumptions. However, the accuracy of the results
obtained is usually adequate for engineering purposes.
833
CHAPTER 15
15-9 - LIQUID COOLING
Liquids normally have much higher thermal conductivities than gases, and thus
much higher heat transfer coefficients associated with them. Therefore, liquid
cooling is far more effective than gas cooling. However, liquid cooling comes
with its own risks and potential problems, such as leakage, corrosion, extra
weight, and condensation. Therefore, liquid cooling is reserved for applications
involving power densities that are too high for safe dissipation by air cooling.
Liquid cooling systems can be classified as direct cooling and indirect cool-
ing systems. In direct cooling systems, the electronic components are in direct
contact with the liquid, and thus the heat generated in the components is trans-
ferred directly to the liquid. In indirect cooling systems, however, there is no
direct contact with the components. The heat generated in this case is first
transferred to a medium such as a cold plate before it is carried away by the liq-
uid. Liquid cooling systems are also classified as closed-loop and open-loop
systems, depending on whether the liquid is discarded or recirculated after it is
heated. In open-loop systems, tap water flows through the cooling system and
is discarded into a drain after it is heated. The heated liquid in closed-loop sys-
tems is cooled in a heat exchanger and recirculated through the system. Closed-
loop systems facilitate better temperature control while conserving water.
The electronic components in direct cooling systems are usually completely
immersed in the liquid. The heat transfer from the components to the liquid
can be by natural or forced convection or boiling, depending on the tempera-
ture levels involved and the properties of the fluids. Immersion cooling of
electronic devices usually involves boiling and thus very high heat transfer co-
efficients, as discussed in the next section. Note that only dielectric fluids can
be used in immersion or direct liquid cooling. This limitation immediately ex-
cludes water from consideration as a prospective fluid in immersion cooling.
cen58 93 3_chl5.qxd 9/9/2002 10:21 AM Page 834
834
HEAT TRANSFER
Expansion
tank
Heat
exchanger Liquid-cooled -
plate
FIGURE 15-55
Schematic of an indirect
liquid cooling system.
Fluorocarbon fluids such as FC75 are well suited for direct cooling and are
commonly used in such applications.
Indirect liquid cooling systems of electronic devices operate just like the
cooling system of a car engine, where the water (actually a mixture of water
and ethylene glycol) circulates through the passages around the cylinders of
the engine block, absorbing heat generated in the cylinders by combustion.
The heated water is then routed by the water pump to the radiator, where it is
cooled by air blown through the radiator coils by the cooling fan. The cooled
water is then rerouted to the engine to transfer more heat. To appreciate the
effectiveness of the cooling system of a car engine, it will suffice to say that
the temperatures encountered in the engine cylinders are typically much
higher than the melting temperatures of the engine blocks.
In an electronic system, the heat is generated in the components instead of
the combustion chambers. The components in this case are mounted on a metal
plate made of a highly conducting material such as copper or aluminum. The
metal plate is cooled by circulating a cooling fluid through tubes attached to it,
as shown in Figure 15-55. The heated liquid is then cooled in a heat exchanger,
usually by air (or sea water in marine applications), and is recirculated by a
pump through the tubes. The expansion and storage tank accommodates any
expansions and contractions of the cooling liquid due to temperature variations
while acting as a liquid reservoir.
The liquids used in the cooling of electronic equipment must meet several
requirements, depending on the specific application. Desirable characteristics
of cooling liquids include high thermal conductivity (yields high heat transfer
coefficients), high specific heat (requires smaller mass flow rate), low vis-
cosity (causes a smaller pressure drop, and thus requires a smaller pump), high
surface tension (less likely to cause leakage problems), high dielectric
strength (a must in direct liquid cooling), chemical inertness (does not react
with surfaces with which it comes into contact), chemical stability (does not
decompose under prolonged use), nontoxic (safe for personnel to handle), low
freezing and high boiling points (extends the useful temperature range), and
low cost. Different fluids may be selected in different applications because of
the different priorities set in the selection process.
The heat sinks or cold plates of an electronic enclosure are usually cooled by
water by passing it through channels made for this purpose or through tubes at-
tached to the cold plate. High heat removal rates can be achieved by circulating
water through these channels or tubes. In high-performance systems, a refrig-
erant can be used in place of water to keep the temperature of the heat sink at
subzero temperatures and thus reduce the junction temperatures of the elec-
tronic components proportionately. The heat transfer and pressure drop calcu-
lations in liquid cooling systems can be performed using appropriate relations.
Liquid cooling can be used effectively to cool clusters of electronic devices
attached to a tubed metal plate (or heat sink), as shown in Figure 15-56. Here
12 TO-3 cases, each dissipating up to 150 W of power, are mounted on a heat
sink equipped with tubes on the back side through which a liquid flows. The
thermal resistance between the case of the devices and the liquid is minimized
in this case, since the electronic devices are mounted directly over the cooling
lines. The case-to-liquid thermal resistance depends on the spacing between
the devices, the quality of the thermal contact between the devices and the
plate, the thickness of the plate, and the flow rate of the liquid, among other
cen5 8 93 3_chl5.qxd 9/9/2002 10:21 AM Page 835
835
CHAPTER 15
FIGURE 15-56
Liquid cooling of TO-3 packages
placed on top of the coolant line
(courtesy of Wakefield Engineering).
things. The tubed metal plate shown is 15.2 cm X 18 cm X 2.5 cm in size and
is capable of dissipating up to 2 kW of power.
The thermal resistance network of a liquid cooling system is shown in Fig-
ure 15-57. The junction temperatures of silicon- based electronic devices are
usually limited to 125°C. The junction-to-case thermal resistance of a device
is provided by the manufacturer. The case-to-liquid thermal resistance can be
determined experimentally by measuring the temperatures of the case and the
liquid, and dividing the difference by the total power dissipated. The liquid-
to-air thermal resistance at the heat exchanger can be determined in a similar
manner. That is,
R.
liquid, device
case-liquid
Q
(°C/W)
(15-25)
^liqu
liquid, hx
quid-air
Q
where r liqllid device and T liquid ta are the inlet temperatures of the liquid to the elec-
tronic device and the heat exchanger, respectively. The required mass flow rate
of the liquid corresponding to a specified temperature rise of the liquid as it
flows through the electronic systems can be determined from Eq. 15-17.
Electronic components mounted on liquid-cooled metal plates should be
provided with good thermal contact in order to minimize the thermal resis-
tance between the components and the plate. The thermal resistance can be
minimized by applying silicone grease or beryllium oxide to the contact sur-
faces and fastening the components tightly to the metal plate. The liquid cool-
ing of a cold plate with a large number of high-power components attached to
it is illustrated in Example 15-17.
EXAMPLE 15-17
Cooling of Power Transistors on a Cold Plate
by Water
A cold plate that supports 20 power transistors, each dissipating 40 W, is to be
cooled with water, as shown in Figure 15-58. Half of the transistors are at-
tached to the back side of the cold plate. It is specified that the temperature
rise of the water is not to exceed 3°C and the velocity of water is to remain un-
der 1 m/s. Assuming that 20 percent of the heat generated is dissipated from
Junction
Case
Liquid
R
case-liquid
R
liquid-ambient
Ambient 1
FIGURE 15-57
Thermal resistance network for a
liquid-cooled electronic device.
_ Water
inlet
_ Water
exit
FIGURE 15-58
Schematic for Example 15-17.
cen58 93 3_chl5.qxd 9/9/2002 10:21 AM Page 836
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HEAT TRANSFER
the components to the surroundings by convection and radiation, and the re-
maining 80 percent is removed by the cooling water, determine the mass flow
rate of water needed and the diameter of the pipe to satisfy the restriction im-
posed on the flow velocity. Also, determine the case temperature of the devices
if the total case-to-liquid thermal resistance is 0.030°C/W and the water enters
the cold plate at 35°C.
SOLUTION A cold plate is to be cooled by water. The mass flow rate of water,
the diameter of the pipe, and the case temperature of the transistors are to be
determined.
Assumptions 1 Steady operating conditions exist. 2 About 20 percent of the
heat generated is dissipated from the components to the surroundings by con-
vection and radiation.
Properties The properties of water at room temperature are p = 1000 kg/m 3
and C p = 4180 J/kg • °C.
Analysis Noting that each of the 20 transistors dissipates 40 W of power and
80 percent of this power must be removed by the water, the amount of heat that
must be removed by the water is
Q = (20 transistors)(40 W/transistor)(0.80) = 640 W
In order to limit the temperature rise of the water to 3°C, the mass flow rate of
water must be no less than
Q
640 J/s
^"p^ * rise
0.051 kg/s = 3.06kg/min
(4180 J/kg • °C)(3°C)
The mass flow rate of a fluid through a circular pipe can be expressed as
irD 2 ,
P A c r
-Y
Then the diameter of the pipe to maintain the velocity of water under 1 m/s is
determined to be
4m 4(0.051 kg/s)
D = i = i = 0.0081 m
' irpT V tt( 1 000 kg/m 3 )( 1 m/s)
0.81 cm
Noting that the total case-to-liquid thermal resistance is 0.030°C/W and the
water enters the cold plate at 35°C, the case temperature of the devices is de-
termined from Eq. 15-25 to be
T
1 liqu
quid, device
+ QR
case-liquid
35°C + (640 W)(0.03°C/W) = 54.2°C
The junction temperature of the device can be determined similarly by using the
junction-to-case thermal resistance of the device supplied by the manufacturer.
