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Contents
Preface xviii
Nomenclature xxvi
CHAPTER ONE
BASICS OF HEAT TRANSFER 1
11 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 SteadyFlow Systems 12
Surface Energy Balance 13
15 Heat Transfer Mechanisms 17
16 Conduction 17
Thermal Conductivity 19
Thermal Diffusivity 23
1 7 Convection 25
1 8 Radiation 27
1 9 Simultaneous Heat Transfer Mechanisms 30
110 ProblemSolving 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
25
26
27
22 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 OneDimensional
Heat Conduction Equation 72
23 General Heat Conduction Equation 74
Rectangular Coordinates 74
Cylindrical Coordinates 75
Spherical Coordinates 76
24 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 OneDimensional
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
31 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
32 Thermal Contact Resistance 138
33 Generalized Thermal Resistance Networks 143
34 Heat Conduction in Cylinders and Spheres 146
Multilayered Cylinders and Spheres 148
35 Critical Radius of Insulation 153
36 Heat Transfer from Finned Surfaces 156
Fin Equation 157
Fin Efficiency 160
Fin Effectiveness 163
Proper Length of a Fin 165
37 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
52 Finite Difference Formulation of
Differential Equations 269
53 One Dimensional Steady Heat Conduction 272
Boundary Conditions 274
54 TwoDimensional
Steady Heat Conduction 282
Boundary Nodes 283
Irregular Boundaries 287
55 Transient Heat Conduction 291
Transient Heat Conduction in a Plane Wall 293
TwoDimensional 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
41 Lumped System Analysis 210
Criteria for Lumped System Analysis 211
Some Remarks on Heat Transfer in Lumped Systems 213
42 Transient Heat Conduction in
Large Plane Walls, Long Cylinders,
and Spheres with Spatial Effects 216
43 Transient Heat Conduction in
SemiInfinite Solids 228
44 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
51 Why Numerical Methods? 266
1 Limitations 267
2 Better Modeling 267
3 Flexibility 268
CHAPTER SIX
FUNDAMENTALS OF CONVECTION
333
61 Physical Mechanism on Convection 334
Nusselt Number 336
62 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 ThreeDimensional Flows 338
63 Velocity Boundary Layer 339
Surface Shear Stress 340
64 Thermal Boundary Layer 341
Prandtl Number 341
65 Laminar and Turbulent Flows 342
Reynolds Number 343
66 Heat and Momentum Transfer
in Turbulent Flow 343
67 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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68 Solutions of Convection Equations
for a Flat Plate 352
The Energy Equation 354
69 Nondimensionalized Convection
Equations and Similarity 356
61 Functional Forms of Friction and
Convection Coefficients 357
61 1 Analogies between Momentum
and Heat Transfer 358
Summary 361
References and Suggested Reading 362
Problems 362
CHAPTER SEVEN
EXTERNAL FORCED CONVECTION
367
CONTENTS
84 General Thermal Analysis 426
Constant Surface Heat Flux [q s = constant) 427
Constant Surface Temperature (T s = constant) 428
85 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
86 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
71 Drag Force and Heat Transfer
in External Flow 368
Friction and Pressure Drag 368
Heat Transfer 370
72 Parallel Flow over Flat Plates 371
Friction Coefficient 372
Heat Transfer Coefficient 373
Flat Plate with Unheated Starting Length 375
Uniform Heat Flux 375
73 Flow across Cylinders and Spheres 380
Effect of Surface Roughness 382
Heat Transfer Coefficient 384
74 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
81 Introduction 420
82 Mean Velocity and Mean Temperature 420
Laminar and Turbulent Flow in Tubes 422
83 The Entrance Region 423
Entry Lengths 425
CHAPTER NINE
NATURAL CONVECTION 459
91 Physical Mechanism of
Natural Convection 460
92 Equation of Motion and
the Grashof Number 463
The Grashof Number 465
93 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
94 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
95 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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96 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
101 Boiling Heat Transfer 516
102 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
103 Flow Boiling 530
1 04 Condensation Heat Transfer 532
1 05 Film Condensation 532
Flow Regimes 534
Heat Transfer Correlations for Film Condensation 535
1 06 Film Condensation Inside
Horizontal Tubes 545
1 07 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
111 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
121 The View Factor 606
1 22 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 CrossedStrings Method 618
123 Radiation Heat Transfer: Black Surfaces 620
1 24 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 TwoSurface Enclosures 627
Radiation Heat Transfer in ThreeSurface Enclosures 629
1 25 Radiation Shields and the Radiation Effect 635
Radiation Effect on Temperature Measurements 637
1 26 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
131 Types of Heat Exchangers 668
1 32 The Overall Heat Transfer Coefficient 67 1
Fouling Factor 674
1 33 Analysis of Heat Exchangers 678
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1 34 The Log Mean Temperature
Difference Method 680
CounterFlow Heat Exchangers 682
Multipass and CrossFlow Heat Exchangers:
Use of a Correction Factor 683
1 35 The EffectivenessNTU Method 690
136 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
141 Introduction 718
142 Analogy between Heat and Mass Transfer 719
Temperature 720
Conduction 720
Heat Generation 720
Convection 721
143 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
144 Boundary Conditions 727
1 45 Steady Mass Diffusion through a Wall 732
146 Water Vapor Migration in Buildings 736
147 Transient Mass Diffusion 740
148 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
149 Mass Convection 754
Analogy between Friction, Heat Transfer, and Mass
Transfer Coefficients 758
Limitation on the HeatMass Convection Analogy 760
Mass Convection Relations 760
CONTENTS
1410 Simultaneous Heat and Mass Transfer 763
Summary 769
References and Suggested Reading 771
Problems 772
CHAPTER FIFTEEN
COOLING OF ELECTRONIC EQUIPMENT
785
151 Introduction and History 786
152 Manufacturing of Electronic Equipment 787
The Chip Carrier 787
Printed Circuit Boards 789
The Enclosure 791
153 Cooling Load of Electronic Equipment 793
1 54 Thermal Environment 794
1 55 Electronics Cooling in
Different Applications 795
1 56 Conduction Cooling 797
Conduction in Chip Carriers 798
Conduction in Printed Circuit Boards 803
Heat Frames 805
The Thermal Conduction Module (TCM) 810
1 57 Air Cooling: Natural Convection
and Radiation 812
158 Air Cooling: Forced Convection 820
Fan Selection 823
Cooling Personal Computers 826
159 Liquid Cooling 833
1510 Immersion Cooling 836
Summary 841
References and Suggested Reading 842
Problems 842
APPENDIX 1
PROPERTY TABLES AND CHARTS
(SI UNITS) 855
Table A1 Molar Mass, Gas Constant, and
CriticalPoint Properties 856
Table A2 Boiling and FreezingPoint
Properties 857
Table A3 Properties of Solid Metals 858
Table A4 Properties of Solid Nonmetals 861
Table A5 Properties of Building Materials 862
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CONTENTS
Table A6 Properties of Insulating Materials 864
Table A7 Properties of Common Foods 865
Table A8 Properties of Miscellaneous
Materials 867
Table A9 Properties of Saturated Water 868
Ta b I e A 1 Properties of S aturated
Refrigerant 134a 869
Table A11 Properties of Saturated Ammonia 870
Ta b I e A 1 2 Properties of S aturated Propane 87 1
TableA13 Properties of Liquids 872
Table A14 Properties of Liquid Metals 873
Table A15 Properties of Air at 1 atm Pressure 874
Table A1 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 A1 9 Solar Radiative Properties of
Materials 880
Figure A20 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 A1 E Molar Mass, Gas Constant, and
CriticalPoint Properties 884
Table A2E Boiling and FreezingPoint
Properties 885
Table A3E Properties of Solid Metals 886
Table A4E Properties of Solid Nonmetals 889
Table A5E Properties of Building Materials 890
Table A6E Properties of Insulating Materials 892
Table A7E Properties of Common Foods 893
Table A8E Properties of Miscellaneous
Materials 895
Table A9E Properties of Saturated Water 896
Table A1 0E Properties of Saturated
Refrigerant 134a 897
Table A1 1 E Properties of Saturated Ammonia 898
Table A1 2E Properties of Saturated Propane 899
Table A1 3E Properties of Liquids 900
Table A1 4E Properties of Liquid Metals 901
Table A1 5E Properties of Air at 1 atm Pressure 902
Table A1 6E Properties of Gases at 1 atm
Pressure 903
Table A1 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 11 Heating of a Copper Ball 10
Example 12 Heating of Water in an
Electric Teapot 14
Example 13 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 19 Radiation Effect on
Thermal Comfort 29
Example 110 Heat Loss from a Person 31
Example 111 Heat Transfer between
Two Isothermal Plates 32
Example 112 Heat Transfer in Conventional
and Microwave Ovens 33
Example 113 Heating of a Plate by
Solar Energy 34
Example 114 Solving a System of Equations
with EES 39
CHAPTER TWO
HEAT CONDUCTION EQUATION 61
Example 21
Heat Gain by a Refrigerator 67
Example 22 Heat Generation in a
Hair Dryer 67
Example 23 Heat Conduction through the
Bottom of a Pan 72
Example 24 Heat Conduction in a
Resistance Heater 72
Exa m p I e 25 Cooling of a Hot Metal B all
in Air 73
Example 26 Heat Conduction in a
Short Cylinder 76
Example 27 Heat Flux Boundary Condition 80
Example 28 Convection and Insulation
Boundary Conditions 82
Example 29 Combined Convection and
Radiation Condition 84
Example 210 Combined Convection, Radiation,
and Heat Flux 85
Exa m p I e 2 1 1 Heat Conduction in a
Plane Wall 86
Example 212 A Wall with Various Sets of
Boundary Conditions 88
Example 21 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 215 Heat Loss through a
Steam Pipe 94
Example 216 Heat Conduction through a
Spherical Shell 96
Example 217 Centerline Temperature of a
Resistance Heater 100
Example 218 Variation of Temperature in a
Resistance Heater 100
Example 219 Heat Conduction in a TwoLayer
Medium 102
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CONTENTS
Example 220 Variation of Temperature in a Wall
with k(T) 105
Example 221 Heat Conduction through a Wall
with k(T) 106
CHAPTER THREE
STEADY HEAT CONDUCTION 1 27
Example 31
Heat Loss through a Wall 134
Example 32
Heat Loss through a
SinglePane Window 135
Example 33
Heat Loss through
DoublePane Windows 136
Example 34
Equivalent Thickness for
Contact Resistance 140
Example 35
Contact Resistance of
Transistors 141
Example 36
Heat Loss through a
Composite Wall 144
Example 37
Heat Transfer to a
Spherical Container 149
:
)
Example 38
Heat Loss through an Insulated
Steam Pipe 151
Example 39
Heat Loss from an Insulated
Electric Wire 154
Example 310
Maximum Power Dissipation of
a Transistor 166
Example 31 1
Selecting a Heat Sink for a
Transistor 167
Example 312
Effect of Fins on Heat Transfer from
Steam Pipes 168
Example 313
Heat Loss from Buried
Steam Pipes 170
Example 314
Heat Transfer between Hot and
Cold Water Pipes 173
Example 315
Cost of Heat Loss through Walls
in Winter 174
Example 316
The i?Value of a Wood
Frame Wall 179
Example 317
The R Value of a Wall with
Rigid Foam 180
Example 318
The #Value of a Masonry Wall 181
Example 319
The R Value of a Pitched Roof 1 82
CHAPTER FOUR
TRANSIENT HEAT CONDUCTION 209
Example 41 Temperature Measurement by
Thermocouples 214
Example 42 Predicting the Time of Death 215
Example 43 Boiling Eggs 224
Example 44 Heating of Large Brass Plates
in an Oven 225
Example 45 Cooling of a Long Stainless Steel
Cylindrical Shaft 226
Example 46 Minimum Burial Depth of Water
Pipes to Avoid Freezing 230
Exa m p I e 47 Cooling of a S hort Brass
Cylinder 234
Example 48 Heat Transfer from a Short
Cylinder 235
Example 49 Cooling of a Long Cylinder
by Water 236
Example 410 Refrigerating Steaks while
Avoiding Frostbite 238
Example 41 1 Chilling of Beef Carcasses in a
Meat Plant 248
CHAPTER FIVE
NUMERICAL METHODS IN
HEAT CONDUCTION 265
Example 51 Steady Heat Conduction in a Large
Uranium Plate 277
Example 52 Heat Transfer from
Triangular Fins 279
Example 53 Steady TwoDimensional Heat
Conduction in LBars 284
Example 54 Heat Loss through Chimneys 287
Example 55 Transient Heat Conduction in a Large
Uranium Plate 296
Example 56 Solar Energy Storage in
Trombe Walls 300
Example 57 Transient TwoDimensional Heat
Conduction in LBars 305
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CHAPTER SIX
FUNDAMENTALS OF CONVECTION
333
Example 86
CONTENTS
Heat Loss from the Ducts of a
Heating System 448
Example 61 Temperature Rise of Oil in a
Journal Bearing 350
Example 62 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 73 Cooling of Plastic Sheets by
Forced Air 378
Example 74 Drag Force Acting on a Pipe
in a River 383
Example 75 Heat Loss from a Steam Pipe
in Windy Air 386
Example 76 Cooling of a Steel Ball by
Forced Air 387
Example 77 Preheating Air by Geothermal Water
in a Tube Bank 393
Example 78 Effect of Insulation on
Surface Temperature 402
Example 79 Optimum Thickness of
Insulation 403
CHAPTER EIGHT
INTERNAL FORCED CONVECTION 41 9
Example 81 Heating of Water in a Tube
by Steam 430
Example 82 Pressure Drop in a Pipe 438
Example 83 Flow of Oil in a Pipeline through
a Lake 439
Example 84 Pressure Drop in a Water Pipe 445
Example 85 Heating of Water by Resistance
Heaters in a Tube 446
CHAPTER NINE
NATURAL CONVECTION 459
Example 91 Heat Loss from Hot
Water Pipes 470
Example 92 Cooling of a Plate in
Different Orientations 47 1
Example 93 Optimum Fin Spacing of a
Heat Sink 476
Example 94 Heat Loss through a DoublePane
Window 482
Example 95 Heat Transfer through a
Spherical Enclosure 483
Example 96 Heating Water in a Tube by
Solar Energy 484
Example 97 [/Factor for CenterofGlass Section
of Windows 496
Example 98 Heat Loss through Aluminum Framed
Windows 497
Example 99 [/Factor of a DoubleDoor
Window 498
CHAPTER TEN
BOILING AND CONDENSATION 515
Example 1 01 Nucleate Boiling Water
in a Pan 526
Example 102 Peak Heat Flux in
Nucleate Boiling 528
Example 1 03 Film Boiling of Water on a
Heating Element 529
Example 104 Condensation of Steam on a
Vertical Plate 541
Example 105 Condensation of Steam on a
Tilted Plate 542
Example 106 Condensation of Steam on
Horizontal Tubes 543
Example 107 Condensation of Steam on
Horizontal Tube Banks 544
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CONTENTS
Example 1 08 Replacing a Heat Pipe by a
Copper Rod 550
CHAPTER ELEVEN
FUNDAMENTALS OF THERMAL RADIATION 561
Example 111 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 121
Example 122
Example 123
Example 124
Example 125
Example 126
Example 127
Example 128
Example 129
Example 1210
Example 1211
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 CrossedStrings 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 1212 Radiation Effect on Temperature
Measurements 639
Example 1213 Effective Emissivity of
Combustion Gases 646
Example 1 214 Radiation Heat Transfer in a
Cylindrical Furnace 647
Example 1215 Effect of Clothing on Thermal
Comfort 652
CHAPTER THIRTEEN
HEAT EXCHANGERS 667
Example 131
Example 132
Example 133
Example 134
Example 135
Example 136
Example 137
Example 138
Example 139
Example 1310
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 CounterFlow
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 141 Determining Mass Fractions from
Mole Fractions 727
Example 142 Mole Fraction of Water Vapor at
the Surface of a Lake 728
Example 143 Mole Fraction of Dissolved Air
in Water 730
Example 1 44 Diffusion of Hydrogen Gas into
a Nickel Plate 732
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Example 145
Example 146
Example 147
Example 148
Example 149
Example 1410
Example 1411
Example 1412
Example 1413
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 51 Predicting the Junction Temperature
of a Transistor 788
Example 152 Determining the JunctiontoCase
Thermal Resistance 789
Example 153 Analysis of Heat Conduction in
a Chip 799
Example 1 54 Predicting the Junction Temperature
of a Device 802
Example 155
Example 156
Example 157
Example 158
Example 159
Example 1510
Example 151 1
Example 1512
Example 1513
Example 1514
Example 1515
Example 1516
Example 1517
Example 1518
Example 1519
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
HollowCore 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
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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 realworld 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 realworld
applications, while deemphasizing 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 endofchapter 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
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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 realworld
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 computerEES 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 14. In a new
section ProblemSolving Technique, the problemsolving 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
SELFINSTRUCTING
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 W0RKED0UT EXAMPLES
Each chapter contains several workedout 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 REALWORLD ENDOFCHAPTER PROBLEMS
The endofchapter 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 safetyrelated 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 wellstructured 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 McGrawHill 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
WisconsinMadison, 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 equationsolving 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 publicationquality 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
builtin 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 builtin, 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 lookup 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 "EESCD" logo in the text). Each
EES solution provides detailed comments and online 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 0072463287). While it does not have the power and functionality of
the professional, commercial packages, HTT uses researchgrade 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.
BOOKSPECIFIC 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
FanBill 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
11 Thermodynamics and
Heat Transfer 2
12 Engineering Heat Transfer 4
13 Heat and Other Forms
of Energy 6
14 The First Law of
Thermodynamics 11
15 Heat Transfer
Mechanisms 17
16 Conduction 17
17 Convection 25
18 Radiation 27
19 Simultaneous Heat Transfer
Mechanism 30
110 ProblemSolving Technique 35
Topic of Special Interest:
Thermal Comfort 40
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HEAT TRANSFER
11  THERMODYNAMICS AND HEAT TRANSFER
Thermos
bottle
TU
Hot
coffee
±
^ Insulation
FIGURE 11
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 12
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. 11).
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. 12). 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 airconditioning system, the refrigerator and freezer, the
water heater, the iron, and even the computer, the TV, and the VCR. Of course,
energyefficient 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. 13).
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
Airconditioning
systems
Water out
oj i n I
Circuit boards
Car radiators
Power plants
Refrigeration systems
FIGURE 13
Some application areas of heat transfer.
cen58933_ch01.qxd 9/10/2002 8:29 AM Page 4
HEAT TRANSFER
FIGURE 14
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 (17431794) 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. 14).
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) (17531814) 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 (18181889) 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.
12  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, timeconsuming, 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 notsovisible 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 notsoaccurate 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. 15). 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 15
Modeling is a powerful engineering
tool that provides great insight and
simplicity at the expense of
some accuracy.
cen58933_ch01.qxd 9/10/2002 8:29 AM Page 6
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.
13  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 wellknown 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. 16).
• Energy = h
Stationary
fluid
Energy ;
FIGURE 16
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
(11)
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 PvT 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. 17).
