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Full text of "heat and mass transfer"

cen58 933_fm.qxd 9/11/2002 10:56 AM Page vii 



Contents 



Preface xviii 
Nomenclature xxvi 



CHAPTER ONE 
BASICS OF HEAT TRANSFER 1 

1-1 Thermodynamics and Heat Transfer 2 

Application Areas of Heat Transfer 3 
Historical Background 3 

1 -2 Engineering Heat Transfer 4 

Modeling in Heat Transfer 5 
1 -3 Heat and Other Forms of Energy 6 

Specific Heats of Gases, Liquids, and Solids 7 
Energy Transfer 9 

1 -4 The First Law of Thermodynamics 1 1 

Energy Balance for Closed Systems (Fixed Mass) 12 
Energy Balance for Steady-Flow Systems 12 
Surface Energy Balance 13 

1-5 Heat Transfer Mechanisms 17 

1-6 Conduction 17 

Thermal Conductivity 19 
Thermal Diffusivity 23 

1 -7 Convection 25 

1 -8 Radiation 27 

1 -9 Simultaneous Heat Transfer Mechanisms 30 

1-10 Problem-Solving Technique 35 

A Remark on Significant Digits 37 

Engineering Software Packages 38 

Engineering Equation Solver (EES) 39 

Heat Transfer Tools (HTT) 39 

Topic of Special Interest: 

Thermal Comfort 40 

Summary 46 

References and Suggested Reading 47 

Problems 47 



CHAPTER TWO 
HEAT CONDUCTION EQUATION 61 



2- 1 Introduction 62 

Steady versus Transient Heat Transfer 
Multidimensional Heat Transfer 64 
Heat Generation 66 



2-5 

2-6 
2-7 






2-2 One -Dimensional 

Heat Conduction Equation 



68 



Heat Conduction Equation in a Large Plane Wall 68 
Heat Conduction Equation in a Long Cylinder 69 
Heat Conduction Equation in a Sphere 71 
Combined One-Dimensional 

Heat Conduction Equation 72 

2-3 General Heat Conduction Equation 74 

Rectangular Coordinates 74 
Cylindrical Coordinates 75 
Spherical Coordinates 76 

2-4 Boundary and Initial Conditions 77 

1 Specified Temperature Boundary Condition 78 

2 Specified Heat Flux Boundary Condition 79 

3 Convection Boundary Condition 81 

4 Radiation Boundary Condition 82 

5 Interface Boundary Conditions 83 

6 Generalized Boundary Conditions 84 

Solution of Steady One-Dimensional 
Heat Conduction Problems 86 



Heat Generation in a Solid 97 

Variable Thermal Conductivity, k(T) 104 

Topic of Special Interest: 

A Brief Review of Differential Equations 107 

Summary 111 

References and Suggested Reading 112 

Problems 113 



CHAPTER THREE 
STEADY HEAT CONDUCTION 1 27 

3-1 Steady Heat Conduction in Plane Walls 128 

The Thermal Resistance Concept 129 



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CONTENTS 



Thermal Resistance Network 131 
Multilayer Plane Walls 133 

3-2 Thermal Contact Resistance 138 

3-3 Generalized Thermal Resistance Networks 143 

3-4 Heat Conduction in Cylinders and Spheres 146 

Multilayered Cylinders and Spheres 148 

3-5 Critical Radius of Insulation 153 

3-6 Heat Transfer from Finned Surfaces 156 

Fin Equation 157 
Fin Efficiency 160 
Fin Effectiveness 163 
Proper Length of a Fin 165 

3-7 Heat Transfer in Common Configurations 169 

Topic of Special Interest: 

Heat Transfer Through Walls and Roofs 175 

Summary 185 

References and Suggested Reading 186 

Problems 187 



4 Complications 268 

5 Human Nature 268 

5-2 Finite Difference Formulation of 
Differential Equations 269 

5-3 One -Dimensional Steady Heat Conduction 272 

Boundary Conditions 274 

5-4 Two-Dimensional 

Steady Heat Conduction 282 

Boundary Nodes 283 
Irregular Boundaries 287 

5-5 Transient Heat Conduction 291 

Transient Heat Conduction in a Plane Wall 293 

Two-Dimensional Transient Heat Conduction 304 

Topic of Special Interest: 

Controlling Numerical Error 309 

Summary 312 

References and Suggested Reading 314 

Problems 314 



CHAPTER FOUR 
TRANSIENT HEAT CONDUCTION 209 

4-1 Lumped System Analysis 210 

Criteria for Lumped System Analysis 211 

Some Remarks on Heat Transfer in Lumped Systems 213 

4-2 Transient Heat Conduction in 

Large Plane Walls, Long Cylinders, 
and Spheres with Spatial Effects 216 

4-3 Transient Heat Conduction in 
Semi-Infinite Solids 228 

4-4 Transient Heat Conduction in 

Multidimensional Systems 231 

Topic of Special Interest: 

Refrigeration and Freezing of Foods 239 

Summary 250 

References and Suggested Reading 251 

Problems 252 



CHAPTER FIVE 

NUMERICAL METHODS 

IN HEAT CONDUCTION 265 

5-1 Why Numerical Methods? 266 

1 Limitations 267 

2 Better Modeling 267 

3 Flexibility 268 



CHAPTER SIX 
FUNDAMENTALS OF CONVECTION 



333 



6-1 Physical Mechanism on Convection 334 

Nusselt Number 336 

6-2 Classification of Fluid Flows 337 

Viscous versus I nviscid Flow 337 
Internal versus External Flow 337 
Compressible versus Incompressible Flow 337 
Laminar versus Turbulent Flow 338 
Natural (or Unforced) versus Forced Flow 338 
Steady versus Unsteady (Transient) Flow 338 
One-, Two-, and Three-Dimensional Flows 338 

6-3 Velocity Boundary Layer 339 

Surface Shear Stress 340 
6-4 Thermal Boundary Layer 341 

Prandtl Number 341 
6-5 Laminar and Turbulent Flows 342 

Reynolds Number 343 

6-6 Heat and Momentum Transfer 
in Turbulent Flow 343 

6-7 Derivation of Differential 

Convection Equations 345 

Conservation of Mass Equation 345 
Conservation of Momentum Equations 346 
Conservation of Energy Equation 348 



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6-8 Solutions of Convection Equations 
for a Flat Plate 352 

The Energy Equation 354 

6-9 Nondimensionalized Convection 
Equations and Similarity 356 

6-1 Functional Forms of Friction and 
Convection Coefficients 357 

6-1 1 Analogies between Momentum 
and Heat Transfer 358 

Summary 361 

References and Suggested Reading 362 

Problems 362 



CHAPTER SEVEN 
EXTERNAL FORCED CONVECTION 



367 



CONTENTS 



8-4 General Thermal Analysis 426 

Constant Surface Heat Flux [q s = constant) 427 
Constant Surface Temperature (T s = constant) 428 

8-5 Laminar Flow in Tubes 431 

Pressure Drop 433 

Temperature Profile and the Nusselt Number 434 

Constant Surface Heat Flux 435 

Constant Surface Temperature 436 

Laminar Flow in Noncircular Tubes 436 

Developing Laminar Flow in the Entrance Region 436 

8-6 Turbulent Flow in Tubes 441 

Rough Surfaces 442 

Developing Turbulent Flow in the Entrance Region 443 

Turbulent Flow in Noncircular Tubes 443 

Flow through Tube Annulus 444 

Heat Transfer Enhancement 444 

Summary 449 

References and Suggested Reading 450 

Problems 452 



7-1 Drag Force and Heat Transfer 
in External Flow 368 

Friction and Pressure Drag 368 
Heat Transfer 370 

7-2 Parallel Flow over Flat Plates 371 

Friction Coefficient 372 

Heat Transfer Coefficient 373 

Flat Plate with Unheated Starting Length 375 

Uniform Heat Flux 375 

7-3 Flow across Cylinders and Spheres 380 

Effect of Surface Roughness 382 
Heat Transfer Coefficient 384 

7-4 Flow across Tube Banks 389 

Pressure Drop 392 

Topic of Special Interest: 

Reducing Heat Transfer through Surfaces 395 

Summary 406 

References and Suggested Reading 407 

Problems 408 



CHAPTER EIGHT 
INTERNAL FORCED CONVECTION 419 

8-1 Introduction 420 

8-2 Mean Velocity and Mean Temperature 420 

Laminar and Turbulent Flow in Tubes 422 
8-3 The Entrance Region 423 

Entry Lengths 425 



CHAPTER NINE 
NATURAL CONVECTION 459 

9-1 Physical Mechanism of 
Natural Convection 460 

9-2 Equation of Motion and 

the Grashof Number 463 

The Grashof Number 465 

9-3 Natural Convection over Surfaces 466 

Vertical Plates (7~ s = constant) 467 

Vertical Plates {q s = constant) 467 

Vertical Cylinders 467 

Inclined Plates 467 

Horizontal Plates 469 

Horizontal Cylinders and Spheres 469 

9-4 Natural Convection from 

Finned Surfaces and PCBs 473 

Natural Convection Cooling of Finned Surfaces 

(T s = constant) 473 
Natural Convection Cooling of Vertical PCBs 

(q s = constant) 474 
Mass Flow Rate through the Space between Plates 475 

9-5 Natural Convection inside Enclosures 477 

Effective Thermal Conductivity 478 

Horizontal Rectangular Enclosures 479 

Inclined Rectangular Enclosures 479 

Vertical Rectangular Enclosures 480 

Concentric Cylinders 480 

Concentric Spheres 481 

Combined Natural Convection and Radiation 481 



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CONTENTS 



9-6 Combined Natural and Forced Convection 486 1 1 -6 Atmospheric and Solar Radiation 586 



Topic of Special Interest: 

Heat Transfer through Windows 489 

Summary 499 

References and Suggested Reading 500 

Problems 501 



Topic of Special Interest: 

Solar Heat Gain through Windows 590 

Summary 597 

References and Suggested Reading 599 

Problems 599 



CHAPTER TEN 
BOILING AND CONDENSATION 



515 



CHAPTER TWELVE 
RADIATION HEAT TRANSFER 605 



10-1 Boiling Heat Transfer 516 

10-2 Pool Boiling 518 

Boiling Regimes and the Boiling Curve 518 
Heat Transfer Correlations in Pool Boiling 522 
Enhancement of Heat Transfer in Pool Boiling 526 

10-3 Flow Boiling 530 

1 0-4 Condensation Heat Transfer 532 

1 0-5 Film Condensation 532 

Flow Regimes 534 

Heat Transfer Correlations for Film Condensation 535 

1 0-6 Film Condensation Inside 
Horizontal Tubes 545 

1 0-7 Dropwise Condensation 545 

Topic of Special Interest: 

Heat Pipes 546 

Summary 551 

References and Suggested Reading 553 

Problems 553 



CHAPTER ELEVEN 
FUNDAMENTALS OF THERMAL RADIATION 561 

11-1 Introduction 562 

1 1 -2 Thermal Radiation 563 

1 1 -3 Blackbody Radiation 565 

1 1 -4 Radiation Intensity 57 1 

Solid Angle 572 

Intensity of Emitted Radiation 573 

Incident Radiation 574 

Radiosity 575 

Spectral Quantities 575 

1 1 -5 Radiative Properties 577 

Emissivity 578 

Absorptivity, Reflectivity, and Transmissivity 582 

Kirchhoffs Law 584 

The Greenhouse Effect 585 



12-1 The View Factor 606 

1 2-2 View Factor Relations 609 

1 The Reciprocity Relation 610 

2 The Summation Rule 613 

3 The Superposition Rule 615 

4 The Symmetry Rule 616 

View Factors between Infinitely Long Surfaces: 
The Crossed-Strings Method 618 

12-3 Radiation Heat Transfer: Black Surfaces 620 

1 2-4 Radiation Heat Transfer: 

Diffuse, Gray Surfaces 623 

Radiosity 623 

Net Radiation Heat Transfer to or from a Surface 623 

Net Radiation Heat Transfer between Any 

Two Surfaces 625 
Methods of Solving Radiation Problems 626 
Radiation Heat Transfer in Two-Surface Enclosures 627 
Radiation Heat Transfer in Three-Surface Enclosures 629 

1 2-5 Radiation Shields and the Radiation Effect 635 

Radiation Effect on Temperature Measurements 637 

1 2-6 Radiation Exchange with Emitting and 
Absorbing Gases 639 

Radiation Properties of a Participating Medium 640 

Emissivity and Absorptivity of Gases and Gas Mixtures 642 

Topic of Special Interest: 

Heat Transfer from the Human Body 649 

Summary 653 

References and Suggested Reading 655 

Problems 655 



CHAPTER THIRTEEN 
HEAT EXCHANGERS 667 

13-1 Types of Heat Exchangers 668 

1 3-2 The Overall Heat Transfer Coefficient 67 1 

Fouling Factor 674 
1 3-3 Analysis of Heat Exchangers 678 



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1 3-4 The Log Mean Temperature 
Difference Method 680 

Counter-Flow Heat Exchangers 682 
Multipass and Cross-Flow Heat Exchangers: 
Use of a Correction Factor 683 

1 3-5 The Effectiveness-NTU Method 690 

13-6 Selection of Heat Exchangers 700 

Heat Transfer Rate 700 

Cost 700 

Pumping Power 701 

Size and Weight 701 

Type 701 

Materials 701 

Other Considerations 702 

Summary 703 

References and Suggested Reading 704 

Problems 705 



CHAPTER FOURTEEN 
MASS TRANSFER 717 

14-1 Introduction 718 

14-2 Analogy between Heat and Mass Transfer 719 

Temperature 720 
Conduction 720 
Heat Generation 720 
Convection 721 

14-3 Mass Diffusion 721 

1 Mass Basis 722 

2 Mole Basis 722 

Special Case: Ideal Gas Mixtures 723 
Fick's Law of Diffusion: Stationary Medium Consisting 
of Two Species 723 

14-4 Boundary Conditions 727 

1 4-5 Steady Mass Diffusion through a Wall 732 

14-6 Water Vapor Migration in Buildings 736 

14-7 Transient Mass Diffusion 740 

14-8 Diffusion in a Moving Medium 743 

Special Case: Gas Mixtures at Constant Pressure 

and Temperature 747 
Diffusion of Vapor through a Stationary Gas: 

Stefan Flow 748 
Equimolar Counterdiffusion 750 

14-9 Mass Convection 754 

Analogy between Friction, Heat Transfer, and Mass 

Transfer Coefficients 758 
Limitation on the Heat-Mass Convection Analogy 760 
Mass Convection Relations 760 



CONTENTS 



14-10 Simultaneous Heat and Mass Transfer 763 

Summary 769 

References and Suggested Reading 771 

Problems 772 



CHAPTER FIFTEEN 
COOLING OF ELECTRONIC EQUIPMENT 



785 



15-1 Introduction and History 786 

15-2 Manufacturing of Electronic Equipment 787 

The Chip Carrier 787 
Printed Circuit Boards 789 
The Enclosure 791 

15-3 Cooling Load of Electronic Equipment 793 

1 5-4 Thermal Environment 794 

1 5-5 Electronics Cooling in 

Different Applications 795 

1 5-6 Conduction Cooling 797 

Conduction in Chip Carriers 798 
Conduction in Printed Circuit Boards 803 
Heat Frames 805 
The Thermal Conduction Module (TCM) 810 

1 5-7 Air Cooling: Natural Convection 
and Radiation 812 

15-8 Air Cooling: Forced Convection 820 

Fan Selection 823 

Cooling Personal Computers 826 

15-9 Liquid Cooling 833 

15-10 Immersion Cooling 836 

Summary 841 

References and Suggested Reading 842 

Problems 842 



APPENDIX 1 
PROPERTY TABLES AND CHARTS 
(SI UNITS) 855 

Table A-1 Molar Mass, Gas Constant, and 
Critical-Point Properties 856 

Table A-2 Boiling- and Freezing-Point 
Properties 857 

Table A-3 Properties of Solid Metals 858 

Table A-4 Properties of Solid Nonmetals 861 

Table A-5 Properties of Building Materials 862 



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CONTENTS 



Table A-6 Properties of Insulating Materials 864 

Table A-7 Properties of Common Foods 865 

Table A-8 Properties of Miscellaneous 
Materials 867 

Table A-9 Properties of Saturated Water 868 

Ta b I e A- 1 Properties of S aturated 
Refrigerant- 134a 869 

Table A-11 Properties of Saturated Ammonia 870 

Ta b I e A- 1 2 Properties of S aturated Propane 87 1 

TableA-13 Properties of Liquids 872 

Table A-14 Properties of Liquid Metals 873 

Table A-15 Properties of Air at 1 atm Pressure 874 

Table A-1 6 Properties of Gases at 1 atm 
Pressure 875 

Ta b I e A- 1 7 Properties of the Atmosphere at 
High Altitude 877 

Ta b I e A- 1 8 Emi ssivitiesofS urf aces 878 

Table A-1 9 Solar Radiative Properties of 
Materials 880 

Figure A-20 The Moody Chart for the Friction 
Factor for Fully Developed Flow 
in Circular Tubes 881 

APPENDIX 2 
PROPERTY TABLES AND CHARTS 
(ENGLISH UNITS) 883 

Table A-1 E Molar Mass, Gas Constant, and 
Critical-Point Properties 884 



Table A-2E Boiling- and Freezing-Point 
Properties 885 

Table A-3E Properties of Solid Metals 886 

Table A-4E Properties of Solid Nonmetals 889 

Table A-5E Properties of Building Materials 890 

Table A-6E Properties of Insulating Materials 892 

Table A-7E Properties of Common Foods 893 

Table A-8E Properties of Miscellaneous 
Materials 895 

Table A-9E Properties of Saturated Water 896 

Table A-1 0E Properties of Saturated 
Refrigerant- 134a 897 

Table A-1 1 E Properties of Saturated Ammonia 898 

Table A-1 2E Properties of Saturated Propane 899 

Table A-1 3E Properties of Liquids 900 

Table A-1 4E Properties of Liquid Metals 901 

Table A-1 5E Properties of Air at 1 atm Pressure 902 

Table A-1 6E Properties of Gases at 1 atm 
Pressure 903 

Table A-1 7E Properties of the Atmosphere at 
High Altitude 905 

APPENDIX 3 
INTRODUCTION TO EES 907 
INDEX 921 



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TABLE OF EXAMPLES: 



CHAPTER ONE 
BASICS OF HEAT TRANSFER 1 

Example 1-1 Heating of a Copper Ball 10 

Example 1-2 Heating of Water in an 
Electric Teapot 14 

Example 1-3 Heat Loss from Heating Ducts 

in a Basement 15 

Example 1 -4 Electric Heating of a House at 
High Elevation 16 

Example 1 -5 The Cost of Heat Loss through 

a Roof 19 

Example 1 -6 Measuring the Thermal Conductivity 
of a Material 23 

Example 1 -7 Conversion between SI and 

English Units 24 

Example 1 -8 Measuring Convection Heat 
Transfer Coefficient 26 

Example 1-9 Radiation Effect on 

Thermal Comfort 29 

Example 1-10 Heat Loss from a Person 31 

Example 1-11 Heat Transfer between 

Two Isothermal Plates 32 

Example 1-12 Heat Transfer in Conventional 
and Microwave Ovens 33 

Example 1-13 Heating of a Plate by 
Solar Energy 34 

Example 1-14 Solving a System of Equations 
with EES 39 



CHAPTER TWO 
HEAT CONDUCTION EQUATION 61 



Example 2-1 



Heat Gain by a Refrigerator 67 



Example 2-2 Heat Generation in a 

Hair Dryer 67 

Example 2-3 Heat Conduction through the 

Bottom of a Pan 72 

Example 2-4 Heat Conduction in a 

Resistance Heater 72 

Exa m p I e 2-5 Cooling of a Hot Metal B all 
in Air 73 

Example 2-6 Heat Conduction in a 
Short Cylinder 76 

Example 2-7 Heat Flux Boundary Condition 80 

Example 2-8 Convection and Insulation 
Boundary Conditions 82 

Example 2-9 Combined Convection and 

Radiation Condition 84 

Example 2-10 Combined Convection, Radiation, 
and Heat Flux 85 

Exa m p I e 2- 1 1 Heat Conduction in a 
Plane Wall 86 

Example 2-12 A Wall with Various Sets of 
Boundary Conditions 88 

Example 2-1 3 Heat Conduction in the Base Plate 
of an Iron 90 

Exa m p I e 2- 1 4 Heat Conduction in a 
Solar Heated Wall 92 

Example 2-15 Heat Loss through a 
Steam Pipe 94 

Example 2-16 Heat Conduction through a 
Spherical Shell 96 

Example 2-17 Centerline Temperature of a 
Resistance Heater 100 

Example 2-18 Variation of Temperature in a 
Resistance Heater 100 

Example 2-19 Heat Conduction in a Two-Layer 
Medium 102 



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CONTENTS 



Example 2-20 Variation of Temperature in a Wall 
with k(T) 105 

Example 2-21 Heat Conduction through a Wall 
with k(T) 106 

CHAPTER THREE 
STEADY HEAT CONDUCTION 1 27 







Example 3-1 


Heat Loss through a Wall 134 




Example 3-2 


Heat Loss through a 
Single-Pane Window 135 




Example 3-3 


Heat Loss through 
Double-Pane Windows 136 




Example 3-4 


Equivalent Thickness for 
Contact Resistance 140 




Example 3-5 


Contact Resistance of 
Transistors 141 




Example 3-6 


Heat Loss through a 
Composite Wall 144 




Example 3-7 


Heat Transfer to a 
Spherical Container 149 


: 


) 


Example 3-8 


Heat Loss through an Insulated 
Steam Pipe 151 






Example 3-9 


Heat Loss from an Insulated 
Electric Wire 154 




Example 3-10 


Maximum Power Dissipation of 
a Transistor 166 




Example 3-1 1 


Selecting a Heat Sink for a 
Transistor 167 




Example 3-12 


Effect of Fins on Heat Transfer from 
Steam Pipes 168 




Example 3-13 


Heat Loss from Buried 
Steam Pipes 170 




Example 3-14 


Heat Transfer between Hot and 
Cold Water Pipes 173 




Example 3-15 


Cost of Heat Loss through Walls 
in Winter 174 




Example 3-16 


The i?-Value of a Wood 
Frame Wall 179 




Example 3-17 


The R -Value of a Wall with 
Rigid Foam 180 




Example 3-18 


The #-Value of a Masonry Wall 181 






Example 3-19 


The R- Value of a Pitched Roof 1 82 



CHAPTER FOUR 
TRANSIENT HEAT CONDUCTION 209 

Example 4-1 Temperature Measurement by 

Thermocouples 214 

Example 4-2 Predicting the Time of Death 215 

Example 4-3 Boiling Eggs 224 

Example 4-4 Heating of Large Brass Plates 

in an Oven 225 

Example 4-5 Cooling of a Long Stainless Steel 

Cylindrical Shaft 226 

Example 4-6 Minimum Burial Depth of Water 
Pipes to Avoid Freezing 230 

Exa m p I e 4-7 Cooling of a S hort Brass 
Cylinder 234 

Example 4-8 Heat Transfer from a Short 
Cylinder 235 

Example 4-9 Cooling of a Long Cylinder 

by Water 236 

Example 4-10 Refrigerating Steaks while 
Avoiding Frostbite 238 

Example 4-1 1 Chilling of Beef Carcasses in a 
Meat Plant 248 



CHAPTER FIVE 
NUMERICAL METHODS IN 
HEAT CONDUCTION 265 

Example 5-1 Steady Heat Conduction in a Large 

Uranium Plate 277 

Example 5-2 Heat Transfer from 

Triangular Fins 279 

Example 5-3 Steady Two-Dimensional Heat 

Conduction in L-Bars 284 

Example 5-4 Heat Loss through Chimneys 287 

Example 5-5 Transient Heat Conduction in a Large 

Uranium Plate 296 

Example 5-6 Solar Energy Storage in 

Trombe Walls 300 

Example 5-7 Transient Two-Dimensional Heat 

Conduction in L-Bars 305 



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CHAPTER SIX 
FUNDAMENTALS OF CONVECTION 



333 



Example 8-6 



CONTENTS 



Heat Loss from the Ducts of a 
Heating System 448 



Example 6-1 Temperature Rise of Oil in a 

Journal Bearing 350 

Example 6-2 Finding Convection Coefficient from 

Drag Measurement 360 



CHAPTER SEVEN 
EXTERNAL FORCED CONVECTION 



367 



Exa m p I e 7 - 1 Flow of Hot Oil over a 
Flat Plate 376 

Exa m p I e 7 -2 Cooling of a Hot B lock by Forced Air 
at High Elevation 377 

Example 7-3 Cooling of Plastic Sheets by 

Forced Air 378 

Example 7-4 Drag Force Acting on a Pipe 
in a River 383 

Example 7-5 Heat Loss from a Steam Pipe 

in Windy Air 386 

Example 7-6 Cooling of a Steel Ball by 
Forced Air 387 

Example 7-7 Preheating Air by Geothermal Water 

in a Tube Bank 393 

Example 7-8 Effect of Insulation on 

Surface Temperature 402 

Example 7-9 Optimum Thickness of 
Insulation 403 



CHAPTER EIGHT 
INTERNAL FORCED CONVECTION 41 9 

Example 8-1 Heating of Water in a Tube 

by Steam 430 

Example 8-2 Pressure Drop in a Pipe 438 

Example 8-3 Flow of Oil in a Pipeline through 
a Lake 439 

Example 8-4 Pressure Drop in a Water Pipe 445 

Example 8-5 Heating of Water by Resistance 

Heaters in a Tube 446 



CHAPTER NINE 
NATURAL CONVECTION 459 

Example 9-1 Heat Loss from Hot 

Water Pipes 470 

Example 9-2 Cooling of a Plate in 

Different Orientations 47 1 

Example 9-3 Optimum Fin Spacing of a 
Heat Sink 476 

Example 9-4 Heat Loss through a Double-Pane 

Window 482 

Example 9-5 Heat Transfer through a 

Spherical Enclosure 483 

Example 9-6 Heating Water in a Tube by 
Solar Energy 484 

Example 9-7 [/-Factor for Center-of-Glass Section 

of Windows 496 

Example 9-8 Heat Loss through Aluminum Framed 
Windows 497 

Example 9-9 [/-Factor of a Double-Door 
Window 498 



CHAPTER TEN 
BOILING AND CONDENSATION 515 

Example 1 0-1 Nucleate Boiling Water 
in a Pan 526 

Example 10-2 Peak Heat Flux in 

Nucleate Boiling 528 

Example 1 0-3 Film Boiling of Water on a 
Heating Element 529 

Example 10-4 Condensation of Steam on a 
Vertical Plate 541 

Example 10-5 Condensation of Steam on a 
Tilted Plate 542 

Example 10-6 Condensation of Steam on 
Horizontal Tubes 543 

Example 10-7 Condensation of Steam on 

Horizontal Tube Banks 544 



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CONTENTS 



Example 1 0-8 Replacing a Heat Pipe by a 
Copper Rod 550 



CHAPTER ELEVEN 
FUNDAMENTALS OF THERMAL RADIATION 561 

Example 11-1 Radiation Emission from a 
Black Ball 568 

Example 1 1 -2 Emission of Radiation from 
aLightbulb 571 

Example 1 1 -3 Radiation Incident on a 
Small Surface 576 

Example 1 1 -4 Emissivity of a Surface 

and Emissive Power 581 

Example 1 1 -5 Selective Absorber and 

Reflective Surfaces 589 

Example 1 1 -6 Installing Reflective Films 
on Windows 596 



CHAPTER TWELVE 
RADIATION HEAT TRANSFER 605 



Example 12-1 
Example 12-2 
Example 12-3 
Example 12-4 
Example 12-5 
Example 12-6 
Example 12-7 
Example 12-8 
Example 12-9 
Example 12-10 
Example 12-11 



View Factors Associated with 
Two Concentric Spheres 614 

Fraction of Radiation Leaving 
through an Opening 615 

View Factors Associated with 
a Tetragon 617 

View Factors Associated with a 
Triangular Duct 617 

The Crossed-Strings Method for 
View Factors 619 

Radiation Heat Transfer in a 
Black Furnace 621 

Radiation Heat Transfer between 
Parallel Plates 627 

Radiation Heat Transfer in a 
Cylindrical Furnace 630 

Radiation Heat Transfer in a 
Triangular Furnace 63 1 

Heat Transfer through a Tubular 
Solar Collector 632 

Radiation Shields 638 



Example 12-12 Radiation Effect on Temperature 
Measurements 639 

Example 12-13 Effective Emissivity of 
Combustion Gases 646 

Example 1 2-14 Radiation Heat Transfer in a 
Cylindrical Furnace 647 

Example 12-15 Effect of Clothing on Thermal 
Comfort 652 



CHAPTER THIRTEEN 
HEAT EXCHANGERS 667 



Example 13-1 
Example 13-2 
Example 13-3 
Example 13-4 
Example 13-5 
Example 13-6 
Example 13-7 
Example 13-8 
Example 13-9 
Example 13-10 



Overall Heat Transfer Coefficient of 
a Heat Exchanger 675 

Effect of Fouling on the Overall Heat 
Transfer Coefficient 677 

The Condensation of Steam in 
a Condenser 685 

Heating Water in a Counter-Flow 
Heat Exchanger 686 

Heating of Glycerin in a Multipass 
Heat Exchanger 687 

Cooling of an 
Automotive Radiator 688 

Upper Limit for Heat Transfer 
in a Heat Exchanger 69 1 

Using the Effectiveness- 
NTU Method 697 

Cooling Hot Oil by Water in a 
Multipass Heat Exchanger 698 

Installing a Heat Exchanger to Save 
Energy and Money 702 



CHAPTER FOURTEEN 
MASS TRANSFER 717 

Example 14-1 Determining Mass Fractions from 
Mole Fractions 727 

Example 14-2 Mole Fraction of Water Vapor at 
the Surface of a Lake 728 

Example 14-3 Mole Fraction of Dissolved Air 
in Water 730 

Example 1 4-4 Diffusion of Hydrogen Gas into 
a Nickel Plate 732 



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Example 14-5 
Example 14-6 
Example 14-7 
Example 14-8 
Example 14-9 
Example 14-10 
Example 14-11 
Example 14-12 
Example 14-13 



Diffusion of Hydrogen through a 
Spherical Container 735 

Condensation and Freezing of 
Moisture in the Walls 738 

Hardening of Steel by the Diffusion 
of Carbon 742 

Venting of Helium in the Atmosphere 
by Diffusion 751 

Measuring Diffusion Coefficient by 
the Stefan Tube 752 

Mass Convection inside a 
Circular Pipe 761 

Analogy between Heat and 
Mass Transfer 762 

Evaporative Cooling of a 
Canned Drink 765 

Heat Loss from Uncovered Hot 
Water Baths 766 



CHAPTER FIFTEEN 
COOLING OF ELECTRONIC EQUIPMENT 785 

Example 1 5-1 Predicting the Junction Temperature 
of a Transistor 788 

Example 15-2 Determining the Junction-to-Case 
Thermal Resistance 789 

Example 15-3 Analysis of Heat Conduction in 
a Chip 799 

Example 1 5-4 Predicting the Junction Temperature 
of a Device 802 



Example 15-5 
Example 15-6 
Example 15-7 
Example 15-8 
Example 15-9 
Example 15-10 
Example 15-1 1 
Example 15-12 
Example 15-13 
Example 15-14 
Example 15-15 
Example 15-16 
Example 15-17 
Example 15-18 
Example 15-19 



CONTENTS 



Heat Conduction along a PCB with 
Copper Cladding 804 

Thermal Resistance of an Epoxy 
Glass Board 805 

Planting Cylindrical Copper Fillings 
in an Epoxy Board 806 

Conduction Cooling of PCB s by a 
Heat Frame 807 

Cooling of Chips by the Thermal 
Conduction Module 812 

Cooling of a Sealed 
Electronic Box 816 

Cooling of a Component by 
Natural Convection 817 

Cooling of a PCB in a Box by 
Natural Convection 818 

Forced- Air Cooling of a 
Hollow-Core PCB 826 

Forced- Air Cooling of a Transistor 
Mounted on a PCB 828 

Choosing a Fan to Cool 
a Computer 830 

Cooling of a Computer 
by a Fan 831 

Cooling of Power Transistors on 
a Cold Plate by Water 835 

Immersion Cooling of 
a Logic Chip 840 

Cooling of a Chip by Boiling 840 



cen5 8 93 3_fm.qxd 9/11/2002 10:56 AM Page xviii 



Preface 



objectives 

Heat transfer is a basic science that deals with the rate of transfer of ther- 
mal energy. This introductory text is intended for use in a first course in 
heat transfer for undergraduate engineering students, and as a reference 
book for practicing engineers. The objectives of this text are 

• To cover the basic principles of heat transfer. 

• To present a wealth of real-world engineering applications to give stu- 
dents a feel for engineering practice. 

• To develop an intuitive understanding of the subject matter by empha- 
sizing the physics and physical arguments. 

Students are assumed to have completed their basic physics and calculus se- 
quence. The completion of first courses in thermodynamics, fluid mechanics, 
and differential equations prior to taking heat transfer is desirable. The rele- 
vant concepts from these topics are introduced and reviewed as needed. 

In engineering practice, an understanding of the mechanisms of heat trans- 
fer is becoming increasingly important since heat transfer plays a crucial role 
in the design of vehicles, power plants, refrigerators, electronic devices, build- 
ings, and bridges, among other things. Even a chef needs to have an intuitive 
understanding of the heat transfer mechanism in order to cook the food "right" 
by adjusting the rate of heat transfer. We may not be aware of it, but we al- 
ready use the principles of heat transfer when seeking thermal comfort. We in- 
sulate our bodies by putting on heavy coats in winter, and we minimize heat 
gain by radiation by staying in shady places in summer. We speed up the cool- 
ing of hot food by blowing on it and keep warm in cold weather by cuddling 
up and thus minimizing the exposed surface area. That is, we already use heat 
transfer whether we realize it or not. 



GENERAL APPROACH 

This text is the outcome of an attempt to have a textbook for a practically ori- 
ented heat transfer course for engineering students. The text covers the stan- 
dard topics of heat transfer with an emphasis on physics and real-world 
applications, while de-emphasizing intimidating heavy mathematical aspects. 
This approach is more in line with students' intuition and makes learning the 
subject matter much easier. 

The philosophy that contributed to the warm reception of the first edition of 
this book has remained unchanged. The goal throughout this project has been 
to offer an engineering textbook that 



cen58 933_fm.qxd 9/11/2002 10:56 AM Page xix 



• Talks directly to the minds of tomorrow's engineers in a simple yet pre- 
cise manner. 

• Encourages creative thinking and development of a deeper understand- 
ing of the subject matter. 

• Is read by students with interest and enthusiasm rather than being used 
as just an aid to solve problems. 

Special effort has been made to appeal to readers' natural curiosity and to help 
students explore the various facets of the exciting subject area of heat transfer. 
The enthusiastic response we received from the users of the first edition all 
over the world indicates that our objectives have largely been achieved. 

Yesterday's engineers spent a major portion of their time substituting values 
into the formulas and obtaining numerical results. However, now formula ma- 
nipulations and number crunching are being left to computers. Tomorrow's 
engineer will have to have a clear understanding and a firm grasp of the basic 
principles so that he or she can understand even the most complex problems, 
formulate them, and interpret the results. A conscious effort is made to em- 
phasize these basic principles while also providing students with a look at 
how modern tools are used in engineering practice. 

NEW IN THIS EDITION 

All the popular features of the previous edition are retained while new ones 
are added. The main body of the text remains largely unchanged except that 
the coverage of forced convection is expanded to three chapters and the cov- 
erage of radiation to two chapters. Of the three applications chapters, only the 
Cooling of Electronic Equipment is retained, and the other two are deleted to 
keep the book at a reasonable size. The most significant changes in this edi- 
tion are highlighted next. 

EXPANDED COVERAGE OF CONVECTION 

Forced convection is now covered in three chapters instead of one. In Chapter 
6, the basic concepts of convection and the theoretical aspects are introduced. 
Chapter 7 deals with the practical analysis of external convection while Chap- 
ter 8 deals with the practical aspects of internal convection. See the Content 
Changes and Reorganization section for more details. 

ADDITIONAL CHAPTER ON RADIATION 

Radiation is now covered in two chapters instead of one. The basic concepts 
associated with thermal radiation, including radiation intensity and spectral 
quantities, are covered in Chapter 11. View factors and radiation exchange be- 
tween surfaces through participating and nonparticipating media are covered 
in Chapter 12. See the Content Changes and Reorganization section for more 
details. 

TOPICS OF SPECIAL INTEREST 

Most chapters now contain a new end-of-chapter optional section called 
"Topic of Special Interest" where interesting applications of heat transfer are 
discussed. Some existing sections such as A Brief Review of Differential 
Equations in Chapter 2, Thermal Insulation in Chapter 7, and Controlling Nu- 
merical Error in Chapter 5 are moved to these sections as topics of special 



cen5 8 93 3_fm.qxd 9/11/2002 10:56 AM Page xx 



interest. Some sections from the two deleted chapters such as the Refrigera- 
tion and Freezing of Foods, Solar Heat Gain through Windows, and Heat 
Transfer through the Walls and Roofs are moved to the relevant chapters as 
special topics. Most topics selected for these sections provide real-world 
applications of heat transfer, but they can be ignored if desired without a loss 
in continuity. 

COMPREHENSIVE PROBLEMS WITH PARAMETRIC STUDIES 

A distinctive feature of this edition is the incorporation of about 130 compre- 
hensive problems that require conducting extensive parametric studies, using 
the enclosed EES (or other suitable) software. Students are asked to study the 
effects of certain variables in the problems on some quantities of interest, to 
plot the results, and to draw conclusions from the results obtained. These 
problems are designated by computer-EES and EES -CD icons for easy recog- 
nition, and can be ignored if desired. Solutions of these problems are given in 
the Instructor's Solutions Manual. 

CONTENT CHANGES AND REORGANIZATION 

With the exception of the changes already mentioned, the main body of the 
text remains largely unchanged. This edition involves over 500 new or revised 
problems. The noteworthy changes in various chapters are summarized here 
for those who are familiar with the previous edition. 

• In Chapter 1, surface energy balance is added to Section 1-4. In a new 
section Problem-Solving Technique, the problem-solving technique is 
introduced, the engineering software packages are discussed, and 
overviews of EES (Engineering Equation Solver) and HTT (Heat Trans- 
fer Tools) are given. The optional Topic of Special Interest in this chap- 
ter is Thermal Comfort. 

• In Chapter 2, the section A Brief Review of Differential Equations is 
moved to the end of chapter as the Topic of Special Interest. 

• In Chapter 3, the section on Thermal Insulation is moved to Chapter 7, 
External Forced Convection, as a special topic. The optional Topic of 
Special Interest in this chapter is Heat Transfer through Walls and 
Roofs. 

• Chapter 4 remains mostly unchanged. The Topic of Special Interest in 
this chapter is Refrigeration and Freezing of Foods. 

• In Chapter 5, the section Solutions Methods for Systems of Algebraic 
Equations and the FORTRAN programs in the margin are deleted, and 
the section Controlling Numerical Error is designated as the Topic of 
Special Interest. 

• Chapter 6, Forced Convection, is now replaced by three chapters: Chap- 
ter 6 Fundamentals of Convection, where the basic concepts of convec- 
tion are introduced and the fundamental convection equations and 
relations (such as the differential momentum and energy equations and 
the Reynolds analogy) are developed; Chapter 7 External Forced Con- 
vection, where drag and heat transfer for flow over surfaces, including 
flow over tube banks, are discussed; and Chapter 8 Internal Forced 
Convection, where pressure drop and heat transfer for flow in tubes are 



cen58 933_fm.qxd 9/11/2002 10:56 AM Page xxi 



presented. Reducing Heat Transfer through Surfaces is added to Chap- 
ter 7 as the Topic of Special Interest. 

• Chapter 7 (now Chapter 9) Natural Convection is completely rewritten. 
The Grashof number is derived from a momentum balance on a differ- 
ential volume element, some Nusselt number relations (especially those 
for rectangular enclosures) are updated, and the section Natural Con- 
vection from Finned Surfaces is expanded to include heat transfer from 
PCBs. The optional Topic of Special Interest in this chapter is Heat 
Transfer through Windows. 

• Chapter 8 (now Chapter 10) Boiling and Condensation remained largely 
unchanged. The Topic of Special Interest in this chapter is Heat Pipes. 

• Chapter 9 is split in two chapters: Chapter 11 Fundamentals of Thermal 
Radiation, where the basic concepts associated with thermal radiation, 
including radiation intensity and spectral quantities, are introduced, and 
Chapter 12 Radiation Heat Transfer, where the view factors and radia- 
tion exchange between surfaces through participating and nonparticipat- 
ing media are discussed. The Topic of Special Interest are Solar Heat 
Gain through Windows in Chapter 11, and Heat Transfer from the Hu- 
man Body in Chapter 12. 

• There are no significant changes in the remaining three chapters of Heat 
Exchangers, Mass Transfer, and Cooling of Electronic Equipment. 

• In the appendices, the values of the physical constants are updated; new 
tables for the properties of saturated ammonia, refrigerant- 134a, and 
propane are added; and the tables on the properties of air, gases, and liq- 
uids (including liquid metals) are replaced by those obtained using EES . 
Therefore, property values in tables for air, other ideal gases, ammonia, 
refrigerant- 134a, propane, and liquids are identical to those obtained 
from EES. 

LEARNING TOOLS 

EMPHASIS ON PHYSICS 

A distinctive feature of this book is its emphasis on the physical aspects of 
subject matter rather than mathematical representations and manipulations. 
The author believes that the emphasis in undergraduate education should re- 
main on developing a sense of underlying physical mechanism and a mastery 
of solving practical problems an engineer is likely to face in the real world. 
Developing an intuitive understanding should also make the course a more 
motivating and worthwhile experience for the students. 

EFFECTIVE USE OF ASSOCIATION 

An observant mind should have no difficulty understanding engineering sci- 
ences. After all, the principles of engineering sciences are based on our every- 
day experiences and experimental observations. A more physical, intuitive 
approach is used throughout this text. Frequently parallels are drawn between 
the subject matter and students' everyday experiences so that they can relate 
the subject matter to what they already know. The process of cooking, for ex- 
ample, serves as an excellent vehicle to demonstrate the basic principles of 
heat transfer. 



cen58 93 3_fm.qxd 9/11/2002 10:56 AM Page xxii 



SELF-INSTRUCTING 

The material in the text is introduced at a level that an average student can 
follow comfortably. It speaks to students, not over students. In fact, it is self- 
instructive. Noting that the principles of sciences are based on experimental 
observations, the derivations in this text are based on physical arguments, and 
thus they are easy to follow and understand. 

EXTENSIVE USE OF ARTWORK 

Figures are important learning tools that help the students "get the picture." 
The text makes effective use of graphics. It contains more figures and illus- 
trations than any other book in this category. Figures attract attention and 
stimulate curiosity and interest. Some of the figures in this text are intended to 
serve as a means of emphasizing some key concepts that would otherwise go 
unnoticed; some serve as paragraph summaries. 

CHAPTER OPENERS AND SUMMARIES 

Each chapter begins with an overview of the material to be covered and its re- 
lation to other chapters. A summary is included at the end of each chapter for 
a quick review of basic concepts and important relations. 

NUMEROUS W0RKED-0UT EXAMPLES 

Each chapter contains several worked-out examples that clarify the material 
and illustrate the use of the basic principles. An intuitive and systematic ap- 
proach is used in the solution of the example problems, with particular atten- 
tion to the proper use of units. 

A WEALTH OF REAL-WORLD END-OF-CHAPTER PROBLEMS 

The end-of-chapter problems are grouped under specific topics in the order 
they are covered to make problem selection easier for both instructors and stu- 
dents. The problems within each group start with concept questions, indicated 
by "C," to check the students' level of understanding of basic concepts. The 
problems under Review Problems are more comprehensive in nature and are 
not directly tied to any specific section of a chapter. The problems under the 
Design and Essay Problems title are intended to encourage students to make 
engineering judgments, to conduct independent exploration of topics of inter- 
est, and to communicate their findings in a professional manner. Several eco- 
nomics- and safety-related problems are incorporated throughout to enhance 
cost and safety awareness among engineering students. Answers to selected 
problems are listed immediately following the problem for convenience to 
students. 

A SYSTEMATIC SOLUTION PROCEDURE 

A well-structured approach is used in problem solving while maintaining an 
informal conversational style. The problem is first stated and the objectives 
are identified, and the assumptions made are stated together with their justifi- 
cations. The properties needed to solve the problem are listed separately. Nu- 
merical values are used together with their units to emphasize that numbers 
without units are meaningless, and unit manipulations are as important as 
manipulating the numerical values with a calculator. The significance of the 
findings is discussed following the solutions. This approach is also used 
consistently in the solutions presented in the Instructor's Solutions Manual. 



cen5 8 93 3_fm.qxd 9/11/2002 10:56 AM Page xxiii 



A CHOICE OF SI ALONE OR SI/ENGLISH UNITS 

In recognition of the fact that English units are still widely used in some in- 
dustries, both SI and English units are used in this text, with an emphasis on 
SI. The material in this text can be covered using combined Si/English units 
or SI units alone, depending on the preference of the instructor. The property 
tables and charts in the appendices are presented in both units, except the ones 
that involve dimensionless quantities. Problems, tables, and charts in English 
units are designated by "E" after the number for easy recognition, and they 
can be ignored easily by the SI users. 

CONVERSION FACTORS 

Frequently used conversion factors and the physical constants are listed on the 
inner cover pages of the text for easy reference. 

SUPPLEMENTS 

These supplements are available to the adopters of the book. 

COSMOS SOLUTIONS MANUAL 

Available to instructors only. 

The detailed solutions for all text problems will be delivered in our 
new electronic Complete Online Solution Manual Organization System 
(COSMOS). COSMOS is a database management tool geared towards as- 
sembling homework assignments, tests and quizzes. No longer do instructors 
need to wade through thick solutions manuals and huge Word files. COSMOS 
helps you to quickly find solutions and also keeps a record of problems as- 
signed to avoid duplication in subsequent semesters. Instructors can contact 
their McGraw-Hill sales representative at http://www.mhhe.com/catalogs/rep/ 
to obtain a copy of the COSMOS solutions manual. 

EES SOFTWARE 

Developed by Sanford Klein and William Beckman from the University of 
Wisconsin-Madison, this software program allows students to solve prob- 
lems, especially design problems, and to ask "what if questions. EES (pro- 
nounced "ease") is an acronym for Engineering Equation Solver. EES is very 
easy to master since equations can be entered in any form and in any order. 
The combination of equation-solving capability and engineering property data 
makes EES an extremely powerful tool for students. 

EES can do optimization, parametric analysis, and linear and nonlinear re- 
gression and provides publication-quality plotting capability. Equations can be 
entered in any form and in any order. EES automatically rearranges the equa- 
tions to solve them in the most efficient manner. EES is particularly useful for 
heat transfer problems since most of the property data needed for solving such 
problems are provided in the program. For example, the steam tables are im- 
plemented such that any thermodynamic property can be obtained from a 
built-in function call in terms of any two properties. Similar capability is pro- 
vided for many organic refrigerants, ammonia, methane, carbon dioxide, and 
many other fluids. Air tables are built-in, as are psychrometric functions and 
JANAF table data for many common gases. Transport properties are also pro- 
vided for all substances. EES also allows the user to enter property data or 
functional relationships with look-up tables, with internal functions written 



cen5 8 93 3_fm.qxd 9/11/2002 10:56 AM Page xxiv 



with EES, or with externally compiled functions written in Pascal, C, C + + , 
or FORTRAN. 

The Student Resources CD that accompanies the text contains the Limited 
Academic Version of the EES program and the scripted EES solutions of about 
30 homework problems (indicated by the "EES-CD" logo in the text). Each 
EES solution provides detailed comments and on-line help, and can easily be 
modified to solve similar problems. These solutions should help students 
master the important concepts without the calculational burden that has been 
previously required. 

HEAT TRANSFER TOOLS (HTT) 

One software package specifically designed to help bridge the gap between 
the textbook fundamentals and commercial software packages is Heat Trans- 
fer Tools, which can be ordered "bundled" with this text (Robert J. Ribando, 
ISBN 0-07-246328-7). While it does not have the power and functionality of 
the professional, commercial packages, HTT uses research-grade numerical 
algorithms behind the scenes and modern graphical user interfaces. Each 
module is custom designed and applicable to a single, fundamental topic in 
heat transfer. 

BOOK-SPECIFIC WEBSITE 

The book website can be found at www.mhhe.com/cengel/. Visit this site for 
book and supplement information, author information, and resources for fur- 
ther study or reference. At this site you will also find PowerPoints of selected 
text figures. 

ACKNOWLEDGMENTS 

I would like to acknowledge with appreciation the numerous and valuable 
comments, suggestions, criticisms, and praise of these academic evaluators: 



Sanjeev Chandra 

University of Toronto, Canada 

Fan-Bill Cheung 

The Pennsylvania State University 

Nicole DeJong 

San Jose State University 

David M. Doner 

West Virginia University Institute of 
Technology 

Mark J. Holowach 

The Pennsylvania State University 

Mehmet Kanoglu 

Gaziantep University, Turkey 

Francis A. Kulacki 

University of Minnesota 



Sai C. Lau 

Texas A&M University 

Joseph Majdalani 

Marquette University 

Jed E. Marquart 

Ohio Northern University 

Robert J. Ribando 

University of Virginia 

Jay M. Ochterbeck 

Clemson University 

James R. Thomas 

Virginia Polytechnic Institute and 
State University 

John D. Wellin 

Rochester Institute of Technology 



cen58 933_fm.qxd 9/11/2002 10:56 AM Page xxv 



Their suggestions have greatly helped to improve the quality of this text. I also 
would like to thank my students who provided plenty of feedback from their 
perspectives. Finally, I would like to express my appreciation to my wife 
Zehra and my children for their continued patience, understanding, and sup- 
port throughout the preparation of this text. 

Yunus A. Cengel 



cen58933_ch01.qxd 9/10/2002 8:29 AM Page 1 



BASICS OF HEAT TRANSFER 



CHAPTER 



The science of thermodynamics deals with the amount of heat transfer as 
a system undergoes a process from one equilibrium state to another, and 
makes no reference to how long the process will take. But in engineer- 
ing, we are often interested in the rate of heat transfer, which is the topic of 
the science of heat transfer. 

We start this chapter with a review of the fundamental concepts of thermo- 
dynamics that form the framework for heat transfer. We first present the 
relation of heat to other forms of energy and review the first law of thermo- 
dynamics. We then present the three basic mechanisms of heat transfer, which 
are conduction, convection, and radiation, and discuss thermal conductivity. 
Conduction is the transfer of energy from the more energetic particles of a 
substance to the adjacent, less energetic ones as a result of interactions be- 
tween the particles. Convection is the mode of heat transfer between a solid 
surface and the adjacent liquid or gas that is in motion, and it involves the 
combined effects of conduction and fluid motion. Radiation is the energy 
emitted by matter in the form of electromagnetic waves (or photons) as a re- 
sult of the changes in the electronic configurations of the atoms or molecules. 
We close this chapter with a discussion of simultaneous heat transfer. 



CONTENTS 

1-1 Thermodynamics and 
Heat Transfer 2 

1-2 Engineering Heat Transfer 4 

1-3 Heat and Other Forms 
of Energy 6 

1-4 The First Law of 

Thermodynamics 11 

1-5 Heat Transfer 

Mechanisms 17 

1-6 Conduction 17 

1-7 Convection 25 

1-8 Radiation 27 

1-9 Simultaneous Heat Transfer 
Mechanism 30 

1-10 Problem-Solving Technique 35 

Topic of Special Interest: 

Thermal Comfort 40 



cen58933_ch01.qxd 9/10/2002 8:29 AM Page 2 



HEAT TRANSFER 



1-1 - THERMODYNAMICS AND HEAT TRANSFER 



Thermos 
bottle 



TU 



Hot 
coffee 



± 



^- Insulation 

FIGURE 1-1 

We are normally interested in how long 
it takes for the hot coffee in a thermos to 
cool to a certain temperature, which 
cannot be determined from a 
thermodynamic analysis alone. 






Cool 

environment 

20°C 




1 Heat 



FIGURE 1-2 

Heat flows in the direction of 
decreasing temperature. 



We all know from experience that a cold canned drink left in a room warms up 
and a warm canned drink left in a refrigerator cools down. This is accom- 
plished by the transfer of energy from the warm medium to the cold one. The 
energy transfer is always from the higher temperature medium to the lower 
temperature one, and the energy transfer stops when the two mediums reach 
the same temperature. 

You will recall from thermodynamics that energy exists in various forms. In 
this text we are primarily interested in heat, which is the form of energy that 
can be transferred from one system to another as a result of temperature dif- 
ference. The science that deals with the determination of the rates of such en- 
ergy transfers is heat transfer. 

You may be wondering why we need to undertake a detailed study on heat 
transfer. After all, we can determine the amount of heat transfer for any sys- 
tem undergoing any process using a thermodynamic analysis alone. The rea- 
son is that thermodynamics is concerned with the amount of heat transfer as a 
system undergoes a process from one equilibrium state to another, and it gives 
no indication about how long the process will take. A thermodynamic analysis 
simply tells us how much heat must be transferred to realize a specified 
change of state to satisfy the conservation of energy principle. 

In practice we are more concerned about the rate of heat transfer (heat trans- 
fer per unit time) than we are with the amount of it. For example, we can de- 
termine the amount of heat transferred from a thermos bottle as the hot coffee 
inside cools from 90°C to 80°C by a thermodynamic analysis alone. But a typ- 
ical user or designer of a thermos is primarily interested in how long it will be 
before the hot coffee inside cools to 80°C, and a thermodynamic analysis can- 
not answer this question. Determining the rates of heat transfer to or from a 
system and thus the times of cooling or heating, as well as the variation of the 
temperature, is the subject of heat transfer (Fig. 1-1). 

Thermodynamics deals with equilibrium states and changes from one equi- 
librium state to another. Heat transfer, on the other hand, deals with systems 
that lack thermal equilibrium, and thus it is a nonequilibrium phenomenon. 
Therefore, the study of heat transfer cannot be based on the principles of 
thermodynamics alone. However, the laws of thermodynamics lay the frame- 
work for the science of heat transfer. The first law requires that the rate of 
energy transfer into a system be equal to the rate of increase of the energy of 
that system. The second law requires that heat be transferred in the direction 
of decreasing temperature (Fig. 1-2). This is like a car parked on an inclined 
road that must go downhill in the direction of decreasing elevation when its 
brakes are released. It is also analogous to the electric current flowing in the 
direction of decreasing voltage or the fluid flowing in the direction of de- 
creasing total pressure. 

The basic requirement for heat transfer is the presence of a temperature dif- 
ference. There can be no net heat transfer between two mediums that are at the 
same temperature. The temperature difference is the driving force for heat 
transfer, just as the voltage difference is the driving force for electric current 
flow and pressure difference is the driving force for fluid flow. The rate of heat 
transfer in a certain direction depends on the magnitude of the temperature 
gradient (the temperature difference per unit length or the rate of change of 



cen58933_ch01.qxd 9/10/2002 8:29 AM Page 3 



CHAPTER 1 



temperature) in that direction. The larger the temperature gradient, the higher 
the rate of heat transfer. 

Application Areas of Heat Transfer 

Heat transfer is commonly encountered in engineering systems and other as- 
pects of life, and one does not need to go very far to see some application ar- 
eas of heat transfer. In fact, one does not need to go anywhere. The human 
body is constantly rejecting heat to its surroundings, and human comfort is 
closely tied to the rate of this heat rejection. We try to control this heat trans- 
fer rate by adjusting our clothing to the environmental conditions. 

Many ordinary household appliances are designed, in whole or in part, by 
using the principles of heat transfer. Some examples include the electric or gas 
range, the heating and air-conditioning system, the refrigerator and freezer, the 
water heater, the iron, and even the computer, the TV, and the VCR. Of course, 
energy-efficient homes are designed on the basis of minimizing heat loss in 
winter and heat gain in summer. Heat transfer plays a major role in the design 
of many other devices, such as car radiators, solar collectors, various compo- 
nents of power plants, and even spacecraft. The optimal insulation thickness 
in the walls and roofs of the houses, on hot water or steam pipes, or on water 
heaters is again determined on the basis of a heat transfer analysis with eco- 
nomic consideration (Fig. 1-3). 

Historical Background 

Heat has always been perceived to be something that produces in us a sensa- 
tion of warmth, and one would think that the nature of heat is one of the first 
things understood by mankind. But it was only in the middle of the nineteenth 




The human body 



Water in 




■ ll 

08 



30D[ 

nnnDn 



Air-conditioning 
systems 




Water out 




oj i n I 

Circuit boards 




Car radiators 



Power plants 



Refrigeration systems 

FIGURE 1-3 

Some application areas of heat transfer. 



cen58933_ch01.qxd 9/10/2002 8:29 AM Page 4 



HEAT TRANSFER 




FIGURE 1-4 

In the early nineteenth century, heat was 
thought to be an invisible fluid called the 
caloric that flowed from warmer bodies 
to the cooler ones. 



century that we had a true physical understanding of the nature of heat, thanks 
to the development at that time of the kinetic theory, which treats molecules 
as tiny balls that are in motion and thus possess kinetic energy. Heat is then 
defined as the energy associated with the random motion of atoms and mole- 
cules. Although it was suggested in the eighteenth and early nineteenth cen- 
turies that heat is the manifestation of motion at the molecular level (called the 
live force), the prevailing view of heat until the middle of the nineteenth cen- 
tury was based on the caloric theory proposed by the French chemist Antoine 
Lavoisier (1743-1794) in 1789. The caloric theory asserts that heat is a fluid- 
like substance called the caloric that is a massless, colorless, odorless, and 
tasteless substance that can be poured from one body into another (Fig. 1-4). 
When caloric was added to a body, its temperature increased; and when 
caloric was removed from a body, its temperature decreased. When a body 
could not contain any more caloric, much the same way as when a glass of 
water could not dissolve any more salt or sugar, the body was said to be satu- 
rated with caloric. This interpretation gave rise to the terms saturated liquid 
and saturated vapor that are still in use today. 

The caloric theory came under attack soon after its introduction. It main- 
tained that heat is a substance that could not be created or destroyed. Yet it 
was known that heat can be generated indefinitely by rubbing one's hands to- 
gether or rubbing two pieces of wood together. In 1798, the American Ben- 
jamin Thompson (Count Rumford) (1753-1814) showed in his papers that 
heat can be generated continuously through friction. The validity of the caloric 
theory was also challenged by several others. But it was the careful experi- 
ments of the Englishman James P. Joule (1818-1889) published in 1843 that 
finally convinced the skeptics that heat was not a substance after all, and thus 
put the caloric theory to rest. Although the caloric theory was totally aban- 
doned in the middle of the nineteenth century, it contributed greatly to the de- 
velopment of thermodynamics and heat transfer. 



1-2 - ENGINEERING HEAT TRANSFER 

Heat transfer equipment such as heat exchangers, boilers, condensers, radia- 
tors, heaters, furnaces, refrigerators, and solar collectors are designed pri- 
marily on the basis of heat transfer analysis. The heat transfer problems 
encountered in practice can be considered in two groups: (1) rating and 
(2) sizing problems. The rating problems deal with the determination of the 
heat transfer rate for an existing system at a specified temperature difference. 
The sizing problems deal with the determination of the size of a system in 
order to transfer heat at a specified rate for a specified temperature difference. 
A heat transfer process or equipment can be studied either experimentally 
(testing and taking measurements) or analytically (by analysis or calcula- 
tions). The experimental approach has the advantage that we deal with the 
actual physical system, and the desired quantity is determined by measure- 
ment, within the limits of experimental error. However, this approach is ex- 
pensive, time-consuming, and often impractical. Besides, the system we are 
analyzing may not even exist. For example, the size of a heating system of 
a building must usually be determined before the building is actually built 
on the basis of the dimensions and specifications given. The analytical ap- 
proach (including numerical approach) has the advantage that it is fast and 



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CHAPTER 1 



inexpensive, but the results obtained are subject to the accuracy of the 
assumptions and idealizations made in the analysis. In heat transfer studies, 
often a good compromise is reached by reducing the choices to just a few by 
analysis, and then verifying the findings experimentally. 



Modeling in Heat Transfer 



The descriptions of most scientific problems involve expressions that relate 
the changes in some key variables to each other. Usually the smaller the 
increment chosen in the changing variables, the more general and accurate 
the description. In the limiting case of infinitesimal or differential changes in 
variables, we obtain differential equations that provide precise mathematical 
formulations for the physical principles and laws by representing the rates of 
changes as derivatives. Therefore, differential equations are used to investi- 
gate a wide variety of problems in sciences and engineering, including heat 
transfer. However, most heat transfer problems encountered in practice can be 
solved without resorting to differential equations and the complications asso- 
ciated with them. 

The study of physical phenomena involves two important steps. In the first 
step, all the variables that affect the phenomena are identified, reasonable as- 
sumptions and approximations are made, and the interdependence of these 
variables is studied. The relevant physical laws and principles are invoked, 
and the problem is formulated mathematically. The equation itself is very in- 
structive as it shows the degree of dependence of some variables on others, 
and the relative importance of various terms. In the second step, the problem 
is solved using an appropriate approach, and the results are interpreted. 

Many processes that seem to occur in nature randomly and without any or- 
der are, in fact, being governed by some visible or not-so-visible physical 
laws. Whether we notice them or not, these laws are there, governing consis- 
tently and predictably what seem to be ordinary events. Most of these laws are 
well defined and well understood by scientists. This makes it possible to pre- 
dict the course of an event before it actually occurs, or to study various aspects 
of an event mathematically without actually running expensive and time- 
consuming experiments. This is where the power of analysis lies. Very accu- 
rate results to meaningful practical problems can be obtained with relatively 
little effort by using a suitable and realistic mathematical model. The prepara- 
tion of such models requires an adequate knowledge of the natural phenomena 
involved and the relevant laws, as well as a sound judgment. An unrealistic 
model will obviously give inaccurate and thus unacceptable results. 

An analyst working on an engineering problem often finds himself or her- 
self in a position to make a choice between a very accurate but complex 
model, and a simple but not-so-accurate model. The right choice depends on 
the situation at hand. The right choice is usually the simplest model that yields 
adequate results. For example, the process of baking potatoes or roasting a 
round chunk of beef in an oven can be studied analytically in a simple way by 
modeling the potato or the roast as a spherical solid ball that has the properties 
of water (Fig. 1-5). The model is quite simple, but the results obtained are suf- 
ficiently accurate for most practical purposes. As another example, when we 
analyze the heat losses from a building in order to select the right size for a 
heater, we determine the heat losses under anticipated worst conditions and 
select a furnace that will provide sufficient heat to make up for those losses. 



Oven 

( Potato ) ■* Actual 

175°C 
Water ■* Ideal 



FIGURE 1-5 

Modeling is a powerful engineering 

tool that provides great insight and 

simplicity at the expense of 

some accuracy. 



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6 
HEAT TRANSFER 



Often we tend to choose a larger furnace in anticipation of some future ex- 
pansion, or just to provide a factor of safety. A very simple analysis will be ad- 
equate in this case. 

When selecting heat transfer equipment, it is important to consider the ac- 
tual operating conditions. For example, when purchasing a heat exchanger 
that will handle hard water, we must consider that some calcium deposits will 
form on the heat transfer surfaces over time, causing fouling and thus a grad- 
ual decline in performance. The heat exchanger must be selected on the basis 
of operation under these adverse conditions instead of under new conditions. 

Preparing very accurate but complex models is usually not so difficult. But 
such models are not much use to an analyst if they are very difficult and time- 
consuming to solve. At the minimum, the model should reflect the essential 
features of the physical problem it represents. There are many significant real- 
world problems that can be analyzed with a simple model. But it should al- 
ways be kept in mind that the results obtained from an analysis are as accurate 
as the assumptions made in simplifying the problem. Therefore, the solution 
obtained should not be applied to situations for which the original assump- 
tions do not hold. 

A solution that is not quite consistent with the observed nature of the prob- 
lem indicates that the mathematical model used is too crude. In that case, a 
more realistic model should be prepared by eliminating one or more of the 
questionable assumptions. This will result in a more complex problem that, of 
course, is more difficult to solve. Thus any solution to a problem should be in- 
terpreted within the context of its formulation. 

1-3 - HEAT AND OTHER FORMS OF ENERGY 

Energy can exist in numerous forms such as thermal, mechanical, kinetic, po- 
tential, electrical, magnetic, chemical, and nuclear, and their sum constitutes 
the total energy E (or e on a unit mass basis) of a system. The forms of energy 
related to the molecular structure of a system and the degree of the molecular 
activity are referred to as the microscopic energy. The sum of all microscopic 
forms of energy is called the internal energy of a system, and is denoted by 
U (or u on a unit mass basis). 

The international unit of energy is joule (J) or kilojoule (1 kJ = 1000 J). 
In the English system, the unit of energy is the British thermal unit (Btu), 
which is defined as the energy needed to raise the temperature of 1 lbm of 
water at 60°F by 1°F. The magnitudes of kJ and Btu are almost identical 
(1 Btu = 1.055056 kJ). Another well-known unit of energy is the calorie 
(1 cal = 4.1868 J), which is defined as the energy needed to raise the temper- 
ature of 1 gram of water at 14.5°C by 1°C. 

Internal energy may be viewed as the sum of the kinetic and potential ener- 
gies of the molecules. The portion of the internal energy of a system asso- 
ciated with the kinetic energy of the molecules is called sensible energy or 
sensible heat. The average velocity and the degree of activity of the mole- 
cules are proportional to the temperature. Thus, at higher temperatures the 
molecules will possess higher kinetic energy, and as a result, the system will 
have a higher internal energy. 

The internal energy is also associated with the intermolecular forces be- 
tween the molecules of a system. These are the forces that bind the molecules 



cen58933_ch01.qxd 9/10/2002 8:29 AM Page 7 



CHAPTER 1 



to each other, and, as one would expect, they are strongest in solids and weak- 
est in gases. If sufficient energy is added to the molecules of a solid or liquid, 
they will overcome these molecular forces and simply break away, turning the 
system to a gas. This is a phase change process and because of this added en- 
ergy, a system in the gas phase is at a higher internal energy level than it is in 
the solid or the liquid phase. The internal energy associated with the phase of 
a system is called latent energy or latent heat. 

The changes mentioned above can occur without a change in the chemical 
composition of a system. Most heat transfer problems fall into this category, 
and one does not need to pay any attention to the forces binding the atoms in 
a molecule together. The internal energy associated with the atomic bonds in 
a molecule is called chemical (or bond) energy, whereas the internal energy 
associated with the bonds within the nucleus of the atom itself is called nu- 
clear energy. The chemical and nuclear energies are absorbed or released dur- 
ing chemical or nuclear reactions, respectively. 

In the analysis of systems that involve fluid flow, we frequently encounter 
the combination of properties u and Pv. For the sake of simplicity and conve- 
nience, this combination is defined as enthalpy /;. That is, h = u + Pv where 
the term Pv represents the//ow energy of the fluid (also called the flow work), 
which is the energy needed to push a fluid and to maintain flow. In the energy 
analysis of flowing fluids, it is convenient to treat the flow energy as part of 
the energy of the fluid and to represent the microscopic energy of a fluid 
stream by enthalpy h (Fig. 1-6). 




• Energy = h 



Stationary 
fluid 



Energy ; 



FIGURE 1-6 

The internal energy u represents the mi- 
croscopic energy of a nonflowing fluid, 
whereas enthalpy h represents the micro- 
scopic energy of a flowing fluid. 



Specific Heats of Gases, Liquids, and Solids 

You may recall that an ideal gas is defined as a gas that obeys the relation 



Pv = RT 



pRT 



(1-1) 



where P is the absolute pressure, v is the specific volume, T is the absolute 
temperature, p is the density, and R is the gas constant. It has been experi- 
mentally observed that the ideal gas relation given above closely approxi- 
mates the P-v-T behavior of real gases at low densities. At low pressures and 
high temperatures, the density of a gas decreases and the gas behaves like an 
ideal gas. In the range of practical interest, many familiar gases such as air, 
nitrogen, oxygen, hydrogen, helium, argon, neon, and krypton and even heav- 
ier gases such as carbon dioxide can be treated as ideal gases with negligible 
error (often less than one percent). Dense gases such as water vapor in 
steam power plants and refrigerant vapor in refrigerators, however, should not 
always be treated as ideal gases since they usually exist at a state near 
saturation. 

You may also recall that specific heat is defined as the energy required to 
raise the temperature of a unit mass of a substance by one degree (Fig. 1-7). 
In general, this energy depends on how the process is executed. In thermo- 
dynamics, we are interested in two kinds of specific heats: specific heat at 
constant volume C, and specific heat at constant pressure C p . The specific 
heat at constant volume C, can be viewed as the energy required to raise the 
temperature of a unit mass of a substance by one degree as the volume is held 
constant. The energy required to do the same as the pressure is held constant 
is the specific heat at constant pressure C p . The specific heat at constant 





m = 


1kg 






AT = 


1°C 




Specific heat 
1 


= 5 kJ/kg- 


°C 



5kJ 

FIGURE 1-7 

Specific heat is the energy required to 

raise the temperature of a unit mass 

of a substance by one degree in a 

specified way. 



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THERMODYNAMICS 



Air 




Air 




in = 1 kg 




m = 1 kg 




300^301 K 
f 




1000 -> 1001 K 
f 




I 

0.718 kJ 


I 

0.855 kJ 


FIGURE 1-8 




The specific heat of a 


substance changes 


with temperatur 


e. 







pressure C p is greater than C r because at constant pressure the system is al- 
lowed to expand and the energy for this expansion work must also be supplied 
to the system. For ideal gases, these two specific heats are related to each 
other by C p = C, + R. 

A common unit for specific heats is kJ/kg • °C or kJ/kg • K. Notice that these 
two units are identical since AT(°C) = AT(K), and 1°C change in temperature 
is equivalent to a change of 1 K. Also, 

1 kJ/kg • °C = 1 J/g • °C = 1 kJ/kg • K = 1 J/g • K 

The specific heats of a substance, in general, depend on two independent 
properties such as temperature and pressure. For an ideal gas, however, they 
depend on temperature only (Fig. 1-8). At low pressures all real gases ap- 
proach ideal gas behavior, and therefore their specific heats depend on tem- 
perature only. 

The differential changes in the internal energy u and enthalpy h of an ideal 
gas can be expressed in terms of the specific heats as 



du = C,,dT 



and 



dh 



C p dT 



(1-2) 



The finite changes in the internal energy and enthalpy of an ideal gas during a 
process can be expressed approximately by using specific heat values at the 
average temperature as 



Am 



C AT 



and 



Ah 



C AT 



(J/g) 



(1-3) 



or 



At/ = mC,. Ar 



and 



AH 



mC pmc AT 



(J) 



(1-4) 




FIGURE 1-9 

The C„ and C p values of incompressible 
substances are identical and are 
denoted by C. 



where m is the mass of the system. 

A substance whose specific volume (or density) does not change with tem- 
perature or pressure is called an incompressible substance. The specific vol- 
umes of solids and liquids essentially remain constant during a process, and 
thus they can be approximated as incompressible substances without sacrific- 
ing much in accuracy. 

The constant- volume and constant-pressure specific heats are identical for 
incompressible substances (Fig. 1-9). Therefore, for solids and liquids the 
subscripts on C r and C p can be dropped and both specific heats can be rep- 
resented by a single symbol, C. That is, C p = C, = C. This result could also 
be deduced from the physical definitions of constant- volume and constant- 
pressure specific heats. Specific heats of several common gases, liquids, and 
solids are given in the Appendix. 

The specific heats of incompressible substances depend on temperature 
only. Therefore, the change in the internal energy of solids and liquids can be 
expressed as 



AU 



mC mc AT 



(J) 



(1-5) 



cen58933_ch01.qxd 9/10/2002 8:29 AM Page 9 



CHAPTER 1 



where C ave is the average specific heat evaluated at the average temperature. 
Note that the internal energy change of the systems that remain in a single 
phase (liquid, solid, or gas) during the process can be determined very easily 
using average specific heats. 



Energy Transfer 



Energy can be transferred to or from a given mass by two mechanisms: heat 
Q and work W. An energy interaction is heat transfer if its driving force is a 
temperature difference. Otherwise, it is work. Arising piston, a rotating shaft, 
and an electrical wire crossing the system boundaries are all associated with 
work interactions. Work done per unit time is called power, and is denoted 
by W. The unit of power is W or hp (1 hp = 746 W). Car engines and hy- 
draulic, steam, and gas turbines produce work; compressors, pumps, and 
mixers consume work. Notice that the energy of a system decreases as it does 
work, and increases as work is done on it. 

In daily life, we frequently refer to the sensible and latent forms of internal 
energy as heat, and we talk about the heat content of bodies (Fig. 1-10). In 
thermodynamics, however, those forms of energy are usually referred to as 
thermal energy to prevent any confusion with heat transfer. 

The term heat and the associated phrases such as heat flow, heat addition, 
heat rejection, heat absorption, heat gain, heat loss, heat storage, heat gener- 
ation, electrical heating, latent heat, body heat, and heat source are in com- 
mon use today, and the attempt to replace heat in these phrases by thermal 
energy had only limited success. These phrases are deeply rooted in our vo- 
cabulary and they are used by both the ordinary people and scientists without 
causing any misunderstanding. For example, the phrase body heat is under- 
stood to mean the thermal energy content of a body. Likewise, heat flow is 
understood to mean the transfer of thermal energy, not the flow of a fluid-like 
substance called heat, although the latter incorrect interpretation, based on the 
caloric theory, is the origin of this phrase. Also, the transfer of heat into a sys- 
tem is frequently referred to as heat addition and the transfer of heat out of a 
system as heat rejection. 

Keeping in line with current practice, we will refer to the thermal energy as 
heat and the transfer of thermal energy as heat transfer. The amount of heat 
transferred during the process is denoted by Q. The amount of heat transferred 
per unit time is called heat transfer rate, and is denoted by Q. The overdot 
stands for the time derivative, or "per unit time." The heat transfer rate Q has 
the unit J/s, which is equivalent to W. 

When the rate of heat transfer Q is available, then the total amount of heat 
transfer Q during a time interval Af can be determined from 









Vapor 
80°C 


1 ^^- Heat 








transfer 




Liquid 




25°C 




80°C 






k_ 




) 





FIGURE 1-10 

The sensible and latent forms of internal 

energy can be transferred as a result of 

a temperature difference, and they are 

referred to as heat or thermal energy. 



Q 



Qdt 



(J) 



(1-6) 



provided that the variation of Q with time is known. For the special case of 
Q = constant, the equation above reduces to 



Q = QM 



(J) 



(1-7) 



cen58933_ch01.qxd 9/10/2002 8:29 AM Page 10 



10 
HEAT TRANSFER 




:24W 

: const. 



3 m 



Q _ 24 W . 
^ 6 m 2 



4 W/m 2 



FIGURE 1-11 

Heat flux is heat transfer per unit 
time and per unit area, and is equal 
to q = QIA when Q is uniform over 
the area A. 




A= kD 2 

FIGURE 1-12 

Schematic for Example 1—1. 



The rate of heat transfer per unit area normal to the direction of heat transfer 
is called heat flux, and the average heat flux is expressed as (Fig. 1-11) 



Q 

A 



(W/m 2 ) 



(1-8) 



where A is the heat transfer area. The unit of heat flux in English units is 
Btu/h • ft 2 . Note that heat flux may vary with time as well as position on a 
surface. 



" EXAMPLE 1-1 Heating of a Copper Ball 

A 10-cm diameter copper ball is to be heated from 100°C to an average tem- 
I perature of 150 C C in 30 minutes (Fig. 1-12). Taking the average density and 

E specific heat of copper in this temperature range to be p = 8950 kg/m 3 and 
C p = 0.395 kJ/kg • C C, respectively, determine (a) the total amount of heat 
transfer to the copper ball, (b) the average rate of heat transfer to the ball, and 
(c) the average heat flux. 



SOLUTION The copper ball is to be heated from 100°C to 150°C. The total 
heat transfer, the average rate of heat transfer, and the average heat flux are to 
be determined. 

Assumptions Constant properties can be used for copper at the average 
temperature. 

Properties The average density and specific heat of copper are given to be 
p = 8950 kg/m 3 and C p = 0.395 kJ/kg ■ °C. 

Analysis (a) The amount of heat transferred to the copper ball is simply the 
change in its internal energy, and is determined from 

Energy transfer to the system = Energy increase of the system 
Q = AU = mC me (T 2 - TO 

where 

m = pV = ^pD 3 = ^(8950 kg/m 3 )(0.1 m) 3 = 4.69 kg 
6 o 

Substituting, 

Q = (4.69 kg)(0.395 kJ/kg • °C)(150 - 100)°C = 92.6 kj 

Therefore, 92.6 kJ of heat needs to be transferred to the copper ball to heat it 
from 100°Cto 150°C. 

(b) The rate of heat transfer normally changes during a process with time. How- 
ever, we can determine the average rate of heat transfer by dividing the total 
amount of heat transfer by the time interval. Therefore, 



e a 



Q = 92.6 kJ 
A? 1800 s 



0.0514 kJ/s = 51.4 W 



cen58933_ch01.qxd 9/10/2002 8:29 AM Page 11 



(c) Heat flux is defined as the heat transfer per unit time per unit area, or the 
rate of heat transfer per unit area. Therefore, the average heat flux in this 
case is 



s^avi 



x£ave 

^D 2 



51.4 W 
tt(0.1 m) 2 



1636 W/m 2 



Discussion Note that heat flux may vary with location on a surface. The value 
calculated above is the average heat flux over the entire surface of the ball. 



1^ ■ THE FIRST LAW OF THERMODYNAMICS 

The first law of thermodynamics, also known as the conservation of energy 

principle, states that energy can neither be created nor destroyed; it can only 
change forms. Therefore, every bit of energy must be accounted for during a 
process. The conservation of energy principle (or the energy balance) for any 
system undergoing any process may be expressed as follows: The net change 
(increase or decrease) in the total energy of the system during a process is 
equal to the difference between the total energy entering and the total energy 
leaving the system during that process. That is, 



(Total energy 
entering the 
system 




Total energy 

leaving the 

system 




Change in the \ 
total energy of 
the system / 



(1-9) 



Noting that energy can be transferred to or from a system by heat, work, and 
mass flow, and that the total energy of a simple compressible system consists 
of internal, kinetic, and potential energies, the energy balance for any system 
undergoing any process can be expressed as 



Net energy transfer 
by heat, work, and mass 



A£„ 



Change in internal, kinetic, 
potential, etc., energies 



(J) 



(1-10) 



n 

CHAPTER 1 



or, in the rate form, as 



Rate of net energy transfer 
by heat, work, and mass 



"^■system'"* 

Rate of change in internal 
kinetic, potential, etc., energies 



(W) 



(1-11) 



Energy is a property, and the value of a property does not change unless the 
state of the system changes. Therefore, the energy change of a system is zero 
(A,E system = 0) if the state of the system does not change during the process, 
that is, the process is steady. The energy balance in this case reduces to 
(Fig. 1-13) 



Steady, rate form: 



(1-12) 



Rate of net energy transfer in 
by heat, work, and mass 



Rate of net energy transfer out 
by heat, work, and mass 



In the absence of significant electric, magnetic, motion, gravity, and surface 
tension effects (i.e., for stationary simple compressible systems), the change 




FIGURE 1-13 

In steady operation, the rate of energy 

transfer to a system is equal to the rate 

of energy transfer from the system. 



cen58933_ch01.qxd 9/10/2002 8:29 AM Page 12 



12 
HEAT TRANSFER 




mC(T.-T„) 



FIGURE 1-14 

In the absence of any work interactions, 
the change in the energy content of a 
closed system is equal to the net 
heat transfer. 



in the total energy of a system during a process is simply the change in its in- 



ternal energy. That is, AE, 



system 



AC/, 



system* 



In heat transfer analysis, we are usually interested only in the forms of en- 
ergy that can be transferred as a result of a temperature difference, that is, heat 
or thermal energy. In such cases it is convenient to write a heat balance and 
to treat the conversion of nuclear, chemical, and electrical energies into ther- 
mal energy as heat generation. The energy balance in that case can be ex- 
pressed as 



Gin " Gou, 

Net heat 
transfer 



AE 



Heat 

generation 



thermal, system 



Change in thermal 
energy of the system 



(J) 



(1-13) 



Energy Balance for Closed Systems (Fixed Mass) 

A closed system consists of a fixed mass. The total energy E for most systems 
encountered in practice consists of the internal energy U. This is especially the 
case for stationary systems since they don't involve any changes in their ve- 
locity or elevation during a process. The energy balance relation in that case 
reduces to 



Stationary closed system: 



AU = mCAT 



(J) 



(1-14) 



where we expressed the internal energy change in terms of mass m, the spe- 
cific heat at constant volume C r , and the temperature change AT of the sys- 
tem. When the system involves heat transfer only and no work interactions 
across its boundary, the energy balance relation further reduces to (Fig. 1-14) 



Stationary closed system, no work: 



Q = mC r AT 



(J) 



(1-15) 



where Q is the net amount of heat transfer to or from the system. This is the 
form of the energy balance relation we will use most often when dealing with 
a fixed mass. 



Energy Balance for Steady-Flow Systems 

A large number of engineering devices such as water heaters and car radiators 
involve mass flow in and out of a system, and are modeled as control volumes. 
Most control volumes are analyzed under steady operating conditions. The 
term steady means no change with time at a specified location. The opposite 
of steady is unsteady or transient. Also, the term uniform implies no change 
with position throughout a surface or region at a specified time. These mean- 
ings are consistent with their everyday usage (steady girlfriend, uniform 
distribution, etc.). The total energy content of a control volume during a 
steady-flow process remains constant (E CY = constant). That is, the change 
in the total energy of the control volume during such a process is zero 
(A£ cv = 0). Thus the amount of energy entering a control volume in all forms 
(heat, work, mass transfer) for a steady-flow process must be equal to the 
amount of energy leaving it. 

The amount of mass flowing through a cross section of a flow device per 
unit time is called the mass flow rate, and is denoted by rh. A fluid may flow 
in and out of a control volume through pipes or ducts. The mass flow rate of a 
fluid flowing in a pipe or duct is proportional to the cross-sectional area A c of 



cen58933_ch01.qxd 9/10/2002 8:29 AM Page 13 



the pipe or duct, the density p, and the velocity T of the fluid. The mass flow 
rate through a differential area dA c can be expressed as 8m = pT„ dA c where 
Y„ is the velocity component normal to dA c . The mass flow rate through the 
entire cross-sectional area is obtained by integration over A c . 

The flow of a fluid through a pipe or duct can often be approximated to be 
one-dimensional. That is, the properties can be assumed to vary in one direc- 
tion only (the direction of flow). As a result, all properties are assumed to be 
uniform at any cross section normal to the flow direction, and the properties 
are assumed to have bulk average values over the entire cross section. Under 
the one-dimensional flow approximation, the mass flow rate of a fluid flow- 
ing in a pipe or duct can be expressed as (Fig. 1-15) 



P TA r 



(kg/s) 



(1-16) 



where p is the fluid density, T is the average fluid velocity in the flow direc- 
tion, and A c is the cross-sectional area of the pipe or duct. 

The volume of a fluid flowing through a pipe or duct per unit time is called 
the volume flow rate V, and is expressed as 



V = TA r 



m 

P 



(m 3 /s) 



(1-17) 



13 
CHAPTER 1 



A = kD 2 /4 — 

c 

for a circular pipe 



m =pVA 



FIGURE 1-15 

The mass flow rate of a fluid at a cross 

section is equal to the product of the 

fluid density, average fluid velocity, 

and the cross-sectional area. 



Note that the mass flow rate of a fluid through a pipe or duct remains constant 
during steady flow. This is not the case for the volume flow rate, however, un- 
less the density of the fluid remains constant. 

For a steady-flow system with one inlet and one exit, the rate of mass flow 
into the control volume must be equal to the rate of mass flow out of it. That 
is, m in = m out = m. When the changes in kinetic and potential energies are 
negligible, which is usually the case, and there is no work interaction, the en- 
ergy balance for such a steady-flow system reduces to (Fig. 1-16) 



Q = ihAh 



rhC p AT 



(kJ/s) 



(1-18) 



where Q is the rate of net heat transfer into or out of the control volume. This 
is the form of the energy balance relation that we will use most often for 
steady-flow systems. 





s- Control volume 




r 


~l 


m 


4 T' 2 


L 


___#_ 



transfer /A 2 V 

FIGURE 1-16 

Under steady conditions, the net rate of 

energy transfer to a fluid in a control 

volume is equal to the rate of increase in 

the energy of the fluid stream flowing 

through the control volume. 



Surface Energy Balance 



As mentioned in the chapter opener, heat is transferred by the mechanisms of 
conduction, convection, and radiation, and heat often changes vehicles as it is 
transferred from one medium to another. For example, the heat conducted to 
the outer surface of the wall of a house in winter is convected away by the 
cold outdoor air while being radiated to the cold surroundings. In such cases, 
it may be necessary to keep track of the energy interactions at the surface, and 
this is done by applying the conservation of energy principle to the surface. 
A surface contains no volume or mass, and thus no energy. Thereore, a sur- 
face can be viewed as a fictitious system whose energy content remains con- 
stant during a process (just like a steady-state or steady-flow system). Then 
the energy balance for a surface can be expressed as 



Surface energy balance: 



(1-19) 



cen58933_ch01.qxd 9/10/2002 8:29 AM Page 14 



14 


HEAT TRANSFER 






WALL 


Control 
i/ surface 




radiation 


conduction 


1^****' Qi 


e, 


N* 




i convection 



FIGURE 1-17 

Energy interactions at the outer wall 
surface of a house. 