15-10 - IMMERSION COOLING
High-power electronic components can be cooled effectively by immersing
them in a dielectric liquid and taking advantage of the very high heat trans-
fer coefficients associated with boiling. Immersion cooling has been used
since the 1940s in the cooling of electronic equipment, but for many years its
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CHAPTER 15
use was largely limited to the electronics of high-power radar systems. The
miniaturization of electronic equipment and the resulting high heat fluxes
brought about renewed interest in immersion cooling, which had been largely
viewed as an exotic cooling technique.
You will recall from thermodynamics that, at a specified pressure, a fluid
boils isothermally at the corresponding saturation temperature. A large amount
of heat is absorbed during the boiling process, essentially in an isothermal
manner. Therefore, immersion cooling also provides a constant-temperature
bath for the electronic components and eliminates hot spots effectively.
The simplest type of immersion cooling system involves an external reser-
voir that supplies liquid continually to the electronic enclosure. The vapor
generated inside is simply allowed to escape to the atmosphere, as shown in
Figure 15-59. A pressure relief valve on the vapor vent line keeps the pressure
and thus the temperature inside at the preset value, just like the petcock of a
pressure cooker. Note that without a pressure relief valve, the pressure inside
the enclosure would be atmospheric pressure and the temperature would have
to be the boiling temperature of the fluid at the atmospheric pressure.
The open-loop-type immersion cooling system described here is simple,
but there are several impracticalities associated with it. First of all, it is heavy
and bulky because of the presence of an external liquid reservoir, and the fluid
lost through evaporation needs to be replenished continually, which adds to
the cost. Further, the release of the vapor into the atmosphere greatly limits the
fluids that can be used in such a system. Therefore, the use of open-loop im-
mersion systems is limited to applications that involve occasional use and thus
have a light duty cycle.
More sophisticated immersion cooling systems operate in a closed loop in
that the vapor is condensed and returned to the electronic enclosure instead of
being purged to the atmosphere. Schematics of two such systems are given in
Figure 15-60. The first system involves a condenser external to the electronics
Liquid
Safety
valve
Vapor
lo
-1
3
:
l
~>
">
j
j
J
6
5
1
1
B
1
J
Dielectric
liquid
FIGURE 15-59
A simple open-loop
immersion cooling system.
Safety
valve
ve Air
U i n n n h r
Air-cooled
condenser
Safety
valve
Vapor
Fins
(C
"b
3
"3
3
*{
I
J
«j
II
P
B
J
Coolant
exit
Coolant
inlet
Dielectric
liquid
(a) System with external condenser
Electronic
components
Dielectric
liquid
(b) System with internal condenser
FIGURE 15-60
The schematics of two closed-loop
immersion cooling systems.
cen58 93 3_chl5.qxd 9/9/2002 10:21 AM Page 838
838
HEAT TRANSFER
FIGURE 15-61
The schematics of two all-liquid
immersion cooling systems.
Expansion
chamber
Pressure
8 relief valve
Expansion
chamber
Electronic
components
(a) System with internal cooling
Pressure
B relief
valve Fins
Electronic Dielectric Fan
components liquid
(b) System with external cooling
enclosure, and the vapor leaving the enclosure is cooled by a cooling fluid such
as air or water outside the enclosure. The condensate is returned to the enclo-
sure for reuse. The condenser in the second system is actually submerged in the
electronic enclosure and is part of the electronic system. The cooling fluid in
this case circulates through the condenser tube, removing heat from the vapor.
The vapor that condenses drips on top of the liquid in the enclosure and con-
tinues to recirculate.
The performance of closed-loop immersion cooling systems is most sus-
ceptible to the presence of noncondensable gases such as air in the vapor
space. An increase of 0.5 percent of air by mass in the vapor can cause the
condensation heat transfer coefficient to drop by a factor of up to 5. Therefore,
the fluid used in immersion cooling systems should be degassed as much as
practical, and care should be taken during the filling process to avoid intro-
ducing any air into the system.
The problems associated with the condensation process and noncondens-
able gases can be avoided by submerging the condenser (actually, heat ex-
changer tubes in this case) in the liquid instead of the vapor in the electronic
enclosure, as shown in Figure 15— 61a. The cooling fluid, such as water, cir-
culating through the tubes absorbs heat from the dielectric liquid at the top
portion of the enclosure and subcools it. The liquid in contact with the elec-
tronic components is heated and may even be vaporized as a result of absorb-
ing heat from the components. But these vapor bubbles collapse as they move
up, as a result of transferring heat to the cooler liquid with which they come
in contact. This system can still remove heat at high rates from the surfaces
of electronic components in an isothermal manner by utilizing the boiling
process, but its overall capacity is limited by the rate of heat that can be
removed by the external cooling fluid in a liquid-to-liquid heat exchanger.
Noting that the heat transfer coefficients associated with forced convection are
far less than those associated with condensation, this all-liquid immersion
cooling system is not suitable for electronic boxes with very high power dis-
sipation rates per unit volume.
A step further in the all-liquid immersion cooling systems is to remove the
heat from the dielectric liquid directly from the outer surface of the electron-
ics enclosure, as shown in Figure 15-61&. In this case, the dielectric liquid
cen58 93 3_chl5.qxd 9/9/2002 10:21 AM Page 839
Air 1-3 atm
Fluorochemical vapor I
Silicone oil
Transformer oil
Fluorochemical liquids
Air 1-3 atm
Fluorochemical vapor I
Transformer oil
Fluorochemical liquids
Boiling fluorochemical liquids
10
izzi
□
> Natural convection
100
W/m 2 -°C
Forced
convection
1000
10,000
839
CHAPTER 15
FIGURE 15-62
Typical heat transfer coefficients for
various dielectric fluids (from
Hwang and Moran, Ref. 11).
inside the sealed enclosure is heated as a result of absorbing the heat dissi-
pated by the electronic components. The heat is then transferred to the walls
of the enclosure, where it is removed by external means. This immersion cool-
ing technique is the most reliable of all since it does not involve any penetra-
tion into the electronics enclosure and the components reside in a completely
sealed liquid environment. However, the use of this system is limited to ap-
plications that involve moderate power dissipation rates. The heat dissipation
is limited by the ability of the system to reject the heat from the outer surface
of the enclosure. To enhance this ability, the outer surfaces of the enclosures
are often finned, especially when the enclosure is cooled by air.
Typical ranges of heat transfer coefficients for various dielectric fluids suit-
able for use in the cooling of electronic equipment are given in Figure 15-62
for natural convection, forced convection, and boiling. Note that extremely
high heat transfer coefficients (from about 1500 to 6000 W/m 2 • °C) can be at-
tained with the boiling of fluorocarbon fluids such as FC78 and FC86 manu-
factured by the 3M company. Fluorocarbon fluids, not to be confused with the
ozone-destroying fluorochloro fluids, are found to be very suitable for im-
mersion cooling of electronic equipment. They have boiling points ranging
from 30°C to 174°C and freezing points below — 50°C. They are nonflamma-
ble, chemically inert, and highly compatible with materials used in electronic
equipment.
Experimental results for the power dissipation of a chip having a heat
transfer area of 0.457 cm 2 and its substrate during immersion cooling in an
FC86 bath are given in Figure 15-63. The FC86 liquid is maintained at a
bulk temperature of 5°C during the experiments by the use of a heat ex-
changer. Heat transfer from the chip is by natural convection in regime A-B,
and bubble formation and thus boiling begins in regime B-C. Note that the
chip surface temperature suddenly drops with the onset of boiling because of
the high heat transfer coefficients associated with boiling. Heat transfer is by
nucleate boiling in regime C-D, and very high heat transfer rates can be
achieved in this regime with relatively small temperature differences.
Chip
power, W
20
liquid
= 5°C
I
/
■
I
f
f
I —
B
Ay
Heat flux,
W/m 2
43.8
21.9
17.5
10
20
13.1
11.0
6.6
4.4
2.2
chip
30 40 50 70 100
J liquid' *-
FIGURE 15-63
Heat transfer from a chip immersed in
the fluorocarbon fluid FC86 (from
Hwang and Moran, Ref. 11).
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840
HEAT TRANSFER
4W
Logic chip
*80°C
Dielectric
fluid
20°C
FIGURE 15-64
Schematic for Example 15-1 i
EXAMPLE 15-18 Immersion Cooling of a Logic Chip
A logic chip used in an IBM 3081 computer dissipates 4 W of power and has a
heat transfer surface area of 0.279 cm 2 , as shown in Figure 15-64. If the sur-
face of the chip is to be maintained at 80°C while being cooled by immersion
in a dielectric fluid at 20°C, determine the necessary heat transfer coefficient
and the type of cooling mechanism that needs to be used to achieve that heat
transfer coefficient.