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 17
Specific heat is the energy required to
raise the temperature of a unit mass
of a substance by one degree in a
specified way.
cen58933_ch01.qxd 9/10/2002 8:29 AM Page 8
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 18
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. 18). 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
(12)
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)
(13)
or
At/ = mC,. Ar
and
AH
mC pmc AT
(J)
(14)
FIGURE 19
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 constantpressure specific heats are identical for
incompressible substances (Fig. 19). 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)
(15)
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. 110). 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 fluidlike
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 110
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)
(16)
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)
(17)
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10
HEAT TRANSFER
:24W
: const.
3 m
Q _ 24 W .
^ 6 m 2
4 W/m 2
FIGURE 111
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 112
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. 111)
Q
A
(W/m 2 )
(18)
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 11 Heating of a Copper Ball
A 10cm diameter copper ball is to be heated from 100°C to an average tem
I perature of 150 C C in 30 minutes (Fig. 112). 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 /
(19)
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)
(110)
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)
(111)
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. 113)
Steady, rate form:
(112)
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 113
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 114
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)
(113)
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)
(114)
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. 114)
Stationary closed system, no work:
Q = mC r AT
(J)
(115)
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 SteadyFlow 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
steadyflow 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 steadyflow 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 crosssectional 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 crosssectional area is obtained by integration over A c .
The flow of a fluid through a pipe or duct can often be approximated to be
onedimensional. 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 onedimensional flow approximation, the mass flow rate of a fluid flow
ing in a pipe or duct can be expressed as (Fig. 115)
P TA r
(kg/s)
(116)
where p is the fluid density, T is the average fluid velocity in the flow direc
tion, and A c is the crosssectional 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)
(117)
13
CHAPTER 1
A = kD 2 /4 —
c
for a circular pipe
m =pVA
FIGURE 115
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 crosssectional 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 steadyflow 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 steadyflow system reduces to (Fig. 116)
Q = ihAh
rhC p AT
(kJ/s)
(118)
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
steadyflow systems.
s Control volume
r
~l
m
4 T' 2
L
___#_
transfer /A 2 V
FIGURE 116
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 steadystate or steadyflow system). Then
the energy balance for a surface can be expressed as
Surface energy balance:
(119)
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14
HEAT TRANSFER
WALL
Control
i/ surface
radiation
conduction
1^****' Qi
e,
N*
i convection
FIGURE 117
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. 117,
for example, can be expressed as
61 = 62 + 63
(120)
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 118
Schematic for Example 12.
EXAMPLE 12 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 1200W electric heating element inside (Fig. 118). 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 1200W 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 13 Heat Loss from Heating Ducts in a Basement
A 5mlong section of an air heating system of a house passes through an un
heated space in the basement (Fig. 119). 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 119
Schematic for Example 13.
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 A15).
Analysis We take the basement section of the heating system as our system,
which is a steadyflow 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 crosssectional 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
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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 120
Schematic for Example 14.

EXAMPLE 14 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. 120). 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 airtight 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 constantvolume and constantpressure 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 A1E and A15E).
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
15  HEAT TRANSFER MECHANISMS
In Section 11 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 highertemperature
medium to the lowertemperature 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 hightemperature medium to a
lowertemperature 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.
16  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
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18
HEAT TRANSFER
FIGURE 121
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 122
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)
(121)
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. 122). In the limiting case of Ax — > 0, the equation above reduces to the
differential form
Oc
kA
(IT
dx
(W)
(122)
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 Tx 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. 122 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 5mlong, 3mhigh, and 25cmthick 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. 123).
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19
CHAPTER 1
EXAMPLE 15 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. 124). 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 10hour 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 123
In heat conduction analysis, A represents
the area normal to the direction
of heat transfer.
Concrete roof
8 m
0.25 m
FIGURE 124
Schematic for Example 15.
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 122 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
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20
HEAT TRANSFER
TABLE
11
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 125
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 1mthick 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. 122 together with other
known quantities give the thermal conductivity (Fig. 125).
The thermal conductivities of materials vary over a wide range, as shown in
Fig. 126. 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
(highertemperature) 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 126
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. 127). 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 127
The mechanisms of heat conduction in
different phases of a substance.
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22
HEAT TRANSFER
TABLE 12
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
13
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 128
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 12. 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 13).
The variation of thermal conductivity over certain temperature ranges is neg
ligible for some materials, but significant for others, as shown in Fig. 128.
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 A3 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)
(123)
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 14. 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 16 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. 129. 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 onedimensional. Two
I thermocouples are embedded into each sample some distance L apart, and a
23
CHAPTER 1
TABLE 14
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 129
Apparatus to measure the thermal
conductivity of a material using two
identical samples and a thin resistance
heater (Example 16).
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 onedimensional 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 crosssectional 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 17 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. 130). 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/hft°F
FIGURE 130
The thermal conductivity value in
English units is obtained by multiplying
the value in SI units by 0.5778.
17 ■ 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 131
Heat transfer from a hot
surface to air by convection.
Natural
convection
Air
\ Xhot egg) v. /
FIGURE 132
The cooling of a boiled egg
by forced and natural convection.
TABLE 15
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
225
101000
25250
5020,000
2500100,000
•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. 131). 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. 132). For example, in the absence of a
fan, heat transfer from the surface of the hot block in Fig. 131 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)
(124)
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 15.
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 18 Measuring Convection Heat Transfer Coefficient
A 2mlong, 0.3cmdiameter electrical wire extends across a room at 15°C, as
shown in Fig. 133. 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 133
Schematic for Example 18.
18  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 xrays, 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 134
Blackbody radiation represents the
maximum amount of radiation that
can be emitted from a surface
at a specified temperature.
TABLE 16
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
0.920.97
Asphalt pavement
0.850.93
Red brick
0.930.96
Human skin
0.95
Wood
0.820.92
Soil
0.930.96
Water
0.96
Vegetation
0.920.96
V*
*^ref ^ "incid
Q ,. = a 6  A ,
*abs ^incident
FIGURE 135
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 StefanBoltzmann law as
£?e
uAJt
(W)
(125)
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
StefanBoltzmann 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. 134). 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)
(126)
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 16.
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. 135)
Q,
a 2incid
(W)
(127)
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. 136)
Q,
BaA,(T}T*^
(W)
(128)
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)
(129)
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 136
Radiation heat transfer between a
surface and the surfaces surrounding it.
EXAMPLE 19 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. 137).
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 16).
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 137
Schematic for Example 19.
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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 138
Although there are three mechanisms of
heat transfer, a medium may involve
only two of them simultaneously.
19  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. 138).
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 110 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. 139).
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 16).
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 139
Heat transfer from the person
described in Example 110.
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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 140
Schematic for Example 1—11.
r 2 = 200K 'EXAMPLE 111 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. 140. 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 Al 1), 0.026 W/m ■ °C for urethane insula
tion (Table A6), 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 141
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. 141 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 112 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. 142). 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 142
A chicken being cooked in a
microwave oven (Example 112).
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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 143
Schematic for Example 113.
I
2 EXAMPLE 113 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. 143). 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
110  PROBLEMSOLVING 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 realworld
problems. Solving such problems, especially complicated ones, requires a sys
tematic approach. By using a stepbystep approach, an engineer can reduce
the solution of a complicated problem into the solution of a series of simple
problems (Fig. 144). 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 144
A stepbystep approach can greatly
simplify problem solving.
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36
HEAT TRANSFER
FIGURE 145
The assumptions made while solving
an engineering problem must be
reasonable and justifiable.
FIGURE 146
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. 145).
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. 146).
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.
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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 3.75L 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. 147). 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 147
A result with more significant digits than
that of given data falsely implies
more accuracy.
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38
HEAT TRANSFER
FIGURE 148
An excellent wordprocessing
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 highsalaried engineers since any person who knows how to
use a wordprocessing 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 wordprocessing 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. 148). 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 builtin 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 researchgrade 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 114 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 doubleclicking on its icon, open a new
file, and type the following on the blank screen that appears:
xy=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 149
Most animals come into this world with
builtin insulation, but human beings
come with a delicate skin.
Thermal Comfort
Unlike animals such as a fox or a bear that are born with builtin 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 doublefloor 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 southfacing houses indicate that the value of solar heating was
recognized early in the history.
The term airconditioning 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 airconditioning 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 airconditioning 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 1kW resis
tance heater (Fig. 150). 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 17 per unit
body surface area. The surface area of a nude body was given by D.
DuBois in 1916 as
FIGURE150
Two fastdancing people supply
more heat to a room than a
1kW resistance heater.
0.202m  425 h
D.425 1,0.725
(m 2 )
(130)
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 17
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
60115
Aircraft, routine
70
Heavy vehicle
185
Miscellaneous Occupational
Activities:
Cooking
95115
Cleaning house
115140
Machine work:
Light
115140
Heavy
235
Handling 50kg bags
235
Pick and shovel work
235280
Miscellaneous Leisure
Activities:
Dancing, social
140255
Calisthenics/exercise
175235
Tennis, singles
210270
Basketball
290440
Wrestling, competitive
410505
*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. 151). 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. 152).
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 windchill 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,
solarheated 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. 153). 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 doublepane 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 70kg
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 handskin
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 151
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 551981).
FIGURE 152
A thermally comfortable environment.
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HEAT TRANSFER
FIGURE 153
Cold surfaces cause excessive heat loss
from the body by radiation, and thus
discomfort on that side of the body.
Shivering
FIGURE 154
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.
154). 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. 155). 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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CHAPTER 1
of sweating, increased heat production by shivering, and a heat stroke with
irreversible and lifethreatening 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 18.
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 155
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 18
Minimum fresh air requirements
in buildings (from ASHRAE
Standard 621989)
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
1.01.5
(per m 2 )
0.20.3
(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 StefanBoltzmann
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 StefanBoltzmann
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.
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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: McGrawHill, 2001 .
3. Y. A. Cengel and M. A. Boles. Thermodynamics — An
Engineering Approach. 4th ed. New York: McGrawHill,
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: PrenticeHall, 1999.
8. M. N. Ozisik. Heat Transfer — A Basic Approach. New
York: McGrawHill, 1985.
9. Robert J. Ribando. Heat Transfer Tools. New York:
McGrawHill, 2002.
10. F M. White. Heat and Mass Transfer. Reading, MA:
AddisonWesley, 1988.
PROBLEMS
Thermodynamics and Heat Transfer
11 C How does the science of heat transfer differ from the
science of thermodynamics?
12C What is the driving force for (a) heat transfer, (b) elec
tric current flow, and (c) fluid flow?
13C 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?
15C What is the difference between the analytical and ex
perimental approach to heat transfer? Discuss the advantages
and disadvantages of each approach.
16C What is the importance of modeling in engineering?
How are the mathematical models for engineering processes
prepared?
17C 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
18C 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 CDEES icon ® are solved using EES, and
complete solutions together with parametric studies are included
on the enclosed CD. Problems with a computerEES icon H are
comprehensive in nature, and are intended to be solved with a
computer, preferably using the EES software that accompanies
this text.
19C What are the mechanisms of energy transfer to a closed
system? How is heat transfer distinguished from the other
forms of energy transfer?
110C How are heat, internal energy, and thermal energy
related to each other?
111C 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?
112 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 24hour period, (b) the heat flux, and (c) the
fraction of heat dissipated from the top and bottom surfaces.
113E 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 eighthour work day, in kWh, and (b) the
heat flux on the surface of the chip, in W/in 2 .
114 Consider a 150W 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
115 A 1200W 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
2hour 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 2hour period. Take the unit cost of
electricity to be $0.07/kWh.
116 A 15cm X 20cm 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 10hour 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
117 A 15cmdiameter 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
118 The average specific heat of the human body is 3.6
kJ/kg • °C. If the body temperature of a 70kg 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.
119 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
120 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.
121E Consider a 60gallon 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
122C 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.
123C 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?
124C Define mass and volume flow rates. How are they re
lated to each other?
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CHAPTER 1
125 Two 800kg cars moving at a velocity of 90 km/h have
a headon 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.
126 A classroom that normally contains 40 people is to be
airconditioned using window airconditioning units of 5kW
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
127E 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
128 A 1m 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.
129 A 4m X 5m X 6m 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
130 A 4m X 5m X 7m 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 100W 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 P131
131 A student living in a 4m X 6m X 6m dormitory
room turns his 150W 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
132E A 10ft 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.
133 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.
134 A 50kg 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.
135 A 20kg 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
136 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 130
FIGURE P136
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HEAT TRANSFER
time, a paddle wheel driven by a 200W 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
137E A 90lbm mass of copper at 160°F and a 50lbm 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.
138 A 5m X 6m X 8m 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 200W 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.
139 A house has an electric heating system that consists of
a 300W 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.
140 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 1200W hair dryer at 100 kPa and 22°C, and leaves at 47°C.
The crosssectional 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 P140
141 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.
142E Air enters the duct of an airconditioning 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
143 Water is heated in an insulated, constant diameter tube
by a 7kW 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 \?
iAWvWVWvWV^n
\70°C
Resistance
heater, 7 kW
FIGURE P1 43
Heat Transfer Mechanisms
144C Define thermal conductivity and explain its signifi
cance in heat transfer.
145C What are the mechanisms of heat transfer? How are
they distinguished from each other?
146C What is the physical mechanism of heat conduction in
a solid, a liquid, and a gas?
147C 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.
148C Write down the expressions for the physical laws that
govern each mode of heat transfer, and identify the variables
involved in each relation.
149C How does heat conduction differ from convection?
150C Does any of the energy of the sun reach the earth by
conduction or convection?
151 C How does forced convection differ from natural
convection?
152C Define emissivity and absorptivity. What is Kirch
hoff's law of radiation?
153C What is a blackbody? How do real bodies differ from
blackbodies?
154C 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.
155C 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.
156C Which is a better heat conductor, diamond or silver?
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CHAPTER 1
157C Consider two walls of a house that are identical ex
cept that one is made of 10cmthick wood, while the other is
made of 25cmthick brick. Through which wall will the house
lose more heat in winter?
158C How do the thermal conductivity of gases and liquids
vary with temperature?
159C Why is the thermal conductivity of superinsulation
orders of magnitude lower than the thermal conductivity of
ordinary insulation?
160C Why do we characterize the heat conduction ability
of insulators in terms of their apparent thermal conductivity
instead of the ordinary thermal conductivity?
161 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
162 The inner and outer surfaces of a 5m X 6m 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 P162
163 The inner and outer surfaces of a 0.5cmthick 2m X
2m 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
164 [JJ^l Reconsider Problem 163. 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.
165 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
166E 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.
167 In a certain experiment, cylindrical samples of diameter
4 cm and length 7 cm are used (see Fig. 129). 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
168 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 0.5cmthick 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
169 Repeat Problem 168 for an electric power consump
tion of 28 W.
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HEAT TRANSFER
170 A heat flux meter attached to the inner surface of a
3cmthick 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
171 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.
172 [7(^1 Reconsider Problem 171. 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.
173 For heat transfer purposes, a standing man can be mod
eled as a 30cmdiameter, 1 70cmlong 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
174 Hot air at 80°C is blown over a 2m X 4m 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
175 rSi'M Reconsider Problem 174. 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.
176 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 0.5mmhigh side surfaces,
determine the temperature difference between the front and
back surfaces of the chip in steady operation.
Silicon
chip
0.5 mm
177 A 50cmlong, 800W 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.
178 A 5 cmexternaldiameter, 10mlong 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
179 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 P176
FIGURE P179
180 VcgM Reconsider Problem 179. 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.
181E The inner and outer glasses of a 6ft X 6ft double
pane window are at 60°F and 42°F, respectively. If the 0.25in.
space between the two glasses is filled with still air, determine
the rate of heat transfer through the window.
Answer: 439 Btu/h
182 Two surfaces of a 2cmthick 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.
183 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.
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184 An ice chest whose outer dimensions are 30 cm X
40 cm X 40 cm is made of 3cmthick 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 40cm X 40cm 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 P184
185 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 P185
186 [Z?vfl Reconsider Problem 185. 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
187E A 200ftlong 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
188 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 4mdiameter 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 thinshelled 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
189 Repeat Problem 188 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.
190 rSi'M Reconsider Problem 188. 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.
191 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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54
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
192 A 0.3cmthick, 12cmhigh, and 18cmlong 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
193 Consider a sealed 20cmhigh 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 P193
194 Using the conversion factors between W and Btu/h, m
and ft, and K and R, express the StefanBoltzmann constant
o = 5.67 X 10 8 W/m 2 • K 4 in the English unit Btu/h • ft 2 ■ R 4 .
195 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
196C Can all three modes of heat transfer occur simultane
ously (in parallel) in a medium?
197C 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.
198C 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.
199C 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.
1100 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
1101 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.
1102 A 1 .4mlong, 0.2cmdiameter 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
1103
Reconsider Problem 1102. 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.
1104E A 2indiameter 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.
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1105 fl&\ A 1000W 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 P1105
1106 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.
1107 A 3minternaldiameter 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
24hour 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
1108 fJb\ The roof of a house consists of a 15cmthick
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
14hour period.
1109E 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 P1109E
Problem Solving Technique and EES
1110C 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
1112
[J3 Solve the following system of two equations
with two unknowns using EES:
x*  y
3xy + y
7.75
3.5
1113
Solve the following system of three equations
with three unknowns using EES:
2x — y + z
3x 2 + 2y
xy + 2z
5
z + 2
1114
[tt3 Solve the following system of three equations
with three unknowns using EES:
1
Special Topic: Thermal Comfort
1115C 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?
1116C 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?
1117C What is asymmetric thermal radiation? How does it
cause thermal discomfort in the occupants of a room?
1118C How do (a) draft and (b) cold floor surfaces cause
discomfort for a room's occupants?
1119C 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?
1120C 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
1121 2.5 kg of liquid water initially at 18°C is to be heated
to 96°C in a teapot equipped with a 1200W 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.
1122 A 4mlong 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 P1122
1123
Reconsider Problem 1122. 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.
1124 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 P11 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 10minute shower as a result of installing this heat exchanger.
Answers: 18.8 kW, 10.8 kW, $0.0113
1125 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.
1126 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 thermostatcontrolled 15kW backup 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
1127 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 calmair
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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 30cm diameter, 170cmlong 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
1128 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.
1129 A 4m X 5m X 6m 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 24hour 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
1130 Consider a 3m X 3m X 3m 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
1131 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 P11 28
FIGURE P1131
1132 A 0.2L 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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HEAT TRANSFER
FIGURE P1132
1133 Tu'M Reconsider Problem 1132. 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.
1134E 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
1135 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.
1136 The hot water needs of a household are met by an
electric 60L hot water tank equipped with a 1 .6kW 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.
1137 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.
1138 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 lowe coated, argonfilled, 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 1.2m X 1.8m 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 P1138
1139
Reconsider Problem 1138. 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
1140 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.
1141 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.
1142 Design a 1200W electric hair dryer such that the air
temperature and velocity in the dryer will not exceed 50°C and
3/ms, respectively.
1143 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.
1144 Conduct this experiment to determine the heat transfer
coefficient between an incandescent lightbulb and the sur
rounding air using a 60W 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 threedimensional 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 onedimensional
when conduction is significant in one dimension only and negligible in the
other two dimensions, twodimensional when conduction in the third dimen
sion is negligible, and threedimensional 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 threedimensional 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
21 Introduction 62
11 OneDimensional Heat
Conduction Equation 68
23 General Heat
Conduction Equation 74
24 Boundary and
Initial Conditions 77
25 Solution of Steady
OneDimensional Heat
Conduction Problems 86
26 Heat Generation in a Solid 97
11 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 21
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 22
Indicating direction for heat transfer
(positive in the positive direction;
negative in the negative direction).