This relation is valid for both steady and transient conditions, and the surface 
energy balance does not involve heat generation since a surface does not have 
a volume. The energy balance for the outer surface of the wall in Fig. 1-17, 
for example, can be expressed as 



61 = 62 + 63 



(1-20) 



where Q , is conduction through the wall to the surface, Q 2 is convection from 
the surface to the outdoor air, and <2 3 is net radiation from the surface to the 
surroundings. 

When the directions of interactions are not known, all energy interactions 
can be assumed to be towards the surface, and the surface energy balance can 
be expressed as 2 E in = 0. Note that the interactions in opposite direction will 
end up having negative values, and balance this equation. 




Electric 
heating 
element 



FIGURE 1-18 

Schematic for Example 1-2. 



EXAMPLE 1-2 Heating of Water in an Electric Teapot 

1.2 kg of liquid water initially at 15°C is to be heated to 95°C in a teapot 
equipped with a 1200-W electric heating element inside (Fig. 1-18). The 
teapot is 0.5 kg and has an average specific heat of 0.7 kJ/kg • °C. Taking the 
specific heat of water to be 4.18 kJ/kg • °C and disregarding any heat loss from 
the teapot, determine how long it will take for the water to be heated. 

SOLUTION Liquid water is to be heated in an electric teapot. The heating time 
is to be determined. 

Assumptions 1 Heat loss from the teapot is negligible. 2 Constant properties 
can be used for both the teapot and the water. 

Properties The average specific heats are given to be 0.7 kJ/kg • °C for the 
teapot and 4.18 kJ/kg • °C for water. 

Analysis We take the teapot and the water in it as the system, which is 
a closed system (fixed mass). The energy balance in this case can be ex- 
pressed as 






AF 

ulJ system 

system ^ ^ water 



AU 



teapot 



Then the amount of energy needed to raise the temperature of water and the 
teapot from 15°C to 95°C is 



E m = (mCAT), 



(mCAT 



teapot 



= (1.2 kg)(4.18 kJ/kg • °C)(95 - 15)°C + (0.5 kg)(0.7 kJ/kg • °C) 
(95 - 15)°C 

= 429.3 kJ 

The 1200-W electric heating unit will supply energy at a rate of 1.2 kW or 
1.2 kJ per second. Therefore, the time needed for this heater to supply 
429.3 kJ of heat is determined from 



At 



Total energy transferred E in 429 3 ^ j 
Rate of energy transfer E 1 .2 kJ/s 

CJ ^ transfer 



358 s = 6.0 min 



cen58933_ch01.qxd 9/10/2002 8:29 AM Page 15 



Discussion In reality, it will take more than 6 minutes to accomplish this heat- 
ing process since some heat loss is inevitable during heating. 



15 
CHAPTER 1 



EXAMPLE 1-3 Heat Loss from Heating Ducts in a Basement 

A 5-m-long section of an air heating system of a house passes through an un- 
heated space in the basement (Fig. 1-19). The cross section of the rectangular 
duct of the heating system is 20 cm X 25 cm. Hot air enters the duct at 
100 kPa and 60°C at an average velocity of 5 m/s. The temperature of the air 
in the duct drops to 54°C as a result of heat loss to the cool space in the base- 
ment. Determine the rate of heat loss from the air in the duct to the basement 
under steady conditions. Also, determine the cost of this heat loss per hour if 
the house is heated by a natural gas furnace that has an efficiency of 80 per- 
cent, and the cost of the natural gas in that area is $0.60/therm (1 therm = 
100,000 Btu = 105,500 kJ). 




54°C 



- loss 

FIGURE 1-19 

Schematic for Example 1-3. 



SOLUTION The temperature of the air in the heating duct of a house drops as 
a result of heat loss to the cool space in the basement. The rate of heat loss 
from the hot air and its cost are to be determined. 

Assumptions 1 Steady operating conditions exist. 2 Air can be treated as an 
ideal gas with constant properties at room temperature. 

Properties The constant pressure specific heat of air at the average tempera- 
ture of (54 + 60)/2 = 57°C is 1.007 kJ/kg • °C (Table A-15). 
Analysis We take the basement section of the heating system as our system, 
which is a steady-flow system. The rate of heat loss from the air in the duct can 
be determined from 



Q 



mC p AT 



where rh is the mass flow rate and A T is the temperature drop. The density of 
air at the inlet conditions is 



P_ 
RT 



100 kPa 



(0.287 kPa ■ m 3 /kg • K)(60 + 273)K 



1.046 kg/m 3 



The cross-sectional area of the duct is 

A c = (0.20 m)(0.25 m) = 0.05 m 2 

Then the mass flow rate of air through the duct and the rate of heat loss 
become 

m = pTA c = (1.046 kg/m 3 )(5 m/s)(0.05 m 2 ) = 0.2615 kg/s 



and 



Q loss ~~ " Z< --p(i m T out ) 

= (0.2615 kg/s)(1.007 kJ/kg ■ °C)(60 - 54)°C 
= 1.580 kj/s 



cen58933_ch01.qxd 9/10/2002 8:29 AM Page 16 



16 
HEAT TRANSFER 



or 5688 kJ/h. The cost of this heat loss to the home owner is 

(Rate of heat loss)(Unit cost of energy input) 



Cost of heat loss 



Furnace efficiency 
(5688 kJ/h)($0.60/therm)/ i therm 



0.80 



105,500 kJ 



= $0.040/h 

Discussion The heat loss from the heating ducts in the basement is costing the 
home owner 4 cents per hour. Assuming the heater operates 2000 hours during 
a heating season, the annual cost of this heat loss adds up to $80. Most of this 
money can be saved by insulating the heating ducts in the unheated areas. 




9 ft 



FIGURE 1-20 

Schematic for Example 1-4. 



- 



EXAMPLE 1-4 Electric Heating of a House at High Elevation 

Consider a house that has a floor space of 2000 ft 2 and an average height of 9 
ft at 5000 ft elevation where the standard atmospheric pressure is 12.2 psia 
(Fig. 1-20). Initially the house is at a uniform temperature of 50°F. Now the 
electric heater is turned on, and the heater runs until the air temperature in the 
house rises to an average value of 70°F. Determine the amount of energy trans- 
ferred to the air assuming (a) the house is air-tight and thus no air escapes dur- 
ing the heating process and (£>) some air escapes through the cracks as the 
heated air in the house expands at constant pressure. Also determine the cost 
of this heat for each case if the cost of electricity in that area is $0.075/kWh. 



SOLUTION The air in the house is heated from 50°F to 70°F by an electric 
heater. The amount and cost of the energy transferred to the air are to be de- 
termined for constant-volume and constant-pressure cases. 
Assumptions 1 Air can be treated as an ideal gas with constant properties at 
room temperature. 2 Heat loss from the house during heating is negligible. 
3 The volume occupied by the furniture and other things is negligible. 
Properties The specific heats of air at the average temperature of (50 + 70)/2 



60°F are C„ = 0.240 Btu/lbm ■ °F and C„ 



C„- R 



0.171 Btu/lbm • °F 



(Tables A-1E and A-15E). 

Analysis The volume and the mass of the air in the house are 



V = (Floor area)(Height) = (2000 ft 2 )(9 ft) 
PV (12.2psia)(18,000ft 3 ) 



18,000 ft 3 



m 



RT (0.3704 psia • ftMbm ■ R)(50 + 460)R 



1162 lbm 



(a) The amount of energy transferred to air at constant volume is simply the 
change in its internal energy, and is determined from 



F — F = AF 

in out system 

^in, constant volume ^^air — ^C,, 111 



= (1162 lbm)(0.171 Btu/lbm • °F)(70 - 50)°F 
= 3974 Btu 

At a unit cost of $0.075/kWh, the total cost of this energy is 



cen58933_ch01.qxd 9/10/2002 8:29 AM Page 17 



Cost of energy = (Amount of energy)(Unit cost of energy) 

1 kWh 



(3974 Btu)($0.075/kWh) 
$0,087 



3412 Btu 



(b) The amount of energy transferred to air at constant pressure is the change 
in its enthalpy, and is determined from 

in, constant pressure air iil^pLxl 

= (1162 lbm)(0.240 Btu/lbm • °F)(70 - 50)°F 
= 5578 Btu 

At a unit cost of $0.075/kWh, the total cost of this energy is 

Cost of energy = (Amount of energy)(Unit cost of energy) 

1 kWh 



(5578 Btu)($0.075/kWh) 
$0,123 



3412 Btu 



Discussion It will cost about 12 cents to raise the temperature of the air in 
this house from 50°F to 70°F. The second answer is more realistic since every 
house has cracks, especially around the doors and windows, and the pressure in 
the house remains essentially constant during a heating process. Therefore, the 
second approach is used in practice. This conservative approach somewhat 
overpredicts the amount of energy used, however, since some of the air will es- 
cape through the cracks before it is heated to 70°F. 



17 
CHAPTER 1 



1-5 - HEAT TRANSFER MECHANISMS 

In Section 1-1 we defined heat as the form of energy that can be transferred 
from one system to another as a result of temperature difference. A thermo- 
dynamic analysis is concerned with the amount of heat transfer as a system 
undergoes a process from one equilibrium state to another. The science that 
deals with the determination of the rates of such energy transfers is the heat 
transfer. The transfer of energy as heat is always from the higher-temperature 
medium to the lower-temperature one, and heat transfer stops when the two 
mediums reach the same temperature. 

Heat can be transferred in three different modes: conduction, convection, 
and radiation. All modes of heat transfer require the existence of a tempera- 
ture difference, and all modes are from the high-temperature medium to a 
lower-temperature one. Below we give a brief description of each mode. A de- 
tailed study of these modes is given in later chapters of this text. 



1-6 - CONDUCTION 

Conduction is the transfer of energy from the more energetic particles of a 
substance to the adjacent less energetic ones as a result of interactions be- 
tween the particles. Conduction can take place in solids, liquids, or gases. In 
gases and liquids, conduction is due to the collisions and diffusion of the 



cen58933_ch01.qxd 9/10/2002 8:29 AM Page 16 



18 
HEAT TRANSFER 




FIGURE 1-21 

Heat conduction through a large plane 
wall of thickness Ax and area A. 



molecules during their random motion. In solids, it is due to the combination 
of vibrations of the molecules in a lattice and the energy transport by free 
electrons. A cold canned drink in a warm room, for example, eventually 
warms up to the room temperature as a result of heat transfer from the room 
to the drink through the aluminum can by conduction. 

The rate of heat conduction through a medium depends on the geometry of 
the medium, its thickness, and the material of the medium, as well as the tem- 
perature difference across the medium. We know that wrapping a hot water 
tank with glass wool (an insulating material) reduces the rate of heat loss from 
the tank. The thicker the insulation, the smaller the heat loss. We also know 
that a hot water tank will lose heat at a higher rate when the temperature of the 
room housing the tank is lowered. Further, the larger the tank, the larger the 
surface area and thus the rate of heat loss. 

Consider steady heat conduction through a large plane wall of thickness 
Ax = L and area A, as shown in Fig. 1—21. The temperature difference across 
the wall is AT = T 2 — T { . Experiments have shown that the rate of heat trans- 
fer Q through the wall is doubled when the temperature difference AT across 
the wall or the area A normal to the direction of heat transfer is doubled, but is 
halved when the wall thickness L is doubled. Thus we conclude that the rate 
of heat conduction through a plane layer is proportional to the temperature 
difference across the layer and the heat transfer area, but is inversely propor- 
tional to the thickness of the layer. That is, 



Rate of heat conduction « 



(Area)(Temperature difference) 
Thickness 



30°C 



lm 



20°C 



4 = 4010W/m 2 



{a) Copper (Jfc = 401 W/m-°C) 



30°C 



lm 



20°C 



g=1480W/m 2 



(b) Silicon (k = 148 W/m-°C) 

FIGURE 1-22 

The rate of heat conduction through a 
solid is directly proportional to 
its thermal conductivity. 



or, 



xl, cond 



kA 



7\ 



Ax 



-kA 



AT 

Ax 



(W) 



(1-21) 



where the constant of proportionality k is the thermal conductivity of the 
material, which is a measure of the ability of a material to conduct heat 
(Fig. 1-22). In the limiting case of Ax — > 0, the equation above reduces to the 
differential form 



Oc 



-kA 



(IT 
dx 



(W) 



(1-22) 



which is called Fourier's law of heat conduction after J. Fourier, who ex- 
pressed it first in his heat transfer text in 1822. Here dT/dx is the temperature 
gradient, which is the slope of the temperature curve on a T-x diagram (the 
rate of change of T with x), at location x. The relation above indicates that the 
rate of heat conduction in a direction is proportional to the temperature gradi- 
ent in that direction. Heat is conducted in the direction of decreasing tem- 
perature, and the temperature gradient becomes negative when temperature 
decreases with increasing x. The negative sign in Eq. 1-22 ensures that heat 
transfer in the positive x direction is a positive quantity. 

The heat transfer area A is always normal to the direction of heat transfer. 
For heat loss through a 5-m-long, 3-m-high, and 25-cm-thick wall, for exam- 
ple, the heat transfer area is A = 15 m 2 . Note that the thickness of the wall has 
no effect on A (Fig. 1-23). 



cen58933_ch01.qxd 9/10/2002 8:29 AM Page 19 



19 
CHAPTER 1 



EXAMPLE 1-5 The Cost of Heat Loss through a Roof 

The roof of an electrically heated home is 6 m long, 8 m wide, and 0.25 m 
thick, and is made of a flat layer of concrete whose thermal conductivity is 
k = 0.8 W/m • °C (Fig. 1-24). The temperatures of the inner and the outer sur- 
faces of the roof one night are measured to be 15°C and 4°C, respectively, for a 
period of 10 hours. Determine (a) the rate of heat loss through the roof that 
night and (b) the cost of that heat loss to the home owner if the cost of elec- 
tricity is $0.08/kWh. 

SOLUTION The inner and outer surfaces of the flat concrete roof of an electri- 
cally heated home are maintained at specified temperatures during a night. The 
heat loss through the roof and its cost that night are to be determined. 
Assumptions 1 Steady operating conditions exist during the entire night since 
the surface temperatures of the roof remain constant at the specified values. 
2 Constant properties can be used for the roof. 

Properties The thermal conductivity of the roof is given to be k = 0.8 
W/m ■ °C. 

Analysis (a) Noting that heat transfer through the roof is by conduction and 
the area of the roof is/4=6mX8m = 48 m 2 , the steady rate of heat trans- 
fer through the roof is determined to be 



Q =kA 



(0.8 W/m • °C)(48 m 2 ) 



(15 - 4)°C 
0.25 m 



1690 W = 1.69 kW 



(b) The amount of heat lost through the roof during a 10-hour period and its 
cost are determined from 

Q = Q At = (1.69 kW)(10 h) = 16.9 kWh 
Cost = (Amount of energy)(Unit cost of energy) 
= (16.9 kWh)($0.08/kWh) = $1.35 

Discussion The cost to the home owner of the heat loss through the roof that 
night was $1.35. The total heating bill of the house will be much larger since 
the heat losses through the walls are not considered in these calculations. 




FIGURE 1-23 

In heat conduction analysis, A represents 

the area normal to the direction 

of heat transfer. 



Concrete roof 
8 m 



0.25 m 




FIGURE 1-24 

Schematic for Example 1-5. 



Thermal Conductivity 



We have seen that different materials store heat differently, and we have de- 
fined the property specific heat C p as a measure of a material's ability to store 
thermal energy. For example, C p = 4.18 kJ/kg • °C for water and C p = 0.45 
kJ/kg • °C for iron at room temperature, which indicates that water can store 
almost 10 times the energy that iron can per unit mass. Likewise, the thermal 
conductivity A; is a measure of a material's ability to conduct heat. For exam- 
ple, k = 0.608 W/m • °C for water and k = 80.2 W/m • °C for iron at room 
temperature, which indicates that iron conducts heat more than 100 times 
faster than water can. Thus we say that water is a poor heat conductor relative 
to iron, although water is an excellent medium to store thermal energy. 

Equation 1-22 for the rate of conduction heat transfer under steady condi- 
tions can also be viewed as the defining equation for thermal conductivity. 
Thus the thermal conductivity of a material can be defined as the rate of 



cen58933_ch01.qxd 9/10/2002 8:29 AM Page 20 



20 


HEAT TRANSFER 




TABLE 


1-1 




The thermal conductivities of some 
materials at room temperature 


Materia 


k, W/m 


°c* 



Diamond 


2300 


Silver 


429 


Copper 


401 


Gold 


317 


Aluminum 


237 


Iron 


80.2 


Mercury (I) 


8.54 


Glass 


0.78 


Brick 


0.72 


Water (I) 


0.613 


Human skin 


0.37 


Wood (oak) 


0.17 


Helium (g) 


0.152 


Soft rubber 


0.13 


Glass fiber 


0.043 


Air (g) 


0.026 


Urethane, rigid foam 


0.026 



♦Multiply by 0.5778 to convert to Btu/h • ft • °F. 




Sample 
material 



■TV) 



Q 



A(T r 

FIGURE 1-25 

A simple experimental setup to 
determine the thermal conductivity 
of a material. 



heat transfer through a unit thickness of the material per unit area per unit 
temperature difference. The thermal conductivity of a material is a measure of 
the ability of the material to conduct heat. A high value for thermal conduc- 
tivity indicates that the material is a good heat conductor, and a low value 
indicates that the material is a poor heat conductor or insulator. The thermal 
conductivities of some common materials at room temperature are given in 
Table 1—1. The thermal conductivity of pure copper at room temperature is 
k = 401 W/m • °C, which indicates that a 1-m-thick copper wall will conduct 
heat at a rate of 401 W per m 2 area per °C temperature difference across the 
wall. Note that materials such as copper and silver that are good electric con- 
ductors are also good heat conductors, and have high values of thermal con- 
ductivity. Materials such as rubber, wood, and styrofoam are poor conductors 
of heat and have low conductivity values. 

A layer of material of known thickness and area can be heated from one side 
by an electric resistance heater of known output. If the outer surfaces of the 
heater are well insulated, all the heat generated by the resistance heater will be 
transferred through the material whose conductivity is to be determined. Then 
measuring the two surface temperatures of the material when steady heat 
transfer is reached and substituting them into Eq. 1-22 together with other 
known quantities give the thermal conductivity (Fig. 1-25). 

The thermal conductivities of materials vary over a wide range, as shown in 
Fig. 1-26. The thermal conductivities of gases such as air vary by a factor of 
10 4 from those of pure metals such as copper. Note that pure crystals and met- 
als have the highest thermal conductivities, and gases and insulating materials 
the lowest. 

Temperature is a measure of the kinetic energies of the particles such as the 
molecules or atoms of a substance. In a liquid or gas, the kinetic energy of the 
molecules is due to their random translational motion as well as their 
vibrational and rotational motions. When two molecules possessing differ- 
ent kinetic energies collide, part of the kinetic energy of the more energetic 
(higher-temperature) molecule is transferred to the less energetic (lower- 
temperature) molecule, much the same as when two elastic balls of the same 
mass at different velocities collide, part of the kinetic energy of the faster 
ball is transferred to the slower one. The higher the temperature, the faster the 
molecules move and the higher the number of such collisions, and the better 
the heat transfer. 

The kinetic theory of gases predicts and the experiments confirm that the 
thermal conductivity of gases is proportional to the square root of the abso- 
lute temperature T, and inversely proportional to the square root of the molar 
mass M. Therefore, the thermal conductivity of a gas increases with increas- 
ing temperature and decreasing molar mass. So it is not surprising that the 
thermal conductivity of helium (M = 4) is much higher than those of air 
(M = 29) and argon (M = 40). 

The thermal conductivities of gases at 1 arm pressure are listed in Table 
A- 16. However, they can also be used at pressures other than 1 atm, since the 
thermal conductivity of gases is independent of pressure in a wide range of 
pressures encountered in practice. 

The mechanism of heat conduction in a liquid is complicated by the fact that 
the molecules are more closely spaced, and they exert a stronger intermolecu- 
lar force field. The thermal conductivities of liquids usually lie between those 



cen58933_ch01.qxd 9/10/2002 8:29 AM Page 21 



NONMETALLIC 
CRYSTALS 



1000 



k, 
W/m°C 



100 



10 



0.1 



0.0 I 

















Diamond 
Graphite 

Silicon 
carbide 

Beryllium 
oxide 

Quartz 






METAL 
ALLOYS 


PURE 
METALS 




Silver 
Copper 

Iron 
Manganese 


Aluminum 
alloys 

Bronze 

Steel 

Nichrome 






NONMETALLIC 
SOLIDS 




Oxides 

Rock 
Food 
Rubber 




LIQUIDS 








Mercury 
Water 

Oils 




NSULATORS 






GASES 


Fibers 
Wood 
Foams 




Hydrogen 
Helium 

Aii- 
Carbon 
dioxide 































21 
CHAPTER 1 



FIGURE 1-26 

The range of thermal conductivity of 
various materials at room temperature. 



of solids and gases. The thermal conductivity of a substance is normally high- 
est in the solid phase and lowest in the gas phase. Unlike gases, the thermal 
conductivities of most liquids decrease with increasing temperature, with wa- 
ter being a notable exception. Like gases, the conductivity of liquids decreases 
with increasing molar mass. Liquid metals such as mercury and sodium have 
high thermal conductivities and are very suitable for use in applications where 
a high heat transfer rate to a liquid is desired, as in nuclear power plants. 

In solids, heat conduction is due to two effects: the lattice vibrational waves 
induced by the vibrational motions of the molecules positioned at relatively 
fixed positions in a periodic manner called a lattice, and the energy trans- 
ported via the free flow of electrons in the solid (Fig. 1-27). The ther- 
mal conductivity of a solid is obtained by adding the lattice and electronic 
components. The relatively high thermal conductivities of pure metals are pri- 
marily due to the electronic component. The lattice component of thermal 
conductivity strongly depends on the way the molecules are arranged. For ex- 
ample, diamond, which is a highly ordered crystalline solid, has the highest 
known thermal conductivity at room temperature. 

Unlike metals, which are good electrical and heat conductors, crystalline 
solids such as diamond and semiconductors such as silicon are good heat con- 
ductors but poor electrical conductors. As a result, such materials find wide- 
spread use in the electronics industry. Despite their higher price, diamond heat 
sinks are used in the cooling of sensitive electronic components because of the 




FIGURE 1-27 

The mechanisms of heat conduction in 
different phases of a substance. 



cen58933_ch01.qxd 9/10/2002 8:29 AM Page 22 



22 

HEAT TRANSFER 



TABLE 1-2 

The thermal conductivity of an 
alloy is usually much lower than 
the thermal conductivity of either 
metal of which it is composed 

Pure metal or k, W/m • °C, 
alloy at 300 K 



Copper 


401 


Nickel 


91 


Constantan 




(55% Cu, 45% Ni) 


23 


Copper 


401 


Aluminum 


237 


Commercial bronze 




(90% Cu, 10% Al) 


52 





TABLE 


1-3 




Thermal conductivities 
vary with temperature 


of materials 


T, K 


Copper 


Aluminum 



100 


482 


200 


413 


300 


401 


400 


393 


600 


379 


800 


366 



302 
237 
237 
240 
231 
218 



FIGURE 1-28 

The variation of the thermal 

conductivity of various solids, 

liquids, and gases with temperature 

(from White, Ref. 10). 



excellent thermal conductivity of diamond. Silicon oils and gaskets are com- 
monly used in the packaging of electronic components because they provide 
both good thermal contact and good electrical insulation. 

Pure metals have high thermal conductivities, and one would think that 
metal alloys should also have high conductivities. One would expect an alloy 
made of two metals of thermal conductivities k { and k 2 to have a conductivity 
k between k [ and k 2 . But this turns out not to be the case. The thermal conduc- 
tivity of an alloy of two metals is usually much lower than that of either metal, 
as shown in Table 1-2. Even small amounts in a pure metal of "foreign" mol- 
ecules that are good conductors themselves seriously disrupt the flow of heat 
in that metal. For example, the thermal conductivity of steel containing just 
1 percent of chrome is 62 W/m • °C, while the thermal conductivities of iron 
and chromium are 83 and 95 W/m • °C, respectively. 

The thermal conductivities of materials vary with temperature (Table 1-3). 
The variation of thermal conductivity over certain temperature ranges is neg- 
ligible for some materials, but significant for others, as shown in Fig. 1-28. 
The thermal conductivities of certain solids exhibit dramatic increases at tem- 
peratures near absolute zero, when these solids become superconductors. For 
example, the conductivity of copper reaches a maximum value of about 
20,000 W/m • °C at 20 K, which is about 50 times the conductivity at room 
temperature. The thermal conductivities and other thermal properties of vari- 
ous materials are given in Tables A-3 to A- 16. 



10,000 

k, 
W/m-°C 



1000 



100 



10 



0.1 



0.01 



^^v^Diamonds 

s / v \ s- ^" v Type Ha 
"\/ ' Type lib 
• Type I 


- Solids 

. Liquids 

Gases 

Silver r>„„„„. 




" Aluminum , 

Tungsten . 


Gold ■ J ■ 


*v^ ■ Platinum 

' Iron 

■ Pyroceram glass 




Aluminum oxide 


Clear fused quartz 

^-- -. Water 


Helium 


— _ Carbon tetrachloride 

""• -. Steam 


Air 
Argon 



200 



400 



600 



800 



1000 



1200 1400 



cen58933_ch01.qxd 9/10/2002 8:29 AM Page 23 



The temperature dependence of thermal conductivity causes considerable 
complexity in conduction analysis. Therefore, it is common practice to evalu- 
ate the thermal conductivity k at the average temperature and treat it as a con- 
stant in calculations. 

In heat transfer analysis, a material is normally assumed to be isotropic; that 
is, to have uniform properties in all directions. This assumption is realistic for 
most materials, except those that exhibit different structural characteristics in 
different directions, such as laminated composite materials and wood. The 
thermal conductivity of wood across the grain, for example, is different than 
that parallel to the grain. 



Thermal Diffusivity 



The product pC p , which is frequently encountered in heat transfer analysis, is 
called the heat capacity of a material. Both the specific heat C p and the heat 
capacity pC p represent the heat storage capability of a material. But C p ex- 
presses it per unit mass whereas pC p expresses it per unit volume, as can be 
noticed from their units J/kg • °C and J/m 3 • °C, respectively. 

Another material property that appears in the transient heat conduction 
analysis is the thermal diffusivity, which represents how fast heat diffuses 
through a material and is defined as 



Heat conducted 
Heat stored 



k 

pC P 



(m 2 /s) 



(1-23) 



Note that the thermal conductivity k represents how well a material con- 
ducts heat, and the heat capacity pC p represents how much energy a material 
stores per unit volume. Therefore, the thermal diffusivity of a material can be 
viewed as the ratio of the heat conducted through the material to the heat 
stored per unit volume. A material that has a high thermal conductivity or a 
low heat capacity will obviously have a large thermal diffusivity. The larger 
the thermal diffusivity, the faster the propagation of heat into the medium. 
A small value of thermal diffusivity means that heat is mostly absorbed by the 
material and a small amount of heat will be conducted further. 

The thermal diffusivities of some common materials at 20°C are given in 
Table 1-4. Note that the thermal diffusivity ranges from a = 0.14 X IO 6 m 2 /s 
for water to 174 X 10~ 6 m 2 /s for silver, which is a difference of more than a 
thousand times. Also note that the thermal diffusivities of beef and water are 
the same. This is not surprising, since meat as well as fresh vegetables and 
fruits are mostly water, and thus they possess the thermal properties of water. 



EXAMPLE 1-6 Measuring the Thermal Conductivity of a Material 

A common way of measuring the thermal conductivity of a material is to sand- 
wich an electric thermofoil heater between two identical samples of the ma- 
terial, as shown in Fig. 1-29. The thickness of the resistance heater, including 
its cover, which is made of thin silicon rubber, is usually less than 0.5 mm. 
A circulating fluid such as tap water keeps the exposed ends of the samples 
at constant temperature. The lateral surfaces of the samples are well insulated 
to ensure that heat transfer through the samples is one-dimensional. Two 
I thermocouples are embedded into each sample some distance L apart, and a 



23 

CHAPTER 1 



TABLE 1-4 

The thermal diffusivities of some 
materials at room temperature 



Material 


a, 


m 2 /s* 


Silver 


149 


X 


10- 5 


Gold 


127 


X 


10- 5 


Copper 


113 


X 


io- 5 


Aluminum 


97.5 


X 


io- 6 


Iron 


22.8 X 


io- 5 


Mercury (I) 


4.7 


X 


io- 5 


Marble 


1.2 


X 


io- 5 


Ice 


1.2 


X 


io- 6 


Concrete 


0.75 


X 


io- 5 


Brick 


0.52 


X 


io- 6 


Heavy soil (dry) 


0.52 


X 


io- 5 


Glass 


0.34 


X 


io- 5 


Glass wool 


0.23 


X 


IO" 5 


Water (I) 


0.14 


X 


IO" 5 


Beef 


0.14 


X 


IO" 5 


Wood (oak) 


0.13 


X 


IO" 5 



*Mu Iti ply by 10.76 to convert to ft 2 /s. 











, Cooling 
r fluid 




Insulation 




Sample 

< 




Thermocouple 
/ 


X 






l }Ar, 


Resistance 
heater 






a 


-z?= 


Sample 






a 








L W, 




1 




, Cooling 




r fluid 



FIGURE 1-29 

Apparatus to measure the thermal 

conductivity of a material using two 

identical samples and a thin resistance 

heater (Example 1-6). 



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24 
HEAT TRANSFER 



differential thermometer reads the temperature drop AT across this distance 
along each sample. When steady operating conditions are reached, the total 
rate of heat transfer through both samples becomes equal to the electric power 
drawn by the heater, which is determined by multiplying the electric current by 
the voltage. 

In a certain experiment, cylindrical samples of diameter 5 cm and length 
10 cm are used. The two thermocouples in each sample are placed 3 cm apart. 
After initial transients, the electric heater is observed to draw 0.4 A at 110 V, 
and both differential thermometers read a temperature difference of 15°C. De- 
termine the thermal conductivity of the sample. 



SOLUTION The thermal conductivity of a material is to be determined by en- 
suring one-dimensional heat conduction, and by measuring temperatures when 
steady operating conditions are reached. 

Assumptions 1 Steady operating conditions exist since the temperature 
readings do not change with time. 2 Heat losses through the lateral surfaces 
of the apparatus are negligible since those surfaces are well insulated, and 
thus the entire heat generated by the heater is conducted through the samples. 
3 The apparatus possesses thermal symmetry. 

Analysis The electrical power consumed by the resistance heater and con- 
verted to heat is 



W. 



VI = (110V)(0.4A) = 44W 



The rate of heat flow through each sample is 

Q = \ W e = \ X (44 W) = 22 W 

since only half of the heat generated will flow through each sample because of 
symmetry. Reading the same temperature difference across the same distance 
in each sample also confirms that the apparatus possesses thermal symmetry. 
The heat transfer area is the area normal to the direction of heat flow, which is 
the cross-sectional area of the cylinder in this case: 

A = \ ttD 2 = \ tt(0.05 m) 2 = 0.00196 m 2 

Noting that the temperature drops by 15 C C within 3 cm in the direction of heat 
flow, the thermal conductivity of the sample is determined to be 



Q =kA 



AT 



QL 



(22 W)(0.03 m) 



A AT (0.00196 m 2 )(15°C) 



22.4 W/m • °C 



Discussion Perhaps you are wondering if we really need to use two samples in 
the apparatus, since the measurements on the second sample do not give any 
additional information. It seems like we can replace the second sample by in- 
sulation. Indeed, we do not need the second sample; however, it enables us to 
verify the temperature measurements on the first sample and provides thermal 
symmetry, which reduces experimental error. 



EXAMPLE 1-7 Conversion between SI and English Units 

An engineer who is working on the heat transfer analysis of a brick building in 
English units needs the thermal conductivity of brick. But the only value he can 



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find from his handbooks is 0.72 W/m • °C, which is in SI units. To make mat- 
i ters worse, the engineer does not have a direct conversion factor between the 
two unit systems for thermal conductivity. Can you help him out? 



25 
CHAPTER 1 



SOLUTION The situation this engineer is facing is not unique, and most engi- 
neers often find themselves in a similar position. A person must be very careful 
during unit conversion not to fall into some common pitfalls and to avoid some 
costly mistakes. Although unit conversion is a simple process, it requires utmost 
care and careful reasoning. 

The conversion factors for W and m are straightforward and are given in con- 
version tables to be 



1 W 
lm 



3.41214 Btu/h 
3.2808 ft 



But the conversion of C C into C F is not so simple, and it can be a source of er- 
ror if one is not careful. Perhaps the first thought that comes to mind is to re- 
place °C by (°F - 32V1.8 since 7"(°C) = [T(°F) - 32]/1.8. But this will be 
wrong since the °C in the unit W/m • °C represents per °C change in tempera- 
ture. Noting that 1°C change in temperature corresponds to 1.8°F, the proper 
conversion factor to be used is 



1°C = 1.8°F 



Substituting, we get 



1 w/m • ° c = S^ = °- 5778 Btu/h ■ ft ■ ° F 

which is the desired conversion factor. Therefore, the thermal conductivity of 
the brick in English units is 



= 0.72 X (0.5778 Btu/h • ft • °F) 
= 0.42 Btu/h • ft • °F 

Discussion Note that the thermal conductivity value of a material in English 
units is about half that in SI units (Fig. 1-30). Also note that we rounded the 
result to two significant digits (the same number in the original value) since ex- 
pressing the result in more significant digits (such as 0.4160 instead of 0.42) 
would falsely imply a more accurate value than the original one. 



k = 0.72 W/m-°C 
= 0.42 Btu/h-ft-°F 




FIGURE 1-30 

The thermal conductivity value in 

English units is obtained by multiplying 

the value in SI units by 0.5778. 



1-7 ■ CONVECTION 

Convection is the mode of energy transfer between a solid surface and the 
adjacent liquid or gas that is in motion, and it involves the combined effects of 
conduction and fluid motion. The faster the fluid motion, the greater the 
convection heat transfer. In the absence of any bulk fluid motion, heat trans- 
fer between a solid surface and the adjacent fluid is by pure conduction. The 
presence of bulk motion of the fluid enhances the heat transfer between the 
solid surface and the fluid, but it also complicates the determination of heat 
transfer rates. 



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26 
HEAT TRANSFER 



Velocity 

variation 

of air 



Temperature 

variation 

of air 




Hot Block 

FIGURE 1-31 

Heat transfer from a hot 
surface to air by convection. 




Natural 
convection 



Air 



\ Xhot egg) v. / 






FIGURE 1-32 

The cooling of a boiled egg 

by forced and natural convection. 



TABLE 1-5 

Typical values of convection heat 
transfer coefficient 



Type of 
convection 



h, W/m 2 



Free convection of 

gases 
Free convection of 

liquids 
Forced convection 

of gases 
Forced convection 

of liquids 
Boiling and 

condensation 



2-25 

10-1000 

25-250 

50-20,000 

2500-100,000 



•Multiply by 0.176 to convert to Btu/h • ft 2 ■ °F. 



Consider the cooling of a hot block by blowing cool air over its top surface 
(Fig. 1-31). Energy is first transferred to the air layer adjacent to the block by 
conduction. This energy is then carried away from the surface by convection, 
that is, by the combined effects of conduction within the air that is due to ran- 
dom motion of air molecules and the bulk or macroscopic motion of the air 
that removes the heated air near the surface and replaces it by the cooler air. 

Convection is called forced convection if the fluid is forced to flow over 
the surface by external means such as a fan, pump, or the wind. In contrast, 
convection is called natural (or free) convection if the fluid motion is caused 
by buoyancy forces that are induced by density differences due to the varia- 
tion of temperature in the fluid (Fig. 1-32). For example, in the absence of a 
fan, heat transfer from the surface of the hot block in Fig. 1-31 will be by nat- 
ural convection since any motion in the air in this case will be due to the rise 
of the warmer (and thus lighter) air near the surface and the fall of the cooler 
(and thus heavier) air to fill its place. Heat transfer between the block and the 
surrounding air will be by conduction if the temperature difference between 
the air and the block is not large enough to overcome the resistance of air to 
movement and thus to initiate natural convection currents. 

Heat transfer processes that involve change of phase of a fluid are also con- 
sidered to be convection because of the fluid motion induced during the 
process, such as the rise of the vapor bubbles during boiling or the fall of the 
liquid droplets during condensation. 

Despite the complexity of convection, the rate of convection heat transfer is 
observed to be proportional to the temperature difference, and is conveniently 
expressed by Newton's law of cooling as 



6 c 



hA s (T s - r„) 



(W) 



(1-24) 



where h is the convection heat transfer coefficient in W/m 2 • °C or Btu/h • ft 2 ■ °F, 
A s is the surface area through which convection heat transfer takes place, T s is 
the surface temperature, and T m is the temperature of the fluid sufficiently far 
from the surface. Note that at the surface, the fluid temperature equals the sur- 
face temperature of the solid. 

The convection heat transfer coefficient h is not a property of the fluid. It is 
an experimentally determined parameter whose value depends on all the vari- 
ables influencing convection such as the surface geometry, the nature of fluid 
motion, the properties of the fluid, and the bulk fluid velocity. Typical values 
of h are given in Table 1-5. 

Some people do not consider convection to be a fundamental mechanism of 
heat transfer since it is essentially heat conduction in the presence of fluid mo- 
tion. But we still need to give this combined phenomenon a name, unless we 
are willing to keep referring to it as "conduction with fluid motion." Thus, it 
is practical to recognize convection as a separate heat transfer mechanism de- 
spite the valid arguments to the contrary. 



EXAMPLE 1-8 Measuring Convection Heat Transfer Coefficient 

A 2-m-long, 0.3-cm-diameter electrical wire extends across a room at 15°C, as 

shown in Fig. 1-33. Heat is generated in the wire as a result of resistance heat- 

■ ing, and the surface temperature of the wire is measured to be 152°C in steady 



cen58933_ch01.qxd 9/10/2002 8:29 AM Page 27 



operation. Also, the voltage drop and electric current through the wire are mea- 
sured to be 60 V and 1.5 A, respectively. Disregarding any heat transfer by 
radiation, determine the convection heat transfer coefficient for heat transfer 
between the outer surface of the wire and the air in the room. 

SOLUTION The convection heat transfer coefficient for heat transfer from an 
electrically heated wire to air is to be determined by measuring temperatures 
when steady operating conditions are reached and the electric power consumed. 
Assumptions 1 Steady operating conditions exist since the temperature read- 
ings do not change with time. 2 Radiation heat transfer is negligible. 
Analysis When steady operating conditions are reached, the rate of heat loss 
from the wire will equal the rate of heat generation in the wire as a result of 
resistance heating. That is, 

Q = generated = W = (60 V)(1.5 A) = 90 W 

The surface area of the wire is 

A s = ttDL = tt(0.003 m)(2 m) = 0.01885 m 2 
Newton's law of cooling for convection heat transfer is expressed as 

Gco„v = hA s (T s - zy 

Disregarding any heat transfer by radiation and thus assuming all the heat loss 
from the wire to occur by convection, the convection heat transfer coefficient is 
determined to be 



h 



Gc 



90 W 



A£T S - r„) (0.01885 m 2 )(152 - 15)°C 



34.9 W/m 2 



Discussion Note that the simple setup described above can be used to deter- 
mine the average heat transfer coefficients from a variety of surfaces in air. 
Also, heat transfer by radiation can be eliminated by keeping the surrounding 
surfaces at the temperature of the wire. 



27 

CHAPTER 1 



T = 15°C 



1.5 A 



152°C 



■60 V- 



FIGURE 1-33 

Schematic for Example 1-8. 



1-8 - RADIATION 

Radiation is the energy emitted by matter in the form of electromagnetic 
waves (or photons) as a result of the changes in the electronic configurations 
of the atoms or molecules. Unlike conduction and convection, the transfer of 
energy by radiation does not require the presence of an intervening medium. 
In fact, energy transfer by radiation is fastest (at the speed of light) and it 
suffers no attenuation in a vacuum. This is how the energy of the sun reaches 
the earth. 

In heat transfer studies we are interested in thermal radiation, which is the 
form of radiation emitted by bodies because of their temperature. It differs 
from other forms of electromagnetic radiation such as x-rays, gamma rays, 
microwaves, radio waves, and television waves that are not related to temper- 
ature. All bodies at a temperature above absolute zero emit thermal radiation. 

Radiation is a volumetric phenomenon, and all solids, liquids, and gases 
emit, absorb, or transmit radiation to varying degrees. However, radiation is 



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28 

HEAT TRANSFER 




FIGURE 1-34 

Blackbody radiation represents the 
maximum amount of radiation that 
can be emitted from a surface 
at a specified temperature. 



TABLE 1-6 

Emissivities of some materials 
at 300 K 



Materia] 


Emissivity 


Aluminum foil 


0.07 


Anodized aluminum 


0.82 


Polished copper 


0.03 


Polished gold 


0.03 


Polished silver 


0.02 


Polished stainless steel 


0.17 


Black paint 


0.98 


White paint 


0.90 


White paper 


0.92-0.97 


Asphalt pavement 


0.85-0.93 


Red brick 


0.93-0.96 


Human skin 


0.95 


Wood 


0.82-0.92 


Soil 


0.93-0.96 


Water 


0.96 


Vegetation 


0.92-0.96 



-V* 



*^ref ^ "incid 



Q ,. = a 6 - A , 

*--abs ^incident 

FIGURE 1-35 

The absorption of radiation incident on 
an opaque surface of absorptivity a. 



usually considered to be a surface phenomenon for solids that are opaque to 
thermal radiation such as metals, wood, and rocks since the radiation emitted 
by the interior regions of such material can never reach the surface, and the 
radiation incident on such bodies is usually absorbed within a few microns 
from the surface. 

The maximum rate of radiation that can be emitted from a surface at an ab- 
solute temperature T s (in K or R) is given by the Stefan-Boltzmann law as 



£?e 



uAJt 



(W) 



(1-25) 



where a = 5.67 X 1(T 8 W/m 2 • K 4 or 0.1714 X 1(T 8 Btu/h • ft 2 • R 4 is the 
Stefan-Boltzmann constant. The idealized surface that emits radiation at this 
maximum rate is called a blackbody, and the radiation emitted by a black- 
body is called blackbody radiation (Fig. 1-34). The radiation emitted by all 
real surfaces is less than the radiation emitted by a blackbody at the same tem- 
perature, and is expressed as 



C?e 



saATj 



(W) 



(1-26) 



where e is the emissivity of the surface. The property emissivity, whose value 
is in the range ^ e < 1, is a measure of how closely a surface approximates 
a blackbody for which e = 1 . The emissivities of some surfaces are given in 
Table 1-6. 

Another important radiation property of a surface is its absorptivity a, 
which is the fraction of the radiation energy incident on a surface that is ab- 
sorbed by the surface. Like emissivity, its value is in the range ^ a < 1. 
A blackbody absorbs the entire radiation incident on it. That is, a blackbody is 
a perfect absorber (a = 1) as it is a perfect emitter. 

In general, both e and a of a surface depend on the temperature and the 
wavelength of the radiation. Kirchhoff 's law of radiation states that the emis- 
sivity and the absorptivity of a surface at a given temperature and wavelength 
are equal. In many practical applications, the surface temperature and the 
temperature of the source of incident radiation are of the same order of mag- 
nitude, and the average absorptivity of a surface is taken to be equal to its av- 
erage emissivity. The rate at which a surface absorbs radiation is determined 
from (Fig. 1-35) 



Q, 



a 2incid 



(W) 



(1-27) 



where 2 incident is the rate at which radiation is incident on the surface and a is 
the absorptivity of the surface. For opaque (nontransparent) surfaces, the 
portion of incident radiation not absorbed by the surface is reflected back. 

The difference between the rates of radiation emitted by the surface and the 
radiation absorbed is the net radiation heat transfer. If the rate of radiation ab- 
sorption is greater than the rate of radiation emission, the surface is said to be 
gaining energy by radiation. Otherwise, the surface is said to be losing energy 
by radiation. In general, the determination of the net rate of heat transfer by ra- 
diation between two surfaces is a complicated matter since it depends on the 
properties of the surfaces, their orientation relative to each other, and the in- 
teraction of the medium between the surfaces with radiation. 



cen58933_ch01.qxd 9/10/2002 8:29 AM Page 29 



When a surface of emissivity e and surface area A s at an absolute tempera- 
ture T s is completely enclosed by a much larger (or black) surface at absolute 
temperature T sun separated by a gas (such as air) that does not intervene with 
radiation, the net rate of radiation heat transfer between these two surfaces is 
given by (Fig. 1-36) 



Q, 



BaA,(T}-T*^ 



(W) 



(1-28) 



In this special case, the emissivity and the surface area of the surrounding sur- 
face do not have any effect on the net radiation heat transfer. 

Radiation heat transfer to or from a surface surrounded by a gas such as air 
occurs parallel to conduction (or convection, if there is bulk gas motion) be- 
tween the surface and the gas. Thus the total heat transfer is determined by 
adding the contributions of both heat transfer mechanisms. For simplicity and 
convenience, this is often done by defining a combined heat transfer co- 
efficient ^combined that includes the effects of both convection and radiation. 
Then the total heat transfer rate to or from a surface by convection and radia- 
tion is expressed as 



fit 



K 



iA s (1 s I a,) 



(W) 



(1-29) 



Note that the combined heat transfer coefficient is essentially a convection 
heat transfer coefficient modified to include the effects of radiation. 

Radiation is usually significant relative to conduction or natural convection, 
but negligible relative to forced convection. Thus radiation in forced convec- 
tion applications is usually disregarded, especially when the surfaces involved 
have low emissivities and low to moderate temperatures. 



29 
CHAPTER 1 




Q mi = £GA S (T A S -Ti m ) 

FIGURE 1-36 

Radiation heat transfer between a 
surface and the surfaces surrounding it. 



EXAMPLE 1-9 Radiation Effect on Thermal Comfort 

It is a common experience to feel "chilly" in winter and "warm" in summer in 
our homes even when the thermostat setting is kept the same. This is due to the 
so called "radiation effect" resulting from radiation heat exchange between our 
bodies and the surrounding surfaces of the walls and the ceiling. 

Consider a person standing in a room maintained at 22°C at all times. The 
inner surfaces of the walls, floors, and the ceiling of the house are observed to 
be at an average temperature of 10 C C in winter and 25°C in summer. Determine 
the rate of radiation heat transfer between this person and the surrounding sur- 
faces if the exposed surface area and the average outer surface temperature of 
the person are 1.4 m 2 and 30°C, respectively (Fig. 1-37). 

SOLUTION The rates of radiation heat transfer between a person and the sur- 
rounding surfaces at specified temperatures are to be determined in summer 
and winter. 

Assumptions 1 Steady operating conditions exist. 2 Heat transfer by convection 
is not considered. 3 The person is completely surrounded by the interior sur- 
faces of the room. 4 The surrounding surfaces are at a uniform temperature. 
Properties The emissivity of a person is e = 0.95 (Table 1-6). 
Analysis The net rates of radiation heat transfer from the body to the sur- 
rounding walls, ceiling, and floor in winter and summer are 




FIGURE 1-37 

Schematic for Example 1-9. 



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30 
HEAT TRANSFER 



rad, winter oun j \ l s -* surr, winter/ 

= (0.95)(5.67 X l(T 8 W/m 2 • K 4 )(1.4nr) 
X [(30 + 273) 4 - (10 + 273) 4 ] K 4 

= 152 W 



and 



= (0.95)(5.67 X 10~ 8 W/m 2 • K 4 )(1.4 m 2 ) 
X [(30 + 273) 4 - (25 + 273) 4 ] K 4 

= 40.9 W 

Discussion Note that we must use absolute temperatures in radiation calcula- 
tions. Also note that the rate of heat loss from the person by radiation is almost 
four times as large in winter than it is in summer, which explains the "chill" we 
feel in winter even if the thermostat setting is kept the same. 



OPAQUE 
SOLID 



Conduction 



1 mode 




Conduction or 
convection 



2 modes 



VACUUM 



Radiation 



1 mode 



FIGURE 1-38 

Although there are three mechanisms of 
heat transfer, a medium may involve 
only two of them simultaneously. 



1-9 - SIMULTANEOUS HEAT TRANSFER 
MECHANISMS 

We mentioned that there are three mechanisms of heat transfer, but not all 
three can exist simultaneously in a medium. For example, heat transfer is 
only by conduction in opaque solids, but by conduction and radiation in 
semitransparent solids. Thus, a solid may involve conduction and radiation 
but not convection. However, a solid may involve heat transfer by convection 
and/or radiation on its surfaces exposed to a fluid or other surfaces. For 
example, the outer surfaces of a cold piece of rock will warm up in a warmer 
environment as a result of heat gain by convection (from the air) and radiation 
(from the sun or the warmer surrounding surfaces). But the inner parts of the 
rock will warm up as this heat is transferred to the inner region of the rock by 
conduction. 

Heat transfer is by conduction and possibly by radiation in a still fluid (no 
bulk fluid motion) and by convection and radiation in a flowing fluid. In the 
absence of radiation, heat transfer through a fluid is either by conduction or 
convection, depending on the presence of any bulk fluid motion. Convection 
can be viewed as combined conduction and fluid motion, and conduction in a 
fluid can be viewed as a special case of convection in the absence of any fluid 
motion (Fig. 1-38). 

Thus, when we deal with heat transfer through & fluid, we have either con- 
duction or convection, but not both. Also, gases are practically transparent to 
radiation, except that some gases are known to absorb radiation strongly at 
certain wavelengths. Ozone, for example, strongly absorbs ultraviolet radia- 
tion. But in most cases, a gas between two solid surfaces does not interfere 
with radiation and acts effectively as a vacuum. Liquids, on the other hand, 
are usually strong absorbers of radiation. 

Finally, heat transfer through a vacuum is by radiation only since conduc- 
tion or convection requires the presence of a material medium. 



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31 
CHAPTER 1 



EXAMPLE 1-10 Heat Loss from a Person 

Consider a person standing in a breezy room at 20°C. Determine the total rate 
of heat transfer from this person if the exposed surface area and the average 
outer surface temperature of the person are 1.6 m 2 and 29°C, respectively, and 
the convection heat transfer coefficient is 6 W/m 2 • °C (Fig. 1-39). 

SOLUTION The total rate of heat transfer from a person by both convection 
and radiation to the surrounding air and surfaces at specified temperatures is to 
be determined. 

Assumptions 1 Steady operating conditions exist. 2 The person is completely 
surrounded by the interior surfaces of the room. 3 The surrounding surfaces are 
at the same temperature as the air in the room. 4 Heat conduction to the floor 
through the feet is negligible. 

Properties The emissivity of a person is e = 0.95 (Table 1-6). 

Analysis The heat transfer between the person and the air in the room will be 
by convection (instead of conduction) since it is conceivable that the air in the 
vicinity of the skin or clothing will warm up and rise as a result of heat transfer 
from the body, initiating natural convection currents. It appears that the exper- 
imentally determined value for the rate of convection heat transfer in this case 
is 6 W per unit surface area (m 2 ) per unit temperature difference (in K or °C) 
between the person and the air away from the person. Thus, the rate of convec- 
tion heat transfer from the person to the air in the room is 

Gconv = hA s (T s - TJ 

= (6 W/m 2 • °C)(1.6 m 2 )(29 - 20)°C 
= 86.4 W 

The person will also lose heat by radiation to the surrounding wall surfaces. 
We take the temperature of the surfaces of the walls, ceiling, and floor to be 
equal to the air temperature in this case for simplicity, but we recognize that 
this does not need to be the case. These surfaces may be at a higher or lower 
temperature than the average temperature of the room air, depending on the 
outdoor conditions and the structure of the walls. Considering that air does not 
intervene with radiation and the person is completely enclosed by the sur- 
rounding surfaces, the net rate of radiation heat transfer from the person to the 
surrounding walls, ceiling, and floor is 



Srad = evA s (Tf- r s 4 urr ) 
= (0.95)(5.67 X 10- 
X [(29 + 273) 4 - 

= 81.7W 



B W/m 2 -K 4 )(1.6m 2 ) 
(20 + 273) 4 ] K 4 



Note that we must use absolute temperatures in radiation calculations. Also 
note that we used the emissivity value for the skin and clothing at room tem- 
perature since the emissivity is not expected to change significantly at a slightly 
higher temperature. 

Then the rate of total heat transfer from the body is determined by adding 
these two quantities: 



fit 



x£ conv x£r 



(86.4 + 81.7) W = 168.1 W 




Q A 

FIGURE 1-39 

Heat transfer from the person 
described in Example 1-10. 



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32 

HEAT TRANSFER 



Discussion The heat transfer would be much higher if the person were not 
dressed since the exposed surface temperature would be higher. Thus, an im- 
portant function of the clothes is to serve as a barrier against heat transfer. 

In these calculations, heat transfer through the feet to the floor by conduc- 
tion, which is usually very small, is neglected. Heat transfer from the skin by 
perspiration, which is the dominant mode of heat transfer in hot environments, 
is not considered here. 



T l = 300 K 



-L=\ cm 



~e = V 



FIGURE 1-40 

Schematic for Example 1—11. 



r 2 = 200K 'EXAMPLE 1-11 Heat Transfer between Two Isothermal Plates 

Consider steady heat transfer between two large parallel plates at constant 
I temperatures of 7"! = 300 K and 7~ z = 200 K that are L = 1 cm apart, as shown 
I in Fig. 1-40. Assuming the surfaces to be black (emissivity e = 1), determine 
I the rate of heat transfer between the plates per unit surface area assuming the 
gap between the plates is (a) filled with atmospheric air, (£>) evacuated, (c) filled 
with urethane insulation, and (d) filled with superinsulation that has an appar- 
ent thermal conductivity of 0.00002 W/m • °C. 



SOLUTION The total rate of heat transfer between two large parallel plates at 
specified temperatures is to be determined for four different cases. 
Assumptions 1 Steady operating conditions exist. 2 There are no natural con- 
vection currents in the air between the plates. 3 The surfaces are black and 
thus e = 1. 

Properties The thermal conductivity at the average temperature of 250 K is 
k = 0.0219 W/m • °C for air (Table A-l 1), 0.026 W/m ■ °C for urethane insula- 
tion (Table A-6), and 0.00002 W/m ■ °C for the superinsulation. 
Analysis (a) The rates of conduction and radiation heat transfer between the 
plates through the air layer are 



6c 



M 



T 2 



(0.0219 W/m -°C)(lm 2 ) 



(300 - 200)°C 
0.01m 



219W 



and 



!2 rad = svMTf - 7\ 4 ) 

= (1)(5.67 X 10- 8 W/m 2 • K 4 )(l m 2 )[(300 K) 4 



(200 K) 4 ] = 368 W 



Therefore, 



Qu 



fico„d + e ra d = 219 + 368 = 587W 



The heat transfer rate in reality will be higher because of the natural convection 
currents that are likely to occur in the air space between the plates. 
(b) When the air space between the plates is evacuated, there will be no con- 
duction or convection, and the only heat transfer between the plates will be by 
radiation. Therefore, 



G t 



G, 



368 W 



(c) An opaque solid material placed between two plates blocks direct radiation 
heat transfer between the plates. Also, the thermal conductivity of an insulating 
material accounts for the radiation heat transfer that may be occurring through 



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33 
CHAPTER 1 



300 K 



200 K 300 K 



200 K 300 K 



2 = 587W 



1 cm 



Q = 368 W 



1 cm 



200 K 300 K 

I 

!j 



Q = 260 W 



1 cm 



200 K 



2 = 0.2W 




4 

1 cm 



(a) Air space (b) Vacuum (c) Insulation (d) Superinsulation 

FIGURE 1-41 

Different ways of reducing heat transfer between two isothermal plates, and their effectiveness. 



the voids in the insulating material. The rate of heat transfer through the ure- 
thane insulation is 



Qu 



Gcond = kA 



. (300 - 200)°C 

(0.026 W/m • °C)(1 m 2 )- — — : — = 260 W 

0.01 m 



Note that heat transfer through the urethane material is less than the heat 
transfer through the air determined in (a), although the thermal conductivity of 
the insulation is higher than that of air. This is because the insulation blocks 
the radiation whereas air transmits it. 

(d) The layers of the superinsulation prevent any direct radiation heat transfer 
between the plates. However, radiation heat transfer between the sheets of su- 
perinsulation does occur, and the apparent thermal conductivity of the super- 
insulation accounts for this effect. Therefore, 



e„ 



kA 



T 2 



(0.00002 W/m •°C)(lm 2 ) 



(300 - 200)°C 
0.01m 



0.2 W 



which is yi_ of the heat transfer through the vacuum. The results of this ex- 
ample are summarized in Fig. 1-41 to put them into perspective. 
Discussion This example demonstrates the effectiveness of superinsulations, 
which are discussed in the next chapter, and explains why they are the insula- 
tion of choice in critical applications despite their high cost. 



EXAMPLE 1-12 Heat Transfer in Conventional 
and Microwave Ovens 

The fast and efficient cooking of microwave ovens made them one of the es- 
sential appliances in modern kitchens (Fig. 1-42). Discuss the heat transfer 
mechanisms associated with the cooking of a chicken in microwave and con- 
ventional ovens, and explain why cooking in a microwave oven is more efficient. 

SOLUTION Food is cooked in a microwave oven by absorbing the electromag- 
netic radiation energy generated by the microwave tube, called the magnetron. 







® 






} 



FIGURE 1-42 

A chicken being cooked in a 
microwave oven (Example 1-12). 



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34 
HEAT TRANSFER 



The radiation emitted by the magnetron is not thermal radiation, since its emis- 
sion is not due to the temperature of the magnetron; rather, it is due to the 
conversion of electrical energy into electromagnetic radiation at a specified 
wavelength. The wavelength of the microwave radiation is such that it is re- 
flected by metal surfaces; transmitted by the cookware made of glass, ceramic, 
or plastic; and absorbed and converted to internal energy by food (especially the 
water, sugar, and fat) molecules. 

In a microwave oven, the radiation that strikes the chicken is absorbed by 
the skin of the chicken and the outer parts. As a result, the temperature of the 
chicken at and near the skin rises. Heat is then conducted toward the inner 
parts of the chicken from its outer parts. Of course, some of the heat absorbed 
by the outer surface of the chicken is lost to the air in the oven by convection. 

In a conventional oven, the air in the oven is first heated to the desired tem- 
perature by the electric or gas heating element. This preheating may take sev- 
eral minutes. The heat is then transferred from the air to the skin of the chicken 
by natural convection in most ovens or by forced convection in the newer con- 
vection ovens that utilize a fan. The air motion in convection ovens increases 
the convection heat transfer coefficient and thus decreases the cooking time. 
Heat is then conducted toward the inner parts of the chicken from its outer 
parts as in microwave ovens. 

Microwave ovens replace the slow convection heat transfer process in con- 
ventional ovens by the instantaneous radiation heat transfer. As a result, micro- 
wave ovens transfer energy to the food at full capacity the moment they are 
turned on, and thus they cook faster while consuming less energy. 




a = 0.6 



25°C 



FIGURE 1-43 

Schematic for Example 1-13. 



I 

2 EXAMPLE 1-13 Heating of a Plate by Solar Energy 

A thin metal plate is insulated on the back and exposed to solar radiation at the 
I front surface (Fig. 1-43). The exposed surface of the plate has an absorptivity 

■ of 0.6 for solar radiation. If solar radiation is incident on the plate at a rate of 

■ 700 W/m 2 and the surrounding air temperature is 25 C C, determine the surface 
temperature of the plate when the heat loss by convection and radiation equals 
the solar energy absorbed by the plate. Assume the combined convection and 
radiation heat transfer coefficient to be 50 W/m 2 ■ C C. 



SOLUTION The back side of the thin metal plate is insulated and the front 
side is exposed to solar radiation. The surface temperature of the plate is to be 
determined when it stabilizes. 

Assumptions 1 Steady operating conditions exist. 2 Heat transfer through the 
insulated side of the plate is negligible. 3 The heat transfer coefficient remains 
constant. 

Properties The solar absorptivity of the plate is given to be a = 0.6. 
Analysis The absorptivity of the plate is 0.6, and thus 60 percent of the solar 
radiation incident on the plate will be absorbed continuously. As a result, the 
temperature of the plate will rise, and the temperature difference between the 
plate and the surroundings will increase. This increasing temperature difference 
will cause the rate of heat loss from the plate to the surroundings to increase. 
At some point, the rate of heat loss from the plate will equal the rate of solar 



cen58933_ch01.qxd 9/10/2002 8:30 AM Page 35 



energy absorbed, and the temperature of the plate will no longer change. The 
temperature of the plate when steady operation is established is deter- 
mined from 



J gained 



OF CLA S q incident, solar ^combined ™s V-* s * °°J 



Solving for 7" s and substituting, the plate surface temperature is determined 
to be 



T„ + a- 



^i incide 



K, 



25°C + 



0.6 X (700 W/m 2 ) 
50 W/m 2 ■ °C 



33.4°C 



Discussion Note that the heat losses will prevent the plate temperature from 
rising above 33.4°C. Also, the combined heat transfer coefficient accounts for 
the effects of both convection and radiation, and thus it is very convenient 
to use in heat transfer calculations when its value is known with reasonable 
accuracy. 



35 
CHAPTER 1 



SOLUTION 



1-10 - PROBLEM-SOLVING TECHNIQUE 

The first step in learning any science is to grasp the fundamentals, and to gain 
a sound knowledge of it. The next step is to master the fundamentals by 
putting this knowledge to test. This is done by solving significant real-world 
problems. Solving such problems, especially complicated ones, requires a sys- 
tematic approach. By using a step-by-step approach, an engineer can reduce 
the solution of a complicated problem into the solution of a series of simple 
problems (Fig. 1-44). When solving a problem, we recommend that you use 
the following steps zealously as applicable. This will help you avoid some of 
the common pitfalls associated with problem solving. 

Step 1: Problem Statement 

In your own words, briefly state the problem, the key information given, and 
the quantities to be found. This is to make sure that you understand the prob- 
lem and the objectives before you attempt to solve the problem. 

Step 2: Schematic 

Draw a realistic sketch of the physical system involved, and list the relevant 
information on the figure. The sketch does not have to be something elaborate, 
but it should resemble the actual system and show the key features. Indicate 
any energy and mass interactions with the surroundings. Listing the given in- 
formation on the sketch helps one to see the entire problem at once. Also, 
check for properties that remain constant during a process (such as tempera- 
ture during an isothermal process), and indicate them on the sketch. 

Step 3: Assumptions 

State any appropriate assumptions made to simplify the problem to make it 
possible to obtain a solution. Justify the questionable assumptions. Assume 
reasonable values for missing quantities that are necessary. For example, in 
the absence of specific data for atmospheric pressure, it can be taken to be 



<& 



4? 



<P 



% 
£ 



PROBLEM 

FIGURE 1-44 

A step-by-step approach can greatly 
simplify problem solving. 



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36 
HEAT TRANSFER 




FIGURE 1-45 

The assumptions made while solving 
an engineering problem must be 
reasonable and justifiable. 




FIGURE 1-46 

The results obtained from 
an engineering analysis must 
be checked for reasonableness. 



1 atm. However, it should be noted in the analysis that the atmospheric pres- 
sure decreases with increasing elevation. For example, it drops to 0.83 atm in 
Denver (elevation 1610 m) (Fig. 1-45). 

Step 4: Physical Laws 

Apply all the relevant basic physical laws and principles (such as the conser- 
vation of energy), and reduce them to their simplest form by utilizing the as- 
sumptions made. However, the region to which a physical law is applied must 
be clearly identified first. For example, the heating or cooling of a canned 
drink is usually analyzed by applying the conservation of energy principle to 
the entire can. 

Step 5: Properties 

Determine the unknown properties at known states necessary to solve the 
problem from property relations or tables. List the properties separately, and 
indicate their source, if applicable. 

Step 6: Calculations 

Substitute the known quantities into the simplified relations and perform the 
calculations to determine the unknowns. Pay particular attention to the units 
and unit cancellations, and remember that a dimensional quantity without a 
unit is meaningless. Also, don't give a false implication of high accuracy by 
copying all the digits from the screen of the calculator — round the results to 
an appropriate number of significant digits. 

Step 7: Reasoning, Verification, and Discussion 

Check to make sure that the results obtained are reasonable and intuitive, and 
verify the validity of the questionable assumptions. Repeat the calculations 
that resulted in unreasonable values. For example, insulating a water heater 
that uses $80 worth of natural gas a year cannot result in savings of $200 a 
year (Fig. 1-46). 

Also, point out the significance of the results, and discuss their implications. 
State the conclusions that can be drawn from the results, and any recommen- 
dations that can be made from them. Emphasize the limitations under which 
the results are applicable, and caution against any possible misunderstandings 
and using the results in situations where the underlying assumptions do not 
apply. For example, if you determined that wrapping a water heater with a 
$20 insulation jacket will reduce the energy cost by $30 a year, indicate that 
the insulation will pay for itself from the energy it saves in less than a year. 
However, also indicate that the analysis does not consider labor costs, and that 
this will be the case if you install the insulation yourself. 

Keep in mind that you present the solutions to your instructors, and any en- 
gineering analysis presented to others is a form of communication. Therefore 
neatness, organization, completeness, and visual appearance are of utmost im- 
portance for maximum effectiveness. Besides, neatness also serves as a great 
checking tool since it is very easy to spot errors and inconsistencies in a neat 
work. Carelessness and skipping steps to save time often ends up costing more 
time and unnecessary anxiety. 



cen58933_ch01.qxd 9/10/2002 8:30 AM Page 37 



The approach just described is used in the solved example problems with- 
out explicitly stating each step, as well as in the Solutions Manual of this text. 
For some problems, some of the steps may not be applicable or necessary. 
However, we cannot overemphasize the importance of a logical and orderly 
approach to problem solving. Most difficulties encountered while solving a 
problem are not due to a lack of knowledge; rather, they are due to a lack of 
coordination. You are strongly encouraged to follow these steps in problem 
solving until you develop your own approach that works best for you. 



37 

CHAPTER 1 



A Remark on Significant Digits 



In engineering calculations, the information given is not known to more than 
a certain number of significant digits, usually three digits. Consequently, the 
results obtained cannot possibly be accurate to more significant digits. Re- 
porting results in more significant digits implies greater accuracy than exists, 
and it should be avoided. 

For example, consider a 3.75-L container filled with gasoline whose density 
is 0.845 kg/L, and try to determine its mass. Probably the first thought that 
comes to your mind is to multiply the volume and density to obtain 3.16875 
kg for the mass, which falsely implies that the mass determined is accurate to 
six significant digits. In reality, however, the mass cannot be more accurate 
than three significant digits since both the volume and the density are accurate 
to three significant digits only. Therefore, the result should be rounded to three 
significant digits, and the mass should be reported to be 3.17 kg instead of 
what appears in the screen of the calculator. The result 3.16875 kg would be 
correct only if the volume and density were given to be 3.75000 L and 
0.845000 kg/L, respectively. The value 3.75 L implies that we are fairly con- 
fident that the volume is accurate within ±0.01 L, and it cannot be 3.74 or 
3.76 L. However, the volume can be 3.746, 3.750, 3.753, etc., since they all 
round to 3.75 L (Fig. 1-47). It is more appropriate to retain all the digits dur- 
ing intermediate calculations, and to do the rounding in the final step since 
this is what a computer will normally do. 

When solving problems, we will assume the given information to be accu- 
rate to at least three significant digits. Therefore, if the length of a pipe is 
given to be 40 m, we will assume it to be 40.0 m in order to justify using three 
significant digits in the final results. You should also keep in mind that all ex- 
perimentally determined values are subject to measurement errors, and such 
errors will reflect in the results obtained. For example, if the density of a sub- 
stance has an uncertainty of 2 percent, then the mass determined using this 
density value will also have an uncertainty of 2 percent. 

You should also be aware that we sometimes knowingly introduce small er- 
rors in order to avoid the trouble of searching for more accurate data. For ex- 
ample, when dealing with liquid water, we just use the value of 1000 kg/m 3 
for density, which is the density value of pure water at 0°C. Using this value 
at 75°C will result in an error of 2.5 percent since the density at this tempera- 
ture is 975 kg/m 3 . The minerals and impurities in the water will introduce ad- 
ditional error. This being the case, you should have no reservation in rounding 
the final results to a reasonable number of significant digits. Besides, having 
a few percent uncertainty in the results of engineering analysis is usually the 
norm, not the exception. 




FIGURE 1-47 

A result with more significant digits than 

that of given data falsely implies 

more accuracy. 



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38 
HEAT TRANSFER 




FIGURE 1-48 

An excellent word-processing 
program does not make a person a good 
writer; it simply makes a good writer 
a better and more efficient writer. 



Engineering Software Packages 

Perhaps you are wondering why we are about to undertake a painstaking study 
of the fundamentals of heat transfer. After all, almost all such problems we are 
likely to encounter in practice can be solved using one of several sophisticated 
software packages readily available in the market today. These software pack- 
ages not only give the desired numerical results, but also supply the outputs in 
colorful graphical form for impressive presentations. It is unthinkable to prac- 
tice engineering today without using some of these packages. This tremen- 
dous computing power available to us at the touch of a button is both a 
blessing and a curse. It certainly enables engineers to solve problems easily 
and quickly, but it also opens the door for abuses and misinformation. In the 
hands of poorly educated people, these software packages are as dangerous as 
sophisticated powerful weapons in the hands of poorly trained soldiers. 

Thinking that a person who can use the engineering software packages 
without proper training on fundamentals can practice engineering is like 
thinking that a person who can use a wrench can work as a car mechanic. If it 
were true that the engineering students do not need all these fundamental 
courses they are taking because practically everything can be done by com- 
puters quickly and easily, then it would also be true that the employers would 
no longer need high-salaried engineers since any person who knows how to 
use a word-processing program can also learn how to use those software pack- 
ages. However, the statistics show that the need for engineers is on the rise, 
not on the decline, despite the availability of these powerful packages. 

We should always remember that all the computing power and the engi- 
neering software packages available today are just tools, and tools have mean- 
ing only in the hands of masters. Having the best word-processing program 
does not make a person a good writer, but it certainly makes the job of a good 
writer much easier and makes the writer more productive (Fig. 1-48). Hand 
calculators did not eliminate the need to teach our children how to add or sub- 
tract, and the sophisticated medical software packages did not take the place 
of medical school training. Neither will engineering software packages re- 
place the traditional engineering education. They will simply cause a shift in 
emphasis in the courses from mathematics to physics. That is, more time will 
be spent in the classroom discussing the physical aspects of the problems in 
greater detail, and less time on the mechanics of solution procedures. 

All these marvelous and powerful tools available today put an extra burden 
on today's engineers. They must still have a thorough understanding of the 
fundamentals, develop a "feel" of the physical phenomena, be able to put the 
data into proper perspective, and make sound engineering judgments, just like 
their predecessors. However, they must do it much better, and much faster, us- 
ing more realistic models because of the powerful tools available today. The 
engineers in the past had to rely on hand calculations, slide rules, and later 
hand calculators and computers. Today they rely on software packages. The 
easy access to such power and the possibility of a simple misunderstanding or 
misinterpretation causing great damage make it more important today than 
ever to have a solid training in the fundamentals of engineering. In this text we 
make an extra effort to put the emphasis on developing an intuitive and phys- 
ical understanding of natural phenomena instead of on the mathematical de- 
tails of solution procedures. 



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39 
CHAPTER 1 



Engineering Equation Solver (EES) 

EES is a program that solves systems of linear or nonlinear algebraic or dif- 
ferential equations numerically. It has a large library of built-in thermody- 
namic property functions as well as mathematical functions, and allows the 
user to supply additional property data. Unlike some software packages, EES 
does not solve thermodynamic problems; it only solves the equations supplied 
by the user. Therefore, the user must understand the problem and formulate it 
by applying any relevant physical laws and relations. EES saves the user con- 
siderable time and effort by simply solving the resulting mathematical equa- 
tions. This makes it possible to attempt significant engineering problems not 
suitable for hand calculations, and to conduct parametric studies quickly and 
conveniently. EES is a very powerful yet intuitive program that is very easy to 
use, as shown in the examples below. The use and capabilities of EES are ex- 
plained in Appendix 3. 

Heat Transfer Tools (HTT) 

One software package specifically designed to help bridge the gap between 
the textbook fundamentals and these powerful software packages is Heat 
Transfer Tools, which may be ordered "bundled" with this text. The software 
included in that package was developed for instructional use only and thus is 
applicable only to fundamental problems in heat transfer. While it does not 
have the power and functionality of the professional, commercial packages, 
HTT uses research-grade numerical algorithms behind the scenes and modern 
graphical user interfaces. Each module is custom designed and applicable to a 
single, fundamental topic in heat transfer to ensure that almost all time at the 
computer is spent learning heat transfer. Nomenclature and all inputs and 
outputs are consistent with those used in this and most other textbooks in 
the field. In addition, with the capability of testing parameters so readily 
available, one can quickly gain a physical feel for the effects of all the non- 
dimensional numbers that arise in heat transfer. 



EXAMPLE 1-14 Solving a System of Equations with EES 

The difference of two numbers is 4, and the sum of the squares of these 
two numbers is equal to the sum of the numbers plus 20. Determine these two 
numbers. 

SOLUTION Relations are given for the difference and the sum of the squares 
of two numbers. They are to be determined. 

Analysis We start the EES program by double-clicking on its icon, open a new 
file, and type the following on the blank screen that appears: 

x-y=4 
x A 2+y A 2=x+y+20 

which is an exact mathematical expression of the problem statement with 
x and y denoting the unknown numbers. The solution to this system of two 



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40 
HEAT TRANSFER 



nonlinear equations with two unknowns is obtained by a single click on the 
"calculator" symbol on the taskbar. It gives 

x=5 and y=l 

Discussion Note that all we did is formulate the problem as we would on pa- 
per; EES took care of all the mathematical details of solution. Also note that 
equations can be linear or nonlinear, and they can be entered in any order with 
unknowns on either side. Friendly equation solvers such as EES allow the user 
to concentrate on the physics of the problem without worrying about the mathe- 
matical complexities associated with the solution of the resulting system of 
equations. 

Throughout the text, problems that are unsuitable for hand calculations and 
are intended to be solved using EES are indicated by a computer icon. 



TOPIC OF SPECIAL INTEREST 




FIGURE 1-49 

Most animals come into this world with 
built-in insulation, but human beings 
come with a delicate skin. 



Thermal Comfort 

Unlike animals such as a fox or a bear that are born with built-in furs, hu- 
man beings come into this world with little protection against the harsh en- 
vironmental conditions (Fig. 1^9). Therefore, we can claim that the search 
for thermal comfort dates back to the beginning of human history. It is be- 
lieved that early human beings lived in caves that provided shelter as well 
as protection from extreme thermal conditions. Probably the first form of 
heating system used was open fire, followed by fire in dwellings through 
the use of a chimney to vent out the combustion gases. The concept of cen- 
tral heating dates back to the times of the Romans, who heated homes by 
utilizing double-floor construction techniques and passing the fire's fumes 
through the opening between the two floor layers. The Romans were also 
the first to use transparent windows made of mica or glass to keep the wind 
and rain out while letting the light in. Wood and coal were the primary en- 
ergy sources for heating, and oil and candles were used for lighting. The ru- 
ins of south-facing houses indicate that the value of solar heating was 
recognized early in the history. 

The term air-conditioning is usually used in a restricted sense to imply 
cooling, but in its broad sense it means to condition the air to the desired 
level by heating, cooling, humidifying, dehumidifying, cleaning, and de- 
odorizing. The purpose of the air-conditioning system of a building is to 
provide complete thermal comfort for its occupants. Therefore, we need to 
understand the thermal aspects of the human body in order to design an ef- 
fective air-conditioning system. 

The building blocks of living organisms are cells, which resemble minia- 
ture factories performing various functions necessary for the survival of 
organisms. The human body contains about 100 trillion cells with an aver- 
age diameter of 0.01 mm. In a typical cell, thousands of chemical reactions 



"This section can be skipped without a loss in continuity. 



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41 
CHAPTER 1 



occur every second during which some molecules are broken down and en- 
ergy is released and some new molecules are formed. The high level of 
chemical activity in the cells that maintain the human body temperature at 
a temperature of 37.0°C (98.6°F) while performing the necessary bodily 
functions is called the metabolism. In simple terms, metabolism refers to 
the burning of foods such as carbohydrates, fat, and protein. The metabo- 
lizable energy content of foods is usually expressed by nutritionists in 
terms of the capitalized Calorie. One Calorie is equivalent to 1 Cal = 1 
kcal = 4.1868 kJ. 

The rate of metabolism at the resting state is called the basal metabolic 
rate, which is the rate of metabolism required to keep a body performing 
the necessary bodily functions such as breathing and blood circulation at 
zero external activity level. The metabolic rate can also be interpreted as 
the energy consumption rate for a body. For an average man (30 years old, 
70 kg, 1.73 m high, 1.8 m 2 surface area), the basal metabolic rate is 84 W. 
That is, the body is converting chemical energy of the food (or of the body 
fat if the person had not eaten) into heat at a rate of 84 J/s, which is then 
dissipated to the surroundings. The metabolic rate increases with the level 
of activity, and it may exceed 10 times the basal metabolic rate when some- 
one is doing strenuous exercise. That is, two people doing heavy exercising 
in a room may be supplying more energy to the room than a 1-kW resis- 
tance heater (Fig. 1-50). An average man generates heat at a rate of 108 W 
while reading, writing, typing, or listening to a lecture in a classroom in a 
seated position. The maximum metabolic rate of an average man is 1250 W 
at age 20 and 730 at age 70. The corresponding rates for women are about 
30 percent lower. Maximum metabolic rates of trained athletes can exceed 
2000 W. 

Metabolic rates during various activities are given in Table 1-7 per unit 
body surface area. The surface area of a nude body was given by D. 
DuBois in 1916 as 




FIGURE1-50 

Two fast-dancing people supply 

more heat to a room than a 

1-kW resistance heater. 



0.202m - 425 h 



D.425 1,0.725 



(m 2 ) 



(1-30) 



where m is the mass of the body in kg and h is the height in m. Clothing in- 
creases the exposed surface area of a person by up to about 50 percent. The 
metabolic rates given in the table are sufficiently accurate for most pur- 
poses, but there is considerable uncertainty at high activity levels. More ac- 
curate values can be determined by measuring the rate of respiratory 
oxygen consumption, which ranges from about 0.25 L/min for an average 
resting man to more than 2 L/min during extremely heavy work. The entire 
energy released during metabolism can be assumed to be released as heat 
(in sensible or latent forms) since the external mechanical work done by the 
muscles is very small. Besides, the work done during most activities such 
as walking or riding an exercise bicycle is eventually converted to heat 
through friction. 

The comfort of the human body depends primarily on three environmen- 
tal factors: the temperature, relative humidity, and air motion. The temper- 
ature of the environment is the single most important index of comfort. 
Extensive research is done on human subjects to determine the "thermal 
comfort zone" and to identify the conditions under which the body feels 



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42 
HEAT TRANSFER 



TABLE 1-7 

Metabolic rates during various 
activities (from ASHRAE 
Handbook of Fundamentals, 
Ret. 1, Chap. 8, Table 4). 





Metabolic 




rate* 


Activity 


W/m 2 


Resting: 




Sleeping 


40 


Reclining 


45 


Seated, quiet 


60 


Standing, relaxed 


70 


Walking (on the level): 




2 mph (0.89 m/s) 


115 


3 mph (1.34 m/s) 


150 


4 mph (1.79 m/s) 


220 


Office Activities: 




Reading, seated 


55 


Writing 


60 


Typing 


65 


Filing, seated 


70 


Filing, standing 


80 


Walking about 


100 


Lifting/packing 


120 


Driving/Flying: 




Car 


60-115 


Aircraft, routine 


70 


Heavy vehicle 


185 


Miscellaneous Occupational 


Activities: 




Cooking 


95-115 


Cleaning house 


115-140 


Machine work: 




Light 


115-140 


Heavy 


235 


Handling 50-kg bags 


235 


Pick and shovel work 


235-280 


Miscellaneous Leisure 


Activities: 


Dancing, social 


140-255 


Calisthenics/exercise 


175-235 


Tennis, singles 


210-270 


Basketball 


290-440 


Wrestling, competitive 


410-505 



*M ultiply by 1.8 m 2 to obtain metabolic rates for 
an average man. Multiply by 0.3171 to convert 
to Btu/h • ft 2 . 



comfortable in an environment. It has been observed that most normally 
clothed people resting or doing light work feel comfortable in the operative 
temperature (roughly, the average temperature of air and surrounding sur- 
faces) range of 23°C to 27°C or 73°C to 80°F (Fig. 1-51). For unclothed 
people, this range is 29°C to 31°C. Relative humidity also has a con- 
siderable effect on comfort since it is a measure of air's ability to absorb 
moisture and thus it affects the amount of heat a body can dissipate by 
evaporation. High relative humidity slows down heat rejection by evapora- 
tion, especially at high temperatures, and low relative humidity speeds it 
up. The desirable level of relative humidity is the broad range of 30 to 
70 percent, with 50 percent being the most desirable level. Most people at 
these conditions feel neither hot nor cold, and the body does not need to 
activate any of the defense mechanisms to maintain the normal body tem- 
perature (Fig. 1-52). 

Another factor that has a major effect on thermal comfort is excessive air 
motion or draft, which causes undesired local cooling of the human body. 
Draft is identified by many as a most annoying factor in work places, auto- 
mobiles, and airplanes. Experiencing discomfort by draft is most common 
among people wearing indoor clothing and doing light sedentary work, and 
least common among people with high activity levels. The air velocity 
should be kept below 9 m/min (30 ft/min) in winter and 15 m/min 
(50 ft/min) in summer to minimize discomfort by draft, especially when the 
air is cool. A low level of air motion is desirable as it removes the warm, 
moist air that builds around the body and replaces it with fresh air. There- 
fore, air motion should be strong enough to remove heat and moisture from 
the vicinity of the body, but gentle enough to be unnoticed. High speed air 
motion causes discomfort outdoors as well. For example, an environment 
at 10°C (50°F) with 48 km/h winds feels as cold as an environment at 
— 7°C (20°F) with 3 km/h winds because of the chilling effect of the air 
motion (the wind-chill factor). 

A comfort system should provide uniform conditions throughout the 
living space to avoid discomfort caused by nonuniformities such as drafts, 
asymmetric thermal radiation, hot or cold floors, and vertical temperature 
stratification. Asymmetric thermal radiation is caused by the cold sur- 
faces of large windows, uninsulated walls, or cold products and the warm 
surfaces of gas or electric radiant heating panels on the walls or ceiling, 
solar-heated masonry walls or ceilings, and warm machinery. Asymmetric 
radiation causes discomfort by exposing different sides of the body to sur- 
faces at different temperatures and thus to different heat loss or gain by 
radiation. A person whose left side is exposed to a cold window, for exam- 
ple, will feel like heat is being drained from that side of his or her body 
(Fig. 1-53). For thermal comfort, the radiant temperature asymmetry 
should not exceed 5°C in the vertical direction and 10°C in the horizontal 
direction. The unpleasant effect of radiation asymmetry can be minimized 
by properly sizing and installing heating panels, using double-pane win- 
dows, and providing generous insulation at the walls and the roof. 

Direct contact with cold or hot floor surfaces also causes localized dis- 
comfort in the feet. The temperature of the floor depends on the way it is 
constructed (being directly on the ground or on top of a heated room, being 
made of wood or concrete, the use of insulation, etc.) as well as the floor 



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43 
CHAPTER 1 



covering used such as pads, carpets, rugs, and linoleum. A floor tempera- 
ture of 23 to 25°C is found to be comfortable to most people. The floor 
asymmetry loses its significance for people with footwear. An effective and 
economical way of raising the floor temperature is to use radiant heating 
panels instead of turning the thermostat up. Another nonuniform condition 
that causes discomfort is temperature stratification in a room that ex- 
poses the head and the feet to different temperatures. For thermal comfort, 
the temperature difference between the head and foot levels should not ex- 
ceed 3°C. This effect can be minimized by using destratification fans. 

It should be noted that no thermal environment will please everyone. No 
matter what we do, some people will express some discomfort. The thermal 
comfort zone is based on a 90 percent acceptance rate. That is, an environ- 
ment is deemed comfortable if only 10 percent of the people are dissatis- 
fied with it. Metabolism decreases somewhat with age, but it has no effect 
on the comfort zone. Research indicates that there is no appreciable differ- 
ence between the environments preferred by old and young people. Exper- 
iments also show that men and women prefer almost the same environment. 
The metabolism rate of women is somewhat lower, but this is compensated 
by their slightly lower skin temperature and evaporative loss. Also, there is 
no significant variation in the comfort zone from one part of the world to 
another and from winter to summer. Therefore, the same thermal comfort 
conditions can be used throughout the world in any season. Also, people 
cannot acclimatize themselves to prefer different comfort conditions. 