SOLUTION A logic chip is to be cooled by immersion in a dielectric fluid. The
minimum heat transfer coefficient and the type of cooling mechanism are to be
determined.
Assumptions Steady operating conditions exist.
Analysis The average heat transfer coefficient over the surface of the chip can
be determined from Newton's law of cooling,
Q = hA s (T t
chip
Tfluid)
Solving for h and substituting the given values, the convection heat transfer co-
efficient is determined to be
Q
4W
A s (T chip - r„ uid ) (0.279 X 10- 4 m 2 )(80 - 20)°C
2390 W/m 2 • °C
which is rather high. Examination of Figure 15-62 reveals that we can obtain
such high heat transfer coefficients with the boiling of fluorocarbon fluids.
Therefore, a suitable cooling technique in this case is immersion cooling in
such a fluid. A viable alternative to immersion cooling is the thermal conduc-
tion module discussed earlier.
FIGURE 15-65
Schematic for Example 15-
19.
EXAMPLE 15-19 Cooling of a Chip by Boiling
An 8-W chip having a surface area of 0.6 cm 2 is cooled by immersing it into
FC86 liquid that is maintained at a temperature of 15°C, as shown in Figure
15-65. Using the boiling curve in Figure 15-63, estimate the temperature of
the chip surface.
SOLUTION A chip is cooled by boiling in a dielectric fluid. The surface tem-
perature of the chip is to be determined.
Assumptions The boiling curve in Figure 15-63 is prepared for a chip having a
surface area of 0.457 cm 2 being cooled in FC86 maintained at 5°C. The chart
can be used for similar cases with reasonable accuracy.
Analysis The heat flux is
Q 8 w
k = T" = , = 13-3 W/cm 2
1 A s 0.6 cm 2
Corresponding to this value on the chart is 7" chip - 7" f | Uid = 60°C. Therefore,
'chip
60 = 15 + 60 = 75°C
That is, the surface of this 8-W chip will be at 75°C as it is cooled by boiling in
the dielectric fluid FC86.
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CHAPTER 15
A liquid- based cooling system brings with it the possibility of leakage and
associated reliability concerns. Therefore, the consideration of immersion
cooling should be limited to applications that require precise temperature con-
trol and those that involve heat dissipation rates that are too high for effective
removal by conduction or air cooling.
SUMMARY
Electric current flow through a resistance is always accompa-
nied by heat generation, and the essence of thermal design is
the safe removal of this internally generated heat by providing
an effective path for heat flow from electronic components to
the surrounding medium. In this chapter, we have discussed
several cooling techniques commonly used in electronic
equipment, such as conduction cooling, natural convection and
radiation cooling, forced-air convection cooling, liquid cool-
ing, immersion cooling, and heat pipes.
In a chip carrier, heat generated at the junction is conducted
along the thickness of the chip, the bonding material, the lead
frame, the case material, and the leads. The junction-to-case
thermal resistance ^ Junction . casc represents the total resistance
to heat transfer between the junction of a component and its
case. This resistance should be as low as possible to minimize
the temperature rise of the junction above the case temperature.
The epoxy board used in PCBs is a poor heat conductor, and so
it is necessary to use copper cladding or to attach the PCB to a
heat frame in conduction-cooled systems.
Low-power electronic systems can be cooled effectively
with natural convection and radiation. The heat transfer from a
surface at temperature T s to a fluid at temperature T Mli by con-
vection is expressed as
Gc
hA s (T s - r fluid )
(W)
where h is the convection heat transfer coefficient and A s is the
heat transfer surface area. The natural convection heat transfer
coefficient for laminar flow of air at atmospheric pressure is
given by a simplified relation of the form
/An 0,25
h = Kl^j-\ (W/m 2 -°C)
where AT = T s — T md is the temperature difference between
the surface and the fluid, L is the characteristic length (the
length of the body along the heat flow path), and K is a con-
stant, whose value is given in Table 15—1. The relations in
Table 15-1 can also be used at pressures other than 1 atm by
multiplying them by y/p, where P is the air pressure in atm.
Radiation heat transfer between a surface at temperature 7^
completely surrounded by a much larger surface at temperature
7^ can be expressed as
where e is the emissivity of the surface, A s is the heat transfer sur-
face area, and cr is the Stefan-Boltzmann constant, whose value
is ct = 5.67 X 1(T 8 W/m 2 ■ K 4 = 0.1714 X 1(T 8 Btu/h • ft 2 • R 4 .
Fluid flow over a body such as a transistor is called external
flow, and flow through a confined space such as inside a tube
or through the parallel passage area between two circuit boards
in an enclosure is called internal flow. Fluid flow is also cate-
gorized as being laminar or turbulent, depending on the
value of the Reynolds number. In convection analysis, the con-
vection heat transfer coefficient is usually expressed in terms
of the dimensionless Nusselt number Nu as
h = jj Nu
(W/m 2 • °C)
where k is thermal conductivity of the fluid and D is the char-
acteristic length of the geometry. Relations for the average
Nusselt number based on experimental data are given in Table
1 5-2 for external flow and in Table 1 5-3 for laminar internal
flow under the uniform heat flux condition, which is closely
approximated by electronic equipment. In forced-air-cooled
systems, the heat transfer can also be expressed as
Q = mCJT 0Ut - TJ
(W)
g rad = eAMT?~ r 4 rr )
(W)
where Q is the rate of heat transfer to the air; C p is the specific
heat of air; T ln and T OM are the average temperatures of air at
the inlet and exit of the enclosure, respectively; and m is the
mass flow rate of air.
The heat transfer coefficients associated with liquids are usu-
ally an order of magnitude higher than those associated with
gases. Liquid cooling systems can be classified as direct cool-
ing and indirect cooling systems. In direct cooling systems,
the electronic components are in direct contact with the liquid,
and thus the heat generated in the components is transferred di-
rectly to the liquid. In indirect cooling systems, however, there
is no direct contact with the components. Liquid cooling
systems are also classified as closed-loop and open-loop sys-
tems, depending on whether the liquid is discharged or recir-
culated after it is heated. Only dielectric fluids can be used in
immersion or direct liquid cooling.
High-power electronic components can be cooled effectively
by immersing them in a dielectric liquid and taking advan-
tage of the very high heat transfer coefficients associated with
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HEAT TRANSFER
boiling. The simplest type of immersion cooling system in-
volves an external reservoir that supplies liquid continually
to the electronic enclosure. This open-loop-type immersion
cooling system is simple but often impractical. Immersion
REFERENCES AND SUGGESTED READING
1. E. P. Black and E. M. Daley. "Thermal Design
Considerations for Electronic Components." ASME Paper
No. 70-DE-17, 1970.
2. R. A. Colclaser, D. A. Neaman, and C. F. Hawkins.
Electronic Circuit Analysis. New York: John Wiley &
Sons, 1984.
3. J. W. Dally. Packaging of Electronic Systems. New
York: McGraw-Hill, 1990.
4. Design Manual of Cooling Methods for Electronic
Equipment. NAVSHIPS 900-190. Department of the
Navy, Bureau of Ships, March 1955.
5. Design Manual of Natural Methods of Cooling
Electronic Equipment. NAVSHIPS 900-192.
Department of the Navy, Bureau of Ships, November
1956.
6. G. N. Ellison. Thermal Computations for Electronic
Equipment. New York: Van Nostrand Reinhold, 1984.
7. J. A. Gardner. "Liquid Cooling Safeguards High-Power
Semiconductors." Electronics, February 24, 1974, p. 103.
8. R. S. Gaugler. "Heat Transfer Devices." U.S. Patent
2350348, 1944.
9. G. M. Grover, T. P. Cotter, and G. F. Erickson. "Structures
of Very High Thermal Conductivity." Journal of
Applied Physics 35 (1964), pp. 1190-1191.
PROBLEMS
cooling systems usually operate in a closed loop, in that the
vapor is condensed and returned to the electronic enclosure in-
stead of being purged to the atmosphere.
10. W. F. Hilbert and F H. Kube. "Effects on Electronic
Equipment Reliability of Temperature Cycling in
Equipment." Report No. EC-69-400. Final Report,
Grumman Aircraft Engineering Corporation, Bethpage,
NY, February 1969.
11. U. P. Hwang and K. P. Moran. "Boiling Heat Transfer of
Silicon Integrated Circuits Chip Mounted on a Substrate."
ASME HTD 20 (1981), pp. 53-59.
12. M. Jakob. Heat Transfer. Vol. 1. New York: John Wiley
& Sons, 1949.
13. R. D. Johnson. "Enclosures — State of the Art." New
Electronics, July 29, 1983, p. 29.
14. A. D. Kraus and A. Bar-Cohen. Thermal Analysis and
Control of Electronic Equipment. New York:
McGraw-Hill/Hemisphere, 1983.