21  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. 21). 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. 22).
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 23. 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
zcoordinates. 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 23
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
xdirection 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. 24). 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 indepth 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) Steadystate
FIGURE 24
Steady and transient heat
conduction in a plane wall.
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64
HEAT TRANSFER
••65°C
FIGURE 25
Twodimensional heat transfer
in a long rectangular bar.
Primary
direction of
heat transfer
FIGURE 26
Heat transfer through the window
of a house can be taken to be
onedimensional.
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 onedimensional, two
dimensional, or threedimensional, 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 threedimensional.
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 twodimensional. 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 zdirection (along the bar) is negligible and
there is no change with time (Fig. 25).
A heat transfer problem is said to be onedimensional 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 onedimensional 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. 26).
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 onedimensional 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 Adirection) 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 onedimensional heat conduction as
2 c
kA
dT
dx
(W)
(21)
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 Tx diagram (Fig. 27). 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
xdirection. The negative sign in Eq. 21 ensures that heat transfer in the posi
tive xdirection 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 threedimensional. Figure
28 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)
(22)
65
CHAPTER 2
FIGURE 27
The temperature gradient dT/dx is
simply the slope of the temperature
curve on a Tx diagram.
In rectangular coordinates, the heat conduction vector can be expressed in
terms of its components as
Qn = QJ +Qyj +QJ
(23)
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 zdirections, 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
(24)
Here A x , A y and A z are heat conduction areas normal to the x, y, and
Zdirections, respectively (Fig. 28).
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 28
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 29
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 210
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. 29). 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. 210). 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)
(25)
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. 25
reduces to G = gV, where g is the constant rate of heat generation per unit
volume.
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67
CHAPTER 2
EXAMPLE 21 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. 211).
In your analysis, would you treat this as a transient or steadystate heat transfer
problem? Also, would you consider the heat transfer to be onedimensional 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 socalled 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 threedimensional 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 onedimensional. 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 21 1
Schematic for Example 21 .
EXAMPLE 22 Heat Generation in a Hair Dryer
The resistance wire of a 1200W hair dryer is 80 cm long and has a diameter of
D = 0.3 cm (Fig. 212). 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 1200W 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 212
Schematic for Example 22.
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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 213
Onedimensional heat conduction
through a volume element
in a large plane wall.
22  ONEDIMENSIONAL
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 onedimensional
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 213. 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
(26)
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
(27)
(28)
Substituting into Equation 26, 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
(29)
(210)
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
(211)
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
(212)
Noting that the area A is constant for a plane wall, the onedimensional tran
sient heat conduction equation in a plane wall becomes
Variable conductivity:
d , dT\ , . n dT
(213)
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
(214)
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. 214):
(1) Steadystate:
(d/dt = 0)
(2) Transient, no heat generation:
(8 = 0)
(3) Steadystate, 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
(215)
(216)
(217)
Note that we replaced the partial derivatives by ordinary derivatives in the
onedimensional 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 215. 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, onedimensional:
d 2 T
dx 2
=
FIGURE 214
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 215
Onedimensional heat conduction
through a volume element
in a long cylinder.
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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
(218)
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,) (219)
= gAAr (220)
Substituting into Eq. 218, we get
Q r Q r + Ar + gAAr = pCAAr
At
(221)
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
(222)
Taking the limit as Ar — > and Ar — > yields
Adr( kA Tr +8 = PC V
(223)
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
(224)
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
(225)
For the case of constant thermal conductivity, the equation above reduces to
1 dT
Constant conductivity:
ld_ dT\ .__
>' drV dr k a dt
(226)
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CHAPTER 2
where again the property a = k/pC is the thermal diffusivity of the material.
Equation 226 reduces to the following forms under specified conditions
(Fig. 216):
(1) Steadystate: 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) Steadystate, no heat generation: d I dT
(d/dt = and g = 0) d~r\ r ~dr
(227)
(228)
(229)
Note that we again replaced the partial derivatives by ordinary derivatives in
the onedimensional 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 onedimensional transient heat conduction equation for a
sphere is determined to be (Fig. 217)
Variable conductivity:
1 d
, r l k —
dr \ dr
P C
dT
dt
(230)
(a) The form that is ready to integrate
my"
(b) The equivalent alternative form
d 2 T dT „
FIGURE 216
Two equivalent forms of the
differential equation for the one
dimensional steady heat conduction in
a cylinder with no heat generation.
FIGURE 217
Onedimensional 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
(231)
where again the property a = k/pC is the thermal diffusivity of the material.
It reduces to the following forms under specified conditions:
(1) Steadystate:
(d/dt = 0)
(2) Transient,
no heat generation:
(g = 0)
(3) Steadystate,
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
(232)
(233)
(234)
where again we replaced the partial derivatives by ordinary derivatives in the
onedimensional steady heat conduction case.
cen58933_ch02.qxd 9/10/2002 8:46 AM Page 72
72
HEAT TRANSFER
Combined OneDimensional
Heat Conduction Equation
An examination of the onedimensional 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
(235)
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 steadystate or no heat generation cases as
described before.
800 W
FIGURE 218
Schematic for Example 23.
EXAMPLE 23 Heat Conduction through the Bottom of a Pan
Consider a steel pan placed on top of an electric range to cook spaghetti (Fig.
218). 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 xaxis, 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. 217,
d 2 T _
dx 2 '
which is the steady onedimensional 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 24
Heat Conduction in a Resistance Heater
A 2kW 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. 219). 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 onedimensional.
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. 227,
1 d I dT
r dr\ dr
which is the steady onedimensional 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 219
Schematic for Example 24.
EXAMPLE 25 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. 220). 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 onedimensional
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. 230 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 220
Schematic for Example 25.
cen58933_ch02.qxd 9/10/2002 8:46 AM Page 74
74
HEAT TRANSFER
which is the onedimensional 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 221
Threedimensional heat conduction
through a rectangular volume element.
23  GENERAL HEAT CONDUCTION EQUATION
In the last section we considered onedimensional 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 221. 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 + QzQx + AxQy
Q.
A£„
At
(236)
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£„
Ei + At
o 'element
E, = mC(T, + A ,
= gAxAyAz
T,) = pCAxAyAz{T, + At T t )
Substituting into Eq. 236, we get
Qx+Qy + QzQx
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
(237)
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
(238)
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 238 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
(239)
where the property a = k/pC is again the thermal dijfusivity of the material.
Equation 239 is known as the FourierBiot equation, and it reduces to these
forms under specified conditions:
(1) Steadystate:
(called the Poisson equation)
(2) Transient, no heat generation:
(called the diffusion equation)
(3) Steadystate, no heat generation:
(called the Laplace equation)
Note that in the special case of onedimensional heat transfer in the
xdirection, 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. 222).
d 2 T  d 2 T  d 2 T  g _
dx 2 dy 2 dz 2 k
(240)
d 2 T d 2 T d 2 T 1 dT
dx 2 3y 2 dz 2 ~ a dt
(241)
d_^ + djT + d^T =0
dx 2 9y 2 dz 2
(242)
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 223, by following the steps just outlined. It can also be ob
tained directly from Eq. 238 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 222
The threedimensional heat
conduction equations reduce to the
onedimensional ones when the
temperature varies in one
dimension only.
FIGURE 223
A differential volume element in
cylindrical coordinates.
cen58933_ch02.qxd 9/10/2002 8:46 AM Page 76
76
HEAT TRANSFER
FIGURE 224
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
(243)
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 224, by following the steps outlined above. It can also be
obtained directly from Eq. 238 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
(244)
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 onedimensional steadystate cases or lumped systems, since they result in
ordinary differential equations.
r
Metal
Heat
loss
600°F
billet
>
T„ = 65°F
FIGURE 225
Schematic for Example 26.
EXAMPLE 26 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 225 is initially at a uniform tempera
ture and is cooled uniformly from the top and bottom surfaces in the zdirection
as well as the lateral surface in the radial rdirection. Also, the temperature at
any point in the ball will change with time during cooling. Therefore, this is a
twodimensional 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. 243 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
2A  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. 226). 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. 227).
Therefore, we need to specify two boundary conditions for onedimensional
problems, four boundary conditions for twodimensional problems, and six
boundary conditions for threedimensional 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
onedimensional. 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 226
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 227
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
secondorder 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)
(245)
150°C
T(x, t)
m
70°C
7X0, = 150°C
T(L, f) = 70°C
FIGURE 228
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. 245 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 onedimensional heat transfer
through a plane wall of thickness L, for example, the specified temperature
boundary conditions can be expressed as (Fig. 228)
7/(0, t) = T x
T(L, t) = T 2
(246)
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 xdirection 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 )
(247)
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. 229).
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
(248)
79
CHAPTER 2
Heat
flux
%■■
Conduction
. 37X0,
dx
Conduction
.dT(L,t)
Heat
flux
9 L
FIGURE 229
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 xdirection,
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 wellinsulated 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. 230)
3T(0, t)
dx
dT(0,
dx
(249)
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 230
A plane wall with insulation
and specified temperature
boundary conditions.
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80
HEAT TRANSFER
L^— Center plane
Zero I
slope
Temperature
distribution
(symmetric
about center
plane)
:()
I —
I 2
dT(LI2, f)
dx
FIGURE 231
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. 231)
dT(L/2, t)
dx
(250)
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 232
Schematic for Example 27.
EXAMPLE 27 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
onedimensional. 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 232.
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 onedimensional heat transfer in the xdirection 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 ]
(251a)
(251 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 233.
In writing Eqs. 25 1 for convection boundary conditions, we have selected
the direction of heat transfer to be the positive xdirection 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. 234). 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 233
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 234
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 235
Schematic for Example 2
EXAMPLE 28 Convection and Insulation Boundary Conditions
Steam flows through a pipe shown in Figure 235 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 onedimensional. 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 /
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For onedimensional heat transfer in the xdirection in a plate of thickness L,
the radiation boundary conditions on both surfaces can be expressed as
(Fig. 236)
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
(252a)
(252b)
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 StefanBoltzmann 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 236
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. 237)
T A (x , t) = T B (x , t)
and
dT A (x , t)
dx
dT B (x Q , t)
' dx
(253)
(254)
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 237
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 /
(255)
This is illustrated in Examples 29 and 210.
T = 78°F
FIGURE 238
Schematic for Example 29.
EXAMPLE 29
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 238. 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 onedimensional 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
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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 210 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 onedimensional, 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 xaxis with
the origin at the inner surface of the wall, as shown in Figure 239. The heat
transfer through the wall is given to be steady and onedimensional, 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 xdirection, 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 xdirection, 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 239
Schematic for Example 210.
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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 onedimensional 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 240
Basic steps involved in the solution of
heat transfer problems.
25  SOLUTION OF STEADY ONEDIMENSIONAL
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
onedimensional 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 onedimensional 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. 240). This is demonstrated below with examples.
Plane
wall
^
T
2
L x
FIGURE 241
Schematic for Example 211.
EXAMPLE 211
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 241. 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
dT _
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. 242). 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. 243),
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 242
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 243
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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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, (256)
which is the desired solution since it satisfies not only the differential equation
but also the two specified boundary conditions. That is, differentiating Eq.
256 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. 256 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
(257)
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 212 A Wall with Various Sets of Boundary Conditions
Consider steady onedimensional 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. 244):
(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 244
Schematic for Example 212.
SOLUTION This is a steady onedimensional 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. 217)
cVT _
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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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 245
A boundaryvalue problem may have a
unique solution, infinitely many
solutions, or no solutions at all.
Resistance heater
1200 W
Base plate
FIGURE 246
Schematic for Example 213.
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 245.
EXAMPLE 213
Heat Conduction in the Base Plate of an Iron
Consider the base plate of a 1200W 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 246.
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 onedimensional 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 xdirection with its origin on
the inner surface, the differential equation for this problem can be expressed as
(Fig. 247)
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 247
The boundary conditions on the base
plate of the iron discussed
in Example 213.
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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 248
Schematic for Example 214.
EXAMPLE 214
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 248. 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 onedimensional 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
closedform 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 249
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 250
Schematic for Example 215.
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 trialanderror method) to give (Fig. 249)
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 215 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
250. 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 onedimensional 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. 251)
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 /7i)
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 {
(258)
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 251
Basic steps involved in the solution
of the steady onedimensional
heat conduction equation in
cylindrical coordinates.
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96
HEAT TRANSFER
x£ cylinder
kA
dT
dr
k(2irrL) ■
2<nkLC, = 2nkL , , , '
ln(r 2 /r,)
(259)
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 252
Schematic for Example 216.
EXAMPLE 216 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 252.
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 onedimensional 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,
(260)
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
(261)
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 253.
26 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. 254). 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 253
During steady onedimensional
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 254
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 )
(262)
h,T,
Q = E
FIGURE 255
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 startup
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. 255)
( Rate of ^
heat transfer
I from the solid /
Rate of 1
energy generation
\ within the solid j
(263)
or
Q =gV
(W)
(264)
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)
(265)
Combining Eqs. 264 and 265 and solving for the surface temperature
T s gives
T=T^ +
gV
hA,
(266)
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. 266 reduces to
h
2h
T — T A
J s, sphere * <» ni.
T = T 4
* 5, plane wall x °° ' /
T = T +
* 5, cylinder L °o ' ^^
(267)
(268)
(269)
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. 256). 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
(270)
99
CHAPTER 2
FIGURE 256
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. 270 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
(271)
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. 257)
T + AT
(272)
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,
(273)
(274)
r* Symmetry
line
FIGURE 257
The maximum temperature in
a symmetrical solid with uniform
heat generation occurs at its center.
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100
HEAT TRANSFER
Again the maximum temperature at the center can be determined from
Eq. 272 by adding the maximum temperature rise to the surface temperature
of the solid.
FIGURE 258
Schematic for Example 217.
EXAMPLE 217 Centerline Temperature of a Resistance Heater
A 2kW 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. 258). 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 onedimensional 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 2kW 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. 271 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 259
Schematic for Example 218.
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
218 and 219.
EXAMPLE 218 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 259. 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
217, 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 onedimensional 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. 227,
ld_( dT\ l
r drV dr k
This is a secondorder 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. 260). 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 260
The thermal symmetry condition at the
centerline of a wire in which heat
is generated uniformly.
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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. 271, which was obtained using an
energy balance on a control volume.
Interface
45°C
Ceramic layer
FIGURE 261
Schematic for Example 219.
EXAMPLE 219 Heat Conduction in a TwoLayer 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. 261). The wire
is embedded in a 0.5cmthick layer of ceramic whose thermal conductivity is
Ceramic =12 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 onedimensional since this twolayer 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 218, 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 215, 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
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104
HEAT TRANSFER
Thus the temperature of the centerline will be slightly above the interface
temperature.
Discussion This example demonstrates how steady onedimensional 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 262
Variation of the thermal conductivity
of some solids with temperature.
27 ■ VARIABLE THERMAL CONDUCTIVITY, k(T)
You will recall from Chapter 1 that the thermal conductivity of a material, in
general, varies with temperature (Fig. 262). 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 temperaturedependent 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 onedimensional 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
(275)
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. 275 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. 257, 259,
and 261 by the k awe expression (or value) from Eq. 275:
2= k A
plane wall ave
Ti  T 2 A rT,
k{T)dT
H cylinder •^' 1 ™ave ^
L
T,  T,
2rrL 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
(276)
(277)
(278)
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)
(279)
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)
(280)
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 onedimensional 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 263.
105
CHAPTER 2
T
Plane wall
k{T) = k Q (l+PT)
V
p>o
0
j3<0
■ T 2
L x
FIGURE 263
The variation of temperature
in a plane wall during steady
onedimensional heat conduction
for the cases of constant and variable
thermal conductivity.
EXAMPLE 220 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 264.
Assuming steady onedimensional 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 onedimensional.
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. 280).
{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 264
Schematic for Example 220.
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 265
Schematic for Example 221.
EXAMPLE 221 Heat Conduction through a Wall with k(T)
Consider a 2mhigh and 0.7mwide 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 265. 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 onedimensional
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 onedimensional.
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.
276 to be
Q ~ kaveA 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. 276 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. 266).
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 266
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 267
The derivative of a function at a point
represents the slope of the tangent
line of the function at that point.
FIGURE 268
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. 267)
y'W
dy(x)
dx
hm —
Ax>0 AX
lim
y(x + Ax)  y(x)
Ax
(281)
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 thirdorder 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. 268)
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
(282)
(283)
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
(284)
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. 284, 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. 284 is
called an indefinite integral since the value of the arbitrary constant C is
indefinite. The described procedure can be extended to higherorder deriv
atives (Fig. 269). 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
(285)
FIGURE 269
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. 284. 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. 270).
(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 270
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 271
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 272
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)
(286)
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. 271). 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. 272).
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
cen58933_ch02.qxd 9/10/2002 8:47 AM Page 111
111
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. 273).
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 273
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 onedimensional when conduction is significant
in one dimension only and negligible in the other two di
mensions. It is said to be twodimensional when conduction in
the third dimension is negligible and threedimensional 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 onedimensional 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,[TM  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 StefanBoltzmann 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:
McGrawHill, 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: McGrawHill, 1985.
6. F. M. White. Heat and Mass Transfer. Reading, MA:
AddisonWesley, 1988.
PROBLEMS
Introduction
21C Is heat transfer a scalar or vector quantity? Explain.
Answer the same question for temperature.
22C How does transient heat transfer differ from steady
heat transfer? How does onedimensional heat transfer differ
from twodimensional heat transfer?
23C Consider a cold canned drink left on a dinner table.
Would you model the heat transfer to the drink as one, two, or
threedimensional? 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.
24C Consider a round potato being baked in an oven.
Would you model the heat transfer to the potato as one, two,
or threedimensional? 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 P24
*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 EESCD icon ® are solved using EES, and
complete solutions together with parametric studies are included
on the enclosed CD. Problems with a computerEES icon H are
comprehensive in nature, and are intended to be solved with a
computer, preferably using the EES software that accompanies
this text.
25C Consider an egg being cooked in boiling water in a
pan. Would you model the heat transfer to the egg as one,
two, or threedimensional? 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.
26C 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 threedimensional? 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 P26
27C 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 threedimensional? Explain.
28C Consider heat loss from a 200L 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 threedimensional? Explain.
29C 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.
210C From a heat transfer point of view, what is the differ
ence between isotropic and unisotropic materials?
211C What is heat generation in a solid? Give examples.
212C Heat generation is also referred to as energy genera
tion or thermal energy generation. What do you think of these
phrases?
213C 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.
214E The resistance wire of a 1000W 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 P214E
ity and heat generation in its simplest form, and indicate what
each variable represents.
220 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.
221 Starting with an energy balance on a rectangular vol
ume element, derive the onedimensional transient heat con
duction equation for a plane wall with constant thermal
conductivity and no heat generation.
222 Starting with an energy balance on a cylindrical shell
volume element, derive the steady onedimensional heat con
duction equation for a long cylinder with constant thermal con
ductivity in which heat is generated at a rate of g.
215E [T^vfl Reconsider Problem 214E. 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.
216 In a nuclear reactor, heat is generated uniformly in the
5cmdiameter 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
217 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 P21 7
218 Consider a large 3cmthick 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
219 Write down the onedimensional transient heat conduc
tion equation for a plane wall with constant thermal conductiv
FIGURE P222
223 Starting with an energy balance on a spherical shell
volume element, derive the onedimensional transient heat
conduction equation for a sphere with constant thermal con
ductivity and no heat generation.