In a cold environment, the rate of heat loss from the body may exceed 
the rate of metabolic heat generation. Average specific heat of the human 
body is 3.49 kJ/kg • °C, and thus each 1°C drop in body temperature corre- 
sponds to a deficit of 244 kJ in body heat content for an average 70-kg 
man. A drop of 0.5°C in mean body temperature causes noticeable but ac- 
ceptable discomfort. A drop of 2.6°C causes extreme discomfort. A sleep- 
ing person will wake up when his or her mean body temperature drops by 
1.3°C (which normally shows up as a 0.5°C drop in the deep body and 3°C 
in the skin area). The drop of deep body temperature below 35°C may dam- 
age the body temperature regulation mechanism, while a drop below 28°C 
may be fatal. Sedentary people reported to feel comfortable at a mean skin 
temperature of 33.3°C, uncomfortably cold at 31°C, shivering cold at 
30°C, and extremely cold at 29°C. People doing heavy work reported to 
feel comfortable at much lower temperatures, which shows that the activity 
level affects human performance and comfort. The extremities of the body 
such as hands and feet are most easily affected by cold weather, and their 
temperature is a better indication of comfort and performance. A hand-skin 
temperature of 20°C is perceived to be uncomfortably cold, 15°C to be 
extremely cold, and 5°C to be painfully cold. Useful work can be per- 
formed by hands without difficulty as long as the skin temperature of fin- 
gers remains above 16°C (ASHRAE Handbook of Fundamentals, Ref. 1, 
Chapter 8). 

The first line of defense of the body against excessive heat loss in a cold 
environment is to reduce the skin temperature and thus the rate of heat loss 
from the skin by constricting the veins and decreasing the blood flow to the 
skin. This measure decreases the temperature of the tissues subjacent to 
the skin, but maintains the inner body temperature. The next preventive 



2.0 



& 1.5 



20 



25 



30 



1.0 



3 0.5 

u 



Sedentary 




50% RH 




v '■.. T < 30 fpm 


Heavy 


>. \ (0.15 m/s) 


clothing 


^ X. 


Winter 


- s \ 
^ X 

N X 

v x. 

^ X 


clothing 


^ X 
^ X 
S X. 


Summer 
■..clothing 


s x_ 
s X 
s 


\ 



64 



72 



76 






Operative temperature 

Upper acceptability limit 

Optimum 

Lower acceptability limit 

FIGURE 1-51 

The effect of clothing on 

the environment temperature 

that feels comfortable (1 clo = 

0.155 m 2 • °C/W = 0.880 ft 2 ■ °F ■ h/Btu) 

(from ASHRAE Standard 55-1981). 




FIGURE 1-52 

A thermally comfortable environment. 



cen58933_ch01.qxd 9/10/2002 8:30 AM Page 44 



44 
HEAT TRANSFER 




FIGURE 1-53 

Cold surfaces cause excessive heat loss 
from the body by radiation, and thus 
discomfort on that side of the body. 



Shivering 




FIGURE 1-54 

The rate of metabolic heat generation 
may go up by six times the resting 
level during total body shivering 
in cold weather. 



measure is increasing the rate of metabolic heat generation in the body by 
shivering, unless the person does it voluntarily by increasing his or her 
level of activity or puts on additional clothing. Shivering begins slowly in 
small muscle groups and may double the rate of metabolic heat production 
of the body at its initial stages. In the extreme case of total body shivering, 
the rate of heat production may reach six times the resting levels (Fig. 
1-54). If this measure also proves inadequate, the deep body temperature 
starts falling. Body parts furthest away from the core such as the hands and 
feet are at greatest danger for tissue damage. 

In hot environments, the rate of heat loss from the body may drop be- 
low the metabolic heat generation rate. This time the body activates the op- 
posite mechanisms. First the body increases the blood flow and thus heat 
transport to the skin, causing the temperature of the skin and the subjacent 
tissues to rise and approach the deep body temperature. Under extreme heat 
conditions, the heart rate may reach 180 beats per minute in order to main- 
tain adequate blood supply to the brain and the skin. At higher heart rates, 
the volumetric efficiency of the heart drops because of the very short time 
between the beats to fill the heart with blood, and the blood supply to the 
skin and more importantly to the brain drops. This causes the person to 
faint as a result of heat exhaustion. Dehydration makes the problem worse. 
A similar thing happens when a person working very hard for a long time 
stops suddenly. The blood that has flooded the skin has difficulty returning 
to the heart in this case since the relaxed muscles no longer force the blood 
back to the heart, and thus there is less blood available for pumping to the 
brain. 

The next line of defense is releasing water from sweat glands and resort- 
ing to evaporative cooling, unless the person removes some clothing and 
reduces the activity level (Fig. 1-55). The body can maintain its core tem- 
perature at 37°C in this evaporative cooling mode indefinitely, even in en- 
vironments at higher temperatures (as high as 200°C during military 
endurance tests), if the person drinks plenty of liquids to replenish his or 
her water reserves and the ambient air is sufficiently dry to allow the sweat 
to evaporate instead of rolling down the skin. If this measure proves inad- 
equate, the body will have to start absorbing the metabolic heat and the 
deep body temperature will rise. A person can tolerate a temperature rise of 
1.4°C without major discomfort but may collapse when the temperature 
rise reaches 2.8°C. People feel sluggish and their efficiency drops consid- 
erably when the core body temperature rises above 39°C. A core tempera- 
ture above 41°C may damage hypothalamic proteins, resulting in cessation 



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45 
CHAPTER 1 



of sweating, increased heat production by shivering, and a heat stroke with 
irreversible and life-threatening damage. Death can occur above 43°C. 

A surface temperature of 46°C causes pain on the skin. Therefore, direct 
contact with a metal block at this temperature or above is painful. How- 
ever, a person can stay in a room at 100°C for up to 30 min without any 
damage or pain on the skin because of the convective resistance at the skin 
surface and evaporative cooling. We can even put our hands into an oven at 
200°C for a short time without getting burned. 

Another factor that affects thermal comfort, health, and productivity is 
ventilation. Fresh outdoor air can be provided to a building naturally by 
doing nothing, ox forcefully by a mechanical ventilation system. In the first 
case, which is the norm in residential buildings, the necessary ventilation is 
provided by infiltration through cracks and leaks in the living space and by 
the opening of the windows and doors. The additional ventilation needed in 
the bathrooms and kitchens is provided by air vents with dampers or ex- 
haust fans. With this kind of uncontrolled ventilation, however, the fresh 
air supply will be either too high, wasting energy, or too low, causing poor 
indoor air quality. But the current practice is not likely to change for resi- 
dential buildings since there is not a public outcry for energy waste or air 
quality, and thus it is difficult to justify the cost and complexity of me- 
chanical ventilation systems. 

Mechanical ventilation systems are part of any heating and air condi- 
tioning system in commercial buildings, providing the necessary amount of 
fresh outdoor air and distributing it uniformly throughout the building. This 
is not surprising since many rooms in large commercial buildings have no 
windows and thus rely on mechanical ventilation. Even the rooms with 
windows are in the same situation since the windows are tightly sealed and 
cannot be opened in most buildings. It is not a good idea to oversize the 
ventilation system just to be on the "safe side" since exhausting the heated 
or cooled indoor air wastes energy. On the other hand, reducing the venti- 
lation rates below the required minimum to conserve energy should also be 
avoided so that the indoor air quality can be maintained at the required lev- 
els. The minimum fresh air ventilation requirements are listed in Table 1-8. 
The values are based on controlling the C0 2 and other contaminants with 
an adequate margin of safety, which requires each person be supplied with 
at least 7.5 L/s (15 ftVmin) of fresh air. 

Another function of the mechanical ventilation system is to clean the air 
by filtering it as it enters the building. Various types of filters are available 
for this purpose, depending on the cleanliness requirements and the allow- 
able pressure drop. 




Evaporation 



FIGURE 1-55 

In hot environments, a body can 

dissipate a large amount of metabolic 

heat by sweating since the sweat absorbs 

the body heat and evaporates. 



TABLE 1-8 

Minimum fresh air requirements 
in buildings (from ASHRAE 
Standard 62-1989) 



Requirement 
(per person) 



Application 


L/s 


ft 3 /min 


Classrooms, 

libraries, 

supermarkets 


8 


15 


Dining rooms, 
conference 
rooms, offices 


10 


20 


Hospital 
rooms 


13 


25 


Hotel rooms 15 

(per room) 


30 
(per room) 


Smoking 
lounges 


30 


60 


Retail stores 


1.0-1.5 
(per m 2 ) 


0.2-0.3 
(per ft 2 ) 



Residential 0.35 air change per 
buildings hour, but not less than 

7.5 L/s (or 15ft 3 /min) 

per person 



cen58933_ch01.qxd 9/10/2002 8:30 AM Page 46 



46 
HEAT TRANSFER 



SUMMARY 



In this chapter, the basics of heat transfer are introduced and 
discussed. The science of thermodynamics deals with the 
amount of heat transfer as a system undergoes a process from 
one equilibrium state to another, whereas the science of heat 
transfer deals with the rate of heat transfer, which is the main 
quantity of interest in the design and evaluation of heat transfer 
equipment. The sum of all forms of energy of a system is called 
total energy, and it includes the internal, kinetic, and potential 
energies. The internal energy represents the molecular energy 
of a system, and it consists of sensible, latent, chemical, and 
nuclear forms. The sensible and latent forms of internal energy 
can be transferred from one medium to another as a result of a 
temperature difference, and are referred to as heat or thermal 
energy. Thus, heat transfer is the exchange of the sensible and 
latent forms of internal energy between two mediums as a re- 
sult of a temperature difference. The amount of heat transferred 
per unit time is called heat transfer rate and is denoted by Q. 
The rate of heat transfer per unit area is called heat flux, q. 

A system of fixed mass is called a closed system and a sys- 
tem that involves mass transfer across its boundaries is called 
an open system or control volume. The first law of thermody- 
namics or the energy balance for any system undergoing any 
process can be expressed as 

When a stationary closed system involves heat transfer only 
and no work interactions across its boundary, the energy bal- 
ance relation reduces to 

Q = mC r AT 

where Q is the amount of net heat transfer to or from the sys- 
tem. When heat is transferred at a constant rate of Q, the 
amount of heat transfer during a time interval At can be deter- 
mined from Q = Q At. 

Under steady conditions and in the absence of any work in- 
teractions, the conservation of energy relation for a control vol- 
ume with one inlet and one exit with negligible changes in 
kinetic and potential energies can be expressed as 



e c 



kA 



Q 



hi C p AT 



where m = p°VA c is the mass flow rate and Q is the rate of net 
heat transfer into or out of the control volume. 

Heat can be transferred in three different modes: conduction, 
convection, and radiation. Conduction is the transfer of energy 
from the more energetic particles of a substance to the adjacent 
less energetic ones as a result of interactions between the parti- 
cles, and is expressed by Fourier's law of heat conduction as 



cIT 
dx 



where k is the thermal conductivity of the material, A is the 
area normal to the direction of heat transfer, and dT/dx is the 
temperature gradient. The magnitude of the rate of heat con- 
duction across a plane layer of thickness L is given by 



6 c 



kA 



AT 



where AT is the temperature difference across the layer. 

Convection is the mode of heat transfer between a solid sur- 
face and the adjacent liquid or gas that is in motion, and in- 
volves the combined effects of conduction and fluid motion. 
The rate of convection heat transfer is expressed by Newton 's 
law of cooling as 



e 



convection 



hA, (T. - TJ 



where h is the convection heat transfer coefficient in W/m 2 • °C 
or Btu/h • ft 2 ■ °F, A s is the surface area through which con- 
vection heat transfer takes place, T s is the surface temperature, 
and T^ is the temperature of the fluid sufficiently far from the 
surface. 

Radiation is the energy emitted by matter in the form of 
electromagnetic waves (or photons) as a result of the changes 
in the electronic configurations of the atoms or molecules. The 
maximum rate of radiation that can be emitted from a surface 
at an absolute temperature T s is given by the Stefan-Boltzmann 
law as 2 crait . raax = uAJ* where <r = 5.67 X 10" 8 W/m 2 • K 4 
or 0.1714 X 10" 8 Btu/h • ft 2 • R 4 is the Stefan-Boltzmann 
constant. 

When a surface of emissivity 8 and surface area A s at an ab- 
solute temperature T s is completely enclosed by a much larger 
(or black) surface at absolute temperature T SUII separated by a 
gas (such as air) that does not intervene with radiation, the net 
rate of radiation heat transfer between these two surfaces is 
given by 

e rad = e aA s (r s 4 -r 8 4 urr ) 

In this case, the emissivity and the surface area of the sur- 
rounding surface do not have any effect on the net radiation 
heat transfer. 

The rate at which a surface absorbs radiation is determined 
from g absorbed = afiinciden, where g lncidcnt is the rate at which ra- 
diation is incident on the surface and a is the absorptivity of 
the surface. 



cen58933_ch01.qxd 9/10/2002 8:30 AM Page 47 



REFERENCES AND SUGGESTED READING 



47 
CHAPTER 1 



1. American Society of Heating, Refrigeration, and Air- 
Conditioning Engineers, Handbook of Fundamentals. 
Atlanta: ASHRAE, 1993. 

2. Y. A. Cengel and R. H. Turner. Fundamentals of Thermal- 
Fluid Sciences. New York: McGraw-Hill, 2001 . 

3. Y. A. Cengel and M. A. Boles. Thermodynamics — An 
Engineering Approach. 4th ed. New York: McGraw-Hill, 
2002. 

4. J. P. Holman. Heat Transfer. 9th ed. New York: McGraw- 
Hill, 2002. 

5. F. P. Incropera and D. P. DeWitt. Introduction to Heat 
Transfer. 4th ed. New York: John Wiley & Sons, 2002. 



6. F. Kreith and M. S. Bohn. Principles of Heat Transfer. 6th 
ed. Pacific Grove, CA: Brooks/Cole, 2001. 

7. A. F. Mills. Basic Heat and Mass Transfer. 2nd ed. 
Upper Saddle River, NJ: Prentice-Hall, 1999. 

8. M. N. Ozisik. Heat Transfer — A Basic Approach. New 
York: McGraw-Hill, 1985. 

9. Robert J. Ribando. Heat Transfer Tools. New York: 
McGraw-Hill, 2002. 

10. F M. White. Heat and Mass Transfer. Reading, MA: 
Addison-Wesley, 1988. 



PROBLEMS 



Thermodynamics and Heat Transfer 

1-1 C How does the science of heat transfer differ from the 
science of thermodynamics? 

1-2C What is the driving force for (a) heat transfer, (b) elec- 
tric current flow, and (c) fluid flow? 

1-3C What is the caloric theory? When and why was it 
abandoned? 

1— 4C How do rating problems in heat transfer differ from the 
sizing problems? 

1-5C What is the difference between the analytical and ex- 
perimental approach to heat transfer? Discuss the advantages 
and disadvantages of each approach. 

1-6C What is the importance of modeling in engineering? 
How are the mathematical models for engineering processes 
prepared? 

1-7C When modeling an engineering process, how is the 
right choice made between a simple but crude and a complex 
but accurate model? Is the complex model necessarily a better 
choice since it is more accurate? 

Heat and Other Forms of Energy 

1-8C What is heat flux? How is it related to the heat trans- 
fer rate? 



*Problems designated by a "C" are concept questions, and 
students are encouraged to answer them all. Problems designated 
by an "E" are in English units, and the SI users can ignore them. 
Problems with a CD-EES icon ® are solved using EES, and 
complete solutions together with parametric studies are included 
on the enclosed CD. Problems with a computer-EES icon H are 
comprehensive in nature, and are intended to be solved with a 
computer, preferably using the EES software that accompanies 
this text. 



1-9C What are the mechanisms of energy transfer to a closed 
system? How is heat transfer distinguished from the other 
forms of energy transfer? 

1-10C How are heat, internal energy, and thermal energy 
related to each other? 

1-11C An ideal gas is heated from 50°C to 80°C (a) at con- 
stant volume and (b) at constant pressure. For which case do 
you think the energy required will be greater? Why? 

1-12 A cylindrical resistor element on a circuit board dis- 
sipates 0.6 W of power. The resistor is 1.5 cm long, and has a 
diameter of 0.4 cm. Assuming heat to be transferred uniformly 
from all surfaces, determine (a) the amount of heat this resistor 
dissipates during a 24-hour period, (b) the heat flux, and (c) the 
fraction of heat dissipated from the top and bottom surfaces. 

1-13E A logic chip used in a computer dissipates 3 W of 
power in an environment at 120°F, and has a heat transfer sur- 
face area of 0.08 in 2 . Assuming the heat transfer from the sur- 
face to be uniform, determine (a) the amount of heat this chip 
dissipates during an eight-hour work day, in kWh, and (b) the 
heat flux on the surface of the chip, in W/in 2 . 

1-14 Consider a 150-W incandescent lamp. The filament 
of the lamp is 5 cm long and has a diameter of 0.5 mm. The 
diameter of the glass bulb of the lamp is 8 cm. Determine the 
heat flux, in W/m 2 , (a) on the surface of the filament and (b) on 
the surface of the glass bulb, and (c) calculate how much it will 
cost per year to keep that lamp on for eight hours a day every 
day if the unit cost of electricity is $0.08/kWh. 

Answers: (a) 1.91 x 10 6 W/m 2 , (b) 7500 W/m 2 , (c) $35.04/yr 

1-15 A 1200-W iron is left on the ironing board with its base 
exposed to the air. About 90 percent of the heat generated 
in the iron is dissipated through its base whose surface area is 
150 cm 2 , and the remaining 10 percent through other surfaces. 
Assuming the heat transfer from the surface to be uniform, 



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48 
HEAT TRANSFER 




8 cm 



FIGURE P1 -14 

determine (a) the amount of heat the iron dissipates during a 
2-hour period, in kWh, (b) the heat flux on the surface of the 
iron base, in W/m 2 , and (c) the total cost of the electrical en- 
ergy consumed during this 2-hour period. Take the unit cost of 
electricity to be $0.07/kWh. 

1-16 A 15-cm X 20-cm circuit board houses on its surface 
120 closely spaced logic chips, each dissipating 0.12 W. If the 
heat transfer from the back surface of the board is negligible, 
determine (a) the amount of heat this circuit board dissipates 
during a 10-hour period, in kWh, and (b) the heat flux on the 
surface of the circuit board, in W/m 2 . 



15 cm 




Chips 



FIGURE P1 -16 

1-17 A 15-cm-diameter aluminum ball is to be heated from 
80°C to an average temperature of 200°C. Taking the average 
density and specific heat of aluminum in this temperature 
range to be p = 2700 kg/m 3 and C p = 0.90 kJ/kg • °C, respec- 
tively, determine the amount of energy that needs to be trans- 
ferred to the aluminum ball. Answer: 515 kJ 



1-18 The average specific heat of the human body is 3.6 
kJ/kg • °C. If the body temperature of a 70-kg man rises from 
37°C to 39°C during strenuous exercise, determine the increase 
in the thermal energy content of the body as a result of this rise 
in body temperature. 

1-19 Infiltration of cold air into a warm house during winter 
through the cracks around doors, windows, and other openings 
is a major source of energy loss since the cold air that enters 
needs to be heated to the room temperature. The infiltration is 
often expressed in terms of ACH (air changes per hour). An 
ACH of 2 indicates that the entire air in the house is replaced 
twice every hour by the cold air outside. 

Consider an electrically heated house that has a floor space 
of 200 m 2 and an average height of 3 m at 1000 m elevation, 
where the standard atmospheric pressure is 89.6 kPa. The 
house is maintained at a temperature of 22°C, and the infiltra- 
tion losses are estimated to amount to 0.7 ACH. Assuming the 
pressure and the temperature in the house remain constant, de- 
termine the amount of energy loss from the house due to infil- 
tration for a day during which the average outdoor temperature 
is 5°C. Also, determine the cost of this energy loss for that day 
if the unit cost of electricity in that area is $0.082/kWh. 

Answers: 53.8 kWh/day, $4.41/day 

1-20 Consider a house with a floor space of 200 m 2 and an 
average height of 3 m at sea level, where the standard atmos- 
pheric pressure is 101 .3 kPa. Initially the house is at a uniform 
temperature of 10°C. Now the electric heater is turned on, and 
the heater runs until the air temperature in the house rises to an 
average value of 22°C. Determine how much heat is absorbed 
by the air assuming some air escapes through the cracks as the 
heated air in the house expands at constant pressure. Also, de- 
termine the cost of this heat if the unit cost of electricity in that 
area is $0.075/kWh. 

1-21E Consider a 60-gallon water heater that is initially 
filled with water at 45°F. Determine how much energy needs to 
be transferred to the water to raise its temperature to 140°F. 
Take the density and specific heat of water to be 62 lbm/ft 3 and 
1 .0 Btu/lbm ■ °F, respectively. 



The First Law of Thermodynamics 

1-22C On a hot summer day, a student turns his fan on when 
he leaves his room in the morning. When he returns in the 
evening, will his room be warmer or cooler than the neighbor- 
ing rooms? Why? Assume all the doors and windows are kept 
closed. 

1-23C Consider two identical rooms, one with a refrigerator 
in it and the other without one. If all the doors and windows are 
closed, will the room that contains the refrigerator be cooler or 
warmer than the other room? Why? 

1-24C Define mass and volume flow rates. How are they re- 
lated to each other? 



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49 
CHAPTER 1 



1-25 Two 800-kg cars moving at a velocity of 90 km/h have 
a head-on collision on a road. Both cars come to a complete 
rest after the crash. Assuming all the kinetic energy of cars is 
converted to thermal energy, determine the average tempera- 
ture rise of the remains of the cars immediately after the crash. 
Take the average specific heat of the cars to be 0.45 kJ/kg • °C. 

1-26 A classroom that normally contains 40 people is to be 
air-conditioned using window air-conditioning units of 5-kW 
cooling capacity. A person at rest may be assumed to dissipate 
heat at a rate of 360 kJ/h. There are 10 lightbulbs in the room, 
each with a rating of 100 W. The rate of heat transfer to the 
classroom through the walls and the windows is estimated to 
be 15,000 kJ/h. If the room air is to be maintained at a constant 
temperature of 21°C, determine the number of window air- 
conditioning units required. Answer: two units 

1-27E A rigid tank contains 20 lbm of air at 50 psia and 
80°F. The air is now heated until its pressure is doubled. Deter- 
mine (a) the volume of the tank and (b) the amount of heat 
transfer. Answers: (a) 80 ft 3 , (b) 2035 Btu 

1-28 A 1-m 3 rigid tank contains hydrogen at 250 kPa and 
420 K. The gas is now cooled until its temperature drops to 300 
K. Determine (a) the final pressure in the tank and (b) the 
amount of heat transfer from the tank. 

1-29 A 4-m X 5-m X 6-m room is to be heated by a base- 
board resistance heater. It is desired that the resistance heater 
be able to raise the air temperature in the room from 7°C to 
25°C within 15 minutes. Assuming no heat losses from the 
room and an atmospheric pressure of 100 kPa, determine the 
required power rating of the resistance heater. Assume constant 
specific heats at room temperature. Answer: 3.01 kW 

1-30 A 4-m X 5-m X 7-m room is heated by the radiator of 
a steam heating system. The steam radiator transfers heat at a 
rate of 10,000 kJ/h and a 100-W fan is used to distribute the 
warm air in the room. The heat losses from the room are esti- 
mated to be at a rate of about 5000 kJ/h. If the initial tempera- 
ture of the room air is 10°C, determine how long it will take for 
the air temperature to rise to 20°C. Assume constant specific 
heats at room temperature. 



5000 kJ/h 



Steam 


Ro 


Jin 
4 n 


i X 5 m x7 m 


»- « 


4 


10,000 kJ/h 


*■ $ 


•< 4 











Room 

4mx6mx6m 




FIGURE P1-31 

1-31 A student living in a 4-m X 6-m X 6-m dormitory 
room turns his 150-W fan on before she leaves her room on a 
summer day hoping that the room will be cooler when she 
comes back in the evening. Assuming all the doors and win- 
dows are tightly closed and disregarding any heat transfer 
through the walls and the windows, determine the temperature 
in the room when she comes back 10 hours later. Use specific 
heat values at room temperature and assume the room to be at 
100 kPa and 15°C in the morning when she leaves. 
Answer: 58.1°C 

1-32E A 10-ft 3 tank contains oxygen initially at 14.7 psia 
and 80°F. A paddle wheel within the tank is rotated until the 
pressure inside rises to 20 psia. During the process 20 Btu of 
heat is lost to the surroundings. Neglecting the energy stored in 
the paddle wheel, determine the work done by the paddle 
wheel. 

1-33 A room is heated by a baseboard resistance heater. 
When the heat losses from the room on a winter day amount to 
7000 kJ/h, it is observed that the air temperature in the room 
remains constant even though the heater operates continuously. 
Determine the power rating of the heater, in kW. 

1-34 A 50-kg mass of copper at 70°C is dropped into an in- 
sulated tank containing 80 kg of water at 25°C. Determine the 
final equilibrium temperature in the tank. 

1-35 A 20-kg mass of iron at 100°C is brought into contact 
with 20 kg of aluminum at 200°C in an insulated enclosure. 
Determine the final equilibrium temperature of the combined 
system. Answer: 168°C 

1-36 An unknown mass of iron at 90 C C is dropped into an 
insulated tank that contains 80 L of water at 20°C. At the same 



Water 




n 



w 



FIGURE P 1-30 



FIGURE P1-36 



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50 
HEAT TRANSFER 



time, a paddle wheel driven by a 200-W motor is activated to 
stir the water. Thermal equilibrium is established after 25 min- 
utes with a final temperature of 27°C. Determine the mass of 
the iron. Neglect the energy stored in the paddle wheel, and 
take the density of water to be 1000 kg/m 3 . Answer: 72.1 kg 

1-37E A 90-lbm mass of copper at 160°F and a 50-lbm mass 
of iron at 200°F are dropped into a tank containing 1 80 lbm of 
water at 70°F. If 600 Btu of heat is lost to the surroundings dur- 
ing the process, determine the final equilibrium temperature. 

1-38 A 5-m X 6-m X 8-m room is to be heated by an elec- 
trical resistance heater placed in a short duct in the room. Ini- 
tially, the room is at 15°C, and the local atmospheric pressure 
is 98 kPa. The room is losing heat steadily to the outside at a 
rate of 200 kJ/min. A 200-W fan circulates the air steadily 
through the duct and the electric heater at an average mass flow 
rate of 50 kg/min. The duct can be assumed to be adiabatic, and 
there is no air leaking in or out of the room. If it takes 15 min- 
utes for the room air to reach an average temperature of 25°C, 
find (a) the power rating of the electric heater and (b) the tem- 
perature rise that the air experiences each time it passes 
through the heater. 

1-39 A house has an electric heating system that consists of 
a 300-W fan and an electric resistance heating element placed 
in a duct. Air flows steadily through the duct at a rate of 0.6 
kg/s and experiences a temperature rise of 5°C. The rate of heat 
loss from the air in the duct is estimated to be 250 W. De- 
termine the power rating of the electric resistance heating 
element. 

1-40 A hair dryer is basically a duct in which a few layers of 
electric resistors are placed. A small fan pulls the air in and 
forces it to flow over the resistors where it is heated. Air enters 
a 1200-W hair dryer at 100 kPa and 22°C, and leaves at 47°C. 
The cross-sectional area of the hair dryer at the exit is 60 cm 2 . 
Neglecting the power consumed by the fan and the heat losses 
through the walls of the hair dryer, determine (a) the volume 
flow rate of air at the inlet and (b) the velocity of the air at the 
exit. Answers: (a) 0.0404 m 3 /s, (b) 7.30 m/s 



T- = 47°C 



: 60 



cm ~ rAVS 



= 100 kPa 

:22°C 



W =1200W 

e 

FIGURE P1-40 

1-41 The ducts of an air heating system pass through an un- 
heated area. As a result of heat losses, the temperature of the air 
in the duct drops by 3°C. If the mass flow rate of air is 120 
kg/min, determine the rate of heat loss from the air to the cold 
environment. 



1-42E Air enters the duct of an air-conditioning system at 1 5 
psia and 50°F at a volume flow rate of 450 ft'/min. The diam- 
eter of the duct is 10 inches and heat is transferred to the air in 
the duct from the surroundings at a rate of 2 Btu/s. Determine 
(a) the velocity of the air at the duct inlet and (b) the tempera- 
ture of the air at the exit. Answers: (a) 825 ft/min, (£>) 64°F 

1-43 Water is heated in an insulated, constant diameter tube 
by a 7-kW electric resistance heater. If the water enters the 
heater steadily at 15°C and leaves at 70°C, determine the mass 
flow rate of water. 



Water M 
LvC \? 



i-AWvWVWvWV^n 



\70°C 



Resistance 
heater, 7 kW 



FIGURE P1 -43 



Heat Transfer Mechanisms 

1-44C Define thermal conductivity and explain its signifi- 
cance in heat transfer. 

1-45C What are the mechanisms of heat transfer? How are 
they distinguished from each other? 

1-46C What is the physical mechanism of heat conduction in 
a solid, a liquid, and a gas? 

1-47C Consider heat transfer through a windowless wall of 
a house in a winter day. Discuss the parameters that affect the 
rate of heat conduction through the wall. 

1-48C Write down the expressions for the physical laws that 
govern each mode of heat transfer, and identify the variables 
involved in each relation. 

1-49C How does heat conduction differ from convection? 

1-50C Does any of the energy of the sun reach the earth by 
conduction or convection? 

1-51 C How does forced convection differ from natural 
convection? 

1-52C Define emissivity and absorptivity. What is Kirch- 
hoff's law of radiation? 

1-53C What is a blackbody? How do real bodies differ from 
blackbodies? 

1-54C Judging from its unit W/m • °C, can we define ther- 
mal conductivity of a material as the rate of heat transfer 
through the material per unit thickness per unit temperature 
difference? Explain. 

1-55C Consider heat loss through the two walls of a house 
on a winter night. The walls are identical, except that one of 
them has a tightly fit glass window. Through which wall will 
the house lose more heat? Explain. 

1-56C Which is a better heat conductor, diamond or silver? 



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51 
CHAPTER 1 



1-57C Consider two walls of a house that are identical ex- 
cept that one is made of 10-cm-thick wood, while the other is 
made of 25-cm-thick brick. Through which wall will the house 
lose more heat in winter? 

1-58C How do the thermal conductivity of gases and liquids 
vary with temperature? 

1-59C Why is the thermal conductivity of superinsulation 
orders of magnitude lower than the thermal conductivity of 
ordinary insulation? 

1-60C Why do we characterize the heat conduction ability 
of insulators in terms of their apparent thermal conductivity 
instead of the ordinary thermal conductivity? 

1-61 C Consider an alloy of two metals whose thermal con- 
ductivities are k t and k 2 . Will the thermal conductivity of the 
alloy be less than k t , greater than k 2 , or between k x and k{l 

1-62 The inner and outer surfaces of a 5-m X 6-m brick wall 
of thickness 30 cm and thermal conductivity 0.69 W/m ■ °C are 
maintained at temperatures of 20°C and 5°C, respectively. 
Determine the rate of heat transfer through the wall, in W. 
Answer: 1035 W 



20°C 



- — 



Brick 



wall 



30 cm 



5°C 



FIGURE P1-62 



1-63 The inner and outer surfaces of a 0.5-cm-thick 2-m X 
2-m window glass in winter are 10°C and 3°C, respectively. If 
the thermal conductivity of the glass is 0.78 W/m • °C, deter- 
mine the amount of heat loss, in kJ, through the glass over a 
period of 5 hours. What would your answer be if the glass were 
1 cm thick? Answers: 78,624 kJ, 39,312 kJ 

1-64 [JJ^l Reconsider Problem 1-63. Using EES (or other) 
b^2 software, plot the amount of heat loss through the 
glass as a function of the window glass thickness in the range 
of 0. 1 cm to 1 .0 cm. Discuss the results. 

1-65 An aluminum pan whose thermal conductivity is 
237 W/m ■ °C has a flat bottom with diameter 20 cm and thick- 
ness 0.4 cm. Heat is transferred steadily to boiling water in the 
pan through its bottom at a rate of 800 W. If the inner surface 
of the bottom of the pan is at 105°C, determine the temperature 
of the outer surface of the bottom of the pan. 



ki (, 



X< 



5°C 



0.4 cm 



I I 1 I I I I III I t 

800 W 

FIGURE P1 -65 



1-66E The north wall of an electrically heated home is 20 ft 
long, 10 ft high, and 1 ft thick, and is made of brick whose 
thermal conductivity is k = 0.42 Btu/h • ft • °F. On a certain 
winter night, the temperatures of the inner and the outer sur- 
faces of the wall are measured to be at about 62°F and 25°F, 
respectively, for a period of 8 hours. Determine (a) the rate of 
heat loss through the wall that night and (b) the cost of that heat 
loss to the home owner if the cost of electricity is $0.07/kWh. 

1-67 In a certain experiment, cylindrical samples of diameter 
4 cm and length 7 cm are used (see Fig. 1-29). The two 
thermocouples in each sample are placed 3 cm apart. After ini- 
tial transients, the electric heater is observed to draw 0.6 A at 
110 V, and both differential thermometers read a temperature 
difference of 10°C. Determine the thermal conductivity of the 
sample. Answer: 78.8 W/m • °C 

1-68 One way of measuring the thermal conductivity of a 
material is to sandwich an electric thermofoil heater between 
two identical rectangular samples of the material and to heavily 
insulate the four outer edges, as shown in the figure. Thermo- 
couples attached to the inner and outer surfaces of the samples 
record the temperatures. 

During an experiment, two 0.5-cm-thick samples 10 cm X 
10 cm in size are used. When steady operation is reached, the 
heater is observed to draw 35 W of electric power, and the tem- 
perature of each sample is observed to drop from 82°C at the 
inner surface to 74°C at the outer surface. Determine the ther- 
mal conductivity of the material at the average temperature. 



Samples 




"- Insulation 



- Insulation 



Source 



FIGURE P1 -68 



0.5 cm 



1-69 Repeat Problem 1-68 for an electric power consump- 
tion of 28 W. 



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52 
HEAT TRANSFER 



1-70 A heat flux meter attached to the inner surface of a 
3-cm-thick refrigerator door indicates a heat flux of 25 W/m 2 
through the door. Also, the temperatures of the inner and the 
outer surfaces of the door are measured to be 7°C and 15°C, 
respectively. Determine the average thermal conductivity of 
the refrigerator door. Answer: 0.0938 W/m • °C 

1-71 Consider a person standing in a room maintained at 
20°C at all times. The inner surfaces of the walls, floors, and 
ceiling of the house are observed to be at an average tempera- 
ture of 12°C in winter and 23°C in summer. Determine the 
rates of radiation heat transfer between this person and the sur- 
rounding surfaces in both summer and winter if the exposed 
surface area, emissivity, and the average outer surface temper- 
ature of the person are 1.6 m 2 , 0.95, and 32°C, respectively. 

1-72 [7(^1 Reconsider Problem 1-71. Using EES (or other) 
t^S software, plot the rate of radiation heat transfer in 
winter as a function of the temperature of the inner surface of 
the room in the range of 8°C to 18°C. Discuss the results. 

1-73 For heat transfer purposes, a standing man can be mod- 
eled as a 30-cm-diameter, 1 70-cm-long vertical cylinder with 
both the top and bottom surfaces insulated and with the side 
surface at an average temperature of 34°C. For a convection 
heat transfer coefficient of 15 W/m 2 ■ °C, determine the rate of 
heat loss from this man by convection in an environment at 
20°C. Answer: 336 W 

1-74 Hot air at 80°C is blown over a 2-m X 4-m flat surface 
at 30 C C. If the average convection heat transfer coefficient is 
55 W/m 2 • °C, determine the rate of heat transfer from the air to 
the plate, in kW. Answer: 22 kW 

1-75 rSi'M Reconsider Problem 1-74. Using EES (or other) 
b^2 software, plot the rate of heat transfer as a func- 
tion of the heat transfer coefficient in the range of 20 W/m 2 ■ °C 
to 100 W/m 2 • °C. Discuss the results. 

1-76 The heat generated in the circuitry on the surface of a 
silicon chip (k = 130 W/m • °C) is conducted to the ceramic 
substrate to which it is attached. The chip is 6 mm X 6 mm in 
size and 0.5 mm thick and dissipates 3 W of power. Disregard- 
ing any heat transfer through the 0.5-mm-high side surfaces, 
determine the temperature difference between the front and 
back surfaces of the chip in steady operation. 



Silicon 
chip 



0.5 mm 




1-77 A 50-cm-long, 800-W electric resistance heating ele- 
ment with diameter 0.5 cm and surface temperature 120°C is 
immersed in 60 kg of water initially at 20°C. Determine how 
long it will take for this heater to raise the water temperature to 
80°C. Also, determine the convection heat transfer coefficients 
at the beginning and at the end of the heating process. 

1-78 A 5 -cm-external-diameter, 10-m-long hot water pipe at 
80°C is losing heat to the surrounding air at 5°C by natural 
convection with a heat transfer coefficient of 25 W/m 2 ■ °C. 
Determine the rate of heat loss from the pipe by natural con- 
vection, in W. Answer: 2945 W 

1-79 A hollow spherical iron container with outer diameter 
20 cm and thickness 0.4 cm is filled with iced water at 0°C. If 
the outer surface temperature is 5°C, determine the approxi- 
mate rate of heat loss from the sphere, in kW, and the rate at 
which ice melts in the container. The heat from fusion of water 
is 333.7 kJ/kg. 



5 C C 




0.4 cm 



Ceramic 
substrate 

FIGURE P1-76 



FIGURE P1-79 

1-80 VcgM Reconsider Problem 1-79. Using EES (or other) 
t£^ software, plot the rate at which ice melts as a 
function of the container thickness in the range of 0.2 cm to 
2.0 cm. Discuss the results. 

1-81E The inner and outer glasses of a 6-ft X 6-ft double- 
pane window are at 60°F and 42°F, respectively. If the 0.25-in. 
space between the two glasses is filled with still air, determine 
the rate of heat transfer through the window. 
Answer: 439 Btu/h 

1-82 Two surfaces of a 2-cm-thick plate are maintained at 
0°C and 80°C, respectively. If it is determined that heat is 
transferred through the plate at a rate of 500 W/m 2 , determine 
its thermal conductivity. 

1-83 Four power transistors, each dissipating 15 W, are 
mounted on a thin vertical aluminum plate 22 cm X 22 cm in 
size. The heat generated by the transistors is to be dissipated by 
both surfaces of the plate to the surrounding air at 25 °C, which 
is blown over the plate by a fan. The entire plate can be as- 
sumed to be nearly isothermal, and the exposed surface area of 
the transistor can be taken to be equal to its base area. If the 
average convection heat transfer coefficient is 25 W/m 2 ■ °C, 
determine the temperature of the aluminum plate. Disregard 
any radiation effects. 



cen58933_ch01.qxd 9/10/2002 8:30 AM Page 53 



1-84 An ice chest whose outer dimensions are 30 cm X 
40 cm X 40 cm is made of 3-cm-thick Styrofoam (k = 0.033 
W/m ■ °C). Initially, the chest is filled with 40 kg of ice at 0°C, 
and the inner surface temperature of the ice chest can be taken 
to be 0°C at all times. The heat of fusion of ice at 0°C is 333.7 
kJ/kg, and the surrounding ambient air is at 30°C. Disregarding 
any heat transfer from the 40-cm X 40-cm base of the ice 
chest, determine how long it will take for the ice in the chest to 
melt completely if the outer surfaces of the ice chest are at 8°C. 
Answer: 32.7 days 



:30°C 



. 




, 


ts « ° 

o Ice chest Q O 

pop O 

Q o 



-3 cm 



Styrofoam 



FIGURE P1-84 



1-85 A transistor with a height of 0.4 cm and a diameter of 
0.6 cm is mounted on a circuit board. The transistor is cooled 
by air flowing over it with an average heat transfer coefficient 
of 30 W/m 2 • °C. If the air temperature is 55°C and the tran- 
sistor case temperature is not to exceed 70°C, determine the 
amount of power this transistor can dissipate safely. Disregard 
any heat transfer from the transistor base. 



Air 
55°C 






Power 
transistor 
T <70°C 



0.6 cm 



0.4 cm - 



FIGURE P1-85 



1-86 [Z?vfl Reconsider Problem 1-85. Using EES (or other) 
b^2 software, plot the amount of power the transistor 
can dissipate safely as a function of the maximum case tem- 
perature in the range of 60°C to 90°C. Discuss the results. 



53 
CHAPTER 1 



1-87E A 200-ft-long section of a steam pipe whose outer di- 
ameter is 4 inches passes through an open space at 50°F. The 
average temperature of the outer surface of the pipe is mea- 
sured to be 280°F, and the average heat transfer coefficient on 
that surface is determined to be 6 Btu/h ■ ft 2 • °F. Determine 
(a) the rate of heat loss from the steam pipe and (b) the annual 
cost of this energy loss if steam is generated in a natural gas 
furnace having an efficiency of 86 percent, and the price of nat- 
ural gas is $0.58/therm (1 therm = 100,000 Btu). 
Answers: (a) 289,000 Btu/h, (b) $17,074/yr 

1-88 The boiling temperature of nitrogen at atmospheric 
pressure at sea level (1 atm) is — 196°C. Therefore, nitrogen is 
commonly used in low temperature scientific studies since the 
temperature of liquid nitrogen in a tank open to the atmosphere 
will remain constant at — 196°C until the liquid nitrogen in 
the tank is depleted. Any heat transfer to the tank will result 
in the evaporation of some liquid nitrogen, which has a heat of 
vaporization of 198 kJ/kg and a density of 810 kg/m 3 at 1 atm. 
Consider a 4-m-diameter spherical tank initially filled 
with liquid nitrogen at 1 atm and — 196°C. The tank is ex- 
posed to 20°C ambient air with a heat transfer coefficient of 
25 W/m 2 • °C. The temperature of the thin-shelled spherical 
tank is observed to be almost the same as the temperature of 
the nitrogen inside. Disregarding any radiation heat exchange, 
determine the rate of evaporation of the liquid nitrogen in the 
tank as a result of the heat transfer from the ambient air. 




FIGURE P1 -88 

1-89 Repeat Problem 1-88 for liquid oxygen, which has 
a boiling temperature of — 183°C, a heat of vaporization of 
213 kJ/kg, and a density of 1 140 kg/m 3 at 1 atm pressure. 

1-90 rSi'M Reconsider Problem 1-88. Using EES (or other) 
1^2 software, plot the rate of evaporation of liquid 
nitrogen as a function of the ambient air temperature in the 
range of 0°C to 35°C. Discuss the results. 

1-91 Consider a person whose exposed surface area is 
1.7 m 2 , emissivity is 0.7, and surface temperature is 32°C. 



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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 

1-92 A 0.3-cm-thick, 12-cm-high, and 18-cm-long circuit 
board houses 80 closely spaced logic chips on one side, each 
dissipating 0.06 W. The board is impregnated with copper fill- 
ings and has an effective thermal conductivity of 16 W/m • °C. 
All the heat generated in the chips is conducted across the cir- 
cuit board and is dissipated from the back side of the board to 
the ambient air. Determine the temperature difference between 
the two sides of the circuit board. Answer-. 0.042°C 

1-93 Consider a sealed 20-cm-high electronic box whose 
base dimensions are 40 cm X 40 cm placed in a vacuum cham- 
ber. The emissivity of the outer surface of the box is 0.95. If the 
electronic components in the box dissipate a total of 100 W of 
power and the outer surface temperature of the box is not to ex- 
ceed 55°C, determine the temperature at which the surrounding 
surfaces must be kept if this box is to be cooled by radiation 
alone. Assume the heat transfer from the bottom surface of the 
box to the stand to be negligible. 




FIGURE P1-93 

1-94 Using the conversion factors between W and Btu/h, m 
and ft, and K and R, express the Stefan-Boltzmann constant 
o- = 5.67 X 10- 8 W/m 2 • K 4 in the English unit Btu/h • ft 2 ■ R 4 . 

1-95 An engineer who is working on the heat transfer analy- 
sis of a house in English units needs the convection heat trans- 
fer coefficient on the outer surface of the house. But the only 
value he can find from his handbooks is 20 W/m 2 • °C, which 
is in SI units. The engineer does not have a direct conversion 
factor between the two unit systems for the convection heat 
transfer coefficient. Using the conversion factors between 
W and Btu/h, m and ft, and °C and °F, express the given con- 
vection heat transfer coefficient in Btu/h • ft 2 ■ °F. 
Answer: 3.52 Btu/h • ft 2 • °F 

Simultaneous Heat Transfer Mechanisms 

1-96C Can all three modes of heat transfer occur simultane- 
ously (in parallel) in a medium? 

1-97C Can a medium involve (a) conduction and con- 
vection, (b) conduction and radiation, or (c) convection and ra- 
diation simultaneously? Give examples for the "yes" answers. 



1-98C The deep human body temperature of a healthy 
person remains constant at 37°C while the temperature and 
the humidity of the environment change with time. Discuss the 
heat transfer mechanisms between the human body and the en- 
vironment both in summer and winter, and explain how a per- 
son can keep cooler in summer and wanner in winter. 

1-99C We often turn the fan on in summer to help us cool. 
Explain how a fan makes us feel cooler in the summer. Also 
explain why some people use ceiling fans also in winter. 

1-100 Consider a person standing in a room at 23°C. Deter- 
mine the total rate of heat transfer from this person if the ex- 
posed surface area and the skin temperature of the person are 
1 .7 m 2 and 32°C, respectively, and the convection heat transfer 
coefficient is 5 W/m 2 ■ °C. Take the emissivity of the skin and 
the clothes to be 0.9, and assume the temperature of the inner 
surfaces of the room to be the same as the air temperature. 
Answer: 161 W 

1-101 Consider steady heat transfer between two large 
parallel plates at constant temperatures of T t = 290 K and 
T 2 = 1 50 K that are L = 2 cm apart. Assuming the surfaces to 
be black (emissivity e = 1 ), determine the rate of heat transfer 
between the plates per unit surface area assuming the gap 
between the plates is (a) filled with atmospheric air, (b) evacu- 
ated, (c) filled with fiberglass insulation, and (rf) filled with 
superinsulation having an apparent thermal conductivity of 
0.00015 W/m - C. 

1-102 A 1 .4-m-long, 0.2-cm-diameter electrical wire extends 
across a room that is maintained at 20°C. Heat is generated in 
the wire as a result of resistance heating, and the surface tem- 
perature of the wire is measured to be 240°C in steady op- 
eration. Also, the voltage drop and electric current through 
the wire are measured to be 110 V and 3 A, respectively. Dis- 
regarding any heat transfer by radiation, determine the con- 
vection heat transfer coefficient for heat transfer between the 
outer surface of the wire and the air in the room. 
Answer: 170.5 W/m 2 • °C 

Room 
20°C 

x- 240°C 



^ Electric resistance heater 

FIGURE P1 -102 



1-103 



Reconsider Problem 1-102. Using EES (or 
other) software, plot the convection heat trans- 
fer coefficient as a function of the wire surface temperature in 
the range of 100°C to 300°C. Discuss the results. 

1-104E A 2-in-diameter spherical ball whose surface is 
maintained at a temperature of 170°F is suspended in the mid- 
dle of a room at 70°F. If the convection heat transfer coefficient 
is 12 Btu/h ■ ft 2 ■ °F and the emissivity of the surface is 0.8, de- 
termine the total rate of heat transfer from the ball. 



cen58933_ch01.qxd 9/10/2002 8:30 AM Page 55 



1-105 fl&\ A 1000-W iron is left on the iron board with its 
Yss£y base exposed to the air at 20°C. The convection 
heat transfer coefficient between the base surface and the sur- 
rounding air is 35 W/m 2 • °C. If the base has an emissivity of 
0.6 and a surface area of 0.02 m 2 , determine the temperature of 
the base of the iron. Answer: 674°C 



20°C 




FIGURE P1-105 

1-106 The outer surface of a spacecraft in space has an emis- 
sivity of 0.8 and a solar absorptivity of 0.3. If solar radiation is 
incident on the spacecraft at a rate of 950 W/m 2 , determine the 
surface temperature of the spacecraft when the radiation emit- 
ted equals the solar energy absorbed. 

1-107 A 3-m-internal-diameter spherical tank made of 1 -cm- 
thick stainless steel is used to store iced water at 0°C. The tank 
is located outdoors at 25°C. Assuming the entire steel tank to 
be at 0°C and thus the thermal resistance of the tank to be neg- 
ligible, determine (a) the rate of heat transfer to the iced water 
in the tank and (b) the amount of ice at 0°C that melts during a 
24-hour period. The heat of fusion of water at atmospheric pres- 
sure is h jf = 333.7 kJ/kg. The emissivity of the outer surface of 
the tank is 0.6, and the convection heat transfer coefficient on 
the outer surface can be taken to be 30 W/m 2 ■ °C. Assume the 
average surrounding surface temperature for radiation ex- 
change to be 15°C. Answer: 5898 kg 

1-108 fJb\ The roof of a house consists of a 15-cm-thick 
W concrete slab (k = 2 W/m • °C) that is 15 m 
wide and 20 m long. The emissivity of the outer surface of the 
roof is 0.9, and the convection heat transfer coefficient on that 
surface is estimated to be 15 W/m 2 • °C. The inner surface of 
the roof is maintained at 15°C. On a clear winter night, the am- 
bient air is reported to be at 10°C while the night sky tempera- 
ture for radiation heat transfer is 255 K. Considering both 
radiation and convection heat transfer, determine the outer sur- 
face temperature and the rate of heat transfer through the roof. 
If the house is heated by a furnace burning natural gas with 
an efficiency of 85 percent, and the unit cost of natural gas is 
$0.60/fherm (1 therm = 105,500 kJ of energy content), de- 
termine the money lost through the roof that night during a 
14-hour period. 

1-109E Consider a flat plate solar collector placed horizon- 
tally on the flat roof of a house. The collector is 5 ft wide and 
15 ft long, and the average temperature of the exposed surface 



55 
CHAPTER 1 



of the collector is 100°F. The emissivity of the exposed sur- 
face of the collector is 0.9. Determine the rate of heat loss from 
the collector by convection and radiation during a calm day 
when the ambient air temperature is 70°F and the effective 
sky temperature for radiation exchange is 50°F. Take the con- 
vection heat transfer coefficient on the exposed surface to be 
2.5 Btu/h ■ ft 2 • °F. 




FIGURE P1-109E 

Problem Solving Technique and EES 

1-110C What is the value of the engineering software pack- 
ages in (a) engineering education and (b) engineering practice? 

| Determine a positive real root of the following 
equation using EES: 

2x 3 - lOx 05 - 3x = -3 



1-112 



[J3 Solve the following system of two equations 
with two unknowns using EES: 



x* - y 

3xy + y 



7.75 
3.5 



1-113 



Solve the following system of three equations 
with three unknowns using EES: 



2x — y + z 

3x 2 + 2y 

xy + 2z 



5 

z + 2 



1-114 



[tt3 Solve the following system of three equations 
with three unknowns using EES: 



1 



Special Topic: Thermal Comfort 

1-115C What is metabolism? What is the range of metabolic 
rate for an average man? Why are we interested in metabolic 



x 2 y - 


- z 


3y ' 5 + 


xz 


x + y - 


- z 



cen58933_ch01.qxd 9/10/2002 8:30 AM Page 56 



56 
HEAT TRANSFER 



rate of the occupants of a building when we deal with heating 
and air conditioning? 

1-116C Why is the metabolic rate of women, in general, 
lower than that of men? What is the effect of clothing on the 
environmental temperature that feels comfortable? 

1-117C What is asymmetric thermal radiation? How does it 
cause thermal discomfort in the occupants of a room? 

1-118C How do (a) draft and (b) cold floor surfaces cause 
discomfort for a room's occupants? 

1-119C What is stratification? Is it likely to occur at places 
with low or high ceilings? How does it cause thermal discom- 
fort for a room's occupants? How can stratification be pre- 
vented? 

1-120C Why is it necessary to ventilate buildings? What is 
the effect of ventilation on energy consumption for heating in 
winter and for cooling in summer? Is it a good idea to keep the 
bathroom fans on all the time? Explain. 

Review Problems 

1-121 2.5 kg of liquid water initially at 18°C is to be heated 
to 96°C in a teapot equipped with a 1200-W electric heating 
element inside. The teapot is 0.8 kg and has an average specific 
heat of 0.6 kJ/kg ■ °C. Taking the specific heat of water to be 
4.18 kJ/kg • °C and disregarding any heat loss from the teapot, 
determine how long it will take for the water to be heated. 

1-122 A 4-m-long section of an air heating system of a house 
passes through an unheated space in the attic. The inner diam- 
eter of the circular duct of the heating system is 20 cm. Hot air 
enters the duct at 100 kPa and 65 °C at an average velocity of 
3 m/s. The temperature of the air in the duct drops to 60°C as a 
result of heat loss to the cool space in the attic. Determine the 
rate of heat loss from the air in the duct to the attic under steady 
conditions. Also, determine the cost of this heat loss per hour if 
the house is heated by a natural gas furnace having an effi- 
ciency of 82 percent, and the cost of the natural gas in that area 
is $0.58/therm (1 therm = 105,500 kJ). 
Answers: 0.488 kJ/s, $0.012/h 



4 m 



65 °C 



3 m/s 



Hot air 




FIGURE P1-122 



1-123 



Reconsider Problem 1-122. Using EES (or 
other) software, plot the cost of the heat loss per 

hour as a function of the average air velocity in the range of 

1 m/s to 10 m/s. Discuss the results. 

1-124 Water flows through a shower head steadily at a rate 
of 10 L/min. An electric resistance heater placed in the water 
pipe heats the water from 16°C to 43°C. Taking the density of 




Resistance 
heater 




FIGURE P1-1 24 

water to be 1 kg/L, determine the electric power input to the 
heater, in kW. 

In an effort to conserve energy, it is proposed to pass the 
drained warm water at a temperature of 39°C through a heat 
exchanger to preheat the incoming cold water. If the heat ex- 
changer has an effectiveness of 0.50 (that is, it recovers only 
half of the energy that can possibly be transferred from the 
drained water to incoming cold water), determine the electric 
power input required in this case. If the price of the electric en- 
ergy is 8.5 0/kWh, determine how much money is saved during 
a 10-minute shower as a result of installing this heat exchanger. 

Answers: 18.8 kW, 10.8 kW, $0.0113 

1-125 It is proposed to have a water heater that consists of an 
insulated pipe of 5 cm diameter and an electrical resistor in- 
side. Cold water at 15°C enters the heating section steadily at a 
rate of 18 L/min. If water is to be heated to 50°C, determine 
(a) the power rating of the resistance heater and (b) the average 
velocity of the water in the pipe. 

1-126 A passive solar house that is losing heat to the out- 
doors at an average rate of 50,000 kJ/h is maintained at 22°C at 
all times during a winter night for 10 hours. The house is to be 
heated by 50 glass containers each containing 20 L of water 
heated to 80°C during the day by absorbing solar energy. 
A thermostat-controlled 15-kW back-up electric resistance 
heater turns on whenever necessary to keep the house at 22°C. 

(a) How long did the electric heating system run that night? 

(b) How long would the electric heater have run that night if 
the house incorporated no solar heating? 

Answers: (a) 4.77 h, (b) 9.26 h 

1-127 It is well known that wind makes the cold air feel 
much colder as a result of the windchill effect that is due to the 
increase in the convection heat transfer coefficient with in- 
creasing air velocity. The windchill effect is usually expressed 
in terms of the windchill factor, which is the difference be- 
tween the actual air temperature and the equivalent calm-air 



cen58933_ch01.qxd 9/10/2002 8:30 AM Page 57 



50,000 kJ/h 




22°C 




FIGURE P1 126 

temperature. For example, a windchill factor of 20°C for an 
actual air temperature of 5°C means that the windy air at 5°C 
feels as cold as the still air at — 15°C. In other words, a person 
will lose as much heat to air at 5°C with a windchill factor of 
20°C as he or she would in calm air at — 15°C. 

For heat transfer purposes, a standing man can be modeled 
as a 30-cm -diameter, 170-cm-long vertical cylinder with both 
the top and bottom surfaces insulated and with the side surface 
at an average temperature of 34 °C. For a convection heat trans- 
fer coefficient of 15 W/m 2 • °C, determine the rate of heat loss 
from this man by convection in still air at 20°C. What would 
your answer be if the convection heat transfer coefficient is in- 
creased to 50 W/m 2 • °C as a result of winds? What is the wind- 
chill factor in this case? Answers: 336 W, 1120 W, 32.7°C 

1-128 A thin metal plate is insulated on the back and ex- 
posed to solar radiation on the front surface. The exposed sur- 
face of the plate has an absorptivity of 0.7 for solar radiation. If 
solar radiation is incident on the plate at a rate of 700 W/m 2 




57 
CHAPTER 1 



and the surrounding air temperature is 10°C, determine the sur- 
face temperature of the plate when the heat loss by convection 
equals the solar energy absorbed by the plate. Take the convec- 
tion heat transfer coefficient to be 30 W/m 2 • °C, and disregard 
any heat loss by radiation. 

1-129 A 4-m X 5-m X 6-m room is to be heated by one ton 
(1000 kg) of liquid water contained in a tank placed in the 
room. The room is losing heat to the outside at an average rate 
of 10,000 kJ/h. The room is initially at 20°C and 100 kPa, and 
is maintained at an average temperature of 20°C at all times. If 
the hot water is to meet the heating requirements of this room 
for a 24-hour period, determine the minimum temperature of 
the water when it is first brought into the room. Assume con- 
stant specific heats for both air and water at room temperature. 
Answer: 77.4°C 

1-130 Consider a 3-m X 3-m X 3-m cubical furnace whose 
top and side surfaces closely approximate black surfaces at a 
temperature of 1200 K. The base surface has an emissivity of 
e = 0.7, and is maintained at 800 K. Determine the net rate 
of radiation heat transfer to the base surface from the top and 
side surfaces. Answer: 594,400 W 

1-131 Consider a refrigerator whose dimensions are 1 .8 m X 
1.2 m X 0.8 m and whose walls are 3 cm thick. The refrigera- 
tor consumes 600 W of power when operating and has a COP 
of 2.5. It is observed that the motor of the refrigerator remains 
on for 5 minutes and then is off for 15 minutes periodically. If 
the average temperatures at the inner and outer surfaces of the 
refrigerator are 6°C and 17°C, respectively, determine the av- 
erage thermal conductivity of the refrigerator walls. Also, de- 
termine the annual cost of operating this refrigerator if the unit 
cost of electricity is $0.08/kWh. 




FIGURE P1-1 28 



FIGURE P1-131 

1-132 A 0.2-L glass of water at 20°C is to be cooled with 
ice to 5°C. Determine how much ice needs to be added to 
the water, in grams, if the ice is at 0°C. Also, determine how 
much water would be needed if the cooling is to be done with 
cold water at 0°C. The melting temperature and the heat of 
fusion of ice at atmospheric pressure are 0°C and 333.7 kJ/kg, 
respectively, and the density of water is 1 kg/L. 



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58 
HEAT TRANSFER 




FIGURE P1-132 



1-133 Tu'M Reconsider Problem 1-132. Using EES (or 
I^S other) software, plot the amount of ice that 
needs to be added to the water as a function of the ice temper- 
ature in the range of — 24°C to 0°C. Discuss the results. 

1-134E In order to cool 1 short ton (2000 lbm) of water at 
70°F in a tank, a person pours 160 lbm of ice at 25°F into the 
water. Determine the final equilibrium temperature in the tank. 
The melting temperature and the heat of fusion of ice at atmo- 
spheric pressure are 32°F and 143.5 Btu/lbm, respectively. 
Answer: 56.3T 

1-135 Engine valves (C p = 440 J/kg • °C and p = 7840 
kg/m 3 ) are to be heated from 40°C to 800°C in 5 minutes in the 
heat treatment section of a valve manufacturing facility. The 
valves have a cylindrical stem with a diameter of 8 mm and a 
length of 10 cm. The valve head and the stem may be assumed 
to be of equal surface area, with a total mass of 0.0788 kg. For 
a single valve, determine (a) the amount of heat transfer, 
(b) the average rate of heat transfer, and (c) the average heat 
flux, (d) the number of valves that can be heat treated per day if 
the heating section can hold 25 valves, and it is used 10 hours 
per day. 

1-136 The hot water needs of a household are met by an 
electric 60-L hot water tank equipped with a 1 .6-kW heating 
element. The tank is initially filled with hot water at 80°C, and 
the cold water temperature is 20°C. Someone takes a shower 
by mixing constant flow rates of hot and cold waters. After a 
showering period of 8 minutes, the average water temperature 
in the tank is measured to be 60°C. The heater is kept on during 
the shower and hot water is replaced by cold water. If the cold 
water is mixed with the hot water stream at a rate of 0.06 kg/s, 
determine the flow rate of hot water and the average tempera- 
ture of mixed water used during the shower. 

1-137 Consider a flat plate solar collector placed at the roof 
of a house. The temperatures at the inner and outer surfaces of 
glass cover are measured to be 28°C and 25°C, respectively. 
The glass cover has a surface area of 2.2. m 2 and a thickness of 



0.6 cm and a thermal conductivity of 0.7 W/m • C. Heat is lost 
from the outer surface of the cover by convection and radiation 
with a convection heat transfer coefficient of 10 W/m 2 • °C and 
an ambient temperature of 15°C. Determine the fraction of heat 
lost from the glass cover by radiation. 

1-138 The rate of heat loss through a unit surface area of 
a window per unit temperature difference between the in- 
doors and the outdoors is called the {/-factor. The value of 
the [/-factor ranges from about 1.25 W/m 2 ■ °C (or 0.22 
Btu/h ■ ft 2 • °F) for low-e coated, argon-filled, quadruple -pane 
windows to 6.25 W/m 2 ■ °C (or 1.1 Btu/h ■ ft 2 • °F) for a single- 
pane window with aluminum frames. Determine the range for 
the rate of heat loss through a 1.2-m X 1.8-m window of a 
house that is maintained at 20°C when the outdoor air temper- 
ature is -8°C. 




Indoors 
20°Cfc 



M. 




Outdoors 



C C 



FIGURE P1-138 



1-139 



Reconsider Problem 1-138. Using EES (or 
other) software, plot the rate of heat loss 

through the window as a function of the [/-factor. Discuss 

the results. 

Design and Essay Problems 

1-140 Write an essay on how microwave ovens work, and 
explain how they cook much faster than conventional ovens. 
Discuss whether conventional electric or microwave ovens 
consume more electricity for the same task. 

1-141 Using information from the utility bills for the coldest 
month last year, estimate the average rate of heat loss from 
your house for that month. In your analysis, consider the con- 
tribution of the internal heat sources such as people, lights, and 
appliances. Identify the primary sources of heat loss from your 
house and propose ways of improving the energy efficiency of 
your house. 

1-142 Design a 1200-W electric hair dryer such that the air 
temperature and velocity in the dryer will not exceed 50°C and 
3/ms, respectively. 

1-143 Design an electric hot water heater for a family of 
four in your area. The maximum water temperature in the tank 



cen58933_ch01.qxd 9/10/2002 8:30 AM Page 59 



and the power consumption are not to exceed 60°C and 4 kW, 
respectively. There are two showers in the house, and the 
flow rate of water through each of the shower heads is about 
10 L/min. Each family member takes a 5 -minute shower every 
morning. Explain why a hot water tank is necessary, and deter- 
mine the proper size of the tank for this family. 

1-144 Conduct this experiment to determine the heat transfer 
coefficient between an incandescent lightbulb and the sur- 
rounding air using a 60-W lightbulb. You will need an indoor- 
outdoor thermometer, which can be purchased for about $ 1 in 



59 
CHAPTER 1 



a hardware store, and a metal glue. You will also need a piece 
of string and a ruler to calculate the surface area of the light- 
bulb. First, measure the air temperature in the room, and then 
glue the tip of the thermocouple wire of the thermometer to the 
glass of the lightbulb. Turn the light on and wait until the tem- 
perature reading stabilizes. The temperature reading will give 
the surface temperature of the lightbulb. Assuming 10 percent 
of the rated power of the bulb is converted to light, calculate 
the heat transfer coefficient from Newton's law of cooling. 



cen58933_ch01.qxd 9/10/2002 8:30 AM Page 60 



cen58933_ch02.qxd 9/10/2002 8:46 AM Page 61 



HEAT CONDUCTION 
EOUATION 



CHAPTER 



Heat transfer has direction as well as magnitude. The rate of heat con- 
duction in a specified direction is proportional to the temperature gra- 
dient, which is the change in temperature per unit length in that 
direction. Heat conduction in a medium, in general, is three-dimensional and 
time dependent. That is, T = T(x, y, z, t) and the temperature in a medium 
varies with position as well as time. Heat conduction in a medium is said to be 
steady when the temperature does not vary with time, and unsteady or tran- 
sient when it does. Heat conduction in a medium is said to be one-dimensional 
when conduction is significant in one dimension only and negligible in the 
other two dimensions, two-dimensional when conduction in the third dimen- 
sion is negligible, and three-dimensional when conduction in all dimensions 
is significant. 

We start this chapter with a description of steady, unsteady, and multi- 
dimensional heat conduction. Then we derive the differential equation that 
governs heat conduction in a large plane wall, a long cylinder, and a sphere, 
and generalize the results to three-dimensional cases in rectangular, cylin- 
drical, and spherical coordinates. Following a discussion of the boundary con- 
ditions, we present the formulation of heat conduction problems and their 
solutions. Finally, we consider heat conduction problems with variable ther- 
mal conductivity. 

This chapter deals with the theoretical and mathematical aspects of heat 
conduction, and it can be covered selectively, if desired, without causing a 
significant loss in continuity. The more practical aspects of heat conduction 
are covered in the following two chapters. 



CONTENTS 

2-1 Introduction 62 

1-1 One-Dimensional Heat 

Conduction Equation 68 

2-3 General Heat 

Conduction Equation 74 

2-4 Boundary and 

Initial Conditions 77 

2-5 Solution of Steady 

One-Dimensional Heat 
Conduction Problems 86 

2-6 Heat Generation in a Solid 97 

1-1 Variable Thermal 

Conductivity k(T) 104 

Topic of Special Interest: 

A Brief Review of 
Differential Equations 107 



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62 
HEAT TRANSFER 




Magnitude of 
temperature 
at a point A 
(no direction) 

80 W/m 2 



Magnitude and 
direction of heat 
flux at the same 
point 



FIGURE 2-1 

Heat transfer has direction as well 
as magnitude, and thus it is 
a vector quantity. 



500 W 













^►•2 = 


Hot 
edium 




Cold 
medium 







L 













■ e=- 


Cold 
tedium 




Hot 
medium 







L 



500 W 



FIGURE 2-2 

Indicating direction for heat transfer 
(positive in the positive direction; 
negative in the negative direction). 



2-1 - INTRODUCTION 

In Chapter 1 heat conduction was defined as the transfer of thermal energy 
from the more energetic particles of a medium to the adjacent less energetic 
ones. It was stated that conduction can take place in liquids and gases as well 
as solids provided that there is no bulk motion involved. 

Although heat transfer and temperature are closely related, they are of a dif- 
ferent nature. Unlike temperature, heat transfer has direction as well as mag- 
nitude, and thus it is a vector quantity (Fig. 2-1). Therefore, we must specify 
both direction and magnitude in order to describe heat transfer completely at 
a point. For example, saying that the temperature on the inner surface of a 
wall is 18°C describes the temperature at that location fully. But saying that 
the heat flux on that surface is 50 W/m 2 immediately prompts the question "in 
what direction?" We can answer this question by saying that heat conduction 
is toward the inside (indicating heat gain) or toward the outside (indicating 
heat loss). 

To avoid such questions, we can work with a coordinate system and indicate 
direction with plus or minus signs. The generally accepted convention is that 
heat transfer in the positive direction of a coordinate axis is positive and in the 
opposite direction it is negative. Therefore, a positive quantity indicates heat 
transfer in the positive direction and a negative quantity indicates heat trans- 
fer in the negative direction (Fig. 2-2). 

The driving force for any form of heat transfer is the temperature difference, 
and the larger the temperature difference, the larger the rate of heat transfer. 
Some heat transfer problems in engineering require the determination of the 
temperature distribution (the variation of temperature) throughout the 
medium in order to calculate some quantities of interest such as the local heat 
transfer rate, thermal expansion, and thermal stress at some critical locations 
at specified times. The specification of the temperature at a point in a medium 
first requires the specification of the location of that point. This can be done 
by choosing a suitable coordinate system such as the rectangular, cylindrical, 
or spherical coordinates, depending on the geometry involved, and a conve- 
nient reference point (the origin). 

The location of a point is specified as (x, y, z) in rectangular coordinates, as 
(r, 4>, z) in cylindrical coordinates, and as (r, <j>, 6) in spherical coordinates, 
where the distances x, y, z, and r and the angles cj) and 8 are as shown in Fig- 
ure 2-3. Then the temperature at a point (x, y, z) at time t in rectangular coor- 
dinates is expressed as T(x, y, z, t). The best coordinate system for a given 
geometry is the one that describes the surfaces of the geometry best. 
For example, a parallelepiped is best described in rectangular coordinates 
since each surface can be described by a constant value of the x-, y-, or 
z-coordinates. A cylinder is best suited for cylindrical coordinates since its lat- 
eral surface can be described by a constant value of the radius. Similarly, the 
entire outer surface of a spherical body can best be described by a constant 
value of the radius in spherical coordinates. For an arbitrarily shaped body, we 
normally use rectangular coordinates since it is easier to deal with distances 
than with angles. 

The notation just described is also used to identify the variables involved 
in a heat transfer problem. For example, the notation T(x, y, z, implies that 
the temperature varies with the space variables x, y, and z as well as time. The 



cen58933_ch02.qxd 9/10/2002 8:46 AM Page 63 



63 
CHAPTER 2 





(a) Rectangular coordinates (b) Cylindrical coordinates 



(c) Spherical coordinates 



FIGURE 2-3 

The various distances 
and angles involved when 
describing the location of a point 
in different coordinate systems. 



notation T{x), on the other hand, indicates that the temperature varies in the 
x-direction only and there is no variation with the other two space coordinates 
or time. 

Steady versus Transient Heat Transfer 

Heat transfer problems are often classified as being steady (also called steady- 
state) or transient (also called unsteady). The term steady implies no change 
with time at any point within the medium, while transient implies variation 
with time or time dependence. Therefore, the temperature or heat flux remains 
unchanged with time during steady heat transfer through a medium at any lo- 
cation, although both quantities may vary from one location to another 
(Fig. 2-4). For example, heat transfer through the walls of a house will be 
steady when the conditions inside the house and the outdoors remain constant 
for several hours. But even in this case, the temperatures on the inner and 
outer surfaces of the wall will be different unless the temperatures inside and 
outside the house are the same. The cooling of an apple in a refrigerator, on 
the other hand, is a transient heat transfer process since the temperature at any 
fixed point within the apple will change with time during cooling. During 
transient heat transfer, the temperature normally varies with time as well as 
position. In the special case of variation with time but not with position, the 
temperature of the medium changes uniformly with time. Such heat transfer 
systems are called lumped systems. A small metal object such as a thermo- 
couple junction or a thin copper wire, for example, can be analyzed as a 
lumped system during a heating or cooling process. 

Most heat transfer problems encountered in practice are transient in nature, 
but they are usually analyzed under some presumed steady conditions since 
steady processes are easier to analyze, and they provide the answers to our 
questions. For example, heat transfer through the walls and ceiling of a typi- 
cal house is never steady since the outdoor conditions such as the temperature, 
the speed and direction of the wind, the location of the sun, and so on, change 
constantly. The conditions in a typical house are not so steady either. There- 
fore, it is almost impossible to perform a heat transfer analysis of a house ac- 
curately. But then, do we really need an in-depth heat transfer analysis? If the 



Time = 2 PM 



Time = 5 PM 



15°C 



7°C 12°C 



5'C 



■e,*e, 



(a) Transient 



15°C 



7°C 15°C 



7°C 



■e, = e, 



(b) Steady-state 

FIGURE 2-4 

Steady and transient heat 
conduction in a plane wall. 



cen58933_ch02.qxd 9/10/2002 8:46 AM Page 64 



64 
HEAT TRANSFER 



••65°C 




FIGURE 2-5 

Two-dimensional heat transfer 
in a long rectangular bar. 




Primary 
direction of 
heat transfer 



FIGURE 2-6 

Heat transfer through the window 
of a house can be taken to be 
one-dimensional. 



purpose of a heat transfer analysis of a house is to determine the proper size of 
a heater, which is usually the case, we need to know the maximum rate of heat 
loss from the house, which is determined by considering the heat loss from the 
house under worst conditions for an extended period of time, that is, during 
steady operation under worst conditions. Therefore, we can get the answer to 
our question by doing a heat transfer analysis under steady conditions. If the 
heater is large enough to keep the house warm under the presumed worst con- 
ditions, it is large enough for all conditions. The approach described above is 
a common practice in engineering. 

Multidimensional Heat Transfer 

Heat transfer problems are also classified as being one-dimensional, two- 
dimensional, or three-dimensional, depending on the relative magnitudes of 
heat transfer rates in different directions and the level of accuracy desired. In 
the most general case, heat transfer through a medium is three-dimensional. 
That is, the temperature varies along all three primary directions within the 
medium during the heat transfer process. The temperature distribution 
throughout the medium at a specified time as well as the heat transfer rate at 
any location in this general case can be described by a set of three coordinates 
such as the x, y, and z in the rectangular (or Cartesian) coordinate system; 
the r, <|>, and z in the cylindrical coordinate system; and the r, <j), and in the 
spherical (or polar) coordinate system. The temperature distribution in this 
case is expressed as T(x, y, z, t), T(r, 4>, z, t), and T(r, 4>, 9, t) in the respective 
coordinate systems. 

The temperature in a medium, in some cases, varies mainly in two primary 
directions, and the variation of temperature in the third direction (and thus 
heat transfer in that direction) is negligible. A heat transfer problem in that 
case is said to be two-dimensional. For example, the steady temperature dis- 
tribution in a long bar of rectangular cross section can be expressed as T(x, y) 
if the temperature variation in the z-direction (along the bar) is negligible and 
there is no change with time (Fig. 2-5). 

A heat transfer problem is said to be one-dimensional if the temperature in 
the medium varies in one direction only and thus heat is transferred in one di- 
rection, and the variation of temperature and thus heat transfer in other direc- 
tions are negligible or zero. For example, heat transfer through the glass of a 
window can be considered to be one-dimensional since heat transfer through 
the glass will occur predominantly in one direction (the direction normal to 
the surface of the glass) and heat transfer in other directions (from one side 
edge to the other and from the top edge to the bottom) is negligible (Fig. 2-6). 
Likewise, heat transfer through a hot water pipe can be considered to be one- 
dimensional since heat transfer through the pipe occurs predominantly in the 
radial direction from the hot water to the ambient, and heat transfer along the 
pipe and along the circumference of a cross section (z- and cjvdirections) is 
typically negligible. Heat transfer to an egg dropped into boiling water is also 
nearly one-dimensional because of symmetry. Heat will be transferred to the 
egg in this case in the radial direction, that is, along straight lines passing 
through the midpoint of the egg. 

We also mentioned in Chapter 1 that the rate of heat conduction through a 
medium in a specified direction (say, in the A-direction) is proportional to the 
temperature difference across the medium and the area normal to the direction 



cen58933_ch02.qxd 9/10/2002 8:46 AM Page 65 



of heat transfer, but is inversely proportional to the distance in that direction. 
This was expressed in the differential form by Fourier's law of heat conduc- 
tion for one-dimensional heat conduction as 



2 c 



-kA 



dT 
dx 



(W) 



(2-1) 



where k is the thermal conductivity of the material, which is a measure of the 
ability of a material to conduct heat, and dT/dx is the temperature gradient, 
which is the slope of the temperature curve on a T-x diagram (Fig. 2-7). The 
thermal conductivity of a material, in general, varies with temperature. But 
sufficiently accurate results can be obtained by using a constant value for 
thermal conductivity at the average temperature. 

Heat is conducted in the direction of decreasing temperature, and thus 
the temperature gradient is negative when heat is conducted in the positive 
x-direction. The negative sign in Eq. 2-1 ensures that heat transfer in the posi- 
tive x-direction is a positive quantity. 

To obtain a general relation for Fourier's law of heat conduction, consider a 
medium in which the temperature distribution is three-dimensional. Figure 
2-8 shows an isothermal surface in that medium. The heat flux vector at a 
point P on this surface must be perpendicular to the surface, and it must point 
in the direction of decreasing temperature. If n is the normal of the isothermal 
surface at point P, the rate of heat conduction at that point can be expressed by 
Fourier's law as 



Qn 



-kA 



BT 
dn 



(W) 



(2-2) 



65 
CHAPTER 2 




FIGURE 2-7 

The temperature gradient dT/dx is 

simply the slope of the temperature 

curve on a T-x diagram. 



In rectangular coordinates, the heat conduction vector can be expressed in 
terms of its components as 



Qn = QJ +Qyj +Q-J 



(2-3) 



where i, j, and k are the unit vectors, and Q x , Q y , and Q z are the magnitudes 
of the heat transfer rates in the x-, y-, and z-directions, which again can be de- 
termined from Fourier's law as 



Q x = -kA x 



dT 

dx' 



Q y = ~kA y 



dT 

By' 



and Q. 



-kA 



BT 
z Bz 



(2-4) 



Here A x , A y and A z are heat conduction areas normal to the x-, y-, and 
Z-directions, respectively (Fig. 2-8). 

Most engineering materials are isotropic in nature, and thus they have the 
same properties in all directions. For such materials we do not need to be con- 
cerned about the variation of properties with direction. But in anisotropic ma- 
terials such as the fibrous or composite materials, the properties may change 
with direction. For example, some of the properties of wood along the grain 
are different than those in the direction normal to the grain. In such cases the 
thermal conductivity may need to be expressed as a tensor quantity to account 
for the variation with direction. The treatment of such advanced topics is be- 
yond the scope of this text, and we will assume the thermal conductivity of a 
material to be independent of direction. 




FIGURE 2-8 

The heat transfer vector is 

always normal to an isothermal 

surface and can be resolved into its 

components like any other vector. 



cen58933_ch02.qxd 9/10/2002 8:46 AM Page 66 



66 
HEAT TRANSFER 




FIGURE 2-9 

Heat is generated in the heating coils 
of an electric range as a result of the 
conversion of electrical energy to heat. 






-^tl Sun 



Solar 
radiation 



X 


' 


Wh 






IV 


- Solar energy 




■T 


absorbed by 




■V 


water 


Water 


1+ 




1 


L «(*) = 


's, absorbed^ ' 



FIGURE 2-10 

The absorption of solar radiation by 
water can be treated as heat 
generation. 



Heat Generation 

A medium through which heat is conducted may involve the conversion of 
electrical, nuclear, or chemical energy into heat (or thermal) energy. In heat 
conduction analysis, such conversion processes are characterized as heat 
generation. 

For example, the temperature of a resistance wire rises rapidly when elec- 
tric current passes through it as a result of the electrical energy being con- 
verted to heat at a rate of I 2 R, where / is the current and R is the electrical 
resistance of the wire (Fig. 2-9). The safe and effective removal of this heat 
away from the sites of heat generation (the electronic circuits) is the subject 
of electronics cooling, which is one of the modern application areas of heat 
transfer. 

Likewise, a large amount of heat is generated in the fuel elements of nuclear 
reactors as a result of nuclear fission that serves as the heat source for the nu- 
clear power plants. The natural disintegration of radioactive elements in nu- 
clear waste or other radioactive material also results in the generation of heat 
throughout the body. The heat generated in the sun as a result of the fusion of 
hydrogen into helium makes the sun a large nuclear reactor that supplies heat 
to the earth. 

Another source of heat generation in a medium is exothermic chemical re- 
actions that may occur throughout the medium. The chemical reaction in this 
case serves as a heat source for the medium. In the case of endothermic reac- 
tions, however, heat is absorbed instead of being released during reaction, and 
thus the chemical reaction serves as a heat sink. The heat generation term be- 
comes a negative quantity in this case. 

Often it is also convenient to model the absorption of radiation such as so- 
lar energy or gamma rays as heat generation when these rays penetrate deep 
into the body while being absorbed gradually. For example, the absorption of 
solar energy in large bodies of water can be treated as heat generation 
throughout the water at a rate equal to the rate of absorption, which varies 
with depth (Fig. 2-10). But the absorption of solar energy by an opaque body 
occurs within a few microns of the surface, and the solar energy that pene- 
trates into the medium in this case can be treated as specified heat flux on the 
surface. 

Note that heat generation is a volumetric phenomenon. That is, it occurs 
throughout the body of a medium. Therefore, the rate of heat generation in a 
medium is usually specified per unit volume and is denoted by g, whose unit 
is W/m 3 or Btu/h • ft 3 . 

The rate of heat generation in a medium may vary with time as well as po- 
sition within the medium. When the variation of heat generation with position 
is known, the total rate of heat generation in a medium of volume V can be de- 
termined from 



l#v 



(W) 



(2-5) 



In the special case of uniform heat generation, as in the case of electric resis- 
tance heating throughout a homogeneous material, the relation in Eq. 2-5 
reduces to G = gV, where g is the constant rate of heat generation per unit 
volume. 



cen58933_ch02.qxd 9/10/2002 8:46 AM Page 67 



67 
CHAPTER 2 



EXAMPLE 2-1 Heat Gain by a Refrigerator 

In order to size the compressor of a new refrigerator, it is desired to determine 
the rate of heat transfer from the kitchen air into the refrigerated space through 
the walls, door, and the top and bottom section of the refrigerator (Fig. 2-11). 
In your analysis, would you treat this as a transient or steady-state heat transfer 
problem? Also, would you consider the heat transfer to be one-dimensional or 
multidimensional? Explain. 

SOLUTION The heat transfer process from the kitchen air to the refrigerated 
space is transient in nature since the thermal conditions in the kitchen and the 
refrigerator, in general, change with time. However, we would analyze this prob- 
lem as a steady heat transfer problem under the worst anticipated conditions 
such as the lowest thermostat setting for the refrigerated space, and the antic- 
ipated highest temperature in the kitchen (the so-called design conditions). If 
the compressor is large enough to keep the refrigerated space at the desired 
temperature setting under the presumed worst conditions, then it is large 
enough to do so under all conditions by cycling on and off. 

Heat transfer into the refrigerated space is three-dimensional in nature since 
heat will be entering through all six sides of the refrigerator. However, heat 
transfer through any wall or floor takes place in the direction normal to the sur- 
face, and thus it can be analyzed as being one-dimensional. Therefore, this 
problem can be simplified greatly by considering the heat transfer to be one- 
dimensional at each of the four sides as well as the top and bottom sections, 
and then by adding the calculated values of heat transfer at each surface. 



Heat transfer 



Hi 



nz\ 



B 




FIGURE 2-1 1 

Schematic for Example 2-1 . 



EXAMPLE 2-2 Heat Generation in a Hair Dryer 

The resistance wire of a 1200-W hair dryer is 80 cm long and has a diameter of 
D = 0.3 cm (Fig. 2-12). Determine the rate of heat generation in the wire per 
unit volume, in W/cm 3 , and the heat flux on the outer surface of the wire as a 
result of this heat generation. 

SOLUTION The power consumed by the resistance wire of a hair dryer is given. 
The heat generation and the heat flux are to be determined. 
Assumptions Heat is generated uniformly in the resistance wire. 
Analysis A 1200-W hair dryer will convert electrical energy into heat in the 
wire at a rate of 1200 W. Therefore, the rate of heat generation in a resistance 
wire is equal to the power consumption of a resistance heater. Then the rate of 
heat generation in the wire per unit volume is determined by dividing the total 
rate of heat generation by the volume of the wire, 



1200 W 



V„ 



(ttD 2 /4)L [tt(0.3 cm) 2 /4](80 cm) 



212 W/cm 3 



Similarly, heat flux on the outer surface of the wire as a result of this heat gen- 
eration is determined by dividing the total rate of heat generation by the surface 
area of the wire, 



1200 W 



A wire ttDL 



tt(0.3 cm)(80 cm) 



15.9 W/cm 2 



Hair dryer 
1200 W 




FIGURE 2-12 

Schematic for Example 2-2. 



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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 2-13 

One-dimensional heat conduction 
through a volume element 
in a large plane wall. 



2-2 - ONE-DIMENSIONAL 

HEAT CONDUCTION EQUATION 

Consider heat conduction through a large plane wall such as the wall of a 
house, the glass of a single pane window, the metal plate at the bottom of 
a pressing iron, a cast iron steam pipe, a cylindrical nuclear fuel element, 
an electrical resistance wire, the wall of a spherical container, or a spherical 
metal ball that is being quenched or tempered. Heat conduction in these 
and many other geometries can be approximated as being one-dimensional 
since heat conduction through these geometries will be dominant in one 
direction and negligible in other directions. Below we will develop the one- 
dimensional heat conduction equation in rectangular, cylindrical, and spheri- 
cal coordinates. 