15. Reliability Prediction of Electronic Equipment. U.S.
Department of Defense, MIL-HDBK-2178B, NTIS,
Springfield, VA, 1974.
16. D. S. Steinberg. Cooling Techniques for Electronic
Equipment. New York: John Wiley & Sons, 1980.
17. A. Zukauskas. "Heat Transfer from Tubes in Cross Flow."
In Advances in Heat Transfer, vol. 8, ed. J. P. Hartnett
and T. F. Irvine, Jr. New York: Academic Press, 1972.
Introduction and History
15-1C What invention started the electronic age? Why did
the invention of the transistor mark the beginning of a revolu-
tion in that age?
15-2C What is an integrated circuit? What is its significance
in the electronics era? What do the initials MSI, LSI, and VLSI
stand for?
*Problems designated by a "C" are concept questions, and
students are encouraged to answer them all. Problems designated
by an "E" are in English units, and the SI users can ignore them.
Problems with an EES-CD icon ® are solved using EES, and
complete solutions together with parametric studies are included
on the enclosed CD. Problems with a computer-EES icon H are
comprehensive in nature, and are intended to be solved with a
computer, preferably using the EES software that accompanies
this text.
15-3C When electric current /passes through an electrical
element having a resistance R, why is heat generated in the el-
ement? How is the amount of heat generation determined?
15-4C Consider a TV that is wrapped in the blankets from
all sides except its screen. Explain what will happen when the
TV is turned on and kept on for a long time, and why. What
will happen if the TV is kept on for a few minutes only?
15-5C Consider an incandescent light bulb that is com-
pletely wrapped in a towel. Explain what will happen when the
light is turned on and kept on. (PS. Do not try this at home!)
15-6C A businessman ties a large cloth advertisement ban-
ner in front of his car such that it completely blocks the airflow
to the radiator. What do you think will happen when he starts
the car and goes on a trip?
cen58 93 3_chl5.qxd 9/9/2002 10:21 AM Page 843
15-7C Which is more likely to break: a car or a TV? Why?
15-8C Why do electronic components fail under prolonged
use at high temperatures?
15-9 The temperature of the case of a power transistor that is
dissipating 12 W is measured to be 60°C. If the junction-to-
case thermal resistance of this transistor is specified by the
manufacturer to be 5°C/W, determine the junction temperature
of the transistor. Answer: 120°C
15-10 Power is supplied to an electronic component from a
12-V source, and the variation in the electric current, the junc-
tion temperature, and the case temperatures with time are ob-
served. When everything is stabilized, the current is observed
to be 0.15 A and the temperatures to be 80°C and 55°C at the
junction and the case, respectively. Calculate the junction-to-
case thermal resistance of this component.
15-11 A logic chip used in a computer dissipates 6 W of
power in an environment at 55°C and has a heat transfer sur-
face area of 0.32 cm 2 . Assuming the heat transfer from the sur-
face to be uniform, determine (a) the amount of heat this chip
dissipates during an 8-h work day, in kWh, and (b) the heat
flux on the surface of the chip, in W/cm 2 .
15-12 A 15-cm X 20-cm circuit board houses 90 closely
spaced logic chips, each dissipating 0.1 W, on its surface. If the
heat transfer from the back surface of the board is negligible,
determine (a) the amount of heat this circuit board dissipates
during a 10-h period, in kWh, and (b) the heat flux on the sur-
face of the circuit board, in W/cm 2 .
15 cm
Chips
FIGURE P1 5-1 2
15-13E A resistor on a circuit board has a total thermal re-
sistance of 130°F7W If the temperature of the resistor is not to
exceed 360°F, determine the power at which it can operate
safely in an ambient at 120°F.
15-14 Consider a 0.1 -kfl resistor whose surface-to-ambient
thermal resistance is 300°C/W. If the voltage drop across the
843
CHAPTER 15
resistor is 7.5 V and its surface temperature is not to exceed
1 50°C, determine the power at which it can operate safely in an
ambient at 30°C. Answer: 0.4 W
15-15 rfi£M Reconsider Problem 15-14. Using EES (or
kS other) software, plot the power at which the re-
sistor can operate safely as a function of the ambient tempera-
ture as the temperature varies from 20°C to 40°C, and discuss
the results.
Manufacturing of Electronic Equipment
15-16C Why is a chip in a chip carrier bonded to a lead
frame instead of the plastic case of the chip carrier?
15-17C Draw a schematic of a chip carrier, and explain how
heat is transferred from the chip to the medium outside of the
chip carrier.
15-18C What does the junction-to-case thermal resistance
represent? On what does it depend for a chip carrier?
15-19C What is a hybrid chip carrier? What are the advan-
tages of hybrid electronic packages?
15-20C What is a PCB? Of what is the board of a PCB
made? What does the "device-to-PCB edge" thermal resistance
in conduction-cooled systems represent? Why is this resistance
relatively high?
15-21C What are the three types of printed circuit boards?
What are the advantages and disadvantages of each type?
15-22C What are the desirable characteristics of the materi-
als used in the fabrication of the circuit boards?
15-23C What is an electronic enclosure? What is the pri-
mary function of the enclosure of an electronic system? Of
what materials are the enclosures made?
Cooling Load of Electronic Equipment and
Thermal Environment
15-24C Consider an electronics box that consumes 120 W of
power when plugged in. How is the heating load of this box
determined?
15-25C Why is the development of superconducting materi-
als generating so much excitement among designers of elec-
tronic equipment?
15-26C How is the duty cycle of an electronic system de-
fined? How does the duty cycle affect the design and selection
of a cooling technique for a system?
15-27C What is temperature cycling? How does it affect the
reliability of electronic equipment?
15-28C What is the ultimate heat sink for (a) a TV, (b) an
airplane, and (c) a ship? For each case, what is the range of
temperature variation of the ambient?
15-29C What is the ultimate heat sink for (a) a VCR, (b) a
spacecraft, and (c) a communication system on top of a moun-
tain? For each case, what is the range of temperature variation
of the ambient?
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HEAT TRANSFER
Electronics Cooling in Different Applications
15-30C How are the electronics of short-range and long-
range missiles cooled?
15-31C What is dynamic temperature? What causes it? How
is it determined? At what velocities is it significant?
15-32C How are the electronics of a ship or submarine
cooled?
15-33C How are the electronics of the communication sys-
tems at remote areas cooled?
15-34C How are the electronics of high-power microwave
equipment such as radars cooled?
15-35C How are the electronics of a space vehicle cooled?
15-36 Consider an airplane cruising in the air at a tempera-
ture of — 25°C at a velocity of 850 km/h. Determine the tem-
perature rise of air at the nose of the airplane as a result of the
ramming effect of the air.
850 km/h
FIGURE P1 5-36
15-37 The temperature of air in high winds is measured by a
thermometer to be 12°C. Determine the true temperature of air
if the wind velocity is 90 km/h. Answer: 11.7°C
15-38 Tu'M Reconsider Problem 15-37. Using EES (or
k^S other) software, plot the true temperature of air
as a function of the wind velocity as the velocity varies from
20 km/h to 120 km/h, and discuss the results.
15-39 Air at 25°C is flowing in a channel at a velocity of
(a) \,{b) 10, (c) 100, and (d) 1000 m/s. Determine the temper-
ature that a stationary probe inserted into the channel will read
for each case.
15—40 An electronic device dissipates 2 W of power and has
a surface area of 5 cm 2 . If the surface temperature of the device
is not to exceed the ambient temperature by more than 50°C,
determine a suitable cooling technique for this device. Use
Figure 15-17.
15— 41E A stand-alone circuit board, 6 in. X 8 in. in size, dis-
sipates 20 W of power. The surface temperature of the board is
not to exceed 165°F in an 85°F environment. Using Figure
15-17 as a guide, select a suitable cooling mechanism.
Conduction Cooling
15-42C What are the major considerations in the selection of
a cooling technique for electronic equipment?
15-43C What is thermal resistance? To what is it analogous
in electrical circuits? Can thermal resistance networks be ana-
lyzed like electrical circuits? Explain.
15-44C If the rate of heat conduction through a medium and
the thermal resistance of the medium are known, how can the
temperature difference between the two sides of the medium be
determined?
15-45C Consider a wire of electrical resistance R, length L,
and cross-sectional area A through which electric current / is
flowing. How is the voltage drop across the wire determined?
What happens to the voltage drop when L is doubled while / is
held constant?
Now consider heat conduction at a rate of Q through the
same wire having a thermal resistance of R. How is the tem-
perature drop across the wire determined? What happens to the
temperature drop when L is doubled while Q is held constant?
^Wire
^
3—
FIGURE P15-45C
15-46C What is a heat frame? How does it enhance heat
transfer along a PCB? Which components on a PCB attached
to a heat frame operate at the highest temperatures: those at the
middle of the PCB or those near the edge?
15-47C What is constriction resistance in heat flow? To what
is it analogous in fluid flow through tubes and electric current
flow in wires?