FIGURE P223
224 Consider a medium in which the heat conduction equa
tion is given in its simplest form as
dx 1
]_dT
a dt
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(«) Is heat transfer steady or transient?
(b) Is heat transfer one, two, or threedimensional?
(c) Is there heat generation in the medium?
(d) Is the thermal conductivity of the medium constant or
variable?
225 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 P229
(a) Is heat transfer steady or transient?
(b) Is heat transfer one, two, or threedimensional?
(c) Is there heat generation in the medium?
(d) Is the thermal conductivity of the medium constant or
variable?
226 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 threedimensional?
Is there heat generation in the medium?
Is the thermal conductivity of the medium constant or
variable?
227 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 threedimensional?
Is there heat generation in the medium?
Is the thermal conductivity of the medium constant or
variable?
228 Starting with an energy balance on a volume element,
derive the twodimensional transient heat conduction equation
in rectangular coordinates for T(x, y, f) for the case of constant
thermal conductivity and no heat generation.
229 Starting with an energy balance on a ringshaped vol
ume element, derive the twodimensional steady heat conduc
tion equation in cylindrical coordinates for T(r, z) for the case
of constant thermal conductivity and no heat generation.
230 Starting with an energy balance on a disk volume ele
ment, derive the onedimensional 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.
231 Consider a medium in which the heat conduction equa
tion is given in its simplest form as
Disk
Insulation
FIGURE P230
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 threedimensional?
(c) Is there heat generation in the medium?
(d) Is the thermal conductivity of the medium constant or
variable?
232 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 threedimensional?
(c) Is there heat generation in the medium?
(d) Is the thermal conductivity of the medium constant or
variable?
233 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 threedimensional?
(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
234C What is a boundary condition? How many boundary
conditions do we need to specify for a twodimensional heat
transfer problem?
235C What is an initial condition? How many initial condi
tions do we need to specify for a twodimensional heat transfer
problem?
236C What is a thermal symmetry boundary condition?
How is it expressed mathematically?
237C How is the boundary condition on an insulated sur
face expressed mathematically?
238C It is claimed that the temperature profile in a medium
must be perpendicular to an insulated surface. Is this a valid
claim? Explain.
239C Why do we try to avoid the radiation boundary con
ditions in heat transfer analysis?
240 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 P240
241 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.
242 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.
243 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.
244 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.
245 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 onedimensional 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 P245
246E A 2kW 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 onedimensional
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.
247 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 P247
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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 temperaturedependent thermal conductivity and
onedimensional 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.
248 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 onedimensional 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 P248
249 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 onedimensional heat transfer, express the mathemat
ical formulation (the differential equation and the boundary
and initial conditions) of this heat conduction problem. Do not
solve.
250 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 onedimensional heat transfer, express the mathemat
ical formulation (the differential equation and the boundary
Radiation
FIGURE P250
and initial conditions) of this heat conduction problem. Do not
solve.
251 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 onedimensional, 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 P251
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HEAT TRANSFER
Solution of Steady OneDimensional
Heat Conduction Problems
252C Consider onedimensional 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?
253C It is stated that the temperature in a plane wall with
constant thermal conductivity and no heat generation varies
linearly during steady onedimensional heat conduction. Will
this still be the case when the wall loses heat by radiation from
its surfaces?
254C 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?
255C 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?
256 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
onedimensional 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
257 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.
258 rSpM Reconsider Problem 257. 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.
259 Consider the base plate of a 800W household iron with
a thickness of L = 0.6 cm, base area of A = 160 cm 2 , and ther
Base
plate
85°C
FIGURE P259
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 onedimensional 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
260 Repeat Problem 259 for a 1200W iron.
261 ta'M Reconsider Problem 259. 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.
262E 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 P262E
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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
263 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.
264 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
onedimensional 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 P264
265 Repeat Problem 264 for a heat flux of 950 W/m 2 and
a surface temperature of 85 °C at the left surface at x = 0.
266E 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 onedimensional 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 P266E
the differential equation, and (c) determine the value of the
lower surface temperature of the plate at x = 0.
267E Repeat Problem 266E by disregarding radiation heat
transfer.
268 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 300W 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 onedimensional 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 P268
269
Reconsider Problem 268. 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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HEAT TRANSFER
the range of r = r x to r = r 2 , and discuss the results. Use the
EES (or other) software.
270 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 500W 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 onedimensional 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 P270
271
Reconsider Problem 270. 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
272C Does heat generation in a solid violate the first law of
thermodynamics, which states that energy cannot be created or
destroyed? Explain.
273C What is heat generation? Give some examples.
274C 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?
275C 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.
276C 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?
277 A 2kW 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 P277
278 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 threedimensional,
and (c) the value of heat flux on the side surface of the cylinder
at r = r .
279 TtPM Reconsider Problem 278. 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.
280E 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 P280E
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 onedimensional 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
281E [?(,">! Reconsider Problem 280E. 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.
282 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 P282
283 Consider a large 3cmthick 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.
284 Consider a large 5cmthick 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.
285 Tu'M Reconsider Problem 284. 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.
286 A 6mlong 2kW electrical resistance wire is made of
0.2cmdiameter 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
287E Heat is generated uniformly at a rate of 3 kW per ft
length in a 0.08in. 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.
288E Repeat Problem 287E for a manganese wire (k =
4.5 Btu/h ■ ft ■ °F).
289 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 P289
290
FIGURE P284
rSi'M Reconsider Problem 289. 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.
cen58933_ch02.qxd 9/10/2002 8:47 AM Page 122
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HEAT TRANSFER
291 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 P291
292 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 onedimensional 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
293 [ft^S Reconsider Problem 292. 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?
299 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 onedimensional heat transfer, obtain a relation for the
heat transfer rate through the wall.
2100 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 P21 00
Variable Thermal Conductivity, k[T)
294C Consider steady onedimensional 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.
295C Is the thermal conductivity of a medium, in general,
constant or does it vary with temperature?
296C Consider steady onedimensional 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.
297C 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?
298C When the thermal conductivity of a medium varies
linearly with temperature, is the average thermal conductivity
2101 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 onedimensional
heat transfer, obtain a relation for (a) the heat transfer rate
through the shell and (b) the temperature distribution T(r) in
the shell.
2102 Consider a 1.5mhigh and 0.6mwide 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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123
CHAPTER 1
2103 PT^ Reconsider Problem 2102. 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
2104C Why do we often utilize simplifying assumptions
when we derive differential equations?
2105C What is a variable? How do you distinguish a de
pendent variable from an independent one in a problem?
2106C Can a differential equation involve more than one
independent variable? Can it involve more than one dependent
variable? Give examples.
2107C What is the geometrical interpretation of a deriva
tive? What is the difference between partial derivatives and or
dinary derivatives?
2108C What is the difference between the degree and the
order of a derivative?
2109C 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?
2110C Consider a function f(x) and its derivative df/dx.
Does this derivative have to be a function of x?
2111C How is integration related to derivation?
2112C What is the difference between an algebraic equa
tion and a differential equation?
2113C What is the difference between an ordinary differen
tial equation and a partial differential equation?
2114C How is the order of a differential equation deter
mined?
2115C How do you distinguish a linear differential equation
from a nonlinear one?
2116C How do you recognize a linear homogeneous differ
ential equation? Give an example and explain why it is linear
and homogeneous.
2117C How do differential equations with constant coeffi
cients differ from those with variable coefficients? Give an ex
ample for each type.
2118C What kind of differential equations can be solved by
direct integration?
2119C Consider a third order linear and homogeneous dif
ferential equation. How many arbitrary constants will its gen
eral solution involve?
Review Problems
2120 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 P21 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.
2121 Consider a long rectangular bar of length a in the
xdirection and width b in the ydirection 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 twodimensional 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 P21 21
2122 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 twodimensional heat transfer, express the mathematical
formulation (the differential equation and the boundary condi
tions) of this heat conduction problem. Do not solve.
2123E 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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124
HEAT TRANSFER
Plate
520 R
. 'solai
FIGURE P2123E
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
2124E Repeat Problem 2123E for the case of no solar
radiation incident on the surface.
2125 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 onedimensional 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 P21 25
2126 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 thickwalled 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
2127 Repeat Problem 2126 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.
2128 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 onedimensional 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 P21 28
cen58933_ch02.qxd 9/10/2002 8:47 AM Page 125
2129 A 1000W 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 onedimensional 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 P21 29
2130 Repeat Problem 21 29 for a 1 500W iron.
2131E The roof of a house consists of a 0.8ftthick 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.
2132 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 0.4cm
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 P21 32
2133 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.
2134 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 P21 31 E
D
Fuel rod
FIGURE P21 34
2135 Consider a 20cmthick 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
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HEAT TRANSFER
radiation, (a) express the differential equations and the bound
ary conditions for steady onedimensional 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.
2136 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 25kW 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.
2137 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 icedwater 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
2138 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.
2139 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 onedimensional
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.
2140 Write an interactive computer program to calculate the
heat transfer rate and the value of temperature anywhere in
the medium for steady onedimensional 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.
2141 Write an interactive computer program to calculate the
heat transfer rate and the value of temperature anywhere in the
medium for steady onedimensional 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 onedimensional 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
31 Steady Heat Conduction
in Plane Walls 128
32 Thermal Contact
Resistance 138
33 Generalized Thermal
Resistance Networks 143
34 Heat Conduction in
Cylinders and Spheres 146
35 Critical Radius
of Insulation 153
36 Heat Transfer from
Finned Surfaces 156
37 Heat Transfer in
Common Configurations 169
Topic of Special Interest:
Heat Transfer Through
Walls and Roofs 175
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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 31
Heat flow through a wall is one
dimensional when the temperature of
the wall varies in one direction only.
31  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. 31).
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 onedimensional. The temperature of the wall in this case
will depend on one direction only (say the xdirection) 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
(31)
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 onedimensional 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)
(32)
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. 32).
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)
(33)
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. 33
by T, and L by x.
The Thermal Resistance Concept
Equation 33 for heat conduction through a plane wall can be rearranged as
Q
r,
cond, wall
(W)
(34)
129
CHAPTER 3
L x
FIGURE 32
Under steady conditions,
the temperature distribution in
a plane wall is a straight line.
where
L_
kA
(°C/W)
(35)
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,
(36)
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. 33).
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)
(37)
• 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 33
Analogy between thermal
and electrical resistance concepts.
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130
HEAT TRANSFER
Solid
r.
AAAAAA^
>T X
R.
1
com M ^
FIGURE 34
Schematic for convection
resistance at a surface.
where
R,,
hA,
(°C/W)
(38)
is the thermal resistance of the surface against heat convection, or simply the
convection resistance of the surface (Fig. 34). 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 38 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)
(39)
where
1
"rad"™j
(K/W)
(310)
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)
(311)
V
T<
r^A/VvW — * T °°
R
Solid
rad
Q=Q +Q ,
FIGURE 35
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. 35, 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)
(312)
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 36
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 onedimensional 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. 36. 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. 37)
Q
R„
(W)
(313)
(314)
(315)
FIGURE 37
A useful mathematical identity.
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132
HEAT TRANSFER
FIGURE 38
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 hiA
(316)
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. 315 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)
(317)
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. 38). 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)
(318)
where U is the overall heat transfer coefficient. A comparison of Eqs. 315
and 318 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 39
The thermal resistance network for
heat transfer through a twolayer
plane wall subjected to
convection on both sides.
UA
R„
(319)
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
(320)
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 twolayer
composite wall can be expressed as (Fig. 39)
Q
(321)
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134
HEAT TRANSFER
To find 7,: G=^r L
K conv,l
To find T 2 : Q=I^zJl
To find Ty Q
«...
FIGURE 310
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
(322)
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 singlelayer 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 twolayer case is analogous to the singlelayer 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
(323)
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 twolayer case shown in Fig. 39 are available and Q is
calculated from Eq. 321, the interface temperature T 2 between the two walls
can be determined from (Fig. 310)
Q
T,
r,
"conv, 1 "T "wall, 1
(324)
16°C
Wall
•2°C
3 m
L = 0.3m
FIGURE 31 1
Schematic for Example 31.
The temperature drop across a layer is easily determined from Eq. 317 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 31 Heat Loss through a Wall
Consider a 3mhigh, 5mwide, and 0.3mthick wall whose thermal con
ductivity is k = 0.9 W/m • °C (Fig. 311). 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. 33 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 32 Heat Loss through a SinglePane Window
Consider a 0.8mhigh and 1.5mwide 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
♦WWWh — WWW^^WWW— ♦
Glass
10°C
h 2 = 40 W/m 2 °C
L = 8 mm
FIGURE 31 2
Schematic for Example 32.
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 onedimensional 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. 312. 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 33 Heat Loss through DoublePane Windows
Consider a 0.8mhigh and 1.5mwide doublepane window consisting of two
4mmthick layers of glass (k = 0.78 W/m • °C) separated by a 10mmwide
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 doublepane 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 doublepane 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 8mmthick window glass is replaced by two 4mmthick glasses that
enclose a 10mmwide 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. 313. 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 onefourth of the result obtained in the previous example. This
explains the popularity of the double and even triplepane 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 doublepane window will rarely get fogged. A doublepane
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 313
Schematic for Example 33.
cen58933_ch03.qxd 9/10/2002 8:59 AM Page 13E
138
HEAT TRANSFER
FIGURE 314
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 315
A typical experimental setup for
the determination of thermal contact
resistance (from Song et al., Ref. 11).
32 ■ 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. 314, 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. 315, 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 crosssectional 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
(325)
It can also be expressed in an analogous manner to Newton's law of cooling as
Q =ft f AAr interfacc (326)
cen58933_ch03.qxd 9/10/2002 8:59 AM Page 139
where A is the apparent interface area (which is the same as the crosssectional
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) (327)
^* 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. 328 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 1cmthick 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 1cmthick 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
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140
HEAT TRANSFER
TABLE 31
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 316
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 31.
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. 316 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 32
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 34 Equivalent Thickness for Contact Resistance
The thermal contact conductance at the interface of two 1cmthick 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. 317).
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 A3).
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
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141
CHAPTER 3
TABLE 32
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
90200
0.32.5
3800
304 Stainless steel
Ground
1.14
20
47
1900
Aluminum
Ground
2.54
150
1.22.5
11,400
Copper
Ground
1.27
20
1.220
143,000
Copper
Milled
3.81
20
15
55,500
Copper (vacuum)
Milled
0.25
30
0.77
11,400
Dissimilar Metal Pairs
Stainless steel
10
2900
Aluminum
2030
20
20
3600
Stainless steel
Aluminum
1.02.0
20
10
20
16,400
20,800
Steel Ct30
Aluminum
Ground
1.42.0
20
10
1535
50,000
59,000
Steel Ct30
Aluminum
Milled
4.57.2
20
10
30
4800
8300
AluminumCopper
Ground
1.31.4
20
5
15
42,000
56,000
AluminumCopper
Milled
4.44.5
20
10
2035
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 2.3cmthick 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 35
Contact Resistance of Transistors
Four identical power transistors with aluminum casing are attached on one side
of a 1cmthick 20cm X 20cm square copper plate (k = 386 W/m • °C) by
screws that exert an average pressure of 6 MPa (Fig. 318). The base area of
each transistor is 8 cm 2 , and each transistor is placed at the center of a 10cm
X 10cm 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 317
Schematic for Example 34.
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142
HEAT TRANSFER
20°C
Copper
plate
FIGURE 3
Schematic
70°C
18
for Example
Plexiglas cover
35.
the transistor is not to exceed 70°C, determine the maximum power each
transistor can dissipate safely, and the temperature jump at the caseplate
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 onedimensional, although it is recognized that heat con
duction in some parts of the plate will be twodimensional 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 32 to be h c =
42,000 W/m 2 • °C, which corresponds to copperaluminum interface for the
case of 1.31.4 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
33  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
seriesparallel arrangements. Although such problems are often two or even
threedimensional, approximate solutions can be obtained by assuming one
dimensional heat transfer and using the thermal resistance network.
Consider the composite wall shown in Fig. 319, 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~,
(329)
(330)
(331)
since the resistances are in parallel.
Now consider the combined seriesparallel arrangement shown in Fig.
320. The total rate of heat transfer through this composite system can again
be expressed as
Q
(332)
143
CHAPTER 3
Insulation
V
,4,
CD *,
© k 2
« L ►
— www
R 2
Q = Q\+Q 2
FIGURE 31 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
(333)
(334)
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 onedimensional (say, in the
MAMAA
Qi
■AWI/VV^
Q
w —
*3 *c
vww vwvw — ♦
conv
FIGURE 320
Thermal resistance network for
combined seriesparallel 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 321
Schematic for Example 36.
xdirection) using the thermal resistance network are (1) any plane wall nor
mal to the xaxis is isothermal (i.e., to assume the temperature to vary in the
xdirection only) and (2) any plane parallel to the xaxis is adiabatic (i.e., to
assume heat transfer to occur in the xdirection 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 36 Heat Loss through a Composite Wall
A 3mhigh and 5mwide wall consists of long 16cm X 22cm cross section
horizontal bricks (k = 0.72 W/m • °C) separated by 3cmthick plaster layers
(k = 0.22 W/m • °C). There are also 2cmthick plaster layers on each side of
the brick and a 3cmthick rigid foam (k = 0.026 W/m ■ °C) on the inner side
of the wall, as shown in Fig. 321. 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 onedimensional 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 onedimensional since
it is predominantly in the xdirection. 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 25cm distance in the vertical direction. There is no variation in the hori
zontal direction. Therefore, we consider a 1mdeep and 0.25mhigh portion of
the wall, since it is representative of the entire wall.
Assuming any cross section of the wall normal to the xdirection to be
isothermal, the thermal resistance network for the representative section of
the wall becomes as shown in Fig. 321. 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 xdirection to be isothermal. We could also
solve this problem by going to the other extreme and assuming the surfaces par
allel to the xdirection to be adiabatic. The thermal resistance network in this
case will be as shown in Fig. 322. 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 322
Alternative thermal resistance
network for Example 36 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 323
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 onedimensional.
FIGURE 324
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. 323). 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
onedimensional. The temperature of the pipe in this case will depend on one
direction only (the radial rdirection) 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. 324). 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 onedimensional 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)
(335)
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
(336)
Substituting A = 2tjtL and performing the integrations give
T — T
Q coai ^ = 2itLk^j^ (W)
(337)
since Q
cond, cyl
constant. This equation can be rearranged as
xl, con
r.
d, cyl
R.
cy I
(W)
(338)
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)
(339)
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. 336. The result can be expressed as
Q
cond, sph
(340)
v sp h
where
Outer radius — Inner radius
sph Aur^k
4Tr(Outer radius)(Inner radius)(Thermal conductivity)
(341)
is the thermal resistance of the spherical layer against heat conduction, or sim
ply the conduction resistance of the spherical layer.