Heat Conduction Equation in a Large Plane Wall 

Consider a thin element of thickness Ax in a large plane wall, as shown in Fig- 
ure 2-13. Assume the density of the wall is p, the specific heat is C, and the 
area of the wall normal to the direction of heat transfer is A. An energy bal- 
ance on this thin element during a small time interval At can be expressed as 



/Rate of heat\ 

conduction 
I atjc / 



or 



/Rate of heat\ 

conduction 
\ at x + Ax J 



\lx x!x + Ax 



/ Rate of heat \ 

generation 

inside the 

element 



/ Rate of change \ 

of the energy 

content of the 

element 



A£„ 



At 



(2-6) 



But the change in the energy content of the element and the rate of heat gen- 
eration within the element can be expressed as 



A£eien,e„t = E, + Af - E, = mC(T, + A , - T,) = pCAAxiT, + Al - T,) 

^element — § 'element — §Al±X 



(2-7) 
(2-8) 



Substituting into Equation 2-6, we get 

Q x -Q I + tu + gAAx = pCAAx- 

Dividing by AAx gives 



1 2.V + A.V Qx , . _ T, + \, 

+ g = pC 



At 



Ax 



At 



(2-9) 



(2-10) 



cen58933_ch02.qxd 9/10/2002 8:46 AM Page 69 



Taking the limit as Ax — > and At — > yields 



1 d Ua dT \ j- ■ r dT 



(2-11) 



69 
CHAPTER 2 



since, from the definition of the derivative and Fourier's law of heat conduc- 
tion, 

e, + A, - e, dQ 



lim 

Ax-> 



Ax 



dx dx \ dx 



(2-12) 



Noting that the area A is constant for a plane wall, the one-dimensional tran- 
sient heat conduction equation in a plane wall becomes 



Variable conductivity: 



d , dT\ , . n dT 



(2-13) 



The thermal conductivity k of a material, in general, depends on the tempera- 
ture T (and therefore x), and thus it cannot be taken out of the derivative. 
However, the thermal conductivity in most practical applications can be as- 
sumed to remain constant at some average value. The equation above in that 
case reduces to 



Constant conductivity: 



PT 
dx 2 



]_dT 
a dt 



(2-14) 



where the property a = k/pC is the thermal diffusivity of the material and 
represents how fast heat propagates through a material. It reduces to the fol- 
lowing forms under specified conditions (Fig. 2-14): 



(1) Steady-state: 
(d/dt = 0) 

(2) Transient, no heat generation: 
(8 = 0) 

(3) Steady-state, no heat generation: 
{d/dt = and g = 0) 



d 2 T 


S 




dx 2 


k 


d 2 T 


1 dT 


dx 2 


a dt 


d 2 T 






= 


dx 2 









(2-15) 



(2-16) 



(2-17) 



Note that we replaced the partial derivatives by ordinary derivatives in the 
one-dimensional steady heat conduction case since the partial and ordinary 
derivatives of a function are identical when the function depends on a single 
variable only [T = T(x) in this case]. 

Heat Conduction Equation in a Long Cylinder 

Now consider a thin cylindrical shell element of thickness Ar in a long cylin- 
der, as shown in Figure 2-15. Assume the density of the cylinder is p, the spe- 
cific heat is C, and the length is L. The area of the cylinder normal to the 
direction of heat transfer at any location is A = 2irrL where r is the value of 
the radius at that location. Note that the heat transfer area A depends on r in 
this case, and thus it varies with location. An energy balance on this thin 
cylindrical shell element during a small time interval A; can be expressed as 



General, one dimensional: 

No Steady- 
generation state 








Steady, one-dimensional: 
d 2 T 



dx 2 



= 



FIGURE 2-14 

The simplification of the one- 
dimensional heat conduction equation 
in a plane wall for the case of constant 
conductivity for steady conduction 
with no heat generation. 




■ Volume element 

FIGURE 2-15 

One-dimensional heat conduction 

through a volume element 

in a long cylinder. 



cen58933_ch02.qxd 9/10/2002 8:46 AM Page 7C 



70 
HEAT TRANSFER 



' Rate of heat \ 

conduction 
\ at r 



( Rate of heat \ 

conduction + 
1 at r + Ar j 



Rate of heat \ 

generation 

inside the 

element 



/ Rate of change \ 

of the energy 

content of the 

element 



or 



Qr ~ Qr + Ar + G c | e 



A£„ 



Ar 



(2-18) 



The change in the energy content of the element and the rate of heat genera- 
tion within the element can be expressed as 



A£„ 



gV A 



E, = mC(T, + A , - T,) = pCAAr(T, + A , - T,) (2-19) 
= gAAr (2-20) 



Substituting into Eq. 2-18, we get 



Q r -Q r + Ar + gAAr = pCAAr 



At 



(2-21) 



where A = 2ittL. You may be tempted to express the area at the middle of the 
element using the average radius as A = 2ir(r + Ar/2)L. But there is nothing 
we can gain from this complication since later in the analysis we will take the 
limit as Ar — > and thus the term Ar/2 will drop out. Now dividing the equa- 
tion above by AAr gives 



1 8 r+ A, ~ Qr 

A Ar 



P C- 



Ar 



(2-22) 



Taking the limit as Ar — > and Ar — > yields 

Adr( kA Tr +8 = PC V 



(2-23) 



since, from the definition of the derivative and Fourier's law of heat 
conduction, 



lim 

Ar->0 



g,. + Ar " Qr dQ 



d 



Ar 



dr dr 



-kA 



dT 
dr 



(2-24) 



Noting that the heat transfer area in this case is A = 2irrL, the one- 
dimensional transient heat conduction equation in a cylinder becomes 



Variable conductivity: 



1 d / , dT\, 
rTr V k Tr ] 



P C 



dT 

dt 



(2-25) 



For the case of constant thermal conductivity, the equation above reduces to 

1 dT 



Constant conductivity: 



ld_ dT\ .__ 

>' drV dr k a dt 



(2-26) 



cen58933_ch02.qxd 9/10/2002 8:46 AM Page 71 



71 
CHAPTER 2 







where again the property a = k/pC is the thermal diffusivity of the material. 

Equation 2-26 reduces to the following forms under specified conditions 
(Fig. 2-16): 

(1) Steady-state: l_d/ dT\ g 
(d/dt = 0) '" dr \ r dr) k 

(2) Transient, no heat generation: Id/ dT\ _ 1 dT 
(g = 0) r dr\ r dr ) ~~ « dt 

(3) Steady-state, no heat generation: d I dT 
(d/dt = and g = 0) d~r\ r ~dr 



(2-27) 
(2-28) 
(2-29) 



Note that we again replaced the partial derivatives by ordinary derivatives in 
the one-dimensional steady heat conduction case since the partial and ordinary 
derivatives of a function are identical when the function depends on a single 
variable only [T = T(r) in this case]. 

Heat Conduction Equation in a Sphere 

Now consider a sphere with density p, specific heat C, and outer radius R. The 
area of the sphere normal to the direction of heat transfer at any location is 
A = 4ttt 2 , where r is the value of the radius at that location. Note that the heat 
transfer area A depends on r in this case also, and thus it varies with location. 
By considering a thin spherical shell element of thickness Ar and repeating 
the approach described above for the cylinder by using A = 4 tit 2 instead of 
A = 2irrL, the one-dimensional transient heat conduction equation for a 
sphere is determined to be (Fig. 2-17) 



Variable conductivity: 



1 d 



, r l k — 
dr \ dr 



P C 



dT 

dt 



(2-30) 



(a) The form that is ready to integrate 

my-" 

(b) The equivalent alternative form 



d 2 T dT „ 



FIGURE 2-16 

Two equivalent forms of the 
differential equation for the one- 
dimensional steady heat conduction in 
a cylinder with no heat generation. 




FIGURE 2-17 

One-dimensional heat conduction 
through a volume element in a sphere. 



which, in the case of constant thermal conductivity, reduces to 

1 dT 



Constant conductivity: 



L± r 2§T + _. 

r 2 dr\ dr k a dt 



(2-31) 



where again the property a = k/pC is the thermal diffusivity of the material. 
It reduces to the following forms under specified conditions: 



(1) Steady-state: 
(d/dt = 0) 

(2) Transient, 

no heat generation: 
(g = 0) 

(3) Steady-state, 

no heat generation: 
(d/dt = and g = 0) 



11 [ r idT 
r 2 dr\ dr 



\_d_l 2 dT 
r 2 dr\ r dr 



d_ 2 dT 
dr \ dr 



~k = 

\_dT 
a dt 



dr 2 dr 



(2-32) 



(2-33) 



(2-34) 



where again we replaced the partial derivatives by ordinary derivatives in the 
one-dimensional steady heat conduction case. 



cen58933_ch02.qxd 9/10/2002 8:46 AM Page 72 



72 

HEAT TRANSFER 



Combined One-Dimensional 
Heat Conduction Equation 

An examination of the one-dimensional transient heat conduction equations 
for the plane wall, cylinder, and sphere reveals that all three equations can be 
expressed in a compact form as 



d_ 
; dr 



r" k 



dT 
dr 



P C 



dT 

dt 



(2-35) 



where n = for a plane wall, n = 1 for a cylinder, and n = 2 for a sphere. In 
the case of a plane wall, it is customary to replace the variable r by x. This 
equation can be simplified for steady-state or no heat generation cases as 
described before. 




800 W 

FIGURE 2-18 

Schematic for Example 2-3. 



EXAMPLE 2-3 Heat Conduction through the Bottom of a Pan 

Consider a steel pan placed on top of an electric range to cook spaghetti (Fig. 
2-18). The bottom section of the pan is L = 0.4 cm thick and has a diameter 
of D = 18 cm. The electric heating unit on the range top consumes 800 W of 
power during cooking, and 80 percent of the heat generated in the heating ele- 
ment is transferred uniformly to the pan. Assuming constant thermal conduc- 
tivity, obtain the differential equation that describes the variation of the 
temperature in the bottom section of the pan during steady operation. 

SOLUTION The bottom section of the pan has a large surface area relative to 
its thickness and can be approximated as a large plane wall. Heat flux is ap- 
plied to the bottom surface of the pan uniformly, and the conditions on the 
inner surface are also uniform. Therefore, we expect the heat transfer through 
the bottom section of the pan to be from the bottom surface toward the top, 
and heat transfer in this case can reasonably be approximated as being one- 
dimensional. Taking the direction normal to the bottom surface of the pan to be 
the x-axis, we will have T = T(x) during steady operation since the temperature 
in this case will depend on x only. 

The thermal conductivity is given to be constant, and there is no heat gener- 
ation in the medium (within the bottom section of the pan). Therefore, the dif- 
ferential equation governing the variation of temperature in the bottom section 
of the pan in this case is simply Eq. 2-17, 



d 2 T _ 
dx 2 ' 







which is the steady one-dimensional heat conduction equation in rectangular 
coordinates under the conditions of constant thermal conductivity and no heat 
generation. Note that the conditions at the surface of the medium have no ef- 
fect on the differential equation. 



EXAMPLE 2-4 



Heat Conduction in a Resistance Heater 



A 2-kW resistance heater wire with thermal conductivity k = 15 W/m ■ °C, di- 
ameter D = 0.4 cm, and length L = 50 cm is used to boil water by immersing 



cen58933_ch02.qxd 9/10/2002 8:46 AM Page 73 



it in water (Fig. 2-19). Assuming the variation of the thermal conductivity of the 
wire with temperature to be negligible, obtain the differential equation that de- 
scribes the variation of the temperature in the wire during steady operation. 

SOLUTION The resistance wire can be considered to be a very long cylinder 
since its length is more than 100 times its diameter. Also, heat is generated 
uniformly in the wire and the conditions on the outer surface of the wire are uni- 
form. Therefore, it is reasonable to expect the temperature in the wire to vary in 
the radial r direction only and thus the heat transfer to be one-dimensional. 
Then we will have T = T(r) during steady operation since the temperature in 
this case will depend on ronly. 

The rate of heat generation in the wire per unit volume can be determined 
from 



G 



G 



2000 W 



v„ 



(ttD 2 /4)L [tt(0.004 m) 2 /4](0.5 cm) 



0.318 X 10"W/m 3 



Noting that the thermal conductivity is given to be constant, the differential 
equation that governs the variation of temperature in the wire is simply 
Eq. 2-27, 



1 d I dT 



r dr\ dr 







which is the steady one-dimensional heat conduction equation in cylindrical co- 
ordinates for the case of constant thermal conductivity. Note again that the con- 
ditions at the surface of the wire have no effect on the differential equation. 



73 
CHAPTER 2 




FIGURE 2-19 

Schematic for Example 2-4. 



EXAMPLE 2-5 Cooling of a Hot Metal Ball in Air 

A spherical metal ball of radius R is heated in an oven to a temperature of 
600°F throughout and is then taken out of the oven and allowed to cool in am- 
bient air at 7" x = 75°F by convection and radiation (Fig. 2-20). The thermal 
conductivity of the ball material is known to vary linearly with temperature. As- 
suming the ball is cooled uniformly from the entire outer surface, obtain the dif- 
ferential equation that describes the variation of the temperature in the ball 
during cooling. 

SOLUTION The ball is initially at a uniform temperature and is cooled uni- 
formly from the entire outer surface. Also, the temperature at any point in the 
ball will change with time during cooling. Therefore, this is a one-dimensional 
transient heat conduction problem since the temperature within the ball will 
change with the radial distance rand the time t. That is, T = T(r, t). 

The thermal conductivity is given to be variable, and there is no heat genera- 
tion in the ball. Therefore, the differential equation that governs the variation of 
temperature in the ball in this case is obtained from Eq. 2-30 by setting the 
heat generation term equal to zero. We obtain 



r 2 dr 



r 2 k 



§T 
dr 



PC 



dT 
dt 



75°F 




FIGURE 2-20 

Schematic for Example 2-5. 



cen58933_ch02.qxd 9/10/2002 8:46 AM Page 74 



74 
HEAT TRANSFER 



which is the one-dimensional transient heat conduction equation in spherical 
coordinates under the conditions of variable thermal conductivity and no heat 
generation. Note again that the conditions at the outer surface of the ball have 
no effect on the differential equation. 




FIGURE 2-21 

Three-dimensional heat conduction 
through a rectangular volume element. 



2-3 - GENERAL HEAT CONDUCTION EQUATION 

In the last section we considered one-dimensional heat conduction and 
assumed heat conduction in other directions to be negligible. Most heat trans- 
fer problems encountered in practice can be approximated as being one- 
dimensional, and we will mostly deal with such problems in this text. 
However, this is not always the case, and sometimes we need to consider heat 
transfer in other directions as well. In such cases heat conduction is said to be 
multidimensional, and in this section we will develop the governing differen- 
tial equation in such systems in rectangular, cylindrical, and spherical coordi- 
nate systems. 

Rectangular Coordinates 

Consider a small rectangular element of length Ax, width Ay, and height Az, 
as shown in Figure 2-21. Assume the density of the body is p and the specific 
heat is C. An energy balance on this element during a small time interval At 
can be expressed as 

\ / Rate of heat \ 
generation 
inside the 



/ Rate of heat \ 
conduction at 

\ x, y, and z J 



Rate of heat 
conduction 
at x + Ax, 

y + Ay, and z + Az I 



element 



Rate of change \ 

of the energy 

content of 

the element 



or 



Qt+Qy + Qz-Qx + Ax-Qy 



Q-. 



A£„ 



At 



(2-36) 



Noting that the volume of the element is V e i ement = AxAyAz, the change in the 
energy content of the element and the rate of heat generation within the ele- 
ment can be expressed as 



A£„ 



E-i + At 
o 'element 



E, = mC(T, + A , 
= gAxAyAz 



T,) = pCAxAyAz{T, + At -T t ) 



Substituting into Eq. 2-36, we get 



Qx+Qy + Qz-Qx 



Qy + . 



Q-. 



g AxAyAz = pCAxAyAz 



T'f + Af 



Ar 



Dividing by AxAyAz gives 

G.Y+A.V _ Qx 1 Gy + Ay 



1 

AyAz 



e, 



Ax 



AxAz 



Av 



1 Q l + Az- 

AxAy Az 



Qz 



PC 



T t +At 



At 
(2-37) 



cen58933_ch02.qxd 9/10/2002 8:46 AM Page 75 



Noting that the heat transfer areas of the element for heat conduction in the 
x, y, and z directions are A x = AyAz, A y = AxAz, and A, = AxAy, respectively, 
and taking the limit as Ax, A_y, Az and At — > yields 



dx\ dx I dy \ dy 



+ l*fH 



pC 



dT 

dt 



(2-38) 



75 
CHAPTER 2 



since, from the definition of the derivative and Fourier's law of heat 
conduction, 





l 


Q 


+ Ax 


Q x 


Ax-» 


AyAz 




Ax 






1 


Q. 


+ A.v ~ 


e, 


Ay-»0 


AxAz 




Ay 






1 


Q 


+ Az — 


Qz 



Az^oAxAy Az 



1 dQ x 

AyAz dx 

1 9Qy 

AxAz dy 
1 dQ z 
AxAy dz AxAy dz 



1 


d 


AyAz 


dx 


1 


d 


AxAz dy 


1 


a 



-kAyAz 
-kAxAz 
-kAxAy 



i)T 
dx 

8T 

dy 

dT 

dz 



dx \ dx 



^- k- 

dy\ dy 

dz r dz 



Equation 2-38 is the general heat conduction equation in rectangular coordi- 
nates. In the case of constant thermal conductivity, it reduces to 



d 2 T d 2 T d 2 T 8 = 1 dT 
dx 2 dy 2 dz 2 k « dt 



(2-39) 



where the property a = k/pC is again the thermal dijfusivity of the material. 
Equation 2-39 is known as the Fourier-Biot equation, and it reduces to these 
forms under specified conditions: 



(1) Steady-state: 

(called the Poisson equation) 

(2) Transient, no heat generation: 
(called the diffusion equation) 

(3) Steady-state, no heat generation: 
(called the Laplace equation) 



Note that in the special case of one-dimensional heat transfer in the 
x-direction, the derivatives with respect to y and z drop out and the equations 
above reduce to the ones developed in the previous section for a plane wall 
(Fig. 2-22). 



d 2 T | d 2 T | d 2 T | g _ 
dx 2 dy 2 dz 2 k 


(2-40) 


d 2 T d 2 T d 2 T 1 dT 
dx 2 3y 2 dz 2 ~ a dt 


(2-41) 


d_^ + djT + d^T =0 
dx 2 9y 2 dz 2 


(2-42) 



Cylindrical Coordinates 



The general heat conduction equation in cylindrical coordinates can be ob- 
tained from an energy balance on a volume element in cylindrical coordinates, 
shown in Figure 2-23, by following the steps just outlined. It can also be ob- 
tained directly from Eq. 2-38 by coordinate transformation using the follow- 
ing relations between the coordinates of a point in rectangular and cylindrical 
coordinate systems: 

x = r cos <f>, y = r sin 4>, and z = Z 




FIGURE 2-22 

The three-dimensional heat 

conduction equations reduce to the 

one-dimensional ones when the 

temperature varies in one 

dimension only. 




FIGURE 2-23 

A differential volume element in 
cylindrical coordinates. 



cen58933_ch02.qxd 9/10/2002 8:46 AM Page 76 



76 
HEAT TRANSFER 




FIGURE 2-24 

A differential volume element in 
spherical coordinates. 



After lengthy manipulations, we obtain 



r dr 



kr 



dT 
dr 



,11 , dT 
r 2 acp \ 5cj> 



d_ 
dz 



dT 
dz 



+ g 



n dT 

P c s7 



(2-43) 



Spherical Coordinates 

The general heat conduction equations in spherical coordinates can be ob- 
tained from an energy balance on a volume element in spherical coordinates, 
shown in Figure 2-24, by following the steps outlined above. It can also be 
obtained directly from Eq. 2-38 by coordinate transformation using the fol- 
lowing relations between the coordinates of a point in rectangular and spheri- 
cal coordinate systems: 



x = r cos 4> sin G, y = r sin 4> sin G, and 



cos G 



Again after lengthy manipulations, we obtain 



r 2 dr \ dr 



1 



d 



dT 
r 2 sin 2 G d<t> V" ^ 



6 96 



k sin 6 



dT 

ae 



P c 



dT 
dt 
(2-44) 



Obtaining analytical solutions to these differential equations requires a 
knowledge of the solution techniques of partial differential equations, which 
is beyond the scope of this introductory text. Here we limit our consideration 
to one-dimensional steady-state cases or lumped systems, since they result in 
ordinary differential equations. 



r 

Metal 



Heat 

loss 



600°F 



billet 



> 



T„ = 65°F 



FIGURE 2-25 

Schematic for Example 2-6. 



EXAMPLE 2-6 Heat Conduction in a Short Cylinder 

A short cylindrical metal billet of radius R and height h is heated in an oven to 
a temperature of 600 C F throughout and is then taken out of the oven and al- 
lowed to cool in ambient air at 7" x = 65°F by convection and radiation. Assum- 
ing the billet is cooled uniformly from all outer surfaces and the variation of the 
thermal conductivity of the material with temperature is negligible, obtain the 
differential equation that describes the variation of the temperature in the bil- 
let during this cooling process. 

SOLUTION The billet shown in Figure 2-25 is initially at a uniform tempera- 
ture and is cooled uniformly from the top and bottom surfaces in the z-direction 
as well as the lateral surface in the radial r-direction. Also, the temperature at 
any point in the ball will change with time during cooling. Therefore, this is a 
two-dimensional transient heat conduction problem since the temperature 
within the billet will change with the radial and axial distances rand zand with 
time t. That is, T= T(r, z, f). 

The thermal conductivity is given to be constant, and there is no heat gener- 
ation in the billet. Therefore, the differential equation that governs the variation 
of temperature in the billet in this case is obtained from Eq. 2-43 by setting 
the heat generation term and the derivatives with respect to 4> equal to zero. We 
obtain 



IA 
r dr 



kr 



+ f 



dT 
dz 



P c 



dT 
dt 



cen58933_ch02.qxd 9/10/2002 8:46 AM Page 77 



In the 


case 


of constc 


nt therma 


cond 


JCt 


vity, 


it reduces 


to 








1A 

r dr 




+ 


dz 2 


_ 1 dT 

a dt 




which 


is the desired 


equation. 













77 
CHAPTER 2 



2-A - BOUNDARY AND INITIAL CONDITIONS 

The heat conduction equations above were developed using an energy balance 
on a differential element inside the medium, and they remain the same re- 
gardless of the thermal conditions on the surfaces of the medium. That is, the 
differential equations do not incorporate any information related to the condi- 
tions on the surfaces such as the surface temperature or a specified heat flux. 
Yet we know that the heat flux and the temperature distribution in a medium 
depend on the conditions at the surfaces, and the description of a heat transfer 
problem in a medium is not complete without a full description of the thermal 
conditions at the bounding surfaces of the medium. The mathematical expres- 
sions of the thermal conditions at the boundaries are called the boundary 
conditions. 

From a mathematical point of view, solving a differential equation is essen- 
tially a process of removing derivatives, or an integration process, and thus 
the solution of a differential equation typically involves arbitrary constants 
(Fig. 2-26). It follows that to obtain a unique solution to a problem, we need 
to specify more than just the governing differential equation. We need to spec- 
ify some conditions (such as the value of the function or its derivatives at 
some value of the independent variable) so that forcing the solution to satisfy 
these conditions at specified points will result in unique values for the arbi- 
trary constants and thus a unique solution. But since the differential equation 
has no place for the additional information or conditions, we need to supply 
them separately in the form of boundary or initial conditions. 

Consider the variation of temperature along the wall of a brick house in 
winter. The temperature at any point in the wall depends on, among other 
things, the conditions at the two surfaces of the wall such as the air tempera- 
ture of the house, the velocity and direction of the winds, and the solar energy 
incident on the outer surface. That is, the temperature distribution in a medium 
depends on the conditions at the boundaries of the medium as well as the heat 
transfer mechanism inside the medium. To describe a heat transfer problem 
completely, two boundary conditions must be given for each direction of 
the coordinate system along which heat transfer is significant (Fig. 2-27). 
Therefore, we need to specify two boundary conditions for one-dimensional 
problems, four boundary conditions for two-dimensional problems, and six 
boundary conditions for three-dimensional problems. In the case of the wall 
of a house, for example, we need to specify the conditions at two locations 
(the inner and the outer surfaces) of the wall since heat transfer in this case is 
one-dimensional. But in the case of a parallelepiped, we need to specify six 
boundary conditions (one at each face) when heat transfer in all three dimen- 
sions is significant. 



The differential equation: 

d 2 T 
dx 2 







C,JE 

T 



c, 



General solution: 
T(x) 



Arbitrary constants 
Some specific solutions: 

T(x) = 2x + 5 
T(x) = -x + 12 
T(x) = -3 
T(x) = 6.2* 



FIGURE 2-26 

The general solution of a typical 

differential equation involves 

arbitrary constants, and thus an 

infinite number of solutions. 



50°C 




Some solutions of 
d 2 T 
dx 2 ' 



:0 



15°C 



The only solution 
that satisfies 
the conditions 
7/(0) = 50°C 
and T(L) = 15°C. 

FIGURE 2-27 

To describe a heat transfer problem 

completely, two boundary conditions 

must be given for each direction along 

which heat transfer is significant. 



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78 
HEAT TRANSFER 



The physical argument presented above is consistent with the mathematical 
nature of the problem since the heat conduction equation is second order (i.e., 
involves second derivatives with respect to the space variables) in all direc- 
tions along which heat conduction is significant, and the general solution of a 
second-order linear differential equation involves two arbitrary constants for 
each direction. That is, the number of boundary conditions that needs to be 
specified in a direction is equal to the order of the differential equation in that 
direction. 

Reconsider the brick wall already discussed. The temperature at any point 
on the wall at a specified time also depends on the condition of the wall at the 
beginning of the heat conduction process. Such a condition, which is usually 
specified at time t = 0, is called the initial condition, which is a mathemati- 
cal expression for the temperature distribution of the medium initially. Note 
that we need only one initial condition for a heat conduction problem regard- 
less of the dimension since the conduction equation is first order in time (it in- 
volves the first derivative of temperature with respect to time). 

In rectangular coordinates, the initial condition can be specified in the gen- 
eral form as 



T(x, y, z, 0) = f(x, y, z) 



(2-45) 



150°C 



T(x, t) 



m 



70°C 



7X0, = 150°C 
T(L, f) = 70°C 

FIGURE 2-28 

Specified temperature boundary 
conditions on both surfaces 
of a plane wall. 



where the function /(x, y, z) represents the temperature distribution throughout 
the medium at time t = 0. When the medium is initially at a uniform tem- 
perature of T h the initial condition of Eq. 2-45 can be expressed as 
T(x, y, z, 0) = Tj. Note that under steady conditions, the heat conduction equa- 
tion does not involve any time derivatives, and thus we do not need to specify 
an initial condition. 

The heat conduction equation is first order in time, and thus the initial con- 
dition cannot involve any derivatives (it is limited to a specified temperature). 
However, the heat conduction equation is second order in space coordinates, 
and thus a boundary condition may involve first derivatives at the boundaries 
as well as specified values of temperature. Boundary conditions most com- 
monly encountered in practice are the specified temperature, specified heat 
flux, convection, and radiation boundary conditions. 



1 Specified Temperature Boundary Condition 

The temperature of an exposed surface can usually be measured directly and 
easily. Therefore, one of the easiest ways to specify the thermal conditions on 
a surface is to specify the temperature. For one-dimensional heat transfer 
through a plane wall of thickness L, for example, the specified temperature 
boundary conditions can be expressed as (Fig. 2-28) 



7/(0, t) = T x 
T(L, t) = T 2 



(2-46) 



where T { and T 2 are the specified temperatures at surfaces at x = and x = L, 
respectively. The specified temperatures can be constant, which is the case for 
steady heat conduction, or may vary with time. 



cen58933_ch02.qxd 9/10/2002 8:46 AM Page 79 



2 Specified Heat Flux Boundary Condition 

When there is sufficient information about energy interactions at a surface, it 
may be possible to determine the rate of heat transfer and thus the heat flux q 
(heat transfer rate per unit surface area, W/m 2 ) on that surface, and this infor- 
mation can be used as one of the boundary conditions. The heat flux in the 
positive x-direction anywhere in the medium, including the boundaries, can be 
expressed by Fourier's law of heat conduction as 



dT = Heat flux in the \ 
dx ^positive ^-direction/ 



(W/m 2 ) 



(2-47) 



Then the boundary condition at a boundary is obtained by setting the specified 
heat flux equal to —k(dT/dx) at that boundary. The sign of the specified heat 
flux is determined by inspection: positive if the heat flux is in the positive di- 
rection of the coordinate axis, and negative if it is in the opposite direction. 
Note that it is extremely important to have the correct sign for the specified 
heat flux since the wrong sign will invert the direction of heat transfer and 
cause the heat gain to be interpreted as heat loss (Fig. 2-29). 

For a plate of thickness L subjected to heat flux of 50 W/m 2 into the medium 
from both sides, for example, the specified heat flux boundary conditions can 
be expressed as 



37(0, t) 
-k—i = 50 

dx 



and 



dT(L, t) 
'' dx 



-50 



(2-48) 



79 
CHAPTER 2 



Heat 
flux 



%■■ 



Conduction 



. 37X0, 



dx 



Conduction 



.dT(L,t) 



Heat 
flux 



9 L 



FIGURE 2-29 

Specified heat flux 

boundary conditions on both 

surfaces of a plane wall. 



Note that the heat flux at the surface at x = L is in the negative x-direction, 
and thus it is —50 W/m 2 . 

Special Case: Insulated Boundary 

Some surfaces are commonly insulated in practice in order to minimize heat 
loss (or heat gain) through them. Insulation reduces heat transfer but does not 
totally eliminate it unless its thickness is infinity. However, heat transfer 
through a properly insulated surface can be taken to be zero since adequate 
insulation reduces heat transfer through a surface to negligible levels. There- 
fore, a well-insulated surface can be modeled as a surface with a specified 
heat flux of zero. Then the boundary condition on a perfectly insulated surface 
(at x = 0, for example) can be expressed as (Fig. 2-30) 



3T(0, t) 

dx 







dT(0, 
dx 







(2-49) 



That is, on an insulated surface, the first derivative of temperature with re- 
spect to the space variable (the temperature gradient) in the direction normal 
to the insulated surface is zero. This also means that the temperature function 
must be perpendicular to an insulated surface since the slope of temperature at 
the surface must be zero. 

Another Special Case: Thermal Symmetry 

Some heat transfer problems possess thermal symmetry as a result of the 
symmetry in imposed thermal conditions. For example, the two surfaces of a 
large hot plate of thickness L suspended vertically in air will be subjected to 



Insulation 



T(x, t) 



97/(0, t) 



60°C 







dx 

T(L, t) = 60°C 

FIGURE 2-30 

A plane wall with insulation 

and specified temperature 

boundary conditions. 



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80 
HEAT TRANSFER 



L^— Center plane 



Zero I 



slope 








Temperature 
distribution 
(symmetric 
about center 
plane) 



:() 



I — 

I 2 
dT(LI2, f) 
dx 

FIGURE 2-31 

Thermal symmetry boundary 
condition at the center plane 
of a plane wall. 



the same thermal conditions, and thus the temperature distribution in one half 
of the plate will be the same as that in the other half. That is, the heat transfer 
problem in this plate will possess thermal symmetry about the center plane at 
x = L/2. Also, the direction of heat flow at any point in the plate will be 
toward the surface closer to the point, and there will be no heat flow across the 
center plane. Therefore, the center plane can be viewed as an insulated sur- 
face, and the thermal condition at this plane of symmetry can be expressed as 
(Fig. 2-31) 



dT(L/2, t) 
dx 







(2-50) 



which resembles the insulation or zero heat flux boundary condition. This 
result can also be deduced from a plot of temperature distribution with a 
maximum, and thus zero slope, at the center plane. 

In the case of cylindrical (or spherical) bodies having thermal symmetry 
about the center line (or midpoint), the thermal symmetry boundary condition 
requires that the first derivative of temperature with respect to r (the radial 
variable) be zero at the centerline (or the midpoint). 



L 



Water 
110°C 




I! 

% 

FIGURE 2-32 

Schematic for Example 2-7. 



EXAMPLE 2-7 Heat Flux Boundary Condition 

Consider an aluminum pan used to cook beef stew on top of an electric range. 
The bottom section of the pan is L = 0.3 cm thick and has a diameter of D = 
20 cm. The electric heating unit on the range top consumes 800 W of power 
during cooking, and 90 percent of the heat generated in the heating element is 
transferred to the pan. During steady operation, the temperature of the inner 
surface of the pan is measured to be 110 C C. Express the boundary conditions 
for the bottom section of the pan during this cooking process. 

SOLUTION The heat transfer through the bottom section of the pan is from the 
bottom surface toward the top and can reasonably be approximated as being 
one-dimensional. We take the direction normal to the bottom surfaces of the 
pan as the x axis with the origin at the outer surface, as shown in Figure 2-32. 
Then the inner and outer surfaces of the bottom section of the pan can be rep- 
resented by x = and x = L, respectively. During steady operation, the tem- 
perature will depend on x only and thus T = T(x). 

The boundary condition on the outer surface of the bottom of the pan at 
x = can be approximated as being specified heat flux since it is stated that 
90 percent of the 800 W (i.e., 720 W) is transferred to the pan at that surface. 
Therefore, 



dT(0) 
dx 



q a 



where 



<?o 



Heat transfer rate 



0.720 kW 



Bottom surface area tt(0.1 m) 2 



22.9 kW/m 2 



cen58933_ch02.qxd 9/10/2002 8:46 AM Page 81 



The temperature at the inner surface of the bottom of the pan is specified to be 
110°C. Then the boundary condition on this surface can be expressed as 

T(L) = 110°C 

where L = 0.003 m. Note that the determination of the boundary conditions 
may require some reasoning and approximations. 



81 
CHAPTER 2 



3 Convection Boundary Condition 

Convection is probably the most common boundary condition encountered 
in practice since most heat transfer surfaces are exposed to an environment at 
a specified temperature. The convection boundary condition is based on a sur- 
face energy balance expressed as 



I Heat conduction \ 

at the surface in a 

\ selected direction/ 



' Heat convection \ 

at the surface in 
Uhe same direction/ 



For one-dimensional heat transfer in the x-direction in a plate of thickness L, 
the convection boundary conditions on both surfaces can be expressed as 



and 



dT(0, t) 
dx 



dT(L, t) 
dx 



hJT^- 7X0, r)] 



h 2 [T(L, t) - r„ 2 ] 



(2-51a) 



(2-51 b) 



where /?, and h 2 are the convection heat transfer coefficients and T rj3l and T^ 2 
are the temperatures of the surrounding mediums on the two sides of the plate, 
as shown in Figure 2-33. 

In writing Eqs. 2-5 1 for convection boundary conditions, we have selected 
the direction of heat transfer to be the positive x-direction at both surfaces. But 
those expressions are equally applicable when heat transfer is in the opposite 
direction at one or both surfaces since reversing the direction of heat transfer 
at a surface simply reverses the signs of both conduction and convection terms 
at that surface. This is equivalent to multiplying an equation by — 1 , which has 
no effect on the equality (Fig. 2-34). Being able to select either direction as 
the direction of heat transfer is certainly a relief since often we do not know 
the surface temperature and thus the direction of heat transfer at a surface in 
advance. This argument is also valid for other boundary conditions such as the 
radiation and combined boundary conditions discussed shortly. 

Note that a surface has zero thickness and thus no mass, and it cannot store 
any energy. Therefore, the entire net heat entering the surface from one side 
must leave the surface from the other side. The convection boundary condi- 
tion simply states that heat continues to flow from a body to the surrounding 
medium at the same rate, and it just changes vehicles at the surface from con- 
duction to convection (or vice versa in the other direction). This is analogous 
to people traveling on buses on land and transferring to the ships at the shore. 



Convection 



Conduction 



. dm t) 

dx 
Conduction 



Convection 



-k d y ( L ' f) = h 2 lT(L, - 7^.,] 



L x 



FIGURE 2-33 

Convection boundary conditions on 
the two surfaces of a plane wall. 











Convection 


Conduction 






hjr^-no, 


o,=-»« 






*,, r-, 








Convection 


Conduction 






h l [m,f)-T 




»i J " * dx 








L 


X 



FIGURE 2-34 

The assumed direction of heat transfer 

at a boundary has no effect on the 

boundary condition expression. 



cen58933_ch02.qxd 9/10/2002 8:46 AM Page 82 



82 
HEAT TRANSFER 



If the passengers are not allowed to wander around at the shore, then the rate 
at which the people are unloaded at the shore from the buses must equal the 
rate at which they board the ships. We may call this the conservation of "peo- 
ple" principle. 

Also note that the surface temperatures T(0, t) and T(L, t) are not known 
(if they were known, we would simply use them as the specified temperature 
boundary condition and not bother with convection). But a surface tempera- 
ture can be determined once the solution T(x, t) is obtained by substituting the 
value of x at that surface into the solution. 



Insulation 




FIGURE 2-35 

Schematic for Example 2- 



EXAMPLE 2-8 Convection and Insulation Boundary Conditions 

Steam flows through a pipe shown in Figure 2-35 at an average temperature of 
Ty, = 200°C. The inner and outer radii of the pipe are r x = 8 cm and r 2 = 
8.5 cm, respectively, and the outer surface of the pipe is heavily insulated. If 
the convection heat transfer coefficient on the inner surface of the pipe is 
h = 65 W/m 2 • °C, express the boundary conditions on the inner and outer sur- 
faces of the pipe during transient periods. 

SOLUTION During initial transient periods, heat transfer through the pipe ma- 
terial will predominantly be in the radial direction, and thus can be approxi- 
mated as being one-dimensional. Then the temperature within the pipe material 
will change with the radial distance r and the time t. That is, T = T(r, t). 

It is stated that heat transfer between the steam and the pipe at the inner 
surface is by convection. Then taking the direction of heat transfer to be the 
positive ^direction, the boundary condition on that surface can be expressed as 



dT(r u t) 
dr 



h[T a - T(r,)] 



The pipe is said to be well insulated on the outside, and thus heat loss through 
the outer surface of the pipe can be assumed to be negligible. Then the bound- 
ary condition at the outer surface can be expressed as 



dT(r 2 , t) 
dr 







That is, the temperature gradient must be zero on the outer surface of the pipe 
at all times. 



4 Radiation Boundary Condition 

In some cases, such as those encountered in space and cryogenic applications, 
a heat transfer surface is surrounded by an evacuated space and thus there is 
no convection heat transfer between a surface and the surrounding medium. In 
such cases, radiation becomes the only mechanism of heat transfer between 
the surface under consideration and the surroundings. Using an energy bal- 
ance, the radiation boundary condition on a surface can be expressed as 



(Heat conduction | 
at the surface in a 
selected direction 



[ Radiation exchange \ 

at the surface in 
\ the same direction / 



cen58933_ch02.qxd 9/10/2002 8:46 AM Page 83 



For one-dimensional heat transfer in the x-direction in a plate of thickness L, 
the radiation boundary conditions on both surfaces can be expressed as 
(Fig. 2-36) 



and 



dT(0, t) 

dx 



dT(L, t) 
dx 



e i CT U SUIT, 1 



e 2 (T[T(L, t) 4 



T(0, f) 4 



(2-52a) 



(2-52b) 



5.67 X 
tr, 2 are the 



where e { and e 2 are the emissivities of the boundary surfaces, a 
10~ 8 W/m 2 • K 4 is the Stefan-Boltzmann constant, and T sum , and T, 
average temperatures of the surfaces surrounding the two sides of the plate, 
respectively. Note that the temperatures in radiation calculations must be ex- 
pressed in K or R (not in °C or °F). 

The radiation boundary condition involves the fourth power of temperature, 
and thus it is a nonlinear condition. As a result, the application of this bound- 
ary condition results in powers of the unknown coefficients, which makes it 
difficult to determine them. Therefore, it is tempting to ignore radiation ex- 
change at a surface during a heat transfer analysis in order to avoid the com- 
plications associated with nonlinearity. This is especially the case when heat 
transfer at the surface is dominated by convection, and the role of radiation is 
minor. 



83 
CHAPTER 2 



Radiation Conduction 




37X0, Q 
dx 



Conduction 



surr, 2 

Radiation 



37X^0 =4 _ r 4 , 
dx 2 K ' surr - 2 



L x 



FIGURE 2-36 

Radiation boundary conditions on 
both surfaces of a plane wall. 



5 Interface Boundary Conditions 

Some bodies are made up of layers of different materials, and the solution of 
a heat transfer problem in such a medium requires the solution of the heat 
transfer problem in each layer. This, in turn, requires the specification of the 
boundary conditions at each interface. 

The boundary conditions at an interface are based on the requirements that 
(1) two bodies in contact must have the same temperature at the area of con- 
tact and (2) an interface (which is a surface) cannot store any energy, and thus 
the heat flux on the two sides of an interface must be the same. The boundary 
conditions at the interface of two bodies A and B in perfect contact at x = x 
can be expressed as (Fig. 2-37) 



T A (x , t) = T B (x , t) 



and 



dT A (x , t) 
dx 



dT B (x Q , t) 

' dx 



(2-53) 



(2-54) 



Material 

A 



Interface 

Material 
B 

T A (x ,t) = T B (x ,t) 
TJx, t) 




dT A (x Q , 1) dT B (x Q ,t) 

K. - - "d 



A dx 



B dx 



L x 



FIGURE 2-37 

Boundary conditions at the interface 
of two bodies in perfect contact. 



where k A and k B are the thermal conductivities of the layers A and B, respec- 
tively. The case of imperfect contact results in thermal contact resistance, 
which is considered in the next chapter. 



cen58933_ch02.qxd 9/10/2002 8:46 AM Page 84 



84 
HEAT TRANSFER 



6 Generalized Boundary Conditions 

So far we have considered surfaces subjected to single mode heat transfer, 
such as the specified heat flux, convection, or radiation for simplicity. In gen- 
eral, however, a surface may involve convection, radiation, and specified heat 
flux simultaneously. The boundary condition in such cases is again obtained 
from a surface energy balance, expressed as 



IHeat transfer \ 
to the surface 
in all modes / 



I Heat transfer \ 

from the surface 

\ in all modes / 



(2-55) 



This is illustrated in Examples 2-9 and 2-10. 




T = 78°F 



FIGURE 2-38 

Schematic for Example 2-9. 



EXAMPLE 2-9 



Combined Convection and Radiation Condition 



A spherical metal ball of radius r is heated in an oven to a temperature of 
600°F throughout and is then taken out of the oven and allowed to cool in am- 
bient air at 7"„ = 78°F, as shown in Figure 2-38. The thermal conductivity of 
the ball material is k = 8.3 Btu/h • ft • °F, and the average convection heat 
transfer coefficient on the outer surface of the ball is evaluated to be h = 4.5 
Btu/h ■ ft 2 • °F. The emissivity of the outer surface of the ball is e = 0.6, and the 
average temperature of the surrounding surfaces is T smr = 525 R. Assuming the 
ball is cooled uniformly from the entire outer surface, express the initial and 
boundary conditions for the cooling process of the ball. 

SOLUTION The ball is initially at a uniform temperature and is cooled uni- 
formly from the entire outer surface. Therefore, this is a one-dimensional tran- 
sient heat transfer problem since the temperature within the ball will change 
with the radial distance rand the time f. That is, T = T(r, t). Taking the mo- 
ment the ball is removed from the oven to be t = 0, the initial condition can be 
expressed as 



T(r, 0) 



600°F 



The problem possesses symmetry about the midpoint (r = 0) since the 
isotherms in this case will be concentric spheres, and thus no heat will be 
crossing the midpoint of the ball. Then the boundary condition at the midpoint 
can be expressed as 



dT(0, t) 
dr 







The heat conducted to the outer surface of the ball is lost to the environment by 
convection and radiation. Then taking the direction of heat transfer to be the 
positive r direction, the boundary condition on the outer surface can be ex- 
pressed as 



, d1\r , t) 
dr 



h[T(r ) - rj + eo-[r(r ) 4 - TU 



cen58933_ch02.qxd 9/10/2002 8:46 AM Page 85 



All the quantities in the above relations are known except the temperatures 
and their derivatives at r = and r . Also, the radiation part of the boundary 
condition is often ignored for simplicity by modifying the convection heat trans- 
fer coefficient to account for the contribution of radiation. The convection coef- 
ficient h in that case becomes the combined heat transfer coefficient. 



85 
CHAPTER 2 



EXAMPLE 2-10 Combined Convection, Radiation, and Heat Flux 

Consider the south wall of a house that is L = 0.2 m thick. The outer surface of 
the wall is exposed to solar radiation and has an absorptivity of a = 0.5 for so- 
lar energy. The interior of the house is maintained at 7" xl = 20 C C, while the am- 
bient air temperature outside remains at T a2 = 5°C. The sky, the ground, and 
the surfaces of the surrounding structures at this location can be modeled as a 
surface at an effective temperature of 7" sky = 255 K for radiation exchange on 
the outer surface. The radiation exchange between the inner surface of the wall 
and the surfaces of the walls, floor, and ceiling it faces is negligible. The con- 
vection heat transfer coefficients on the inner and the outer surfaces of the wall 
are h 1 = 6 W/m 2 • °C and h 2 = 25 W/m 2 • °C, respectively. The thermal con- 
ductivity of the wall material is k = 0.7 W/m • °C, and the emissivity of the 
outer surface is e z = 0.9. Assuming the heat transfer through the wall to be 
steady and one-dimensional, express the boundary conditions on the inner and 
the outer surfaces of the wall. 

SOLUTION We take the direction normal to the wall surfaces as the x-axis with 
the origin at the inner surface of the wall, as shown in Figure 2-39. The heat 
transfer through the wall is given to be steady and one-dimensional, and thus 
the temperature depends on x only and not on time. That is, T = T(x). 

The boundary condition on the inner surface of the wall at x = is a typical 
convection condition since it does not involve any radiation or specified heat 
flux. Taking the direction of heat transfer to be the positive x-direction, the 
boundary condition on the inner surface can be expressed as 



dT(0) 
dx 



AiLT., - 7X0)] 



The boundary condition on the outer surface at x = is quite general as it in- 
volves conduction, convection, radiation, and specified heat flux. Again taking 
the direction of heat transfer to be the positive x-direction, the boundary condi- 
tion on the outer surface can be expressed as 



dT{L) 
dx 



h 2 [T(L) - KJ + e 2 o-[r(L) 4 - TLA - a# solar 



where q solar is the incident solar heat flux. Assuming the opposite direction for 
heat transfer would give the same result multiplied by -1, which is equivalent 
to the relation here. All the quantities in these relations are known except the 
temperatures and their derivatives at the two boundaries. 




FIGURE 2-39 

Schematic for Example 2-10. 



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86 
HEAT TRANSFER 



Note that a heat transfer problem may involve different kinds of boundary 
conditions on different surfaces. For example, a plate may be subject to heat 
flux on one surface while losing or gaining heat by convection from the other 
surface. Also, the two boundary conditions in a direction may be specified at 
the same boundary, while no condition is imposed on the other boundary. For 
example, specifying the temperature and heat flux at x = of a plate of thick- 
ness L will result in a unique solution for the one-dimensional steady temper- 
ature distribution in the plate, including the value of temperature at the surface 
x = L. Although not necessary, there is nothing wrong with specifying more 
than two boundary conditions in a specified direction, provided that there is 
no contradiction. The extra conditions in this case can be used to verify the 
results. 



S- 



Heat transfer problem \ 

I 

Mathematical formulation 

(Differential equation and 

boundary conditions) 

I 
General solution of differential equation 

I 
Application of boundary conditions 

I 
Solution of the problem 




FIGURE 2-40 

Basic steps involved in the solution of 
heat transfer problems. 



2-5 - SOLUTION OF STEADY ONE-DIMENSIONAL 
HEAT CONDUCTION PROBLEMS 

So far we have derived the differential equations for heat conduction in 
various coordinate systems and discussed the possible boundary conditions. 
A heat conduction problem can be formulated by specifying the applicable 
differential equation and a set of proper boundary conditions. 

In this section we will solve a wide range of heat conduction problems in 
rectangular, cylindrical, and spherical geometries. We will limit our attention 
to problems that result in ordinary differential equations such as the steady 
one-dimensional heat conduction problems. We will also assume constant 
thermal conductivity, but will consider variable conductivity later in this chap- 
ter. If you feel rusty on differential equations or haven't taken differential 
equations yet, no need to panic. Simple integration is all you need to solve the 
steady one-dimensional heat conduction problems. 

The solution procedure for solving heat conduction problems can be sum- 
marized as (1) formulate the problem by obtaining the applicable differential 
equation in its simplest form and specifying the boundary conditions, (2) ob- 
tain the general solution of the differential equation, and (3) apply the bound- 
ary conditions and determine the arbitrary constants in the general solution 
(Fig. 2-40). This is demonstrated below with examples. 





Plane 








wall 




^ 








T 
2 






L x 



FIGURE 2-41 

Schematic for Example 2-11. 



EXAMPLE 2-11 



Heat Conduction in a Plane Wall 



Consider a large plane wall of thickness L = 0.2 m, thermal conductivity k = 
1.2 W/m ■ °C, and surface area A = 15 m 2 . The two sides of the wall are main- 
tained at constant temperatures of 7", = 120 C C and T 2 = 50°C, respectively, as 
shown in Figure 2-41. Determine (a) the variation of temperature within the 
wall and the value of temperature at x = 0.1 m and (b) the rate of heat con- 
duction through the wall under steady conditions. 

SOLUTION A plane wall with specified surface temperatures is given. The vari- 
ation of temperature and the rate of heat transfer are to be determined. 
Assumptions 1 Heat conduction is steady. 2 Heat conduction is one- 
dimensional since the wall is large relative to its thickness and the thermal 



cen58933_ch02.qxd 9/10/2002 8:46 AM Page 87 



conditions on both sides are uniform. 3 Thermal conductivity is constant. 
4 There is no heat generation. 

Properties The thermal conductivity is given to be k = 1.2 W/m • °C. 

Analysis (a) Taking the direction normal to the surface of the wall to be the 
^-direction, the differential equation for this problem can be expressed as 



d-T _ 
dx 2 ' 







with boundary conditions 



7\0) 

T(L) = T 2 = 50°C 

The differential equation is linear and second order, and a quick inspection of 
it reveals that it has a single term involving derivatives and no terms involving 
the unknown function 7"as a factor. Thus, it can be solved by direct integration. 
Noting that an integration reduces the order of a derivative by one, the general 
solution of the differential equation above can be obtained by two simple suc- 
cessive integrations, each of which introduces an integration constant. 
Integrating the differential equation once with respect to x yields 



dT 

dx 



C, 



where C 1 is an arbitrary constant. Notice that the order of the derivative went 
down by one as a result of integration. As a check, if we take the derivative of 
this equation, we will obtain the original differential equation. This equation is 
not the solution yet since it involves a derivative. 
Integrating one more time, we obtain 



T(x) 



Co 



which is the general solution of the differential equation (Fig. 2-42). The gen- 
eral solution in this case resembles the general formula of a straight line whose 
slope is C x and whose value at x = is C 2 . This is not surprising since the sec- 
ond derivative represents the change in the slope of a function, and a zero sec- 
ond derivative indicates that the slope of the function remains constant. 
Therefore, any straight line is a solution of this differential equation. 

The general solution contains two unknown constants Cj and C 2 , and thus we 
need two equations to determine them uniquely and obtain the specific solu- 
tion. These equations are obtained by forcing the general solution to satisfy the 
specified boundary conditions. The application of each condition yields one 
equation, and thus we need to specify two conditions to determine the con- 
stants C 1 and C 2 . 

When applying a boundary condition to an equation, all occurrences of the 
dependent and independent variables and any derivatives are replaced by the 
specified values. Thus the only unknowns in the resulting equations are the ar- 
bitrary constants. 

The first boundary condition can be interpreted as in the general solution, re- 
place all the x's by zero and T(x) by T x . That is (Fig. 2-43), 



7X0) = C, X + c 2 



c, 



87 
CHAPTER 2 



Differential equation: 






dx 1 


= 


Integrate: 








dT 
dx 


= c, 


Integrate again: 






T(x) = 
General 


C t x + C, 
Arbitrary 


solution 




constants 



FIGURE 2-42 

Obtaining the general solution of a 

simple second order differential 

equation by integration. 



Boundary condition: 

T(0) = r, 
General solution: 

T(x) = C,x + C 2 
Applying the boundary condition: 
C, 



T(x) 

1 




C,x 

1 




Substituting: 

T, = C, X + C, -> 



c, 



It cannot involve x or T(x) after the 
boundary condition is applied. 



FIGURE 2-43 

When applying a boundary condition 

to the general solution at a specified 

point, all occurrences of the dependent 

and independent variables should be 

replaced by their specified values 

at that point. 



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88 
HEAT TRANSFER 



The second boundary condition can be interpreted as in the general solution, re- 
place all the x's by L and T{x) by T 2 . That is, 



T(L) = C X L + C 2 



C,L + T, 



C, 



r, 



Substituting the C x and C 2 expressions into the general solution, we obtain 

T — T 
T(x)= 2 L - x + T, (2-56) 

which is the desired solution since it satisfies not only the differential equation 
but also the two specified boundary conditions. That is, differentiating Eq. 
2-56 with respect to x twice will give d 2 Tldx 2 , which is the given differential 
equation, and substituting x = and x = L into Eq. 2-56 gives 7(0) = T x and 
T(L) = T z , respectively, which are the specified conditions at the boundaries. 
Substituting the given information, the value of the temperature at x = 0.1 m 
is determined to be 

(50 - 120)°C 
T(0.l m) = — — (0.1 m) + 120°C = 85°C 

(b) The rate of heat conduction anywhere in the wall is determined from 
Fourier's law to be 



Gv 



-kA 



dT 

dx 



-kAC, 



-kA- 



r, 



kA- 



T 2 



(2-57) 



The numerical value of the rate of heat conduction through the wall is deter- 
mined by substituting the given values to be 

T,-T 2 „ (120 - 50)°C 
Q = kA ' = (1.2W/m-°C)(15m 2 ) — — = 6300 W 

Discussion Note that under steady conditions, the rate of heat conduction 
through a plane wall is constant. 



EXAMPLE 2-12 A Wall with Various Sets of Boundary Conditions 

Consider steady one-dimensional heat conduction in a large plane wall of thick- 
ness L and constant thermal conductivity k with no heat generation. Obtain ex- 
pressions for the variation of temperature within the wall for the following pairs 
of boundary conditions (Fig. 2-44): 



(a) -k 

(b) -k 

(c) -k 



dT(0) 

dx 
dT(0) 

dx 
dT(0) 

dx 



q = 40 W/cm 2 


and 


7T[0) = T 


= 15°C 


4 = 40 W/cm 2 


and 


k dT(L) 
dx 


= q L = -25 W/cm 


q = 40 W/cm 2 


and 


dT(L) 
k ^ 


= q a = 40 W/cm 2 



cen58933_ch02.qxd 9/10/2002 8:46 AM Page 89 



15°C 



40 W/cm 7 



(a) 



Plane 
wall 

T(x) 



I W/cm 7 





Plane 




.. — 




wall 




- — 


■" 


T(x) 




- — 2f 


0< 






- — 




L 


X 



(l>) 



40 W/cm 7 
25 W/cm 2 



(c) 



89 
CHAPTER 2 



Plane 
wall 



T(x) 



40 W/cnr 



FIGURE 2-44 

Schematic for Example 2-12. 



SOLUTION This is a steady one-dimensional heat conduction problem with 
constant thermal conductivity and no heat generation in the medium, and the 
heat conduction equation in this case can be expressed as (Eq. 2-17) 



cV-T _ 

dx 2 ' 







whose general solution was determined in the previous example by direct inte- 
gration to be 



T(x) 



C, 



where C 1 and C 2 are two arbitrary integration constants. The specific solutions 
corresponding to each specified pair of boundary conditions are determined as 
follows. 

(a) In this case, both boundary conditions are specified at the same boundary 
at x = 0, and no boundary condition is specified at the other boundary at x = L. 
Noting that 



dT 
dx 



C, 



the application of the boundary conditions gives 
rf7T0) 



dx 



<?o 



and 



ZY0) 



-kCi = q 



c,xo + c, 



c, 



<7o 
' k 



Co 



Substituting, the specific solution in this case is determined to be 

T(x) =~t+T 



Therefore, the two boundary conditions can be specified at the same boundary, 
and it is not necessary to specify them at different locations. In fact, the fun- 
damental theorem of linear ordinary differential equations guarantees that a 



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90 
HEAT TRANSFER 



Differential equation: 






T"(x) = 






General solution: 






T{x) = C,x + C 


i 




(a) Unique solution: 






-kT'(0) = q } T . ._ _ 

r(0) = r J K ' 


la 
k 


x + T 


(b) No solution: 






-kT'(0) = q o \ 
-kT(L) = q L \ 1(Xy 


= None 


(c) Multiple solutions: 






-kT (L) = q \ 


la 
k 


x + C, 
f 
Arbitrary 



FIGURE 2-45 

A boundary-value problem may have a 
unique solution, infinitely many 
solutions, or no solutions at all. 



Resistance heater 
1200 W 



Base plate 




FIGURE 2-46 

Schematic for Example 2-13. 



unique solution exists when both conditions are specified at the same location. 
But no such guarantee exists when the two conditions are specified at different 
boundaries, as you will see below. 

(b) In this case different heat fluxes are specified at the two boundaries. The 
application of the boundary conditions gives 



and 



dT(0) 

dx 



dT{L) 

dx 



Qo 



1l 



-kC x = q 



-kCi = 4l 



c, 



c, 



k 



Ik 

k 



Since q + q L and the constant Cj cannot be equal to two different things at 
the same time, there is no solution in this case. This is not surprising since this 
case corresponds to supplying heat to the plane wall from both sides and ex- 
pecting the temperature of the wall to remain steady (not to change with time). 
This is impossible. 

(c) In this case, the same values for heat flux are specified at the two bound- 
aries. The application of the boundary conditions gives 



and 



dT(0) 
dx 



dT(L) 
dx 



<7o 



9o 



-kC x = q 



kC x — q 



C, 



c, 



<7o 
' k 



<7o 
' k 



Thus, both conditions result in the same value for the constant C 1 , but no value 
for C z . Substituting, the specific solution in this case is determined to be 



T(x) 



4o 



C, 



which is not a unique solution since C 2 is arbitrary. This solution represents a 
family of straight lines whose slope is -q /k. Physically, this problem corre- 
sponds to requiring the rate of heat supplied to the wall at x = be equal to the 
rate of heat removal from the other side of the wall at x = L. But this is a con- 
sequence of the heat conduction through the wall being steady, and thus the 
second boundary condition does not provide any new information. So it is not 
surprising that the solution of this problem is not unique. The three cases dis- 
cussed above are summarized in Figure 2-45. 



EXAMPLE 2-13 



Heat Conduction in the Base Plate of an Iron 



Consider the base plate of a 1200-W household iron that has a thickness of 
L = 0.5 cm, base area of A = 300 cm 2 , and thermal conductivity of k = 
15 W/m ■ C C. The inner surface of the base plate is subjected to uniform heat 
flux generated by the resistance heaters inside, and the outer surface loses 
heat to the surroundings at 7~ x = 20°C by convection, as shown in Figure 2-46. 



cen58933_ch02.qxd 9/10/2002 8:46 AM Page 91 



Taking the convection heat transfer coefficient to be h = 80 W/m 2 • °C and 
disregarding heat loss by radiation, obtain an expression for the variation of 
temperature in the base plate, and evaluate the temperatures at the inner and 
the outer surfaces. 

SOLUTION The base plate of an iron is considered. The variation of tempera- 
ture in the plate and the surface temperatures are to be determined. 
Assumptions 1 Heat transfer is steady since there is no change with time. 
2 Heat transfer is one-dimensional since the surface area of the base plate is 
large relative to its thickness, and the thermal conditions on both sides are uni- 
form. 3 Thermal conductivity is constant. 4 There is no heat generation in the 
medium. 5 Heat transfer by radiation is negligible. 6 The upper part of the iron 
is well insulated so that the entire heat generated in the resistance wires is 
transferred to the base plate through its inner surface. 
Properties The thermal conductivity is given to be k = 15 W/m • °C. 
Analysis The inner surface of the base plate is subjected to uniform heat flux 
at a rate of 



1o 



Qo 



1200 W 
0.03 m 2 



40,000 W/m 2 



The outer side of the plate is subjected to the convection condition. Taking the 
direction normal to the surface of the wall as the x-direction with its origin on 
the inner surface, the differential equation for this problem can be expressed as 
(Fig. 2-47) 



d 2 T 

dx 2 ' 







with the boundary conditions 

dlXP) 

dx 



q = 40,000 W/m 2 



dT(L) 

_,_ = /![r(L) _ rj 

The general solution of the differential equation is again obtained by two suc- 
cessive integrations to be 



dT 

dx 



C, 



and 



T(x) = C x x + C 2 (a) 

where C 1 and C 2 are arbitrary constants. Applying the first boundary condition, 



dx 



qo 



-kC x = q Q 



C, 



Noting that dT/dx = C 1 and T(L) = C 1 L + C 2 , the application of the second 
boundary condition gives 



91 
CHAPTER 2 



Heat flux 



%-- 



Base plate 
Conduction 



rf7\0) 
dx 



Conduction 



h 
T-, 



Convection 



-k^&-=h[T(L)-T a , 



FIGURE 2-47 

The boundary conditions on the base 

plate of the iron discussed 

in Example 2-13. 



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92 
HEAT TRANSFER 



dT(L) 



kC x = h[(C t L + C 2 ) - rj 



Substituting C 1 = - q /k and solving for C 2 , we obtain 



Qo <7o 
h k 



Now substituting Cj and C 2 into the general solution (a) gives 

'L- jc , V 



T(x) = T m + 4, 



h 



(W 



which is the solution for the variation of the temperature in the plate. The tem- 
peratures at the inner and outer surfaces of the plate are determined by substi- 
tuting x = and x = L, respectively, into the relation (£>): 



7X0) = T^ + q 



20°C + (40,000 W/m 2 ) 



1 



0.005 m 



15W/m-°C 80W/m 2 -°C 



533°C 



and 



T(L) 



«.0 + 



20°C + 



40,000 W/m 2 
80 W/m 2 • °C 



520°C 



Discussion Note that the temperature of the inner surface of the base plate 
will be 13°C higher than the temperature of the outer surface when steady op- 
erating conditions are reached. Also note that this heat transfer analysis enables 
us to calculate the temperatures of surfaces that we cannot even reach. This ex- 
ample demonstrates how the heat flux and convection boundary conditions are 
applied to heat transfer problems. 




FIGURE 2-48 

Schematic for Example 2-14. 



EXAMPLE 2-14 



Heat Conduction in a Solar Heated Wall 



Consider a large plane wall of thickness L = 0.06 m and thermal conductivity 
k = 1.2 W/m • °C in space. The wall is covered with white porcelain tiles that 
have an emissivity of e = 0.85 and a solar absorptivity of a = 0.26, as shown 
in Figure 2-48. The inner surface of the wall is maintained at T x = 300 K at all 
times, while the outer surface is exposed to solar radiation that is incident at a 



rate of q s 



800 W/m 2 . The outer surface is also losing heat by radiation to 



deep space at K. Determine the temperature of the outer surface of the wall 
and the rate of heat transfer through the wall when steady operating conditions 
are reached. What would your response be if no solar radiation was incident on 
the surface? 

SOLUTION A plane wall in space is subjected to specified temperature on one 
side and solar radiation on the other side. The outer surface temperature and 
the rate of heat transfer are to be determined. 



cen58933_ch02.qxd 9/10/2002 8:46 AM Page 93 



Assumptions 1 Heat transfer is steady since there is no change with time. 
2 Heat transfer is one-dimensional since the wall is large relative to its 
thickness, and the thermal conditions on both sides are uniform. 3 Thermal 
conductivity is constant. 4 There is no heat generation. 
Properties The thermal conductivity is given to be k = 1.2 W/m • °C. 
Analysis Taking the direction normal to the surface of the wall as the 
^-direction with its origin on the inner surface, the differential equation for this 
problem can be expressed as 



d 2 T _ 

dx 2 ' 







with boundary conditions 



7X0) 



dT(L) 
-k—j- L = su[T(Lf 



300 K 



spaceJ ^s( 



where 7" space = 0. The general solution of the differential equation is again ob- 
tained by two successive integrations to be 



T(x) 



C, 



(a) 



where C x and C 2 are arbitrary constants. Applying the first boundary condition 
yields 



7X0) 



C, 



C, 



Noting that dT/dx = C x and T(L) = C^ + C 2 = CjL + T lt the application of 
the second boundary conditions gives 



dT(L) 
dx 



euT(L) 4 - aq st 



kC [ = s(j{C x L + T { ) 4 — <xq si 



Although C 1 is the only unknown in this equation, we cannot get an explicit ex- 
pression for it because the equation is nonlinear, and thus we cannot get a 
closed-form expression for the temperature distribution. This should explain 
why we do our best to avoid nonlinearities in the analysis, such as those asso- 
ciated with radiation. 

Let us back up a little and denote the outer surface temperature by T{L) = T L 
instead of T[L) = C X L + T v The application of the second boundary condition 
in this case gives 



dT(L) 

~k—H L = saTiLf 
dx 



a( 7solar 



-kC, 



atfs, 



Solving for C x gives 



C, 



a^soiar ~ eoT L 4 



Now substituting Cj and C 2 into the general solution (a), we obtain 

K^solar ~ «t7 l 4 



T(x) 



■x + T, 



(« 



(c) 



93 
CHAPTER 2 



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94 
HEAT TRANSFER 



(1) Rearrange the equation to be solved: 

T L = 310.4 - 0.240975(— £-) 
\100/ 

The equation is in the proper form since the 
left side consists of T L only. 

(2) Guess the value ofT D say 300 K, and 
substitute into the right side of the equation. 
It gives 

T L = 290.2 K 

(3) Now substitute this value ofT L into the 
right side of the equation and get 

T L = 293.1 K 

(4) Repeat step (3) until convergence to 
desired accuracy is achieved. The 
subsequent iterations give 

T L = 292.6 K 
T L = 292.7 K 
T L = 292.7 K 

Therefore, the solution is T L = 292.7 K. The 
result is independent of the initial guess. 



FIGURE 2-49 

A simple method of solving a 
nonlinear equation is to arrange the 
equation such that the unknown is 
alone on the left side while everything 
else is on the right side, and to iterate 
after an initial guess until 
convergence. 




FIGURE 2-50 

Schematic for Example 2-15. 



which is the solution for the variation of the temperature in the wall in terms of 
the unknown outer surface temperature T L . At x = £ it becomes 



°"?solar 



suTt 



L+T, 



(d) 



which is an implicit relation for the outer surface temperature T L . Substituting 
the given values, we get 



0.26 X (800 W/m 2 ) - 0.85 X (5.67 X 10- 8 W/m 2 ■ K 4 ) Tt 



1.2 W/m • K 



(0.06 m) + 300 K 



which simplifies to 



0.240975 



100/ 



This equation can be solved by one of the several nonlinear equation solvers 
available (or by the old fashioned trial-and-error method) to give (Fig. 2-49) 

T L = 292.7 K 

Knowing the outer surface temperature and knowing that it must remain con- 
stant under steady conditions, the temperature distribution in the wall can be 
determined by substituting the T L value above into Eq. (c): 



m 



0.26 X (800 W/m 2 ) - 0.85 X (5.67 X 10~ 8 W/m 2 • K 4 )(292.7 K) 4 



which simplifies to 



1.2 W/m ■ K 



T(x) = (-121.5 K/m)x + 300 K 



300 K 



Note that the outer surface temperature turned out to be lower than the inner 
surface temperature. Therefore, the heat transfer through the wall will be toward 
the outside despite the absorption of solar radiation by the outer surface. Know- 
ing both the inner and outer surface temperatures of the wall, the steady rate of 
heat conduction through the wall can be determined from 



(1.2 W/m -K) 



(300 - 292.7) K 
0.06 m 



146 W/m 2 



Discussion In the case of no incident solar radiation, the outer surface tem- 
perature, determined from Eq. (d) by setting <j S0 | ar = 0, will be T L = 284.3 K. It 
is interesting to note that the solar energy incident on the surface causes the 
surface temperature to increase by about 8 K only when the inner surface tem- 
perature of the wall is maintained at 300 K. 



EXAMPLE 2-15 Heat Loss through a Steam Pipe 

Consider a steam pipe of length L = 20 m, inner radius r x = 6 cm, outer radius 
r 2 = 8 cm, and thermal conductivity k = 20 W/m • °C, as shown in Figure 
2-50. The inner and outer surfaces of the pipe are maintained at average tem- 
peratures of 7*1 = 150°C and T 2 = 60°C, respectively. Obtain a general relation 



cen58933_ch02.qxd 9/10/2002 8:46 AM Page 95 



95 
CHAPTER 2 



for the temperature distribution inside the pipe under steady conditions, and 
determine the rate of heat loss from the steam through the pipe. 

SOLUTION A steam pipe is subjected to specified temperatures on its 

surfaces. The variation of temperature and the rate of heat transfer are to be 

determined. 

Assumptions 1 Heat transfer is steady since there is no change with time. 

2 Heat transfer is one-dimensional since there is thermal symmetry about the 

centerline and no variation in the axial direction, and thus T = T(r). 3 Thermal 

conductivity is constant. 4 There is no heat generation. 

Properties The thermal conductivity is given to be k = 20 W/m • °C. 

Analysis The mathematical formulation of this problem can be expressed as 



d_ dT 

dr \ dr 



with boundary conditions 



T(r 2 ) 







150°C 
60°C 



Integrating the differential equation once with respect to rgives 

dT „ 

where Cj is an arbitrary constant. We now divide both sides of this equation by 
a - to bring it to a readily integrable form, 



dT 

dr 



C, 



Again integrating with respect to /"gives (Fig. 2-51) 

T(r) = C, In r + C 2 



(a) 



We now apply both boundary conditions by replacing all occurrences of rand 
T(r) in Eq. (a) with the specified values at the boundaries. We get 



T(A) = r, 
T(r 2 ) = T 2 



C l In r t + C 2 = Tj 

C, In r, + C = T 2 



which are two equations in two unknowns, C : and C 2 . Solving them simultane- 
ously gives 



C, 



ln(r 2 /7-i) 



and 



C, 



t 2 - r, 

ln(r 2 /r{) 



Substituting them into Eq. (a) and rearranging, the variation of temperature 
within the pipe is determined to be 



\ln(r 2 'rij " 



T { ) + T { 



(2-58) 



The rate of heat loss from the steam is simply the total rate of heat conduction 
through the pipe, and is determined from Fourier's law to be 



Differential equation: 



Integrate: 



m=° 



dT 
dr 



Divide by r (r + 0): 

dT = C 1 
dr '' 
Integrate again: 

T(r) = C, In r + C 2 
which is the general solution. 



FIGURE 2-51 

Basic steps involved in the solution 

of the steady one-dimensional 

heat conduction equation in 

cylindrical coordinates. 



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96 
HEAT TRANSFER 



x£ cylinder 



-kA 



dT 

dr 



-k(2irrL) ■ 



-2<nkLC, = 2-nkL , , , ' 
ln(r 2 /r,) 



(2-59) 



The numerical value of the rate of heat conduction through the pipe is deter- 
mined by substituting the given values 

(150 - 60)°C 
Q = 2.(20 W/m ■ °C)(20 m) ln(Q Q8/Q Q6) = 786 kW 

DISCUSSION Note that the total rate of heat transfer through a pipe is con- 
stant, but the heat flux is not since it decreases in the direction of heat trans- 
fer with increasing radius since q = Q/(2nrL). 




FIGURE 2-52 

Schematic for Example 2-16. 



EXAMPLE 2-16 Heat Conduction through a Spherical Shell 

Consider a spherical container of inner radius r x = 8 cm, outer radius r z = 
10 cm, and thermal conductivity k = 45 W/m • °C, as shown in Figure 2-52. 
The inner and outer surfaces of the container are maintained at constant tem- 
peratures of 7i = 200°C and T 2 = 80°C, respectively, as a result of some chem- 
ical reactions occurring inside. Obtain a general relation for the temperature 
distribution inside the shell under steady conditions, and determine the rate of 
heat loss from the container. 

SOLUTION A spherical container is subjected to specified temperatures on its 

surfaces. The variation of temperature and the rate of heat transfer are to be 

determined. 

Assumptions 1 Heat transfer is steady since there is no change with time. 

2 Heat transfer is one-dimensional since there is thermal symmetry about the 

midpoint, and thus T = T[r). 3 Thermal conductivity is constant. 4 There is no 

heat generation. 

Properties The thermal conductivity is given to be k = 45 W/m • °C. 

Analysis The mathematical formulation of this problem can be expressed as 



d_l 2 dT 
dr \ dr 







with boundary conditions 



T{r0 = T { = 200°C 
T(r 2 ) = T 2 = 80°C 

Integrating the differential equation once with respect to /-yields 

,dT 



dr 



C, 



where C x is an arbitrary constant. We now divide both sides of this equation by 
r 2 to bring it to a readily integrable form, 

dT = C± 
dr r 2 



cen58933_ch02.qxd 9/10/2002 8:46 AM Page 97 



Again integrating with respect to r gives 

T(r) = - -y- + C 2 



(a) 



We now apply both boundary conditions by replacing all occurrences of rand 
T(r) in the relation above by the specified values at the boundaries. We get 



T(r 2 ) = T 2 



C, 



r 2 



Co 



which are two equations in two unknowns, C, and C 2 . Solving them simultane- 
ously gives 



C, 



r, 



■ (Tj — T 2 ) and C 2 



r 2 T 2 - r,T, 



Substituting into Eq. (a), the variation of temperature within the spherical shell 
is determined to be 



T(r) 



r(r 2 - /-,) 



(T t 



r 2 T 2 - r.r, 



(2-60) 



The rate of heat loss from the container is simply the total rate of heat conduc- 
tion through the container wall and is determined from Fourier's law 



Sc spher 



sphere 



-kA 



dT 
dr 



Q 

-fe(4irr 2 ) — = -\iskC x = A^kr x r 2 ■ 
r- 



(2-61) 



The numerical value of the rate of heat conduction through the wall is deter- 
mined by substituting the given values to be 

(200 - 80)°C 
Q = 4tt(45 W/m • °C)(0.08 m)(0.10 m) ( q 10 _ g } m = 27,140 W 

Discussion Note that the total rate of heat transfer through a spherical shell is 
constant, but the heat flux, q = QIAur 2 , is not since it decreases in the direc- 
tion of heat transfer with increasing radius as shown in Figure 2-53. 



2-6 HEAT GENERATION IN A SOLID 

Many practical heat transfer applications involve the conversion of some form 
of energy into thermal energy in the medium. Such mediums are said to in- 
volve internal heat generation, which manifests itself as a rise in temperature 
throughout the medium. Some examples of heat generation are resistance 
heating in wires, exothermic chemical reactions in a solid, and nuclear reac- 
tions in nuclear fuel rods where electrical, chemical, and nuclear energies are 
converted to heat, respectively (Fig. 2-54). The absorption of radiation 
throughout the volume of a semitransparent medium such as water can also be 
considered as heat generation within the medium, as explained earlier. 



97 
CHAPTER 2 




«2- 



Q 2 



27.14kW 



27.14kW 



: 337.5 kW/m- 



: 216.0 kW/m- 



\ 2 4rc(0.10mr 

FIGURE 2-53 

During steady one-dimensional 

heat conduction in a spherical (or 

cylindrical) container, the total rate 

of heat transfer remains constant, 

but the heat flux decreases with 

increasing radius. 




Electric 

resistance 

wires 

FIGURE 2-54 

Heat generation in solids is 
commonly encountered in practice. 



cen58933_ch02.qxd 9/10/2002 8:46 AM Page 98 



98 
HEAT TRANSFER 



Heat generation is usually expressed per unit volume of the medium, and is 
denoted by g, whose unit is W/m 3 . For example, heat generation in an electri- 
cal wire of outer radius r and length L can be expressed as 



V„ 



I 2 R e 
vr}L 



(W/m 3 ) 



(2-62) 



h,T, 




Q = E 



FIGURE 2-55 

At steady conditions, the entire heat 
generated in a solid must leave the 
solid through its outer surface. 



where / is the electric current and R e is the electrical resistance of the wire. 

The temperature of a medium rises during heat generation as a result of the 
absorption of the generated heat by the medium during transient start-up 
period. As the temperature of the medium increases, so does the heat transfer 
from the medium to its surroundings. This continues until steady operating 
conditions are reached and the rate of heat generation equals the rate of heat 
transfer to the surroundings. Once steady operation has been established, the 
temperature of the medium at any point no longer changes. 

The maximum temperature T mm in a solid that involves uniform heat gener- 
ation will occur at a location farthest away from the outer surface when the 
outer surface of the solid is maintained at a constant temperature T s . For ex- 
ample, the maximum temperature occurs at the midplane in a plane wall, at 
the centerline in a long cylinder, and at the midpoint in a sphere. The temper- 
ature distribution within the solid in these cases will be symmetrical about the 
center of symmetry. 

The quantities of major interest in a medium with heat generation are the 
surface temperature T s and the maximum temperature T max that occurs in the 
medium in steady operation. Below we develop expressions for these two 
quantities for common geometries for the case of uniform heat generation 
(g = constant) within the medium. 

Consider a solid medium of surface area A s , volume V, and constant thermal 
conductivity k, where heat is generated at a constant rate of g per unit volume. 
Heat is transferred from the solid to the surrounding medium at T„, with a 
constant heat transfer coefficient of h. All the surfaces of the solid are main- 
tained at a common temperature T s . Under steady conditions, the energy bal- 
ance for this solid can be expressed as (Fig. 2-55) 



( Rate of ^ 

heat transfer 

I from the solid / 



Rate of 1 

energy generation 
\ within the solid j 



(2-63) 



or 



Q =gV 



(W) 



(2-64) 



Disregarding radiation (or incorporating it in the heat transfer coefficient h), 
the heat transfer rate can also be expressed from Newton's law of cooling as 



Q = hA s (T, - r„) 



(W) 



(2-65) 



Combining Eqs. 2-64 and 2-65 and solving for the surface temperature 
T s gives 



T=T^ + 



gV 

hA, 



(2-66) 



cen58933_ch02.qxd 9/10/2002 8:46 AM Page 99 



For a large plane wall of thickness 2L (A s = 2j4 wall and V = 2LA wall ), a long 
solid cylinder of radius r (A s = 2irr L and V = Tir^L), and a solid sphere of 



radius r (A s = 4ttt„ and V = |irr|), Eq. 2-66 reduces to 



h 

2h 

T — T A- 

J s, sphere * <» ni. 



T = T 4- 

-* 5, plane wall x °° ' / 



T = T + 

-* 5, cylinder L °o ' ^^ 



(2-67) 
(2-68) 
(2-69) 



Note that the rise in surface temperature T s is due to heat generation in the 
solid. 

Reconsider heat transfer from a long solid cylinder with heat generation. 
We mentioned above that, under steady conditions, the entire heat generated 
within the medium is conducted through the outer surface of the cylinder. 
Now consider an imaginary inner cylinder of radius r within the cylinder 
(Fig. 2-56). Again the heat generated within this inner cylinder must be equal 
to the heat conducted through the outer surface of this inner cylinder. That is, 
from Fourier's law of heat conduction, 



-kA 



dT 

' dr ' 



gV r 



(2-70) 



99 
CHAPTER 2 




FIGURE 2-56 

Heat conducted through a cylindrical 

shell of radius r is equal to the heat 

generated within a shell. 



where A,. = 2irrL and V r = irr 2 L at any location r. Substituting these expres- 
sions into Eq. 2-70 and separating the variables, we get 



-k{2TtrL) =£■ = e(Trr 2 L) 
ar 



dT 



2k 



rdr 



Integrating from r = where T(0) = T to r = r where T(r ) = T s yields 



AT = T 

max, cylinder o 



8[l 
Ak 



(2-71) 



where T is the centerline temperature of the cylinder, which is the maximum 
temperature, and AT max is the difference between the centerline and the sur- 
face temperatures of the cylinder, which is the maximum temperature rise in 
the cylinder above the surface temperature. Once Ar max is available, the cen- 
terline temperature can easily be determined from (Fig. 2-57) 



T + AT 



(2-72) 



The approach outlined above can also be used to determine the maximum tem- 
perature rise in a plane wall of thickness 2L and a solid sphere of radius r , 
with these results: 



AT, 



max, plane wall r\ t 

max, sphere sik. 



AT, 



(2-73) 
(2-74) 




r* Symmetry 
line 

FIGURE 2-57 

The maximum temperature in 

a symmetrical solid with uniform 

heat generation occurs at its center. 



cen58933_ch02.qxd 9/10/2002 8:46 AM Page 1C 



100 
HEAT TRANSFER 



Again the maximum temperature at the center can be determined from 
Eq. 2-72 by adding the maximum temperature rise to the surface temperature 
of the solid. 




FIGURE 2-58 

Schematic for Example 2-17. 



EXAMPLE 2-17 Centerline Temperature of a Resistance Heater 

A 2-kW resistance heater wire whose thermal conductivity is k = 15 W/m • °C 
has a diameter of D = 4 mm and a length of L = 0.5 m, and is used to boil 
water (Fig. 2-58). If the outer surface temperature of the resistance wire is T s = 
105°C, determine the temperature at the center of the wire. 

SOLUTION The surface temperature of a resistance heater submerged in water 
is to be determined. 

Assumptions 1 Heat transfer is steady since there is no change with time. 
2 Heat transfer is one-dimensional since there is thermal symmetry about the 
centerline and no change in the axial direction. 3 Thermal conductivity is con- 
stant. 4 Heat generation in the heater is uniform. 
Properties The thermal conductivity is given to be k = 15 W/m • °C. 
Analysis The 2-kW resistance heater converts electric energy into heat at a rate 
of 2 kW. The heat generation per unit volume of the wire is 






G« 



2000 W 



irrJL ir(0.002 m) 2 (0.5 m) 



0.318 X 10 9 W/m 3 



Then the center temperature of the wire is determined from Eq. 2-71 to be 
T = T 



4k 



(0.318 X 10 9 W/m 3 )(0.002m) 2 
105°C + - ttttt^tt^^ = 126°C 



4 X (15 W/m • °C) 



Discussion Note that the temperature difference between the center and the 
surface of the wire is 21°C. 




FIGURE 2-59 

Schematic for Example 2-18. 



We have developed these relations using the intuitive energy balance ap- 
proach. However, we could have obtained the same relations by setting up the 
appropriate differential equations and solving them, as illustrated in Examples 
2-18 and 2-19. 



EXAMPLE 2-18 Variation of Temperature in a Resistance Heater 

A long homogeneous resistance wire of radius r = 0.2 in. and thermal con- 
ductivity k = 7.8 Btu/h • ft • °F is being used to boil water at atmospheric pres- 
sure by the passage of electric current, as shown in Figure 2-59. Heat is 
generated in the wire uniformly as a result of resistance heating at a rate of g = 
2400 Btu/h ■ in 3 . If the outer surface temperature of the wire is measured to be 
T s = 226 C F, obtain a relation for the temperature distribution, and determine 
the temperature at the centerline of the wire when steady operating conditions 
are reached. 



cen58933_ch02.qxd 9/10/2002 8:46 AM Page 101 



SOLUTION This heat transfer problem is similar to the problem in Example 
2-17, except that we need to obtain a relation for the variation of temperature 
within the wire with r. Differential equations are well suited for this purpose. 
Assumptions 1 Heat transfer is steady since there is no change with time. 
2 Heat transfer is one-dimensional since there is no thermal symmetry about 
the centerline and no change in the axial direction. 3 Thermal conductivity is 
constant. 4 Heat generation in the wire is uniform. 

Properties The thermal conductivity is given to be k = 7.8 Btu/h • ft ■ C F. 
Analysis The differential equation which governs the variation of temperature 
in the wire is simply Eq. 2-27, 



ld_( dT\ l 
r drV dr k 







This is a second-order linear ordinary differential equation, and thus its general 
solution will contain two arbitrary constants. The determination of these con- 
stants requires the specification of two boundary conditions, which can be 
taken to be 



T(r ) = T S = 226°F 



and 



dT(0) 
dr 







The first boundary condition simply states that the temperature of the outer 
surface of the wire is 226°F. The second boundary condition is the symmetry 
condition at the centerline, and states that the maximum temperature in the 
wire will occur at the centerline, and thus the slope of the temperature at r = 
must be zero (Fig. 2-60). This completes the mathematical formulation of the 
problem. 

Although not immediately obvious, the differential equation is in a form that 
can be solved by direct integration. Multiplying both sides of the equation by r 
and rearranging, we obtain 

d_l dT\_l 
dry dr k r 



Integrating with respect to r gives 

dT 
r Tr = 



k 2 



C, 



(a) 



since the heat generation is constant, and the integral of a derivative of a func- 
tion is the function itself. That is, integration removes a derivative. It is conve- 
nient at this point to apply the second boundary condition, since it is related to 
the first derivative of the temperature, by replacing all occurrences of rand 
dT/dr in Eq. (a) by zero. It yields 



X 



dT(0) 
dr 



2k 



X0 + C, 



C, = 



101 
CHAPTER 2 



T J(r) 



dT(0) 
dr 



FIGURE 2-60 

The thermal symmetry condition at the 

centerline of a wire in which heat 

is generated uniformly. 



cen58933_ch02.qxd 9/10/2002 8:47 AM Page 102 



102 
HEAT TRANSFER 



Thus C x cancels from the solution. We now divide Eq. (a) by rto bring it to a 
readily integrable form, 



dT = 
dr 

Again integrating with respect to r gives 

g_ 
4k 



T{r) 



2k 



r 2 + C 7 



m 



We now apply the first boundary condition by replacing all occurrences of r by 
r and all occurrences of T by 7" s . We get 



Jk r ° + C > 



c, 



\k rs 



Substituting this C 2 relation into Eq. (b) and rearranging give 



T{r) 



T - + -hM-'*> 



(c) 



which is the desired solution for the temperature distribution in the wire as a 
function of r. The temperature at the centerline (r = 0) is obtained by replacing 
r in Eq. (c) by zero and substituting the known quantities, 

„ m rr , 8 2 „„, ot , . 2400 Btu/h - in 3 /l2inA . , .,„„ 

HO) = T s + - ^ = 226 F + 4x(78Btu/h . ft .o F) (T7TJ (a2 m ' } = 263 F 

Discussion The temperature of the centerline will be 37 C F above the tempera- 
ture of the outer surface of the wire. Note that the expression above for the cen- 
terline temperature is identical to Eq. 2-71, which was obtained using an 
energy balance on a control volume. 