15-48C What does the junction-to-case thermal resistance of
an electronic component represent? In practice, how is this
value determined? How can the junction temperature of a com-
ponent be determined when the case temperature, the power
dissipation of the component, and the junction-to-case thermal
resistance are known?
15-49C What does the case-to-ambient thermal resistance of
an electronic component represent? In practice, how is this
value determined? How can the case temperature of a compo-
nent be determined when the ambient temperature, the power
dissipation of the component, and the case-to-ambient thermal
resistance are known?
15-50C Consider an electronic component whose junction-to-
case thermal resistance ^j unction . C! , se is provided by the manufac-
turer and whose case-to-ambient thermal resistance -/? casc . am bient is
determined by the thermal designer. When the power dissipation
of the component and the ambient temperature are known,
explain how the junction temperature can be determined. When
^ junction-case is greater than tf case _ arabient , will the case temperature
be closer to the junction or ambient temperature?
15-51C Why is the rate of heat conduction along a PCB very
low? How can heat conduction from the mid-parts of a PCB to
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CHAPTER 15
its outer edges be improved? How can heat conduction across
the thickness of the PCB be improved?
15-52C Why is the warping of epoxy boards that are copper-
cladded on one side a major concern? What is the cause of this
warping? How can the warping of PCBs be avoided?
15-53C Why did the thermal conduction module receive so
much attention from thermal designers of electronic equip-
ment? How does the design of TCM differ from traditional
chip carrier design? Why is the cavity in the TCM filled with
helium instead of air?
15-54 Consider a chip dissipating 0.8 W of power in a DIP
with 18 pin leads. The materials and the dimensions of various
sections of this electronic device are given in the table. If the
temperature of the leads is 50°C, estimate the temperature at
the junction of the chip.
Thermal
Section and Conductivity, Thickness, Heat Transfer
Material W/m • °C mm Surface Area
Junction
constriction
—
—
Diameter 0.5 mm
Silicon chip
120
0.5
4 mm x 4 mm
Eutectic bond
296
0.05
4 mm x 4 mm
Copper lead
frame
386
0.25
4 mm x 4 mm
Plastic
separator
1
0.3
18 x 1 mm x 0.25 mm
Copper leads
386
6
18 x 1 mm x 0.25 mm
Junction
Air gap
Bond wires
Case
Leads
Lead frame Bond
IGUREP15-54
15-55 A fan blows air at 25°C over a 2-W plastic DIP with
16 leads mounted on a PCB at a velocity of 300 m/min. Using
data from Figure 15-23, determine the junction temperature of
the electronic device. What would the junction temperature be
if the fan were to fail?
15-56 Heat is to be conducted along a PCB with copper
cladding on one side. The PCB is 12 cm long and 12 cm wide,
Epoxy
Copper
FIGURE P1 5-56
and the thicknesses of the copper and epoxy layers are
0.06 mm and 0.5 mm, respectively. Disregarding heat transfer
from the side surfaces, determine the percentages of heat con-
duction along the copper (k = 386 W/m ■ °C) and epoxy (k =
0.26 W/m ■ °C) layers. Also, determine the effective thermal
conductivity of the PCB.
Answers: 0.6 percent, 99.4 percent, 41.6 W/m • °C
15-57 rfi£M Reconsider Problem 15-56. Using EES (or
)SZ2 other) software, investigate the effect of the
thickness of the copper layer on the percentage of heat con-
ducted along the copper layer and the effective thermal con-
ductivity of the PCB. Let the thickness vary from 0.02 mm to
0.10 mm. Plot the percentage of heat conducted along the cop-
per layer and the effective thermal conductivity as a function of
the thickness of the copper layer, and discuss the results.
15-58 The heat generated in the circuitry on the surface of a
silicon chip (k = 130 W/m • °C) is conducted to the ceramic
substrate to which it is attached. The chip is 6 mm X 6 mm in
size and 0.5-mm thick and dissipates 3 W of power. Determine
the temperature difference between the front and back surfaces
of the chip in steady operation.
15-59E Consider a 6-in. X 7-in. glass-epoxy laminate (k =
0.15 Btu/h • ft • °F) whose thickness is 0.05 in. Determine the
thermal resistance of this epoxy layer for heat flow (a) along
the 7-in. -long side and (£>) across its thickness.
15-60 Consider a 15-cm X 18-cm glass-epoxy laminate
(k = 0.26 W/m • °C) whose thickness is 1.4 mm. In order to re-
duce the thermal resistance across its thickness, cylindrical
copper fillings (k = 386 W/m ■ °C) of diameter 1 mm are to be
planted throughout the board with a center-to-center distance
of 3 mm. Determine the new value of the thermal resistance of
the epoxy board for heat conduction across its thickness as a
result of this modification.
15-61 [ct^S Reconsider Problem 15-60. Using EES (or
1^13 other) software, investigate the effects of the
thermal conductivity and the diameter of the filling material on
the thermal resistance of the epoxy board. Let the thermal con-
ductivity vary from 10 W/m ■ 0C to 400 W/m • 0C and the di-
ameter from 0.5 mm to 2.0 mm. Plot the thermal resistance as
cen58 93 3_chl5.qxd 9/9/2002 10:21 AM Page 846
846
HEAT TRANSFER
functions of the thermal conductivity and the diameter, and dis-
cuss the results.
15-62 A 12-cm X 15-cm circuit board dissipating 45 W of
heat is to be conduction-cooled by a 1.5-mm-fhick copper heat
frame (k = 386 W/m • °C) 12 cm X 17 cm in size. The epoxy
laminate (k = 0.26 W/m • °C) has a thickness of 2 mm and is
attached to the heat frame with conductive epoxy adhesive
(k = 1.8 W/m ■ °C) of thickness . 1 2 mm . The PCB is attached
to a heat sink by clamping a 5-mm-wide portion of the edge to
the heat sink from both ends. The temperature of the heat frame
at this point is 30°C. Heat is uniformly generated on the PCB
at a rate of 3 W per 1-cm X 12-cm strip. Considering only one-
half of the PCB board because of symmetry, determine the
maximum surface temperature on the PCB and the temperature
distribution along the heat frame.
Electronic
components
PCB
QtO.
'mvi'cmh-i
Cold
plate
FIGURE P1 5-62
Heat
frame
Epoxy
adhesive
15-63 Consider a 15-cm X 20-cm double-sided circuit
board dissipating a total of 30 W of heat. The board consists of
a 3-mm-thick epoxy layer (k = 0.26 W/m • °C) with 1-mm-di-
ameter aluminum wires (k = 237 W/m • °C) inserted along the
20-cm-long direction, as shown in Figure P15-63. The dis-
tance between the centers of the aluminum wires is 2 mm. The
circuit board is attached to a heat sink from both ends, and the
temperature of the board at both ends is 30°C. Heat is consid-
ered to be uniformly generated on both sides of the epoxy layer
Double-sided
PCB
o
o
o
o
GO
ai 2
Electronic
components
Aluminum
wire
3 mm
FIGURE P1 5-63
of the PCB. Considering only a portion of the PCB because of
symmetry, determine the magnitude and location of the maxi-
mum temperature that occurs in the PCB. Answer: 88.7°C
15-64 Repeat Problem 15-63, replacing the aluminum wires
by copper wires (k = 386 W/m • °C).
15-65 Repeat Problem 15-63 for a center-to-center distance
of 4 mm instead of 2 mm between the wires.
15-66 Consider a thermal conduction module with 80 chips,
each dissipating 4 W of power. The module is cooled by water
at 18°C flowing through the cold plate on top of the module.
The thermal resistances in the path of heat flow are 7? chlp =
12°C/W between the junction and the surface of the chip,
R mt = 9°C/W between the surface of the chip and the outer
surface of the thermal conduction module, and 7? cxt = 7°C/W
between the outer surface of the module and the cooling water.
Determine the junction temperature of the chip.
15-67 Consider a 0. 3-mm-thick epoxy board (k = 0.26 W/m
• °C) that is 15 cm X 20 cm in size. Now a 0.1 -mm -thick layer
of copper (k = 386 W/m • °C) is attached to the back surface of
the PCB. Determine the effective thermal conductivity of the
PCB along its 20-cm-long side. What fraction of the heat con-
ducted along that side is conducted through copper?
15-68 A 0.5-mm-thick copper plate (k = 386 W/m • °C) is
sandwiched between two 3-mm-thick epoxy boards (k = 0.26
W/m ■ °C) that are 12 cm X 18 cm in size. Determine the ef-
fective thermal conductivity of the PCB along its 1 8-cm-long
side. What fraction of the heat conducted along that side is con-
ducted through copper?