Now consider steady onedimensional 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. 325. 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
(342)
: ^conv,l +S cyI +R conv,:
FIGURE 325
The thermal resistance network
for a cylindrical (or spherical)
shell subjected to convection from
both the inner and the outer sides.
cen58933_ch03.qxd 9/10/2002 8:59 AM Page 14E
148
HEAT TRANSFER
where
Motel ^conv, 1
l
v cyl Jv conv, 2
( ln(r 2 /r,) 
1
(2nTL)/i 2ttL& (2nr 2 L)/7 2
(343)
for a cylindrical layer, and
^,,
#
conv, 1
l
v sph
1
(4irr 1 2 )/ !l
4m\r 2 k (4TTr?)h 2
(344)
/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 threelayered composite cylinder
of length L shown in Fig. 326 with convection on both sides can be ex
pressed as
Q
(345)
R
cyl,3
WWW < L A/WWV ♦ T^
R
conv, 2
FIGURE 326
The thermal resistance network for heat transfer through a threelayered 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
(346)
where A x = litr^L and A 4 = 2irr 4 L. Equation 346 can also be used for a
threelayered 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. 327). 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
(347)
149
CHAPTER 3
", T l T 2 T 3 T^
«JVWvVv^» J VWWv^^ J WVWV^^ J VWvW^»
R 2
Tv,
ir,
conv, 1
r„
ir 2
R conv,l +R l
r,
T,
*i
+ R,
h
T 3
«2
h
T„ 2
Rt + R„,
FIGURE 327
The ratio AT/R across any layer is
equal to Q , which remains constant in
onedimensional 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)
(348)
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 37 Heat Transfer to a Spherical Container
A 3m internal diameter spherical tank made of 2cmthick 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 24h 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.
cen58933_ch03.qxd 9/10/2002 8:59 AM Page 15C
150
HEAT TRANSFER
'Vad
I — WWW 1
* — WWW — ♦ — www
Rj R,
i — WWW — i
FIGURE 328
Schematic for Example 37.
Assumptions 1 Heat transfer is steady since the specified thermal conditions at
the boundaries do not change with time. 2 Heat transfer is onedimensional
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. 328. 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 24h 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 24h 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 38 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 3cmthick 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
cen58933_ch03.qxd 9/10/2002 8:59 AM Page 152
152
HEAT TRANSFER
Insulation
T l T 2 T 3
FIGURE 329
Schematic for Example 38.
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 onedimensional 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. 329. 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. 317 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
35  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. 330). 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. 331)
Q
S;„ + R n
1
lnf/j/ri)
2nLk h(2nr 2 L)
(349)
The variation of Q with the outer radius of the insulation r 2 is plotted in
Fig. 331. 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)
(350)
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 330
An insulated cylindrical pipe
exposed to convection from the outer
surface and the thermal resistance
network associated with it.
FIGURE 331
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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.
(351)
where k is the thermal conductivity of the insulation and h is the convection
heat transfer coefficient on the outer surface.
EXAMPLE 39
Heat Loss from an Insulated Electric Wire
A 3mmdiameter and 5mlong electric wire is tightly wrapped with a 2mm
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 onedimensional 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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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. 332. 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
2nkL 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. 350 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 4mm
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 332
Schematic for Example 39.
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HEAT TRANSFER
FIGURE 333
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).
36  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 334
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. 333 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. 334).
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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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. 335. 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)(TT.J
Substituting and dividing by Ax, we obtain
xi cond. X + A.Y x£ cond, X
Volume
element
FIGURE 335
Volume element of a fin at location x
having a length of Ax, crosssectional
area of A c , and perimeter of p.
Ax
hp(T  7/J =
(352)
Taking the limit as Ax — > gives
"Q cond
dx
hp(T  T„) =
(353)
From Fourier's law of heat conduction we have
dT
« cond
kA,.
dx
(354)
where A c is the crosssectional area of the fin at location x. Substitution of this
relation into Eq. 353 gives the differential equation governing heat transfer
in fins,
d_
dx
kA
dT
dx
hp(T  T a ) =
(355)
In general, the crosssectional 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 355 reduces to
dx 1
a 2 e = o
(356)
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158
HEAT TRANSFER
where
hp
~kA~.
(357)
and = 7 — r„ is the temperature excess. At the fin base we have
0/, — T b — T<n
Equation 356 is a linear, homogeneous, secondorder 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. 356 yields
zero. Therefore, the general solution of the differential equation Eq. 356 is
Q(x)
CV
(358)
~ Specified
temperature
(a) Specified temperature
(b) Negligible heat loss
(c) Convection
(d) Convection and radiation
FIGURE 336
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)
(359)
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. 336). 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)
(360)
Note that the temperature along the fin in this case decreases exponentially
from T b to T m as shown in Fig. 337. 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.)
(361)
where p is the perimeter, A c is the crosssectional 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
(362)
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. 338).
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
(363)
159
CHAPTER 3
(p = kD, A c = kD /4 for a cylindrical fin)
FIGURE 337
A long circular fin of uniform cross
section and the variation of
temperature along it.
i e fin
I / / > / 4 1
^base *"iin
FIGURE 338
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. 359. The
application of these two conditions on the general solution (Eq. 358) yields,
after some manipulations, this relation for the temperature distribution:
Adiabatic fin tip:
T(x)
cosh a(L — x)
cosh ah
(364)
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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
(365)
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 339
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. 339)
Corrected fin length:
L +
(366)
where A c is the crosssectional 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. 364 and 365.
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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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 crosssectional area A c = A b and length
L that is attached to the surface with a perfect contact (Fig. 340). 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 crosssectional 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.)
(367)
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. 341. 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
(368)
or
Q fin — "Hfin Q fin, r
%.n hA fm (T b  7/„)
(369)
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
(370)
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
(371)
since A fm = pL for fins with constant cross section. Equation 371 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 340
Fins enhance heat transfer from
a surface by enhancing surface area.
(a) Ideal
80°C
(b) Actual
56°C
FIGURE 341
Ideal and actual
temperature distribution in a fin.
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HEAT TRANSFER
FIGURE 342
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. 342 for fins on a plain surface and in Fig. 343 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 343
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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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
nofin case. The performance of fins expressed in terms of the/w effectiveness
e fin is defined as (Fig. 344)
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
(372)
Here, A b is the crosssectional 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 344
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
(373)
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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. 361. Substituting this relation into
Eq. 372, 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
(374)
^^
 3 x (r x w)
A fln =2XLXw + tXw (one fin)
~2x Lxw
FIGURE 345
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 crosssectional 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 liquidtogas 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„)
(375)
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")
(376)
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. 345). 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. 346.
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 346
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
(377)
Using a hand calculator, the values of tanh aL are evaluated for some values
of aL and the results are given in Table 33. 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 onedimensional 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 onedimensional fin analysis is negligi
ble (less than about 1 percent) when
TABLE 33
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 oneofakind complex
geometries, as shown in Table 34. 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„)
(378)
A small value of thermal resistance indicates a small temperature drop across
the heat sink, and thus a high fin efficiency.
FIGURE 347
Schematic for Example 310.
EXAMPLE 31 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 casetoambient thermal resistance, which accounts for both the natural
convection and radiation heat transfers.
The casetoambient 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 casetoambient 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. 347. 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 6063T5, are black anodized,
and are 76 mm (3 in.) long (courtesy of Vemaline Products, Inc.).
HS 5030
R= 0.9°C/W (vertical)
R = 1.2C/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= 1.8C/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 311 Selecting a Heat Sink for a Transistor
A 60W power transistor is to be cooled by attaching it to one of the commer
cially available heat sinks shown in Table 34. 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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168
HEAT TRANSFER
SOLUTION A commercially available heat sink from Table 34 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 60W 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 34 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 348
Schematic for Example 312.
EXAMPLE 312 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. 348. 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.
343. 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
37 ■ 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 onedimensional, and simple analytical
solutions can be obtained easily. But many problems encountered in practice
are two or threedimensional and involve rather complicated geometries for
which no simple solutions are available.
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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 )
(379)
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 35 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 34 and 379 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 313 and 314.
sT 2 = 10°C
Z = 0.5m
I_
 ^  D = 10 cm )
[' ■' ■ ■■' L = 30 m  — 4
FIGURE 349
Schematic for Example 313.
EXAMPLE 313 Heat Loss from Buried Steam Pipes
A 30mlong, 10cmdiameter hot water pipe of a district heating system is
buried in the soil 50 cm below the ground surface, as shown in Figure 349.
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 35 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
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171
CHAPTER 3
TABLE 35
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 semiinfinite 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 semiinfinite 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 semiinfinite 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)
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HEAT TRANSFER
TABLE 35 (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 semiinfinite
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
semiinfinite medium
^
2kD
1  0.25D/Z
(16) Isothermal sphere buried
in a semiinfinite 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 314 Heat Transfer between Hot and Cold Water Pipes
A 5mlong section of hot and cold water pipes run parallel to each other in a
thick concrete layer, as shown in Figure 350. 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 35 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 350
Schematic for Example 314.
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 315.
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174
HEAT TRANSFER
75°F
Wall, R= 13
■ 45°F
7/cc. ♦^vwvw^ i — ww < ^vwwv^^ 7^ 9
FIGURE 351
Schematic for Example 315.
EXAMPLE 315 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 thicknesstothermal 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 R13 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 onedimensional 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. 351.
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 24h period and
its cost to the home owner are
Q = Q At = (2712 Btu/h)(24h/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
(380)
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 Rvalue). 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 36 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 37 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 37
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.
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HEAT TRANSFER
TABLE 36
Unit thermal resistance (the ffvalue) 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 surfaceair 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 38. The effective emissivity of a
planeparallel air space is given by
Effective e l e 2
1
(381)
where s { and e 2 are the emissivities of the surfaces of the air space. Table
38 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 39 lists the thermal resistances of 20mm, 40mm, and 90mm
(0.75in., 1.5in., and 3.5in.) 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 lowemissivity surface reduces radiation heat transfer across
an air space and thus significantly increases the thermal resistance. The
thermal effectiveness of a lowemissivity 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 Ryalue 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 38
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 39
Unit thermal resistances (ffvalues) of wellsealed 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
20mm Air Space
40mm Air Space
90mm 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
0.75in.
Air Space
1.5in. A
ir Space
3.5in. 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 316
The /? Value of a Wood Frame Wall
Determine the overall unit thermal resistance (the ffvalue) and the overall heat
transfer coefficient (the L/factor) of a wood frame wall that is built around
38mm X 90mm (2x4 nominal) wood studs with a centertocenter distance
of 400 mm. The 90mmwide cavity between the studs is filled with glass fiber
insulation. The inside is finished with 13mm gypsum wallboard and the out
side with 13mm wood fiberboard and 13mm X 200mm 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 ffvalue 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 onedimensional. 3 Thermal properties of the wall and the heat transfer
coefficients are constant.
Properties The ffvalues of different materials are given in Table 36.
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 ffvalues from Table 36 and
calculating others, the total ffvalues 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 ffvalue of 2.23 X 5.68 = 12.7 (or nearly ff13) 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 Ufactor 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 Ufactor: 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 1kW 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 317 The ffValue of a Wall with Rigid Foam
The 13mmthick wood fiberboard sheathing of the wood stud wall discussed in
the previous example is replaced by a 25mmthick 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 ffvalue of the existing wall was determined in Example
316 to be 2.23 m 2 • °C/W. Noting that the ffvalues 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 ffvalue 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 ffvalue of the wall. This example demonstrated how to evaluate the new
ffvalue of a structure when some structural members are added or removed.
EXAMPLE 318 The ffValue of a Masonry Wall
Determine the overall unit thermal resistance (the ffvalue) and the overall heat
transfer coefficient (the ^/factor) of a masonry cavity wall that is built around
6in. 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 4in. face
brick with iin. cement mortar between the bricks and concrete blocks. The in
side finish consists of  in. gypsum wallboard separated from the concrete block
by fin.thick (1in. X 3in. nominal) vertical furring (ff = 4.2 h • ft 2 • °F/Btu)
whose centertocenter 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 ffvalue 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 ffvalue and the Ufactor of a masonry cavity wall are to be
determined.
Assumptions 1 Steady operating conditions exist. 2 Heat transfer through the
wall is onedimensional. 3 Thermal properties of the wall and the heat transfer
coefficients are constant.
Properties The ffvalues of different materials are given in Table 36.
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 ffvalues from Table 36, the overall ffvalue 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
flva
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 Ufactor 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 319
The ffValue 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
2in. X 4in. wood studs with a centertocenter distance of 16 in. The 3.5in.
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 ffvalue 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 ffvalue 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 onedimensional. 3 Thermal properties of the roof and the heat transfer
coefficients are constant.
Properties The ffvalues of different materials are given in Table 36.
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 ffvalues from Table 36, the
overall R
/alue of
the roof can be determined in
the table here.
Schematic
flva
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 Ui 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 tZfactor 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 2in. X
6in. joists on 400mm (16in.) or 600mm (24in.) centers. The fraction of
framing is usually taken to be 0.10 for joists on 400mm centers and 0.07
for joists on 600mm 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 roofattic
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. 352). 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 352
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/CE0335P,
U.S. Dept. of Energy).
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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 353
Three possible locations for an attic radiant barrier (from DOE/CE0335P, 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 354
Thermal resistance network for a
pitched roofatticceiling 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
aluminumcoated 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
353. 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. 354). Therefore, the overall
Kvalue of the roofceiling combination with an unvented attic depends on
the combined effects of the 7?value of the ceiling and the Kvalue 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 stillair case for the attic surfaces). Then it can be shown that the over
all i?value of the ceilingroof 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
(382)
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
Onedimensional 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
seriesparallel 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 Rvalue, 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.
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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 nofin 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 crosssectional 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 ElectricalAnalogue Shape Factors."
Chemical Engineering Progress 5 (1955), p. 67.
3. R. Barron. Cryogenic Systems. New York: McGrawHill,
1967.
4. L. S. Fletcher. "Recent Developments in Contact
Conductance Heat Transfer." Journal of Heat Transfer
110, no. 4B (1988), pp. 105979.
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. 62131.
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: McGrawHill, 1972.
cen58933_ch03.qxd 9/10/2002 9:00 AM Page 187
9. M. N. Ozisik. Heat Transfer — A Basic Approach. New
York: McGrawHill, 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. 4551.
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. 23741.
13. N. V. Suryanarayana. Engineering Heat Transfer. St. Paul,
MN: West Publishing, 1995.
Steady Heat Conduction in Plane Walls
31 C Consider onedimensional 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?
32C 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.
33C 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?
34C What does the thermal resistance of a medium
represent?
35C How is the combined heat transfer coefficient defined?
What convenience does it offer in heat transfer calculations?
36C Can we define the convection resistance per unit
surface area as the inverse of the convection heat transfer
coefficient?
37C Why are the convection and the radiation resistances at
a surface in parallel instead of being in series?
38C 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.
39C How does the thermal resistance network associated
with a singlelayer plane wall differ from the one associated
with a fivelayer 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 EESCD icon ® are solved using EES, and
complete solutions together with parametric studies are included
on the enclosed CD. Problems with a computerEES icon H are
comprehensive in nature, and are intended to be solved with a
computer, preferably using the EES software that accompanies
this text.
310C Consider steady onedimensional 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.
311C Consider steady onedimensional 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.
312C 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?
313C Consider a window glass consisting of two 4mm
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 8mmthick glass sheet under identical con
ditions.
314C 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.
31 5C The bottom of a pan is made of a 4mmthick 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 3mmthick copper layer sand
wiched between two 2mmthick aluminum layers. Will the
new design conduct heat better? Explain. Assume perfect con
tact between the layers.
316C 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?
317 Consider a 4mhigh, 6mwide, and 0.3mthick 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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188
HEAT TRANSFER
Aluminum
FIGURE P315C
Copper
318 Consider a 1.2mhigh and 2mwide 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.
319 Consider a 1 .2mhigh and 2mwide doublepane win
dow consisting of two 3mmthick layers of glass (k = 0.78
W/m • °C) separated by a 12mmwide stagnant air space (k =
0.026 W/m • °C). Determine the steady rate of heat transfer
through this doublepane 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
320 Repeat Problem 319, assuming the space between the
two glass layers is evacuated.
321 Ta'M Reconsider Problem 319. 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.
322E 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 P31 9
FIGURE P322E
323 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 24h 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.
324 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 24h 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.
325 A 12cm X 18cm 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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189
CHAPTER 3
30°C
Power
transistor
0.2 W
0.5 cm
— 0.4 cm
IGURE P324
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.
326 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
327 Water is boiling in a 25cmdiameter aluminum pan {k =
237 W/m ■ °C) at 95°C. Heat is transferred steadily to the boil
ing water in the pan through its 0.5cmthick 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.
328E A wall is constructed of two layers of 0.5inthick
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.
329 The roof of a house consists of a 3cmthick 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 14h
period.
r sky = iooK
T. = 10°C
15 cm
FIGURE P328E
FIGURE P329
330 A 2m X 1 .5m 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.
331 Repeat Problem. 330 for expanded perlite insulation
assuming conductivity is k = 0.052 W/m • °C.
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190
HEAT TRANSFER
332 [?(.*)! Reconsider Problem 330. 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.
333E 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 0.25in.
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 P333E
334 Consider a house that has a 10m X 20m base and a
4mhigh wall. All four walls of the house have an /{value of
2.31 m 2 • °C/W. The two 10m X 4m walls have no windows.
The third wall has five windows made of 0.5cmthick 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 1.5cmthick stagnant air space (k = 0.026
W/m • °C) enclosed between two 0.5cmthick 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 sevenmonth
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 singlepane win
dows to double pane windows.
335 The wall of a refrigerator is constructed of fiberglass in
sulation (k = 0.035 W/m • °C) sandwiched between two layers
of 1mmthick 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 P335
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.
336 rSi'M Reconsider Problem 335. 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.
337 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
338E A 0.03inthick 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 9in.long
side. What fraction of the heat conducted along that side is con
ducted through copper?
Thermal Contact Resistance
339C What is thermal contact resistance? How is it related
to thermal contact conductance?
340C 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
341C 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?
342C 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?
343C 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?
344C Explain how the thermal contact resistance can be
minimized.
3—45 The thermal contact conductance at the interface of two
1cmthick 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.
346 Six identical power transistors with aluminum casing
are attached on one side of a 1 .2cmthick 20cm X 30cm
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 10cm X
10cm 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 caseplate interface.
Copper
Transistor plate
15°C
1.2 cm
FIGURE P3^6
347 Two 5cmdiameter, 15cmlong 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 twobar 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
348 A 1 mmthick copper plate (k = 386 W/m ■ °C) is sand
wiched between two 5mmthick 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 P348
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HEAT TRANSFER
Generalized Thermal Resistance Networks
349C 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.
350C 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?
351 C What are the two approaches used in the develop
ment of the thermal resistance network for twodimensional
problems?
352 A 4mhigh and 6mwide wall consists of a long
18cm X 30cm cross section of horizontal bricks (k = 0.72
W/m ■ °C) separated by 3cmthick plaster layers (k = 0.22
W/m • °C). There are also 2cmthick plaster layers on each
side of the wall, and a 2cmthick 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 onedimensional heat transfer and disregarding radi
ation, determine the rate of heat transfer through the wall.
Foam
Brick
2 2
FIGURE P352
18 cm
1.5 cm
30 cm
1.5 cm
353
Reconsider Problem 352. 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.
354 A 10cmthick wall is to be constructed with 2.5m
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 2.5mlong 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.
355 A 12mlong and 5mhigh wall is constructed of two
layers of 1 cmthick 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.
356E A 10in. thick, 30ftlong, and 10fthigh wall is
to be constructed using 9in.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 0.5in. 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 P356E
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357 Consider a 5mhigh, 8mlong, and 0.22mthick 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 onedimensional, 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 P357
358 Repeat Problem 357 assuming that the thermal contact
resistance at the interfaces DF and EF is 0.00012 m 2 • °C/W.