Interface 




45°C 



Ceramic layer 



FIGURE 2-61 

Schematic for Example 2-19. 



EXAMPLE 2-19 Heat Conduction in a Two-Layer Medium 

Consider a long resistance wire of radius r x = 0.2 cm and thermal conductivity 
/f wire = 15 W/m • °C in which heat is generated uniformly as a result of re- 
sistance heating at a constant rate of g = 50 W/cm 3 (Fig. 2-61). The wire 
is embedded in a 0.5-cm-thick layer of ceramic whose thermal conductivity is 
Ceramic =1-2 W/m • °C. If the outer surface temperature of the ceramic layer 
is measured to be 7" s = 45°C, determine the temperatures at the center of the 
resistance wire and the interface of the wire and the ceramic layer under steady 
conditions. 



SOLUTION The surface and interface temperatures of a resistance wire cov- 
ered with a ceramic layer are to be determined. 

Assumptions 1 Heat transfer is steady since there is no change with time. 
2 Heat transfer is one-dimensional since this two-layer heat transfer problem 
possesses symmetry about the centerline and involves no change in the axial di- 
rection, and thus T = T(r). 3 Thermal conductivities are constant. 4 Heat gen- 
eration in the wire is uniform. 



Properties It is given that /c wire =15 W/m • °C and k a 



= 1.2 W/m • ° C. 



cen58933_ch02.qxd 9/10/2002 8:47 AM Page 103 



Analysis Letting 7, denote the unknown interface temperature, the heat trans- 
fer problem in the wire can be formulated as 



]_Cl_ I rf?w,re 

r drV dr 







with 



-* wire( r l) ~~ Tj 

rfr wirc (0) 



dr 







This problem was solved in Example 2-18, and its solution was determined 
to be 



* wirev) ~~ -'/ 



4fc» 



W " r 2 ) 



(a) 



Noting that the ceramic layer does not involve any heat generation and its 
outer surface temperature is specified, the heat conduction problem in that 
layer can be expressed as 



d I dT a 



dr 



with 



dr 



to = T, 

■ to = T s 







45°C 



This problem was solved in Example 2-15, and its solution was determined 
to be 



ln(r/r t ) 



(W 



We have already utilized the first interface condition by setting the wire and ce- 
ramic layer temperatures equal to T, at the interface r = r x . The interface tem- 
perature 7", is determined from the second interface condition that the heat flux 
in the wire and the ceramic layer at r = r 1 must be the same: 



gmre to 

dr 



dT r , 



,ic to 



dr 



9 



T >(l 



ln(r 2 to 



Solving for 7, and substituting the given values, the interface temperature is de- 
termined to be 



gr{ 



2k, 



ceramic 



r 2 

In t + T s 

M 



(50 X 10 6 W/m 3 )(0.002m) 2 0.007 m 



2(1.2 W/m • °C) 



In 



0.002 m 



+ 45° C = 149.4°C 



Knowing the interface temperature, the temperature at the centerline (r = 0) is 
obtained by substituting the known quantities into Eq. (a), 



:(0) 



grr 



149.4°C 



(50 X 10 6 W/m 3 )(0.002 m) 2 
4 X (15 W/m • °C) 



152.7°C 



103 
CHAPTER 2 



cen58933_ch02.qxd 9/10/2002 8:47 AM Page 104 



104 
HEAT TRANSFER 



Thus the temperature of the centerline will be slightly above the interface 
temperature. 

Discussion This example demonstrates how steady one-dimensional heat con- 
duction problems in composite media can be solved. We could also solve this 
problem by determining the heat flux at the interface by dividing the total heat 
generated in the wire by the surface area of the wire, and then using this value 
as the specifed heat flux boundary condition for both the wire and the ceramic 
layer. This way the two problems are decoupled and can be solved separately. 



500 
400 
300 

200 



£ 



100 



50 



-g 20 



£ 



10 



































,J^ Copper _ 
'T^Gold 


























T 


in 


?s 


ten 
















































































































I 










































-s 


tainles 
AISI 


s St 
30 


ee 
4 


,- 












































































































































































Fuse 


d quart 


z 









1 

100 300 500 1000 2000 4000 

Temperature (K) 

FIGURE 2-62 

Variation of the thermal conductivity 
of some solids with temperature. 



2-7 ■ VARIABLE THERMAL CONDUCTIVITY, k(T) 

You will recall from Chapter 1 that the thermal conductivity of a material, in 
general, varies with temperature (Fig. 2-62). However, this variation is mild 
for many materials in the range of practical interest and can be disregarded. In 
such cases, we can use an average value for the thermal conductivity and treat 
it as a constant, as we have been doing so far. This is also common practice for 
other temperature-dependent properties such as the density and specific heat. 

When the variation of thermal conductivity with temperature in a specified 
temperature interval is large, however, it may be necessary to account for this 
variation to minimize the error. Accounting for the variation of the thermal 
conductivity with temperature, in general, complicates the analysis. But in the 
case of simple one-dimensional cases, we can obtain heat transfer relations in 
a straightforward manner. 

When the variation of thermal conductivity with temperature k(T) is known, 
the average value of the thermal conductivity in the temperature range be- 
tween T, and T 2 can be determined from 



k(T)dT 



T 2 



(2-75) 



This relation is based on the requirement that the rate of heat transfer through 
a medium with constant average thermal conductivity & ave equals the rate of 
heat transfer through the same medium with variable conductivity k(T). Note 
that in the case of constant thermal conductivity k(T) = k, Eq. 2-75 reduces to 
£ ave = k, as expected. 

Then the rate of steady heat transfer through a plane wall, cylindrical layer, 
or spherical layer for the case of variable thermal conductivity can be deter- 
mined by replacing the constant thermal conductivity k in Eqs. 2-57, 2-59, 
and 2-61 by the k awe expression (or value) from Eq. 2-75: 



2= k A 
plane wall ave 



Ti - T 2 A rT, 

k{T)dT 



H cylinder •^' 1 ™ave ^ 



L 

T, - T, 



2-rrL rr, 



Q 



sphere 



4TrL vp r,r. 



ln(r 2 /r,) ln(r 2 /r,) J r , 
T, - To 4m-,/-., rr, 



k(T)dT 



ave' V 2 f. 






k{T)dT 



(2-76) 
(2-77) 
(2-78) 



cen58933_ch02.qxd 9/10/2002 8:47 AM Page 105 



The variation in thermal conductivity of a material with temperature in the 
temperature range of interest can often be approximated as a linear function 
and expressed as 



k{T) = *o(l + $T) 



(2-79) 



where (3 is called the temperature coefficient of thermal conductivity. The 

average value of thermal conductivity in the temperature range T x to T 2 in this 
case can be determined from 



*o(l + $T)dT 



k 1 + ■ 



To + T, 



k(lme) 



(2-80) 



Note that the average thermal conductivity in this case is equal to the thermal 
conductivity value at the average temperature. 

We have mentioned earlier that in a plane wall the temperature varies 
linearly during steady one-dimensional heat conduction when the thermal 
conductivity is constant. But this is no longer the case when the thermal con- 
ductivity changes with temperature, even linearly, as shown in Figure 2-63. 



105 
CHAPTER 2 



T 


Plane wall 
k{T) = k Q (l+PT) 




V 


p>o 




0- 


j3<0 


■ T 2 




L x 



FIGURE 2-63 

The variation of temperature 

in a plane wall during steady 

one-dimensional heat conduction 

for the cases of constant and variable 

thermal conductivity. 



EXAMPLE 2-20 Variation of Temperature in a Wall with k[T) 

Consider a plane wall of thickness L whose thermal conductivity varies linearly 
in a specified temperature range as k(T) = k (l + p7") where k and p are con- 
stants. The wall surface at x = is maintained at a constant temperature of 7"! 
while the surface at x = £ is maintained at T z , as shown in Figure 2-64. 
Assuming steady one-dimensional heat transfer, obtain a relation for (a) the 
heat transfer rate through the wall and (b) the temperature distribution T{x) in 
the wall. 

SOLUTION A plate with variable conductivity is subjected to specified tem- 
peratures on both sides. The variation of temperature and the rate of heat trans- 
fer are to be determined. 

Assumptions 1 Heat transfer is given to be steady and one-dimensional. 
2 Thermal conductivity varies linearly. 3 There is no heat generation. 
Properties The thermal conductivity is given to be k(T) = k (\ + p7"). 
Analysis (a) The rate of heat transfer through the wall can be determined from 



Q 



If A — 

^ave ^ J 



where A is the heat conduction area of the wall and 



£(r ave ) = k 1 + (3 



is the average thermal conductivity (Eq. 2-80). 



{b) To determine the temperature distribution in the wall, we begin with 
Fourier's law of heat conduction, expressed as 



Q = -k(T)A 



dT 

dx 



k(T) = kJi+pr> 





,nyi 


J-* Y 




Plane 






wall 




()• 




T 




L x 



FIGURE 2-64 

Schematic for Example 2-20. 



cen58933_ch02.qxd 9/10/2002 8:47 AM Page 106 



106 
HEAT TRANSFER 



where the rate of conduction heat transfer Q and the area A are constant. 
Separating variables and integrating from x = where 7~(0) = 7"! to any x where 
T(x) = T, we get 

" Qdx = -A ( k{T)dT 

Substituting k(T) = k Q (\ + fJ7") and performing the integrations we obtain 

Qx = -Ak [(T - 7\) + P(J- - T?)/2] 
Substituting the Q expression from part (a) and rearranging give 



t 2 + |r- 



2^ave £ 

P^o L 



(r, - r 2 ) - r, 2 



P 



o 



which is a quadratic equation in the unknown temperature T. Using the qua- 
dratic formula, the temperature distribution T(x) in the wall is determined to be 



T(x) 



1 
P 



1 



2k 



P 2 P*o L 



(Xi 



T 2 ) + n + -r l 



The proper sign of the square root term (+ or -) is determined from the re- 
quirement that the temperature at any point within the medium must remain 
between 7"! and T z . This result explains why the temperature distribution in a 
plane wall is no longer a straight line when the thermal conductivity varies with 
temperature. 



.k(T) = k n (l+pT) 



Bronze 
plate 



FIGURE 2-65 

Schematic for Example 2-21. 



EXAMPLE 2-21 Heat Conduction through a Wall with k(T) 

Consider a 2-m-high and 0.7-m-wide bronze plate whose thickness is 0.1 m. 
One side of the plate is maintained at a constant temperature of 600 K while 
the other side is maintained at 400 K, as shown in Figure 2-65. The thermal 
conductivity of the bronze plate can be assumed to vary linearly in that temper- 
ature range as k(T) = k (\ + (57) where k = 38 W/m • K and (5 = 9.21 X 
10~ 4 K _1 . Disregarding the edge effects and assuming steady one-dimensional 
heat transfer, determine the rate of heat conduction through the plate. 

SOLUTION A plate with variable conductivity is subjected to specified tem- 
peratures on both sides. The rate of heat transfer is to be determined. 
Assumptions 1 Heat transfer is given to be steady and one-dimensional. 
2 Thermal conductivity varies linearly. 3 There is no heat generation. 
Properties The thermal conductivity is given to be k(T) = k (l + p7"). 
Analysis The average thermal conductivity of the medium in this case is sim- 
ply the value at the average temperature and is determined from 

/ T 2 + r, 

^ave — k(T ave) — *o( 1 + P 9 



(38 W/m -K) 
55.5 W/m ■ K 



1 + (9.21 X lO^Kr 1 ) 



(600 + 400) K 



cen58933_ch02.qxd 9/10/2002 8:47 AM Page 107 



Then the rate of heat conduction through the plate can be determined from Eq. 
2-76 to be 

Q ~ k-aveA j 

(600 - 400)K 

= (55.5 W/m • K)(2 m X 0.7 m) — : — = 155,400 W 

0.1 m 

Discussion We would have obtained the same result by substituting the given 
k{T) relation into the second part of Eq. 2-76 and performing the indicated 
integration. 



107 
CHAPTER 2 



TOPIC OF SPECIAL INTEREST 



A Brief Review of Differential Equations* 

As we mentioned in Chapter 1, the description of most scientific problems 
involves relations that involve changes in some key variables with respect 
to each other. Usually the smaller the increment chosen in the changing 
variables, the more general and accurate the description. In the limiting 
case of infinitesimal or differential changes in variables, we obtain differ- 
ential equations, which provide precise mathematical formulations for the 
physical principles and laws by representing the rates of change as deriva- 
tives. Therefore, differential equations are used to investigate a wide vari- 
ety of problems in science and engineering, including heat transfer. 

Differential equations arise when relevant physical laws and principles 
are applied to a problem by considering infinitesimal changes in the vari- 
ables of interest. Therefore, obtaining the governing differential equation 
for a specific problem requires an adequate knowledge of the nature of the 
problem, the variables involved, appropriate simplifying assumptions, and 
the applicable physical laws and principles involved, as well as a careful 
analysis (Fig. 2-66). 

An equation, in general, may involve one or more variables. As the name 
implies, a variable is a quantity that may assume various values during a 
study. A quantity whose value is fixed during a study is called a constant. 
Constants are usually denoted by the earlier letters of the alphabet such as 
a, b, c, and d, whereas variables are usually denoted by the later ones such 
as t, x, y, and z- A variable whose value can be changed arbitrarily is called 
an independent variable (or argument). A variable whose value depends 
on the value of other variables and thus cannot be varied independently is 
called a dependent variable (or a function). 

A dependent variable y that depends on a variable x is usually denoted as 
y(x) for clarity. However, this notation becomes very inconvenient and 
cumbersome when y is repeated several times in an expression. In such 
cases it is desirable to denote y(x) simply as y when it is clear that y is a 
function of x. This shortcut in notation improves the appearance and the 





Physical problem 




Identify 
important 
variables 




Make 
reasonable 


Apply 
relevant 




assumptions and 
approximations 


jhysical laws 








' ' 






A differential equation 




Apply 

applicable 

solution 

technique 


Boundary 
and initial 
conditions 




Solution of the problem 





FIGURE 2-66 

Mathematical modeling 
of physical problems. 



*This section can be skipped if desired without a loss in continuity. 



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108 
HEAT TRANSFER 




X x + Ax 

FIGURE 2-67 

The derivative of a function at a point 
represents the slope of the tangent 
line of the function at that point. 




FIGURE 2-68 

Graphical representation of 
partial derivative dzldx. 



readability of the equations. The value of y at a fixed number a is denoted 
by y(a). 

The derivative of a function y(x) at a point is equivalent to the slope of 
the tangent line to the graph of the function at that point and is defined as 
(Fig. 2-67) 



y'W 



dy(x) 
dx 



hm — 

Ax->0 AX 



lim 



y(x + Ax) - y(x) 
Ax 



(2-81) 



Here Ax represents a (small) change in the independent variable x and is 
called an increment of x. The corresponding change in the function y is 
called an increment of y and is denoted by Ay. Therefore, the derivative of 
a function can be viewed as the ratio of the increment Ay of the function to 
the increment Ax of the independent variable for very small Ax. Note that 
Ay and thus y'(x) will be zero if the function y does not change with x. 

Most problems encountered in practice involve quantities that change 
with time t, and their first derivatives with respect to time represent the rate 
of change of those quantities with time. For example, if N(t) denotes the 
population of a bacteria colony at time /, then the first derivative N' = 
dNIdt represents the rate of change of the population, which is the amount 
the population increases or decreases per unit time. 

The derivative of the first derivative of a function y is called the second 
derivative of y, and is denoted by y" or d 2 y/dx 2 . In general, the derivative of 
the (n — l)st derivative of y is called the nth derivative of y and is denoted 
by y (,,) or d n y/dx". Here, n is a positive integer and is called the order of the 
derivative. The order n should not be confused with the degree of a deriva- 
tive. For example, y'" is the third-order derivative of y, but (y') 3 is the third 
degree of the first derivative of y. Note that the first derivative of a function 
represents the slope or the rate of change of the function with the indepen- 
dent variable, and the second derivative represents the rate of change of the 
slope of the function with the independent variable. 

When a function y depends on two or more independent variables such 
as x and t, it is sometimes of interest to examine the dependence of the 
function on one of the variables only. This is done by taking the derivative 
of the function with respect to that variable while holding the other vari- 
ables constant. Such derivatives are called partial derivatives. The first 
partial derivatives of the function y(x, t) with respect to x and t are defined 
as (Fig. 2-68) 



dy y(x + Ax, t) - y(x, t) 

— = hm -. 

OX Ax -^ o Ax 

y(x, t + At) - y(x, t) 



dy 

T^ = llm 

at \t^o 



At 



(2-82) 
(2-83) 



Note that when finding dy/dx we treat (asa constant and differentiate y 
with respect to x. Likewise, when finding dy/dt we treat x as a constant and 
differentiate y with respect to /. 

Integration can be viewed as the inverse process of differentiation. Inte- 
gration is commonly used in solving differential equations since solving a 
differential equation is essentially a process of removing the derivatives 



cen58933_ch02.qxd 9/10/2002 8:47 AM Page 109 



from the equation. Differentiation is the process of finding y'(x) when a 
function y(x) is given, whereas integration is the process of finding the 
function y(x) when its derivative y'(x) is given. The integral of this deriva- 
tive is expressed as 



J y'(x)dx = J dy = y(x) + C 



(2-84) 



since y'(x)dx = dy and the integral of the differential of a function is the 
function itself (plus a constant, of course). In Eq. 2-84, x is the integration 
variable and C is an arbitrary constant called the integration constant. 

The derivative of y(x) + C is y'(x) no matter what the value of the con- 
stant C is. Therefore, two functions that differ by a constant have the same 
derivative, and we always add a constant C during integration to recover 
this constant that is lost during differentiation. The integral in Eq. 2-84 is 
called an indefinite integral since the value of the arbitrary constant C is 
indefinite. The described procedure can be extended to higher-order deriv- 
atives (Fig. 2-69). For example, 



109 
CHAPTER 2 



I 


dy- 


= y + C 


.(>' 


dx = 


-y + C 


J" 


dx = 


= y' + c 


\' 


dx = 


= y" + C 


I" 


dx = 


--y<"-» + C 



y"{x)dx = y'(x) + C 



(2-85) 



FIGURE 2-69 

Some indefinite integrals 
that involve derivatives. 



This can be proved by defining a new variable u(x) = y'(x), differentiating 
it to obtain u'(x) = y"{x), and then applying Eq. 2-84. Therefore, the order 
of a derivative decreases by one each time it is integrated. 



Classification of Differential Equations 

A differential equation that involves only ordinary derivatives is called an 
ordinary differential equation, and a differential equation that involves 
partial derivatives is called a partial differential equation. Then it follows 
that problems that involve a single independent variable result in ordinary 
differential equations, and problems that involve two or more independent 
variables result in partial differential equations. A differential equation may 
involve several derivatives of various orders of an unknown function. The 
order of the highest derivative in a differential equation is the order of the 
equation. For example, the order of /" + (y") 4 = 7x 5 is 3 since it contains 
no fourth or higher order derivatives. 

You will remember from algebra that the equation 3x — 5 = is much 
easier to solve than the equation x 4 + 3x — 5 = because the first equation 
is linear whereas the second one is nonlinear. This is also true for differen- 
tial equations. Therefore, before we start solving a differential equation, we 
usually check for linearity. A differential equation is said to be linear if the 
dependent variable and all of its derivatives are of the first degree and their 
coefficients depend on the independent variable only. In other words, a dif- 
ferential equation is linear if it can be written in a form that does not in- 
volve (1) any powers of the dependent variable or its derivatives such as y 3 
or (y') 2 , (2) any products of the dependent variable or its derivatives such 
as yy' or y'y'", and (3) any other nonlinear functions of the dependent vari- 
able such as sin y or e>. If any of these conditions apply, it is nonlinear 
(Fig. 2-70). 



(a) A nonlinear equation: 
30") 2 -4vv' + e 2 



Power Product 



-6x l 



Other 
nonlinear 
functions 



(b) A linear equation: 

3x y" - 4xy' + e y = 6x 



FIGURE 2-70 

A differential equation that is 

(a) nonlinear and (b) linear. When 

checking for linearity, we examine the 

dependent variable only. 



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110 
HEAT TRANSFER 



(a) 


With constant coefficients: 








y" + 6y' — 2y = xe 


-2x 






v 








V 

Constant 




(b) 


With 


variable coefficients: 






y 


■> i 2 


xe 2 * 


AT X — 1 

V 






V 
Variable 





FIGURE 2-71 

A differential equation with 

(a) constant coefficients and 

(b) variable coefficients. 



(a) An algebraic equation: 

y 2 - 7v - 10 = 
Solution: y = 2 and y = 5 



(b) A differential equation: 

y' - v.v = o 

Solution: y = e 7 ' 



FIGURE 2-72 

Unlike those of algebraic equations, 
the solutions of differential equations 
are typically functions instead of 
discrete values. 



A linear differential equation, however, may contain (1) powers or non- 
linear functions of the independent variable, such as x 2 and cos x and 



(2) products of the dependent variable (or its derivatives) and functions of 
the independent variable, such as x 3 y' , x 2 y, and e^^y". A linear differential 
equation of order n can be expressed in the most general form as 



yM +f l (x)yf» -'>+■•■ +f„_ l (x)y' +f„(x)y = R(x) 



(2-86) 



A differential equation that cannot be put into this form is nonlinear. A 
linear differential equation in y is said to be homogeneous as well if 
R(x) = 0. Otherwise, it is nonhomogeneous. That is, each term in a linear 
homogeneous equation contains the dependent variable or one of its deriv- 
atives after the equation is cleared of any common factors. The term R(x) is 
called the nonhomogeneous term. 

Differential equations are also classified by the nature of the coefficients 
of the dependent variable and its derivatives. A differential equation is said 
to have constant coefficients if the coefficients of all the terms that involve 
the dependent variable or its derivatives are constants. If, after clearing any 
common factors, any of the terms with the dependent variable or its deriv- 
atives involve the independent variable as a coefficient, that equation is 
said to have variable coefficients (Fig. 2-71). Differential equations with 
constant coefficients are usually much easier to solve than those with vari- 
able coefficients. 

Solutions of Differential Equations 

Solving a differential equation can be as easy as performing one or more 
integrations; but such simple differential equations are usually the excep- 
tion rather than the rule. There is no single general solution method appli- 
cable to all differential equations. There are different solution techniques, 
each being applicable to different classes of differential equations. Some- 
times solving a differential equation requires the use of two or more tech- 
niques as well as ingenuity and mastery of solution methods. Some 
differential equations can be solved only by using some very clever tricks. 
Some cannot be solved analytically at all. 

In algebra, we usually seek discrete values that satisfy an algebraic equa- 
tion such as x 1 — Ix —10 = 0. When dealing with differential equations, 
however, we seek functions that satisfy the equation in a specified interval. 
For example, the algebraic equation x 2 — Ix — 10 = is satisfied by two 
numbers only: 2 and 5. But the differential equation y' — ly = is satis- 
fied by the function e lx for any value of x (Fig. 2-72). 

Consider the algebraic equation x 3 — 6x 2 + llx — 6 = 0. Obviously, 
x = 1 satisfies this equation, and thus it is a solution. However, it is not the 
only solution of this equation. We can easily show by direct substitution 
that x = 2 and x = 3 also satisfy this equation, and thus they are solutions 
as well. But there are no other solutions to this equation. Therefore, we 
say that the set 1, 2, and 3 forms the complete solution to this algebraic 
equation. 

The same line of reasoning also applies to differential equations. Typi- 
cally, differential equations have multiple solutions that contain at least one 
arbitrary constant. Any function that satisfies the differential equation on an 



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. 2-73). 



Function:/ = 3e~ 2> 








Differential equation: y" — 


4y = 





Derivatives off: 








r -- 


= -6e- 


2.v 




/" = 


= 12e- 2 






Substituting into y" 


-4y = 


0: 






f"- 


4/i 





He- 2 * - 


4X 3e 


-2x ? 









= 





Therefore, the function 3e~ 


^is £ 


solution of 


the differential equation y" 


-4y 


= 0. 



FIGURE 2-73 

Verifying that a given function is a 
solution of a differential equation. 



SUMMARY 



In this chapter we have studied the heat conduction equation 
and its solutions. Heat conduction in a medium is said to be 
steady when the temperature does not vary with time and un- 
steady or transient when it does. Heat conduction in a medium 
is said to be one-dimensional when conduction is significant 
in one dimension only and negligible in the other two di- 
mensions. It is said to be two-dimensional when conduction in 
the third dimension is negligible and three-dimensional when 
conduction in all dimensions is significant. In heat transfer 
analysis, the conversion of electrical, chemical, or nuclear 
energy into heat (or thermal) energy is characterized as heat 
generation. 

The heat conduction equation can be derived by performing 
an energy balance on a differential volume element. The one- 
dimensional heat conduction equation in rectangular, cylindri- 
cal, and spherical coordinate systems for the case of constant 
thermal conductivities are expressed as 

d 2 T , 8 1 dT 



i a 



d.\ 2 

dT 



r dr \ dr 

1 d I , dT 



dr 



dr 



k « dt 

8_ = }_dT 
k a dt 

8_ = }_dT 
k a dt 



where the property a = k/pC is the thermal diffusivity of the 
material. 

The solution of a heat conduction problem depends on the 
conditions at the surfaces, and the mathematical expressions 
for the thermal conditions at the boundaries are called the 



boundary conditions. The solution of transient heat conduction 
problems also depends on the condition of the medium at the 
beginning of the heat conduction process. Such a condition, 
which is usually specified at time t = 0, is called the initial 
condition, which is a mathematical expression for the temper- 
ature distribution of the medium initially. Complete mathemat- 
ical description of a heat conduction problem requires the 
specification of two boundary conditions for each dimension 
along which heat conduction is significant, and an initial con- 
dition when the problem is transient. The most common 
boundary conditions are the specified temperature, specified 
heat flux, convection, and radiation boundary conditions. A 
boundary surface, in general, may involve specified heat flux, 
convection, and radiation at the same time. 

For steady one-dimensional heat transfer through a plate of 
thickness L, the various types of boundary conditions at the 
surfaces at x = and x = L can be expressed as 



Specified temperature: 
T(0) = T { 



and 



T(L) 



where T { and T 2 are the specified temperatures at surfaces at 
x = and x = L. 



Specified heat flux: 

,_dT(0) 
k dx 



4o 



and 



dT(L) 
dx 



1l 



where q and q L are the specified heat fluxes at surfaces at 
x = and x = L. 



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112 
HEAT TRANSFER 



Insulation or thermal symmetry: 

dT(0) dT(L) 



dx 







and 



Convection 
dT(0) 



dx 



h,[T-M - T(0)] and 



dx 



-k-^ = h 2 [T(L)-T oc2 ] 



where h x and h 2 are the convection heat transfer coefficients 
and T^, and T„ 2 are the temperatures of the surrounding medi- 
ums on the two sides of the plate. 



Radiation: 



-k^ 1 = sMTL, l -T(0)' i ] and 

dT(L) 
-k^^ = ^MT(L) 4 -Ti urr J 



where h is the convection heat transfer coefficient. The maxi- 
mum temperature rise between the surface and the midsection 
of a medium is given by 



AT, 



= *kL 

max, plane wall n y 



AT, 



max, cylinder a k 



max, sphere /- r. 

When the variation of thermal conductivity with temperature 
k(T) is known, the average value of the thermal conductivity in 
the temperature range between T { and T 2 can be determined 

from 



f : k(T)dT 
T 2 -T t 



where s t and e 2 are the emissivities of the boundary surfaces, 
o- = 5.67 X 1CT 8 W/m 2 • K 4 is the Stefan-Boltzmann constant, 
and r surr , and r slllT 2 are the average temperatures of the sur- 
faces surrounding the two sides of the plate. In radiation calcu- 
lations, the temperatures must be in K or R. 

Interface of two bodies A and B in perfect contact at x = x : 



T A (x ) = T B (x ) and 



AT A (*o) 
dx 



dT B (jc ) 

' dx 



Then the rate of steady heat transfer through a plane wall, 
cylindrical layer, or spherical layer can be expressed as 

r, - T 2 _ A f T < 

G plane wall ~~ K ave A 7 — T I k(T)dT 



Q cylinder ^'"'^ave-^ 



Q 



4irk„„ r r,r 



T, ~ T 2 2ttL f 7 ' 
\r\{r 2 lr{) ln(r 2 /r,) ) T , 
r, - T 2 4<ir/-,r 2 f r, 



sphere ^'"' v ave'l'2 f„ — }• y 2 ~ T 



Fi \ T k(T)dT 



where k A and k B are the thermal conductivities of the layers 
A and B. 

Heat generation is usually expressed per unit volume of the 
medium and is denoted by g, whose unit is W/m 3 . Under steady 
conditions, the surface temperature T s of a plane wall of thick- 
ness 2L, a cylinder of outer radius r , and a sphere of radius r 
in which heat is generated at a constant rate of g per unit vol- 
ume in a surrounding medium at T„ can be expressed as 



Ik 

ii 

gr a 
2h 
gr 

s, sphere <» o/. 



7* = T + 

5, plane wall <» j^ 



T = T -f- 

5, cylinder ro ^ /^ 



The variation of thermal conductivity of a material with 
temperature can often be approximated as a linear function and 
expressed as 

k{T) = k (l + p7) 

where (3 is called the temperature coefficient of thermal 
conductivity. 



REFERENCES AND SUGGESTED READING 



1. W. E. Boyce and R. C. Diprima. Elementary Differential 
Equations and Boundary Value Problems. 4th ed. 
New York: John Wiley & Sons, 1986. 



2. J. R Holman. Heat Transfer. 9th ed. New York: 
McGraw-Hill, 2002. 



cen58933_ch02.qxd 9/10/2002 8:47 AM Page 113 



3. F. P. Incropera and D. P. DeWitt. Introduction to Heat 
Transfer. 4th ed. New York: John Wiley & Sons, 2002. 

4. S. S. Kutateladze. Fundamentals of Heat Transfer. 
New York: Academic Press, 1963. 



113 
CHAPTER 1 



5. M. N. Ozisik. Heat Transfer — A Basic Approach. 
New York: McGraw-Hill, 1985. 

6. F. M. White. Heat and Mass Transfer. Reading, MA: 
Addison-Wesley, 1988. 



PROBLEMS 



Introduction 

2-1C Is heat transfer a scalar or vector quantity? Explain. 
Answer the same question for temperature. 

2-2C How does transient heat transfer differ from steady 
heat transfer? How does one-dimensional heat transfer differ 
from two-dimensional heat transfer? 

2-3C Consider a cold canned drink left on a dinner table. 
Would you model the heat transfer to the drink as one-, two-, or 
three-dimensional? Would the heat transfer be steady or tran- 
sient? Also, which coordinate system would you use to analyze 
this heat transfer problem, and where would you place the ori- 
gin? Explain. 

2-4C Consider a round potato being baked in an oven. 
Would you model the heat transfer to the potato as one-, two-, 
or three-dimensional? Would the heat transfer be steady or 
transient? Also, which coordinate system would you use to 
solve this problem, and where would you place the origin? 
Explain. 




FIGURE P2-4 



*Problems designated by a "C" are concept questions, and 
students are encouraged to answer them all. Problems designated 
by an "E" are in English units, and the SI users can ignore them. 
Problems with an EES-CD icon ® are solved using EES, and 
complete solutions together with parametric studies are included 
on the enclosed CD. Problems with a computer-EES icon H are 
comprehensive in nature, and are intended to be solved with a 
computer, preferably using the EES software that accompanies 
this text. 



2-5C Consider an egg being cooked in boiling water in a 
pan. Would you model the heat transfer to the egg as one-, 
two-, or three-dimensional? Would the heat transfer be steady 
or transient? Also, which coordinate system would you use to 
solve this problem, and where would you place the origin? 
Explain. 

2-6C Consider a hot dog being cooked in boiling water in a 
pan. Would you model the heat transfer to the hot dog as one-, 
two-, or three-dimensional? Would the heat transfer be steady 
or transient? Also, which coordinate system would you use to 
solve this problem, and where would you place the origin? 
Explain. 




FIGURE P2-6 

2-7C Consider the cooking process of a roast beef in an 
oven. Would you consider this to be a steady or transient heat 
transfer problem? Also, would you consider this to be one-, 
two-, or three-dimensional? Explain. 

2-8C Consider heat loss from a 200-L cylindrical hot water 
tank in a house to the surrounding medium. Would you con- 
sider this to be a steady or transient heat transfer problem? 
Also, would you consider this heat transfer problem to be one-, 
two-, or three-dimensional? Explain. 

2-9C Does a heat flux vector at a point P on an isothermal 
surface of a medium have to be perpendicular to the surface at 
that point? Explain. 

2-10C From a heat transfer point of view, what is the differ- 
ence between isotropic and unisotropic materials? 

2-11C What is heat generation in a solid? Give examples. 

2-12C Heat generation is also referred to as energy genera- 
tion or thermal energy generation. What do you think of these 
phrases? 

2-13C In order to determine the size of the heating element 
of a new oven, it is desired to determine the rate of heat trans- 
fer through the walls, door, and the top and bottom section of 
the oven. In your analysis, would you consider this to be a 



cen58933_ch02.qxd 9/10/2002 8:47 AM Page 114 



114 
HEAT TRANSFER 



steady or transient heat transfer problem? Also, would you con- 
sider the heat transfer to be one -dimensional or multidimen- 
sional? Explain. 

2-14E The resistance wire of a 1000-W iron is 15 in. long 
and has a diameter of D = 0.08 in. Determine the rate of heat 
generation in the wire per unit volume, in Btu/h • ft 3 , and the 
heat flux on the outer surface of the wire, in Btu/h ■ ft 2 , as a re- 
sult of this heat generation. 



FIGURE P2-14E 



ity and heat generation in its simplest form, and indicate what 
each variable represents. 

2-20 Write down the one -dimensional transient heat conduc- 
tion equation for a long cylinder with constant thermal con- 
ductivity and heat generation, and indicate what each variable 
represents. 

2-21 Starting with an energy balance on a rectangular vol- 
ume element, derive the one-dimensional transient heat con- 
duction equation for a plane wall with constant thermal 
conductivity and no heat generation. 

2-22 Starting with an energy balance on a cylindrical shell 
volume element, derive the steady one-dimensional heat con- 
duction equation for a long cylinder with constant thermal con- 
ductivity in which heat is generated at a rate of g. 



2-15E [T^vfl Reconsider Problem 2-14E. Using EES (or 
h^2 other) software, evaluate and plot the surface 
heat flux as a function of wire diameter as the diameter varies 
from 0.02 to 0.20 in. Discuss the results. 

2-16 In a nuclear reactor, heat is generated uniformly in the 
5-cm-diameter cylindrical uranium rods at a rate of 7 X 10 7 
W/m 3 . If the length of the rods is 1 m, determine the rate of 
heat generation in each rod. Answer: 137 A kW 

2-17 In a solar pond, the absorption of solar energy can be 
modeled as heat generation and can be approximated by g = 
g e~ bx , where g Q is the rate of heat absorption at the top surface 
per unit volume and b is a constant. Obtain a relation for the to- 
tal rate of heat generation in a water layer of surface area A and 
thickness L at the top of the pond. 



Radiation 

beam being 

absorbed 




FIGURE P2-1 7 

2-18 Consider a large 3-cm-thick stainless steel plate in 
which heat is generated uniformly at a rate of 5 X 10 6 W/m 3 . 
Assuming the plate is losing heat from both sides, determine 
the heat flux on the surface of the plate during steady opera- 
tion. Answer: 75,000 W/m 2 

Heat Conduction Equation 

2-19 Write down the one-dimensional transient heat conduc- 
tion equation for a plane wall with constant thermal conductiv- 




FIGURE P2-22 

2-23 Starting with an energy balance on a spherical shell 
volume element, derive the one-dimensional transient heat 
conduction equation for a sphere with constant thermal con- 
ductivity and no heat generation. 




FIGURE P2-23 

2-24 Consider a medium in which the heat conduction equa- 
tion is given in its simplest form as 



dx 1 



]_dT 
a dt 



cen58933_ch02.qxd 9/10/2002 8:47 AM Page 115 



(«) Is heat transfer steady or transient? 

(b) Is heat transfer one-, two-, or three-dimensional? 

(c) Is there heat generation in the medium? 

(d) Is the thermal conductivity of the medium constant or 
variable? 

2-25 Consider a medium in which the heat conduction equa- 
tion is given in its simplest form as 



\d_ 

r dr 



rk 



dT 
dr 







115 
CHAPTER 1 




FIGURE P2-29 



(a) Is heat transfer steady or transient? 

(b) Is heat transfer one-, two-, or three-dimensional? 

(c) Is there heat generation in the medium? 

(d) Is the thermal conductivity of the medium constant or 
variable? 

2-26 Consider a medium in which the heat conduction equa- 
tion is given in its simplest form as 



r 2 dr 



_ 2 ar 

dr 



\_dT 
a dt 



(a) 
(b) 
(c) 

(d) 



Is heat transfer steady or transient? 

Is heat transfer one-, two-, or three-dimensional? 

Is there heat generation in the medium? 

Is the thermal conductivity of the medium constant or 

variable? 



2-27 Consider a medium in which the heat conduction equa- 
tion is given in its simplest form as 



d 2 T dT 
' dr 2 dr 







(fl) 
(b) 
(c) 
(d) 



Is heat transfer steady or transient? 

Is heat transfer one-, two-, or three-dimensional? 

Is there heat generation in the medium? 

Is the thermal conductivity of the medium constant or 

variable? 



2-28 Starting with an energy balance on a volume element, 
derive the two-dimensional transient heat conduction equation 
in rectangular coordinates for T(x, y, f) for the case of constant 
thermal conductivity and no heat generation. 

2-29 Starting with an energy balance on a ring-shaped vol- 
ume element, derive the two-dimensional steady heat conduc- 
tion equation in cylindrical coordinates for T(r, z) for the case 
of constant thermal conductivity and no heat generation. 

2-30 Starting with an energy balance on a disk volume ele- 
ment, derive the one-dimensional transient heat conduction 
equation for T(z, t) in a cylinder of diameter D with an insu- 
lated side surface for the case of constant thermal conductivity 
with heat generation. 

2-31 Consider a medium in which the heat conduction equa- 
tion is given in its simplest form as 



Disk 



Insulation 




FIGURE P2-30 



d 2 T d 2 T = 1 dT 
dx 2 dy 2 ~ a dt 



(a) Is heat transfer steady or transient? 

(b) Is heat transfer one-, two-, or three-dimensional? 

(c) Is there heat generation in the medium? 

(d) Is the thermal conductivity of the medium constant or 
variable? 

2-32 Consider a medium in which the heat conduction equa- 
tion is given in its simplest form as 



r dr 



kr 



dT 

dr 



d_ 
dz 



dT 

dz 







(a) Is heat transfer steady or transient? 

(b) Is heat transfer one-, two-, or three-dimensional? 

(c) Is there heat generation in the medium? 

(d) Is the thermal conductivity of the medium constant or 
variable? 

2-33 Consider a medium in which the heat conduction equa- 
tion is given in its simplest form as 



j_d_ 
r 2 dr 



dT 
dt 



1 



d 2 T 



sin 2 6 d§ 2 



\_dT 
a dt 



(a) Is heat transfer steady or transient? 

(b) Is heat transfer one-, two-, or three-dimensional? 

(c) Is there heat generation in the medium? 

(d) Is the thermal conductivity of the medium constant or 
variable? 



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116 
HEAT TRANSFER 



Boundary and Initial Conditions; 
Formulation of Heat Conduction Problems 

2-34C What is a boundary condition? How many boundary 
conditions do we need to specify for a two-dimensional heat 
transfer problem? 

2-35C What is an initial condition? How many initial condi- 
tions do we need to specify for a two-dimensional heat transfer 
problem? 

2-36C What is a thermal symmetry boundary condition? 
How is it expressed mathematically? 

2-37C How is the boundary condition on an insulated sur- 
face expressed mathematically? 

2-38C It is claimed that the temperature profile in a medium 
must be perpendicular to an insulated surface. Is this a valid 
claim? Explain. 

2-39C Why do we try to avoid the radiation boundary con- 
ditions in heat transfer analysis? 

2-40 Consider a spherical container of inner radius r x , outer 
radius r 2 , and thermal conductivity k. Express the boundary 
condition on the inner surface of the container for steady one- 
dimensional conduction for the following cases: (a) specified 
temperature of 50°C, {b) specified heat flux of 30 W/m 2 toward 
the center, (c) convection to a medium at 7V_ with a heat trans- 
fer coefficient of h. 



Spherical container 




FIGURE P2-40 

2-41 Heat is generated in a long wire of radius r at a con- 
stant rate of g per unit volume. The wire is covered with a 
plastic insulation layer. Express the heat flux boundary condi- 
tion at the interface in terms of the heat generated. 

2-42 Consider a long pipe of inner radius r lt outer radius r 2 , 
and thermal conductivity k. The outer surface of the pipe is 
subjected to convection to a medium at T a with a heat transfer 
coefficient of h, but the direction of heat transfer is not known. 
Express the convection boundary condition on the outer sur- 
face of the pipe. 

2-43 Consider a spherical shell of inner radius r u outer ra- 
dius r 2 , thermal conductivity k, and emissivity e. The outer sur- 
face of the shell is subjected to radiation to surrounding 



Express the radiation boundary condition on the outer surface 
of the shell. 

2-44 A container consists of two spherical layers, A and B, 
that are in perfect contact. If the radius of the interface is r , 
express the boundary conditions at the interface. 

2-45 Consider a steel pan used to boil water on top of an 
electric range. The bottom section of the pan is L = 0.5 cm 
thick and has a diameter of D = 20 cm. The electric heating 
unit on the range top consumes 1000 W of power during cook- 
ing, and 85 percent of the heat generated in the heating element 
is transferred uniformly to the pan. Heat transfer from the top 
surface of the bottom section to the water is by convection with 
a heat transfer coefficient of h. Assuming constant thermal 
conductivity and one-dimensional heat transfer, express the 
mathematical formulation (the differential equation and the 
boundary conditions) of this heat conduction problem during 
steady operation. Do not solve. 



Steel pan 




FIGURE P2-45 

2-46E A 2-kW resistance heater wire whose thermal con- 
ductivity is k = 10.4 Btu/h ■ ft • °F has a radius of r = 0.06 in. 
and a length of L = 15 in., and is used for space heating. As- 
suming constant thermal conductivity and one-dimensional 
heat transfer, express the mathematical formulation (the differ- 
ential equation and the boundary conditions) of this heat con- 
duction problem during steady operation. Do not solve. 

2-47 Consider an aluminum pan used to cook stew on top of 
an electric range. The bottom section of the pan is L = 0.25 cm 
thick and has a diameter of D = 18 cm. The electric heating 
unit on the range top consumes 900 W of power during cook- 
ing, and 90 percent of the heat generated in the heating element 



Aluminum pan 




FIGURE P2-47 



cen58933_ch02.qxd 9/10/2002 8:47 AM Page 117 



117 
CHAPTER 1 



is transferred to the pan. During steady operation, the temper- 
ature of the inner surface of the pan is measured to be 108°C. 
Assuming temperature-dependent thermal conductivity and 
one-dimensional heat transfer, express the mathematical for- 
mulation (the differential equation and the boundary condi- 
tions) of this heat conduction problem during steady operation. 
Do not solve. 

2-48 Water flows through a pipe at an average temperature 
of T„ = 50°C. The inner and outer radii of the pipe are r { = 
6 cm and r 2 = 6.5 cm, respectively. The outer surface of 
the pipe is wrapped with a thin electric heater that consumes 
300 W per m length of the pipe. The exposed surface of the 
heater is heavily insulated so that the entire heat generated in 
the heater is transferred to the pipe. Heat is transferred from the 
inner surface of the pipe to the water by convection with a heat 
transfer coefficient of h = 55 W/m 2 ■ °C. Assuming constant 
thermal conductivity and one-dimensional heat transfer, ex- 
press the mathematical formulation (the differential equation 
and the boundary conditions) of the heat conduction in the pipe 
during steady operation. Do not solve. 



Insulation 




Electric heater 



FIGURE P2-48 



2-49 A spherical metal ball of radius r is heated in an oven 
to a temperature of T, throughout and is then taken out of the 
oven and dropped into a large body of water at T„ where it is 
cooled by convection with an average convection heat transfer 
coefficient of h. Assuming constant thermal conductivity and 
transient one-dimensional heat transfer, express the mathemat- 
ical formulation (the differential equation and the boundary 
and initial conditions) of this heat conduction problem. Do not 
solve. 

2-50 A spherical metal ball of radius r is heated in an oven 
to a temperature of T, throughout and is then taken out of the 
oven and allowed to cool in ambient air at T„ by convection 
and radiation. The emissivity of the outer surface of the cylin- 
der is e, and the temperature of the surrounding surfaces is 
r surr . The average convection heat transfer coefficient is esti- 
mated to be h. Assuming variable thermal conductivity and 
transient one-dimensional heat transfer, express the mathemat- 
ical formulation (the differential equation and the boundary 



Radiation 




FIGURE P2-50 

and initial conditions) of this heat conduction problem. Do not 
solve. 

2-51 Consider the north wall of a house of thickness L. The 
outer surface of the wall exchanges heat by both convection 
and radiation. The interior of the house is maintained at T al , 
while the ambient air temperature outside remains at T^ 2 - The 
sky, the ground, and the surfaces of the surrounding structures 
at this location can be modeled as a surface at an effective tem- 
perature of T sky for radiation exchange on the outer surface. 
The radiation exchange between the inner surface of the wall 
and the surfaces of the walls, floor, and ceiling it faces is neg- 
ligible. The convection heat transfer coefficients on the inner 
and outer surfaces of the wall are h { and h 2 , respectively. The 
thermal conductivity of the wall material is k and the emissiv- 
ity of the outer surface is e 2 . Assuming the heat transfer 
through the wall to be steady and one-dimensional, express the 
mathematical formulation (the differential equation and the 
boundary and initial conditions) of this heat conduction prob- 
lem. Do not solve. 



Wall 




h 2 

T 



FIGURE P2-51 



cen58933_ch02.qxd 9/10/2002 8:47 AM Page HE 



118 
HEAT TRANSFER 



Solution of Steady One-Dimensional 
Heat Conduction Problems 

2-52C Consider one-dimensional heat conduction through a 
large plane wall with no heat generation that is perfectly insu- 
lated on one side and is subjected to convection and radiation 
on the other side. It is claimed that under steady conditions, the 
temperature in a plane wall must be uniform (the same every- 
where). Do you agree with this claim? Why? 

2-53C It is stated that the temperature in a plane wall with 
constant thermal conductivity and no heat generation varies 
linearly during steady one-dimensional heat conduction. Will 
this still be the case when the wall loses heat by radiation from 
its surfaces? 

2-54C Consider a solid cylindrical rod whose ends are main- 
tained at constant but different temperatures while the side sur- 
face is perfectly insulated. There is no heat generation. It is 
claimed that the temperature along the axis of the rod varies 
linearly during steady heat conduction. Do you agree with this 
claim? Why? 

2-55C Consider a solid cylindrical rod whose side surface is 
maintained at a constant temperature while the end surfaces are 
perfectly insulated. The thermal conductivity of the rod mater- 
ial is constant and there is no heat generation. It is claimed that 
the temperature in the radial direction within the rod will not 
vary during steady heat conduction. Do you agree with this 
claim? Why? 

2-56 Consider a large plane wall of thickness L = 0.4 m, 
thermal conductivity k = 2.3 W/m • °C, and surface area A = 
20 m 2 . The left side of the wall is maintained at a constant tem- 
perature of T { = 80°C while the right side loses heat by con- 
vection to the surrounding air at T„ = 15°C with a heat transfer 
coefficient of h = 24 W/m 2 • °C. Assuming constant thermal 
conductivity and no heat generation in the wall, (a) express the 
differential equation and the boundary conditions for steady 
one-dimensional heat conduction through the wall, (b) obtain a 
relation for the variation of temperature in the wall by solving 
the differential equation, and (c) evaluate the rate of heat trans- 
fer through the wall. Answer-, (c) 6030 W 

2-57 Consider a solid cylindrical rod of length 0.15 m and 
diameter 0.05 m. The top and bottom surfaces of the rod are 
maintained at constant temperatures of 20°C and 95°C, re- 
spectively, while the side surface is perfectly insulated. Deter- 
mine the rate of heat transfer through the rod if it is made of 
(a) copper, k = 380 W/m • °C, (b) steel, k = 18 W/m • °C, and 
(c) granite, k = 1.2 W/m ■ °C. 

2-58 rSpM Reconsider Problem 2-57. Using EES (or other) 
b^2 software, plot the rate of heat transfer as a func- 
tion of the thermal conductivity of the rod in the range of 
1 W/m • °C to 400 W/m • °C. Discuss the results. 

2-59 Consider the base plate of a 800-W household iron with 
a thickness of L = 0.6 cm, base area of A = 160 cm 2 , and ther- 



Base 
plate 



-85°C 



FIGURE P2-59 



mal conductivity of k = 20 W/m ■ °C. The inner surface of the 
base plate is subjected to uniform heat flux generated by the re- 
sistance heaters inside. When steady operating conditions are 
reached, the outer surface temperature of the plate is measured 
to be 85°C. Disregarding any heat loss through the upper part 
of the iron, (a) express the differential equation and the bound- 
ary conditions for steady one-dimensional heat conduction 
through the plate, (b) obtain a relation for the variation of tem- 
perature in the base plate by solving the differential equation, 
and (c) evaluate the inner surface temperature. 
Answer: (c) 100°C 

2-60 Repeat Problem 2-59 for a 1200-W iron. 

2-61 ta'M Reconsider Problem 2-59. Using the relation ob- 
1^2 tained for the variation of temperature in the base 
plate, plot the temperature as a function of the distance x in the 
range of x = to x = L, and discuss the results. Use the EES 
(or other) software. 

2-62E Consider a steam pipe of length L = 1 5 ft, inner ra- 
dius r x = 2 in., outer radius r 2 = 2.4 in., and thermal conduc- 
tivity k = 7.2 Btu/h • ft • °F. Steam is flowing through the pipe 
at an average temperature of 250°F, and the average convection 
heat transfer coefficient on the inner surface is given to be h = 
1 .25 Btu/h • ft 2 • °F . If the average temperature on the outer 




FIGURE P2-62E 



cen58933_ch02.qxd 9/10/2002 8:47 AM Page 119 



119 
CHAPTER 1 



surfaces of the pipe is T 2 = 160°F, (a) express the differential 
equation and the boundary conditions for steady one- 
dimensional heat conduction through the pipe, (b) obtain a re- 
lation for the variation of temperature in the pipe by solving the 
differential equation, and (c) evaluate the rate of heat loss from 
the steam through the pipe. Answer: (c) 16,800 Btu/h 

2-63 A spherical container of inner radius r { = 2 m, outer ra- 
dius r 2 = 2.1 m, and thermal conductivity k = 30 W/m • °C is 
filled with iced water at 0°C. The container is gaining heat by 
convection from the surrounding air at T-^ = 25 °C with a heat 
transfer coefficient of h = 18 W/m 2 ■ °C. Assuming the inner 
surface temperature of the container to be 0°C, (a) express the 
differential equation and the boundary conditions for steady 
one -dimensional heat conduction through the container, (b) ob- 
tain a relation for the variation of temperature in the container 
by solving the differential equation, and (c) evaluate the rate of 
heat gain to the iced water. 

2-64 Consider a large plane wall of thickness L = 0.3 m, 
thermal conductivity k = 2.5 W/m • °C, and surface area A = 
12 m 2 . The left side of the wall at x = is subjected to a net 
heat flux of q = 700 W/m 2 while the temperature at that sur- 
face is measured to be T x = 80°C. Assuming constant thermal 
conductivity and no heat generation in the wall, (a) express the 
differential equation and the boundary conditions for steady 
one-dimensional heat conduction through the wall, (b) obtain a 
relation for the variation of temperature in the wall by solving 
the differential equation, and (c) evaluate the temperature of 
the right surface of the wall at x = L. Answer-, (c) -4°C 



V 



FIGURE P2-64 

2-65 Repeat Problem 2-64 for a heat flux of 950 W/m 2 and 
a surface temperature of 85 °C at the left surface at x = 0. 

2-66E A large steel plate having a thickness of L = 4 in., 
thermal conductivity of k = 7.2 Btu/h • ft • °F, and an emissiv- 
ity of e = 0.6 is lying on the ground. The exposed surface of 
the plate at x = L is known to exchange heat by convection 
with the ambient air at !T„ = 90°F with an average heat transfer 
coefficient of h = 12 Btu/h ■ ft 2 • °F as well as by radiation with 
the open sky with an equivalent sky temperature of r sky = 
5 1 R. Also, the temperature of the upper surface of the plate is 
measured to be 75°F. Assuming steady one-dimensional heat 
transfer, (a) express the differential equation and the boundary 
conditions for heat conduction through the plate, (b) obtain a 
relation for the variation of temperature in the plate by solving 



Radiation 



75°F, 



jU 



h, T x 
Convection 



Plate 



Ground 



FIGURE P2-66E 



the differential equation, and (c) determine the value of the 
lower surface temperature of the plate at x = 0. 

2-67E Repeat Problem 2-66E by disregarding radiation heat 
transfer. 

2-68 When a long section of a compressed air line passes 
through the outdoors, it is observed that the moisture in the 
compressed air freezes in cold weather, disrupting and even 
completely blocking the air flow in the pipe. To avoid this 
problem, the outer surface of the pipe is wrapped with electric 
strip heaters and then insulated. 

Consider a compressed air pipe of length L = 6m, inner ra- 
dius /•[ = 3.7 cm, outer radius r 2 = 4.0 cm, and thermal con- 
ductivity k = 14 W/m • °C equipped with a 300-W strip heater. 
Air is flowing through the pipe at an average temperature of 
— 10°C, and the average convection heat transfer coefficient on 
the inner surface is h = 30 W/m 2 • °C. Assuming 15 percent of 
the heat generated in the strip heater is lost through the insula- 
tion, (a) express the differential equation and the boundary 
conditions for steady one-dimensional heat conduction through 
the pipe, (b) obtain a relation for the variation of temperature in 
the pipe material by solving the differential equation, and 
(c) evaluate the inner and outer surface temperatures of the 
pipe. Answers: (c) -3.91°C, -3.87°C 



Electric heater 



Compressed air ■ 



-10°C 



I 



Insulation 



FIGURE P2-68 



2-69 



Reconsider Problem 2-68. Using the relation ob- 
tained for the variation of temperature in the pipe 
material, plot the temperature as a function of the radius r in 



cen58933_ch02.qxd 9/10/2002 8:47 AM Page 12C 



120 
HEAT TRANSFER 



the range of r = r x to r = r 2 , and discuss the results. Use the 
EES (or other) software. 

2-70 In a food processing facility, a spherical container of 
inner radius r x = 40 cm, outer radius r 2 = 41 cm, and thermal 
conductivity k = 1 .5 W/m • °C is used to store hot water and to 
keep it at 100°C at all times. To accomplish this, the outer sur- 
face of the container is wrapped with a 500-W electric strip 
heater and then insulated. The temperature of the inner surface 
of the container is observed to be nearly 100°C at all times. As- 
suming 10 percent of the heat generated in the heater is lost 
through the insulation, (a) express the differential equation and 
the boundary conditions for steady one-dimensional heat con- 
duction through the container, (b) obtain a relation for the vari- 
ation of temperature in the container material by solving the 
differential equation, and (c) evaluate the outer surface tem- 
perature of the container. Also determine how much water at 
100°C this tank can supply steadily if the cold water enters 
at 20°C. 



Insulation 




Spherical 
container 

FIGURE P2-70 



2-71 



Reconsider Problem 2-70. Using the relation ob- 
tained for the variation of temperature in the con- 
tainer material, plot the temperature as a function of the radius 
r in the range of r = r { to r = r 2 , and discuss the results. Use 
the EES (or other) software. 

Heat Generation in a Solid 

2-72C Does heat generation in a solid violate the first law of 
thermodynamics, which states that energy cannot be created or 
destroyed? Explain. 

2-73C What is heat generation? Give some examples. 

2-74C An iron is left unattended and its base temperature 
rises as a result of resistance heating inside. When will the rate 
of heat generation inside the iron be equal to the rate of heat 
loss from the iron? 

2-75C Consider the uniform heating of a plate in an envi- 
ronment at a constant temperature. Is it possible for part of the 
heat generated in the left half of the plate to leave the plate 
through the right surface? Explain. 



2-76C Consider uniform heat generation in a cylinder and a 
sphere of equal radius made of the same material in the same 
environment. Which geometry will have a higher temperature 
at its center? Why? 

2-77 A 2-kW resistance heater wire with thermal conductiv- 
ity of k = 20 W/m ■ °C, a diameter of D = 5 mm, and a length 
of L = 0.7 m is used to boil water. If the outer surface temper- 
ature of the resistance wire is T s = 110°C, determine the tem- 
perature at the center of the wire. 



110°C 



^D^ 




I 

FIGURE P2-77 

2-78 Consider a long solid cylinder of radius r = 4 cm and 
thermal conductivity k = 25 W/m ■ °C. Heat is generated in the 
cylinder uniformly at a rate of g = 35 W/cm 3 . The side surface 
of the cylinder is maintained at a constant temperature of T s = 
80°C. The variation of temperature in the cylinder is given by 



T(r) 



grp 
k 



1 



+ T, 



Based on this relation, determine (a) if the heat conduction is 
steady or transient, (b) if it is one-, two-, or three-dimensional, 
and (c) the value of heat flux on the side surface of the cylinder 
at r = r . 

2-79 TtPM Reconsider Problem 2-78. Using the relation 
fc^S obtained for the variation of temperature in the 
cylinder, plot the temperature as a function of the radius r in 
the range of r = to r = r , and discuss the results. Use the 
EES (or other) software. 

2-80E A long homogeneous resistance wire of radius r = 
0.25 in. and thermal conductivity k = 8.6 Btu/h ■ ft • °F is being 
used to boil water at atmospheric pressure by the passage of 



Water 




h 




_ _, 


► 





I" 








^ Resistance 
heater 



FIGURE P2-80E 



cen58933_ch02.qxd 9/10/2002 8:47 AM Page 121 



electric current. Heat is generated in the wire uniformly as a 
result of resistance heating at a rate of g = 1800 Btu/h • in 3 . 
The heat generated is transferred to water at 212°F by con- 
vection with an average heat transfer coefficient of h = 820 
Btu/h • ft 2 ■ °F. Assuming steady one-dimensional heat transfer, 
(a) express the differential equation and the boundary condi- 
tions for heat conduction through the wire, (b) obtain a relation 
for the variation of temperature in the wire by solving the dif- 
ferential equation, and (c) determine the temperature at the 
centerline of the wire. Answer: (c) 290. 8°F 

2-81E [?(,">! Reconsider Problem 2-80E. Using the relation 
I^S obtained for the variation of temperature in the 
wire, plot the temperature at the centerline of the wire as a 
function of the heat generation g in the range of 400 Btu/h • in 3 
to 2400 Btu/h • in 3 , and discuss the results. Use the EES (or 
other) software. 

2-82 In a nuclear reactor, 1 -cm -diameter cylindrical uranium 
rods cooled by water from outside serve as the fuel. Heat is 
generated uniformly in the rods (k = 29.5 W/m • °C) at a rate 
of 7 X 10 7 W/m 3 . If the outer surface temperature of rods is 
175°C, determine the temperature at their center. 

s- 175°C 



Uranium rod 



FIGURE P2-82 

2-83 Consider a large 3-cm-thick stainless steel plate (k = 
15.1 W/m • °C) in which heat is generated uniformly at a rate 
of 5 X 10 5 W/m 3 . Both sides of the plate are exposed to an en- 
vironment at 30°C with a heat transfer coefficient of 60 W/m 2 
■ °C. Explain where in the plate the highest and the lowest tem- 
peratures will occur, and determine their values. 

2-84 Consider a large 5-cm-thick brass plate (k = 111 
W/m • °C) in which heat is generated uniformly at a rate of 
2 X 10 5 W/m 3 . One side of the plate is insulated while the other 
side is exposed to an environment at 25°C with a heat transfer 



Brass 
plate 



h 

T 



0" 



121 
CHAPTER 1 



coefficient of 44 W/m 2 ■ °C. Explain where in the plate the 
highest and the lowest temperatures will occur, and determine 
their values. 

2-85 Tu'M Reconsider Problem 2-84. Using EES (or other) 
k^^ software, investigate the effect of the heat trans- 
fer coefficient on the highest and lowest temperatures in the 
plate. Let the heat transfer coefficient vary from 20 W/m 2 ■ °C 
to 100 W/m 2 ■ °C. Plot the highest and lowest temperatures as 
a function of the heat transfer coefficient, and discuss the 
results. 

2-86 A 6-m-long 2-kW electrical resistance wire is made of 
0.2-cm-diameter stainless steel (k = 15.1 W/m • °C). The re- 
sistance wire operates in an environment at 30°C with a heat 
transfer coefficient of 140 W/m 2 • °C at the outer surface. De- 
termine the surface temperature of the wire (a) by using the ap- 
plicable relation and (b) by setting up the proper differential 
equation and solving it. Answers: (a) 409°C, (b) 409°C 

2-87E Heat is generated uniformly at a rate of 3 kW per ft 
length in a 0.08-in. -diameter electric resistance wire made of 
nickel steel {k = 5.8 Btu/h • ft • °F). Determine the temperature 
difference between the centerline and the surface of the wire. 

2-88E Repeat Problem 2-87E for a manganese wire (k = 
4.5 Btu/h ■ ft ■ °F). 

2-89 Consider a homogeneous spherical piece of radioactive 
material of radius r = 0.04 m that is generating heat at a con- 
stant rate of g = 4 X 10 7 W/m 3 . The heat generated is dissi- 
pated to the environment steadily. The outer surface of the 
sphere is maintained at a uniform temperature of 80°C and 
the thermal conductivity of the sphere is k = 15 W/m ■ °C. As- 
suming steady one -dimensional heat transfer, (a) express the 
differential equation and the boundary conditions for heat con- 
duction through the sphere, (b) obtain a relation for the varia- 
tion of temperature in the sphere by solving the differential 
equation, and (c) determine the temperature at the center of the 
sphere. 




FIGURE P2-89 



2-90 



FIGURE P2-84 



rSi'M Reconsider Problem 2-89. Using the relation ob- 
1^2 tained for the variation of temperature in the 
sphere, plot the temperature as a function of the radius r in the 
range of r = to r = r . Also, plot the center temperature of 
the sphere as a function of the thermal conductivity in the 
range of 10 W/m ■ °C to 400 W/m • °C. Discuss the results. Use 
the EES (or other) software. 



cen58933_ch02.qxd 9/10/2002 8:47 AM Page 122 



122 
HEAT TRANSFER 



2-91 A long homogeneous resistance wire of radius r = 
5 mm is being used to heat the air in a room by the passage of 
electric current. Heat is generated in the wire uniformly at a 
rate of g = 5 X 10 7 W/m 3 as a result of resistance heating. If 
the temperature of the outer surface of the wire remains at 
1 80°C, determine the temperature at r = 2 mm after steady op- 
eration conditions are reached. Take the thermal conductivity 
of the wire to be k = 8 W/m ■ °C. Answer: 212. 8°C 



180°C 






FIGURE P2-91 



2-92 Consider a large plane wall of thickness L = 0.05 m. 
The wall surface at x = is insulated, while the surface at x = 
L is maintained at a temperature of 30°C. The thermal conduc- 
tivity of the wall is k = 30 W/m • °C, and heat is generated in 
the wall at a rate of g = g <?~°' 5,/L W/m 3 where j = 8X 10 6 
W/m 3 . Assuming steady one-dimensional heat transfer, (a) ex- 
press the differential equation and the boundary conditions for 
heat conduction through the wall, (b) obtain a relation for the 
variation of temperature in the wall by solving the differential 
equation, and (c) determine the temperature of the insulated 
surface of the wall. Answer: (c) 314°C 

2-93 [ft^S Reconsider Problem 2-92. Using the relation 
H^«2 given for the heat generation in the wall, plot the 
heat generation as a function of the distance x in the range of 
x = to x = L, and discuss the results. Use the EES (or other) 
software. 



always equivalent to the conductivity value at the average tem- 
perature? 

2-99 Consider a plane wall of thickness L whose thermal 
conductivity varies in a specified temperature range as k(T) = 
k (l + (3T 2 ) where k and (3 are two specified constants. The 
wall surface at x = is maintained at a constant temperature of 
7*!, while the surface at x = L is maintained at T 2 . Assuming 
steady one-dimensional heat transfer, obtain a relation for the 
heat transfer rate through the wall. 

2-100 Consider a cylindrical shell of length L, inner radius 
r u and outer radius r 2 whose thermal conductivity varies 
linearly in a specified temperature range as k(T) = k (l + (37) 
where k and (3 are two specified constants. The inner surface 
of the shell is maintained at a constant temperature of 7\, while 
the outer surface is maintained at T 2 . Assuming steady one- 
dimensional heat transfer, obtain a relation for (a) the heat 
transfer rate through the wall and (b) the temperature distribu- 
tion T(r) in the shell. 



Cylindrical 




FIGURE P2-1 00 



Variable Thermal Conductivity, k[T) 

2-94C Consider steady one-dimensional heat conduction in 
a plane wall, long cylinder, and sphere with constant thermal 
conductivity and no heat generation. Will the temperature in 
any of these mediums vary linearly? Explain. 

2-95C Is the thermal conductivity of a medium, in general, 
constant or does it vary with temperature? 

2-96C Consider steady one-dimensional heat conduction in 
a plane wall in which the thermal conductivity varies linearly. 
The error involved in heat transfer calculations by assuming 
constant thermal conductivity at the average temperature is 
(a) none, {b) small, or (c) significant. 

2-97C The temperature of a plane wall during steady one- 
dimensional heat conduction varies linearly when the thermal 
conductivity is constant. Is this still the case when the ther- 
mal conductivity varies linearly with temperature? 

2-98C When the thermal conductivity of a medium varies 
linearly with temperature, is the average thermal conductivity 



2-101 Consider a spherical shell of inner radius r x and outer 
radius r 2 whose thermal conductivity varies linearly in a speci- 
fied temperature range as k(T) = k (l + (37) where k and (3 
are two specified constants. The inner surface of the shell is 
maintained at a constant temperature of 7\ while the outer sur- 
face is maintained at T 2 . Assuming steady one-dimensional 
heat transfer, obtain a relation for (a) the heat transfer rate 
through the shell and (b) the temperature distribution T(r) in 
the shell. 

2-102 Consider a 1.5-m-high and 0.6-m-wide plate whose 
thickness is 0.15 m. One side of the plate is maintained at a 
constant temperature of 500 K while the other side is main- 
tained at 350 K. The thermal conductivity of the plate can be 
assumed to vary linearly in that temperature range as k(T) = 
k {\ + 07) where k = 25 W/m • K and = 8.7 X 10~ 4 K" 1 . 
Disregarding the edge effects and assuming steady one- 
dimensional heat transfer, determine the rate of heat conduc- 
tion through the plate. Answer: 30,800 W 



cen58933_ch02.qxd 9/10/2002 8:47 AM Page 123 



123 
CHAPTER 1 



2-103 PT^| Reconsider Problem 2-102. Using EES (or 
1^13 other) software, plot the rate of heat conduction 
through the plate as a function of the temperature of the hot 
side of the plate in the range of 400 K to 700 K. Discuss the 
results. 

Special Topic: Review of Differential Equations 

2-104C Why do we often utilize simplifying assumptions 
when we derive differential equations? 

2-105C What is a variable? How do you distinguish a de- 
pendent variable from an independent one in a problem? 

2-106C Can a differential equation involve more than one 
independent variable? Can it involve more than one dependent 
variable? Give examples. 

2-107C What is the geometrical interpretation of a deriva- 
tive? What is the difference between partial derivatives and or- 
dinary derivatives? 

2-108C What is the difference between the degree and the 
order of a derivative? 

2-109C Consider a function /(x, y) and its partial derivative 
df/dx. Under what conditions will this partial derivative be 
equal to the ordinary derivative df/dx? 

2-110C Consider a function f(x) and its derivative df/dx. 
Does this derivative have to be a function of x? 

2-111C How is integration related to derivation? 

2-112C What is the difference between an algebraic equa- 
tion and a differential equation? 

2-113C What is the difference between an ordinary differen- 
tial equation and a partial differential equation? 

2-114C How is the order of a differential equation deter- 
mined? 

2-115C How do you distinguish a linear differential equation 
from a nonlinear one? 

2-116C How do you recognize a linear homogeneous differ- 
ential equation? Give an example and explain why it is linear 
and homogeneous. 

2-117C How do differential equations with constant coeffi- 
cients differ from those with variable coefficients? Give an ex- 
ample for each type. 

2-118C What kind of differential equations can be solved by 
direct integration? 

2-119C Consider a third order linear and homogeneous dif- 
ferential equation. How many arbitrary constants will its gen- 
eral solution involve? 

Review Problems 

2-120 Consider a small hot metal object of mass m and spe- 
cific heat C that is initially at a temperature of T t . Now the ob- 
ject is allowed to cool in an environment at r„ by convection 




FIGURE P2-1 20 

with a heat transfer coefficient of h. The temperature of the 
metal object is observed to vary uniformly with time during 
cooling. Writing an energy balance on the entire metal object, 
derive the differential equation that describes the variation of 
temperature of the ball with time, Tit). Assume constant ther- 
mal conductivity and no heat generation in the object. Do not 
solve. 

2-121 Consider a long rectangular bar of length a in the 
x-direction and width b in the y-direction that is initially at a 
uniform temperature of T t . The surfaces of the bar at x = and 
y = are insulated, while heat is lost from the other two sur- 
faces by convection to the surrounding medium at temperature 
r„ with a heat transfer coefficient of h. Assuming constant 
thermal conductivity and transient two-dimensional heat trans- 
fer with no heat generation, express the mathematical formula- 
tion (the differential equation and the boundary and initial 
conditions) of this heat conduction problem. Do not solve. 




FIGURE P2-1 21 

2-122 Consider a short cylinder of radius r and height H in 
which heat is generated at a constant rate of g . Heat is lost 
from the cylindrical surface at r = r by convection to the sur- 
rounding medium at temperature T„ with a heat transfer coeffi- 
cient of h. The bottom surface of the cylinder at z = is 
insulated, while the top surface at z = H is subjected to uni- 
form heat flux q h . Assuming constant thermal conductivity and 
steady two-dimensional heat transfer, express the mathematical 
formulation (the differential equation and the boundary condi- 
tions) of this heat conduction problem. Do not solve. 

2-123E Consider a large plane wall of thickness L = 0.5 ft 
and thermal conductivity k = 1.2 Btu/h ■ ft • °F. The wall 
is covered with a material that has an emissivity of e = 0.80 
and a solar absorptivity of a = 0.45. The inner surface of the 
wall is maintained at T x = 520 R at all times, while the outer 
surface is exposed to solar radiation that is incident at a rate of 
<7soiar = 300 Btu/h • ft 2 . The outer surface is also losing heat by 



cen58933_ch02.qxd 9/10/2002 8:47 AM Page 124 



124 
HEAT TRANSFER 



Plate 



520 R 




. 'solai 



FIGURE P2-123E 

radiation to deep space at K. Determine the temperature of 
the outer surface of the wall and the rate of heat transfer 
through the wall when steady operating conditions are reached. 

Answers: 530.9 R, 26.2 Btu/h • ft 2 

2-124E Repeat Problem 2-123E for the case of no solar 
radiation incident on the surface. 

2-125 Consider a steam pipe of length L, inner radius r u 
outer radius r 2 , and constant thermal conductivity k. Steam 
flows inside the pipe at an average temperature of T, with a 
convection heat transfer coefficient of h t . The outer surface of 
the pipe is exposed to convection to the surrounding air at a 
temperature of T with a heat transfer coefficient of h B . Assum- 
ing steady one-dimensional heat conduction through the pipe, 
(a) express the differential equation and the boundary condi- 
tions for heat conduction through the pipe material, (b) obtain 
a relation for the variation of temperature in the pipe material 
by solving the differential equation, and (c) obtain a relation 
for the temperature of the outer surface of the pipe. 




FIGURE P2-1 25 

2-126 The boiling temperature of nitrogen at atmospheric 
pressure at sea level (1 atm pressure) is — 196°C. Therefore, ni- 
trogen is commonly used in low temperature scientific studies 



since the temperature of liquid nitrogen in a tank open to the at- 
mosphere will remain constant at — 196°C until the liquid ni- 
trogen in the tank is depleted. Any heat transfer to the tank will 
result in the evaporation of some liquid nitrogen, which has a 
heat of vaporization of 198 kJ/kg and a density of 810 kg/m 3 at 

1 atm. 

Consider a thick-walled spherical tank of inner radius r x = 

2 m, outer radius r 2 = 2.1 m , and constant thermal conductiv- 
ity k = 18 W/m • °C. The tank is initially filled with liquid 
nitrogen at 1 atm and — 196°C, and is exposed to ambient air 
at T^ = 20°C with a heat transfer coefficient of h = 25 
W/m 2 • °C. The inner surface temperature of the spherical tank 
is observed to be almost the same as the temperature of the ni- 
trogen inside. Assuming steady one -dimensional heat transfer, 
(a) express the differential equation and the boundary condi- 
tions for heat conduction through the tank, (b) obtain a relation 
for the variation of temperature in the tank material by solving 
the differential equation, and (c) determine the rate of evapora- 
tion of the liquid nitrogen in the tank as a result of the heat 
transfer from the ambient air. Answer: (c) 1.32 kg/s 

2-127 Repeat Problem 2-126 for liquid oxygen, which has 
a boiling temperature of — 183°C, a heat of vaporization of 
213 kJ/kg, and a density of 1 140 kg/m 3 at 1 atm. 

2-128 Consider a large plane wall of thickness L = 0.4 m 
and thermal conductivity k = 8.4 W/m • °C. There is no access 
to the inner side of the wall at x = and thus the thermal con- 
ditions on that surface are not known. However, the outer sur- 
face of the wall at x = L, whose emissivity is e = 0.7, is known 
to exchange heat by convection with ambient air at T m = 25°C 
with an average heat transfer coefficient of h = 14 W/m 2 • °C 
as well as by radiation with the surrounding surfaces at an av- 
erage temperature of T smT = 290 K. Further, the temperature of 
the outer surface is measured to be T 2 = 45 C C. Assuming 
steady one-dimensional heat transfer, (a) express the differen- 
tial equation and the boundary conditions for heat conduction 
through the plate, (b) obtain a relation for the temperature of 
the outer surface of the plate by solving the differential equa- 
tion, and (c) evaluate the inner surface temperature of the wall 
at x = 0. Answer: (c) 64.3°C 



Plane 
wall 



45°C 



h 

T 




FIGURE P2-1 28 



cen58933_ch02.qxd 9/10/2002 8:47 AM Page 125 



2-129 A 1000-W iron is left on the iron board with its base 
exposed to ambient air at 20°C. The base plate of the iron has 
a thickness of L = 0.5 cm, base area of A = 150 cm 2 , and ther- 
mal conductivity of k = 18 W/m • °C. The inner surface of the 
base plate is subjected to uniform heat flux generated by the re- 
sistance heaters inside. The outer surface of the base plate 
whose emissivity is e = 0.7, loses heat by convection to ambi- 
ent air at 7U = 22° C with an average heat transfer coefficient 
of h = 30 W/m 2 • °C as well as by radiation to the surrounding 
surfaces at an average temperature of T mn = 290 K. Dis- 
regarding any heat loss through the upper part of the iron, 
(a) express the differential equation and the boundary con- 
ditions for steady one-dimensional heat conduction through 
the plate, (b) obtain a relation for the temperature of the outer 
surface of the plate by solving the differential equation, and 
(c) evaluate the outer surface temperature. 



Iron 
base 
plate 




/; 
T 



FIGURE P2-1 29 

2-130 Repeat Problem 2-1 29 for a 1 500-W iron. 

2-131E The roof of a house consists of a 0.8-ft-thick con- 
crete slab (k = 1.1 Btu/h • ft • °F) that is 25 ft wide and 35 ft 
long. The emissivity of the outer surface of the roof is 0.8, and 
the convection heat transfer coefficient on that surface is esti- 
mated to be 3.2 Btu/h • ft 2 • °F. On a clear winter night, the am- 
bient air is reported to be at 50°F, while the night sky 
temperature for radiation heat transfer is 310 R. If the inner 



125 
CHAPTER 1 



surface temperature of the roof is T t = 62°F, determine the 
outer surface temperature of the roof and the rate of heat loss 
through the roof when steady operating conditions are reached. 

2-132 Consider a long resistance wire of radius r x = 0.3 cm 
and thermal conductivity A; wire =18 W/m • °C in which heat is 
generated uniformly at a constant rate of g = 1.5 W/cm 3 as a 
result of resistance heating. The wire is embedded in a 0.4-cm- 
thick layer of plastic whose thermal conductivity is A: plastic =1.8 
W/m ■ °C. The outer surface of the plastic cover loses heat by 
convection to the ambient air at T a = 25°C with an average 
combined heat transfer coefficient of h = 14 W/m 2 ■ °C. As- 
suming one -dimensional heat transfer, determine the tempera- 
tures at the center of the resistance wire and the wire -plastic 
layer interface under steady conditions. 
/lnswers;97.1°C, 97.3°C 




- Plastic cover 

FIGURE P2-1 32 

2-133 Consider a cylindrical shell of length L, inner radius 
r u and outer radius r 2 whose thermal conductivity varies in 
a specified temperature range as k(T) = fc (l + pr 2 ) where 
k and p are two specified constants. The inner surface of the 
shell is maintained at a constant temperature of 7", while 
the outer surface is maintained at T 2 . Assuming steady one- 
dimensional heat transfer, obtain a relation for the heat transfer 
rate through the shell. 

2-134 In a nuclear reactor, heat is generated in 1 -cm- 
diameter cylindrical uranium fuel rods at a rate of 4 X 
10 7 W/m 3 . Determine the temperature difference between the 
center and the surface of the fuel rod. Answer: 9.0°C 



• v t 



h 



Concrete 



FIGURE P2-1 31 E 



D 



Fuel rod 



FIGURE P2-1 34 

2-135 Consider a 20-cm-thick large concrete plane wall 
(k = 0.77 W/m ■ °C) subjected to convection on both sides with 
r„, = 27°C and h x = 5 W/m 2 • °C on the inside, and T x2 = 8°C 
and h 2 = 12 W/m 2 • °C on the outside. Assuming constant 
thermal conductivity with no heat generation and negligible 



cen58933_ch02.qxd 9/10/2002 8:47 AM Page 126 



126 
HEAT TRANSFER 



radiation, (a) express the differential equations and the bound- 
ary conditions for steady one-dimensional heat conduction 
through the wall, {b) obtain a relation for the variation of tem- 
perature in the wall by solving the differential equation, and 
(c) evaluate the temperatures at the inner and outer surfaces of 
the wall. 

2-136 Consider a water pipe of length L = 12 m, inner ra- 
dius r, = 15 cm, outer radius r 2 = 20 cm, and thermal conduc- 
tivity k = 20 W/m ■ °C. Heat is generated in the pipe material 
uniformly by a 25-kW electric resistance heater. The inner and 
outer surfaces of the pipe are at T t = 60°C and T 2 = 80°C, re- 
spectively. Obtain a general relation for temperature distribu- 
tion inside the pipe under steady conditions and determine the 
temperature at the center plane of the pipe. 

2-137 Heat is generated uniformly at a rate of 2.6 X 10 6 
W/m 3 in a spherical ball (k = 45 W/m • °C) of diameter 30 cm. 
The ball is exposed to iced-water at 0°C with a heat transfer co- 
efficient of 1200 W/m 2 • °C. Determine the temperatures at the 
center and the surface of the ball. 

Computer, Design, and Essay Problems 

2-138 Write an essay on heat generation in nuclear fuel rods. 
Obtain information on the ranges of heat generation, the varia- 
tion of heat generation with position in the rods, and the ab- 
sorption of emitted radiation by the cooling medium. 



2-139 f^tb Write an interactive computer program to calcu- 
xifv7 late the heat transfer rate and the value of tem- 
perature anywhere in the medium for steady one-dimensional 
heat conduction in a long cylindrical shell for any combination 
of specified temperature, specified heat flux, and convection 
boundary conditions. Run the program for five different sets of 
specified boundary conditions. 

2-140 Write an interactive computer program to calculate the 
heat transfer rate and the value of temperature anywhere in 
the medium for steady one-dimensional heat conduction in 
a spherical shell for any combination of specified tempera- 
ture, specified heat flux, and convection boundary conditions. 
Run the program for five different sets of specified boundary 
conditions. 

2-141 Write an interactive computer program to calculate the 
heat transfer rate and the value of temperature anywhere in the 
medium for steady one-dimensional heat conduction in a plane 
wall whose thermal conductivity varies linearly as k(T) = 
k (l + (37) where the constants k and p are specified by the 
user for specified temperature boundary conditions. 



cen58933_ch03.qxd 9/10/2002 8:58 AM Page 127 



STEADY HEAT CONDUCTION 



CHAPTER 



In heat transfer analysis, we are often interested in the rate of heat transfer 
through a medium under steady conditions and surface temperatures. Such 
problems can be solved easily without involving any differential equations 
by the introduction of thermal resistance concepts in an analogous manner to 
electrical circuit problems. In this case, the thermal resistance corresponds 
to electrical resistance, temperature difference corresponds to voltage, and the 
heat transfer rate corresponds to electric current. 

We start this chapter with one-dimensional steady heat conduction in 
a plane wall, a cylinder, and a sphere, and develop relations for thermal resis- 
tances in these geometries. We also develop thermal resistance relations for 
convection and radiation conditions at the boundaries. We apply this concept 
to heat conduction problems in multilayer plane walls, cylinders, and spheres 
and generalize it to systems that involve heat transfer in two or three dimen- 
sions. We also discuss the thermal contact resistance and the overall heat 
transfer coefficient and develop relations for the critical radius of insulation 
for a cylinder and a sphere. Finally, we discuss steady heat transfer from 
finned surfaces and some complex geometries commonly encountered in 
practice through the use of conduction shape factors. 



CONTENTS 

3-1 Steady Heat Conduction 
in Plane Walls 128 

3-2 Thermal Contact 
Resistance 138 

3-3 Generalized Thermal 

Resistance Networks 143 

3-4 Heat Conduction in 

Cylinders and Spheres 146 

3-5 Critical Radius 

of Insulation 153 

3-6 Heat Transfer from 

Finned Surfaces 156 

3-7 Heat Transfer in 

Common Configurations 169 

Topic of Special Interest: 

Heat Transfer Through 
Walls and Roofs 175 



cen58933_ch03.qxd 9/10/2002 8:58 AM Page 12E 



128 
HEAT TRANSFER 



20°C 



20°C 



20°C 



20°C 



20°C 



20°C 



20°C, 



11°C 



20= 



11°C 



f ♦ < 

ire 

♦ < 


• 3°C 

• 3°C 






ire 

♦ < 

11°C 

L ♦ * 

\j(x) 
11°C\ 

11°C 

' ♦ » < 


• 3°C 

• 3°C 
3°C 

At 



♦ 3°C 



3°C 



3°C 



3'C 



3'C 



3'C 



3'C 




+ Q 



3'C 



3'C 



FIGURE 3-1 

Heat flow through a wall is one- 
dimensional when the temperature of 
the wall varies in one direction only. 



3-1 - STEADY HEAT CONDUCTION IN PLANE WALLS 

Consider steady heat conduction through the walls of a house during a winter 
day. We know that heat is continuously lost to the outdoors through the wall. 
We intuitively feel that heat transfer through the wall is in the normal direc- 
tion to the wall surface, and no significant heat transfer takes place in the wall 
in other directions (Fig. 3-1). 

Recall that heat transfer in a certain direction is driven by the temperature 
gradient in that direction. There will be no heat transfer in a direction in which 
there is no change in temperature. Temperature measurements at several loca- 
tions on the inner or outer wall surface will confirm that a wall surface is 
nearly isothermal. That is, the temperatures at the top and bottom of a wall 
surface as well as at the right or left ends are almost the same. Therefore, there 
will be no heat transfer through the wall from the top to the bottom, or from 
left to right, but there will be considerable temperature difference between the 
inner and the outer surfaces of the wall, and thus significant heat transfer in 
the direction from the inner surface to the outer one. 

The small thickness of the wall causes the temperature gradient in that 
direction to be large. Further, if the air temperatures in and outside the house 
remain constant, then heat transfer through the wall of a house can be modeled 
as steady and one-dimensional. The temperature of the wall in this case 
will depend on one direction only (say the x-direction) and can be expressed 
as T(x). 