18 cm
Epoxy
boards
Copper
plate
0.5 mm
FIGURE P1 5-68
15-69E A 6-in. X 8-in. X 0.06-in. copper heat frame is
used to conduct 20 W of heat generated in a PCB along the
cen58 93 3_chl5.qxd 9/9/2002 10:21 AM Page 847
8-in.-long side toward the ends. Determine the temperature dif-
ference between the midsection and either end of the heat
frame. Answer: 12.8°F
15-70 A 12-W power transistor is cooled by mounting it on
an aluminum bracket (k = 231 W/m • °C) that is attached to a
liquid-cooled plate by 0.2-mm-thick epoxy adhesive (k = 1.8
W/m ■ °C), as shown in Figure PI 5-70. The thermal resistance
of the plastic washer is given as 2.5°C/W. Preliminary calcula-
tions show that about 20 percent of the heat is dissipated by
convection and radiation, and the rest is conducted to the liq-
uid-cooled plate. If the temperature of the cold plate is 50°C,
determine the temperature of the transistor case.
Liquid
channels
I Transistor
Aluminum
1 cm / bracket
FIGURE P1 5-70
Air Cooling: Natural Convection and Radiation
15-71C A student puts his books on top of a VCR, com-
pletely blocking the air vents on the top surface. What do you
think will happen as the student watches a rented movie played
by that VCR?
15-72C Can a low-power electronic system in space be
cooled by natural convection? Can it be cooled by radiation?
Explain.
15-73C Why are there several openings on the various sur-
faces of a TV, VCR, and other electronic enclosures? What
happens if a TV or VCR is enclosed in a cabinet with no free
air space around?
15-74C Why should radiation heat transfer always be con-
sidered in the analysis of natural convection-cooled electronic
equipment?
15-75C How does atmospheric pressure affect natural con-
vection heat transfer? What are the best and worst orientations
for heat transfer from a square surface?
15-76C What is view factor? How does it affect radiation
heat transfer between two surfaces?
15-77C What is emissivity? How does it affect radiation
heat transfer between two surfaces?
847
CHAPTER 15
15-78C For most effective natural convection cooling of a
PCB array, should the PCBs be placed horizontally or verti-
cally? Should they be placed close to each other or far from
each other?
15-79C Why is radiation heat transfer from the components
on the PCBs in an enclosure negligible?
15-80 Consider a sealed 20-cm-high electronic box whose
base dimensions are 35 cm X 50 cm that is placed on top of a
stand in a room at 30°C. The emissivity of the outer surface
of the box is 0.85. If the electronic components in the box
dissipate a total of 100 W of power and the outer surface tem-
perature of the box is not to exceed 65°C, determine if this
box can be cooled by natural convection and radiation alone.
Assume the heat transfer from the bottom surface of the
box to the stand to be negligible, and the temperature of the
surrounding surfaces to be the same as the air temperature of
the room.
Electronic box
FIGURE P1 5-80
15-81 Repeat Problem 15-80, assuming the box is mounted
on a wall instead of a stand such that it is 0.5 m high. Again, as-
sume heat transfer from the bottom surface to the wall to be
negligible.
15-82E A 0.1 5-W small cylindrical resistor mounted on a
circuit board is 0.5 in. long and has a diameter of 0.15 in. The
view of the resistor is largely blocked by the circuit board fac-
ing it, and the heat transfer from the connecting wires is neg-
ligible. The air is free to flow through the parallel flow
passages between the PCBs as a result of natural convection
currents. If the air temperature in the vicinity of the resistor is
130°F, determine the surface temperature of the resistor.
Answer: 194°F
15-83 A 14-cm X 20-cm PCB has electronic components on
one side, dissipating a total of 7 W. The PCB is mounted in a
rack vertically (height 14 cm) together with other PCBs. If the
surface temperature of the components is not to exceed 90°C,
determine the maximum temperature of the environment in
which this PCB can operate safely at sea level. What would
your answer be if this rack is located at a location at 3000 m al-
titude where the atmospheric pressure is 70.12 kPa?
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HEAT TRANSFER
FIGURE P1 5-83
15-84 A cylindrical electronic component whose diameter is
2 cm and length is 4 cm is mounted on a board with its axis in
the vertical direction and is dissipating 3 W of power. The
emissivity of the surface of the component is 0.8, and the tem-
perature of the ambient air is 30°C. Assuming the temperature
of the surrounding surfaces to be 20°C, determine the average
surface temperature of the component under combined natural
convection and radiation cooling.
15-85 Repeat Problem 15-84, assuming the component is
oriented horizontally.
15-86 Tu'M Reconsider Problem 15-84. Using EES (or
k^S other) software, investigate the effects of sur-
face emissivity and ambient temperature on the average sur-
face temperature of the component. Let the emissivity vary
from 0.1 to 1.0 and the ambient temperature from 15°C to
35°C. Take the temperature of the surrounding surfaces to be
10°C smaller than the ambient air temperature. Plot the average
surface temperature as functions of the emissivity and the am-
bient air temperature, and discuss the results.
15-87 Consider a power transistor that dissipates 0.1 W of
power in an environment at 30°C. The transistor is 0.4 cm long
and has a diameter of 0.4 cm. Assuming heat to be transferred
uniformly from all surfaces, determine (a) the heat flux on the
surface of the transistor, in W/cm 2 , and (b) the surface temper-
ature of the transistor for a combined convection and radiation
heat transfer coefficient of 18 W/m 2 • °C.
15-88 The components of an electronic system dissipating
150 W are located in a 1-m-long horizontal duct whose cross
section is 15 cm X 15 cm. The components in the duct are
cooled by forced air, which enters at 30°C at a rate of 0.4
m 3 /min and leaves at 45°C. The surfaces of the sheet metal duct
are not painted, and thus radiation heat transfer from the outer
surfaces is negligible. If the ambient air temperature is 25°C,
determine (a) the heat transfer from the outer surfaces of the
duct to the ambient air by natural convection and (b) the aver-
age temperature of the duct. Answers: (a) 31.7 W, (b) 40°C
. 45°C
FIGURE P1 5-88
15-89 Repeat Problem 15— S
of diameter 10 cm.
15-90
for a circular horizontal duct
Reconsider Problem 15-88. Using EES (or
other) software, investigate the effects of the
volume flow rate of air and the side-length of the duct on heat
transfer by natural convection and the average temperature of
the duct. Let the flow rate vary from 0.1 mVmin to 0.5 m 3 /min,
and the side-length from 10 cm to 20 cm. Plot the heat transfer
rate by natural convection and the average duct temperature as
functions of flow rate and side-length, and discuss the results.
15-91 Repeat Problem 15-88, assuming that the fan fails and
thus the entire heat generated inside the duct must be rejected
to the ambient air by natural convection from the outer surfaces
of the duct.
15-92 A 20-cm X 20-cm circuit board containing 81 square
chips on one side is to be cooled by combined natural convection
20 cm
FIGURE P1 5-92
cen58 93 3_chl5.qxd 9/9/2002 10:21 AM Page 849
and radiation by mounting it on a vertical surface in a room at
25°C. Each chip dissipates 0.08 W of power, and the emissiv-
ity of the chip surfaces is 0.65. Assuming the heat transfer from
the back side of the circuit board to be negligible and the tem-
perature of the surrounding surfaces to be the same as the air
temperature of the room, determine the surface temperature of
the chips.
15-93 Repeat Problem 15-92, assuming the circuit board
to be positioned horizontally with (a) chips facing up and
(b) chips facing down.
Air Cooling: Forced Convection
15-94C Why is radiation heat transfer in forced-air-cooled
systems disregarded?
15-95C If an electronic system can be cooled adequately by
either natural convection or forced-air convection, which
would you prefer? Why?
15-96C Why is forced convection cooling much more effec-
tive than natural convection cooling?
15-97C Consider a forced-air-cooled electronic system dis-
sipating a fixed amount of power. How will increasing the flow
rate of air affect the surface temperature of the components?
Explain. How will it affect the exit temperature of the air?
15-98C To what do internal and external flow refer in forced
convection cooling? Give an example of a forced-air-cooled
electronic system that involves both types of flow.
15-99C For a specified power dissipation and air inlet tem-
perature, how does the convection heat transfer coefficient af-
fect the surface temperature of the electronic components?
Explain.
15-100C How does high altitude affect forced convection
heat transfer? How would you modify your forced-air cooling
system to operate at high altitudes safely?
15-101C What are the advantages and disadvantages of
placing the cooling fan at the inlet or at the exit of an electronic
box?
15-102C How is the volume flow rate of air in a forced-air-
cooled electronic system that has a constant-speed fan estab-
lished? If a few more PCBs are added to the box while keeping
the fan speed constant, will the flow rate of air through the sys-
tem change? Explain.
15-103C What happens if we attempt to cool an electronic
system with an undersized fan? What about if we do that with
an oversized fan?
15-104 Consider a hollow -core PCB that is 15 cm high and
20 cm long, dissipating a total of 30 W. The width of the air
gap in the middle of the PCB is 0.25 cm. The cooling air enters
the core at 30°C at a rate of 1 L/s. Assuming the heat generated
to be uniformly distributed over the two side surfaces of the
849
CHAPTER 15
PCB, determine (a) the temperature at which the air leaves the
hollow core and (b) the highest temperature on the inner sur-
face of the core. Answers: (a) 56.4°C, (£>) 67.6°C
15 cm
Hollow core
for air flow
Electronic
components
FIGURE P1 5-1 04
15-105 Repeat Problem 15-104 for a hollow-core PCB dis-
sipating 45 W.