359 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 oldfashioned coats.
Consider a jacket made of five layers of 0.1mmthick syn
thetic fabric {k = 0.13 W/m • °C) with 1.5mmthick 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 0.5mmthick 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 fivelayer ski jacket?
360 Repeat Problem 359 assuming the layers of the jacket
are made of cotton fabric (k = 0.06 W/m • °C).
361 A 5mwide, 4mhigh, and 40mlong kiln used to cure
concrete pipes is made of 20cmthick 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 3mmthick sheet metal covered with
2cmthick 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 P359
4 m
5 in
FIGURE P361
362 TtTM Reconsider Problem 361. 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.
363E Consider a 6in. X 8in. 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 centertocenter
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 P363E
Heat Conduction in Cylinders and Spheres
364C What is an infinitely long cylinder? When is it proper
to treat an actual cylinder as being infinitely long, and when
is it not?
365C 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.
366C Can the thermal resistance concept be used for a solid
cylinder or sphere in steady operation? Explain.
367 A 5minternaldiameter spherical tank made of
1.5cm 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
vQOIQ'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 24h
period. The heat of fusion of water at atmospheric pressure is
h if = 333.7 kJ/kg.
368 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 3cmthick 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.
369 fitt'M Reconsider Problem 368. 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.
370 (Jb\ A 50mlong 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 P370
FIGURE P367
371 Consider a 2mhigh 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 P371
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 46cmdiameter 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 3cmthick 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
372 rfigM Reconsider Problem 371. 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.
373 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 cmthick 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.
3C
:25°C
12.5 cm
FIGURE P373
374 Repeat Problem 3
resistance of 0.00008 m 2
insulation.
6 cm
73, assuming a thermal contact
°C/W between the can and the
375E 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 2in. 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 P375E
376 Hot water at an average temperature of 90°C is flowing
through a 15m 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.
377 Repeat Problem 376 for a pipe made of copper (k =
386 W/m ■ °C) instead of cast iron.
378E 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 P378E
379E Repeat Problem 378E, 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.
380
Reconsider Problem 378E. 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.
381 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 P381
Consider a 3mdiameter 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 thinshelled 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 5cmthick fiberglass insu
lation (k = 0.035 W/m ■ °C), and (c) insulated with 2cmthick
superinsulation which has an effective thermal conductivity of
0.00005 W/m • °C.
382 Repeat Problem 381 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
383C What is the critical radius of insulation? How is it
defined for a cylindrical layer?
384C 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?
385C 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?
386C 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?
387C 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?
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388 A 2mmdiameter and 1 0mlong electric wire is tightly
wrapped with a 1mmthick 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 P388
389E A 0.083in. diameter electrical wire at 115°F is
covered by 0.02in. 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
390E Repeat Problem 389E, assuming a thermal contact
resistance of 0.001 h • ft 2 • °F/Btu at the interface of the wire
and the insulation.
391 A 5mmdiameter spherical ball at 50°C is covered by a
1mmthick 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 P391
392
Reconsider Problem 391. 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
393C What is the reason for the widespread use of fins on
surfaces?
197
CHAPTER 3
394C What is the difference between the fin effectiveness
and the fin efficiency?
395C 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?
396C 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.
397C How does the overall effectiveness of a finned sur
face differ from the effectiveness of a single fin?
398C 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?
399C 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?
3100C 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.
3101C 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?
3102C Does the (a) efficiency and (b) effectiveness of a fin
increase or decrease as the fin length is increased?
3103C 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.
3104C 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.
3105C 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.
3106 Obtain a relation for the fin efficiency for a fin of con
stant crosssectional 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 P31 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.
3107 The casetoambient 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.
3108 A 40W power transistor is to be cooled by attaching
it to one of the commercially available heat sinks shown in
Table 3A. 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 P31 08
3109 A 30W power transistor is to be cooled by attaching
it to one of the commercially available heat sinks shown in
Table 3A. 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.
3110 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 2024T6 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
3111E 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 P31 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 P3111E
VR*.
3112E Repeat Problem 3111 for a silver spoon (k = 247
Btu/h • ft • °F).
3113E fu'M Reconsider Problem 31 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.
3114 A 0.3cmthick, 12cmhigh, and 18cmlong 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 0.2cmthick, 12cmhigh, and
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CHAPTER 3
18cmlong aluminum plate (k = 237 W/m • °C) with 864
2cmlong aluminum pin fins of diameter 0.25 cm is attached
to the back side of the circuit board with a 0.02 cmthick epoxy
adhesive (k = 1.8 W/m • °C). Determine the new temperatures
on the two sides of the circuit board.
3115 Repeat Problem 31 14 using a copper plate with cop
per fins (k = 386 W/m • °C) instead of aluminum ones.
3116 A hot surface at 100°C is to be cooled by attach
ing 3cmlong, 0.25cmdiameter aluminum pin fins (k =
237 W/m • °C) to it, with a centertocenter 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
1m X 1m section of the plate. Also determine the overall
effectiveness of the fins.
FIGURE P31 16
386
3117 Repeat Problem 3116 using copper fins (k
W/m • °C) instead of aluminum ones.
3118 fitt'M Reconsider Problem 3116. Using EES (or
KS other) software, investigate the effect of the cen
tertocenter distance of the fins on the rate of heat transfer
from the surface and the overall effectiveness of the fins. Let
the centertocenter 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 centertocenter distance, and discuss the results.
3119 Two 3mlong and 0.4cmthick cast iron (k = 52
W/m • °C) steam pipes of outer diameter 10 cm are connected
to each other through two 1 cmthick 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 P31 19
Heat Transfer in Common Configurations
3120C What is a conduction shape factor? How is it related
to the thermal resistance?
3121C What is the value of conduction shape factors in
engineering?
3122 A 20mlong and 8cmdiameter 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 P31 22
3123
Reconsider Problem 3122. 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.
3124 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
3125 [?(,■>! Reconsider Problem 3124. 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.
3126E A row of 3ftlong and 1 in. diameter used uranium
fuel rods that are still radioactive are buried in the ground par
allel to each other with a centertocenter 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"p8 in.^j
FIGURE P31 26
3127 Hot water at an average temperature of 60°C and an
average velocity of 0.6 m/s is flowing through a 5m section
of a thinwalled hot water pipe that has an outer diameter of
2.5 cm. The pipe passes through the center of a 14cmfhick
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
5mlong 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 P31 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
25mlong section of the pipe.
3129 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 20cm
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?
3130 Consider a 10mlong thickwalled concrete duct (k =
0.75 W/m • °C) of square crosssection. 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 P31 27
16 cm
20 cm
FIGURE P31 30
3128 Hot water at an average temperature of 80°C and an
average velocity of 1.5 m/s is flowing through a 25m 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
3131 A 3mdiameter 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.
3132 fitt'M Reconsider Problem 3131. 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.
3133 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 P31 39
Special Topics:
Heat Transfer through the Walls and Roofs
3134C What is the TJvalue of a wall? How does it differ
from the unit thermal resistance of the wall? How is it related
to the [/factor?
3135C What is effective emissivity for a planeparallel air
space? How is it determined? How is radiation heat transfer
through the air space determined when the effective emissivity
is known?
3136C The unit thermal resistances (/{values) of both
40mm and 90mm vertical air spaces are given in Table 39 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.
3137C 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?
3138C 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.
3139 Determine the summer R value and the [/factor of a
wood frame wall that is built around 38mm X 140mm wood
studs with a centertocenter distance of 400 mm. The 140
mmwide cavity between the studs is filled with mineral fiber
batt insulation. The inside is finished with 13mm gypsum
wallboard and the outside with 13mm wood fiberboard and
13mm X 200mm 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
3140 The 13mmthick wood fiberboard sheathing of the
wood stud wall in Problem 3139 is replaced by a 25mm
thick rigid foam insulation. Determine the percent increase in
the R value of the wall as a result.
3141E Determine the winter R value and the [/factor of a
masonry cavity wall that is built around 4in. thick concrete
blocks made of lightweight aggregate. The outside is finished
with 4in. 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 (1in. by 3in. nominal) vertical furring whose centerto
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 P31 41 E
3142 Consider a flat ceiling that is built around 38mm X
90mm wood studs with a centertocenter distance of 400 mm.
The lower part of the ceiling is finished with 13mm gypsum
wallboard, while the upper part consists of a wood subfloor
(R = 0.166 m 2 • °C/W), a 13mm 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 P31 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 90mm
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.
3143 Determine the winter lvalue and the {/factor of a
masonry cavity wall that consists of 100mm common bricks,
a 90mm air space, 100mm concrete blocks made of light
weight aggregate, 20mm air space, and 13mm gypsum wall
board separated from the concrete block by 20mmthick
(1in. X 3 in. nominal) vertical furring whose centertocenter
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
3144 Repeat Problem 3143 assuming one side of both air
spaces is coated with a reflective film of e = 0.05.
3145 Determine the winter Rvahie and the [/factor of a
masonry wall that consists of the following layers: 100mm
face bricks, 100mm common bricks, 25mm urethane rigid
foam insulation, and 13mm gypsum wallboard.
Answers: 1.404 m 2 • °C/W, 0.712 W/m 2 • °C
3146 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.
3147 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 100mm face brick is added to the outside,
leaving a 20mm 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 3mhigh, 7mlong section of the
wall after modification when the indoor and outdoor tempera
tures are 22°C and — 5°C, respectively.
FIGURE P31 47
FIGURE P31 43
3148 Determine the summer and winter ^values, in
m 2 • °C/W, of a masonry wall that consists of 100mm face
bricks, 13mm of cement mortar, 100mm lightweight concrete
block, 40mm air space, and 20mm plasterboard.
Answers: 0.809 and 0.795 m 2 ■ °C/W
3149E 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
3150 Two homes are identical, except that the walls of one
house consist of 200mm lightweight concrete blocks, 20mm
air space, and 20mm plasterboard, while the walls of the other
house involve the standard R2.4 m 2 • °C/W frame wall con
struction. Which house do you think is more energy efficient?
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3151 Determine the /Jvalue of a ceiling that consists of a
layer of 19mm 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 P31 51
Review Problems
3152E 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.
3153E Repeat Problem 3152E, assuming that a 0.01in.
thick limestone layer has formed on both the inner and outer
surfaces of the tube.
3154 A 1.2mdiameter and 6mlong 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 7.5cmthick glass wool
insulation (k = 0.038 W/m • °C).
3155 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 15mlong 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
3156 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 10m 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 10h curing
process that night.
Steam is supplied by a gasfired 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 P31 54
FIGURE P31 56
3157 Consider an 18cm X 18cm multilayer circuit board
dissipating 27 W of heat. The board consists of four layers
of 0.2mmthick copper (k = 386 W/m • °C) and three layers of
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HEAT TRANSFER
Copper
FIGURE P31 57
1.5mmthick 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 18cm epoxy laminate strip (or 1.5 W per 1cm X 18cm
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.
3158 The plumbing system of a house involves a 0.5m 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 P31 58
3159 Repeat Problem 3158 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.
31 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
0.12in.thick towel (k = 0.035 Btu/h • ft ■ °F). You may use the
properties of water for potato.
3161E Repeat Problem 3160E assuming there is a 0.02
in. thick air space (k = 0.015 Btu/h • ft • °F) between the potato
and the towel.
3162 An ice chest whose outer dimensions are 30 cm X
40 cm X 50 cm is made of 3cmthick 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 40cm X
50cm 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 P31 62
3163 A 4mhigh and 6mlong wall is constructed of two
large 2cmthick steel plates (k = 15 W/m • °C) separated by
1cm thick and 20cmwide steel bars placed 99 cm apart. The
Steel plates
2 cm
20 cm
Fiberglass
 insulation
99 cm
1 cm
2 cm
FIGURE P31 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?
3164 A 0.2cmthick, 10cmhigh, and 15cmlong 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 cmthick, 10cmhigh, and 15cmlong alumi
num plate (k = 237 W/m • °C) with 20 0.2cmthick, 2cm
long, and 15cmwide aluminum fins of rectangular profile are
attached to the back side of the circuit board with a 0.03cm
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 P31 64
3165 Repeat Problem 3164 using a copper plate with cop
per fins (k = 386 W/m • °C) instead of aluminum ones.
3166 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 10m X
5m 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 P31 66
3167 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 1mmthick leather (k = 0.159
W/m ■ °C) that covers half of the body while the second one is
wearing clothes made of 1mmthick 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
25cmdiameter 1.7mlong cylinder, determine the fractions
of heat lost from each person by perspiration.
3168 A 6mwide 2.8mhigh wall is constructed of one
layer of common brick (k = 0.72 W/m • °C) of thickness
20 cm, one inside layer of lightweight 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.
3169 Reconsider Problem 31 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).
3170 Cold conditioned air at 12°C is flowing inside a
1.5cmthick 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
convectionradiation 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.
3171 When analyzing heat transfer through windows, it
is important to consider the frame as well as the glass area.
Consider a 2mwide 1.5mhigh woodframed window with
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HEAT TRANSFER
85 percent of the area covered by 3mmthick singlepane 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.
3172 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.
3173 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 6mdiameter 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 5cmthick 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.
3174 A 15cm X 20cm hot surface at 85°C is to be cooled
by attaching 4cmlong aluminum (k = 237 W/m ■ °C) fins of
2mm X 2mm 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.
3175 [?(,■>! Reconsider Problem 3174. 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?
3176 A 1 .4mdiameter 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?
3177 A 0. 6mdiameter 1.9mlong cylindrical tank con
taining liquefied natural gas (LNG) at — 160°C is placed at the
center of a 1 .9mlong 1 .4m X 1 .4m 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
3178 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.
3179 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?
3180 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.
3181 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.
3182 Repeat Problem 3181 for a submarine with a crew of
60 people.
3183 A house with 200m 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 singlepaned windows in the house that are 1.2 m wide
and 1.8 m high. The house has RX9 (in h • ft 2 • °F/Btu) insula
tion in the walls and R30 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.
3184 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.
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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 onedimensional heat conduction problems such as those asso
ciated with a large plane wall, a long cylinder, a sphere, and a semiinfinite
medium using transient temperature charts and analytical solutions. Finally,
we consider transient heat conduction in multidimensional systems by uti
lizing the product solution.
CONTENTS
41 Lumped Systems Analysis 210
42 Transient Heat Conduction
in Large Plane Walls, Long
Cylinders, and Spheres
with Spatial Effects 216
43 Transient Heat Conduction
in SemiInfinite Solids 228
44 Transient Heat Conduction in
Multidimensional Systems 231
Topic of Special Interest:
Refrigeration and
Freezing of Foods 239
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210
HEAT TRANSFER
(a) Copper ball
(b) Roast beef
FIGURE 41
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 42
The geometry and parameters
involved in the lumped
system analysis.
41 ■ 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. 41). 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. 42).
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
(41)
Noting that m = pV and dT = d(T — r„) since T„ = constant, Eq. 41 can be
rearranged as
d(T TJ)
TT a
hA s
pvc„
dt
(42)
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 '
(43)
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)
(44)
(45)
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 44 is plotted
in Fig. 43 for different values of b. There are two observations that can be
made from this figure and the relation above:
1. Equation 44 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. AA, 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)
(46)
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)
(47)
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. 44)
e„
mC p {T a  r,)
(kJ)
(48)
We could also obtain this equation by substituting the T(t) relation from Eq.
44 into the Q(t) relation in Eq. 46 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 43
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 44
Heat transfer to
or from
a body
reaches its maximum value
when the body reaches
the environment temperature.
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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
(49)
Bi
heat convection
heat conduction
FIGURE 45
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. 45)
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
wellrounded 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.
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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. 46).
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. 47). 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. 48). 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 46
Small bodies with high thermal
conductivities and low convection
coefficients are most likely
to satisfy the criterion for
lumped system analysis.
FIGURE 47
Analogy between heat transfer to a
solid and passenger traffic
to an island.
Convection
/? = 2000W/m 2 °C
FIGURE 48
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
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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 49
Schematic for Example 41.
EXAMPLE 41 Temperature Measurement by Thermocouples
The temperature of a gas stream is to be measured by a thermocouple whose
junction can be approximated as a 1mmdiameter sphere, as shown in Fig.
49. 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. 44 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 junctiongas 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 42 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
30cmdiameter, 1.70mlong cylinder, estimate the time of death of that per
son (Fig. 410).
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 30cmdiameter, 1.70mlong
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 A9).
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 + 2nrj 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 410
Schematic for Example 42.
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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. 44,
T(t)  r»
which yields
25 20
3720
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.
42  TRANSIENT HEAT CONDUCTION IN LARGE
PLANE WALLS, LONG CYLINDERS, AND
SPHERES WITH SPATIAL EFFECTS
In Section, 41, 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 onedimensional
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.
411. 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. 412. 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 41 1
Schematic of the simple
geometries in which heat
transfer is onedimensional.
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 timeconsuming 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 41 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 halfthickness L by the outer radius r . Note that
the characteristic length in the definition of the Biot number is taken to be the
cen58933_ch04.qxd 9/10/2002 9:12 AM Page 21E
218
HEAT TRANSFER
halfthickness L for the plane wall, and the radius r for the long cylinder and
sphere instead of VIA used in lumped system analysis.
The onedimensional 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
(410)
(411)
(412)
where the constants A, and \ l are functions of the Bi number only, and their
values are listed in Table 41 against the Bi number for all three geometries.
The function J is the zerothorder Bessel function of the first kind, whose
value can be determined from Table 42. 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
(413)
(414)
(415)
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 oneterm 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. 413, 414, and 415 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 oneterm 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 firstorder
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. 416).
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)
(416)
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
SJ , . 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
PsNJS^^""^ ^^t^T^vV
.^^"I^zlt^U^^
5§§§fffipfc====;
i \— i ^yh ri —  i a r, h— ii
JuTtlmH^ mt^
+_T E!?^ >c + + L L T3"ZdiJ 
S \ \ ^JTOH — ^_.^
+ t*'tg*4iI—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^VrlX^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)
TT„
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' /
/ '
ttt
<tty
1 t
/
/ /
tjt 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
ttjl
1111/
SV O'j O'JJ 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
At 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 413
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
ZLLl
f4
J 1 _
^^ :
iiii ! 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
41
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)
TT 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.
64
*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 414
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 "44=
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\^ \ liO A Vl\^
■ifo\rt \ N \ \ v vV
4 HP V V \ Mlfe V A^ ^
a^ \ v \ \ M \\
pU \ V v v V \ \\
LpjII lJ t V\ i_i_i_Li_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)
TT 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 415
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 413c, 414c, and 415c
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. 417).
The fraction of heat transfer can also be determined from these relations,
which are based on the oneterm 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
(417)
(418)
(419)
The use of the Heisler/Grober charts and the oneterm 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. 418)
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
(420)
223
CHAPTER 4
(a) Finite convection coefficient
(b) Infinite convection coefficient
FIGURE 416
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 oneterm 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 417
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 418
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 43 Boiling Eggs
An ordinary egg can be approximated as a 5cmdiameter sphere (Fig. 419).
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 onedimensional 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 oneterm
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 A9).
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 oneterm 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 41,
3.0753,
1.9958
Substituting these and other values into Eq. 415 and solving for t gives
r  7L
T; ~ T„
A,e~V
70 — 95
>  „: = 1.9958e' 30753 ^
595
> t = 0.209
which is greater than 0.2, and thus the oneterm 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 44 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. 420). 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 onedimensional 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 oneterm 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 A3).