Noting that heat transfer is the only energy interaction involved in this case 
and there is no heat generation, the energy balance for the wall can be ex- 
pressed as 



( Rate of \ 

heat transfer 

linto the wall/ 



1 Rate of \ 

heat transfer 
\ out of the wall/ 



/Rate of change] 

of the energy 
\ of the wall J 



or 



i£in t^out 



dE^ 



dt 



(3-1) 



But dE mU /dt = for steady operation, since there is no change in the temper- 
ature of the wall with time at any point. Therefore, the rate of heat transfer into 
the wall must be equal to the rate of heat transfer out of it. In other words, the 
rate of heat transfer through the wall must be constant, Q cond wall = constant. 
Consider a plane wall of thickness L and average thermal conductivity k. 
The two surfaces of the wall are maintained at constant temperatures of 
T { and T 2 . For one-dimensional steady heat conduction through the wall, 
we have T(x). Then Fourier's law of heat conduction for the wall can be 
expressed as 



fie 



-kA 



dT 

dx 



(W) 



(3-2) 



where the rate of conduction heat transfer Q cond wall and the wall area A are 
constant. Thus we have dTldx = constant, which means that the temperature 



cen58933_ch03.qxd 9/10/2002 8:58 AM Page 129 



through the wall varies linearly with x. That is, the temperature distribution in 
the wall under steady conditions is a straight line (Fig. 3-2). 

Separating the variables in the above equation and integrating from x = 0, 
where T(0) = T h tox = L, where T(L) = T 2 , we get 



[I 



Gc 



dx 



kAdT 



Performing the integrations and rearranging gives 



xl- cond 



kA- 



(W) 



(3-3) 



which is identical to Eq. 3—1. Again, the rate of heat conduction through 
a plane wall is proportional to the average thermal conductivity, the wall 
area, and the temperature difference, but is inversely proportional to the 
wall thickness. Also, once the rate of heat conduction is available, the tem- 
perature T(x) at any location x can be determined by replacing T 2 in Eq. 3-3 
by T, and L by x. 

The Thermal Resistance Concept 

Equation 3-3 for heat conduction through a plane wall can be rearranged as 



Q 



r, 



cond, wall 



(W) 



(3-4) 



129 
CHAPTER 3 




L x 

FIGURE 3-2 

Under steady conditions, 

the temperature distribution in 

a plane wall is a straight line. 



where 



L_ 

kA 



(°C/W) 



(3-5) 



is the thermal resistance of the wall against heat conduction or simply the 
conduction resistance of the wall. Note that the thermal resistance of a 
medium depends on the geometry and the thermal properties of the medium. 
The equation above for heat flow is analogous to the relation for electric 
current flow I, expressed as 



R, 



(3-6) 



where R e = Llu e A is the electric resistance and \ l — V 2 is the voltage differ- 
ence across the resistance (<j e is the electrical conductivity). Thus, the rate of 
heat transfer through a layer corresponds to the electric current, the thermal 
resistance corresponds to electrical resistance, and the temperature difference 
corresponds to voltage difference across the layer (Fig. 3-3). 

Consider convection heat transfer from a solid surface of area A s and tem- 
perature T s to a fluid whose temperature sufficiently far from the surface is !T m , 
with a convection heat transfer coefficient h. Newton's law of cooling for con- 
vection heat transfer rate Q com = hA s (T s — T^) can be rearranged as 



2c 



r« 



(W) 



(3-7) 



• T, - T. 
Q = — 

T i — -WWW — ■ T 2 

R 

V -V 

. M 12 

R 



(a) Heat flow 



v i« VWW\A 



-v, 



(b) Electric current flow 

FIGURE 3-3 

Analogy between thermal 
and electrical resistance concepts. 



cen58933_ch03.qxd 9/10/2002 8:59 AM Page 13C 



130 
HEAT TRANSFER 



Solid 



r. 



AAAAAA^ 



->T X 



R. 



1 



com M ^ 

FIGURE 3-4 

Schematic for convection 
resistance at a surface. 



where 



R,, 



hA, 



(°C/W) 



(3-8) 



is the thermal resistance of the surface against heat convection, or simply the 
convection resistance of the surface (Fig. 3-4). Note that when the convec- 
tion heat transfer coefficient is very large (h — > °°), the convection resistance 
becomes zero and T s ~ T m . That is, the surface offers no resistance to convec- 
tion, and thus it does not slow down the heat transfer process. This situation is 
approached in practice at surfaces where boiling and condensation occur. Also 
note that the surface does not have to be a plane surface. Equation 3-8 for 
convection resistance is valid for surfaces of any shape, provided that the as- 
sumption of h = constant and uniform is reasonable. 

When the wall is surrounded by a gas, the radiation effects, which we have 
ignored so far, can be significant and may need to be considered. The rate of 
radiation heat transfer between a surface of emissivity e and area A s at tem- 
perature T s and the surrounding surfaces at some average temperature T smr can 
be expressed as 



Q md = suAAT? - r«J = h mi A s (T s - T m ) = T ' Tsm 

"rad 



(W) 



(3-9) 



where 



1 



"rad"™j 



(K/W) 



(3-10) 



is the thermal resistance of a surface against radiation, or the radiation re- 
sistance, and 



Q, 



-"■A-* s -*■ surr/ 



eCT (r s 2 + r s 2 urr )(7; + r surr ) 



(W/m 2 • K) 



(3-11) 



V 



T<- 



r^A/VvW — * T °° 

R 



Solid 

rad 

Q=Q +Q , 

FIGURE 3-5 

Schematic for convection and 
radiation resistances at a surface. 



is the radiation heat transfer coefficient. Note that both T s and T^ must be 
in K in the evaluation of h md . The definition of the radiation heat transfer co- 
efficient enables us to express radiation conveniently in an analogous manner 
to convection in terms of a temperature difference. But h lad depends strongly 
on temperature while /i conv usually does not. 

A surface exposed to the surrounding air involves convection and radiation 
simultaneously, and the total heat transfer at the surface is determined by 
adding (or subtracting, if in the opposite direction) the radiation and convec- 
tion components. The convection and radiation resistances are parallel to each 
other, as shown in Fig. 3-5, and may cause some complication in the thermal 
resistance network. When T smr ~ T m , the radiation effect can properly be ac- 
counted for by replacing h in the convection resistance relation by 



K, 



(W/m 2 • K) 



(3-12) 



where /z combined is the combined heat transfer coefficient. This way all the 
complications associated with radiation are avoided. 



cen58933_ch03.qxd 9/10/2002 8:59 AM Page 131 



R , + R „ + R -, 

conv, 1 wall conv, 2 



T".,- 



conv, 1 

AWvW 



wall 

AAA/VW^ 



131 
CHAPTER 3 




A/VWW^ 



-*T, 



Thermal 
^2 network 



T, -T, 



R , +R ,+fi , 

e, 1 e, 2 (*, 3 



Tj- 



=1 AAAAAA^ 



«,2 

AAAAAAA 



AAAAMA 



-.T, 



Electrical 



analogy 

FIGURE 3-6 

The thermal resistance network for heat transfer through a plane wall subjected to convection on both sides, 

and the electrical analogy. 



Thermal Resistance Network 

Now consider steady one-dimensional heat flow through a plane wall of thick- 
ness L, area A, and thermal conductivity k that is exposed to convection on 
both sides to fluids at temperatures T m \ and T m2 with heat transfer coefficients 
hi and h 2 , respectively, as shown in Fig. 3-6. Assuming T m2 < T m{ , the varia- 
tion of temperature will be as shown in the figure. Note that the temperature 
varies linearly in the wall, and asymptotically approaches T^ and T„, 2 in the 
fluids as we move away from the wall. 
Under steady conditions we have 

/ Rate of \ / Rate of \ / Rate of \ 

heat convection = heat conduction = heat convection 

\ into the wall / \ through the wall / \ from the wall / 



or 



Q =h t A(T xl -T l ) = kA- 



T, -T 7 



h 2 A(T 2 - T x2 ) 



which can be rearranged as 
Q 



l/h,A LlkA 



\lh 2 A 



r„i - r, _ r, -t 2 _t 2 - t^ 2 

D 

**rnnv 1 



"wall ^xonv, 2 



Adding the numerators and denominators yields (Fig. 3-7) 



Q 



R„ 



(W) 



(3-13) 



(3-14) 



(3-15) 




FIGURE 3-7 

A useful mathematical identity. 



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132 
HEAT TRANSFER 



FIGURE 3-8 

The temperature drop across a layer is 
proportional to its thermal resistance. 



►e= iow 




conv, 1 I 

»i* — VWWV^ 

2°C/W 



wall 

15°C/W 
AT=QR 



Y conv, 2 

-^^WVVW — ♦ r »2 

3°C/W 



where 

"total 



7T7 + 7T + 7^7 (° C/W ) 

/?,A kA h-iA 



(3-16) 



Note that the heat transfer area A is constant for a plane wall, and the rate of 
heat transfer through a wall separating two mediums is equal to the tempera- 
ture difference divided by the total thermal resistance between the mediums. 
Also note that the thermal resistances are in series, and the equivalent thermal 
resistance is determined by simply adding the individual resistances, just like 
the electrical resistances connected in series. Thus, the electrical analogy still 
applies. We summarize this as the rate of steady heat transfer between two 
surfaces is equal to the temperature difference divided by the total thermal re- 
sistance between those two surfaces. 

Another observation that can be made from Eq. 3-15 is that the ratio of the 
temperature drop to the thermal resistance across any layer is constant, and 
thus the temperature drop across any layer is proportional to the thermal 
resistance of the layer. The larger the resistance, the larger the temperature 
drop. In fact, the equation Q = AT/R can be rearranged as 



AT = QR 



(°C) 



(3-17) 



which indicates that the temperature drop across any layer is equal to the rate 
of heat transfer times the thermal resistance across that layer (Fig. 3-8). You 
may recall that this is also true for voltage drop across an electrical resistance 
when the electric current is constant. 

It is sometimes convenient to express heat transfer through a medium in an 
analogous manner to Newton's law of cooling as 



Q = UA AT 



(W) 



(3-18) 



where U is the overall heat transfer coefficient. A comparison of Eqs. 3-15 
and 3-18 reveals that 



cen58933_ch03.qxd 9/10/2002 8:59 AM Page 133 




r »i« WWW WWW 



R, = L l 

1 k t A 



-A/WWW 

R 1= L 2 



-AA/WW » r «2 



conv ' 2 h 2 A 



133 
CHAPTER 3 



FIGURE 3-9 

The thermal resistance network for 
heat transfer through a two-layer 
plane wall subjected to 
convection on both sides. 



UA 



R„ 



(3-19) 



Therefore, for a unit area, the overall heat transfer coefficient is equal to the 
inverse of the total thermal resistance. 

Note that we do not need to know the surface temperatures of the wall in or- 
der to evaluate the rate of steady heat transfer through it. All we need to know 
is the convection heat transfer coefficients and the fluid temperatures on both 
sides of the wall. The surface temperature of the wall can be determined as 
described above using the thermal resistance concept, but by taking the 
surface at which the temperature is to be determined as one of the terminal 
surfaces. For example, once Q is evaluated, the surface temperature T x can be 
determined from 



Q 



l//i, A 



(3-20) 



Multilayer Plane Walls 



In practice we often encounter plane walls that consist of several layers of dif- 
ferent materials. The thermal resistance concept can still be used to determine 
the rate of steady heat transfer through such composite walls. As you may 
have already guessed, this is done by simply noting that the conduction resis- 
tance of each wall is LlkA connected in series, and using the electrical analogy. 
That is, by dividing the temperature difference between two surfaces at known 
temperatures by the total thermal resistance between them. 

Consider a plane wall that consists of two layers (such as a brick wall with 
a layer of insulation). The rate of steady heat transfer through this two-layer 
composite wall can be expressed as (Fig. 3-9) 



Q 



(3-21) 



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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 3-10 

The evaluation of the surface and 
interface temperatures when T„ { and 
7^2 are given and Q is calculated. 



where i? tota i is the total thermal resistance, expressed as 



R„- 



R 



conv, 

l 



+ R* 



"wall, 2 ' R* 



conv. 2 






(3-22) 



The subscripts 1 and 2 in the R waU relations above indicate the first and the 
second layers, respectively. We could also obtain this result by following the 
approach used above for the single-layer case by noting that the rate of steady 
heat transfer Q through a multilayer medium is constant, and thus it must be 
the same through each layer. Note from the thermal resistance network that 
the resistances are in series, and thus the total thermal resistance is simply the 
arithmetic sum of the individual thermal resistances in the path of heat flow. 

This result for the two-layer case is analogous to the single-layer case, ex- 
cept that an additional resistance is added for the additional layer. This result 
can be extended to plane walls that consist of three or more layers by adding 
an additional resistance for each additional layer. 

Once Q is known, an unknown surface temperature Tj at any surface or in- 
terface j can be determined from 



Q 



v total, i—j 



total, ( — j 



(3-23) 



is the total thermal 



where T t is a known temperature at location i and R 
resistance between locations i and j. For example, when the fluid temperatures 
T^ and T m2 for the two-layer case shown in Fig. 3-9 are available and Q is 
calculated from Eq. 3-21, the interface temperature T 2 between the two walls 
can be determined from (Fig. 3-10) 



Q 



T-, 



r, 



-"conv, 1 "T "wall, 1 



(3-24) 




16°C 



Wall 



•2°C 



3 m 




L = 0.3m 

FIGURE 3-1 1 

Schematic for Example 3-1. 



The temperature drop across a layer is easily determined from Eq. 3-17 by 
multiplying Q by the thermal resistance of that layer. 

The thermal resistance concept is widely used in practice because it is intu- 
itively easy to understand and it has proven to be a powerful tool in the solu- 
tion of a wide range of heat transfer problems. But its use is limited to systems 
through which the rate of heat transfer Q remains constant; that is, to systems 
involving steady heat transfer with no heat generation (such as resistance 
heating or chemical reactions) within the medium. 



EXAMPLE 3-1 Heat Loss through a Wall 

Consider a 3-m-high, 5-m-wide, and 0.3-m-thick wall whose thermal con- 
ductivity is k = 0.9 W/m • °C (Fig. 3-11). On a certain day, the temperatures of 
the inner and the outer surfaces of the wall are measured to be 16°C and 2 C C, 
respectively. Determine the rate of heat loss through the wall on that day. 



cen58933_ch03.qxd 9/10/2002 8:59 AM Page 135 



SOLUTION The two surfaces of a wall are maintained at specified tempera- 
tures. The rate of heat loss through the wall is to be determined. 
Assumptions 1 Heat transfer through the wall is steady since the surface 
temperatures remain constant at the specified values. 2 Heat transfer is one- 
dimensional since any significant temperature gradients will exist in the direc- 
tion from the indoors to the outdoors. 3 Thermal conductivity is constant. 
Properties The thermal conductivity is given to be k = 0.9 W/m ■ C C. 
Analysis Noting that the heat transfer through the wall is by conduction and 
the area of the wall is/! = 3mX5m=15 m 2 , the steady rate of heat transfer 
through the wall can be determined from Eq. 3-3 to be 



Q=kA- 



T 2 , (16 

— = (0.9 W/m • °C)(15 m 2 ) 



2)°C 



0.3 m 



630 W 



We could also determine the steady rate of heat transfer through the wall by 
making use of the thermal resistance concept from 



Q 



Ar„ 



where 



ft„ 



0.3 m 



lwa " kA (0.9 W/m • °C)(15m 2 ) 
Substituting, we get 

Q 



0.02222°C/W 



(16 - 2)°C 



0.02222°C/W 



630 W 



Discussion This is the same result obtained earlier. Note that heat conduction 
through a plane wall with specified surface temperatures can be determined 
directly and easily without utilizing the thermal resistance concept. However, 
the thermal resistance concept serves as a valuable tool in more complex heat 
transfer problems, as you will see in the following examples. 



EXAMPLE 3-2 Heat Loss through a Single-Pane Window 

Consider a 0.8-m-high and 1.5-m-wide glass window with a thickness of 8 mm 
and a thermal conductivity of k = 0.78 W/m • °C. Determine the steady rate of 
heat transfer through this glass window and the temperature of its inner surface 
for a day during which the room is maintained at 20 C C while the temperature of 
the outdoors is — 10°C. Take the heat transfer coefficients on the inner and 
outer surfaces of the window to be h x = 10 W/m 2 • °C and h 2 = 40 W/m 2 • °C, 
which includes the effects of radiation. 

SOLUTION Heat loss through a window glass is considered. The rate of 
heat transfer through the window and the inner surface temperature are to be 
determined. 



135 
CHAPTER 3 



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136 
HEAT TRANSFER 



20°C- 



h, = 10 W/m 2 -°C 



glass 



♦-WWW-h — WWW-^^WWW— ♦ 



Glass 



-10°C 



h 2 = 40 W/m 2 -°C 



L = 8 mm 



FIGURE 3-1 2 

Schematic for Example 3-2. 



Assumptions 1 Heat transfer through the window is steady since the surface 
temperatures remain constant at the specified values. 2 Heat transfer through 
the wall is one-dimensional since any significant temperature gradients will ex- 
ist in the direction from the indoors to the outdoors. 3 Thermal conductivity is 
constant. 

Properties The thermal conductivity is given to be k = 0.78 W/m • °C. 
Analysis This problem involves conduction through the glass window and con- 
vection at its surfaces, and can best be handled by making use of the thermal 
resistance concept and drawing the thermal resistance network, as shown in 
Fig. 3-12. Noting that the area of the window is A = 0.8 m X 1.5 m = 1.2 m 2 , 
the individual resistances are evaluated from their definitions to be 



1 



1 



R 



h { A (10W/m 2 • °C)(1.2m 2 ) 
0.008 m 



0.08333°C/W 



slass kA (0.78 W/m • °C)( 1.2 m 2 ) 
1 1 



Rn — R n 



• 2 h 2 A (40 W/m 2 • °C)(1.2m 2 ) 



0.00855°C/W 

0.02083°C/W 



Noting that all three resistances are in series, the total resistance is 



glass 

0.1127°C/W 



R„ 



0.08333 + 0.00855 + 0.02083 



Then the steady rate of heat transfer through the window becomes 
T^-T-^ [20-(-10)]°C 

Q = -^ " = L TTT77^7^7- = 266 W 



Knowing the rate of heat transfer, the inner surface temperature of the window 
glass can be determined from 



Q 



T m , - T, 



*conv, 1 



-> Tj - r„] G^conv, 1 

= 20°C - (266 W)(0.08333°C/W) 
= -2.2°C 



Discussion Note that the inner surface temperature of the window glass will be 
-2.2°C even though the temperature of the air in the room is maintained at 
20°C. Such low surface temperatures are highly undesirable since they cause 
the formation of fog or even frost on the inner surfaces of the glass when the 
humidity in the room is high. 



EXAMPLE 3-3 Heat Loss through Double-Pane Windows 

Consider a 0.8-m-high and 1.5-m-wide double-pane window consisting of two 
4-mm-thick layers of glass (k = 0.78 W/m • °C) separated by a 10-mm-wide 
stagnant air space (k = 0.026 W/m • °C). Determine the steady rate of heat 



cen58933_ch03.qxd 9/10/2002 8:59 AM Page 137 



transfer through this double-pane window and the temperature of its inner sur- 
face for a day during which the room is maintained at 20°C while the tempera- 
ture of the outdoors is - 10°C. Take the convection heat transfer coefficients on 
the inner and outer surfaces of the window to be /?, = 10 W/m 2 • °C and h z = 
40 W/m 2 • °C, which includes the effects of radiation. 

SOLUTION A double-pane window is considered. The rate of heat transfer 
through the window and the inner surface temperature are to be determined. 
Analysis This example problem is identical to the previous one except that 
the single 8-mm-thick window glass is replaced by two 4-mm-thick glasses that 
enclose a 10-mm-wide stagnant air space. Therefore, the thermal resistance 
network of this problem will involve two additional conduction resistances cor- 
responding to the two additional layers, as shown in Fig. 3-13. Noting that the 
area of the window is again A = 0.8 m X 1.5 m = 1.2 m 2 , the individual re- 
sistances are evaluated from their definitions to be 



1 



1 



R ' ' Rconv • 1 h y A (10 W/m 2 • °C)(1.2m 2 ) 



R t — R 3 — R^ 



0.004 m 



M (0.78 W/m ■ °C)(1.2m 2 ) 
0.01 m 



0.08333°C/W 



0.00427°C/W 



Rn — Rr. 



k 2 A (0.026 W/m ■ °C)(1 .2 m 2 ) 

1 1 

'■ h 2 A (40 W/m 2 • °C)(1.2m 2 ) 



0.3205°C/W 
0.02083°C/W 



Noting that all three resistances are in series, the total resistance is 



R. 



R,„- + R. 



Rr, 



glass, 1 air glass, 2 conv, 2 

= 0.08333 + 0.00427 + 0.3205 + 0.00427 + 0.02083 
= 0.4332°C/W 

Then the steady rate of heat transfer through the window becomes 



Q 



R„ 



r„ 2 [2o-(-io)]°c 



0.4332°C/W 



69.2 W 



which is about one-fourth of the result obtained in the previous example. This 
explains the popularity of the double- and even triple-pane windows in cold 
climates. The drastic reduction in the heat transfer rate in this case is due to 
the large thermal resistance of the air layer between the glasses. 
The inner surface temperature of the window in this case will be 



QRc 



20°C - (69.2 W)(0.08333°C/W) = 14.2°C 



which is considerably higher than the -2.2°C obtained in the previous ex- 
ample. Therefore, a double-pane window will rarely get fogged. A double-pane 
window will also reduce the heat gain in summer, and thus reduce the air- 
conditioning costs. 



137 
CHAPTER 3 



Glass Glass 



20°C 



• '\AAAAA/- 




10°C 



AAAAAV • 



3 o ^-2 

FIGURE 3-13 

Schematic for Example 3-3. 



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138 
HEAT TRANSFER 



FIGURE 3-14 

Temperature distribution and heat flow 

lines along two solid plates pressed 

against each other for the case of 

perfect and imperfect contact. 





(a) Ideal (perfect) thermal contact 



(b) Actual (imperfect) thermal contact 



Applied load 




Loading shaft 
Alignment collar 
Top plate 
Steel ball 
Pencil heaters 
Heaters block 

Upper test specimen — 
Lower test specimen — 
Lower heat flux meter 



-SC 



Thermocouples 
Interface 



y= 



-t 



Cold 
fluid 



ZP* 



Cold plate 

Load cell - 

Steel ball - 

Bottom plate — H x I 

Bell jar * 

base plate 

FIGURE 3-15 

A typical experimental setup for 
the determination of thermal contact 
resistance (from Song et al., Ref. 11). 



3-2 ■ THERMAL CONTACT RESISTANCE 

In the analysis of heat conduction through multilayer solids, we assumed 
"perfect contact" at the interface of two layers, and thus no temperature drop 
at the interface. This would be the case when the surfaces are perfectly smooth 
and they produce a perfect contact at each point. In reality, however, even flat 
surfaces that appear smooth to the eye turn out to be rather rough when ex- 
amined under a microscope, as shown in Fig. 3-14, with numerous peaks and 
valleys. That is, a surface is microscopically rough no matter how smooth it 
appears to be. 

When two such surfaces are pressed against each other, the peaks will form 
good material contact but the valleys will form voids filled with air. As a re- 
sult, an interface will contain numerous air gaps of varying sizes that act as 
insulation because of the low thermal conductivity of air. Thus, an interface 
offers some resistance to heat transfer, and this resistance per unit interface 
area is called the thermal contact resistance, R c . The value of R c is deter- 
mined experimentally using a setup like the one shown in Fig. 3-15, and as 
expected, there is considerable scatter of data because of the difficulty in char- 
acterizing the surfaces. 

Consider heat transfer through two metal rods of cross-sectional area A that 
are pressed against each other. Heat transfer through the interface of these two 
rods is the sum of the heat transfers through the solid contact spots and the 
gaps in the noncontact areas and can be expressed as 



q =e t 



+ e E 



(3-25) 



It can also be expressed in an analogous manner to Newton's law of cooling as 

Q =ft f AAr interfacc (3-26) 



cen58933_ch03.qxd 9/10/2002 8:59 AM Page 139 



where A is the apparent interface area (which is the same as the cross-sectional 
area of the rods) and AT interface is the effective temperature difference at the 
interface. The quantity h c , which corresponds to the convection heat transfer 
coefficient, is called the thermal contact conductance and is expressed as 

QIA 

K = ^= (W/m 2 • °C) (3-27) 

^-* interface 

It is related to thermal contact resistance by 

h c QIA 

That is, thermal contact resistance is the inverse of thermal contact conduc- 
tance. Usually, thermal contact conductance is reported in the literature, but 
the concept of thermal contact resistance serves as a better vehicle for ex- 
plaining the effect of interface on heat transfer. Note that R c represents ther- 
mal contact resistance per unit area. The thermal resistance for the entire 
interface is obtained by dividing R c by the apparent interface area A. 

The thermal contact resistance can be determined from Eq. 3-28 by 
measuring the temperature drop at the interface and dividing it by the heat 
flux under steady conditions. The value of thermal contact resistance depends 
on the surface roughness and the material properties as well as the tem- 
perature and pressure at the interface and the type of fluid trapped at the 
interface. The situation becomes more complex when plates are fastened by 
bolts, screws, or rivets since the interface pressure in this case is nonuniform. 
The thermal contact resistance in that case also depends on the plate thick- 
ness, the bolt radius, and the size of the contact zone. Thermal contact 
resistance is observed to decrease with decreasing surface roughness 
and increasing interface pressure, as expected. Most experimentally deter- 
mined values of the thermal contact resistance fall between 0.000005 and 
0.0005 m 2 • °C/W (the corresponding range of thermal contact conductance 
is 2000 to 200,000 W/m 2 • °C). 

When we analyze heat transfer in a medium consisting of two or more lay- 
ers, the first thing we need to know is whether the thermal contact resistance 
is significant or not. We can answer this question by comparing the magni- 
tudes of the thermal resistances of the layers with typical values of thermal 
contact resistance. For example, the thermal resistance of a 1-cm-thick layer 
of an insulating material per unit surface area is 

r, £j U.U1 111 „ __ 2 Of~< f\\J 

K c , insulation ~ J ~ 0.04 W/m ■ °C ~ ^ ' 

whereas for a 1-cm-thick layer of copper, it is 

^•»«- = t = 386W/m m °C = °- 000026 m ' ■ ° C/W 

Comparing the values above with typical values of thermal contact resistance, 
we conclude that thermal contact resistance is significant and can even domi- 
nate the heat transfer for good heat conductors such as metals, but can be 



139 
CHAPTER 3 



cen58933_ch03.qxd 9/10/2002 8:59 AM Page 14C 



140 
HEAT TRANSFER 



TABLE 3-1 

Thermal contact conductance 
for aluminum plates with different 
fluids at the interface for a surface 
roughness of 10 |xm and interface 
pressure of 1 atm (from Fried, 
Ref. 5) 





Contact 


Fluid at the 


Conductance, h c , 


Interface 


W/m 2 • °C 


Air 


3640 


Helium 


9520 


Hydrogen 


13,900 


Silicone oil 


19,000 


Glycerin 


37,700 



LU- 



LU" 



10-- 



Contact pressure (psi) 

10 2 10 3 



Coated with 
tin/nickel alloy 



Bronze 



Coated with 
nickel alloy 




Coated 
aluminum 
alloy 



Jji-^T 



10 4 



10-' 



io- 



10 J 



10 J 10' 

Contact pressure (kN/irr) 



■ Uncoated 

■ Coated 



FIGURE 3-16 

Effect of metallic coatings on 
thermal contact conductance 
(from Peterson, Ref. 10). 



disregarded for poor heat conductors such as insulations. This is not surpris- 
ing since insulating materials consist mostly of air space just like the inter- 
face itself. 

The thermal contact resistance can be minimized by applying a thermally 
conducting liquid called a thermal grease such as silicon oil on the surfaces 
before they are pressed against each other. This is commonly done when at- 
taching electronic components such as power transistors to heat sinks. The 
thermal contact resistance can also be reduced by replacing the air at the in- 
terface by a better conducting gas such as helium or hydrogen, as shown in 
Table 3-1. 

Another way to minimize the contact resistance is to insert a soft metallic 
foil such as tin, silver, copper, nickel, or aluminum between the two surfaces. 
Experimental studies show that the thermal contact resistance can be reduced 
by a factor of up to 7 by a metallic foil at the interface. For maximum effec- 
tiveness, the foils must be very thin. The effect of metallic coatings on thermal 
contact conductance is shown in Fig. 3-16 for various metal surfaces. 

There is considerable uncertainty in the contact conductance data reported 
in the literature, and care should be exercised when using them. In Table 3-2 
some experimental results are given for the contact conductance between sim- 
ilar and dissimilar metal surfaces for use in preliminary design calculations. 
Note that the thermal contact conductance is highest (and thus the contact re- 
sistance is lowest) for soft metals with smooth surfaces at high pressure. 



EXAMPLE 3-4 Equivalent Thickness for Contact Resistance 

The thermal contact conductance at the interface of two 1-cm-thick aluminum 
plates is measured to be 11,000 W/m 2 ■ °C. Determine the thickness of the alu- 
minum plate whose thermal resistance is equal to the thermal resistance of the 
interface between the plates (Fig. 3-17). 

SOLUTION The thickness of the aluminum plate whose thermal resistance 
is equal to the thermal contact resistance is to be determined. 
Properties The thermal conductivity of aluminum at room temperature is 
k = 237 W/m • °C (Table A-3). 

Analysis Noting that thermal contact resistance is the inverse of thermal con- 
tact conductance, the thermal contact resistance is 



R, 



1 



1 



h c 1 1,000 W/nr 



0.909 X 10- 4 m 2 -°C/W 



For a unit surface area, the thermal resistance of a flat plate is defined as 

R 



L 
k 



where L is the thickness of the plate and k is the thermal conductivity. Setting 
R = R cl the equivalent thickness is determined from the relation above to be 



kR r 



(237 W/m • °C)(0.909 X 10~ 4 nr • °C/W) = 0.0215 m = 2.15 cm 



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141 
CHAPTER 3 



TABLE 3-2 


Thermal contact conductance of some metal surfaces in air (from various sources) 


Surface Rough- Tempera- 
Material Condition ness, |xm ture, °C 


Pressure, 
MPa 


W/m 2 


°C 



Identical Metal Pairs 












416 Stainless steel 


Ground 


2.54 


90-200 


0.3-2.5 


3800 


304 Stainless steel 


Ground 


1.14 


20 


4-7 


1900 


Aluminum 


Ground 


2.54 


150 


1.2-2.5 


11,400 


Copper 


Ground 


1.27 


20 


1.2-20 


143,000 


Copper 


Milled 


3.81 


20 


1-5 


55,500 


Copper (vacuum) 


Milled 


0.25 


30 


0.7-7 


11,400 


Dissimilar Metal Pairs 












Stainless steel- 








10 


2900 


Aluminum 




20-30 


20 


20 


3600 



Stainless steel- 
Aluminum 




1.0-2.0 


20 


10 
20 


16,400 
20,800 


Steel Ct-30- 
Aluminum 


Ground 


1.4-2.0 


20 


10 
15-35 


50,000 
59,000 


Steel Ct-30- 
Aluminum 


Milled 


4.5-7.2 


20 


10 
30 


4800 
8300 


Aluminum-Copper 


Ground 


1.3-1.4 


20 


5 
15 


42,000 
56,000 


Aluminum-Copper 


Milled 


4.4-4.5 


20 


10 
20-35 


12,000 
22,000 



*Divide the given values by 5.678 to convert to Btu/h • ft 2 • °F. 



Discussion Note that the interface between the two plates offers as much re- 
sistance to heat transfer as a 2.3-cm-thick aluminum plate. It is interesting 
that the thermal contact resistance in this case is greater than the sum of the 
thermal resistances of both plates. 



EXAMPLE 3-5 



Contact Resistance of Transistors 



Four identical power transistors with aluminum casing are attached on one side 
of a 1-cm-thick 20-cm X 20-cm square copper plate (k = 386 W/m • °C) by 
screws that exert an average pressure of 6 MPa (Fig. 3-18). The base area of 
each transistor is 8 cm 2 , and each transistor is placed at the center of a 10-cm 
X 10-cm quarter section of the plate. The interface roughness is estimated to 
be about 1.5 |xm. All transistors are covered by a thick Plexiglas layer, which is 
a poor conductor of heat, and thus all the heat generated at the junction of the 
transistor must be dissipated to the ambient at 20°C through the back surface 
of the copper plate. The combined convection/radiation heat transfer coefficient 
at the back surface can be taken to be 25 W/m 2 • °C. If the case temperature of 



Plate 

1 



1 cm 



Plate 

2 



1 cm 



Interface 



Plate 
1 

1 cm 


Equivalent ] Plate 
aluminum i 2 
layer 

2.15 cm i 1 cm 







FIGURE 3-17 

Schematic for Example 3-4. 



cen58933_ch03.qxd 9/10/2002 8:59 AM Page 142 



142 
HEAT TRANSFER 



20°C 




Copper 
plate 



FIGURE 3- 

Schematic 



70°C 
18 
for Example 



Plexiglas cover 



3-5. 



the transistor is not to exceed 70°C, determine the maximum power each 
transistor can dissipate safely, and the temperature jump at the case-plate 
interface. 

SOLUTION Four identical power transistors are attached on a copper plate. For 
a maximum case temperature of 70°C, the maximum power dissipation and the 
temperature jump at the interface are to be determined. 

Assumptions 1 Steady operating conditions exist. 2 Heat transfer can be ap- 
proximated as being one-dimensional, although it is recognized that heat con- 
duction in some parts of the plate will be two-dimensional since the plate area 
is much larger than the base area of the transistor. But the large thermal con- 
ductivity of copper will minimize this effect. 3 All the heat generated at the 
junction is dissipated through the back surface of the plate since the transistors 
are covered by a thick Plexiglas layer. 4 Thermal conductivities are constant. 

Properties The thermal conductivity of copper is given to be k = 386 
W/m ■ °C. The contact conductance is obtained from Table 3-2 to be h c = 
42,000 W/m 2 • °C, which corresponds to copper-aluminum interface for the 
case of 1.3-1.4 |xm roughness and 5 MPa pressure, which is sufficiently close 
to what we have. 

Analysis The contact area between the case and the plate is given to be 8 cm 2 , 
and the plate area for each transistor is 100 cm 2 . The thermal resistance net- 
work of this problem consists of three resistances in series (interface, plate, and 
convection), which are determined to be 



R 



1 



I 



aterface ^ (42,000 W/m 2 • °C)(8 X 10- 4 m 2 ) 
L 0.01 m 



0.030°C/W 



plate kA (386 W/m ■ °C)(0.01 m 2 ) 
1 1 



h B A (25 W/m 2 • °C)(0.01 m 2 ) 



0.0026°C/W 
4.0°C/W 



The total thermal resistance is then 



R„ 



D _(_ p _i_ p 

"•interface ' Opiate ' "~a\ 



0.030 + 0.0026 + 4.0 = 4.0326°C/W 



Note that the thermal resistance of a copper plate is very small and can be 
ignored altogether. Then the rate of heat transfer is determined to be 



Q 



AT (70 - 20)°C 



R,. 



4.0326°C/W 



12.4 W 



Therefore, the power transistor should not be operated at power levels greater 
than 12.4 W if the case temperature is not to exceed 70°C. 
The temperature jump at the interface is determined from 



AT 



QRm 



(12.4 W)(0.030°C/W) = 0.37°C 



which is not very large. Therefore, even if we eliminate the thermal contact re- 
sistance at the interface completely, we will lower the operating temperature of 
the transistor in this case by less than 0.4°C. 



cen58933_ch03.qxd 9/10/2002 8:59 AM Page 143 



3-3 - GENERALIZED THERMAL RESISTANCE 
NETWORKS 

The thermal resistance concept or the electrical analogy can also be used to 
solve steady heat transfer problems that involve parallel layers or combined 
series-parallel arrangements. Although such problems are often two- or even 
three-dimensional, approximate solutions can be obtained by assuming one- 
dimensional heat transfer and using the thermal resistance network. 

Consider the composite wall shown in Fig. 3-19, which consists of two par- 
allel layers. The thermal resistance network, which consists of two parallel re- 
sistances, can be represented as shown in the figure. Noting that the total heat 
transfer is the sum of the heat transfers through each layer, we have 



Q=Q l + Qi 



Utilizing electrical analogy, we get 



R 7 



(7, - T 2 ) 



-1 + 1 
R, R. 



where 



1 

^total 



Q 



R, R, 



T 2 



"> K 



RiR 2 

r7Tr~, 



(3-29) 



(3-30) 



(3-31) 



since the resistances are in parallel. 

Now consider the combined series-parallel arrangement shown in Fig. 
3-20. The total rate of heat transfer through this composite system can again 
be expressed as 



Q 



(3-32) 



143 
CHAPTER 3 



Insulation 



-V 



,4, 





CD *, 


© k 2 




« L ► 



— www 

R 2 

Q = Q\+Q 2 

FIGURE 3-1 9 

Thermal resistance 
network for two parallel layers. 

Insulation 



<D *i 



fc. 



L, = L n 



h, r„ 



where 



and 



^12 + ^3 + ^conv 



R 2 



L 2 



R t R 2 

/?! + R 2 



L 3 
k 3 A 3 ' 



R 3 + R D 



1 



(3-33) 



(3-34) 



Once the individual thermal resistances are evaluated, the total resistance and 
the total rate of heat transfer can easily be determined from the relations 
above. 

The result obtained will be somewhat approximate, since the surfaces of the 
third layer will probably not be isothermal, and heat transfer between the first 
two layers is likely to occur. 

Two assumptions commonly used in solving complex multidimensional 
heat transfer problems by treating them as one-dimensional (say, in the 



-MAMAA- 



Qi- 



■AWI/VV^ 



Q 

w — 

*3 *c 



-vww vwvw — ♦ 

conv 



FIGURE 3-20 

Thermal resistance network for 
combined series-parallel arrangement. 



cen58933_ch03.qxd 9/10/2002 8:59 AM Page 144 



144 
HEAT TRANSFER 



Foam 



Plaster ■ 




Brick 



1.5 cm 



22 cm 



1.5 cm 




FIGURE 3-21 

Schematic for Example 3-6. 



x-direction) using the thermal resistance network are (1) any plane wall nor- 
mal to the x-axis is isothermal (i.e., to assume the temperature to vary in the 
x-direction only) and (2) any plane parallel to the x-axis is adiabatic (i.e., to 
assume heat transfer to occur in the x-direction only). These two assumptions 
result in different resistance networks, and thus different (but usually close) 
values for the total thermal resistance and thus heat transfer. The actual result 
lies between these two values. In geometries in which heat transfer occurs pre- 
dominantly in one direction, either approach gives satisfactory results. 



EXAMPLE 3-6 Heat Loss through a Composite Wall 

A 3-m-high and 5-m-wide wall consists of long 16-cm X 22-cm cross section 
horizontal bricks (k = 0.72 W/m • °C) separated by 3-cm-thick plaster layers 
(k = 0.22 W/m • °C). There are also 2-cm-thick plaster layers on each side of 
the brick and a 3-cm-thick rigid foam (k = 0.026 W/m ■ °C) on the inner side 
of the wall, as shown in Fig. 3-21. The indoor and the outdoor temperatures are 
20°C and -10°C, and the convection heat transfer coefficients on the inner 
and the outer sides are h 1 = 10 W/m 2 • °C and h 2 = 25 W/m 2 • °C, respectively. 
Assuming one-dimensional heat transfer and disregarding radiation, determine 
the rate of heat transfer through the wall. 

SOLUTION The composition of a composite wall is given. The rate of heat 
transfer through the wall is to be determined. 

Assumptions 1 Heat transfer is steady since there is no indication of change 
with time. 2 Heat transfer can be approximated as being one-dimensional since 
it is predominantly in the x-direction. 3 Thermal conductivities are constant. 
4 Heat transfer by radiation is negligible. 

Properties The thermal conductivities are given to be k = 0.72 W/m • °C 
for bricks, k = 0.22 W/m • °C for plaster layers, and k = 0.026 W/m • °C for the 
rigid foam. 

Analysis There is a pattern in the construction of this wall that repeats itself 
every 25-cm distance in the vertical direction. There is no variation in the hori- 
zontal direction. Therefore, we consider a 1-m-deep and 0.25-m-high portion of 
the wall, since it is representative of the entire wall. 

Assuming any cross section of the wall normal to the x-direction to be 
isothermal, the thermal resistance network for the representative section of 
the wall becomes as shown in Fig. 3-21. The individual resistances are eval- 
uated as: 



1 



1 



Ri ' Rconv ' 1 h y A (10 W/m 2 • °C)(0.25 X 1 m 2 ) 
= L__ 0.03 m 

1 - foam - M - (Q Q26 w/m . o C)( q 25 xlm 2) 

L 0.02 m 



: 0.4°C/W 

4.6°C/W 



Rl Rb ****«"» kA (0.22 W/m • °C)(0.25 X 1 m 2 ) 
= 0.36°C/W 

L 0.16m 

K } - K 5 - K 



5 plaster, center ^ (0.22 W/m ' °C)(0.015 X 1 III 2 ) 

48.48°C/W 



cen58933_ch03.qxd 9/10/2002 8:59 AM Page 145 



^4 — ^h, 



0.16m 



kA (0.72 W/m ■ °C)(0.22 X 1 m 2 



1 



1 

h 2 A (25 W/m 2 • °C)(0.25 X 1 m 2 ) 



1.01°C/W 



0.16°C/W 



The three resistances R 3 , ff 4 , and R 5 in the middle are parallel, and their equiv- 
alent resistance is determined from 



R mi<l R 3 R A R 5 48.48 1.01 48.48 



1.03 W/°C 



which gives 



tfmid = 0.97°C/W 



Now all the resistances are in series, and the total resistance is 

^total = Ri + Rl + R 2 + R m U + ^6 + Ro 

= 0.4 + 4.6 + 0.36 + 0.97 + 0.36 + 0.16 
= 6.85°C/W 

Then the steady rate of heat transfer through the wall becomes 



Q 



[20 - (-10)]°C 
6.85°C/W 



4.38 W (per 0.25 m 2 surface area) 



or 4.38/0.25 = 17.5 W per m 2 area. The total area of the wall is A = 3 m X 5 
m = 15 m 2 . Then the rate of heat transfer through the entire wall becomes 



fit, 



(17.5 W/m 2 )(15 m 2 ) = 263 W 



Of course, this result is approximate, since we assumed the temperature within 
the wall to vary in one direction only and ignored any temperature change (and 
thus heat transfer) in the other two directions. 

Discussion In the above solution, we assumed the temperature at any cross 
section of the wall normal to the x-direction to be isothermal. We could also 
solve this problem by going to the other extreme and assuming the surfaces par- 
allel to the x-direction to be adiabatic. The thermal resistance network in this 
case will be as shown in Fig. 3-22. By following the approach outlined above, 
the total thermal resistance in this case is determined to be ff tota | = 6.97°C/W, 
which is very close to the value 6.85°C/W obtained before. Thus either ap- 
proach would give roughly the same result in this case. This example demon- 
strates that either approach can be used in practice to obtain satisfactory 
results. 



145 
CHAPTER 3 



T 

V 



1, 



Adiabatic 
lines 



R j 
■>!♦— VW- 



-<wv — vw — vw — vw vw— ♦ 



FIGURE 3-22 

Alternative thermal resistance 

network for Example 3-6 for the 

case of surfaces parallel to the 

primary direction of heat 

transfer being adiabatic. 



cen58933_ch03.qxd 9/10/2002 8:59 AM Page 146 



146 
HEAT TRANSFER 




FIGURE 3-23 

Heat is lost from a hot water pipe to 
the air outside in the radial direction, 
and thus heat transfer from a long 
pipe is one-dimensional. 




FIGURE 3-24 

A long cylindrical pipe (or spherical 
shell) with specified inner and outer 
surface temperatures T l and T 2 . 



3^ - HEAT CONDUCTION IN 

CYLINDERS AND SPHERES 

Consider steady heat conduction through a hot water pipe. Heat is continu- 
ously lost to the outdoors through the wall of the pipe, and we intuitively feel 
that heat transfer through the pipe is in the normal direction to the pipe surface 
and no significant heat transfer takes place in the pipe in other directions 
(Fig. 3-23). The wall of the pipe, whose thickness is rather small, separates 
two fluids at different temperatures, and thus the temperature gradient in the 
radial direction will be relatively large. Further, if the fluid temperatures in- 
side and outside the pipe remain constant, then heat transfer through the pipe 
is steady. Thus heat transfer through the pipe can be modeled as steady and 
one-dimensional. The temperature of the pipe in this case will depend on one 
direction only (the radial r-direction) and can be expressed as T = T(r). The 
temperature is independent of the azimuthal angle or the axial distance. This 
situation is approximated in practice in long cylindrical pipes and spherical 
containers. 

In steady operation, there is no change in the temperature of the pipe with 
time at any point. Therefore, the rate of heat transfer into the pipe must be 
equal to the rate of heat transfer out of it. In other words, heat transfer through 
the pipe must be constant, Q cond cyl = constant. 

Consider a long cylindrical layer (such as a circular pipe) of inner radius r x , 
outer radius r 2 , length L, and average thermal conductivity k (Fig. 3-24). The 
two surfaces of the cylindrical layer are maintained at constant temperatures 
Ti and T 2 . There is no heat generation in the layer and the thermal conductiv- 
ity is constant. For one-dimensional heat conduction through the cylindrical 
layer, we have T(r). Then Fourier's law of heat conduction for heat transfer 
through the cylindrical layer can be expressed as 



e 



cond, cyl 



-kA 



dT 
dr 



(W) 



(3-35) 



where A = 2ittL is the heat transfer area at location r. Note that A depends on 
r, and thus it varies in the direction of heat transfer. Separating the variables 
in the above equation and integrating from r = r u where T(r t ) = 7\, to r = r 2 , 
where T{r 2 ) = T 2 , gives 



r 2 Q cond. 



L 



ey I 



dr : 



kdT 



(3-36) 



Substituting A = 2tjtL and performing the integrations give 

T — T 

Q coai ^ = 2itLk^j^ (W) 



(3-37) 



since Q 



cond, cyl 



constant. This equation can be rearranged as 



xl, con 



r. 



d, cyl 



R. 



cy I 



(W) 



(3-38) 



cen58933_ch03.qxd 9/10/2002 8:59 AM Page 147 



where 



147 
CHAPTER 3 



v cyl 



ln(r 2 lr x ) ln(Outer radius/Inner radius) 

2irLk 2tt X (Length) X (Thermal conductivity) 



(3-39) 



is the thermal resistance of the cylindrical layer against heat conduction, or 
simply the conduction resistance of the cylinder layer. 

We can repeat the analysis above for a spherical layer by taking A = 4irr 2 
and performing the integrations in Eq. 3-36. The result can be expressed as 



Q 



cond, sph 



(3-40) 



v sp h 



where 



Outer radius — Inner radius 



sph A-ur^k 



4Tr(Outer radius)(Inner radius)(Thermal conductivity) 



(3-41) 



is the thermal resistance of the spherical layer against heat conduction, or sim- 
ply the conduction resistance of the spherical layer. 

Now consider steady one-dimensional heat flow through a cylindrical or 
spherical layer that is exposed to convection on both sides to fluids at temper- 
atures r„[ and T m2 with heat transfer coefficients h { and h 2 , respectively, as 
shown in Fig. 3-25. The thermal resistance network in this case consists of 
one conduction and two convection resistances in series, just like the one for 
the plane wall, and the rate of heat transfer under steady conditions can be ex- 
pressed as 



Q 



(3-42) 




: ^conv,l +S cyI +R conv,: 



FIGURE 3-25 

The thermal resistance network 

for a cylindrical (or spherical) 

shell subjected to convection from 

both the inner and the outer sides. 



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 



(2-nT|L)/i| 2ttL& (2-nr 2 L)/7 2 



(3-43) 



for a cylindrical layer, and 



^,, 



# 



conv, 1 

l 



v sph 



1 



(4irr 1 2 )/ !l 



4m\r 2 k (4TTr?)h 2 



(3-44) 



/or a spherical layer. Note that A in the convection resistance relation R com = 
1/liA is the surface area at which convection occurs. It is equal to A = 2vrL 
for a cylindrical surface and A = 4irr 2 for a spherical surface of radius r. Also 
note that the thermal resistances are in series, and thus the total thermal resis- 
tance is determined by simply adding the individual resistances, just like the 
electrical resistances connected in series. 

Multilayered Cylinders and Spheres 

Steady heat transfer through multilayered cylindrical or spherical shells can be 
handled just like multilayered plane walls discussed earlier by simply add- 
ing an additional resistance in series for each additional layer. For example, 
the steady heat transfer rate through the three-layered composite cylinder 
of length L shown in Fig. 3-26 with convection on both sides can be ex- 
pressed as 



Q 



(3-45) 





R 



cyl,3 



WWW <- L A/WWV ♦ T^ 



R 



conv, 2 



FIGURE 3-26 

The thermal resistance network for heat transfer through a three-layered composite cylinder 
subjected to convection on both sides. 



cen58933_ch03.qxd 9/10/2002 8:59 AM Page 149 



where i? tota i is the total thermal resistance, expressed as 



R„ 



R 



conv, 

l 



+ R, 



cyl, 1 



^cyl.2 + R 



cyl, 3 



R 



conv, 2 



ln(r 2 /r,) ln(r 3 /r 2 ) ln(r 4 /r 3 ) i 



h,A, 



2TxLk, 



2 r nLk 1 



2 r nLk i h 2 A 4 



(3-46) 



where A x = litr^L and A 4 = 2irr 4 L. Equation 3-46 can also be used for a 
three-layered spherical shell by replacing the thermal resistances of cylindri- 
cal layers by the corresponding spherical ones. Again, note from the thermal 
resistance network that the resistances are in series, and thus the total thermal 
resistance is simply the arithmetic sum of the individual thermal resistances in 
the path of heat flow. 

Once Q is known, we can determine any intermediate temperature T- } by ap- 
plying the relation Q = (T t — T)/R total j_j across any layer or layers such that 
Tj is a known temperature at location i and 7? total , ■ _ ■ is the total thermal resis- 
tance between locations i andy (Fig. 3-27). For example, once Q has been 
calculated, the interface temperature T 2 between the first and second cylindri- 
cal layers can be determined from 



Q 



T x 



"conv, 1 "r °cyl, 1 



1 



\n(r 2 lr x ) 

h x (2ixr x L) 2nLk x 



(3-47) 



149 
CHAPTER 3 



"-, T l T 2 T 3 T^ 

«-JVWvVv^-»- J VWWv^-^ J WVWV^^ J VWvW^» 



R 2 



Tv, 


i-r, 


conv, 1 


r„ 


i-r 2 


R conv,l +R l 


r, 


-T, 


*i 


+ R, 


h 


-T 3 




«2 


h 


-T„ 2 



Rt + R„, 



FIGURE 3-27 

The ratio AT/R across any layer is 

equal to Q , which remains constant in 

one-dimensional steady conduction. 



We could also calculate T 2 from 

t 2 - r«2 






r 2 



R 7 + R, 



R,, 



ln(r 3 /r 2 ) to(r 4 /r 3 ) 



1 



2irLk ? 



2irLk 3 h (2irr 4 L) 



(3-48) 



Although both relations will give the same result, we prefer the first one since 
it involves fewer terms and thus less work. 

The thermal resistance concept can also be used for other geometries, pro- 
vided that the proper conduction resistances and the proper surface areas in 
convection resistances are used. 



EXAMPLE 3-7 Heat Transfer to a Spherical Container 

A 3-m internal diameter spherical tank made of 2-cm-thick stainless steel 
(k = 15 W/m • °C) is used to store iced water at 7"^ = 0°C. The tank is located 
in a room whose temperature is T m2 = 22°C. The walls of the room are also at 
22°C. The outer surface of the tank is black and heat transfer between the outer 
surface of the tank and the surroundings is by natural convection and radiation. 
The convection heat transfer coefficients at the inner and the outer surfaces of 
the tank are h 1 = 80 W/m 2 • °C and h 2 = 10 W/m 2 • °C, respectively. Determine 
(a) the rate of heat transfer to the iced water in the tank and (b) the amount of 
ice at C C that melts during a 24-h period. 

SOLUTION A spherical container filled with iced water is subjected to convec- 
tion and radiation heat transfer at its outer surface. The rate of heat transfer 
and the amount of ice that melts per day are to be determined. 



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 3-28 

Schematic for Example 3-7. 



Assumptions 1 Heat transfer is steady since the specified thermal conditions at 
the boundaries do not change with time. 2 Heat transfer is one-dimensional 
since there is thermal symmetry about the midpoint. 3 Thermal conductivity is 
constant. 

Properties The thermal conductivity of steel is given to be k = 15 W/m • °C. 
The heat of fusion of water at atmospheric pressure is h :f = 333.7 kJ/kg. The 
outer surface of the tank is black and thus its emissivity is e = 1. 

Analysis (a) The thermal resistance network for this problem is given in 
Fig. 3-28. Noting that the inner diameter of the tank is D x = 3 m and the outer 
diameter is D, = 3.04 m, the inner and the outer surface areas of the tank are 



A, = irD? = tt(3 m) 2 = 28.3 m 2 
A 2 = ttD 2 2 = ir(3.04 m) 2 = 29.0 m 2 



Also, the radiation heat transfer coefficient is given by 
h mi = eu(Ti+ T* 2 )(T 2 + T a2 ) 

But we do not know the outer surface temperature T 2 of the tank, and thus we 
cannot calculate /? rad . Therefore, we need to assume a T 2 value now and check 
the accuracy of this assumption later. We will repeat the calculations if neces- 
sary using a revised value for T 2 . 

We note that T 2 must be between 0°C and 22 C C, but it must be closer 
to 0°C, since the heat transfer coefficient inside the tank is much larger. Taking 
7", = 5 C C = 278 K, the radiation heat transfer coefficient is determined to be 



h mi = (1)(5.67 X 10- 8 W/m 2 • K 4 )[(295 K) 2 
= 5.34 W/m 2 ■ K = 5.34 W/m 2 • °C 

Then the individual thermal resistances become 
1 1 



(278 K) 2 ][(295 + 278) K] 



Ri ~ ^conv, 1 



R, — R s , 



hi A, (80 W/m 2 • °C)(28.3 m 2 ) 



0.000442°C/W 



r 2 - r { 



(1.52 - 1.50) m 



1 - sphere ~ ^^^ ~ 4^ (15 W / m . °Q(1.52 m)(1.50 Hi) 

0.000047°C/W 

1 1 



R ° Rc ° m ' 2 h 2 A 2 (10 W/m 2 • °C)(29.0 m 2 ) 



0.00345°C/W 



1 



1 



^^2 (5.34 W/m 2 • °C)(29.0m 2 ) 



0.00646°C/W 



The two parallel resistances R and ff rad can be replaced by an equivalent resis- 



tance ff equiv determined from 



111 1 



#equiv Ro -Rrad 0.00345 0.00646 



444.7 W/°C 



which gives 



flequiv = 0.00225°C/W 



cen58933_ch03.qxd 9/10/2002 8:59 AM Page 151 



Now all the resistances are in series, and the total resistance is determined 
to be 

fltotai = Ri + Ri + fleqiuv = 0.000442 + 0.000047 + 0.00225 = 0.00274°C/W 
Then the steady rate of heat transfer to the iced water becomes 



Q 



7^ - r„ 



(22 - 0)°C 
0.00274°C/W 



8029 W (or Q = 8.027 kJ/s) 



To check the validity of our original assumption, we now determine the outer 
surface temperature from 



Q 



T«a - T 2 



*2 — T^ 2 QR cquiv 

= 22°C - (8029 W)(0.00225°C/W) = 4°C 



which is sufficiently close to the 5°C assumed in the determination of the radi- 
ation heat transfer coefficient. Therefore, there is no need to repeat the calcu- 
lations using 4°C for T z . 
(b) The total amount of heat transfer during a 24-h period is 



Q = QAt 



.029 kJ/s)(24 X 3600 s) = 673,700 kJ 



Noting that it takes 333.7 kJ of energy to melt 1 kg of ice at 0°C, the amount 
of ice that will melt during a 24-h period is 



Q _ 673,700 kJ 
~h iS ~ 333.7 kJ/kg 



2079 kg 



Therefore, about 2 metric tons of ice will melt in the tank every day. 
Discussion An easier way to deal with combined convection and radiation at a 
surface when the surrounding medium and surfaces are at the same tempera- 
ture is to add the radiation and convection heat transfer coefficients and to treat 
the result as the convection heat transfer coefficient. That is, to take h = 10 + 
5.34 = 15.34 W/m 2 • °C in this case. This way, we can ignore radiation since 
its contribution is accounted for in the convection heat transfer coefficient. The 
convection resistance of the outer surface in this case would be 



1 



1 



Combined ^2 (15.34 W/m 2 • °C)(29.0 m 2 ) 



0.00225°C/W 



which is identical to the value obtained for the equivalent resistance for the par- 
allel convection and the radiation resistances. 



151 
CHAPTER 3 



EXAMPLE 3-8 Heat Loss through an Insulated Steam Pipe 

Steam at 7" xl = 320°C flows in a cast iron pipe (k = 80 W/m • °C) whose inner 
and outer diameters are Dj = 5 cm and D 2 = 5.5 cm, respectively. The pipe is 
covered with 3-cm-thick glass wool insulation with k = 0.05 W/m • °C. Heat is 
lost to the surroundings at 7" x2 = 5°C by natural convection and radiation, with 



cen58933_ch03.qxd 9/10/2002 8:59 AM Page 152 



152 
HEAT TRANSFER 



Insulation 




T l T 2 T 3 



FIGURE 3-29 

Schematic for Example 3-8. 



a combined heat transfer coefficient of h 2 = 18 W/m 2 • C C. Taking the heat 
transfer coefficient inside the pipe to be h 1 = 60 W/m 2 • °C, determine the rate 
of heat loss from the steam per unit length of the pipe. Also determine the tem- 
perature drops across the pipe shell and the insulation. 

SOLUTION A steam pipe covered with glass wool insulation is subjected to 
convection on its surfaces. The rate of heat transfer per unit length and the 
temperature drops across the pipe and the insulation are to be determined. 
Assumptions 1 Heat transfer is steady since there is no indication of any 
change with time. 2 Heat transfer is one-dimensional since there is thermal 
symmetry about the centerline and no variation in the axial direction. 3 Thermal 
conductivities are constant. 4 The thermal contact resistance at the interface is 
negligible. 

Properties The thermal conductivities are given to be k = 80 W/m • °C for cast 
iron and k = 0.05 W/m • °C for glass wool insulation. 

Analysis The thermal resistance network for this problem involves four resis- 
tances in series and is given in Fig. 3-29. Taking L = 1 m, the areas of the 
surfaces exposed to convection are determined to be 



A: 
A 3 



2vr x L = 2tt(0.025 m)(l m) = 0.157 m 2 
2irr 3 L = 2tt(0.0575 m)(l m) = 0.361 m 2 



Then the individual thermal resistances become 



1 



1 



R t = R 

"2 — "insulati' 



h,A (60 W/m 2 • °C)(0.157m 2 ) 
ln(r 2 //-,)_ ln(2.75/2.5) 

pipe ~ IvkyL ~ 2ir(80 W/m ■ °C)(1 m) = 
m(r 3 /r 2 ) ln(5.75/2.75) 



2iTk 2 L 2tt(0.05 W/m ■ °C)(1 m) 
1 1 



0.106°C/W 
0.0002°C/W 
2.35°C/W 



R " /?conv ' 2 h 2 A 3 (18 W/m 2 • °C)(0.361m 2 ) 



0.154°C/W 



Noting that all resistances are in series, the total resistance is determined to be 
fltotai = Ri + Ri + R 2 + Ro = 0.106 + 0.0002 + 2.35 + 0.154 = 2.61°C/W 

Then the steady rate of heat loss from the steam becomes 



Q 



R„ 



(320 - 5)°C 
2.61 °C/W 



121 W (per m pipe length) 



The heat loss for a given pipe length can be determined by multiplying the 
above quantity by the pipe length L. 

The temperature drops across the pipe and the insulation are determined 
from Eq. 3-17 to be 

A7/ pipe = gflpipe = (121 W)(0.0002°C/W) = 0.02°C 
Ar insulation = efl insulalion = (121 W)(2.35°C/W) = 284°C 

That is, the temperatures between the inner and the outer surfaces of the pipe 
differ by 0.02°C, whereas the temperatures between the inner and the outer 
surfaces of the insulation differ by 284°C. 



cen58933_ch03.qxd 9/10/2002 8:59 AM Page 153 



Discussion Note that the thermal resistance of the pipe is too small relative to 
the other resistances and can be neglected without causing any significant 
error. Also note that the temperature drop across the pipe is practically zero, 
and thus the pipe can be assumed to be isothermal. The resistance to heat flow 
in insulated pipes is primarily due to insulation. 



153 
CHAPTER 3 



3-5 - CRITICAL RADIUS OF INSULATION 

We know that adding more insulation to a wall or to the attic always decreases 
heat transfer. The thicker the insulation, the lower the heat transfer rate. This 
is expected, since the heat transfer area A is constant, and adding insulation 
always increases the thermal resistance of the wall without increasing the 
convection resistance. 

Adding insulation to a cylindrical pipe or a spherical shell, however, is a dif- 
ferent matter. The additional insulation increases the conduction resistance of 
the insulation layer but decreases the convection resistance of the surface be- 
cause of the increase in the outer surface area for convection. The heat trans- 
fer from the pipe may increase or decrease, depending on which effect 
dominates. 

Consider a cylindrical pipe of outer radius r, whose outer surface tempera- 
ture T x is maintained constant (Fig. 3-30). The pipe is now insulated with a 
material whose thermal conductivity is k and outer radius is r 2 . Heat is lost 
from the pipe to the surrounding medium at temperature T^, with a convection 
heat transfer coefficient h. The rate of heat transfer from the insulated pipe to 
the surrounding air can be expressed as (Fig. 3-31) 



Q 



S;„ + R n 



1 



lnf/j/ri) 

2-nLk h(2-nr 2 L) 



(3-49) 



The variation of Q with the outer radius of the insulation r 2 is plotted in 
Fig. 3-31. The value of r 2 at which Q reaches a maximum is determined from 
the requirement that dQldr 2 = (zero slope). Performing the differentiation 
and solving for r 2 yields the critical radius of insulation for a cylindrical 
body to be 



' cr, cylinder 



/; 



(m) 



(3-50) 



Note that the critical radius of insulation depends on the thermal conductivity 
of the insulation k and the external convection heat transfer coefficient h. 
The rate of heat transfer from the cylinder increases with the addition of insu- 
lation for r 2 < r a , reaches a maximum when r 2 = r cr , and starts to decrease for 
r 2 > r a . Thus, insulating the pipe may actually increase the rate of heat trans- 
fer from the pipe instead of decreasing it when r 2 < r cr 

The important question to answer at this point is whether we need to be con- 
cerned about the critical radius of insulation when insulating hot water pipes 
or even hot water tanks. Should we always check and make sure that the outer 



Insulation 




FIGURE 3-30 

An insulated cylindrical pipe 

exposed to convection from the outer 

surface and the thermal resistance 

network associated with it. 




FIGURE 3-31 



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154 
HEAT TRANSFER 



radius of insulation exceeds the critical radius before we install any insula- 
tion? Probably not, as explained here. 

The value of the critical radius r a . will be the largest when k is large and h is 
small. Noting that the lowest value of h encountered in practice is about 
5 W/m 2 • °C for the case of natural convection of gases, and that the thermal 
conductivity of common insulating materials is about 0.05 W/m 2 • °C, the 
largest value of the critical radius we are likely to encounter is 



"W, insulation __ 0.05 W/m ■ °C 

~h~ 5 W/m 2 • °C 



0.01 m = 1 cm 



This value would be even smaller when the radiation effects are considered. 
The critical radius would be much less in forced convection, often less than 
1 mm, because of much larger h values associated with forced convection. 
Therefore, we can insulate hot water or steam pipes freely without worrying 
about the possibility of increasing the heat transfer by insulating the pipes. 

The radius of electric wires may be smaller than the critical radius. There- 
fore, the plastic electrical insulation may actually enhance the heat transfer 
from electric wires and thus keep their steady operating temperatures at lower 
and thus safer levels. 

The discussions above can be repeated for a sphere, and it can be shown in 
a similar manner that the critical radius of insulation for a spherical shell is 



= 2k 

*cr, sphere r. 



(3-51) 



where k is the thermal conductivity of the insulation and h is the convection 
heat transfer coefficient on the outer surface. 



EXAMPLE 3-9 



Heat Loss from an Insulated Electric Wire 



A 3-mm-diameter and 5-m-long electric wire is tightly wrapped with a 2-mm- 
thick plastic cover whose thermal conductivity is k = 0.15 W/m • C C. Electrical 
measurements indicate that a current of 10 A passes through the wire and there 
is a voltage drop of 8 V along the wire. If the insulated wire is exposed to a 
medium at 7" x = 30°C with a heat transfer coefficient of h = 12 W/m 2 • °C, de- 
termine the temperature at the interface of the wire and the plastic cover in 
steady operation. Also determine whether doubling the thickness of the plastic 
cover will increase or decrease this interface temperature. 

SOLUTION An electric wire is tightly wrapped with a plastic cover. The inter- 
face temperature and the effect of doubling the thickness of the plastic cover 
on the interface temperature are to be determined. 

Assumptions 1 Heat transfer is steady since there is no indication of any 
change with time. 2 Heat transfer is one-dimensional since there is thermal 
symmetry about the centerline and no variation in the axial direction. 3 Thermal 
conductivities are constant. 4 The thermal contact resistance at the interface is 
negligible. 5 Heat transfer coefficient incorporates the radiation effects, if any. 
Properties The thermal conductivity of plastic is given to be k = 0.15 
W/m • °C. 



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155 
CHAPTER 3 



Analysis Heat is generated in the wire and its temperature rises as a result of 
resistance heating. We assume heat is generated uniformly throughout the wire 
and is transferred to the surrounding medium in the radial direction. In steady 
operation, the rate of heat transfer becomes equal to the heat generated within 
the wire, which is determined to be 

Q = W e = VI = (8 V)(10 A) = 80 W 

The thermal resistance network for this problem involves a conduction resis- 
tance for the plastic cover and a convection resistance for the outer surface in 
series, as shown in Fig. 3-32. The values of these two resistances are deter- 
mined to be 

A 2 = (2irr 2 )L = 2tt(0.0035 m)(5 m) = 0.110 m 2 
1 1 



R, 



hA 2 (12W/m 2 • °C)(0.110m 2 ) 
ln(r 2 /r,) ln(3.5/1.5) 



plastic 



2-nkL 2tt(0.15 W/m ■ °C)(5 m) 



0.76°C/W 
0.18°C/W 



and therefore 



#,o, a I = plastic + *co„v = 0.76 + 0.18 = 0.94°C/W 



Then the interface temperature can be determined from 



Q 



T, - 71 



= 30°C + (80 W)(0.94°C/W) = 105°C 



Note that we did not involve the electrical wire directly in the thermal resistance 
network, since the wire involves heat generation. 

To answer the second part of the question, we need to know the critical radius 
of insulation of the plastic cover. It is determined from Eq. 3-50 to be 



k = 0.15 W/m ■ °C 
h 12 W/m 2 ■ °C 



0.0125 m = 12.5 mm 



which is larger than the radius of the plastic cover. Therefore, increasing the 
thickness of the plastic cover will enhance heat transfer until the outer radius 
of the cover reaches 12.5 mm. As a result, the rate of heat transfer Q will in- 
crease when the interface temperature T x is held constant, or T x will decrease 
when Q is held constant, which is the case here. 

Discussion It can be shown by repeating the calculations above for a 4-mm- 
thick plastic cover that the interface temperature drops to 90.6°C when the 
thickness of the plastic cover is doubled. It can also be shown in a similar man- 
ner that the interface reaches a minimum temperature of 83°C when the outer 
radius of the plastic cover equals the critical radius. 




T 2 

^vwvw — • — www » T„ 

D D 

plastic conv 



FIGURE 3-32 

Schematic for Example 3-9. 



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156 
HEAT TRANSFER 



FIGURE 3-33 

The thin plate fins of a car radiator 

greatly increase the rate of 

heat transfer to the air (photo by 

Yunus £engel and James Kleiser). 





3-6 - HEAT TRANSFER FROM FINNED SURFACES 

The rate of heat transfer from a surface at a temperature T s to the surrounding 
medium at T m is given by Newton's law of cooling as 



6 c 



hA s (T s -T x ) 



FIGURE 3-34 

Some innovative fin designs. 



where A s is the heat transfer surface area and h is the convection heat transfer 
coefficient. When the temperatures T s and T„ are fixed by design considera- 
tions, as is often the case, there are two ways to increase the rate of heat trans- 
fer: to increase the convection heat transfer coefficient h or to increase the 
surface area A s . Increasing h may require the installation of a pump or fan, or 
replacing the existing one with a larger one, but this approach may or may not 
be practical. Besides, it may not be adequate. The alternative is to increase the 
surface area by attaching to the surface extended surfaces called fins made of 
highly conductive materials such as aluminum. Finned surfaces are manu- 
factured by extruding, welding, or wrapping a thin metal sheet on a surface. 
Fins enhance heat transfer from a surface by exposing a larger surface area to 
convection and radiation. 

Finned surfaces are commonly used in practice to enhance heat transfer, and 
they often increase the rate of heat transfer from a surface severalfold. The car 
radiator shown in Fig. 3-33 is an example of a finned surface. The closely 
packed thin metal sheets attached to the hot water tubes increase the surface 
area for convection and thus the rate of convection heat transfer from the tubes 
to the air many times. There are a variety of innovative fin designs available 
in the market, and they seem to be limited only by imagination (Fig. 3-34). 

In the analysis of fins, we consider steady operation with no heat generation 
in the fin, and we assume the thermal conductivity k of the material to remain 
constant. We also assume the convection heat transfer coefficient h to be con- 
stant and uniform over the entire surface of the fin for convenience in the 
analysis. We recognize that the convection heat transfer coefficient h, in gen- 
eral, varies along the fin as well as its circumference, and its value at a point 
is a strong function of the fluid motion at that point. The value of h is usually 
much lower at the fin base than it is at the fin tip because the fluid is sur- 
rounded by solid surfaces near the base, which seriously disrupt its motion to 



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157 
CHAPTER 3 



the point of "suffocating" it, while the fluid near the fin tip has little contact 
with a solid surface and thus encounters little resistance to flow. Therefore, 
adding too many fins on a surface may actually decrease the overall heat 
transfer when the decrease in h offsets any gain resulting from the increase in 
the surface area. 



Fin Equation 

Consider a volume element of a fin at location x having a length of Ax, cross- 
sectional area of A c , and a perimeter of/?, as shown in Fig. 3-35. Under steady 
conditions, the energy balance on this volume element can be expressed as 

/ Rate of heat \ / Rate of heat \ / Rate of heat \ 

conduction into = conduction from the + convection from 

I the element at xj I element at x + Ax / I the element / 



or 



where 



tjcond.x t- cond, X + Ax ' Qc 



Q conv = h(pAx)(T-T.J 



Substituting and dividing by Ax, we obtain 

xi cond. X + A.Y x£ cond, X 



Volume 
element 




FIGURE 3-35 

Volume element of a fin at location x 

having a length of Ax, cross-sectional 

area of A c , and perimeter of p. 



Ax 



hp(T - 7/J = 



(3-52) 



Taking the limit as Ax — > gives 

"Q cond 

dx 



hp(T - T„) = 



(3-53) 



From Fourier's law of heat conduction we have 

dT 



«- cond 



-kA,. 



dx 



(3-54) 



where A c is the cross-sectional area of the fin at location x. Substitution of this 
relation into Eq. 3-53 gives the differential equation governing heat transfer 
in fins, 



d_ 
dx 



kA 



dT 
dx 



hp(T - T a ) = 



(3-55) 



In general, the cross-sectional area A c and the perimeter/; of a fin vary with x, 
which makes this differential equation difficult to solve. In the special case of 
constant cross section and constant thermal conductivity, the differential 
equation 3-55 reduces to 



dx 1 



a 2 e = o 



(3-56) 



cen58933_ch03.qxd 9/10/2002 8:59 AM Page 15E 



158 
HEAT TRANSFER 



where 



hp 
~kA~. 



(3-57) 



and = 7 — r„ is the temperature excess. At the fin base we have 

0/, — T b — T<n- 

Equation 3-56 is a linear, homogeneous, second-order differential equation 
with constant coefficients. A fundamental theory of differential equations 
states that such an equation has two linearly independent solution functions, 
and its general solution is the linear combination of those two solution func- 
tions. A careful examination of the differential equation reveals that subtract- 
ing a constant multiple of the solution function from its second derivative 
yields zero. Thus we conclude that the function and its second derivative 
must be constant multiples of each other. The only functions whose deriva- 
tives are constant multiples of the functions themselves are the exponential 
functions (or a linear combination of exponential functions such as sine and 
cosine hyperbolic functions). Therefore, the solution functions of the dif- 
ferential equation above are the exponential functions e~" x or e ax or constant 
multiples of them. This can be verified by direct substitution. For example, 
the second derivative of e~ ax is a 2 e~ ax , and its substitution into Eq. 3-56 yields 
zero. Therefore, the general solution of the differential equation Eq. 3-56 is 



Q(x) 



CV 



(3-58) 



~- Specified 
temperature 

(a) Specified temperature 

(b) Negligible heat loss 

(c) Convection 

(d) Convection and radiation 

FIGURE 3-36 

Boundary conditions at the 
fin base and the fin tip. 



where C { and C 2 are arbitrary constants whose values are to be determined 
from the boundary conditions at the base and at the tip of the fin. Note that we 
need only two conditions to determine C\ and C 2 uniquely. 

The temperature of the plate to which the fins are attached is normally 
known in advance. Therefore, at the fin base we have a specified temperature 
boundary condition, expressed as 



Boundary condition at fin base: 



6(0) 



(3-59) 



At the fin tip we have several possibilities, including specified temperature, 
negligible heat loss (idealized as an insulated tip), convection, and combined 
convection and radiation (Fig. 3-36). Next, we consider each case separately. 

1 Infinitely Long Fin (r fintip = 7" J 

For a sufficiently long fin of uniform cross section (A c = constant), the tem- 
perature of the fin at the fin tip will approach the environment temperature 7^ 
and thus will approach zero. That is, 



Boundary condition at fin tip: 0(L) = T(L) — T m 







This condition will be satisfied by the function e~ ax , but not by the other 
prospective solution function e ax since it tends to infinity as x gets larger. 
Therefore, the general solution in this case will consist of a constant multiple 
of e ax . The value of the constant multiple is determined from the require- 
ment that at the fin base where x = the value of will be Q b . Noting that 



cen58933_ch03.qxd 9/10/2002 8:59 AM Page 159 



e -ax — e o — ^ tne p r0 p er value of the constant is Q b , and the solution function 
we are looking for is 0(x) = Q b e~ ax . This function satisfies the differential 
equation as well as the requirements that the solution reduce to Q b at the fin 
base and approach zero at the fin tip for large x. Noting that = T — T m and 
a = \JhplkA c , the variation of temperature along the fin in this case can be 
expressed as 



Very long fin: 



T(x) 



(3-60) 



Note that the temperature along the fin in this case decreases exponentially 
from T b to T m as shown in Fig. 3-37. The steady rate of heat transfer from the 
entire fin can be determined from Fourier's law of heat conduction 



Very long fin: 



do 



ng fin 



-kA. 



dT 
c dx 



y/hpkA c (T b - r.) 



(3-61) 



where p is the perimeter, A c is the cross-sectional area of the fin, and x is the 
distance from the fin base. Alternatively, the rate of heat transfer from the fin 
could also be determined by considering heat transfer from a differential 
volume element of the fin and integrating it over the entire surface of the fin. 
That is, 



Q 



,„ = f h[T(x 



) - rj dA fm 



f A8(. 

JA,„ 



x) dA fln 



(3-62) 



The two approaches described are equivalent and give the same result since, 
under steady conditions, the heat transfer from the exposed surfaces of the fin 
is equal to the heat transfer to the fin at the base (Fig. 3-38). 



2 Negligible Heat Loss from the Fin Tip 
(Insulated fin tip, Q fin tip = 0) 

Fins are not likely to be so long that their temperature approaches the sur- 
rounding temperature at the tip. A more realistic situation is for heat transfer 
from the fin tip to be negligible since the heat transfer from the fin is propor- 
tional to its surface area, and the surface area of the fin tip is usually a negli- 
gible fraction of the total fin area. Then the fin tip can be assumed to be 
insulated, and the condition at the fin tip can be expressed as 



Boundary condition at fin tip: 



dx 



(3-63) 



159 
CHAPTER 3 




(p = kD, A c = kD /4 for a cylindrical fin) 

FIGURE 3-37 

A long circular fin of uniform cross 

section and the variation of 

temperature along it. 



i e fin 

I / / > / 4 1 



^base *"iin 

FIGURE 3-38 

Under steady conditions, heat transfer 

from the exposed surfaces of the 

fin is equal to heat conduction 

to the fin at the base. 



The condition at the fin base remains the same as expressed in Eq. 3-59. The 
application of these two conditions on the general solution (Eq. 3-58) yields, 
after some manipulations, this relation for the temperature distribution: 



Adiabatic fin tip: 



T(x) 



cosh a(L — x) 
cosh ah 



(3-64) 



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160 
HEAT TRANSFER 



The rate of heat transfer from the fin can be determined again from Fourier's 
law of heat conduction: 



Adiabatic fin tip: 



Q 



insulated tip 



-M, 



dT 
dx 



VhpkA c (T b - T r _) tanh ah 



(3-65) 



Note that the heat transfer relations for the very long fin and the fin with 
negligible heat loss at the tip differ by the factor tanh ah, which approaches 1 
as L becomes very large. 



e fin 



\- L H 



Convection 



(a) Actual fin with 
convection at the tip 



^ 



Insulated 



(b) Equivalent fin with insulated tip 

FIGURE 3-39 

Corrected fin length L c is defined such 
that heat transfer from a fin of length 
L c with insulated tip is equal to heat 
transfer from the actual fin of length L 
with convection at the fin tip. 



3 Convection (or Combined Convection and Radiation) 
from Fin Tip 

The fin tips, in practice, are exposed to the surroundings, and thus the proper 
boundary condition for the fin tip is convection that also includes the effects 
of radiation. The fin equation can still be solved in this case using the convec- 
tion at the fin tip as the second boundary condition, but the analysis becomes 
more involved, and it results in rather lengthy expressions for the temperature 
distribution and the heat transfer. Yet, in general, the fin tip area is a small 
fraction of the total fin surface area, and thus the complexities involved can 
hardly justify the improvement in accuracy. 

A practical way of accounting for the heat loss from the fin tip is to replace 
the fin length L in the relation for the insulated tip case by a corrected length 
defined as (Fig. 3-39) 



Corrected fin length: 



L + 



(3-66) 



where A c is the cross-sectional area and/7 is the perimeter of the fin at the tip. 
Multiplying the relation above by the perimeter gives A corrected = A fm n atera i) + 
A tip , which indicates that the fin area determined using the corrected length is 
equivalent to the sum of the lateral fin area plus the fin tip area. 

The corrected length approximation gives very good results when the vari- 
ation of temperature near the fin tip is small (which is the case when ah > 1) 
and the heat transfer coefficient at the fin tip is about the same as that at the 
lateral surface of the fin. Therefore, fins subjected to convection at their tips 
can be treated as fins with insulated tips by replacing the actual fin length by 
the corrected length in Eqs. 3-64 and 3-65. 

Using the proper relations for A c and p, the corrected lengths for rectangu- 
lar and cylindrical fins are easily determined to be 



T = f 

c, rectangular fin 



and 



D 



*^c, cylindrical fin I 



where t is the thickness of the rectangular fins and D is the diameter of the 
cylindrical fins. 



Fin Efficiency 

Consider the surface of a plane wall at temperature T b exposed to a medium at 
temperature T^. Heat is lost from the surface to the surrounding medium by 



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161 
CHAPTER 3 



convection with a heat transfer coefficient of h. Disregarding radiation or 
accounting for its contribution in the convection coefficient /;, heat transfer 
from a surface area A s is expressed as Q = hA s (T s — T^). 

Now let us consider a fin of constant cross-sectional area A c = A b and length 
L that is attached to the surface with a perfect contact (Fig. 3-40). This time 
heat will flow from the surface to the fin by conduction and from the fin to the 
surrounding medium by convection with the same heat transfer coefficient h. 
The temperature of the fin will be T b at the fin base and gradually decrease to- 
ward the fin tip. Convection from the fin surface causes the temperature at any 
cross section to drop somewhat from the midsection toward the outer surfaces. 
However, the cross-sectional area of the fins is usually very small, and thus 
the temperature at any cross section can be considered to be uniform. Also, the 
fin tip can be assumed for convenience and simplicity to be insulated by using 
the corrected length for the fin instead of the actual length. 

In the limiting case of zero thermal resistance or infinite thermal conduc- 
tivity (& — > oo), the temperature of the fin will be uniform at the base value of 
T b . The heat transfer from the fin will be maximum in this case and can be 
expressed as 




(a) Surface without fins 



Q fin. r 



hA fm (T b - r.) 



(3-67) 




In reality, however, the temperature of the fin will drop along the fin, and 
thus the heat transfer from the fin will be less because of the decreasing tem- 
perature difference T(x) — T«, toward the fin tip, as shown in Fig. 3-41. To ac- 
count for the effect of this decrease in temperature on heat transfer, we define 
a fin efficiency as 



%in 



e, 



2fin, r 



Actual heat transfer rate from the fin 

Ideal heat transfer rate from the fin 

if the entire fin were at base temperature 



(3-68) 



or 



Q fin — "Hfin Q fin, r 



%.n hA fm (T b - 7/„) 



(3-69) 



where A Fm is the total surface area of the fin. This relation enables us to deter- 
mine the heat transfer from a fin when its efficiency is known. For the cases 
of constant cross section of very long fins and fins with insulated tips, the fin 
efficiency can be expressed as 



Mlong fin 



6 fin 

*£■ fin, m£ 



Vtip~kA c (T b - r.) _ i jkAc 
M fln (T b - rj Lihp 



aL 



(3-70) 



and 



'linsulated t 



Q An \ / hpkA c (T b - r„) tanh aL t anh aL 



Q fin, r 



M fm (T b - T x ) 



aL 



(3-71) 



since A fm = pL for fins with constant cross section. Equation 3-71 can also be 
used for fins subjected to convection provided that the fin length L is replaced 
by the corrected length L c . 



(b) Surface with a fin 

: 2 X w X L 
s 2 X w X L 

FIGURE 3-40 

Fins enhance heat transfer from 
a surface by enhancing surface area. 




(a) Ideal 



80°C 




(b) Actual 



56°C 



FIGURE 3-41 

Ideal and actual 



temperature distribution in a fin. 



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162 
HEAT TRANSFER 



FIGURE 3-42 

Efficiency of circular, rectangular, and 

triangular fins on a plain surface of 

width w (from Gardner, Ref. 6). 



Fin efficiency relations are developed for fins of various profiles and are 
plotted in Fig. 3-42 for fins on a plain surface and in Fig. 3-43 for circular 
fins of constant thickness. The fin surface area associated with each profile is 
also given on each figure. For most fins of constant thickness encountered in 
practice, the fin thickness t is too small relative to the fin length L, and thus 
the fin tip area is negligible. 



100 




100 



FIGURE 3-43 

Efficiency of circular fins of length L 

and constant thickness t (from 

Gardner, Ref. 6). 



60 



40 



20 

















tmmmtL. 






r 2 + ±t 








^ h 










\2 








y\ 








:4 

— h 

^2 










^Jt 








h\ A fm = 2K(rl-,j) + 27ir 2 t 






2_ 









0.5 



1.0 1.5 



2.0 



2.5 



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163 
CHAPTER 3 



Note that fins with triangular and parabolic profiles contain less material 
and are more efficient than the ones with rectangular profiles, and thus are 
more suitable for applications requiring minimum weight such as space appli- 
cations. 

An important consideration in the design of finned surfaces is the selection 
of the proper fin length L. Normally the longer the fin, the larger the heat 
transfer area and thus the higher the rate of heat transfer from the fin. But also 
the larger the fin, the bigger the mass, the higher the price, and the larger the 
fluid friction. Therefore, increasing the length of the fin beyond a certain 
value cannot be justified unless the added benefits outweigh the added cost. 
Also, the fin efficiency decreases with increasing fin length because of the de- 
crease in fin temperature with length. Fin lengths that cause the fin efficiency 
to drop below 60 percent usually cannot be justified economically and should 
be avoided. The efficiency of most fins used in practice is above 90 percent. 



Fin Effectiveness 

Fins are used to enhance heat transfer, and the use of fins on a surface cannot 
be recommended unless the enhancement in heat transfer justifies the added 
cost and complexity associated with the fins. In fact, there is no assurance that 
adding fins on a surface will enhance heat transfer. The performance of the 
fins is judged on the basis of the enhancement in heat transfer relative to the 
no-fin case. The performance of fins expressed in terms of the/w effectiveness 
e fin is defined as (Fig. 3-44) 



gf,n 

x£ no fin 



e, 



hA b (T b - ly 



Heat transfer rate from 
the fin of base area A b 

Heat transfer rate from 
the surface of area A h 



(3-72) 



Here, A b is the cross-sectional area of the fin at the base and Q no fin represents 
the rate of heat transfer from this area if no fins are attached to the surface. 
An effectiveness of e fin = 1 indicates that the addition of fins to the surface 
does not affect heat transfer at all. That is, heat conducted to the fin through 
the base area A b is equal to the heat transferred from the same area A b to the 
surrounding medium. An effectiveness of e fln < 1 indicates that the fin actu- 
ally acts as insulation, slowing down the heat transfer from the surface. This 
situation can occur when fins made of low thermal conductivity materials are 
used. An effectiveness of e fln > 1 indicates that fins are enhancing heat trans- 
fer from the surface, as they should. However, the use of fins cannot be justi- 
fied unless e fin is sufficiently larger than 1. Finned surfaces are designed on 
the basis of maximizing effectiveness for a specified cost or minimizing cost 
for a desired effectiveness. 