15-106 FEa| Reconsider Problem 15-104. Using EES (or
I^S other) software, investigate the effects of the
power rating of the PCB and the volume flow rate of the air on
the exit temperature of the air and the maximum temperature
on the inner surface of the core. Let the power vary from 20 W
to 60 W and the flow rate from 0.5 L/s to 2.5 L/s. Plot the
air exit temperature and the maximum surface temperature of
the core as functions of power and flow rate, and discuss the
results.
15-107E A transistor with a height of 0.25 in. and a diameter
of 0.2 in. is mounted on a circuit board. The transistor is cooled
by air flowing over it at a velocity of 400 ft/min. If the air tem-
perature is 140°F and the transistor case temperature is not to
exceed 175°F, determine the amount of power this transistor
can dissipate safely. Answer: 0.15 W
15-108 A desktop computer is to be cooled by a fan. The
electronic components of the computer consume 75 W of
power under full-load conditions. The computer is to operate in
environments at temperatures up to 45°C and at elevations up
to 3400 m where the atmospheric pressure is 66.63 kPa. The
exit temperature of air is not to exceed 60°C to meet reliability
requirements. Also, the average velocity of air is not to exceed
110 m/min at the exit of the computer case, where the fan is in-
stalled to keep the noise level down. Determine the flow rate of
the fan that needs to be installed and the diameter of the casing
of the fan.
15-109 Repeat Problem 15-108 for a computer that con-
sumes 100 W of power.
15-110 A computer cooled by a fan contains eight PCBs,
each dissipating 12 W of power. The height of the PCBs is
12 cm and the length is 18 cm. The clearance between the tips
of the components on the PCB and the back surface of the
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HEAT TRANSFER
adjacent PCB is 0.3 cm. The cooling air is supplied by a 15-W
fan mounted at the inlet. If the temperature rise of air as it
flows through the case of the computer is not to exceed 15°C,
determine (a) the flow rate of the air that the fan needs to de-
liver, (b) the fraction of the temperature rise of air due to the
heat generated by the fan and its motor, and (c) the highest al-
lowable inlet air temperature if the surface temperature of the
components is not to exceed 90°C anywhere in the system.
PCB, 12 W
FIGURE P1 5-1 10
15-111 An array of power transistors, each dissipating 2 W
of power, is to be cooled by mounting them on a 20-cm X
20-cm square aluminum plate and blowing air over the plate
with a fan at 30°C with a velocity of 3 m/s. The average tem-
perature of the plate is not to exceed 60°C. Assuming the heat
transfer from the back side of the plate to be negligible, de-
termine the number of transistors that can be placed on this
plate. Answer: 9
Aluminum Power
plate /* transistors
\ (
00000
00000
00000
Air
FIGURE P1 5-1 11
15-112 Repeat Problem 1 5-1 11 for a location at an elevation
of 1610 m where the atmospheric pressure is 83.4 kPa.
15-113 fifii'M Reconsider Problem 15—111. Using EES (or
si3 other) software, investigate the effects of air
velocity and the maximum plate temperature on the number of
transistors. Let the air velocity vary from 1 rn/s to 8 m/s and
the maximum plate temperature from 40°C to 80°C. Plot the
number of transistors as functions of air velocity and maximum
plate temperature, and discuss the results.
15-114 An enclosure contains an array of circuit boards, 15
cm high and 20 cm long. The clearance between the tips of the
components on the PCB and the back surface of the adjacent
PCB is 0.3 cm. Each circuit board contains 75 square chips on
one side, each dissipating 0.15 W of power. Air enters the
space between the boards through the 0.3-cm X 15-cm cross
section at 40°C with a velocity of 300 m/min. Assuming the
heat transfer from the back side of the circuit board to be neg-
ligible, determine the exit temperature of the air and the high-
est surface temperature of the chips.
15-115 The components of an electronic system dissipating
120 W are located in a 1-m-long horizontal duct whose cross
section is 20 cm X 20 cm. The components in the duct are
cooled by forced air, which enters at 30°C at a rate of 0.5
m 3 /min. Assuming 80 percent of the heat generated inside is
transferred to air flowing through the duct and the remaining
20 percent is lost through the outer surfaces of the duct, deter-
mine (a) the exit temperature of air and (b) the highest compo-
nent surface temperature in the duct.
15-116 Repeat Problem 15-115 for a circular horizontal duct
of diameter 10 cm.
Liquid Cooling
15-117C If an electronic system can be cooled adequately by
either forced-air cooling or liquid cooling, which one would
you prefer? Why?
15-118C Explain how direct and indirect liquid cooling sys-
tems differ from each other.
15-119C Explain how closed-loop and open-loop liquid
cooling systems operate.
15-120C What are the properties of a liquid ideally suited
for cooling electronic equipment?
15-121 A cold plate that supports 10 power transistors, each
dissipating 40 W, is to be cooled with water. It is specified that
the temperature rise of the water not exceed 4°C and the veloc-
ity of water remain under 0.5 m/s. Assuming 25 percent of the
heat generated is dissipated from the components to the sur-
roundings by convection and radiation, and the remaining 75
percent is removed by the cooling water, determine the mass
Cold plate
_ Water
inlet
_ Water
exit
FIGURE P1 5-1 21
cen58 93 3_chl5.qxd 9/9/2002 10:21 AM Page 851
flow rate of water needed and the diameter of the pipe to sat-
isfy the restriction imposed on the flow velocity. Also, deter-
mine the case temperature of the devices if the total
case -to-liquid thermal resistance is 0.04°C/W and the water
enters the cold plate at 25°C.
15-122 Tu'M Reconsider Problem 15-121. Using EES (or
|^i£ other) software, investigate the effect of the
maximum temperature rise of the water on the mass flow rate
of water, the diameter of the pipe, and the case temperature.
Let the maximum temperature rise vary from 1°C to 10°C. Plot
the mass flow rate, the diameter, and the case temperature as a
function of the temperature rise, and discuss the results.
15-123E Water enters the tubes of a cold plate at 95°F with
an average velocity of 60 ft/min and leaves at 105°F. The di-
ameter of the tubes is 0.25 in. Assuming 15 percent of the heat
generated is dissipated from the components to the surround-
ings by convection and radiation, and the remaining 85 percent
is removed by the cooling water, determine the amount of heat
generated by the electronic devices mounted on the cold plate.
Answer: 263 W
15-124 A sealed electronic box is to be cooled by tap water
flowing through channels on two of its sides. It is specified that
the temperature rise of the water not exceed 3°C. The power
dissipation of the box is 2 kW, which is removed entirely by
water. If the box operates 24 h a day, 365 days a year,
determine the mass flow rate of water flowing through the box
and the amount of cooling water used per year.
15-125 Repeat Problem 15-124 for a power dissipation of
3kW.
Immersion Cooling
15-126C What are the desirable characteristics of a liquid
used in immersion cooling of electronic devices?
15-127C How does an open-loop immersion cooling system
operate? How does it differ from closed-loop cooling systems?
15-128C How do immersion cooling systems with internal
and external cooling differ? Why are externally cooled systems
limited to relatively low-power applications?
15-129C Why is boiling heat transfer used in the cooling of
very high-power electronic devices instead of forced air or liq-
uid cooling?
15-130 A logic chip used in a computer dissipates 4 W of
power and has a heat transfer surface area of 0.3 cm 2 . If the
surface of the chip is to be maintained at 70°C while being
cooled by immersion in a dielectric fluid at 20°C, determine
the necessary heat transfer coefficient and the type of cooling
mechanism that needs to be used to achieve that heat transfer
coefficient.
15-131 A 6-W chip having a surface area of 0.5 cm 2 is
cooled by immersing it into FC86 liquid that is maintained at a
851
CHAPTER 15
temperature of 25°C. Using the boiling curve in Figure 15-63,
estimate the temperature of the chip surface. Answer: 82°C
15-132 A logic chip cooled by immersing it in a dielectric
liquid dissipates 3.5 W of power in an environment at 50°C and
has a heat transfer surface area of 0.8 cm 2 . The surface temper-
ature of the chip is measured to be 95°C. Assuming the heat
transfer from the surface to be uniform, determine (a) the heat
flux on the surface of the chip, in W/cm 2 ; (b) the heat transfer
coefficient on the surface of the chip, in W/m 2 • °C; and (c) the
thermal resistance between the surface of the chip and the cool-
ing medium, in °C/W.
;
;
;
3.5 W
,-, Logic
-
'"' chip
:_
^-— 95°C
:
Dielectric
liquid,
^ 50°C
FIGURE P1 5-1 32
15-133
Reconsider Problem 15-132. Using EES (or
other) software, investigate the effect of chip
power on the heat flux, the heat transfer coefficient, and the con-
vection resistance on chip surface. Let the power vary from 2 W
to 10W. Plot the heat flux, the heat transfer coefficient, and the
thermal resistance as a function of power dissipated, and discuss
the results.