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 oneterm solutions. Here we will use the
charts to demonstrate their use. Noting that the halfthickness of the plate is
L = 0.02 m, from Fig. 413 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 41 9
Schematic for Example 43.
T rr _ = 500°C
h = 120 W/m 2 °C
/
2L = 4cm
Brass
plate
J, = 20°C
FIGURE 420
Schematic for Example 44.
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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 halfthickness 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 421
Schematic for Example 45.
EXAMPLE 45 Cooling of a Long
Stainless Steel Cylindrical Shaft
A long 20cmdiameter cylindrical shaft made of stainless steel 304 comes out
of an oven at a uniform temperature of 600°C (Fig. 421). 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 onedimensional 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 oneterm 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 A3). 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. 414 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. 414cfor 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 oneterm
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 41 to be
Substituting these values into Eq. 414 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 42 to be 0.430.
Then the fractiona
1 heat tran
sfer is
determined from Eq. 418 to be
Q
*max
= l26
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 422
Schematic of a semiinfinite body.
43  TRANSIENT HEAT CONDUCTION
IN SEMIINFINITE SOLIDS
A semiinfinite solid is an idealized body that has a single plane surface and
extends to infinity in all directions, as shown in Fig. 422. 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
semiinfinite medium in determining the variation of temperature near its sur
face. Also, a thick wall can be modeled as a semiinfinite 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 semiinfinite 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. 423 graphically for the nondimensionalized temperature
defined as
1  6(x, f) = 1
T(x, t)
T(x,t)T t
(421)
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 semiinfinite 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 423
Variation of temperature with position and time in a semiinfinite 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 onedimensional heat conduction prob
lem in a semiinfinite 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
(422)
where the quantity erfc (£) is the complementary error function, defined as
erfc(£) = 1 ^= f e" 2 du
Vtt Jo
(423)
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 43. For the special
case of h — > °o, the surface temperature T s becomes equal to the fluid temper
ature r„, and Eq. 422 reduces to
T(x, f)
erfc
(424)
cen58933_ch04.qxd 9/10/2002 9:12 AM Page 23C
230
HEAT TRANSFER
TABLE 43
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. 423 is a plot of the exact analytical solution given by Eq. 423, it is sub
ject to reading errors, and thus is of limited accuracy.
,T. = 10°C
Soil
Water pipe
r ; =i5°c
FIGURE 424
Schematic for Example 46.
EXAMPLE 46 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. 424). 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. 423,
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. 424:
T(x,t)
erfc
2Vat,
015
10  15
erfc
iVat
0.60
The argument that corresponds to this value of the complementary error func
tion is determined from Table 43 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
44  TRANSIENT HEAT CONDUCTION IN
MULTIDIMENSIONAL SYSTEMS
The transient temperature charts presented earlier can be used to determine the
temperature distribution and heat transfer in onedimensional heat conduction
problems associated with a large plane wall, a long cylinder, a sphere, and a
semiinfinite 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 semiinfinite cylinder or
plate, and even threedimensional problems associated with geometries such
as a rectangular prism or a semiinfinite 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 (twodimensional)
FIGURE 425
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.
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232
HEAT TRANSFER
Plane wall
^1
Long
cylinder
FIGURE 426
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 427
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. 425). The solution in such multidimensional geometries can
be expressed as the product of the solutions for the onedimensional 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 twodimensional transient heat conduction
problem. When the properties are assumed to be constant, it can be shown that
the solution of this twodimensional problem can be expressed as
T(r, x, t)  T a
short
cylinder
T(x, t)
plane
w;ill
T(r, t)
infinite
cylinder
(425)
That is, the solution for the twodimensional short cylinder of height a and
radius r is equal to the product of the nondimensionalized solutions for the
onedimensional 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. 426. We generalize this as follows: the solution for a multi
dimensional geometry is the product of the solutions of the onedimensional
geometries whose intersection is the multidimensional body.
For convenience, the onedimensional 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
(426)
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. 427, and thus the transient temperature distribution
for this rectangular bar can be expressed as
T(x, y, t)
W*. OQwaiiCy.
(427)
The proper forms of the product solutions for some other geometries are given
in Table 4—4. It is important to note that the xcoordinate is measured from the
surface in a semiinfinite 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 twodimensional problem involves the product of
two onedimensional solutions, whereas the solution of a threedimensional
problem involves the product of three onedimensional 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 onedimensional 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 44
Multidimensional solutions expressed as products of onedimensional 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
Semiinfinite cylinder
Hx,r,t) = e cyl (r,t)() wM (x,f)
Short cylinder
Semiinfinite medium
K.v,v.O = e semWnf (x,f)e semWnf ^o
Quarterinfinite medium
B(x,y,z,t) =
9 semiinf (' V > f )8 scmi i„f & e s emiinf 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
Semiinfinite plate
Hx,y,z,t) =
^wall^Oe^.^O'.Oe^.infCZ.O
Quarterinfinite 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
Semiinfinite rectangular bar
Q(x,y,z,t) =
e „an^oe wall (.v,oe wall ( Z ,o
Rectangular parallelepiped
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HEAT TRANSFER
transient heat transfer for a twodimensional geometry formed by the inter
section of two onedimensional geometries 1 and 2 is
_e
*iraa
total, 2D
_Q_
Cm.
_Q_
_Q_
(428)
Transient heat transfer for a threedimensional body formed by the inter
section of three onedimensional bodies 1, 2, and 3 is given by
_Q_
Sc m a
_Q_
iima
_Q_
i/ma
_Q_
fcraa
_Q_
*£ma
_Q_
sima
(429)
The use of the product solution in transient two and threedimensional heat
conduction problems is illustrated in the following examples.
r„ = 25°C
h = 60W/m 2 °C
T t = 120 D C
FIGURE 428
Schematic for Example 47.
EXAMPLE 47 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 twodimensional, and
thus the temperature varies in both the axial x and the radial rdirections. 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 oneterm 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 A3). 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. 428. The dimensionless temperature at the center of the
plane wall is determined from Figure 413a 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,or„\
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
TiTn
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 48 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 47.
(P =
= 8530
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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. 413c 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. 428,
^ 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 49 Cooling of a Long Cylinder by Water
A semiinfinite 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 semiinfinite 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 semiinfinite cylinder is two
dimensional, and thus the temperature varies in both the axial x and the radial
rdirections. 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 oneterm
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 A3). More accurate results can be
obtained by using properties at average temperature.
Analysis This semiinfinite cylinder can physically be formed by the inter
section of an infinite cylinder of radius r = 10 cm and a semiinfinite medium,
as shown in Fig. 429.
We will solve this problem using the oneterm solution relation for the cylin
der and the analytic solution for the semiinfinite 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 41 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 oneterm approximation is applicable. Substituting these values
into Eq. 414 gives
' = B_ 1 (0,0 =A x e
1.0124e<° 3126 > 2 ( 21)1 > = 0.762
The solution for the semiinfinite solid can be determined from
hx , h 2 at
1  e s emiinftM) = 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 43,
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
semiinfinite
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 429
Schematic for Example 49.
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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 430
Schematic for Example 410.
EXAMPLE 410 Refrigerating Steaks while Avoiding Frostbite
In a meat processing plant, 1in. thick steaks initially at 75°F are to be cooled
in the racks of a large refrigerator that is maintained at 5°F (Fig. 430). The
steaks are placed close to each other, so that heat transfer from the 1in. 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 onedimensional 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 oneterm 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. 4136, we obtain
1.5
x _ 0.5 in. _
L ~ 0.5 in. ~
T(L,t)T„ 355
'  M  75
T T m 455 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 offflavors 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 nogrowth 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. 431). 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. 432).
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 431
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 432
The factors that affect the rate of
growth of microorganisms.
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240
HEAT TRANSFER
Rate of
growth
Temperature
FIGURE 433
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 makeup 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. 433). 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 lowoxygen 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. 434). 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. 435).
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 poorquality 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 434
Freezing may stop the growth of
microorganisms, but it may not
necessarily kill them.
Freezer
18to35°C
o
Frozen
foods
''
Refrigerator
1 to 4°C
ft
y
Fresh
foods
FIGURE 435
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 436
Typical freezing curve of a food item.
TABLE 45
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 436.
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 45). 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 437
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 437. 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
midbackbone 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. 438). 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 moistheat cooking such as boiling. The
Chuck
Sirloin
Brisket Flank Round
Foreshank Short plate
FIGURE 438
Various cuts of beef (from National
Livestock and Meat Board).
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HEAT TRANSFER
Time in days
FIGURE 439
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 440
The freezing time of meat can be
reduced considerably by using low
temperature air at high velocity.
TABLE 46
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°C23°C
Beef
412
618 1224
Lamb
38
616 1218
Veal
34
414 8
Pork
26
412 815
Chopped
beef
34
46 8
Cooked foods
23
24
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 439. 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 firstclass 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 longterm 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 4cmdeep 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. 440). 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 longterm storage. The lower the storage tempera
ture, the longer the storage life of meat products, as shown in Table 46.
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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 coolingroom 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
loadout door is often adequate for such cases.
A refrigerated shipping dock, as shown in Figure 441, 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 441
A refrigerated truck dock for loading
frozen items to a refrigerated truck.
Poultry Products
Poultry products can be preserved by icechilling to 1 to 2°C or deep chill
ing to about — 2°C for shortterm storage, or by freezing them to — 18°C or
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HEAT TRANSFER
FIGURE 442
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 longterm 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 120s 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. 442).
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 flowtype
immersion slush ice chillers. Continuous slush icechillers 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
featherpicking 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
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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. 443). 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 offodor 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 airblast 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 crosssection 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 444. Thermal properties of poultry are given in
Table 47.
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 lowtemperature 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. 445). 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 belowfreezing) temperatures. Significant changes
Storage life (days)
15 20 25
Storage temperature, °C
FIGURE 443
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 444
The variation of freezing time of
poultry with air temperature (from
van der Berg and Lentz, Ref. 11).
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248
HEAT TRANSFER
FIGURE 445
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 47
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 oxygenfree 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 45 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 A9). The density and specific heat of air at
0°C are 1.292 kg/m 3 and 1.006 kJ/kg ■ °C (Table A15). Also, the specific heat
of beef carcass is determined from the relation in Table A7b to be
Analysis (a) A sketch of the chilling room is given in Figure 446. 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 = (0080 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
tevap
FIGURE 446
Schematic for Example 45.
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 = (0080 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.
413, 414, 415, and 423 for a large plane wall, a long cylin
der, a sphere, and a semiinfinite medium, respectively. These
charts are applicable for onedimensional 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 oneterm 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
Atem 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 41 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 oneterm 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 onedimensional transient heat
conduction in a semiinfinite 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 twodimensional transient heat conduction problems en
countered in geometries such as a short cylinder, a long rectan
gular bar, or a semiinfinite cylinder or plate, and even
threedimensional problems associated with geometries such
as a rectangular prism or a semiinfinite 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
onedimensional 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 onedimensional val
ues. The transient heat transfer for a twodimensional geometry
formed by the intersection of two onedimensional geometries
1 and 2 is
_Q_
Q_
!cma
_Q_
Transient heat transfer for a threedimensional 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 AirConditioning Engineers, Inc., 1993.
2. ASHRAE. Handbook of Fundamentals. SI version.
Atlanta, GA: American Society of Heating, Refrigerating,
and AirConditioning 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: McGrawHill, 1961.
5. M. P. Heisler. "Temperature Charts for Induction and
Constant Temperature Heating." ASME Transactions 69
(1947), pp. 22736.
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.
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HEAT TRANSFER
8. L. S. Langston. "Heat Transfer from Multidimensional
Objects Using OneDimensional Solutions for Heat
Loss." International Journal of Heat and Mass Transfer
25(1982), pp. 14950.
9. M. N. Ozisik, Heat Transfer — A Basic Approach. New
York: McGrawHill, 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
41 C What is lumped system analysis? When is it
applicable?
42C 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?
43C 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?
44C 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 P44C
45C 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?
46C 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 EESCD icon ® are solved using EES, and
complete solutions together with parametric studies are included
on the enclosed CD. Problems with a computerEES icon H are
comprehensive in nature, and are intended to be solved with a
computer, preferably using the EES software that accompanies
this text.
47C Consider two identical 4kg 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?
48C 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?
49C In what medium is the lumped system analysis more
likely to be applicable: in water or in air? Why?
410C 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?
41 1C For which kind of bodies made of the same material
is the lumped system analysis more likely to be applicable:
slender ones or wellrounded ones of the same volume? Why?
412 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 .
413 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.
414 The temperature of a gas stream is to be measured by
a thermocouple whose junction can be approximated as a
1 .2mmdiameter 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
415E 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
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250°F
oo<
/
Brass
balls
.OOO
> 120°F r
0000000
V
Water
bath
FIGURE P415E
416E Repeat Problem 41 5E for aluminum balls.
417 To warm up some milk for a baby, a mother pours milk
into a thinwalled 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
418 Repeat Problem 417 for the case of water also
being stirred, so that the heat transfer coefficient is doubled to
240 W/m 2 • °C.
41 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
12fluidoz 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
420 Consider a 1000W iron whose base plate is made of
0.5cmthick aluminum alloy 2024T6 (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 P420
421
Reconsider Problem 420. 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.
422 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
423
C p = 0.465 kJ/kg
Furnace
,900°C
o
Air, 35°C
, Steel ball 100 o C
OOO
FIGURE P41 9E
FIGURE P423
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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.
424 rSpM Reconsider Problem 423. 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.
425 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
426C 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.
427C Can the transient temperature charts in Fig. 413 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.
428C 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?
429C 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?
430C 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?
431 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?
432C 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?
433 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?
434 An ordinary egg can be approximated as a 5.5cm
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 P434
435
Reconsider Problem 434. 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.
436 In a production facility, 3cmthick 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 P436
cen58933_ch04.qxd 9/10/2002 9:13 AM Page 255
437 ScJU Reconsider Problem 436. 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.
438 A long 3 5 cmdiameter 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
439 rfigM Reconsider Problem 438. 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.
440E Long cylindrical AISI stainless steel rods (k = 7.74
Btu/h ■ ft • °F and a = 0.135 ft 2 /h) of 4in. diameter are heat
treated by drawing them at a velocity of 10 ft/min through
a 30ftlong 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 P440E
441 In a meat processing plant, 2cmthick 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.
442 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.
443 In Betty Crocker 's Cookbook, it is stated that it takes 2 h
45 min to roast a 3.2kg 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 P443
444 Repeat Problem 4^3 for a roast rib that is to be "well
done" instead of "rare." A rib is considered to be welldone
when its center temperature reaches 77°C, and the roasting in
this case takes about 4 h 15 min.
445 For heat transfer purposes, an egg can be considered to
be a 5. 5cm 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
446 Repeat Problem 4^5 for a location at 1610m eleva
tion such as Denver, Colorado, where the boiling temperature
of water is 94.4°C.
447 The author and his 6yearold 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 P447
448 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.
449E 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 P449E
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.
450 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 9cmdiameter 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.
451 [ft^S Reconsider Problem 450. 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.
452 Citrus fruits are very susceptible to cold weather, and
extended exposure to subfreezing temperatures can destroy
them. Consider an 8cmdiameter orange that is initially at
Orange
T ; = 15°C
FIGURE P452
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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.
453 An 8cmdiameter 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.
454 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 P454
455E Oranges of 2.5in. 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.
456 A 65kg 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 P456
457 Layers of 23cmthick 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.
458E Layers of 6in. 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
459 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 P459
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HEAT TRANSFER
Transient Heat Conduction in SemiInfinite Solids
460C What is a semiinfinite medium? Give examples of
solid bodies that can be treated as semiinfinite mediums for
heat transfer purposes.
461 C Under what conditions can a plane wall be treated as
a semiinfinite medium?
462C Consider a hot semiinfinite 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 .
463 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.
464 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 10h period.
► Winds,
" 10°C
Soil
10°C
FIGURE P464
465
Reconsider Problem 464. 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.
466E The walls of a furnace are made of 1.5ftthick 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
467 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.
468 A large cast iron container (k = 52 W/m • °C and a =
1.70 X 10~ 5 m 2 /s) with 5cmthick 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 1.2mwide and 2mhigh 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 P468
Transient Heat Conduction in Multidimensional Systems
469C What is the product solution method? How is it used
to determine the transient temperature distribution in a two
dimensional system?
470C How is the product solution used to determine
the variation of temperature with time and position in three
dimensional systems?
471 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 /.
472C 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
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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.
473 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 P473
474
Reconsider Problem 473. 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.
475 A semiinfinite 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.
476E 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
477E Repeat Problem 476E for a location at 5300 ft
elevation such as Denver, Colorado, where the boiling temper
ature of water is 202 °F.
478 A 5cmhigh 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 P478
479
Reconsider Problem 478. 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.
480 A 2cmhigh 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.
481 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.
482 Repeat Problem 481 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 P481
483 A 20cmlong 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.
484 Repeat Problem 483 for the case where the aluminum
block is inserted into the furnace on a lowconductivity mater
ial so that the heat transfer to or from the bottom surface of the
block is negligible.
485 rfigM Reconsider Problem 483. 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
486C What are the common kinds of microorganisms?
What undesirable changes do microorganisms cause in foods?
487C How does refrigeration prevent or delay the spoilage
of foods? Why does freezing extend the storage life of foods
for months?
488C What are the environmental factors that affect the
growth rate of microorganisms in foods?
489C 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?
490C 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?
491 C How does (a) the air motion and (b) the relative hu
midity of the environment affect the growth of microorganisms
in foods?
492C 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?
493C 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?
494C How does the rate of freezing affect the tenderness,
color, and the drip of meat during thawing?
495C 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.
496C What is a refrigerated shipping dock? How does it
reduce the refrigeration load of the cold storage rooms?
497C How does immersion chilling of poultry compare to
forcedair chilling with respect to (a) cooling time, (b) mois
ture loss of poultry, and (c) microbial growth.
498C What is the proper storage temperature of frozen
poultry? What are the primary methods of freezing for poultry?
499C What are the factors that affect the quality of
frozen fish?
4100 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 .
4101 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 P41 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
4102E 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?
4103 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 continuousflowtype 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.
4104 In a meat processing plant, 10cmthick 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 10cmthick 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 P41 04
Review Problems
4105 Consider two 2cmthick 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
4106 Consider a curing kiln whose walls are made of
30cmthick 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 P41 06
4107 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
4108 A hot dog can be considered to be a 12cmlong 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 P41 08
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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.
4109 A long roll of 2mwide and 0.5cmthick 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 P41 09
4110E In Betty Crocker 's Cookbook, it is stated that it takes
5 h to roast a 14lb 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 onethird 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?
4111 (~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 P41 1 1
4112 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 30cmdiameter 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 P41 10
Watermelon, 25°C
FIGURE P41 12
4113 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 2cmdiameter 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 P41 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.
4114 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.
4115E 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 P41 15
4116E Repeat Problem 41 15E for cast iron geometries
(k = 29 Btu/h • ft ■ °F and a = 0.61 ft 2 /h).
4117E rra Reconsider Problem 41 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
4118 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.
4119 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 20cm 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.
4120 10cmthick 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).
4121 A 30cmdiameter, 3.5mhigh 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.
4122 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 P41 22
4123 Repeat Problem 4122 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).
4124 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 P41 24
4125 A 40cmthick 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.
4126 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.
4127 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 28cm diameter, 1.80mlong 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
4128 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.