Note that both the fin efficiency and fin effectiveness are related to the per- 
formance of the fin, but they are different quantities. However, they are 
related to each other by 




-no fin 

FIGURE 3-44 

The effectiveness of a fin. 



Qi 



G fi „ 



%,„ hA fm (T b - r„) A fl 



6, 



hA„ (T b - r„) hA b (T b - T m ) 



A, 



■ %,n 



(3-73) 



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164 
HEAT TRANSFER 



Therefore, the fin effectiveness can be determined easily when the fin effi- 
ciency is known, or vice versa. 

The rate of heat transfer from a sufficiently long fin of uniform cross section 
under steady conditions is given by Eq. 3-61. Substituting this relation into 
Eq. 3-72, the effectiveness of such a long fin is determined to be 



-"long fin 



6 fin _ Vh^kA c (T b -T x 
fi„„ fin hA b {T b -T a ) 



kp 

\n~A c 



(3-74) 




^^ 



- 3 x (r x w) 
A fln =2XLXw + tXw (one fin) 
~2x Lxw 

FIGURE 3-45 

Various surface areas associated with a 
rectangular surface with three fins. 



since A c = A b in this case. We can draw several important conclusions from 
the fin effectiveness relation above for consideration in the design and selec- 
tion of the fins: 

• The thermal conductivity k of the fin material should be as high as 
possible. Thus it is no coincidence that fins are made from metals, with 
copper, aluminum, and iron being the most common ones. Perhaps the 
most widely used fins are made of aluminum because of its low cost and 
weight and its resistance to corrosion. 

• The ratio of the perimeter to the cross-sectional area of the fin p/A c 
should be as high as possible. This criterion is satisfied by thin plate fins 
and slender pin fins. 

• The use of fins is most effective in applications involving a low 
convection heat transfer coefficient. Thus, the use of fins is more easily 
justified when the medium is a gas instead of a liquid and the heat 
transfer is by natural convection instead of by forced convection. 
Therefore, it is no coincidence that in liquid-to-gas heat exchangers such 
as the car radiator, fins are placed on the gas side. 

When determining the rate of heat transfer from a finned surface, we must 
consider the unfinned portion of the surface as well as the fins. Therefore, the 
rate of heat transfer for a surface containing n fins can be expressed as 



Q total, fin — 2unfin + Q fin 

= ^u„r,n (T b ~ 7V) + r\ {m hA Sm (T b 

= KKniin + THfin^finXTft ~ TJ) 



r„) 



(3-75) 



We can also define an overall effectiveness for a finned surface as the ratio 
of the total heat transfer from the finned surface to the heat transfer from the 
same surface if there were no fins, 



Q total, fin h(A mfw + % in A fm )(7; - F„) 



J fin. overall 



Q 



total, no fin 



^nofin^i- J"-) 



(3-76) 



where A no fin is the area of the surface when there are no fins, A fln is the total 
surface area of all the fins on the surface, and A unfin is the area of the unfinned 
portion of the surface (Fig. 3-45). Note that the overall fin effectiveness 
depends on the fin density (number of fins per unit length) as well as the 
effectiveness of the individual fins. The overall effectiveness is a better mea- 
sure of the performance of a finned surface than the effectiveness of the indi- 
vidual fins. 



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165 
CHAPTER 3 



Proper Length of a Fin 



An important step in the design of a fin is the determination of the appropriate 
length of the fin once the fin material and the fin cross section are specified. 
You may be tempted to think that the longer the fin, the larger the surface area 
and thus the higher the rate of heat transfer. Therefore, for maximum heat 
transfer, the fin should be infinitely long. However, the temperature drops 
along the fin exponentially and reaches the environment temperature at some 
length. The part of the fin beyond this length does not contribute to heat trans- 
fer since it is at the temperature of the environment, as shown in Fig. 3-46. 
Therefore, designing such an "extra long" fin is out of the question since it 
results in material waste, excessive weight, and increased size and thus in- 
creased cost with no benefit in return (in fact, such a long fin will hurt perfor- 
mance since it will suppress fluid motion and thus reduce the convection heat 
transfer coefficient). Fins that are so long that the temperature approaches the 
environment temperature cannot be recommended either since the little in- 
crease in heat transfer at the tip region cannot justify the large increase in the 
weight and cost. 

To get a sense of the proper length of a fin, we compare heat transfer from 
a fin of finite length to heat transfer from an infinitely long fin under the same 
conditions. The ratio of these two heat transfers is 




High 

heat 

transfer 



r^t-+-t 



% % 1 



FIGURE 3-46 

Because of the gradual temperature 

drop along the fin, the region 

near the fin tip makes little or 

no contribution to heat transfer. 



Heat transfer 
ratio: 



g fln VhJkA.(T b -T^tanhaL 



G, 



VhpkA-(T b -TJ 



tanh ah 



(3-77) 



Using a hand calculator, the values of tanh aL are evaluated for some values 
of aL and the results are given in Table 3-3. We observe from the table that 
heat transfer from a fin increases with aL almost linearly at first, but the curve 
reaches a plateau later and reaches a value for the infinitely long fin at about 
aL = 5. Therefore, a fin whose length is L = ^a can be considered to be an 
infinitely long fin. We also observe that reducing the fin length by half in that 
case (from aL = 5 to aL = 2.5) causes a drop of just 1 percent in heat trans- 
fer. We certainly would not hesitate sacrificing 1 percent in heat transfer per- 
formance in return for 50 percent reduction in the size and possibly the cost of 
the fin. In practice, a fin length that corresponds to about aL = 1 will transfer 
76.2 percent of the heat that can be transferred by an infinitely long fin, and 
thus it should offer a good compromise between heat transfer performance 
and the fin size. 

A common approximation used in the analysis of fins is to assume the fin 
temperature varies in one direction only (along the fin length) and the tem- 
perature variation along other directions is negligible. Perhaps you are won- 
dering if this one-dimensional approximation is a reasonable one. This is 
certainly the case for fins made of thin metal sheets such as the fins on a car 
radiator, but we wouldn't be so sure for fins made of thick materials. Studies 
have shown that the error involved in one-dimensional fin analysis is negligi- 
ble (less than about 1 percent) when 



TABLE 3-3 

The variation of heat transfer from 
a fin relative to that from an 
infinitely long fin 



aL 


- tanh aL 

"Hong fin 


0.1 


0.100 


0.2 


0.197 


0.5 


0.462 


1.0 


0.762 


1.5 


0.905 


2.0 


0.964 


2.5 


0.987 


3.0 


0.995 


4.0 


0.999 


5.0 


1.000 



hh 



<0.2 



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166 
HEAT TRANSFER 



where 8 is the characteristic thickness of the fin, which is taken to be the plate 
thickness t for rectangular fins and the diameter D for cylindrical ones. 

Specially designed finned surfaces called heat sinks, which are commonly 
used in the cooling of electronic equipment, involve one-of-a-kind complex 
geometries, as shown in Table 3-4. The heat transfer performance of heat 
sinks is usually expressed in terms of their thermal resistances R in °C/W, 
which is defined as 



G, 



R 



hA {m T| fin {T b - r„) 



(3-78) 



A small value of thermal resistance indicates a small temperature drop across 
the heat sink, and thus a high fin efficiency. 




FIGURE 3-47 

Schematic for Example 3-10. 



EXAMPLE 3-1 Maximum Power Dissipation of a Transistor 

Power transistors that are commonly used in electronic devices consume large 
amounts of electric power. The failure rate of electronic components increases 
almost exponentially with operating temperature. As a rule of thumb, the failure 
rate of electronic components is halved for each 10°C reduction in the junction 
operating temperature. Therefore, the operating temperature of electronic com- 
ponents is kept below a safe level to minimize the risk of failure. 

The sensitive electronic circuitry of a power transistor at the junction is pro- 
tected by its case, which is a rigid metal enclosure. Heat transfer characteris- 
tics of a power transistor are usually specified by the manufacturer in terms of 
the case-to-ambient thermal resistance, which accounts for both the natural 
convection and radiation heat transfers. 

The case-to-ambient thermal resistance of a power transistor that has a max- 
imum power rating of 10 W is given to be 20°C/W. If the case temperature of 
the transistor is not to exceed 85°C, determine the power at which this transis- 
tor can be operated safely in an environment at 25°C. 

SOLUTION The maximum power rating of a transistor whose case temperature 
is not to exceed 85°C is to be determined. 

Assumptions 1 Steady operating conditions exist. 2 The transistor case is iso- 
thermal at 85°C. 

Properties The case-to-ambient thermal resistance is given to be 20°C/W. 
Analysis The power transistor and the thermal resistance network associated 
with it are shown in Fig. 3-47. We notice from the thermal resistance network 
that there is a single resistance of 20°C/W between the case at T c = 85°C and 
the ambient at 7" x = 25°C, and thus the rate of heat transfer is 



6 



AT] 

R / 

/ c 



R, 



(85 - 25)°C 
20°C/W 



3W 



Therefore, this power transistor should not be operated at power levels above 
3 W if its case temperature is not to exceed 85°C. 

Discussion This transistor can be used at higher power levels by attaching it to 
a heat sink (which lowers the thermal resistance by increasing the heat transfer 
surface area, as discussed in the next example) or by using a fan (which lowers 
the thermal resistance by increasing the convection heat transfer coefficient). 



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167 
CHAPTER 3 



TABLE 3- 



Combined natural convection and radiation thermal resistance of various 
heat sinks used in the cooling of electronic devices between the heat sink and 
the surroundings. All fins are made of aluminum 6063T-5, are black anodized, 
and are 76 mm (3 in.) long (courtesy of Vemaline Products, Inc.). 



HS 5030 




R= 0.9°C/W (vertical) 
R = 1.2-C/W (horizontal) 

Dimensions: 76 mm X 105 mm X 44 mm 
Surface area: 677 cm 2 



HS606S 




R= 5°C/W 

Dimensions: 76 mm X 38 mm X 24 mm 
Surface area: 387 cm 2 



HS 6071 




R = 1.4°C/W (vertical) 
R= 1.8-C/W (horizontal) 

Dimensions: 76 mm X 92 mm X 26 mm 
Surface area: 968 cm 2 



HS6105 




R= 1.8°C/W (vertical) 
R= 2. 1°C/W (horizontal) 

Dimensions: 76 mm X 127 mm X 91 mm 
Surface area: 677 cm 2 



HS6115 



R= 1.1°C/W (vertical) 
R= 1.3°C/W (horizontal) 

Dimensions: 76 mm X 102 mm X 25 mm 
Surface area: 929 cm 2 



MS 7030 



R= 2.9°C/W (vertical) 
R= 3. 1°C/W (horizontal) 

Dimensions: 76 mm X 97 mm X 19 mm 
Surface area: 290 cm 2 



EXAMPLE 3-11 Selecting a Heat Sink for a Transistor 

A 60-W power transistor is to be cooled by attaching it to one of the commer- 
cially available heat sinks shown in Table 3-4. Select a heat sink that will allow 
the case temperature of the transistor not to exceed 90°C in the ambient air 
at 30°C. 



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168 
HEAT TRANSFER 



SOLUTION A commercially available heat sink from Table 3-4 is to be se- 
lected to keep the case temperature of a transistor below 90°C. 
Assumptions 1 Steady operating conditions exist. 2 The transistor case is iso- 
thermal at 90°C. 3 The contact resistance between the transistor and the heat 
sink is negligible. 

Analysis The rate of heat transfer from a 60-W transistor at full power is 
Q = 60 W. The thermal resistance between the transistor attached to the heat 
sink and the ambient air for the specified temperature difference is determined 
to be 



Q 



AT 
R 



AT 

Q 



(90 - 30)°C 
60 W 



1.0°C/W 



Therefore, the thermal resistance of the heat sink should be below 1.0°C/W. 
An examination of Table 3-4 reveals that the HS 5030, whose thermal resis- 
tance is 0.9°C/W in the vertical position, is the only heat sink that will meet 
this requirement. 




FIGURE 3-48 

Schematic for Example 3-12. 



EXAMPLE 3-12 Effect of Fins on Heat Transfer from Steam Pipes 

Steam in a heating system flows through tubes whose outer diameter is 
D 1 = 3 cm and whose walls are maintained at a temperature of 120°C. Circu- 
lar aluminum fins (k = 180 W/m ■ °C) of outer diameter D 2 = 6 cm and con- 
stant thickness f = 2 mm are attached to the tube, as shown in Fig. 3-48. The 
space between the fins is 3 mm, and thus there are 200 fins per meter length 
of the tube. Heat is transferred to the surrounding air at 7" x = 25°C, with a com- 
bined heat transfer coefficient of h = 60 W/m 2 • °C. Determine the increase in 
heat transfer from the tube per meter of its length as a result of adding fins. 

SOLUTION Circular aluminum fins are to be attached to the tubes of a heating 
system. The increase in heat transfer from the tubes per unit length as a result 
of adding fins is to be determined. 

Assumptions 1 Steady operating conditions exist. 2 The heat transfer coeffi- 
cient is uniform over the entire fin surfaces. 3 Thermal conductivity is constant. 
4 Heat transfer by radiation is negligible. 

Properties The thermal conductivity of the fins is given to be k = 180 
W/m • °C. 

Analysis In the case of no fins, heat transfer from the tube per meter of its 
length is determined from Newton's law of cooling to be 



A 



2 no fin ~~ hA nofin (T b 



- -nDiL = tt(0.03 m)(l m) = 0.0942 m 2 

(60 W/m 2 ■ °C)(0.0942 m 2 )(120 - 25)°C 
537 W 



The efficiency of the circular fins attached to a circular tube is plotted in Fig. 
3-43. Noting that L = \{D 2 - D x ) = 1(0.06 - 0.03) = 0.015 m in this case, 
we have 



cen58933_ch03.qxd 9/10/2002 8:59 AM Page 169 



r 2 + if (0.03 + i X 0.002) m 



2.07 



(L + U) 



0.015 m 

(0.015 +i X 0.002) m X 



60W/m 2 -°C 
V (180 W/m • °C)(0.002m) 



'%.„ 



0.207 



Aa, = 2ir(r 2 2 - /f) + 2irr 2 f 

= 2tt[(0.03 m) 2 - (0.015 m) 2 ] + 2tt(0.03 m)(0.002 m) 
= 0.00462 m 2 

6 fin = THfinGfin, max = T >fin' J Afin (T b — Tj) 

= 0.95(60 W/m 2 • °C)(0.00462 m 2 )(120 - 25)°C 
= 25.0 W 

Heat transfer from the unfinned portion of the tube is 

A unfln = ttDiS = ir(0.03 m)(0.003 m) = 0.000283 m 2 

Gunfin = hA nn f m {T b — 7„) 

= (60 W/m 2 ■ °C)(0.000283 m 2 )(120 - 25)°C 
= 1.60 W 

Noting that there are 200 fins and thus 200 interfin spacings per meter length 
of the tube, the total heat transfer from the finned tube becomes 



Q 



total, fin 



«(2f,„ + Gu„fi„) = 200(25.0 + 1.6) W = 5320 W 



Therefore, the increase in heat transfer from the tube per meter of its length as 
a result of the addition of fins is 

Q ia^e = Q total, fin " 6 „o fin = 5320 - 537 = 4783 W (per m tube length) 
Discussion The overall effectiveness of the finned tube is 

G total, fin 5320 W 



Q 



total, no fin 



537 W 



9.9 



That is, the rate of heat transfer from the steam tube increases by a factor of 
almost 10 as a result of adding fins. This explains the widespread use of finned 
surfaces. 



169 
CHAPTER 3 



0.95 



3-7 ■ HEAT TRANSFER IN 

COMMON CONFIGURATIONS 

So far, we have considered heat transfer in simple geometries such as large 
plane walls, long cylinders, and spheres. This is because heat transfer in such 
geometries can be approximated as one-dimensional, and simple analytical 
solutions can be obtained easily. But many problems encountered in practice 
are two- or three-dimensional and involve rather complicated geometries for 
which no simple solutions are available. 



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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 ) 



(3-79) 



where S is the conduction shape factor, which has the dimension of length, 
and k is the thermal conductivity of the medium between the surfaces. The 
conduction shape factor depends on the geometry of the system only. 

Conduction shape factors have been determined for a number of configura- 
tions encountered in practice and are given in Table 3-5 for some common 
cases. More comprehensive tables are available in the literature. Once the 
value of the shape factor is known for a specific geometry, the total steady 
heat transfer rate can be determined from the equation above using the speci- 
fied two constant temperatures of the two surfaces and the thermal conductiv- 
ity of the medium between them. Note that conduction shape factors are 
applicable only when heat transfer between the two surfaces is by conduction. 
Therefore, they cannot be used when the medium between the surfaces is a 
liquid or gas, which involves natural or forced convection currents. 

A comparison of Equations 3-4 and 3-79 reveals that the conduction shape 
factor S is related to the thermal resistance R by R = 1/kS or S = 1/kR. Thus, 
these two quantities are the inverse of each other when the thermal conduc- 
tivity of the medium is unity. The use of the conduction shape factors is illus- 
trated with examples 3-13 and 3-14. 



s-T 2 = 10°C 



Z = 0.5m 



I_ 



- ^ | D = 10 cm ) 

[' ■' ■ ■■' -L = 30 m - — -4 



FIGURE 3-49 

Schematic for Example 3-13. 



EXAMPLE 3-13 Heat Loss from Buried Steam Pipes 

A 30-m-long, 10-cm-diameter hot water pipe of a district heating system is 
buried in the soil 50 cm below the ground surface, as shown in Figure 3-49. 
The outer surface temperature of the pipe is 80°C. Taking the surface tempera- 
ture of the earth to be 10°C and the thermal conductivity of the soil at that lo- 
cation to be 0.9 W/m • °C, determine the rate of heat loss from the pipe. 

SOLUTION The hot water pipe of a district heating system is buried in the soil. 
The rate of heat loss from the pipe is to be determined. 

Assumptions 1 Steady operating conditions exist. 2 Heat transfer is two- 
dimensional (no change in the axial direction). 3 Thermal conductivity of the 
soil is constant. 

Properties The thermal conductivity of the soil is given to be k = 0.9 W/m • °C. 
Analysis The shape factor for this configuration is given in Table 3-5 to be 



S 



2ttL 
ln(4z/D) 



since z > 1.5D, where z is the distance of the pipe from the ground surface, 
and D is the diameter of the pipe. Substituting, 



2ir X (30 m) 
ln(4 X 0.5/0.1) 



62.9 m 



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171 
CHAPTER 3 



TABLE 3-5 

Conduction shape factors S for several configurations for use in Q = kS{Ti - T 2 ) to determine the steady rate of heat 
transfer through a medium of thermal conductivity k between the surfaces at temperatures 7"! and T 2 



( 1 ) Isothermal cylinder of length L 
buried in a semi-infinite medium 
(L»D and z>l.5D) 



rl 



2%L 
" In (4z/D) 



G 



m 



> ) ■■ 



(2) Vertical isothermal cylinder of length L 
buried in a semi-infinite medium 
(L»D) 



2%L 
"ln(4L/D) 



IX 



(3) Two parallel isothermal cylinders 
placed in an infinite medium 
(L»D l ,D 1 ,z) 



2kL 



cosh 



Az 2 ~D] 



~D\ 




(4) A row of equally spaced parallel isothermal 
cylinders buried in a semi-infinite medium 
(L»D, z andw>1.5D) 



j£± 



2kL 



ln|^ sinh 2^Z 
(per cylinder) 




-j-^-W ; >|» '- w ■ ' »]■■ ; 'Wr»|- 



(5) Circular isothermal cylinder of length L 
in the midplane of an infinite wall 
(Z > 0.5£>) 



2kL 



ln(8ztoD) 




(6) Circular isothermal cylinder of length L 
at the center of a square solid bar of the 
same length 



2%L 



In (1.08 wID) 




(7) Eccentric circular isothermal cylinder 
of length L in a cylinder of the same 
length (L > £>,) j 



2kL 



cosh 



D]+D\-4z 2 




(8) Large plane wall 



-T, 



(continued) 



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172 
HEAT TRANSFER 



TABLE 3-5 (CONCLUDED) 



(9) A long cylindrical layer 



2kL 



In (D/D.) 




(10) A square flow passage 
(a) For alb > 1.4, 

„ 2%L 


T 2\/\ 1 


J 


0.93 In (0.948 alb) 








(b) For alb < 1.41, 


/A 




c 2%L 




L 


0.785 In (alb) 


\^b^ 
-> a 







(1 1) A spherical layer 



2%D X D 1 



(12) Disk buried parallel to 

the surface in a semi-infinite 
medium (z » D) 



D^-D 



1 -u x 




^h_ 



S = 4D 

(5 = 2£)whenz = 0) 



& 



(13) The edge of two adjoining 
walls of equal thickness 



5 = 0.54 w 




(14) Corner of three walls 
of equal thickness 



5 = 0.15L 




(15) Isothermal sphere buried in a 
semi-infinite medium 



^ 



2kD 



1 - 0.25D/Z 




(16) Isothermal sphere buried 

in a semi-infinite medium at T 2 
whose surface is insulated 



2kD 



Insulated 



1 + 0.25D/Z 




cen58933_ch03.qxd 9/10/2002 8:59 AM Page 173 



Then the steady rate of heat transfer from the pipe becomes 

Q = Sk(T, - T 2 ) = (62.9 m)(0.9 W/m • °C)(80 - 10)°C = 3963 W 

Discussion Note that this heat is conducted from the pipe surface to the sur- 
face of the earth through the soil and then transferred to the atmosphere by 
convection and radiation. 



173 
CHAPTER 3 



EXAMPLE 3-14 Heat Transfer between Hot and Cold Water Pipes 

A 5-m-long section of hot and cold water pipes run parallel to each other in a 
thick concrete layer, as shown in Figure 3-50. The diameters of both pipes are 
5 cm, and the distance between the centerline of the pipes is 30 cm. The sur- 
face temperatures of the hot and cold pipes are 70°C and 15°C, respectively. 
Taking the thermal conductivity of the concrete to be k = 0.75 W/m • °C, de- 
termine the rate of heat transfer between the pipes. 

SOLUTION Hot and cold water pipes run parallel to each other in a thick con- 
crete layer. The rate of heat transfer between the pipes is to be determined. 
Assumptions 1 Steady operating conditions exist. 2 Heat transfer is two- 
dimensional (no change in the axial direction). 3 Thermal conductivity of the 
concrete is constant. 

Properties The thermal conductivity of concrete is given to be k = 0.75 
W/m • °C. 
Analysis The shape factor for this configuration is given in Table 3-5 to be 

2ttL 



4z 2 -D]-D} 



cosh 



where z is the distance between the centerlines of the pipes and L is their 
length. Substituting, 



2tt X (5 m) 



4 X 0.3 2 - 0.05 2 - 0.05 2 



cosh 



2 X 0.05 X 0.05 



6.34 m 



Then the steady rate of heat transfer between the pipes becomes 

Q = Sk(T, - T 2 ) = (6.34 m)(0.75 W/m ■ °C)(70 - 15°)C = 262 W 

Discussion We can reduce this heat loss by placing the hot and cold water 
pipes further away from each other. 



15°C 




Z = 30 cm 

FIGURE 3-50 

Schematic for Example 3-14. 



It is well known that insulation reduces heat transfer and saves energy and 
money. Decisions on the right amount of insulation are based on a heat trans- 
fer analysis, followed by an economic analysis to determine the "monetary 
value" of energy loss. This is illustrated with Example 3-15. 



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174 
HEAT TRANSFER 



75°F- 



Wall, R= 13 



■ 45°F 



7/cc. ♦-^vwvw^ i — ww- < -^vwwv^-^ 7^ 9 

FIGURE 3-51 

Schematic for Example 3-15. 



EXAMPLE 3-15 Cost of Heat Loss through Walls in Winter 

Consider an electrically heated house whose walls are 9 ft high and have an 
R-\ia\ue of insulation of 13 (i.e., a thickness-to-thermal conductivity ratio of 
L/k = 13 h • ft 2 • °F/Btu). Two of the walls of the house are 40 ft long and the 
others are 30 ft long. The house is maintained at 75°F at all times, while 
the temperature of the outdoors varies. Determine the amount of heat lost 
through the walls of the house on a certain day during which the average tem- 
perature of the outdoors is 45°F. Also, determine the cost of this heat loss to the 
homeowner if the unit cost of electricity is $0.075/kWh. For combined convec- 
tion and radiation heat transfer coefficients, use the ASHRAE (American Soci- 
ety of Heating, Refrigeration, and Air Conditioning Engineers) recommended 
values of h, = 1.46 Btu/h • ft 2 • °F for the inner surface of the walls and h = 
4.0 Btu/h ■ ft 2 • °F for the outer surface of the walls under 15 mph wind condi- 
tions in winter. 

SOLUTION An electrically heated house with R-13 insulation is considered. 
The amount of heat lost through the walls and its cost are to be determined. 
Assumptions 1 The indoor and outdoor air temperatures have remained at the 
given values for the entire day so that heat transfer through the walls is steady. 
2 Heat transfer through the walls is one-dimensional since any significant 
temperature gradients in this case will exist in the direction from the indoors 
to the outdoors. 3 The radiation effects are accounted for in the heat transfer 
coefficients. 

Analysis This problem involves conduction through the wall and convection at 
its surfaces and can best be handled by making use of the thermal resistance 
concept and drawing the thermal resistance network, as shown in Fig. 3-51. 
The heat transfer area of the walls is 

A = Circumference X Height = (2 X 30 ft + 2 X 40 ft)(9 ft) = 1260 ft 2 
Then the individual resistances are evaluated from their definitions to be 

0.00054 h ■ °F/Btu 



R=R = J_ = 1 

conv '' h,A (1.46 Btu/h ■ ft 2 ■ °F)( 1260 ft 2 ) 



R» 



L /{-value 13 h ■ ft 2 ■ °F/Btu 



kA A 



1260 ft 2 



1 



0.01032 h-°F/Btu 

0.00020 h • °F/Btu 



1 

h c A (4.0 Btu/h • ft 2 ■ °F)(1260 ft 2 ) 

Noting that all three resistances are in series, the total resistance is 

fltotai = R i + #waii + R = 0.00054 + 0.01032 + 0.00020 = 0.01106 h • °F/Btu 

Then the steady rate of heat transfer through the walls of the house becomes 



Q 



*Mntnl 



(75 - 45)°F 
0.01 106 h • °F/Btu 



2712 Btu/h 



Finally, the total amount of heat lost through the walls during a 24-h period and 
its cost to the home owner are 

Q = Q At = (2712 Btu/h)(24-h/day) = 65,099 Btu/day = 19.1 kWh/day 



cen58933_ch03.qxd 9/10/2002 8:59 AM Page 175 



since 1 kWh = 3412 Btu, and 

Heating cost = (Energy lost)(Cost of energy) = (19.1 kWh/day)($0.075/kWh) 
= $1.43/day 

Discussion The heat losses through the walls of the house that day will cost 
the home owner $1.43 worth of electricity. 



175 
CHAPTER 3 



TOPIC OF SPECIAL INTEREST 



Heat Transfer Through Walls and Roofs 

Under steady conditions, the rate of heat transfer through any section of a 
building wall or roof can be determined from 



Q = UA(Ti - T ) 



MTj ~ T ) 
R 



(3-80) 



where T t and T are the indoor and outdoor air temperatures, A is the heat 
transfer area, U is the overall heat transfer coefficient (the [/-factor), and 
R = l/U is the overall unit thermal resistance (the R-value). Walls and 
roofs of buildings consist of various layers of materials, and the structure 
and operating conditions of the walls and the roofs may differ significantly 
from one building to another. Therefore, it is not practical to list the 
^-values (or [/-factors) of different kinds of walls or roofs under different 
conditions. Instead, the overall 7?-value is determined from the thermal 
resistances of the individual components using the thermal resistance net- 
work. The overall thermal resistance of a structure can be determined most 
accurately in a lab by actually assembling the unit and testing it as a whole, 
but this approach is usually very time consuming and expensive. The ana- 
lytical approach described here is fast and straightforward, and the results 
are usually in good agreement with the experimental values. 

The unit thermal resistance of a plane layer of thickness L and thermal 
conductivity k can be determined from R = Llk. The thermal conductivity 
and other properties of common building materials are given in the appen- 
dix. The unit thermal resistances of various components used in building 
structures are listed in Table 3-6 for convenience. 

Heat transfer through a wall or roof section is also affected by the con- 
vection and radiation heat transfer coefficients at the exposed surfaces. The 
effects of convection and radiation on the inner and outer surfaces of walls 
and roofs are usually combined into the combined convection and radiation 
heat transfer coefficients (also called surface conductances) h t and h a , 
respectively, whose values are given in Table 3-7 for ordinary surfaces 
(e = 0.9) and reflective surfaces (e = 0.2 or 0.05). Note that surfaces hav- 
ing a low emittance also have a low surface conductance due to the reduc- 
tion in radiation heat transfer. The values in the table are based on a surface 



TABLE 3-7 

Combined convection and radiation 
heat transfer coefficients at window, 
wall, or roof surfaces (from ASHRAE 
Handbook of Fundamentals, Ref. 1, 
Chap. 22, Table 1). 



Direc- 
tion of 
Posi- Heat 


h, W/m 2 • c 

Surface 
Emittance 


C* 
e 


tion Flow 


0.90 


0.20 


0.05 


Still air (both indoors and 


outdoors) 


Horiz. UpT 


9.26 


5.17 


4.32 


Horiz. Down I 


6.13 


2.10 


1.25 


45° slope UpT 


9.09 


5.00 


4.15 


45° slope Down I 


7.50 


3.41 


2.56 


Vertical Horiz. — > 


8.29 


4.20 


3.35 


Moving air (any position, a 


ny direction) 


Winter condition 








(winds at 15 mpr 








or 24 km/h) 


34.0 


— 


— 


Summer condition 








(winds at 7.5 mph 






or 12 km/h) 


22.7 


— 


— 



*This section can be skipped without a loss of continuity. 



♦Multiply by 0.176 to convert to Btu/h • ft 2 ■ °F. 
Surface resistance can be obtained from Ft = llh. 



cen58933_ch03.qxd 9/10/2002 8:59 AM Page 176 



176 
HEAT TRANSFER 



TABLE 3-6 

Unit thermal resistance (the ff-value) of common components used in buildings 





R- 


value 






R- value 


Component 






-n 2 • °C/W ft 2 


• h • °F/Btu 


Component 


-n 2 • °C/W ft 2 • h • °F/Btu 


Outside surface (winter) 






0.030 


0.17 


Wood stud, nominal 2 in. X 






Outside surface (summer) 




0.044 


0.25 


6 in. (5.5 in. or 140 mm wide) 


0.98 


5.56 


Inside surface, still air 






0.12 


0.68 


Clay tile, 100 mm (4 in.) 


0.18 


1.01 


Plane air space, vertical, 


orc 


inary 


surfaces (e eff = 


0.82): 


Acoustic tile 


0.32 


1.79 


13 mm (1 in.) 






0.16 


0.90 


Asphalt shingle roofing 


0.077 


0.44 


20 mm (| in.) 






0.17 


0.94 


Building paper 


0.011 


0.06 


40 mm (1.5 in.) 






0.16 


0.90 


Concrete block, 100 mm (4 in.): 






90 mm (3.5 in.) 






0.16 


0.91 


Lightweight 


0.27 


1.51 


Insulation, 25 mm (1 in 


) 








Heavyweight 


0.13 


0.71 


Glass fiber 






0.70 


4.00 


Plaster or gypsum board, 






Mineral fiber batt 






0.66 


3.73 


13 mm (i in.) 


0.079 


0.45 


Urethane rigid foam 






0.98 


5.56 


Wood fiberboard, 13 mm (i in.) 


0.23 


1.31 


Stucco, 25 mm (1 in.) 






0.037 


0.21 


Plywood, 13 mm (| in.) 


0.11 


0.62 


Face brick, 100 mm (4 in.) 




0.075 


0.43 


Concrete, 200 mm (8 in.): 






Common brick, 100 mm 


(4 


n.) 


0.12 


0.79 


Lightweight 


1.17 


6.67 


Steel siding 






0.00 


0.00 


Heavyweight 


0.12 


0.67 


Slag, 13 mm (1 in.) 






0.067 


0.38 


Cement mortar, 13 mm (1/2 in.) 


0.018 


0.10 


Wood, 25 mm (1 in.) 






0.22 


1.25 


Wood bevel lapped siding, 






Wood stud, nominal 2 in 


. X 








13 mm X 200 mm 






4 in. (3.5 in. or 90 mm w 


de) 


0.63 


3.58 


(1/2 in. X 8 in.) 


0.14 


0.81 



temperature of 21°C (72°F) and a surface-air temperature difference of 
5.5°C (10°F). Also, the equivalent surface temperature of the environment 
is assumed to be equal to the ambient air temperature. Despite the conve- 
nience it offers, this assumption is not quite accurate because of the addi- 
tional radiation heat loss from the surface to the clear sky. The effect of sky 
radiation can be accounted for approximately by taking the outside tem- 
perature to be the average of the outdoor air and sky temperatures. 

The inner surface heat transfer coefficient h t remains fairly constant 
throughout the year, but the value of h varies considerably because of its 
dependence on the orientation and wind speed, which can vary from less 
than 1 km/h in calm weather to over 40 km/h during storms. The com- 
monly used values of ft, and h for peak load calculations are 



h, = 8.29 W/m 2 • °C = 1 .46 Btu/h • ft 2 ■ °F 
r 34.0 W/m 2 ■ °C = 6.0 Btu/h ■ ft 2 ■ °F 



//„ 



22.7 W/m 2 ■ °C = 4.0 Btu/h ■ ft 2 



(winter and summer) 

(winter) 

(summer) 



which correspond to design wind conditions of 24 km/h (15 mph) for win- 
ter and 12 km/h (7.5 mph) for summer. The corresponding surface thermal 
resistances (^-values) are determined from Rj = l/hj and R = l/h a . The 
surface conductance values under still air conditions can be used for inte- 
rior surfaces as well as exterior surfaces in calm weather. 



cen58933_ch03.qxd 9/10/2002 8:59 AM Page 177 



177 
CHAPTER 3 



Building components often involve trapped air spaces between various 
layers. Thermal resistances of such air spaces depend on the thickness of 
the layer, the temperature difference across the layer, the mean air temper- 
ature, the emissivity of each surface, the orientation of the air layer, and the 
direction of heat transfer. The emissivities of surfaces commonly encoun- 
tered in buildings are given in Table 3-8. The effective emissivity of a 
plane-parallel air space is given by 



Effective e l e 2 



1 



(3-81) 



where s { and e 2 are the emissivities of the surfaces of the air space. Table 
3-8 also lists the effective emissivities of air spaces for the cases where 
(1) the emissivity of one surface of the air space is e while the emissivity 
of the other surface is 0.9 (a building material) and (2) the emissivity of 
both surfaces is e. Note that the effective emissivity of an air space between 
building materials is 0.82/0.03 = 27 times that of an air space between sur- 
faces covered with aluminum foil. For specified surface temperatures, ra- 
diation heat transfer through an air space is proportional to effective 
emissivity, and thus the rate of radiation heat transfer in the ordinary sur- 
face case is 27 times that of the reflective surface case. 

Table 3-9 lists the thermal resistances of 20-mm-, 40-mm-, and 90-mm- 
(0.75-in., 1.5-in., and 3.5-in.) thick air spaces under various conditions. The 
thermal resistance values in the table are applicable to air spaces of uniform 
thickness bounded by plane, smooth, parallel surfaces with no air leakage. 
Thermal resistances for other temperatures, emissivities, and air spaces can 
be obtained by interpolation and moderate extrapolation. Note that the 
presence of a low-emissivity surface reduces radiation heat transfer across 
an air space and thus significantly increases the thermal resistance. The 
thermal effectiveness of a low-emissivity surface will decline, however, if 
the condition of the surface changes as a result of some effects such as con- 
densation, surface oxidation, and dust accumulation. 

The 7?-value of a wall or roof structure that involves layers of uniform 
thickness is determined easily by simply adding up the unit thermal re- 
sistances of the layers that are in series. But when a structure involves 
components such as wood studs and metal connectors, then the ther- 
mal resistance network involves parallel connections and possible two- 
dimensional effects. The overall R-yalue in this case can be determined by 
assuming (1) parallel heat flow paths through areas of different construc- 
tion or (2) isothermal planes normal to the direction of heat transfer. The 
first approach usually overpredicts the overall thermal resistance, whereas 
the second approach usually underpredicts it. The parallel heat flow path 
approach is more suitable for wood frame walls and roofs, whereas the 
isothermal planes approach is more suitable for masonry or metal frame 
walls. 

The thermal contact resistance between different components of building 
structures ranges between 0.01 and 0.1 m 2 ■ °C/W, which is negligible in 
most cases. However, it may be significant for metal building components 
such as steel framing members. 



TABLE 3-8 

Emissivities s of various surfaces 
and the effective emissivity of air 
spaces (from ASHRAE Handbook 
of Fundamentals, Ref. 1, Chap. 22, 
Table 3). 







Effective 






EmissK 


ity of 




? 


Air Space 




1 = s 


e 1 = e 


Surface 


E £ 


2 = 0.9 


£ 2 = E 


Aluminum foil, 






bright 


0.05* 


0.05 


0.03 


Aluminum 








sheet 


0.12 


0.12 


0.06 


Aluminum- 








coated 








paper, 








polished 


0.20 


0.20 


0.11 


Steel, galvan 


ized, 






bright 


0.25 


0.24 


0.15 


Aluminum 








paint 


0.50 


0.47 


0.35 


Building materials: 






Wood, paper, 






masonry, 


nonmetallic 




paints 


0.90 


0.82 


0.82 


Ordinary glass 0.84 


0.77 


0.72 



*Surface emissivity of aluminum foil 
increases to 0.30 with barely visible 
condensation, and to 0.70 with clearly 
visible condensation. 



cen58933_ch03.qxd 9/10/2002 8:59 AM Page 17E 



TABLE 3-9 

Unit thermal resistances (ff-values) of well-sealed plane air spaces (from ASHRAE Handbook of Fundamentals, Ref. 1, 

Chap. 22, Table 2) 

(a) SI units (in m 2 • °C/W) 





Position Direction 
of Air of Heat 
Space Flow 


Mean 

Temp., 

°C 


Temp. 

Diff., 

°C 


20-mm Air Space 


40-mm Air Space 


90-mm Air Spa 

Effective 
Emissivity, e e 


ce 






Effective 
Emissivity, e ef 






Effective 
Emissivity, e ef 




t 




0.03 


0.05 


0.5 


0.82 


0.03 


0.05 


0.5 


0.82 


0.03 


0.05 


0.5 


0.82 






32.2 


5.6 


0.41 


0.39 


0.18 


0.13 


0.45 


0.42 


0.19 


0.14 


0.50 


0.47 


0.20 


0.14 






10.0 


16.7 


0.30 


0.29 


0.17 


0.14 


0.33 


0.32 


0.18 


0.14 


0.27 


0.35 


0.19 


0.15 




Horizontal Up T 


10.0 


5.6 


0.40 


0.39 


0.20 


0.15 


0.44 


0.42 


0.21 


0.16 


0.49 


0.47 


0.23 


0.16 






-17.8 


11.1 


0.32 


0.32 


0.20 


0.16 


0.35 


0.34 


0.22 


0.17 


0.40 


0.38 


0.23 


0.18 






32.2 


5.6 


0.52 


0.49 


0.20 


0.14 


0.51 


0.48 


0.20 


0.14 


0.56 


0.52 


0.21 


0.14 






10.0 


16.7 


0.35 


0.34 


0.19 


0.14 


0.38 


0.36 


0.20 


0.15 


0.40 


0.38 


0.20 


0.15 




45° slope UpT 


10.0 


5.6 


0.51 


0.48 


0.23 


0.17 


0.51 


0.48 


0.23 


0.17 


0.55 


0.52 


0.24 


0.17 






-17.8 


11.1 


0.37 


0.36 


0.23 


0.18 


0.40 


0.39 


0.24 


0.18 


0.43 


0.41 


0.24 


0.19 






32.2 


5.6 


0.62 


0.57 


0.21 


0.15 


0.70 


0.64 


0.22 


0.15 


0.65 


0.60 


0.22 


0.15 






10.0 


16.7 


0.51 


0.49 


0.23 


0.17 


0.45 


0.43 


0.22 


0.16 


0.47 


0.45 


0.22 


0.16 




Vertical Horizontal 


-> 10.0 


5.6 


0.65 


0.61 


0.25 


0.18 


0.67 


0.62 


0.26 


0.18 


0.64 


0.60 


0.25 


0.18 






-17.8 


11.1 


0.55 


0.53 


0.28 


0.21 


0.49 


0.47 


0.26 


0.20 


0.51 


0.49 


0.27 


0.20 






32.2 


5.6 


0.62 


0.58 


0.21 


0.15 


0.89 


0.80 


0.24 


0.16 


0.85 


0.76 


0.24 


0.16 






10.0 


16.7 


0.60 


0.57 


0.24 


0.17 


0.63 


0.59 


0.25 


0.18 


0.62 


0.58 


0.25 


0.18 




45° slope Down i 


10.0 


5.6 


0.67 


0.63 


0.26 


0.18 


0.90 


0.82 


0.28 


0.19 


0.83 


0.77 


0.28 


0.19 






-17.8 


11.1 


0.66 


0.63 


0.30 


0.22 


0.68 


0.64 


0.31 


0.22 


0.67 


0.64 


0.31 


0.22 






32.2 


5.6 


0.62 


0.58 


0.21 


0.15 


1.07 


0.94 


0.25 


0.17 


1.77 


1.44 


0.28 


0.18 








10.0 


16.7 


0.66 


0.62 


0.25 


0.18 


1.10 


0.99 


0.30 


0.20 


1.69 


1.44 


0.33 


0.21 






Horizontal Down i 


10.0 


5.6 


0.68 


0.63 


0.26 


0.18 


1.16 


1.04 


0.30 


0.20 


1.96 


1.63 


0.34 


0.22 


V 


J 




-17.8 


11.1 


0.74 


0.70 


0.32 


0.23 


1.24 


1.13 


0.39 


0.26 


1.92 


1.68 


0.43 


0.29 




(b) English units (in h 


■ ft 2 ■ °F/Btu) 




























Position Direction 
of Air of Heat 
Space Flow 


Mean 

Temp., 

°F 


Temp. 

Diff., 

°F 


0.75-in. 


Air Space 


1.5-in. A 


ir Space 


3.5-in. Air Space 






Effective 
Emissivity, e ef 






Effective 
Emissivity, e ef 






Effective 
Emissivity, e eff 




0.03 


0.05 


0.5 


0.82 


0.03 


0.05 


0.5 


0.82 


0.03 


0.05 


0.5 


0.82 






90 


10 


2.34 


2.22 


1.04 


0.75 


2.55 


2.41 


1.08 


0.77 


2.84 


2.66 


1.13 


0.80 






50 


30 


1.71 


1.66 


0.99 


0.77 


1.87 


1.81 


1.04 


0.80 


2.09 


2.01 


1.10 


0.84 




Horizontal Up T 


50 


10 


2.30 


2.21 


1.16 


0.87 


2.50 


2.40 


1.21 


0.89 


2.80 


2.66 


1.28 


0.93 









20 


1.83 


1.79 


1.16 


0.93 


2.01 


1.95 


1.23 


0.97 


2.25 


2.18 


1.32 


1.03 






90 


10 


2.96 


2.78 


1.15 


0.81 


2.92 


2.73 


1.14 


0.80 


3.18 


2.96 


1.18 


0.82 






50 


30 


1.99 


1.92 


1.08 


0.82 


2.14 


2.06 


1.12 


0.84 


2.26 


2.17 


1.15 


0.86 




45° slope UpT 


50 


10 


2.90 


2.75 


1.29 


0.94 


2.88 


2.74 


1.29 


0.94 


3.12 


2.95 


1.34 


0.96 









20 


2.13 


2.07 


1.28 


1.00 


2.30 


2.23 


1.34 


1.04 


2.42 


2.35 


1.38 


1.06 






90 


10 


3.50 


3.24 


1.22 


0.84 


3.99 


3.66 


1.27 


0.87 


3.69 


3.40 


1.24 


0.85 






50 


30 


2.91 


2.77 


1.30 


0.94 


2.58 


2.46 


1.23 


0.90 


2.67 


2.55 


1.25 


0.91 




Vertical Horizontal 


-* 50 


10 


3.70 


3.46 


1.43 


1.01 


3.79 


3.55 


1.45 


1.02 


3.63 


3.40 


1.42 


1.01 









20 


3.14 


3.02 


1.58 


1.18 


2.76 


2.66 


1.48 


1.12 


2.88 


2.78 


1.51 


1.14 






90 


10 


3.53 


3.27 


1.22 


0.84 


5.07 


4.55 


1.36 


0.91 


4.81 


4.33 


1.34 


0.90 






50 


30 


3.43 


3.23 


1.39 


0.99 


3.58 


3.36 


1.42 


1.00 


3.51 


3.30 


1.40 


1.00 




45° slope Down i 


50 


10 


3.81 


3.57 


1.45 


1.02 


5.10 


4.66 


1.60 


1.09 


4.74 


4.36 


1.57 


1.08 









20 


3.75 


3.57 


1.72 


1.26 


3.85 


3.66 


1.74 


1.27 


3.81 


3.63 


1.74 


1.27 






90 


10 


3.55 


3.29 


1.22 


0.85 


6.09 


5.35 


1.43 


0.94 


10.07 


8.19 


1.57 


1.00 






50 


30 


3.77 


3.52 


1.44 


1.02 


6.27 


5.63 


1.70 


1.14 


9.60 


8.17 


1.88 


1.22 




Horizontal Down i 


50 


10 


3.84 


3.59 


1.45 


1.02 


6.61 


5.90 


1.73 


1.15 


11.15 


9.27 


1.93 


1.24 











20 


4.18 


3.96 


1.81 


1.30 


7.03 


6.43 


2.19 


1.49 


10.90 


9.52 


2.47 


1.62 



178 



cen58933_ch03.qxd 9/10/2002 8:59 AM Page 179 



EXAMPLE 3-16 



The /?- Value of a Wood Frame Wall 



Determine the overall unit thermal resistance (the ff-value) and the overall heat 
transfer coefficient (the L/-factor) of a wood frame wall that is built around 
38-mm X 90-mm (2x4 nominal) wood studs with a center-to-center distance 
of 400 mm. The 90-mm-wide cavity between the studs is filled with glass fiber 
insulation. The inside is finished with 13-mm gypsum wallboard and the out- 
side with 13-mm wood fiberboard and 13-mm X 200-mm wood bevel lapped 
siding. The insulated cavity constitutes 75 percent of the heat transmission 
area while the studs, plates, and sills constitute 21 percent. The headers con- 
stitute 4 percent of the area, and they can be treated as studs. 

Also, determine the rate of heat loss through the walls of a house whose 
perimeter is 50 m and wall height is 2.5 m in Las Vegas, Nevada, whose winter 
design temperature is -2°C. Take the indoor design temperature to be 22°C 
and assume 20 percent of the wall area is occupied by glazing. 



179 
CHAPTER 3 



SOLUTION The ff-value and the ^/-factor of a wood frame wall as well as the 
rate of heat loss through such a wall in Las Vegas are to be determined. 

Assumptions 1 Steady operating conditions exist. 2 Heat transfer through the 
wall is one-dimensional. 3 Thermal properties of the wall and the heat transfer 
coefficients are constant. 



Properties The ff-values of different materials are given in Table 3-6. 

Analysis The schematic of the wall as well as the different elements used in its 
construction are shown here. Heat transfer through the insulation and through 
the studs will meet different resistances, and thus we need to analyze the ther- 
mal resistance for each path separately. Once the unit thermal resistances and 
the L/-factors for the insulation and stud sections are available, the overall av- 
erage thermal resistance for the entire wall can be determined from 

''nv era 1 1 1 / L/ rivers 1 1 



where 



overall V ./area/insi 



(U X/ arca ) stud 



and the value of the area fraction f area is 0.75 for the insulation section and 
0.25 for the stud section since the headers that constitute a small part of the 
wall are to be treated as studs. Using the available ff-values from Table 3-6 and 
calculating others, the total ff-values for each section can be determined in a 
systematic manner in the table in this sample. 

We conclude that the overall unit thermal resistance of the wall is 2.23 
m 2 • °C/W, and this value accounts for the effects of the studs and headers. It 
corresponds to an ff-value of 2.23 X 5.68 = 12.7 (or nearly ff-13) in English 
units. Note that if there were no wood studs and headers in the wall, the over- 
all thermal resistance would be 3.05 m 2 ■ °C/W, which is 37 percent greater 
than 2.23 m 2 • °C/W. Therefore, the wood studs and headers in this case serve 
as thermal bridges in wood frame walls, and their effect must be considered in 
the thermal analysis of buildings. 



cen58933_ch03.qxd 9/10/2002 8:59 AM Page IE 



180 
HEAT TRANSFER 



Schematic 



ft- value, 
m 2 • °C/W 




Construction 



Between At 

Studs Studs 



1. 


Outside surface, 








24 km/h wind 


0.030 


0.030 


2. 


Wood bevel lapped 








siding 


0.14 


0.14 


3. 


Wood fiberboard 








sheeting, 13 mm 


0.23 


0.23 


4a. 


Glass fiber 








insulation, 90 mm 


2.45 


— 


4b. 


Wood stud, 38 mm X 






6 


90 mm 


— 


0.63 


5. 


Gypsum wall board, 








13 mm 


0.079 


0.079 


6. 


Inside surface, still air 


0.12 


0.12 



Total unit thermal resistance of each section, R (in m 2 • °C/W) 3.05 1.23 
The U-factor of each section, U = 1/ff, in W/m 2 • °C 0.328 0.813 

Area fraction of each section, 4rea 0.75 0.25 

Overall U-factor: U= 2f area , U, = 0.75 X 0.328 + 0.25 X 0.813 

= 0.449 W/m 2 • °C 
Overall unit thermal resistance: R = l/U= 2.23 m 2 • °C/W 

The perimeter of the building is 50 m and the height of the walls is 2.5 m. 
Noting that glazing constitutes 20 percent of the walls, the total wall area is 

A wa „ = 0.80(Perimeter)(Height) = 0.80(50 m)(2.5 m) = 100 m 2 

Then the rate of heat loss through the walls under design conditions becomes 

Swan = (HA) wall {T i - T ) 

= (0.449 W/m 2 • °C)(100 m 2 )[22 - (-2)°C] 
= 1078 W 

Discussion Note that a 1-kW resistance heater in this house will make up al- 
most all the heat lost through the walls, except through the doors and windows, 
when the outdoor air temperature drops to -2°C. 



EXAMPLE 3-17 The ff-Value of a Wall with Rigid Foam 

The 13-mm-thick wood fiberboard sheathing of the wood stud wall discussed in 
the previous example is replaced by a 25-mm-thick rigid foam insulation. De- 
termine the percent increase in the R -value of the wall as a result. 



cen58933_ch03.qxd 9/10/2002 8:59 AM Page 181 



SOLUTION The overall ff-value of the existing wall was determined in Example 
3-16 to be 2.23 m 2 • °C/W. Noting that the ff-values of the fiberboard and the 
foam insulation are 0.23 m z • °C/W and 0.98 m 2 • °C/W, respectively, and 
the added and removed thermal resistances are in series, the overall ff-value of 
the wall after modification becomes 



Rr, 



R„ 



old removed added 

2.23 - 0.23 + 0.98 
2.98 m 2 • °C/W 



This represents an increase of (2.98 - 2.23)/2.23 = 0.34 or 34 percent in 
the ff-value of the wall. This example demonstrated how to evaluate the new 
ff-value of a structure when some structural members are added or removed. 



EXAMPLE 3-18 The ff-Value of a Masonry Wall 

Determine the overall unit thermal resistance (the ff-value) and the overall heat 
transfer coefficient (the ^/-factor) of a masonry cavity wall that is built around 
6-in. -thick concrete blocks made of lightweight aggregate with 3 cores filled 
with perlite (ff = 4.2 h • ft 2 • °F/Btu). The outside is finished with 4-in. face 
brick with i-in. cement mortar between the bricks and concrete blocks. The in- 
side finish consists of | in. gypsum wallboard separated from the concrete block 
by f-in.-thick (1-in. X 3-in. nominal) vertical furring (ff = 4.2 h • ft 2 • °F/Btu) 
whose center-to-center distance is 16 in. Both sides of the |-in. -thick air space 
between the concrete block and the gypsum board are coated with reflective 
aluminum foil (e = 0.05) so that the effective emissivity of the air space is 
0.03. For a mean temperature of 50°F and a temperature difference of 30°F, 
the ff-value of the air space is 2.91 h • ft 2 ■ °F/Btu. The reflective air space con- 
stitutes 80 percent of the heat transmission area, while the vertical furring con- 
stitutes 20 percent. 

SOLUTION The ff-value and the U-factor of a masonry cavity wall are to be 
determined. 

Assumptions 1 Steady operating conditions exist. 2 Heat transfer through the 
wall is one-dimensional. 3 Thermal properties of the wall and the heat transfer 
coefficients are constant. 

Properties The ff-values of different materials are given in Table 3-6. 
Analysis The schematic of the wall as well as the different elements used in its 
construction are shown below. Following the approach described here and using 
the available ff-values from Table 3-6, the overall ff-value of the wall is deter- 
mined in this table. 



181 
CHAPTER 3 



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182 
HEAT TRANSFER 



Schematic 




l 







fl-va 


ue, 






h • ft 2 • 


°F/Btu 






Between 


At 




Construction 


Furring 


Furring 


1. 


Outside surface, 








15 mph wind 


0.17 


0.17 


2. 


Face brick, 4 in. 


0.43 


0.43 


3. 


Cement mortar, 








0.5 in. 


0.10 


0.10 


4. 


Concrete block, 6 in 


. 4.20 


4.20 


5a. 


Reflective air space, 








fin- 


2.91 


— 


^ 5b. 


Nominal 1x3 






<\ 


vertical furring 


— 


0.94 


Gypsum wall board, 








0.5 in. 


0.45 


0.45 


7. 


Inside surface, 








still air 


0.68 


0.68 



Total unit thermal resistance of each section, R 8.94 6.97 

The U-factor of each section, U = 1/ff, in Btu/h ■ ft 2 • °F 0.112 0.143 

Area fraction of each section, 4rea 0.80 0.20 

Overall £/-f actor: U = 2f area ,U, = 0.80 X 0.112 + 0.20 X 0.143 

= 0.1 18 Btu/h- ft 2 -°F 
Overall unit thermal resistance: R= IIU= 8.46 h • ft 2 • °F/Btu 

Therefore, the overall unit thermal resistance of the wall is 8.46 h • ft 2 • C F/Btu 
and the overall l/-factor is 0.118 Btu/h • ft 2 • °F. These values account for the 
effects of the vertical furring. 



EXAMPLE 3-19 



The ff-Value of a Pitched Roof 



Determine the overall unit thermal resistance (the ff -value) and the overall heat 
transfer coefficient (the ^/-factor) of a 45° pitched roof built around nominal 
2-in. X 4-in. wood studs with a center-to-center distance of 16 in. The 3.5-in.- 
wide air space between the studs does not have any reflective surface and thus 
its effective emissivity is 0.84. For a mean temperature of 90°F and a temper- 
ature difference of 30°F, the ff-value of the air space is 0.86 h • ft 2 • °F/Btu. 
The lower part of the roof is finished with |-in. gypsum wallboard and the upper 
part with |-in. plywood, building paper, and asphalt shingle roofing. The air 
space constitutes 75 percent of the heat transmission area, while the studs and 
headers constitute 25 percent. 

SOLUTION The ff-value and the ^/-factor of a 45° pitched roof are to be 

determined. 

Assumptions 1 Steady operating conditions exist. 2 Heat transfer through the 

roof is one-dimensional. 3 Thermal properties of the roof and the heat transfer 

coefficients are constant. 

Properties The ff-values of different materials are given in Table 3-6. 



cen58933_ch03.qxd 9/10/2002 9:00 AM Page 183 



183 
CHAPTER 3 



Analysis The schematic 


of the pitched roof as well as the 


different elements 


used in its construction 


are 


shown below. Following the a 


oproach described 


above and using the avai 


able ff-values from Table 3-6, the 


overall R- 


/alue of 


the roof can be determined in 


the table here. 






Schematic 






fl-va 
h • ft 2 • 

Between 


ue, 
°F/Btu 

At 


. . 




Construction 


Studs 


Studs 


1. 


Outside surface, 






/ " V \ X \ X \^ 




15 mph wind 


0.17 


0.17 


/ "^Cv^O^N^/ 


2. 


Asphalt shingle 






/ /Q/^f^^w/ 




roofing 


0.44 


0.44 


/ /r^j^^^A 4 / 


3. 


Building paper 


0.10 


0.10 


TrCyA 


4. 


Plywood deck, | in. 


0.78 


0.78 


I I \ \ \ \ C ) 


5a 


Nonreflective air 






1 2 3 4 5a 5b 6 7 




space, 3.5 in. 


0.86 


— 




5b 


Wood stud, 2 in. by 4 in. 


— 


3.58 




6. 


Gypsum wallboard, 0.5 in. 


0.45 


0.45 




7. 


Inside surface, 










45° slope, still air 


0.63 


0.63 


Total unit thermal resista 


ice of each section, R 


3.43 


6.15 


The Ofactor of each sect 


ion, 


U= 1/fl, in Btu/h -ft 2 - °F 


0.292 


0.163 


Area fraction of each seel 


ion, 


'area 


0.75 


0.25 


Overall U-i actor: U= 2f area 


iUr- 


= 0.75 X 0.292 + 0.25 X 0.163 




= 0.260 Btu/h • ft 2 • °F 










Overall unit thermal resistance: R = 1/U = 


3.85 h • fl : 


• °F/Btu 


Therefore, the overall 


unit 


thermal resistance of this 


pitched 


roof is 


3.85 h • ft 2 • °F/Btu and the overall tZ-factor is 0.260 Btu/h 


• ft 2 • °F. Note that 


the wood studs offer much larger thermal resistance to heat flow than 


the air 


space between the studs. 











The construction of wood frame flat ceilings typically involve 2-in. X 
6-in. joists on 400-mm (16-in.) or 600-mm (24-in.) centers. The fraction of 
framing is usually taken to be 0.10 for joists on 400-mm centers and 0.07 
for joists on 600-mm centers. 

Most buildings have a combination of a ceiling and a roof with an attic 
space in between, and the determination of the 7?-value of the roof-attic- 
ceiling combination depends on whether the attic is vented or not. For ad- 
equately ventilated attics, the attic air temperature is practically the same as 
the outdoor air temperature, and thus heat transfer through the roof is gov- 
erned by the /lvalue of the ceiling only. However, heat is also transferred 
between the roof and the ceiling by radiation, and it needs to be considered 
(Fig. 3-52). The major function of the roof in this case is to serve as a ra- 
diation shield by blocking off solar radiation. Effectively ventilating the at- 
tic in summer should not lead one to believe that heat gain to the building 
through the attic is greatly reduced. This is because most of the heat trans- 
fer through the attic is by radiation. 



Aii- 
exhaust 

.^ppRadiant 
s^' barrier 

K/\/\/w\/w\/\/\/w\/\/\/y 



K 




n 



Ail- 
intake 



Air 
intake 

FIGURE 3-52 

Ventilation paths for a naturally 

ventilated attic and the appropriate 

size of the flow areas around the 

radiant barrier for proper air 

circulation (from DOE/CE-0335P, 

U.S. Dept. of Energy). 



cen58933_ch03.qxd 9/10/2002 9:00 AM Page 184 



184 
HEAT TRANSFER 



- Roof decking 



Air space 



Roof decking 



- Roof decking 




Joist - 



Insulation 



Joist — Insulation - 

(b) At the bottom of rafters 



(a) Under the roof deck 

FIGURE 3-53 

Three possible locations for an attic radiant barrier (from DOE/CE-0335P, U.S. Dept. of Energy). 



Joist — Insulation - 

(c) On top of attic floor insulation 



Shingles 



Deck 



^Rafter/^ %\ 
y^ Attic < " 


A 

r ceiling 
attic \ 


Ceiling joist > 


' ^ceiling 








?i 





FIGURE 3-54 

Thermal resistance network for a 
pitched roof-attic-ceiling combination 
for the case of an unvented attic. 



Radiation heat transfer between the ceiling and the roof can be mini- 
mized by covering at least one side of the attic (the roof or the ceiling side) 
by a reflective material, called radiant barrier, such as aluminum foil or 
aluminum-coated paper. Tests on houses with R- 19 attic floor insulation 
have shown that radiant barriers can reduce summer ceiling heat gains by 
16 to 42 percent compared to an attic with the same insulation level and no 
radiant barrier. Considering that the ceiling heat gain represents about 15 to 
25 percent of the total cooling load of a house, radiant barriers will reduce 
the air conditioning costs by 2 to 10 percent. Radiant barriers also reduce 
the heat loss in winter through the ceiling, but tests have shown that the 
percentage reduction in heat losses is less. As a result, the percentage 
reduction in heating costs will be less than the reduction in the air- 
conditioning costs. Also, the values given are for new and undusted radiant 
barrier installations, and percentages will be lower for aged or dusty radi- 
ant barriers. 

Some possible locations for attic radiant barriers are given in Figure 
3-53. In whole house tests on houses with R-\9 attic floor insulation, radi- 
ant barriers have reduced the ceiling heat gain by an average of 35 percent 
when the radiant barrier is installed on the attic floor, and by 24 percent 
when it is attached to the bottom of roof rafters. Test cell tests also demon- 
strated that the best location for radiant barriers is the attic floor, provided 
that the attic is not used as a storage area and is kept clean. 

For unvented attics, any heat transfer must occur through (1) the ceiling, 
(2) the attic space, and (3) the roof (Fig. 3-54). Therefore, the overall 
K-value of the roof-ceiling combination with an unvented attic depends on 
the combined effects of the 7?-value of the ceiling and the K-value of the 
roof as well as the thermal resistance of the attic space. The attic space can 
be treated as an air layer in the analysis. But a more practical way of ac- 
counting for its effect is to consider surface resistances on the roof and ceil- 
ing surfaces facing each other. In this case, the ^-values of the ceiling and 
the roof are first determined separately (by using convection resistances for 
the still-air case for the attic surfaces). Then it can be shown that the over- 
all i?-value of the ceiling-roof combination per unit area of the ceiling can 
be expressed as 



cen58933_ch03.qxd 9/10/2002 9:00 AM Page 185 



/? = /? H- /? 

"■ "ceiling "roof 



^ceiling 
^roof 



(3-82) 



185 
CHAPTER 3 



where i4 cei | in g and A mof are the ceiling and roof areas, respectively. The area 
ratio is equal to 1 for flat roofs and is less than 1 for pitched roofs. For a 45° 
pitched roof, the area ratio is ^ ce iiin g /^roof = l/\/2 = 0.707. Note that the 
pitched roof has a greater area for heat transfer than the flat ceiling, and the 
area ratio accounts for the reduction in the unit lvalue of the roof when 
expressed per unit area of the ceiling. Also, the direction of heat flow is up 
in winter (heat loss through the roof) and down in summer (heat gain 
through the roof). 

The 7?-value of a structure determined by analysis assumes that the ma- 
terials used and the quality of workmanship meet the standards. Poor work- 
manship and substandard materials used during construction may result in 
^-values that deviate from predicted values. Therefore, some engineers use 
a safety factor in their designs based on experience in critical applications. 



SUMMARY 



One-dimensional heat transfer through a simple or composite 
body exposed to convection from both sides to mediums at 
temperatures T^ t and T rjJ1 can be expressed as 



Q 



(W) 



mediums. For a plane wall exposed to convection on both 
sides, the total resistance is expressed as 



R„ 



R,, 



-"wiiii *~ Rr, 



1 



1 



This relation can be extended to plane walls that consist of two 
or more layers by adding an additional resistance for each ad- 
ditional layer. The elementary thermal resistance relations can 
be expressed as follows: 



where h c is the thermal contact conductance, R c is the thermal 
contact resistance, and the radiation heat transfer coefficient is 
defined as 



/>,, 



s(j(T; + Tl Tr )(T s + T sulr ) 



Once the rate of heat transfer is available, the temperature drop 
across any layer can be determined from 

AT= QR 

The thermal resistance concept can also be used to solve steady 
heat transfer problems involving parallel layers or combined 
series-parallel arrangements. 

Adding insulation to a cylindrical pipe or a spherical shell 
will increase the rate of heat transfer if the outer radius of 
the insulation is less than the critical radius of insulation, 
defined as 



Interface resistance: 
Radiation resistance: 



v cyl 



Conduction resistance (plane wall): R wa[[ 
Conduction resistance (cylinder): R 
Conduction resistance (sphere): 
Convection resistance: 



kA 

Info//-)) 
2nLk 



R 



sph 



R 



R 

■"rad 



4ir r x r 2 k 

J_ 

: onv hA 

j_ = Rc 

nterface ^ £ 

1 



"rad "■ 



cr, cylinder 


h 




2k, 


* cr, sphere 


h 



The effectiveness of an insulation is often given in terms of 
its R-value, the thermal resistance of the material per unit sur- 
face area, expressed as 



lvalue 



L 



(flat insulation) 



where L is the thickness and k is the thermal conductivity of the 
material. 



cen58933_ch03.qxd 9/10/2002 9:00 AM Page 186 



186 
HEAT TRANSFER 



Finned surfaces are commonly used in practice to enhance 
heat transfer. Fins enhance heat transfer from a surface by ex- 
posing a larger surface area to convection. The temperature 
distribution along the fin for very long fins and for fins with 
negligible heat transfer at the fin are given by 



Very long fin: 
Adiabatic fin tip: 



T(x) - T a 



T(x) 



- T a 



p—x\ hplkA 

cosh a(L — x) 
cosh ah 



where a = \/hplkA c , p is the perimeter, and A c is the cross 
sectional area of the fin. The rates of heat transfer for both 
cases are given to be 



Very 

long ei ongfin = ~kA c 
fin: 

Adiabatic 

Jm ^c insulated tip — ~~ ^c 

tip: 



VhpkAc(T b -TJ 



\/kpkA c (T b - r„) tanh ah 



Fins exposed to convection at their tips can be treated as fins 
with insulated tips by using the corrected length L c = L + AJp 
instead of the actual fin length. 

The temperature of a fin drops along the fin, and thus the 
heat transfer from the fin will be less because of the decreasing 
temperature difference toward the fin tip. To account for the ef- 
fect of this decrease in temperature on heat transfer, we define 
fin efficiency as 



%i„ 



Q< 



Q 



fin, max 



Actual heat transfer rate from the fin 
Ideal heat transfer rate from the fin if 
the entire fin were at base temperature 



The performance of the fins is judged on the basis of the en- 
hancement in heat transfer relative to the no-fin case and is ex- 
pressed in terms of the fin effectiveness s fln , defined as 



e f ,» 



g f ,„ 

e» of ,„ hA t {T t -TJ 



Heat transfer rate from 
the fin of base area A b 

Heat transfer rate from 
the surface of area A h 



Here, A h is the cross-sectional area of the fin at the base and 
Q no fin represents the rate of heat transfer from this area if no 
fins are attached to the surface. The overall effectiveness for a 
finned surface is defined as the ratio of the total heat transfer 
from the finned surface to the heat transfer from the same sur- 
face if there were no fins, 

Q total, fin h(A unfm + TlflnAaJfTj, - TJ) 



Q 



total, no fin 



"A no fin (T b T„) 



Fin efficiency and fin effectiveness are related to each other by 

Afl„ 



A, 



%,„ 



Certain multidimensional heat transfer problems involve two 
surfaces maintained at constant temperatures T { and T 2 . The 
steady rate of heat transfer between these two surfaces is ex- 
pressed as 

Q = Sk(T { - T 2 ) 

where S is the conduction shape factor that has the dimen- 
sion of length and k is the thermal conductivity of the medium 
between the surfaces. 



When the fin efficiency is available, the rate of heat transfer 
from a fin can be determined from 



Gfi„ = infi„2f 



■nfin^ fin (T b - T„) 



REFERENCES AND SUGGESTED READING 



1. American Society of Heating, Refrigeration, and Air 
Conditioning Engineers. Handbook of Fundamentals. 
Atlanta: ASHRAE, 1993. 

2. R. V. Andrews. "Solving Conductive Heat Transfer 
Problems with Electrical-Analogue Shape Factors." 
Chemical Engineering Progress 5 (1955), p. 67. 

3. R. Barron. Cryogenic Systems. New York: McGraw-Hill, 
1967. 

4. L. S. Fletcher. "Recent Developments in Contact 
Conductance Heat Transfer." Journal of Heat Transfer 
110, no. 4B (1988), pp. 1059-79. 



5. E. Fried. "Thermal Conduction Contribution to Heat 
Transfer at Contacts." Thermal Conductivity, vol. 2, ed. 
R. R Tye. London: Academic Press, 1969. 

6. K. A. Gardner. "Efficiency of Extended Surfaces." Trans. 
ASME 67 (1945), pp. 621-31. 

7. F. P. Incropera and D. P. DeWitt. Introduction to Heat 
Transfer. 4th ed. New York: John Wiley & Sons, 2002. 

8. D. Q. Kern and A. D. Kraus. Extended Surface Heat 
Transfer. New York: McGraw-Hill, 1972. 



cen58933_ch03.qxd 9/10/2002 9:00 AM Page 187 



9. M. N. Ozisik. Heat Transfer — A Basic Approach. New 
York: McGraw-Hill, 1985. 

10. G. P. Peterson. "Thermal Contact Resistance in Waste 
Heat Recovery Systems." Proceedings of the 18th 
ASME/ETCE Hydrocarbon Processing Symposium. 
Dallas, TX, 1987, pp. 45-51. 

11. S. Song, M. M. Yovanovich, and F. O. Goodman. 
"Thermal Gap Conductance of Conforming Surfaces in 
Contact." Journal of Heat Transfer 115 (1993), p. 533. 



PROBLEMS 



187 
CHAPTER 3 



12. J. E. Sunderland and K. R. Johnson. "Shape Factors for 
Heat Conduction through Bodies with Isothermal or 
Convective Boundary Conditions," Trans. ASME 10 
(1964), pp. 237-41. 

13. N. V. Suryanarayana. Engineering Heat Transfer. St. Paul, 
MN: West Publishing, 1995. 



Steady Heat Conduction in Plane Walls 

3-1 C Consider one-dimensional heat conduction through 
a cylindrical rod of diameter D and length L. What is the 
heat transfer area of the rod if (a) the lateral surfaces of the rod 
are insulated and (b) the top and bottom surfaces of the rod are 
insulated? 

3-2C Consider heat conduction through a plane wall. Does 
the energy content of the wall change during steady heat con- 
duction? How about during transient conduction? Explain. 

3-3C Consider heat conduction through a wall of thickness L 
and area A. Under what conditions will the temperature distri- 
butions in the wall be a straight line? 

3-4C What does the thermal resistance of a medium 
represent? 

3-5C How is the combined heat transfer coefficient defined? 
What convenience does it offer in heat transfer calculations? 

3-6C Can we define the convection resistance per unit 
surface area as the inverse of the convection heat transfer 
coefficient? 

3-7C Why are the convection and the radiation resistances at 
a surface in parallel instead of being in series? 

3-8C Consider a surface of area A at which the convection 
and radiation heat transfer coefficients are h com and /z rad , re- 
spectively. Explain how you would determine (a) the single 
equivalent heat transfer coefficient, and (b) the equivalent ther- 
mal resistance. Assume the medium and the surrounding sur- 
faces are at the same temperature. 

3-9C How does the thermal resistance network associated 
with a single-layer plane wall differ from the one associated 
with a five-layer composite wall? 

*Problems designated by a "C" are concept questions, and 
students are encouraged to answer them all. Problems designated 
by an "E" are in English units, and the SI users can ignore them. 
Problems with an EES-CD icon ® are solved using EES, and 
complete solutions together with parametric studies are included 
on the enclosed CD. Problems with a computer-EES icon H are 
comprehensive in nature, and are intended to be solved with a 
computer, preferably using the EES software that accompanies 
this text. 



3-10C Consider steady one-dimensional heat transfer 
through a multilayer medium. If the rate of heat transfer Q is 
known, explain how you would determine the temperature 
drop across each layer. 

3-11C Consider steady one-dimensional heat transfer 
through a plane wall exposed to convection from both sides to 
environments at known temperatures T rA and T^ 2 with known 
heat transfer coefficients h { and h 2 . Once the rate of heat trans- 
fer Q has been evaluated, explain how you would determine 
the temperature of each surface. 

3-12C Someone comments that a microwave oven can be 
viewed as a conventional oven with zero convection resistance 
at the surface of the food. Is this an accurate statement? 

3-13C Consider a window glass consisting of two 4-mm- 
thick glass sheets pressed tightly against each other. Compare 
the heat transfer rate through this window with that of one con- 
sisting of a single 8-mm-thick glass sheet under identical con- 
ditions. 

3-14C Consider steady heat transfer through the wall of a 
room in winter. The convection heat transfer coefficient at the 
outer surface of the wall is three times that of the inner surface 
as a result of the winds. On which surface of the wall do you 
think the temperature will be closer to the surrounding air tem- 
perature? Explain. 

3-1 5C The bottom of a pan is made of a 4-mm-thick alu- 
minum layer. In order to increase the rate of heat transfer 
through the bottom of the pan, someone proposes a design for 
the bottom that consists of a 3-mm-thick copper layer sand- 
wiched between two 2-mm-thick aluminum layers. Will the 
new design conduct heat better? Explain. Assume perfect con- 
tact between the layers. 

3-16C Consider two cold canned drinks, one wrapped in a 
blanket and the other placed on a table in the same room. 
Which drink will warm up faster? 

3-17 Consider a 4-m-high, 6-m-wide, and 0.3-m-thick brick 
wall whose thermal conductivity is k = 0.8 W/m ■ °C . On a 
certain day, the temperatures of the inner and the outer surfaces 
of the wall are measured to be 14°C and 6°C, respectively. De- 
termine the rate of heat loss through the wall on that day. 



cen58933_ch03.qxd 9/10/2002 9:00 AM Page IE 



188 
HEAT TRANSFER 




Aluminum 

FIGURE P3-15C 



Copper 



3-18 Consider a 1.2-m-high and 2-m-wide glass window 
whose thickness is 6 mm and thermal conductivity is k = 0.78 
W/m ■ °C. Determine the steady rate of heat transfer through 
this glass window and the temperature of its inner surface for a 
day during which the room is maintained at 24°C while the 
temperature of the outdoors is — 5°C. Take the convection heat 
transfer coefficients on the inner and outer surfaces of the win- 
dow to be hi = 10 W/m 2 • °C and h 2 = 25 W/m 2 • °C, and dis- 
regard any heat transfer by radiation. 

3-19 Consider a 1 .2-m-high and 2-m-wide double-pane win- 
dow consisting of two 3-mm-thick layers of glass (k = 0.78 
W/m • °C) separated by a 12-mm-wide stagnant air space (k = 
0.026 W/m • °C). Determine the steady rate of heat transfer 
through this double-pane window and the temperature of its 
inner surface for a day during which the room is maintained 
at 24°C while the temperature of the outdoors is — 5°C. Take 
the convection heat transfer coefficients on the inner and outer 



Glass 




surfaces of the window to be h { = 10 W/m 2 • °C and h 2 = 
25 W/m 2 • °C, and disregard any heat transfer by radiation. 

Answers: 114 W, 19.2°C 

3-20 Repeat Problem 3-19, assuming the space between the 
two glass layers is evacuated. 

3-21 Ta'M Reconsider Problem 3-19. Using EES (or other) 
1^2 software, plot the rate of heat transfer through the 
window as a function of the width of air space in the range of 
2 mm to 20 mm, assuming pure conduction through the air. 
Discuss the results. 

3-22E Consider an electrically heated brick house (k = 0.40 
Btu/h • ft • °F) whose walls are 9 ft high and 1 ft thick. Two of 
the walls of the house are 40 ft long and the others are 30 ft 
long. The house is maintained at 70°F at all times while the 
temperature of the outdoors varies. On a certain day, the tem- 
perature of the inner surface of the walls is measured to be at 
55°F while the average temperature of the outer surface is ob- 
served to remain at 45°F during the day for 10 h and at 35°F at 
night for 14 h. Determine the amount of heat lost from the 
house that day. Also determine the cost of that heat loss to the 
homeowner for an electricity price of $0.09/kWh. 




FIGURE P3-1 9 



FIGURE P3-22E 

3-23 A cylindrical resistor element on a circuit board dissi- 
pates 0. 1 5 W of power in an environment at 40°C. The resistor 
is 1 .2 cm long, and has a diameter of 0.3 cm. Assuming heat to 
be transferred uniformly from all surfaces, determine (a) the 
amount of heat this resistor dissipates during a 24-h period, 

(b) the heat flux on the surface of the resistor, in W/m 2 , and 

(c) the surface temperature of the resistor for a combined con- 
vection and radiation heat transfer coefficient of 9 W/m 2 • °C. 

3-24 Consider a power transistor that dissipates 0.2 W of 
power in an environment at 30°C. The transistor is 0.4 cm long 
and has a diameter of 0.5 cm. Assuming heat to be transferred 
uniformly from all surfaces, determine (a) the amount of heat 
this resistor dissipates during a 24-h period, in kWh; (b) the 
heat flux on the surface of the transistor, in W/m 2 ; and (c) the 
surface temperature of the resistor for a combined convection 
and radiation heat transfer coefficient of 18 W/m 2 ■ °C. 

3-25 A 12-cm X 18-cm circuit board houses on its surface 
100 closely spaced logic chips, each dissipating 0.07 W in an 
environment at 40°C. The heat transfer from the back surface 
of the board is negligible. If the heat transfer coefficient on the 



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189 
CHAPTER 3 



30°C 



Power 

transistor 

0.2 W 



0.5 cm 



— 0.4 cm 

IGURE P3-24 



surface of the board is 10 W/m 2 • °C, determine (a) the heat 
flux on the surface of the circuit board, in W/m 2 ; (b) the surface 
temperature of the chips; and (c) the thermal resistance be- 
tween the surface of the circuit board and the cooling medium, 
in °C/W. 

3-26 Consider a person standing in a room at 20°C with an 
exposed surface area of 1.7 m 2 . The deep body temperature of 
the human body is 37°C, and the thermal conductivity of the 
human tissue near the skin is about 0.3 W/m • °C. The body is 
losing heat at a rate of 150 W by natural convection and radia- 
tion to the surroundings. Taking the body temperature 0.5 cm 
beneath the skin to be 37°C, determine the skin temperature of 
the person. Answer: 35.5° C 

3-27 Water is boiling in a 25-cm-diameter aluminum pan {k = 
237 W/m ■ °C) at 95°C. Heat is transferred steadily to the boil- 
ing water in the pan through its 0.5-cm-thick flat bottom at a 
rate of 800 W. If the inner surface temperature of the bottom of 
the pan is 108°C, determine (a) the boiling heat transfer coeffi- 
cient on the inner surface of the pan, and (b) the outer surface 
temperature of the bottom of the pan. 

3-28E A wall is constructed of two layers of 0.5-in-thick 
sheetrock (k = 0.10 Btu/h • ft • °F), which is a plasterboard 
made of two layers of heavy paper separated by a layer of 
gypsum, placed 5 in. apart. The space between the sheetrocks 



Fiberglass 
insulation 



Sheetrock 



0.5 in. 



5 in. 



0.5 in. 



is filled with fiberglass insulation (k = 0.020 Btu/h • ft • °F). 
Determine (a) the thermal resistance of the wall, and (b) its 
R- value of insulation in English units. 

3-29 The roof of a house consists of a 3-cm-thick concrete 
slab (k = 2 W/m ■ °C) that is 15 m wide and 20 m long. The 
convection heat transfer coefficients on the inner and outer sur- 
faces of the roof are 5 and 12 W/m 2 • °C, respectively. On a 
clear winter night, the ambient air is reported to be at 10°C, 
while the night sky temperature is 100 K. The house and the in- 
terior surfaces of the wall are maintained at a constant temper- 
ature of 20 C C. The emissivity of both surfaces of the concrete 
roof is 0.9. Considering both radiation and convection heat 
transfers, determine the rate of heat transfer through the roof, 
and the inner surface temperature of the roof. 

If the house is heated by a furnace burning natural gas with 
an efficiency of 80 percent, and the price of natural gas is 
$0.60/therm (1 therm = 105,500 kJ of energy content), deter- 
mine the money lost through the roof that night during a 14-h 
period. 



r sky = iooK 



T. = 10°C 



15 cm 




FIGURE P3-28E 



FIGURE P3-29 

3-30 A 2-m X 1 .5-m section of wall of an industrial furnace 
burning natural gas is not insulated, and the temperature at the 
outer surface of this section is measured to be 80°C. The tem- 
perature of the furnace room is 30°C, and the combined con- 
vection and radiation heat transfer coefficient at the surface of 
the outer furnace is 10 W/m 2 • °C. It is proposed to insulate this 
section of the furnace wall with glass wool insulation (k = 
0.038 W/m • °C) in order to reduce the heat loss by 90 percent. 
Assuming the outer surface temperature of the metal section 
still remains at about 80°C, determine the thickness of the in- 
sulation that needs to be used. 

The furnace operates continuously and has an efficiency of 
78 percent. The price of the natural gas is $0.55/therm (1 therm 
= 105,500 kJ of energy content). If the installation of the insu- 
lation will cost $250 for materials and labor, determine how 
long it will take for the insulation to pay for itself from the en- 
ergy it saves. 

3-31 Repeat Problem. 3-30 for expanded perlite insulation 
assuming conductivity is k = 0.052 W/m • °C. 



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190 
HEAT TRANSFER 



3-32 [?(.*)! Reconsider Problem 3-30. Using EES (or other) 
|^£^ software, investigate the effect of thermal con- 
ductivity on the required insulation thickness. Plot the thick- 
ness of insulation as a function of the thermal conductivity of 
the insulation in the range of 0.02 W/m ■ °C to 0.08 W/m • °C, 
and discuss the results. 

3-33E Consider a house whose walls are 12 ft high and 40 ft 
long. Two of the walls of the house have no windows, while 
each of the other two walls has four windows made of 0.25-in.- 
thick glass (k = 0.45 Btu/h ■ ft ■ °F), 3 ft X 5 ft in size. The 
walls are certified to have an /{-value of 19 (i.e., an Llk value of 
19 h • ft 2 • °F/Btu). Disregarding any direct radiation gain or 
loss through the windows and taking the heat transfer coef- 
ficients at the inner and outer surfaces of the house to be 2 and 
4 Btu/h • ft 2 • °F, respectively, determine the ratio of the heat 
transfer through the walls with and without windows. 



Attic 
space 



12 ft 



Sheet metal 




40 ft 

FIGURE P3-33E 



3-34 Consider a house that has a 10-m X 20-m base and a 
4-m-high wall. All four walls of the house have an /{-value of 
2.31 m 2 • °C/W. The two 10-m X 4-m walls have no windows. 
The third wall has five windows made of 0.5-cm-thick glass 
(k = 0.78 W/m • °C), 1.2 m X 1.8 m in size. The fourth wall 
has the same size and number of windows, but they are double- 
paned with a 1.5-cm-thick stagnant air space (k = 0.026 
W/m • °C) enclosed between two 0.5-cm-thick glass layers. 
The thermostat in the house is set at 22°C and the average tem- 
perature outside at that location is 8°C during the seven-month- 
long heating season. Disregarding any direct radiation gain or 
loss through the windows and taking the heat transfer coeffi- 
cients at the inner and outer surfaces of the house to be 7 and 
15 W/m 2 ■ °C, respectively, determine the average rate of heat 
transfer through each wall. 

If the house is electrically heated and the price of electricity 
is $0.08/kWh, determine the amount of money this household 
will save per heating season by converting the single-pane win- 
dows to double -pane windows. 

3-35 The wall of a refrigerator is constructed of fiberglass in- 
sulation (k = 0.035 W/m • °C) sandwiched between two layers 
of 1-mm-thick sheet metal (k = 15.1 W/m • °C). The refriger- 
ated space is maintained at 3°C, and the average heat transfer 
coefficients at the inner and outer surfaces of the wall are 



Kitchen 
air 

25°C 



10°C 



1 mm] 



Insulation 



Refrigerated 
space 
3°C 



1 mm 



FIGURE P3-35 

4 W/m 2 • °C and 9 W/m 2 • °C, respectively. The kitchen tem- 
perature averages 25°C. It is observed that condensation occurs 
on the outer surfaces of the refrigerator when the temperature 
of the outer surface drops to 20°C. Determine the minimum 
thickness of fiberglass insulation that needs to be used in the 
wall in order to avoid condensation on the outer surfaces. 

3-36 rSi'M Reconsider Problem 3-35. Using EES (or other) 
1^2 software, investigate the effects of the thermal 
conductivities of the insulation material and the sheet metal on 
the thickness of the insulation. Let the thermal conductivity 
vary from 0.02 W/m • °C to 0.08 W/m • °C for insulation and 
10 W/m • °C to 400 W/m • °C for sheet metal. Plot the thick- 
ness of the insulation as the functions of the thermal con- 
ductivities of the insulation and the sheet metal, and discuss 
the results. 