15-134 A computer chip dissipates 5 W of power and has a
heat transfer surface area of 0.4 cm 2 . If the surface of the chip
is to be maintained at 55°C while being cooled by immersion
in a dielectric fluid at 10°C, determine the necessary heat trans-
fer coefficient and the type of cooling mechanism that needs to
be used to achieve that heat transfer coefficient.
15-135 A 3-W chip having a surface area of 0.2 cm 2 is
cooled by immersing it into FC86 liquid that is maintained at a
temperature of 45 °C. Using the boiling curve in Figure 15-63,
estimate the temperature of the chip surface. Answer: 108°C
15-136 A logic chip having a surface area of 0.3 cm 2 is to be
cooled by immersing it into FC86 liquid that is maintained at a
temperature of 35°C. The surface temperature of the chip is not
to exceed 60°C. Using the boiling curve in Figure 15-63, esti-
mate the maximum power that the chip can dissipate safely.
15-137 A 2-kW electronic device that has a surface area of
120 cm 2 is to be cooled by immersing it in a dielectric fluid
with a boiling temperature of 60°C contained in a 1-m X 1-m
cen58 93 3_chl5.qxd 9/9/2002 10:22 AM Page 852
852
HEAT TRANSFER
X 1-m cubic enclosure. Noting that the combined natural con-
vection and the radiation heat transfer coefficients in air are
typically about 10 W/m 2 • °C, determine if the heat generated
inside can be dissipated to the ambient air at 20°C by natural
convection and radiation. If it cannot, explain what modifica-
tion you could make to allow natural convection cooling.
Also, determine the heat transfer coefficients at the surface
of the electronic device for a surface temperature of 80°C. As-
sume the liquid temperature remains constant at 60°C through-
out the enclosure.
Air
20°C
Dielectric
liquid
1 m
lra-
- Electronic
device
FIGURE P1 5-1 37
Review Problems
15-138C Several power transistors are to be cooled by
mounting them on a water-cooled metal plate. The total power
dissipation, the mass flow rate of water through the tube, and
the water inlet temperature are fixed. Explain what you would
do for the most effective cooling of the transistors.
15-139C Consider heat conduction along a vertical copper
bar whose sides are insulated. One person claims that the bar
should be oriented such that the hot end is at the bottom and the
cold end is at the top for better heat transfer, since heat rises.
Another person claims that it makes no differences to heat con-
duction whether heat is conducted downward or upward, and
thus the orientation of the bar is irrelevant. With which person
do you agree?
15-140 Consider a 15-cm X 15-cm multilayer circuit board
dissipating 22.5 W of heat. The board consists of four layers of
0.1-mm-fhick copper (k = 386 W/m ■ °C) and three layers of
0.5-mm-thick glass-epoxy (k = 0.26 W/m ■ °C) sandwiched to-
gether, as shown in Figure PI 5- 140. The circuit board is at-
tached to a heat sink from both ends, and the temperature of the
board at those ends is 35°C. Heat is considered to be uniformly
generated in the epoxy layers of the PCB at a rate of 0.5 W per
1-cm X 15-cm epoxy laminate strip (or 1.5 W per 1-cm X
1 5-cm strip of the board). Considering only a portion of the PCB
because of symmetry, determine the magnitude and location of
the maximum temperature that occurs in the PCB . Assume heat
transfer from the top and bottom faces of the PCB to be negligi-
ble. Answer: 55.5°C
Copper
15 cm
FIGURE P1 5-1 40
15-141 Repeat Problem 15-140, assuming that the board
consists of a single 1 .5-mm-thick layer of glass-epoxy, with no
copper layers.
15-142 The components of an electronic system that is dissi-
pating 150 W are located in a 1-m-long horizontal duct whose
cross section is 10 cm X 10 cm. The components in the duct
are cooled by forced air, which enters at 30°C and 50 m/min
and leaves at 45°C. The surfaces of the sheet metal duct are not
painted, and so radiation heat transfer from the outer surfaces
is negligible. If the ambient air temperature is 30°C, determine
(a) the heat transfer from the outer surfaces of the duct to the
ambient air by natural convection, {b) the average temperature
of the duct, and (c) the highest component surface temperature
in the duct.
15-143 Two 10-W power transistors are cooled by mounting
them on the two sides of an aluminum bracket (k = 237 W/m •
°C) that is attached to a liquid-cooled plate by 0.2-mm-thick
epoxy adhesive (k = 1.8 W/m ■ °C), as shown in Figure
P15-143. The thermal resistance of each plastic washer is
Liquid-cooled
plate
Transistor
1 cm
IGURE P15-143
0.3 cm
cen58 93 3_chl5.qxd 9/9/2002 10:22 AM Page 853
given as 2°C/W, and the temperature of the cold plate is 40°C.
The surface of the aluminum plate is untreated, and thus radia-
tion heat transfer from it is negligible because of the low emis-
sivity of aluminum surfaces. Disregarding heat transfer from
the 0.3-cm-wide edges of the aluminum plate, determine the
surface temperature of the transistor case. Also, determine the
fraction of heat dissipation to the ambient air by natural con-
vection and to the cold plate by conduction. Take the ambient
temperature to be 25°C.
15-144E A fan blows air at 70°F and a velocity of 500 ft/min
over a 1 .5-W plastic DIP with 24 leads mounted on a PCB . Us-
ing data from Figure 15-23, determine the junction tempera-
ture of the electronic device. What would the junction
temperature be if the fan were to fail?
15-145 flk\ A 15-cm X 18-cm double-sided circuit board
xM>y dissipating a total of 18 W of heat is to be
conduction-cooled by a 1.2-mm-thick aluminum core plate
(k = 237 W/m • °C) sandwiched between two epoxy laminates
{k = 0.26 W/m • °C). Each epoxy layer has a thickness of
0.5 mm and is attached to the aluminum core plate with con-
ductive epoxy adhesive (k = 1.8 W/m ■ °C) of thickness
0.1 mm. Heat is uniformly generated on each side of the PCB
at a rate of 0.5 W per 1 -cm X 15-cm epoxy laminate strip. All
of the heat is conducted along the 1 8-cm side since the PCB is
cooled along the two 15-cm-long edges. Considering only part
of the PCB board because of symmetry, determine the maxi-
mum temperature rise across the 9-cm distance between the
center and the sides of the PCB. Answer: 10.1°C
15-146 Ten power transistors, each dissipating 2 W, are at-
tached to a 7-cm X 7-cm X 0.2-cm aluminum plate with a
square cutout in the middle in a symmetrical arrangement, as
shown in Figure P15-146. The aluminum plate is cooled from
two sides by liquid at 40°C. If 70 percent of the heat generated
by the transistors is estimated to be conducted through the alu-
Transistors
40°C
Aluminum
plate
40°C
Cut out
section
4 cm X 4 cm
1 cm 5 cm
FIGURE P15-146
1 cm
853
CHAPTER 15
minum plate, determine the temperature rise across the 1 -cm-
wide section of the aluminum plate between the transistors
and the heat sink.
15-147 The components of an electronic system are located
in a 1.2-m-long horizontal duct whose cross section is 10 cm X
20 cm. The components in the duct are not allowed to come
into direct contact with cooling air, and so are cooled by air
flowing over the duct at 30°C with a velocity of 250 m/min.
The duct is oriented such that air strikes the 10-cm-high side of
the duct normally. If the surface temperature of the duct is not
to exceed 60°C, determine the total power rating of the elec-
tronic devices that can be mounted in the duct. What would
your answer be if the duct is oriented such that air strikes the
20-cm-high side normally? Answers: 481 W, 384 W
15-148 Repeat Problem 15-147 for a location at an altitude
of 5000 m, where the atmospheric pressure is 54.05 kPa.
15-149E A computer that consumes 65 W of power is cooled
by a fan blowing air into the computer enclosure. The dimen-
sions of the computer case are 6 in. X 20 in. X 24 in., and all
surfaces of the case are exposed to the ambient, except for the
base surface. Temperature measurements indicate that the case
is at an average temperature of 95°F when the ambient temper-
ature and the temperature of the surrounding walls are 80°F If
the emissivity of the outer surface of the case is 0.85, deter-
mine the fraction of heat lost from the outer surfaces of the
computer case.
FIGURE P15-149E
Computer, Design, and Essay Problems
15-150 Bring an electronic device that is cooled by heat
sinking to class and discuss how the heat sink enhances heat
transfer.
15-151 Obtain a catalog from a heat sink manufacturer and
select a heat sink model that can cool a 10-W power transistor
safely under (a) natural convection and radiation and (b)
forced convection conditions.
15-152 Take the cover off a PC or another electronic box and
identify the individual sections and components. Also, identify
the cooling mechanisms involved and discuss how the current
design facilitates effective cooling.
cen58 93 3_chl5.qxd 9/9/2002 10:22 AM Page 854
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