4129 Repeat Problem 4128 using a banana instead of an
apple. The thermal properties of bananas are practically the
same as those of apples.
4130 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 bvalue, 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?
4131 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
OneDimensional
Steady Heat Conduction 272
TwoDimensional
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 51
The analytical solution of a problem
requires solving the governing
differential equation and applying
the boundary conditions.
51 ■ WHY NUMERICAL METHODS?
The ready availability of highspeed computers and easytouse 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 onedimensional 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. 51)
\_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 {rir^)
(51)
(52)
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
rcoordinate 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' :
(53)
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. 52). 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 onedimensional 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 realworld 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. 53).
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 52
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 53
The approximate numerical solution
of a realworld problem may be more
accurate than the exact (analytical)
solution of an oversimplified
model of that problem.
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268
HEAT TRANSFER
T(r, z)
Analytical solution:
T(r,z)T^ ~ J (Ar) sinhA„(Lz)
where A 's are roots of 7 n (A r n ) =
FIGURE 54
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 "whatif " questions. This is an iter
ative process that is extremely tedious and timeconsuming 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 onedimensional 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. 54). Therefore, even when the solu
tions are available in some handbooks, they are intimidating enough to scare
prospective users away.
FIGURE 55
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 highpowered computers with
sophisticated software packages, with impressive presentationstyle colorful
output in graphical and tabular form (Fig. 55). 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,
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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.
52  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 secondorder
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 yearend 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
(54)
where i is the interest rate for the compounding period and n is the number of
periods. Using the same formula, the yearend 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 yearend 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 51
Yearend balance of a $100 account
earning interest at an annual rate
of 18 percent for various
compounding periods
Number
Compounding
of
YearEnd
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
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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 56
The derivative of a function at a point
represents the slope of the function
at that point.
FIGURE 57
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 56. 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
(55)
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
(56)
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)
(57)
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 onedimensional 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 xdirection, 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 57. The xcoordinate 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. 56, 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
(58)
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
(59)
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
(510)
which is the governing equation for steady onedimensional heat transfer in a
plane wall with heat generation and constant thermal conductivity, can be ex
pressed in the finite difference form as (Fig. 58)
2T,„
Ax 2
 m + 1 . 6 »;
0,
m = 1,2,3,
,M  1
(511)
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
threedimensional heat transfer problems by replacing each second derivative
by a difference equation in that direction. For example, the finite difference
formulation for steady twodimensional 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 58
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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HEAT TRANSFER
n+1
FIGURE 59
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. 59)
jj, IT rp
m,n m—l,n m, n + 1
~Tx 1 +
2T
Ay 2
 m, n — 1 6m, n
(512)
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 xdirection and N equal subregions in the ydirection has a total of
(M + 1)(N + 1) nodes, and Eq. 512 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 m1
m
m + 1
M
X
U ►! 1
r A.v n Ax
Ax
FIGURE 510
The nodal points and volume
elements for the finite difference
formulation of onedimensional
conduction in a plane wall.
53  ONEDIMENSIONAL
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 onedimensional 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 xdirection, 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 510. The xcoordinate 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
fullsize elements (they have a thickness of Ax), whereas the two elements at
the boundaries are halfsized.
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
(513)
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
(514)
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
(515)
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. 511). 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
(516)
Substituting Eqs. 514 and 516 into Eq. 513 gives
kA
Ax
kA
Ax
, AAx =
(517)
which simplifies to
TT + T
Ax 2
m = 1,2,3,
,M~ 1
(518)
Linear
T ,T f
kAa^ *i
Ax
kA
A.v
FIGURE 51 1
In finite difference formulation, the
temperature is assumed to vary
linearly between the nodes.
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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 512
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. 511) 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 512. (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 onedimensional 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 onedimensional 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. 513)
7"(0) = T Q = Specified value
T{L) = T M = Specified value
(519)
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
(520)
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 onedimensional heat conduction can be expressed as (Fig.
514)
Qv
kA
Ax
g (AAx/2) =
(521)
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. 521 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 513
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 514
Schematic for the finite difference
formulation of the left boundary
node of a plane wall.
(joA + kA
Ax
+ g (AAx/2) =
(522)
Special case: Insulated Boundary (q = 0)
T, T,
. ^j^ + £ (AAx/2) =
(523)
2. Convection Boundary Condition
TtT
hA{T.„  T ) + kA
Ax
+ g (AAx/2) =
(524)
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276
HEAT TRANSFER
FIGURE 515
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 516
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) =
(525)
4. Combined Convection and Radiation Boundary Condition
(Fig. 515)
hA(T m  T ) + euA(T* an  T+) + kA ^j^ + g (AAx/2) =
or
,A(T x T a ) + kA
Ax
+ g a {AAxl2) =
(526)
(527)
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) = (528)
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. 516). 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) = (529)
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. 523. 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. 517). 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. 518) 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
(530)
which is equivalent to Eq. 523 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 517
A node on an insulated boundary
can be treated as an interior node by
replacing the insulation by a mirror.
EXAMPLE 51 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 518.
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 onedimensional 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 518
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 519
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. 518 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. 519). Therefore, highly accurate results can be obtained with numerical
methods by using a limited number of nodes.
cen58 93 3_ch05.qxd 9/4/2002 11:41 AM Page 279
279
CHAPTER 5
EXAMPLE 52 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 520. 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 xdirection, 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 520
Schematic for Example 52 and the
volume element of a general
interior node of the fin.
cen58933_ch05.qxd 9/4/2002 11:41 AM Page 2i
280
HEAT TRANSFER
Ax/2
FIGURE 521
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. 521):
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 51). 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 welldefined 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. 522). 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 welldefined steps.
Iterative methods:
Start with an initial guess for the solution,
and iterate until solution converges.
FIGURE 522
Two general categories of solution
methods for solving systems
of algebraic equations.
cen58 93 3_ch05.qxd 9/4/2002 11:41 AM Page 282
282
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)
nl
\y
2
1
Ax
Ar
1 2
 « 1 • • • M x
m — 1 m + 1
FIGURE 523
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
nl
Ax
m, n + 1
m, n
+
I
m, nl
Ax
m + 1 , n
I
y t m  1
m + 1
FIGURE 524
The volume element of a general
interior node (m, n) for two
dimensional conduction in
rectangular coordinates.
5^  TWODIMENSIONAL
STEADY HEAT CONDUCTION
In Section 53 we considered onedimensional 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 onedimensional,
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 twodimensional steady heat conduction in rectangular coordinates
using the finite difference method. The approach presented below can be ex
tended to threedimensional cases.
Consider a rectangular region in which heat conduction is significant in the
x and ydirections. Now divide the xy plane of the region into a rectangular
mesh of nodal points spaced Ax and Ay apart in the x and ydirections,
respectively, as shown in Figure 523, and consider a unit depth of Az = 1
in the zdirection. 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
twodimensional problems is the double subscript notation (m, n) where
m = 0,1,2, ... ,M is the node count in the xdirection and n = 0, 1, 2, . . . ,N
is the node count in the _ydirection. 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 524. 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
(531)
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 ydirection,
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 = (532)
Dividing each term by Ax X Ay and simplifying gives
(533)
* 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. 512 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 xdirection and N equally spaced nodes
in the ydirection has a total of (M + 1)(7V + 1) nodes, and Eq. 533 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
ydirections 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  = (534)
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  = (535)
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 threedimensional 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 twodimensional case involve heat transfer in the ydirection as well as
the xdirection. Insulated surfaces can still be viewed as "mirrors, " and the
283
CHAPTER 5
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284
HEAT TRANSFER
Volume element
of node 2
Boundary
subjected
/ j to convection
e,ef, + e top + Q nght + e bollom + — = o
FIGURE 525
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 526
Schematic for Example 53 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 527
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. 525)
S Q + SKl
(536)
whether the problem is one, two, or threedimensional. 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 53 for various boundary
conditions.
EXAMPLE 53 Steady TwoDimensional Heat Conduction
in LBars
Consider steady heat transfer in an Lshaped solid body whose cross section is
given in Figure 526. Heat transfer in the direction normal to the plane of the
paper is negligible, and thus heat transfer in the body is twodimensional. 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 Lshaped 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 twodimensional, 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. 536 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. 527a]
cen58 93 3_ch05.qxd 9/4/2002 11:41 AM Page 285
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. 5276]
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. 528a]
^♦£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 twodimensional case (Eq. 535) gives [Fig. 5286]
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. 535 to be
[Fig. 529a]
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 528
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 529
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 530
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 Lshaped exposed surface and to conduction at other surfaces.
An energy balance on this element gives [Fig. 5296]
# + 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. 530a]
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. 5306]
, 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
twodimensional 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 531.
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 54 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 531
Approximating an irregular
boundary with a rectangular mesh.
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288
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 1mlong section of the chimney.
Symmetry lines
(Equivalent to insulation)
Representative
7" k section of chimney
FIGURE 532
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 533
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 twodimensional 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
onedimensional, 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 532. 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
oneeighth 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 533a
Ax AyTjr,
+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
5336
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. 534)
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
535)
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 ) + eoAx(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
535)
Av r 8  r, a r Ax
I I
(4) i Minor i
Mirror
FIGURE 534
Converting the boundary
nodes 3 and 5 on symmetry lines to
interior nodes by using mirror images.
Insulation
h, T„
sky
FIGURE 535
Schematics for energy balances on the
volume elements of nodes 7 and 9.
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290
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 536
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 GaussSeidel 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 536.
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 1mlong 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
55  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 537. 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 m1
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 537
Finite difference formulation of time
dependent problems involves discrete
points in time as well as space.
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292
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
(537)
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 *
(538)
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 538
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
(539)
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. 538).
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 threedimensional. 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.
539. 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. 539)
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
(540)
(541)
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. 540 and
541 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 twodimensional cases to keep the complexities at a man
ageable level, but the analysis can readily be extended to threedimensional
cases and other coordinate systems.
Transient Heat Conduction in a Plane Wall
Consider transient onedimensional 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 xdirection, as shown in Figure 540. 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. 539 as
kA
Ax
T T
*■ m , . , in
1 kA —
Ax
+ g„,AAx = pAAxC :
At
(542)
293
CHAPTER 5
If expressed at i + 1 : Implicit method
py A
in in_
At
If expressed at i: Explicit method
FIGURE 539
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
m1
, Volume
element
of node m
1
l kA _m±L
Ax
Ax
m+1 M1
M *
FIGURE 540
The nodal points and volume elements
for the transient finite difference
formulation of onedimensional
conduction in a plane wall.
cen58 93 3_ch05.qxd 9/4/2002 11:42 AM Page 294
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 "
(543)
where a = k/pC is the thermal dijfusivity of the wall material. We now define
a dimensionless mesh Fourier number as
ocAf
Ajc 2
(544)
Then Eq. 543 reduces to
2T + T
g„Ax 2
(545)
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)
(546)
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
(547)
A
M(r.r^)
Ax
2
pA Ax C ° °
y 2 m
W^kA^ °
Ax
Ax Ax
1 2 •••
L x
FIGURE 541
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. 545 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) (548)
which can be rearranged as
x7;;+\(i + 2T)7r i + T7;;,V 1 + T
g^Ax 2
+ Ti =
(549)
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. 5^41)
hA(T x  Ti) + kA ■
Ax
+ &A
Ax
. Ajc £o
PA r C : —
v 2 At
(550)
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
(551)
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 =
(Tmi + 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. 542). 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 onedimensional 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. 547. 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 542
The stability criterion of the
explicit method requires all primary
coefficients to be positive or zero.
ctAf __, J_ /interior nodes, onedimensional heat
Ajc 2 ~ 2 \ transfer in rectangular coordinates
(552)
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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 543
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 544
Schematic for Example 55.
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 ) + (i2t)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. 543). 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 55 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 544. 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 51 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 onedimensional 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. 547 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. 545):
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 545
Schematic for the explicit
finite difference formulation of the
convection condition at the right
boundary of a plane wall.
cen58 93 3_ch05.qxd 9/4/2002 11:42 AM Page 298
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HEAT TRANSFER
TABLE 52
The variation of the nodal
temperatures in Example 55 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? (125 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 52. 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. 549 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
cen58 93 3_ch05.qxd 9/4/2002 11:42 AM Page 299
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 53, 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 53
The variation of the nodal
temperatures in Example 55 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 546
Schematic of a Trombe wall
(Example 56).
TABLE 54
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 AM10 AM
33
114
10 AM1 PM
43
242
1 PM4 PM
45
178
4 PM7 PM
37
7 PM10 PM
32
10 PM1 AM
27
1 AM4 AM
26
4 AM7 AM
25
EXAMPLE 56 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. 546). 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 1ftthick
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 southfacing vertical sur
face throughout the day for a typical day in January is given in Table 54 in 3h
intervals. The Trombe wall has single glazing with an absorptivitytransmissivity
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 12h intervals and the
amount of heat transfer during the first and second days are to be determined.
Assumptions 1 Heat transfer is onedimensional 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
547. Nodes 1 through 4 are interior nodes, and the explicit finite difference
formulations of these nodes are obtained directly from Eq. 547 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. 551 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
(553)
which simplifies to
j.j+1
1  2t  2t ■
h m „ Ax
Ti + 2tTI + 2t 
h„„, Ax
K qj olm . Ax
2t : (554)
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/hft°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 547
The nodal network for the Trombe
wall discussed in Example 56.
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 548
The variation of temperatures in
the Trombe wall discussed
in Example 56.
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 (478 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 / = 112, q' solar = 242 for / = 1324,
<7^ ar = 178 for / = 2536, and q' solar = for / = 3796.
The results after 6, 12, 18, 24, 30, 36, 42, and 48 h are given in Table 55
and are plotted in Figure 548 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 55
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 15min 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 (555)
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 startup 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
ml,n I "••"
Ay
«l
+
Ax
m, n  1
Ax
TwoDimensional Transient Heat Conduction
m + 1, n
y t m  1
m + 1
FIGURE 549
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 ydirections, and consider a unit depth of Az = 1 in the zdirection.
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 xyplane of the region into a rectangu
lar mesh of nodal points spaced Ax and A_y apart in the x and ydirections,
respectively, and consider a general interior node (m, ri) whose coordinates are
x = mAx and y = nAy, as shown in Figure 549. 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. 539 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
(556)
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 :
(557)
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 easytoremember
form:
, J, __ rp
1 1 top ' J right
+ T h ,
4r„,
,/ :
(558)
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'
(559)
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^ (560)
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. 550).
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 threedimensional cases and severely limits the size of the time step At that
can be used with the explicit method. In the case of transient twodimensional
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, twodimensional heat
I 2 ~4 transfer in rectangular coordinates)
(561)
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. 560 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 onedimensional 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. 539. This is illustrated in Example 57.
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 550
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 57 Transient TwoDimensional Heat Conduction
in LBars
Consider twodimensional transient heat transfer in an Lshaped solid body that
is initially at a uniform temperature of 90°C and whose cross section is given
in Figure 551. 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 551
Schematic and nodal network for
Example 57.
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 552
Schematics for energy balances on the
volume elements of nodes 1 and 2.
SOLUTION This is a transient twodimensional heat transfer problem in rec
tangular coordinates, and it was solved in Example 53 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.
552a)
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. 5526)
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. 553a)
AyTlTi . 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. 5536). Noting that T 10 = 90°C, Eq. 560 gives
U 1
(1  4t) T\ + t T[ + 2Tj, + 90 +
(e) Node 5. (Interior node, Fig. 554a). Noting that T n = 90°C, Eq. 560
gives
Ti, + ' = (1  4t) Ti + t I H + r^ + Ti + 90 +
&/ 2
(f) Node 6. (Boundary node subjected to convection on two sides, Fig. 5546)
(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
l4 T 4x]7
hi,
2Tj + 4Ti + 2Tj + 4 X 90 + 4 f T w + 3 ^
[g) Node 7. (Boundary node subjected to convection, Fig. 555)
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 4T2Ty ]
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 553
Schematics for energy balances on the
volume elements of nodes 3 and 4.
m
ii
12
(a) Node 5
(7?) Node 6
FIGURE 554
Schematics for energy balances on the
volume elements of nodes 5 and 6.
h, r„
1\ 1
13 1 ^ ^
1r
• 15
FIGURE 555
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. 555)
,Ax,_ _,, , . Ay A.t Tj 5  Tj
h T (T«,T$) + q R — k—^
kAyTi
Ti , . Ax^J AxAy T<'
+
2 Ax 6J 2 2
Dividing by klA, simplifying, and solving for 7"g +1 gives
p TT c "
At
T' + l
*9
hl\ I 4rI hi
i  4t  2t j r 9 " + 2t m + 90 + ^+ ^r„
2fe
This completes the finite difference formulation of the problem. Next we need
to determine the upper limit of the time step Af from the stability criterion,
which requires the coefficient of T m in the 7~^ +1 expression (the primary coeffi
cient) to be greater than or equal to zero for all nodes. The smallest primary co
efficient in the nine equations here is the coefficient of Ti in the expression,
and thus the stability criterion for this problem can be expressed as
1  4t  4t
///.
1
4(1 + hllk)
A?:
/ :
4a(l + hllk)
since t = aAf// 2 . Substituting the given quantities, the maximum allowable
value of the time step is determined to be
At:
(0.012 m) 2
4(3.2 X 10 6 m 2 /s)[l + (80 W/m 2 • °C)(0.012 m)/(15 W/m ■ °C)]
10.6 s
Therefore, any time step less than 10.6 s can be used to solve this problem. For
convenience, let us choose the time step to be At = 10 s. Then the mesh
Fourier number becomes
aAt _ (32 X 10 6 m 2 /s)(10s)
l r ~ (0.012 m) 2
0.222 (for At = 10 s)
Substituting this value of t and other given quantities, the developed transient
finite difference equations simplify to
T i+
0.08367,' + 0.444(7^ + 7] + 11.2)
0.0S36Ti + 0.222(7/ + Tj + ITi, + 22.4;
0.05527J + 0.444(7; + T b + 12.8)
0.1127] + 0.222(7/ + 27J + 109.2)
0.1127/ + 0.222(7] + 7j + 7 6 ' + 109.2)
cen58 93 3_ch05.qxd 9/4/2002 11:42 AM Page 309
0.093 17^ + 0.074(2^ + 47^' + 2Tj + 424)
0.08367V + 0.222(7,; + TJ + 202.4)
0.08367V + 0.222(Tj + T$ + 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 53. 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 roundoff 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
556. 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 556
The local and global discretization
errors of the finite difference
method at the third time step
at a specified nodal point.
cen58 93 3_ch05.qxd 9/4/2002 11:42 AM Page 31C
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 +■■■ (562)
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
(563)
or
dT(x m , tf)
T(x m , t, + At) = T(x m , t,) + At (564)
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.
cen58 93 3_ch05.qxd 9/4/2002 11:42 AM Page 311
311
CHAPTER 5
Roundoff 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 roundoff 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 557. 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 roundoff 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 roundoff 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 roundoff 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 roundoff 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
roundoff error, which increases with decreasing step size, as shown in Fig
ure 558. 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 557
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 558
As the mesh or time step size
decreases, the discretization error
decreases but the roundoff
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 roundoff error. We should be careful not to
let roundoff 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
roundoff 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 roundoff 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 builtin 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
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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
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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 threedimensional. For
convenience in formulation, we always assume all heat trans
fer to be into the volume element from all surfaces toward the
node under