3-37 Heat is to be conducted along a circuit board that has a 
copper layer on one side. The circuit board is 15 cm long and 
15 cm wide, and the thicknesses of the copper and epoxy lay- 
ers are 0.1 mm and 1.2 mm, respectively. Disregarding heat 
transfer from side surfaces, determine the percentages of heat 
conduction along the copper (k = 386 W/m • °C) and epoxy 
(k = 0.26 W/m ■ °C) layers. Also determine the effective ther- 
mal conductivity of the board. 

Answers: 0.8 percent, 99.2 percent, and 29.9 W/m • °C 

3-38E A 0.03-in-thick copper plate (k = 223 Btu/h • ft ■ °F) 
is sandwiched between two 0.1 -in. -thick epoxy boards (k = 
0.15 Btu/h • ft • °F) that are 7 in. X 9 in. in size. Determine the 
effective thermal conductivity of the board along its 9-in.-long 
side. What fraction of the heat conducted along that side is con- 
ducted through copper? 

Thermal Contact Resistance 

3-39C What is thermal contact resistance? How is it related 
to thermal contact conductance? 

3-40C Will the thermal contact resistance be greater for 
smooth or rough plain surfaces? 



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191 
CHAPTER 3 



Epoxy 
boards 



Copper 
plate 




Plexiglas 



9 in. 



0.03 in. 

IGURI 38E 



3-41C A wall consists of two layers of insulation pressed 
against each other. Do we need to be concerned about the ther- 
mal contact resistance at the interface in a heat transfer analy- 
sis or can we just ignore it? 

3-42C A plate consists of two thin metal layers pressed 
against each other. Do we need to be concerned about the 
thermal contact resistance at the interface in a heat transfer 
analysis or can we just ignore it? 

3-43C Consider two surfaces pressed against each other. 
Now the air at the interface is evacuated. Will the thermal con- 
tact resistance at the interface increase or decrease as a result? 

3-44C Explain how the thermal contact resistance can be 
minimized. 

3—45 The thermal contact conductance at the interface of two 
1-cm-thick copper plates is measured to be 18,000 W/m 2 • °C. 
Determine the thickness of the copper plate whose thermal 
resistance is equal to the thermal resistance of the interface 
between the plates. 

3-46 Six identical power transistors with aluminum casing 
are attached on one side of a 1 .2-cm-thick 20-cm X 30-cm 
copper plate (k = 386 W/m • °C) by screws that exert an aver- 
age pressure of 10 MPa. The base area of each transistor is 
9 cm 2 , and each transistor is placed at the center of a 10-cm X 
10-cm section of the plate. The interface roughness is esti- 
mated to be about 1 .4 jjim. All transistors are covered by a thick 
Plexiglas layer, which is a poor conductor of heat, and thus all 
the heat generated at the junction of the transistor must be dis- 
sipated to the ambient at 15°C through the back surface of the 
copper plate. The combined convection/radiation heat transfer 
coefficient at the back surface can be taken to be 30 W/m 2 ■ °C. 
If the case temperature of the transistor is not to exceed 85°C, 
determine the maximum power each transistor can dissipate 
safely, and the temperature jump at the case-plate interface. 



Copper 
Transistor plate 



15°C 




1.2 cm 



FIGURE P3^6 



3-47 Two 5-cm-diameter, 15-cm-long aluminum bars (k = 
176 W/m ■ °C) with ground surfaces are pressed against each 
other with a pressure of 20 atm. The bars are enclosed in an in- 
sulation sleeve and, thus, heat transfer from the lateral surfaces 
is negligible. If the top and bottom surfaces of the two-bar sys- 
tem are maintained at temperatures of 150°C and 20°C, re- 
spectively, determine (a) the rate of heat transfer along the 
cylinders under steady conditions and (b) the temperature drop 
at the interface. Answers: (a) 142.4 W, (b) 6.4°C 

3-48 A 1 -mm-thick copper plate (k = 386 W/m ■ °C) is sand- 
wiched between two 5-mm-thick epoxy boards (k = 0.26 
W/m • °C) that are 15 cm X 20 cm in size. If the thermal con- 
tact conductance on both sides of the copper plate is estimated 
to be 6000 W/m • °C, determine the error involved in the total 
thermal resistance of the plate if the thermal contact conduc- 
tances are ignored. 



Copper 



Epoxy 




FIGURE P3-48 



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192 
HEAT TRANSFER 



Generalized Thermal Resistance Networks 

3-49C When plotting the thermal resistance network associ- 
ated with a heat transfer problem, explain when two resistances 
are in series and when they are in parallel. 

3-50C The thermal resistance networks can also be used 
approximately for multidimensional problems. For what kind 
of multidimensional problems will the thermal resistance 
approach give adequate results? 

3-51 C What are the two approaches used in the develop- 
ment of the thermal resistance network for two-dimensional 
problems? 

3-52 A 4-m-high and 6-m-wide wall consists of a long 
18-cm X 30-cm cross section of horizontal bricks (k = 0.72 
W/m ■ °C) separated by 3-cm-thick plaster layers (k = 0.22 
W/m • °C). There are also 2-cm-thick plaster layers on each 
side of the wall, and a 2-cm-thick rigid foam (k = 
0.026 W/m • °C) on the inner side of the wall. The indoor and 
the outdoor temperatures are 22°C and — 4°C, and the convec- 
tion heat transfer coefficients on the inner and the outer sides 
are /z, = 10 W/m 2 • °C and h 2 = 20 W/m 2 • °C, respectively. 
Assuming one-dimensional heat transfer and disregarding radi- 
ation, determine the rate of heat transfer through the wall. 



Foam 




Brick 



2 2 

FIGURE P3-52 



18 cm 



1.5 cm 



30 cm 



1.5 cm 



3-53 



Reconsider Problem 3-52. Using EES (or other) 
software, plot the rate of heat transfer through 

the wall as a function of the thickness of the rigid foam in the 

range of 1 cm to 10 cm. Discuss the results. 

3-54 A 10-cm-thick wall is to be constructed with 2.5-m- 
long wood studs (k = 0.11 W/m • °C) that have a cross section 
of 10 cm X 10 cm. At some point the builder ran out of those 
studs and started using pairs of 2.5-m-long wood studs that 
have a cross section of 5 cm X 10 cm nailed to each other 



instead. The manganese steel nails (k = 50 W/m • °C) are 
10 cm long and have a diameter of 0.4 cm. A total of 50 nails 
are used to connect the two studs, which are mounted to the 
wall such that the nails cross the wall. The temperature differ- 
ence between the inner and outer surfaces of the wall is 8°C. 
Assuming the thermal contact resistance between the two 
layers to be negligible, determine the rate of heat transfer 

(a) through a solid stud and (b) through a stud pair of equal 
length and width nailed to each other, (c) Also determine the 
effective conductivity of the nailed stud pair. 

3-55 A 12-m-long and 5-m-high wall is constructed of two 
layers of 1 -cm-thick sheetrock (k = 0.17 W/m • °C) spaced 
12 cm by wood studs (k = 0.11 W/m • °C) whose cross section 
is 12 cm X 5 cm. The studs are placed vertically 60 cm apart, 
and the space between them is filled with fiberglass insulation 
(k = 0.034 W/m • °C). The house is maintained at 20°C and the 
ambient temperature outside is — 5°C. Taking the heat transfer 
coefficients at the inner and outer surfaces of the house to be 
8.3 and 34 W/m 2 • °C, respectively, determine (a) the thermal 
resistance of the wall considering a representative section of it 
and (b) the rate of heat transfer through the wall. 

3-56E A 10-in. -thick, 30-ft-long, and 10-ft-high wall is 
to be constructed using 9-in.-long solid bricks (k = 0.40 
Btu/h ■ ft • °F) of cross section 7 in. X 7 in., or identical size 
bricks with nine square air holes (k = 0.015 Btu/h • ft ■ °F) that 
are 9 in. long and have a cross section of 1.5 in. X 1.5 in. There 
is a 0.5-in. -thick plaster layer (k = 0.10 Btu/h ■ ft ■ °F) between 
two adjacent bricks on all four sides and on both sides of the 
wall. The house is maintained at 80°F and the ambient tem- 
perature outside is 30°F. Taking the heat transfer coefficients 
at the inner and outer surfaces of the wall to be 1.5 and 
4 Btu/h • ft 2 • °F, respectively, determine the rate of heat 
transfer through the wall constructed of (a) solid bricks and 

(b) bricks with air holes. 



Air channels 
1.5 in. X 1.5 in. x9 in. 




Brick 



0.5 in. 



FIGURE P3-56E 



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3-57 Consider a 5-m-high, 8-m-long, and 0.22-m-thick wall 
whose representative cross section is as given in the figure. The 
thermal conductivities of various materials used, in W/m • °C, 



are k. 



2, k B = 8, k c = 20, k D = 15, and k E = 35. The 



left and right surfaces of the wall are maintained at uniform 
temperatures of 300°C and 100°C, respectively. Assuming heat 
transfer through the wall to be one-dimensional, determine 
(a) the rate of heat transfer through the wall; (b) the tem- 
perature at the point where the sections B, D, and E meet; and 
(c) the temperature drop across the section F. Disregard any 
contact resistances at the interfaces. 



100°C 



300°C 




1 cm 



FIGURE P3-57 



3-58 Repeat Problem 3-57 assuming that the thermal contact 
resistance at the interfaces D-F and E-F is 0.00012 m 2 • °C/W. 

3-59 Clothing made of several thin layers of fabric with 
trapped air in between, often called ski clothing, is commonly 
used in cold climates because it is light, fashionable, and a very 
effective thermal insulator. So it is no surprise that such cloth- 
ing has largely replaced thick and heavy old-fashioned coats. 
Consider a jacket made of five layers of 0.1-mm-thick syn- 
thetic fabric {k = 0.13 W/m • °C) with 1.5-mm-thick air space 
(k = 0.026 W/m • °C) between the layers. Assuming the inner 
surface temperature of the jacket to be 28°C and the surface 
area to be 1.1 m 2 , determine the rate of heat loss through the 
jacket when the temperature of the outdoors is — 5°C and the 
heat transfer coefficient at the outer surface is 25 W/m 2 • °C. 



Multilayered 
ski jacket 




193 
CHAPTER 3 



What would your response be if the jacket is made of a sin- 
gle layer of 0.5-mm-thick synthetic fabric? What should be the 
thickness of a wool fabric (k = 0.035 W/m • °C) if the person 
is to achieve the same level of thermal comfort wearing a thick 
wool coat instead of a five-layer ski jacket? 

3-60 Repeat Problem 3-59 assuming the layers of the jacket 
are made of cotton fabric (k = 0.06 W/m • °C). 

3-61 A 5-m-wide, 4-m-high, and 40-m-long kiln used to cure 
concrete pipes is made of 20-cm-thick concrete walls and ceil- 
ing (k = 0.9 W/m • °C). The kiln is maintained at 40°C by in- 
jecting hot steam into it. The two ends of the kiln, 4 m X 5 m 
in size, are made of a 3-mm-thick sheet metal covered with 
2-cm-thick Styrofoam (k = 0.033 W/m • °C). The convection 
heat transfer coefficients on the inner and the outer surfaces of 
the kiln are 3000 W/m 2 • °C and 25 W/m 2 • °C, respectively. 
Disregarding any heat loss through the floor, determine the rate 
of heat loss from the kiln when the ambient air is at — 4°C. 



-4°C 




FIGURE P3-59 



4 m 



5 in 

FIGURE P3-61 



3-62 TtTM Reconsider Problem 3-61. Using EES (or other) 
1^2 software, investigate the effects of the thickness 
of the wall and the convection heat transfer coefficient on the 
outer surface of the rate of heat loss from the kiln. Let the 
thickness vary from 10 cm to 30 cm and the convection heat 
transfer coefficient from 5 W/m 2 • °C to 50 W/m 2 ■ °C. Plot the 
rate of heat transfer as functions of wall thickness and the con- 
vection heat transfer coefficient, and discuss the results. 

3-63E Consider a 6-in. X 8-in. epoxy glass laminate (k = 
0.10 Btu/h ■ ft • °F) whose thickness is 0.05 in. In order to re- 
duce the thermal resistance across its thickness, cylindrical 
copper fillings (k = 223 Btu/h ■ ft • °F) of 0.02 in. diameter are 
to be planted throughout the board, with a center-to-center 
distance of 0.06 in. Determine the new value of the thermal 
resistance of the epoxy board for heat conduction across its 
thickness as a result of this modification. 
Answer: 0.00064 h ■ °F/Btu 



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194 
HEAT TRANSFER 



0.02 in. 



0.06 in.- 




Copper filling Epoxy board 

FIGURE P3-63E 



Heat Conduction in Cylinders and Spheres 

3-64C What is an infinitely long cylinder? When is it proper 
to treat an actual cylinder as being infinitely long, and when 
is it not? 

3-65C Consider a short cylinder whose top and bottom sur- 
faces are insulated. The cylinder is initially at a uniform tem- 
perature Tj and is subjected to convection from its side surface 
to a medium at temperature T^, with a heat transfer coefficient 
of /?. Is the heat transfer in this short cylinder one- or two- 
dimensional? Explain. 

3-66C Can the thermal resistance concept be used for a solid 
cylinder or sphere in steady operation? Explain. 

3-67 A 5-m-internal-diameter spherical tank made of 
1.5-cm -thick stainless steel (k = 15 W/m • °C) is used to store 
iced water at 0°C. The tank is located in a room whose temper- 
ature is 30°C. The walls of the room are also at 30°C. The outer 
surface of the tank is black (emissivity e = 1), and heat trans- 
fer between the outer surface of the tank and the surroundings 
is by natural convection and radiation. The convection heat 



O Iced water VJ 

■o?.£W Q 



O 



1.5 cm 



vQ-OIQ'PIO 



transfer coefficients at the inner and the outer surfaces of the 
tank are 80 W/m 2 ■ °C and 10 W/m 2 ■ °C, respectively. Deter- 
mine (a) the rate of heat transfer to the iced water in the tank 
and (b) the amount of ice at 0°C that melts during a 24-h 
period. The heat of fusion of water at atmospheric pressure is 
h if = 333.7 kJ/kg. 

3-68 Steam at 320°C flows in a stainless steel pipe (k = 
1 5 W/m • °C) whose inner and outer diameters are 5 cm and 
5.5 cm, respectively. The pipe is covered with 3-cm-thick glass 
wool insulation (k = 0.038 W/m • °C). Heat is lost to the sur- 
roundings at 5°C by natural convection and radiation, with 
a combined natural convection and radiation heat transfer co- 
efficient of 15 W/m 2 • °C. Taking the heat transfer coefficient 
inside the pipe to be 80 W/m 2 • °C, determine the rate of heat 
loss from the steam per unit length of the pipe. Also determine 
the temperature drops across the pipe shell and the insulation. 

3-69 fitt'M Reconsider Problem 3-68. Using EES (or other) 
1^2 software, investigate the effect of the thickness of 
the insulation on the rate of heat loss from the steam and the 
temperature drop across the insulation layer. Let the insulation 
thickness vary from 1 cm to 10 cm. Plot the rate of heat loss 
and the temperature drop as a function of insulation thickness, 
and discuss the results. 

3-70 (Jb\ A 50-m-long section of a steam pipe whose outer 
^<UP diameter is 10 cm passes through an open space 
at 15°C. The average temperature of the outer surface of the 
pipe is measured to be 150°C. If the combined heat transfer co- 
efficient on the outer surface of the pipe is 20 W/m 2 • °C, de- 
termine (a) the rate of heat loss from the steam pipe, (b) the 
annual cost of this energy lost if steam is generated in a natural 
gas furnace that has an efficiency of 75 percent and the price of 
natural gas is $0.52/therm (1 therm = 105,500 kJ), and (c) the 
thickness of fiberglass insulation (k = 0.035 W/m • °C) needed 
in order to save 90 percent of the heat lost. Assume the pipe 
temperature to remain constant at 150°C. 



150°C 




\ Q 



J^C ^ 



Fiberglass 
insulation 

FIGURE P3-70 



FIGURE P3-67 



3-71 Consider a 2-m-high electric hot water heater that has a 
diameter of 40 cm and maintains the hot water at 55°C. The 
tank is located in a small room whose average temperature is 



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195 
CHAPTER 3 



3 cm 



27°C 



Foam 
insulation 




40 cm 



T = 55°C 




2 m 



FIGURE P3-71 



27°C, and the heat transfer coefficients on the inner and outer 
surfaces of the heater are 50 and 12 W/m 2 ■ °C, respectively. 
The tank is placed in another 46-cm-diameter sheet metal tank 
of negligible thickness, and the space between the two tanks is 
filled with foam insulation (k = 0.03 W/m ■ °C). The thermal 
resistances of the water tank and the outer thin sheet metal 
shell are very small and can be neglected. The price of elec- 
tricity is $0.08/kWh, and the home owner pays $280 a year for 
water heating. Determine the fraction of the hot water energy 
cost of this household that is due to the heat loss from the tank. 
Hot water tank insulation kits consisting of 3-cm-thick fiber- 
glass insulation (k = 0.035 W/m ■ °C) large enough to wrap the 
entire tank are available in the market for about $30. If such an 
insulation is installed on this water tank by the home owner 
himself, how long will it take for this additional insulation to 
pay for itself? Answers-. 17.5 percent, 1.5 years 

3-72 rfigM Reconsider Problem 3-71. Using EES (or other) 
b^2 software, plot the fraction of energy cost of hot 
water due to the heat loss from the tank as a function of the 
hot water temperature in the range of 40°C to 90°C. Discuss 
the results. 

3-73 Consider a cold aluminum canned drink that is initially 
at a uniform temperature of 3°C. The can is 12.5 cm high and 
has a diameter of 6 cm. If the combined convection/radiation 
heat transfer coefficient between the can and the surrounding 
air at 25 °C is 10 W/m 2 • °C, determine how long it will take for 
the average temperature of the drink to rise to 10°C. 

In an effort to slow down the warming of the cold drink, a 
person puts the can in a perfectly fitting 1 -cm-thick cylindrical 
rubber insulation (k = 0.13 W/m • °C). Now how long will it 
take for the average temperature of the drink to rise to 10°C? 
Assume the top of the can is not covered. 



3-C 



:25°C 




12.5 cm 



FIGURE P3-73 

3-74 Repeat Problem 3- 
resistance of 0.00008 m 2 
insulation. 



6 cm 



73, assuming a thermal contact 
°C/W between the can and the 



3-75E Steam at 450°F is flowing through a steel pipe (k = 8.7 
Btu/h • ft ■ °F) whose inner and outer diameters are 3.5 in. and 
4.0 in., respectively, in an environment at 55°F. The pipe is 
insulated with 2-in. -thick fiberglass insulation (k = 0.020 
Btu/h • ft • °F). If the heat transfer coefficients on the inside and 
the outside of the pipe are 30 and 5 Btu/h • ft 2 • °F, respectively, 
determine the rate of heat loss from the steam per foot length of 
the pipe. What is the error involved in neglecting the thermal 
resistance of the steel pipe in calculations? 



Steel pipe 




Insulation 

FIGURE P3-75E 

3-76 Hot water at an average temperature of 90°C is flowing 
through a 15-m section of a cast iron pipe (k = 52 W/m • °C) 
whose inner and outer diameters are 4 cm and 4.6 cm, respec- 
tively. The outer surface of the pipe, whose emissivity is 0.7, is 
exposed to the cold air at 10°C in the basement, with a heat 
transfer coefficient of 15 W/m 2 • °C. The heat transfer coeffi- 
cient at the inner surface of the pipe is 120 W/m 2 ■ °C. Taking 
the walls of the basement to be at 10°C also, determine the rate 
of heat loss from the hot water. Also, determine the average 



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196 
HEAT TRANSFER 



velocity of the water in the pipe if the temperature of the water 
drops by 3°C as it passes through the basement. 

3-77 Repeat Problem 3-76 for a pipe made of copper (k = 
386 W/m ■ °C) instead of cast iron. 

3-78E Steam exiting the turbine of a steam power plant at 
100°F is to be condensed in a large condenser by cooling water 
flowing through copper pipes (k = 223 Btu/h • ft • °F) of inner 
diameter 0.4 in. and outer diameter 0.6 in. at an average 
temperature of 70°F. The heat of vaporization of water at 
100°F is 1037 Btu/lbm. The heat transfer coefficients are 1500 
Btu/h • ft 2 • °F on the steam side and 35 Btu/h • ft 2 • °F on the 
water side. Determine the length of the tube required to con- 
dense steam at a rate of 120 lbm/h. Answer: 1 148 ft 



Steam, 100°F 
120 lbm/h 




Liquid water 

FIGURE P3-78E 

3-79E Repeat Problem 3-78E, assuming that a 0.01 -in. -thick 
layer of mineral deposit (k = 0.5 Btu/h • ft ■ °F) has formed on 
the inner surface of the pipe. 



3-80 



Reconsider Problem 3-78E. Using EES (or 



other) software, investigate the effects of the 
thermal conductivity of the pipe material and the outer di- 
ameter of the pipe on the length of the tube required. Let 
the thermal conductivity vary from 10 Btu/h • ft • °F to 400 
Btu/h • ft • °F and the outer diameter from 0.5 in. to 1.0 in. Plot 
the length of the tube as functions of pipe conductivity and the 
outer pipe diameter, and discuss the results. 

3-81 The boiling temperature of nitrogen at atmospheric 
pressure at sea level (1 atm pressure) is — 196°C. Therefore, ni- 
trogen is commonly used in low -temperature scientific studies 
since the temperature of liquid nitrogen in a tank open to the at- 
mosphere will remain constant at — 196°C until it is depleted. 
Any heat transfer to the tank will result in the evaporation of 
some liquid nitrogen, which has a heat of vaporization of 198 
kJ/kg and a density of 810 kg/m 3 at 1 atm. 



N, vapor 




Insulation 



FIGURE P3-81 



Consider a 3-m-diameter spherical tank that is initially filled 
with liquid nitrogen at 1 atm and — 196°C. The tank is exposed 
to ambient air at 15°C, with a combined convection and radia- 
tion heat transfer coefficient of 35 W/m 2 ■ °C. The temperature 
of the thin-shelled spherical tank is observed to be almost the 
same as the temperature of the nitrogen inside. Determine 
the rate of evaporation of the liquid nitrogen in the tank as a 
result of the heat transfer from the ambient air if the tank is 
(a) not insulated, (b) insulated with 5-cm-thick fiberglass insu- 
lation (k = 0.035 W/m ■ °C), and (c) insulated with 2-cm-thick 
superinsulation which has an effective thermal conductivity of 
0.00005 W/m • °C. 

3-82 Repeat Problem 3-81 for liquid oxygen, which has 
a boiling temperature of — 183°C, a heat of vaporization of 
213 kJ/kg, and a density of 1 140 kg/m 3 at 1 atm pressure. 

Critical Radius of Insulation 

3-83C What is the critical radius of insulation? How is it 
defined for a cylindrical layer? 

3-84C A pipe is insulated such that the outer radius of the 
insulation is less than the critical radius. Now the insulation is 
taken off. Will the rate of heat transfer from the pipe increase 
or decrease for the same pipe surface temperature? 

3-85C A pipe is insulated to reduce the heat loss from it. 
However, measurements indicate that the rate of heat loss 
has increased instead of decreasing. Can the measurements 
be right? 

3-86C Consider a pipe at a constant temperature whose ra- 
dius is greater than the critical radius of insulation. Someone 
claims that the rate of heat loss from the pipe has increased 
when some insulation is added to the pipe. Is this claim valid? 

3-87C Consider an insulated pipe exposed to the atmo- 
sphere. Will the critical radius of insulation be greater on calm 
days or on windy days? Why? 



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3-88 A 2-mm-diameter and 1 0-m-long electric wire is tightly 
wrapped with a 1-mm-thick plastic cover whose thermal con- 
ductivity is k = 0.15 W/m ■ °C. Electrical measurements indi- 
cate that a current of 10 A passes through the wire and there is 
a voltage drop of 8 V along the wire. If the insulated wire is ex- 
posed to a medium at T„ = 30°C with a heat transfer coeffi- 
cient of h = 24 W/m 2 ■ °C, determine the temperature at the 
interface of the wire and the plastic cover in steady operation. 
Also determine if doubling the thickness of the plastic cover 
will increase or decrease this interface temperature. 



Electrical 
wire 



T. = 30°C 



- Insulation 



2> 



■10m 



FIGURE P3-88 

3-89E A 0.083-in. -diameter electrical wire at 115°F is 
covered by 0.02-in. -thick plastic insulation (k = 0.075 
Btu/h • ft • °F). The wire is exposed to a medium at 50°F, with 
a combined convection and radiation heat transfer coefficient 
of 2.5 Btu/h • ft 2 • °F. Determine if the plastic insulation on the 
wire will increase or decrease heat transfer from the wire. 
Answer: It helps 

3-90E Repeat Problem 3-89E, assuming a thermal contact 
resistance of 0.001 h • ft 2 • °F/Btu at the interface of the wire 
and the insulation. 

3-91 A 5-mm-diameter spherical ball at 50°C is covered by a 
1-mm-thick plastic insulation (k = 0.13 W/m • °C). The ball is 
exposed to a medium at 15°C, with a combined convection and 
radiation heat transfer coefficient of 20 W/m 2 ■ °C. Determine 
if the plastic insulation on the ball will help or hurt heat trans- 
fer from the ball. 




FIGURE P3-91 



3-92 



Reconsider Problem 3-91. Using EES (or other) 
software, plot the rate of heat transfer from the 

ball as a function of the plastic insulation thickness in the range 

of 0.5 mm to 20 mm. Discuss the results. 



Heat Transfer from Finned Surfaces 

3-93C What is the reason for the widespread use of fins on 
surfaces? 



197 
CHAPTER 3 



3-94C What is the difference between the fin effectiveness 
and the fin efficiency? 

3-95C The fins attached to a surface are determined to have 
an effectiveness of 0.9. Do you think the rate of heat transfer 
from the surface has increased or decreased as a result of the 
addition of these fins? 

3-96C Explain how the fins enhance heat transfer from a 
surface. Also, explain how the addition of fins may actually 
decrease heat transfer from a surface. 

3-97C How does the overall effectiveness of a finned sur- 
face differ from the effectiveness of a single fin? 

3-98C Hot water is to be cooled as it flows through the tubes 
exposed to atmospheric air. Fins are to be attached in order to 
enhance heat transfer. Would you recommend attaching the 
fins inside or outside the tubes? Why? 

3-99C Hot air is to be cooled as it is forced to flow through 
the tubes exposed to atmospheric air. Fins are to be added in 
order to enhance heat transfer. Would you recommend attach- 
ing the fins inside or outside the tubes? Why? When would you 
recommend attaching fins both inside and outside the tubes? 

3-100C Consider two finned surfaces that are identical 
except that the fins on the first surface are formed by casting 
or extrusion, whereas they are attached to the second surface 
afterwards by welding or tight fitting. For which case do you 
think the fins will provide greater enhancement in heat trans- 
fer? Explain. 

3-101C The heat transfer surface area of a fin is equal to the 
sum of all surfaces of the fin exposed to the surrounding 
medium, including the surface area of the fin tip. Under what 
conditions can we neglect heat transfer from the fin tip? 

3-102C Does the (a) efficiency and (b) effectiveness of a fin 
increase or decrease as the fin length is increased? 

3-103C Two pin fins are identical, except that the diameter 
of one of them is twice the diameter of the other. For which fin 
will the (a) fin effectiveness and (b) fin efficiency be higher? 
Explain. 

3-104C Two plate fins of constant rectangular cross section 
are identical, except that the thickness of one of them is twice 
the thickness of the other. For which fin will the (a) fin effec- 
tiveness and (b) fin efficiency be higher? Explain. 

3-105C Two finned surfaces are identical, except that the 
convection heat transfer coefficient of one of them is twice that 
of the other. For which finned surface will the (a) fin effective- 
ness and (b) fin efficiency be higher? Explain. 

3-106 Obtain a relation for the fin efficiency for a fin of con- 
stant cross-sectional area A c , perimeter p, length L, and thermal 
conductivity k exposed to convection to a medium at r„ with a 
heat transfer coefficient h. Assume the fins are sufficiently long 
so that the temperature of the fin at the tip is nearly T x . Take 
the temperature of the fin at the base to be T b and neglect heat 



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198 
HEAT TRANSFER 



h, T m 



n{ k 


I" 


^ b =K 




p = kD, A c = 


= TdfilA 


FIGURE P3-1 06 





transfer from the fin tips. Simplify the relation for (a) a circu- 
lar fin of diameter D and (b) rectangular fins of thickness t. 

3-107 The case-to-ambient thermal resistance of a power 
transistor that has a maximum power rating of 1 5 W is given to 
be 25°C/W. If the case temperature of the transistor is not to 
exceed 80°C, determine the power at which this transistor can 
be operated safely in an environment at 40°C. 

3-108 A 40-W power transistor is to be cooled by attaching 
it to one of the commercially available heat sinks shown in 
Table 3-A. Select a heat sink that will allow the case tempera- 
ture of the transistor not to exceed 90° in the ambient air at 20°. 



= 20°C 



90°C 




FIGURE P3-1 08 

3-109 A 30-W power transistor is to be cooled by attaching 
it to one of the commercially available heat sinks shown in 
Table 3-A. Select a heat sink that will allow the case tempera- 
ture of the transistor not to exceed 80°C in the ambient air at 
35°C. 

3-110 Steam in a heating system flows through tubes whose 
outer diameter is 5 cm and whose walls are maintained at a 
temperature of 180°C. Circular aluminum alloy 2024-T6 fins 
(k = 1 86 W/m ■ °C) of outer diameter 6 cm and constant thick- 
ness 1 mm are attached to the tube. The space between the fins 
is 3 mm, and thus there are 250 fins per meter length of the 
tube. Heat is transferred to the surrounding air at T rj -_ = 25°C, 
with a heat transfer coefficient of 40 W/m 2 • °C. Determine the 
increase in heat transfer from the tube per meter of its length as 
a result of adding fins. Answer: 2639 W 

3-111E Consider a stainless steel spoon (k = 8.7 
Btu/h ■ ft • °F) partially immersed in boiling water at 200°F in 
a kitchen at 75°F. The handle of the spoon has a cross section 
of 0.08 in. X 0.5 in., and extends 7 in. in the air from the free 



2.5 cm 



= 25°C * 



180°C 




FIGURE P3-1 10 

surface of the water. If the heat transfer coefficient at the ex- 
posed surfaces of the spoon handle is 3 Btu/h ■ ft 2 • °F, deter- 
mine the temperature difference across the exposed surface of 
the spoon handle. State your assumptions. Answer-. 124. 6°F 

Spoon 



:75°F 




Boiling 
water 
200°F 

FIGURE P3-111E 



VR*. 



3-112E Repeat Problem 3-111 for a silver spoon (k = 247 
Btu/h • ft • °F). 

3-113E fu'M Reconsider Problem 3-1 HE. Using EES (or 
Ei3 other) software, investigate the effects of the 
thermal conductivity of the spoon material and the length of its 
extension in the air on the temperature difference across the 
exposed surface of the spoon handle. Let the thermal conduc- 
tivity vary from 5 Btu/h ■ ft • °F to 225 Btu/h • ft ■ °F and the 
length from 5 in. to 12 in. Plot the temperature difference as the 
functions of thermal conductivity and length, and discuss 
the results. 

3-114 A 0.3-cm-thick, 12-cm-high, and 18-cm-long circuit 
board houses 80 closely spaced logic chips on one side, each 
dissipating 0.04 W. The board is impregnated with copper fill- 
ings and has an effective thermal conductivity of 20 W/m • °C. 
All the heat generated in the chips is conducted across the cir- 
cuit board and is dissipated from the back side of the board 
to a medium at 40°C, with a heat transfer coefficient of 50 
W/m 2 • °C. (a) Determine the temperatures on the two sides 
of the circuit board, (b) Now a 0.2-cm-thick, 12-cm-high, and 



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CHAPTER 3 



18-cm-long aluminum plate (k = 237 W/m • °C) with 864 
2-cm-long aluminum pin fins of diameter 0.25 cm is attached 
to the back side of the circuit board with a 0.02 -cm-thick epoxy 
adhesive (k = 1.8 W/m • °C). Determine the new temperatures 
on the two sides of the circuit board. 

3-115 Repeat Problem 3-1 14 using a copper plate with cop- 
per fins (k = 386 W/m • °C) instead of aluminum ones. 

3-116 A hot surface at 100°C is to be cooled by attach- 
ing 3-cm-long, 0.25-cm-diameter aluminum pin fins (k = 
237 W/m • °C) to it, with a center-to-center distance of 0.6 cm. 
The temperature of the surrounding medium is 30°C, and the 
heat transfer coefficient on the surfaces is 35 W/m 2 • °C. 
Determine the rate of heat transfer from the surface for a 
1-m X 1-m section of the plate. Also determine the overall 
effectiveness of the fins. 




FIGURE P3-1 16 



386 



3-117 Repeat Problem 3-116 using copper fins (k 
W/m • °C) instead of aluminum ones. 

3-118 fitt'M Reconsider Problem 3-116. Using EES (or 
KS other) software, investigate the effect of the cen- 
ter-to-center distance of the fins on the rate of heat transfer 
from the surface and the overall effectiveness of the fins. Let 
the center-to-center distance vary from 0.4 cm to 2.0 cm. Plot 
the rate of heat transfer and the overall effectiveness as a func- 
tion of the center-to-center distance, and discuss the results. 

3-119 Two 3-m-long and 0.4-cm-thick cast iron (k = 52 
W/m • °C) steam pipes of outer diameter 10 cm are connected 
to each other through two 1 -cm-thick flanges of outer diameter 
20 cm. The steam flows inside the pipe at an average tempera- 
ture of 200°C with a heat transfer coefficient of 180 W/m 2 • °C. 
The outer surface of the pipe is exposed to an ambient at 12°C, 
with a heat transfer coefficient of 25 W/m 2 • °C. (a) Disregard- 
ing the flanges, determine the average outer surface tempera- 
ture of the pipe, (b) Using this temperature for the base of the 
flange and treating the flanges as the fins, determine the fin ef- 
ficiency and the rate of heat transfer from the flanges, (c) What 
length of pipe is the flange section equivalent to for heat trans- 
fer purposes? 




FIGURE P3-1 19 



Heat Transfer in Common Configurations 

3-120C What is a conduction shape factor? How is it related 
to the thermal resistance? 

3-121C What is the value of conduction shape factors in 
engineering? 

3-122 A 20-m-long and 8-cm-diameter hot water pipe of a 
district heating system is buried in the soil 80 cm below the 
ground surface. The outer surface temperature of the pipe is 
60°C. Taking the surface temperature of the earth to be 5°C 
and the thermal conductivity of the soil at that location to be 
0.9 W/m • °C, determine the rate of heat loss from the pipe. 



JL 



5°C 



80 cm 



-60°C 



m 



,20 a- 



FIGURE P3-1 22 



3-123 



Reconsider Problem 3-122. Using EES (or 
other) software, plot the rate of heat loss from 

the pipe as a function of the burial depth in the range of 20 cm 

to 2.0 m. Discuss the results. 

3-124 Hot and cold water pipes 8 m long run parallel to each 
other in a thick concrete layer. The diameters of both pipes are 
5 cm, and the distance between the centerlines of the pipes is 
40 cm. The surface temperatures of the hot and cold pipes are 
60°C and 15°C, respectively. Taking the thermal conductivity 
of the concrete to be k = 0.75 W/m • °C, determine the rate of 
heat transfer between the pipes. Answer: 306 W 



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200 
HEAT TRANSFER 



3-125 [?(,■>! Reconsider Problem 3-124. Using EES (or 
I^S other) software, plot the rate of heat transfer 
between the pipes as a function of the distance between the 
centerlines of the pipes in the range of 10 cm to 1 .0 m. Discuss 
the results. 

3-126E A row of 3-ft-long and 1 -in. -diameter used uranium 
fuel rods that are still radioactive are buried in the ground par- 
allel to each other with a center-to-center distance of 8 in. at a 
depth 15 ft from the ground surface at a location where the 
thermal conductivity of the soil is 0.6 Btu/h • ft • °F. If the sur- 
face temperature of the rods and the ground are 350°F and 
60°F, respectively, determine the rate of heat transfer from the 
fuel rods to the atmosphere through the soil. 



60°F 




. •'; 1^8 in. ; »[»'8 in.r"p-8 in.^j 

FIGURE P3-1 26 

3-127 Hot water at an average temperature of 60°C and an 
average velocity of 0.6 m/s is flowing through a 5-m section 
of a thin-walled hot water pipe that has an outer diameter of 
2.5 cm. The pipe passes through the center of a 14-cm-fhick 
wall filled with fiberglass insulation (k = 0.035 W/m ■ °C). If 
the surfaces of the wall are at 1 8°C, determine (a) the rate of 
heat transfer from the pipe to the air in the rooms and (b) the 
temperature drop of the hot water as it flows through this 
5-m-long section of the wall. Answers: 23.5 W, 0.02°C 




Wall 



Hot water pipe 



8°C 



X 



0°C 



3 m 



20 m: ■ f 



FIGURE P3-1 28 

(k = 1.5 W/m ■ °C) vertically for 3 m, and continues horizon- 
tally at this depth for 20 m more before it enters the next build- 
ing. The first section of the pipe is exposed to the ambient air 
at 8°C, with a heat transfer coefficient of 22 W/m 2 • °C. If the 
surface of the ground is covered with snow at 0°C, determine 
(a) the total rate of heat loss from the hot water and (b) the 
temperature drop of the hot water as it flows through this 
25-m-long section of the pipe. 

3-129 Consider a house with a flat roof whose outer dimen- 
sions are 12 m X 12 m. The outer walls of the house are 6 m 
high. The walls and the roof of the house are made of 20-cm- 
thick concrete (k = 0.75 W/m • °C). The temperatures of the in- 
ner and outer surfaces of the house are 15°C and 3°C, 
respectively. Accounting for the effects of the edges of adjoin- 
ing surfaces, determine the rate of heat loss from the house 
through its walls and the roof. What is the error involved in ig- 
noring the effects of the edges and corners and treating the roof 
asal2mX 12m surface and the walls as 6 m X 12 m surfaces 
for simplicity? 

3-130 Consider a 10-m-long thick-walled concrete duct (k = 
0.75 W/m • °C) of square cross-section. The outer dimensions 
of the duct are 20 cm X 20 cm, and the thickness of the duct 
wall is 2 cm. If the inner and outer surfaces of the duct are at 
100°C and 15°C, respectively, determine the rate of heat trans- 
fer through the walls of the duct. Answer: 22.9 kW 




FIGURE P3-1 27 



16 cm- 
20 cm 

FIGURE P3-1 30 



3-128 Hot water at an average temperature of 80°C and an 
average velocity of 1.5 m/s is flowing through a 25-m section 
of a pipe that has an outer diameter of 5 cm. The pipe extends 
2 m in the ambient air above the ground, dips into the ground 



3-131 A 3-m-diameter spherical tank containing some radio- 
active material is buried in the ground (k = 1 .4 W/m • °C). The 
distance between the top surface of the tank and the ground 
surface is 4 m. If the surface temperatures of the tank and the 



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201 
CHAPTER 3 



ground are 140°C and 15°C, respectively, determine the rate of 
heat transfer from the tank. 

3-132 fitt'M Reconsider Problem 3-131. Using EES (or 
k^^ other) software, plot the rate of heat transfer 
from the tank as a function of the tank diameter in the range of 
0.5 m to 5.0 m. Discuss the results. 

3-133 Hot water at an average temperature of 85°C passes 
through a row of eight parallel pipes that are 4 m long and have 
an outer diameter of 3 cm, located vertically in the middle of a 
concrete wall (k = 0.75 W/m ■ °C) that is 4 m high, 8 m long, 
and 15 cm thick. If the surfaces of the concrete walls are 
exposed to a medium at 32°C, with a heat transfer coefficient 
of 12 W/m 2 • °C, determine the rate of heat loss from the hot 
water and the surface temperature of the wall. 




FIGURE P3-1 39 



Special Topics: 

Heat Transfer through the Walls and Roofs 

3-134C What is the TJ-value of a wall? How does it differ 
from the unit thermal resistance of the wall? How is it related 
to the [/-factor? 

3-135C What is effective emissivity for a plane-parallel air 
space? How is it determined? How is radiation heat transfer 
through the air space determined when the effective emissivity 
is known? 

3-136C The unit thermal resistances (/{-values) of both 
40-mm and 90-mm vertical air spaces are given in Table 3-9 to 
be 0.22 m 2 • °C/W, which implies that more than doubling the 
thickness of air space in a wall has no effect on heat transfer 
through the wall. Do you think this is a typing error? Explain. 

3-137C What is a radiant barrier? What kind of materials are 
suitable for use as radiant barriers? Is it worthwhile to use ra- 
diant barriers in the attics of homes? 

3-138C Consider a house whose attic space is ventilated ef- 
fectively so that the air temperature in the attic is the same as 
the ambient air temperature at all times. Will the roof still have 
any effect on heat transfer through the ceiling? Explain. 

3-139 Determine the summer R- value and the [/-factor of a 
wood frame wall that is built around 38-mm X 140-mm wood 
studs with a center-to-center distance of 400 mm. The 140- 
mm-wide cavity between the studs is filled with mineral fiber 
batt insulation. The inside is finished with 13-mm gypsum 
wallboard and the outside with 13-mm wood fiberboard and 
13-mm X 200-mm wood bevel lapped siding. The insulated 
cavity constitutes 80 percent of the heat transmission area, 
while the studs, headers, plates, and sills constitute 20 percent. 
Answers: 3.213 m 2 • °C/W, 0.311 W/m 2 ■ °C 

3-140 The 13-mm-thick wood fiberboard sheathing of the 
wood stud wall in Problem 3-139 is replaced by a 25-mm- 
thick rigid foam insulation. Determine the percent increase in 
the R- value of the wall as a result. 



3-141E Determine the winter R- value and the [/-factor of a 
masonry cavity wall that is built around 4-in. -thick concrete 
blocks made of lightweight aggregate. The outside is finished 
with 4-in. face brick with ^-in. cement mortar between the 
bricks and concrete blocks. The inside finish consists of ^-in. 
gypsum wallboard separated from the concrete block by | -in.- 
thick (1-in. by 3-in. nominal) vertical furring whose center-to- 
center distance is 16 in. Neither side of the | -in. -thick air space 
between the concrete block and the gypsum board is coated 
with any reflective film. When determining the R- value of the 
air space, the temperature difference across it can be taken to 
be 30°F with a mean air temperature of 50°F. The air space 
constitutes 80 percent of the heat transmission area, while the 
vertical furring and similar structures constitute 20 percent. 




FIGURE P3-1 41 E 

3-142 Consider a flat ceiling that is built around 38-mm X 
90-mm wood studs with a center-to-center distance of 400 mm. 
The lower part of the ceiling is finished with 13-mm gypsum 
wallboard, while the upper part consists of a wood subfloor 
(R = 0.166 m 2 • °C/W), a 13-mm plywood, a layer of felt (R = 
0.011 m 2 ■ °C/W), and linoleum (R = 0.009 m 2 ■ °C/W). Both 



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HEAT TRANSFER 




12 3 4 5 

FIGURE P3-1 42 

sides of the ceiling are exposed to still air. The air space con- 
stitutes 82 percent of the heat transmission area, while the 
studs and headers constitute 1 8 percent. Determine the winter 
R- value and the {/-factor of the ceiling assuming the 90-mm- 
wide air space between the studs (a) does not have any reflec- 
tive surface, (b) has a reflective surface with s = 0.05 on one 
side, and (c) has reflective surfaces with e = 0.05 on both 
sides. Assume a mean temperature of 10°C and a temperature 
difference of 5.6°C for the air space. 

3-143 Determine the winter lvalue and the {/-factor of a 
masonry cavity wall that consists of 100-mm common bricks, 
a 90-mm air space, 100-mm concrete blocks made of light- 
weight aggregate, 20-mm air space, and 13-mm gypsum wall- 
board separated from the concrete block by 20-mm-thick 
(1-in. X 3 -in. nominal) vertical furring whose center-to-center 
distance is 400 mm. Neither side of the two air spaces is coated 
with any reflective films. When determining the ^-value of the 
air spaces, the temperature difference across them can be taken 
to be 16.7°C with a mean air temperature of 10°C. The air 
space constitutes 84 percent of the heat transmission area, 



while the vertical furring and similar structures constitute 
16 percent. Answers: 1.02 m 2 • °C/W, 0.978 W/m 2 • °C 

3-144 Repeat Problem 3-143 assuming one side of both air 
spaces is coated with a reflective film of e = 0.05. 

3-145 Determine the winter R-vahie and the [/-factor of a 
masonry wall that consists of the following layers: 100-mm 
face bricks, 100-mm common bricks, 25-mm urethane rigid 
foam insulation, and 13-mm gypsum wallboard. 
Answers: 1.404 m 2 • °C/W, 0.712 W/m 2 • °C 

3-146 The overall heat transfer coefficient (the [/-value) of a 
wall under winter design conditions is U = 1.55 W/m 2 ■ °C. 
Determine the [/-value of the wall under summer design 
conditions. 

3-147 The overall heat transfer coefficient (the [/-value) of a 
wall under winter design conditions is U = 2.25 W/m 2 ■ °C. 
Now a layer of 100-mm face brick is added to the outside, 
leaving a 20-mm air space between the wall and the bricks. De- 
termine the new [/-value of the wall. Also, determine the rate 
of heat transfer through a 3-m-high, 7-m-long section of the 
wall after modification when the indoor and outdoor tempera- 
tures are 22°C and — 5°C, respectively. 




FIGURE P3-1 47 




FIGURE P3-1 43 



3-148 Determine the summer and winter ^-values, in 
m 2 • °C/W, of a masonry wall that consists of 100-mm face 
bricks, 13-mm of cement mortar, 100-mm lightweight concrete 
block, 40-mm air space, and 20-mm plasterboard. 
Answers: 0.809 and 0.795 m 2 ■ °C/W 

3-149E The overall heat transfer coefficient of a wall is 
determined to be U = 0.09 Btu/h • ft 2 ■ °F under the conditions 
of still air inside and winds of 7.5 mph outside. What will the 
[/-factor be when the wind velocity outside is doubled? 
Answer: 0.0907 Btu/h ■ ft 2 • °F 

3-150 Two homes are identical, except that the walls of one 
house consist of 200-mm lightweight concrete blocks, 20-mm 
air space, and 20-mm plasterboard, while the walls of the other 
house involve the standard R-2.4 m 2 • °C/W frame wall con- 
struction. Which house do you think is more energy efficient? 



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3-151 Determine the /J-value of a ceiling that consists of a 
layer of 19-mm acoustical tiles whose top surface is covered 
with a highly reflective aluminum foil for winter conditions. 
Assume still air below and above the tiles. 



Highly 

reflective 

foil 




FIGURE P3-1 51 

Review Problems 

3-152E Steam is produced in the copper tubes (k = 223 
Btu/h • ft • °F) of a heat exchanger at a temperature of 250°F by 
another fluid condensing on the outside surfaces of the tubes at 
350°F. The inner and outer diameters of the tube are 1 in. and 
1.3 in., respectively. When the heat exchanger was new, the 
rate of heat transfer per foot length of the tube was 2 X 10 4 
Btu/h. Determine the rate of heat transfer per foot length 
of the tube when a 0.01 -in. -thick layer of limestone (k = 
1 .7 Btu/h ■ ft ■ °F) has formed on the inner surface of the tube 
after extended use. 

3-153E Repeat Problem 3-152E, assuming that a 0.01-in.- 
thick limestone layer has formed on both the inner and outer 
surfaces of the tube. 

3-154 A 1.2-m-diameter and 6-m-long cylindrical propane 
tank is initially filled with liquid propane whose density is 581 
kg/m 3 . The tank is exposed to the ambient air at 30°C, with a 
heat transfer coefficient of 25 W/m 2 • °C. Now a crack devel- 
ops at the top of the tank and the pressure inside drops to 1 atm 
while the temperature drops to — 42°C, which is the boiling 
temperature of propane at 1 atm. The heat of vaporization of 







Propane 




T 

f 


ir =30°C 


vapor 


"\ 




PROPANE TANK 


1.2 m 


T = -42°C 


P = 1 atm / 




4 


6 m ► 





203 
CHAPTER 3 



propane at 1 atm is 425 kJ/kg. The propane is slowly vaporized 
as a result of the heat transfer from the ambient air into the 
tank, and the propane vapor escapes the tank at — 42°C through 
the crack. Assuming the propane tank to be at about the same 
temperature as the propane inside at all times, determine how 
long it will take for the propane tank to empty if the tank is 
(a) not insulated and (b) insulated with 7.5-cm-thick glass wool 
insulation (k = 0.038 W/m • °C). 

3-155 Hot water is flowing at an average velocity of 1 .5 m/s 
through a cast iron pipe (k = 52 W/m • °C) whose inner and 
outer diameters are 3 cm and 3.5 cm, respectively. The pipe 
passes through a 15-m-long section of a basement whose 
temperature is 15°C. If the temperature of the water drops 
from 70°C to 67°C as it passes through the basement and the 
heat transfer coefficient on the inner surface of the pipe is 400 
W/m 2 ■ °C, determine the combined convection and radiation 
heat transfer coefficient at the outer surface of the pipe. 
Answer: 272.5 W/m 2 • °C 

3-156 Newly formed concrete pipes are usually cured first 
overnight by steam in a curing kiln maintained at a temperature 
of 45°C before the pipes are cured for several days outside. The 
heat and moisture to the kiln is provided by steam flowing in a 
pipe whose outer diameter is 12 cm. During a plant inspection, 
it was noticed that the pipe passes through a 10-m section that 
is completely exposed to the ambient air before it reaches the 
kiln. The temperature measurements indicate that the average 
temperature of the outer surface of the steam pipe is 82°C 
when the ambient temperature is 8°C. The combined convec- 
tion and radiation heat transfer coefficient at the outer surface 
of the pipe is estimated to be 25 W/m 2 • °C. Determine the 
amount of heat lost from the steam during a 10-h curing 
process that night. 

Steam is supplied by a gas-fired steam generator that has 
an efficiency of 80 percent, and the plant pays $0.60/therm of 
natural gas (1 therm = 105,500 kJ). If the pipe is insulated and 
90 percent of the heat loss is saved as a result, determine the 
amount of money this facility will save a year as a result of 
insulating the steam pipes. Assume that the concrete pipes are 
cured 110 nights a year. State your assumptions. 



8 "C 



Furnace 




FIGURE P3-1 54 



FIGURE P3-1 56 

3-157 Consider an 18-cm X 18-cm multilayer circuit board 
dissipating 27 W of heat. The board consists of four layers 
of 0.2-mm-thick copper (k = 386 W/m • °C) and three layers of 



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204 
HEAT TRANSFER 



Copper 




FIGURE P3-1 57 

1.5-mm-thick epoxy glass (k = 0.26 W/m ■ °C) sandwiched 
together, as shown in the figure. The circuit board is attached to 
a heat sink from both ends, and the temperature of the board at 
those ends is 35°C. Heat is considered to be uniformly gener- 
ated in the epoxy layers of the board at a rate of 0.5 W per 1 -cm 
X 18-cm epoxy laminate strip (or 1.5 W per 1-cm X 18-cm 
strip of the board). Considering only a portion of the board be- 
cause of symmetry, determine the magnitude and location of 
the maximum temperature that occurs in the board. Assume 
heat transfer from the top and bottom faces of the board to be 
negligible. 

3-158 The plumbing system of a house involves a 0.5-m sec- 
tion of a plastic pipe {k = 0.16 W/m • °C) of inner diameter 
2 cm and outer diameter 2.4 cm exposed to the ambient air. 
During a cold and windy night, the ambient air temperature re- 
mains at about — 5°C for a period of 14 h. The combined con- 
vection and radiation heat transfer coefficient on the outer 
surface of the pipe is estimated to be 40 W/m 2 • °C, and the 
heat of fusion of water is 333.7 kJ/kg. Assuming the pipe to 
contain stationary water initially at 0°C, determine if the water 
in that section of the pipe will completely freeze that night. 



Exposed 
- water pipe 



AIR 




SOIL- 



FIGURE P3-1 58 

3-159 Repeat Problem 3-158 for the case of a heat transfer 
coefficient of 10 W/m 2 • °C on the outer surface as a result of 
putting a fence around the pipe that blocks the wind. 

3-1 60E The surface temperature of a 3 -in. -diameter baked 
potato is observed to drop from 300°F to 200°F in 5 minutes in 
an environment at 70°F. Determine the average heat transfer 
coefficient between the potato and its surroundings. Using this 



heat transfer coefficient and the same surface temperature, 
determine how long it will take for the potato to experience 
the same temperature drop if it is wrapped completely in a 
0.12-in.-thick towel (k = 0.035 Btu/h • ft ■ °F). You may use the 
properties of water for potato. 

3-161E Repeat Problem 3-160E assuming there is a 0.02- 
in. -thick air space (k = 0.015 Btu/h • ft • °F) between the potato 
and the towel. 

3-162 An ice chest whose outer dimensions are 30 cm X 
40 cm X 50 cm is made of 3-cm-thick Styrofoam (k = 0.033 
W/m • °C). Initially, the chest is filled with 45 kg of ice at 0°C, 
and the inner surface temperature of the ice chest can be taken 
to be 0°C at all times. The heat of fusion of ice at 0°C is 333.7 
kJ/kg, and the heat transfer coefficient between the outer 
surface of the ice chest and surrounding air at 35°C is 18 
W/m 2 ■ °C. Disregarding any heat transfer from the 40-cm X 
50-cm base of the ice chest, determine how long it will take for 
the ice in the chest to melt completely. 



:35°C 









^ ° 

a Ice chest ° 

~ _ 0°C ° 

Q o 



3 cm 



Styrofoam 



FIGURE P3-1 62 



3-163 A 4-m-high and 6-m-long wall is constructed of two 
large 2-cm-thick steel plates (k = 15 W/m • °C) separated by 
1-cm -thick and 20-cm-wide steel bars placed 99 cm apart. The 



Steel plates 



2 cm 




20 cm 



Fiberglass 
- insulation 



99 cm 



1 cm 



2 cm 



FIGURE P3-1 63 



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205 
CHAPTER 3 



remaining space between the steel plates is filled with fiber- 
glass insulation (k = 0.035 W/m ■ °C). If the temperature dif- 
ference between the inner and the outer surfaces of the walls is 
22°C, determine the rate of heat transfer through the wall. Can 
we ignore the steel bars between the plates in heat transfer 
analysis since they occupy only 1 percent of the heat transfer 
surface area? 

3-164 A 0.2-cm-thick, 10-cm-high, and 15-cm-long circuit 
board houses electronic components on one side that dissipate 
a total of 15 W of heat uniformly. The board is impregnated 
with conducting metal fillings and has an effective thermal 
conductivity of 12 W/m • °C. All the heat generated in the com- 
ponents is conducted across the circuit board and is dissipated 
from the back side of the board to a medium at 37°C, with a 
heat transfer coefficient of 45 W/m 2 • °C. (a) Determine the 
surface temperatures on the two sides of the circuit board. 
(b) Now a 0.1 -cm-thick, 10-cm-high, and 15-cm-long alumi- 
num plate (k = 237 W/m • °C) with 20 0.2-cm-thick, 2-cm- 
long, and 15-cm-wide aluminum fins of rectangular profile are 
attached to the back side of the circuit board with a 0.03-cm- 
thick epoxy adhesive (k = 1.8 W/m • °C). Determine the new 
temperatures on the two sides of the circuit board. 



Electronic 
components 




10 cm // 



0.3 cm 



0.2 cm 



^2 mm 

FIGURE P3-1 64 



3-165 Repeat Problem 3-164 using a copper plate with cop- 
per fins (k = 386 W/m • °C) instead of aluminum ones. 

3-166 A row of 10 parallel pipes that are 5 m long and have 
an outer diameter of 6 cm are used to transport steam at 150°C 
through the concrete floor (k = 0.75 W/m • °C) of a 10-m X 
5-m room that is maintained at 25°C. The combined con- 
vection and radiation heat transfer coefficient at the floor is 
12 W/m 2 • °C. If the surface temperature of the concrete floor 
is not to exceed 40°C, determine how deep the steam pipes 
should be buried below the surface of the concrete floor. 







Room 




25°C 






,40°C 


0OOO 


oooooo 

^ Steam pipes 


*D = 6cm 


Concrete floor 



FIGURE P3-1 66 

3-167 Consider two identical people each generating 60 W 
of metabolic heat steadily while doing sedentary work, and dis- 
sipating it by convection and perspiration. The first person 
is wearing clothes made of 1-mm-thick leather (k = 0.159 
W/m ■ °C) that covers half of the body while the second one is 
wearing clothes made of 1-mm-thick synthetic fabric (k = 0.13 
W/m ■ °C) that covers the body completely. The ambient air is 
at 30°C, the heat transfer coefficient at the outer surface is 
1 5 W/m 2 ■ °C, and the inner surface temperature of the clothes 
can be taken to be 32°C. Treating the body of each person as a 
25-cm-diameter 1.7-m-long cylinder, determine the fractions 
of heat lost from each person by perspiration. 

3-168 A 6-m-wide 2.8-m-high wall is constructed of one 
layer of common brick (k = 0.72 W/m • °C) of thickness 
20 cm, one inside layer of light-weight plaster (k = 0.36 
W/m • °C) of thickness 1 cm, and one outside layer of cement 
based covering (k = 1 .40 W/m • °C) of thickness 2 cm. The in- 
ner surface of the wall is maintained at 23 °C while the outer 
surface is exposed to outdoors at 8°C with a combined convec- 
tion and radiation heat transfer coefficient of 17 W/m 2 ■ °C. 
Determine the rate of heat transfer through the wall and tem- 
perature drops across the plaster, brick, covering, and surface- 
ambient air. 

3-169 Reconsider Problem 3-1 68. It is desired to insulate the 
wall in order to decrease the heat loss by 85 percent. For the 
same inner surface temperature, determine the thickness of in- 
sulation and the outer surface temperature if the wall is insu- 
lated with (a) polyurethane foam {k = 0.025 W/m • °C) and 
(b) glass fiber (k = 0.036 W/m • °C). 

3-170 Cold conditioned air at 12°C is flowing inside a 
1.5-cm-thick square aluminum (k = 237 W/m ■ °C) duct of 
inner cross section 22 cm X 22 cm at a mass flow rate of 
0.8 kg/s. The duct is exposed to air at 33°C with a combined 
convection-radiation heat transfer coefficient of 8 W/m 2 • °C. 
The convection heat transfer coefficient at the inner surface is 
75 W/m 2 • °C. If the air temperature in the duct should not 
increase by more than 1 °C determine the maximum length of 
the duct. 

3-171 When analyzing heat transfer through windows, it 
is important to consider the frame as well as the glass area. 
Consider a 2-m-wide 1.5-m-high wood-framed window with 



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206 
HEAT TRANSFER 



85 percent of the area covered by 3-mm-thick single-pane glass 
(k = 0.7 W/m ■ °C). The frame is 5 cm thick, and is made of 
pine wood (k = 0.12 W/m • °C). The heat transfer coefficient is 
7 W/m 2 • °C inside and 13 W/m 2 • °C outside. The room is 
maintained at 24°C, and the temperature outdoors is 40°C. De- 
termine the percent error involved in heat transfer when the 
window is assumed to consist of glass only. 

3-172 Steam at 235°C is flowing inside a steel pipe (k = 
61 W/m • °C) whose inner and outer diameters are 10 cm and 
12 cm, respectively, in an environment at 20°C. The heat trans- 
fer coefficients inside and outside the pipe are 1 05 W/m 2 • °C 
and 14 W/m 2 • °C, respectively. Determine (a) the thickness of 
the insulation (k = 0.038 W/m ■ °C) needed to reduce the heat 
loss by 95 percent and (b) the thickness of the insulation 
needed to reduce the exposed surface temperature of insulated 
pipe to 40°C for safety reasons. 

3-173 When the transportation of natural gas in a pipeline is 
not feasible for economic or other reasons, it is first liquefied at 
about — 160°C, and then transported in specially insulated 
tanks placed in marine ships. Consider a 6-m-diameter spheri- 
cal tank that is filled with liquefied natural gas (LNG) at 
— 160°C. The tank is exposed to ambient air at 18°C with a 
heat transfer coefficient of 22 W/m 2 • °C. The tank is thin- 
shelled and its temperature can be taken to be the same as the 
LNG temperature. The tank is insulated with 5-cm-thick super 
insulation that has an effective thermal conductivity of 0.00008 
W/m • °C. Taking the density and the specific heat of LNG to 
be 425 kg/m 3 and 3.475 kJ/kg • °C, respectively, estimate how 
long it will take for the LNG temperature to rise to — 150°C. 

3-174 A 15-cm X 20-cm hot surface at 85°C is to be cooled 
by attaching 4-cm-long aluminum (k = 237 W/m ■ °C) fins of 
2-mm X 2-mm square cross section. The temperature of sur- 
rounding medium is 25°C and the heat transfer coefficient on 
the surfaces can be taken to be 20 W/m 2 • °C. If it is desired to 
triple the rate of heat transfer from the bare hot surface, deter- 
mine the number of fins that needs to be attached. 

3-175 [?(,■>! Reconsider Problem 3-174. Using EES (or 
1^2 other) software, plot the number of fins as a 
function of the increase in the heat loss by fins relative to no 
fin case (i.e., overall effectiveness of the fins) in the range of 
1.5 to 5. Discuss the results. Is it realistic to assume the heat 
transfer coefficient to remain constant? 

3-176 A 1 .4-m-diameter spherical steel tank filled with iced 
water at 0°C is buried underground at a location where the 
thermal conductivity of the soil is k = 0.55 W/m • °C. The dis- 
tance between the tank center and the ground surface is 2.4 m. 
For ground surface temperature of 18°C, determine the rate of 
heat transfer to the iced water in the tank. What would your 
answer be if the soil temperature were 1 8°C and the ground 
surface were insulated? 



3-177 A 0. 6-m-diameter 1.9-m-long cylindrical tank con- 
taining liquefied natural gas (LNG) at — 160°C is placed at the 
center of a 1 .9-m-long 1 .4-m X 1 .4-m square solid bar made of 
an insulating material with k = 0.0006 W/m • °C. If the outer 
surface temperature of the bar is 20°C, determine the rate of 
heat transfer to the tank. Also, determine the LNG temperature 
after one month. Take the density and the specific heat of LNG 
to be 425 kg/m 3 and 3.475 kJ/kg ■ °C, respectively. 

Design and Essay Problems 

3-178 The temperature in deep space is close to absolute 
zero, which presents thermal challenges for the astronauts who 
do space walks. Propose a design for the clothing of the astro- 
nauts that will be most suitable for the thermal environment in 
space. Defend the selections in your design. 

3-179 In the design of electronic components, it is very de- 
sirable to attach the electronic circuitry to a substrate material 
that is a very good thermal conductor but also a very effective 
electrical insulator. If the high cost is not a major concern, what 
material would you propose for the substrate? 

3-180 Using cylindrical samples of the same material, devise 
an experiment to determine the thermal contact resistance. 
Cylindrical samples are available at any length, and the thermal 
conductivity of the material is known. 

3-181 Find out about the wall construction of the cabins of 
large commercial airplanes, the range of ambient conditions 
under which they operate, typical heat transfer coefficients on 
the inner and outer surfaces of the wall, and the heat generation 
rates inside. Determine the size of the heating and air- 
conditioning system that will be able to maintain the cabin 
at 20°C at all times for an airplane capable of carrying 
400 people. 

3-182 Repeat Problem 3-181 for a submarine with a crew of 
60 people. 

3-183 A house with 200-m 2 floor space is to be heated with 
geothermal water flowing through pipes laid in the ground 
under the floor. The walls of the house are 4 m high, and there 
are 10 single-paned windows in the house that are 1.2 m wide 
and 1.8 m high. The house has R-X9 (in h • ft 2 • °F/Btu) insula- 
tion in the walls and R-30 on the ceiling. The floor temperature 
is not to exceed 40°C. Hot geothermal water is available at 
90°C, and the inner and outer diameter of the pipes to be used 
are 2.4 cm and 3.0 cm. Design such a heating system for this 
house in your area. 

3-184 Using a timer (or watch) and a thermometer, conduct 
this experiment to determine the rate of heat gain of your 
refrigerator. First, make sure that the door of the refrigerator 
is not opened for at least a few hours to make sure that steady 
operating conditions are established. Start the timer when the 
refrigerator stops running and measure the time At t it stays off 



cen58933_ch03.qxd 9/10/2002 9:00 AM Page 207 



before it kicks in. Then measure the time At 2 it stays on. Not- 
ing that the heat removed during At 2 is equal to the heat gain of 
the refrigerator during At { + At 2 and using the power con- 
sumed by the refrigerator when it is running, determine the av- 
erage rate of heat gain for your refrigerator, in watts. Take the 
COP (coefficient of performance) of your refrigerator to be 1 .3 
if it is not available. 



207 
CHAPTER 3 



Now, clean the condenser coils of the refrigerator and re- 
move any obstacles on the way of airflow through the coils By 
replacing these measurements, determine the improvement in 
the COP of the refrigerator. 



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cen58933_ch04.qxd 9/10/2002 9:12 AM Page 209 



TRANSIENT HEAT 
CONDUCTION 



CHAPTER 



The temperature of a body, in general, varies with time as well 
as position. In rectangular coordinates, this variation is expressed as 
T(x, y, z, t), where (x, y, z) indicates variation in the x, y, and z directions, 
respectively, and t indicates variation with time. In the preceding chapter, we 
considered heat conduction under steady conditions, for which the tempera- 
ture of a body at any point does not change with time. This certainly simpli- 
fied the analysis, especially when the temperature varied in one direction only, 
and we were able to obtain analytical solutions. In this chapter, we consider 
the variation of temperature with time as well as position in one- and multi- 
dimensional systems. 

We start this chapter with the analysis of lumped systems in which the tem- 
perature of a solid varies with time but remains uniform throughout the solid 
at any time. Then we consider the variation of temperature with time as well 
as position for one-dimensional heat conduction problems such as those asso- 
ciated with a large plane wall, a long cylinder, a sphere, and a semi-infinite 
medium using transient temperature charts and analytical solutions. Finally, 
we consider transient heat conduction in multidimensional systems by uti- 
lizing the product solution. 



CONTENTS 

4-1 Lumped Systems Analysis 210 

4-2 Transient Heat Conduction 
in Large Plane Walls, Long 
Cylinders, and Spheres 
with Spatial Effects 216 

4-3 Transient Heat Conduction 
in Semi-Infinite Solids 228 

4-4 Transient Heat Conduction in 

Multidimensional Systems 231 

Topic of Special Interest: 

Refrigeration and 
Freezing of Foods 239 



cen58933_ch04.qxd 9/10/2002 9:12 AM Page 21C 



210 
HEAT TRANSFER 




(a) Copper ball 




(b) Roast beef 

FIGURE 4-1 

A small copper ball can be modeled 
as a lumped system, but a roast 
beef cannot. 




SOLID BODY 



m = mass 
V = volume 



: density 

= initial temperature 



T = T(t) 



Q = hA s [T„-T(t)] 

FIGURE 4-2 

The geometry and parameters 
involved in the lumped 
system analysis. 



4-1 ■ LUMPED SYSTEM ANALYSIS 

In heat transfer analysis, some bodies are observed to behave like a "lump" 
whose interior temperature remains essentially uniform at all times during a 
heat transfer process. The temperature of such bodies can be taken to be a 
function of time only, T(t). Heat transfer analysis that utilizes this idealization 
is known as lumped system analysis, which provides great simplification 
in certain classes of heat transfer problems without much sacrifice from 
accuracy. 

Consider a small hot copper ball coming out of an oven (Fig. 4-1). Mea- 
surements indicate that the temperature of the copper ball changes with time, 
but it does not change much with position at any given time. Thus the tem- 
perature of the ball remains uniform at all times, and we can talk about the 
temperature of the ball with no reference to a specific location. 

Now let us go to the other extreme and consider a large roast in an oven. If 
you have done any roasting, you must have noticed that the temperature dis- 
tribution within the roast is not even close to being uniform. You can easily 
verify this by taking the roast out before it is completely done and cutting it in 
half. You will see that the outer parts of the roast are well done while the cen- 
ter part is barely warm. Thus, lumped system analysis is not applicable in this 
case. Before presenting a criterion about applicability of lumped system 
analysis, we develop the formulation associated with it. 

Consider a body of arbitrary shape of mass m, volume V, surface area A s , 
density p, and specific heat C p initially at a uniform temperature T { (Fig. 4-2). 
At time t = 0, the body is placed into a medium at temperature r„, and heat 
transfer takes place between the body and its environment, with a heat trans- 
fer coefficient h. For the sake of discussion, we will assume that T m > T t , but 
the analysis is equally valid for the opposite case. We assume lumped system 
analysis to be applicable, so that the temperature remains uniform within the 
body at all times and changes with time only, T = T(t). 

During a differential time interval dt, the temperature of the body rises by a 
differential amount dT. An energy balance of the solid for the time interval dt 
can be expressed as 



/Heat transfer into the body\ 






during dt 



( The increase in the | 

energy of the body 

\ during dt j 



or 



hA s {T a -T)dt = mC„ dT 



(4-1) 



Noting that m = pV and dT = d(T — r„) since T„ = constant, Eq. 4-1 can be 
rearranged as 



d(T- TJ) 
T-T a 



hA s 

pvc„ 



dt 



(4-2) 



Integrating from t = 0, at which T = T t , to any time t, at which T = T(t), gives 



In 



Tit) - T„ 

T; ~ fZ 



hA s 

pvc p ' 



(4-3) 



cen58933_ch04.qxd 9/10/2002 9:12 AM Page 211 



Taking the exponential of both sides and rearranging, we obtain 

T(t) - T m _„, 



where 



pvc. 



(1/s) 



(4-4) 



(4-5) 



is a positive quantity whose dimension is (time) -1 . The reciprocal of b has 
time unit (usually s), and is called the time constant. Equation 4-4 is plotted 
in Fig. 4-3 for different values of b. There are two observations that can be 
made from this figure and the relation above: 

1. Equation 4-4 enables us to determine the temperature T(t) of a body at 
time t, or alternatively, the time t required for the temperature to reach 
a specified value T(t). 

2. The temperature of a body approaches the ambient temperature T^ 
exponentially. The temperature of the body changes rapidly at the 
beginning, but rather slowly later on. A large value of b indicates that 
the body will approach the environment temperature in a short time. 
The larger the value of the exponent b, the higher the rate of decay in 
temperature. Note that b is proportional to the surface area, but inversely 
proportional to the mass and the specific heat of the body. This is not 
surprising since it takes longer to heat or cool a larger mass, especially 
when it has a large specific heat. 

Once the temperature T(t) at time t is available from Eq. A-A, the rate of con- 
vection heat transfer between the body and its environment at that time can be 
determined from Newton's law of cooling as 



Q(t) = hA s [T(t) - rj 



(W) 



(4-6) 



The total amount of heat transfer between the body and the surrounding 
medium over the time interval t = to t is simply the change in the energy 
content of the body: 



Q = mC.[T(!) ~ n 



(kJ) 



(4-7) 



The amount of heat transfer reaches its upper limit when the body reaches the 
surrounding temperature T„. Therefore, the maximum heat transfer between 
the body and its surroundings is (Fig. 4-4) 



e„ 



mC p {T a - r,) 



(kJ) 



(4-8) 



We could also obtain this equation by substituting the T(t) relation from Eq. 
4-4 into the Q(t) relation in Eq. 4-6 and integrating it from t = to t — > °°. 



Criteria for Lumped System Analysis 

The lumped system analysis certainly provides great convenience in heat 
transfer analysis, and naturally we would like to know when it is appropriate 



211 
CHAPTER 4 



Tin 




FIGURE 4-3 

The temperature of a lumped 

system approaches the environment 

temperature as time gets larger. 





1 = 


h 


t 


— > 00 


T t 


T, 




T a 


T„ 


T> 






r«, 


T T rJ -_ 


r, 


< 




T a 


r» 




Q-- 


tsraax — 


">C p (T r 


-TJ 








FIGURE 4-4 


Heat transfer to 


or from 


a body 



reaches its maximum value 

when the body reaches 

the environment temperature. 



cen58933_ch04.qxd 9/10/2002 9:12 AM Page 212 



212 
HEAT TRANSFER 



Convection 



» Conduction . 



SOLID 
BODY 



to use it. The first step in establishing a criterion for the applicability of the 
lumped system analysis is to define a characteristic length as 



and a Biot number Bi as 



Bi 



hL c 



(4-9) 



Bi 



heat convection 



heat conduction 

FIGURE 4-5 

The Biot number can be viewed as the 
ratio of the convection at the surface 
to conduction within the body. 



It can also be expressed as (Fig. 4-5) 
Bi 



h AT Convection at the surface of the body 
~klL,.~AT = ~ 



or 



Bi 



LJk 



Conduction within the body 



Conduction resistance within the body 



\lh Convection resistance at the surface of the body 



When a solid body is being heated by the hotter fluid surrounding it (such as 
a potato being baked in an oven), heat is first convected to the body and 
subsequently conducted within the body. The Biot number is the ratio of the 
internal resistance of a body to heat conduction to its external resistance to 
heat convection. Therefore, a small Biot number represents small resistance 
to heat conduction, and thus small temperature gradients within the body. 

Lumped system analysis assumes a uniform temperature distribution 
throughout the body, which will be the case only when the thermal resistance 
of the body to heat conduction (the conduction resistance) is zero. Thus, 
lumped system analysis is exact when Bi = and approximate when Bi > 0. 
Of course, the smaller the Bi number, the more accurate the lumped system 
analysis. Then the question we must answer is, How much accuracy are we 
willing to sacrifice for the convenience of the lumped system analysis? 

Before answering this question, we should mention that a 20 percent 
uncertainty in the convection heat transfer coefficient h in most cases is con- 
sidered "normal" and "expected." Assuming h to be constant and uniform is 
also an approximation of questionable validity, especially for irregular geome- 
tries. Therefore, in the absence of sufficient experimental data for the specific 
geometry under consideration, we cannot claim our results to be better than 
±20 percent, even when Bi = 0. This being the case, introducing another 
source of uncertainty in the problem will hardly have any effect on the over- 
all uncertainty, provided that it is minor. It is generally accepted that lumped 
system analysis is applicable if 

Bi<0.1 



When this criterion is satisfied, the temperatures within the body relative to 
the surroundings (i.e., T — T m ) remain within 5 percent of each other even for 
well-rounded geometries such as a spherical ball. Thus, when Bi < 0.1, the 
variation of temperature with location within the body will be slight and can 
reasonably be approximated as being uniform. 



cen58933_ch04.qxd 9/10/2002 9:12 AM Page 213 



213 
CHAPTER 4 



The first step in the application of lumped system analysis is the calculation 
of the Biot number, and the assessment of the applicability of this approach. 
One may still wish to use lumped system analysis even when the criterion 
Bi < 0.1 is not satisfied, if high accuracy is not a major concern. 

Note that the Biot number is the ratio of the convection at the surface to con- 
duction within the body, and this number should be as small as possible for 
lumped system analysis to be applicable. Therefore, small bodies with high 
thermal conductivity are good candidates for lumped system analysis, es- 
pecially when they are in a medium that is a poor conductor of heat (such as 
air or another gas) and motionless. Thus, the hot small copper ball placed in 
quiescent air, discussed earlier, is most likely to satisfy the criterion for 
lumped system analysis (Fig. 4-6). 

Some Remarks on Heat Transfer in Lumped Systems 

To understand the heat transfer mechanism during the heating or cooling of a 
solid by the fluid surrounding it, and the criterion for lumped system analysis, 
consider this analogy (Fig. 4-7). People from the mainland are to go by boat 
to an island whose entire shore is a harbor, and from the harbor to their desti- 
nations on the island by bus. The overcrowding of people at the harbor de- 
pends on the boat traffic to the island and the ground transportation system on 
the island. If there is an excellent ground transportation system with plenty of 
buses, there will be no overcrowding at the harbor, especially when the boat 
traffic is light. But when the opposite is true, there will be a huge overcrowd- 
ing at the harbor, creating a large difference between the populations at the 
harbor and inland. The chance of overcrowding is much lower in a small is- 
land with plenty of fast buses. 

In heat transfer, a poor ground transportation system corresponds to poor 
heat conduction in a body, and overcrowding at the harbor to the accumulation 
of heat and the subsequent rise in temperature near the surface of the body 
relative to its inner parts. Lumped system analysis is obviously not applicable 
when there is overcrowding at the surface. Of course, we have disregarded 
radiation in this analogy and thus the air traffic to the island. Like passengers 
at the harbor, heat changes vehicles at the surface from convection to conduc- 
tion. Noting that a surface has zero thickness and thus cannot store any energy, 
heat reaching the surface of a body by convection must continue its journey 
within the body by conduction. 

Consider heat transfer from a hot body to its cooler surroundings. Heat will 
be transferred from the body to the surrounding fluid as a result of a tempera- 
ture difference. But this energy will come from the region near the surface, 
and thus the temperature of the body near the surface will drop. This creates a 
temperature gradient between the inner and outer regions of the body and ini- 
tiates heat flow by conduction from the interior of the body toward the outer 
surface. 

When the convection heat transfer coefficient h and thus convection heat 
transfer from the body are high, the temperature of the body near the surface 
will drop quickly (Fig. 4-8). This will create a larger temperature difference 
between the inner and outer regions unless the body is able to transfer heat 
from the inner to the outer regions just as fast. Thus, the magnitude of the 
maximum temperature difference within the body depends strongly on the 
ability of a body to conduct heat toward its surface relative to the ability of 




15W/m 2 -°C 



V k n ° 3 , 

— -^ 4-£> = 0.02 m 



1 



f A s kD 2 6 

Bi = jfc = j5xap2 = o,ooo 75 < p.] 
* 401 

FIGURE 4-6 

Small bodies with high thermal 

conductivities and low convection 

coefficients are most likely 

to satisfy the criterion for 

lumped system analysis. 




FIGURE 4-7 

Analogy between heat transfer to a 

solid and passenger traffic 

to an island. 




Convection 



/? = 2000W/m 2 -°C 

FIGURE 4-8 

When the convection coefficient h is 

high and k is low, large temperature 

differences occur between the inner 

and outer regions of a large solid. 



cen58933_ch04.qxd 9/10/2002 9:12 AM Page 214 



214 
HEAT TRANSFER 



the surrounding medium to convect this heat away from the surface. The 
Biot number is a measure of the relative magnitudes of these two competing 
effects. 

Recall that heat conduction in a specified direction n per unit surface area is 
expressed as q = —k dT/dn, where dT/dn is the temperature gradient and k is 
the thermal conductivity of the solid. Thus, the temperature distribution in the 
body will be uniform only when its thermal conductivity is infinite, and no 
such material is known to exist. Therefore, temperature gradients and thus 
temperature differences must exist within the body, no matter how small, in 
order for heat conduction to take place. Of course, the temperature gradient 
and the thermal conductivity are inversely proportional for a given heat flux. 
Therefore, the larger the thermal conductivity, the smaller the temperature 
gradient. 



Thermocouple 



Gas 
T ,h *" ^> Junction 



~D = 1 mm 

T(t) 

FIGURE 4-9 

Schematic for Example 4-1. 



EXAMPLE 4-1 Temperature Measurement by Thermocouples 

The temperature of a gas stream is to be measured by a thermocouple whose 
junction can be approximated as a 1-mm-diameter sphere, as shown in Fig. 
4-9. The properties of the junction are k = 35 W/m • °C, p = 8500 kg/m 3 , and 
C p = 320 J/kg • C C, and the convection heat transfer coefficient between the 
junction and the gas is h = 210 W/m 2 • °C. Determine how long it will take for 
the thermocouple to read 99 percent of the initial temperature difference. 

SOLUTION The temperature of a gas stream is to be measured by a thermo- 
couple. The time it takes to register 99 percent of the initial A T is to be 
determined. 

Assumptions 1 The junction is spherical in shape with a diameter of D = 
0.001 m. 2 The thermal properties of the junction and the heat transfer coeffi- 
cient are constant. 3 Radiation effects are negligible. 

Properties The properties of the junction are given in the problem statement. 
Analysis The characteristic length of the junction is 



V_ 

A, 



ttD 2 



I 



D 



(0.001 m) = 1.67 X 10- 



Then the Biot number becomes 

hL c (210 W/m 2 ■ °C)(1.67 X 10- 4 m) 
Bi = X = 35W/m.°C = 0.001 <0.1 



Therefore, lumped system analysis is applicable, and the error involved in this 
approximation is negligible. 

In order to read 99 percent of the initial temperature difference 7" - T, M 
between the junction and the gas, we must have 



T(t) - 7V„ 



0.01 



For example, when 7" = C C and T„ = 100 C C, a thermocouple is considered to 
have read 99 percent of this applied temperature difference when its reading 
indicates T(t) = 99°C. 



cen58933_ch04.qxd 9/10/2002 9:12 AM Page 215 



The value of the exponent b is 



hA, 



h 



210 W/m 2 • °C 



pC p V pC p L c (8500 kg/m 3 )(320 J/kg • °C)( 1.67 X l(T 4 m) 

We now substitute these values into Eq. 4-4 and obtain 
T(t) ~ T a _ kt 



0.462 s" 



T, 



-> 0.01 



(0.462 s~> 



which yields 



t = 10 s 



Therefore, we must wait at least 10 s for the temperature of the thermocouple 
junction to approach within 1 percent of the initial junction-gas temperature 
difference. 

Discussion Note that conduction through the wires and radiation exchange 
with the surrounding surfaces will affect the result, and should be considered in 
a more refined analysis. 



215 
CHAPTER 4 



EXAMPLE 4-2 Predicting the Time of Death 

A person is found dead at 5 pm in a room whose temperature is 20°C. The tem- 
perature of the body is measured to be 25°C when found, and the heat trans- 
fer coefficient is estimated to be ft = 8 W/m 2 • °C. Modeling the body as a 
30-cm-diameter, 1.70-m-long cylinder, estimate the time of death of that per- 
son (Fig. 4-10). 



SOLUTION A body is found while still warm. The time of death is to be 
estimated. 

Assumptions 1 The body can be modeled as a 30-cm-diameter, 1.70-m-long 
cylinder. 2 The thermal properties of the body and the heat transfer coefficient 
are constant. 3 The radiation effects are negligible. 4 The person was healthy(l) 
when he or she died with a body temperature of 37°C. 

Properties The average human body is 72 percent water by mass, and thus we 
can assume the body to have the properties of water at the average temperature 



of (37 + 25)/2 = 31°C; k 
J/kg • °C (Table A-9). 



0.617 W/m • °C, p = 996 kg/m 3 , and C. = 4178 



Analysis The characteristic length of the body is 



9 T 

TTK- L 



ir(0.15m) 2 (1.7m) 



L =^ = 
c A s 2irr D L + 2-nrj 2ir(0.15 m)(1.7 m) + 2ir(0.15 m) 2 

Then the Biot number becomes 

hL c (8 W/m 2 • °C)(0.0689 m) 



0.0689 m 



Bi 



0.617 W/m 



0.89 > 0.1 




FIGURE 4-10 

Schematic for Example 4-2. 



cen58933_ch04.qxd 9/10/2002 9:12 AM Page 216 



216 
HEAT TRANSFER 



Therefore, lumped system analysis is not applicable. However, we can still use 
it to get a "rough" estimate of the time of death. The exponent b in this case is 



hA s h 



8 W/m 2 ■ °C 



pC p V pC p L c (996kg/m 3 )(4178J/kg • °C)(0.0689 m) 
= 2.79 X 10- 5 s-' 

We now substitute these values into Eq. 4-4, 



T(t) - r» 



which yields 



25 -20 
37-20 



t = 43,860 s = 12.2 h 



Therefore, as a rough estimate, the person died about 12 h before the body was 
found, and thus the time of death is 5 am. This example demonstrates how to 
obtain "ball park" values using a simple analysis. 



4-2 - TRANSIENT HEAT CONDUCTION IN LARGE 
PLANE WALLS, LONG CYLINDERS, AND 
SPHERES WITH SPATIAL EFFECTS 

In Section, 4-1, we considered bodies in which the variation of temperature 
within the body was negligible; that is, bodies that remain nearly isothermal 
during a process. Relatively small bodies of highly conductive materials ap- 
proximate this behavior. In general, however, the temperature within a body 
will change from point to point as well as with time. In this section, we con- 
sider the variation of temperature with time and position in one-dimensional 
problems such as those associated with a large plane wall, a long cylinder, and 
a sphere. 

Consider a plane wall of thickness 2L, a long cylinder of radius r , and 
a sphere of radius r initially at a uniform temperature T t , as shown in Fig. 
4-11. At time t = 0, each geometry is placed in a large medium that is at a 
constant temperature T m and kept in that medium for t > 0. Heat transfer takes 
place between these bodies and their environments by convection with a uni- 
form and constant heat transfer coefficient h. Note that all three cases possess 
geometric and thermal symmetry: the plane wall is symmetric about its center 
plane (x = 0), the cylinder is symmetric about its centerline (r = 0), and the 
sphere is symmetric about its center point (r = 0). We neglect radiation heat 
transfer between these bodies and their surrounding surfaces, or incorporate 
the radiation effect into the convection heat transfer coefficient h. 

The variation of the temperature profile with time in the plane wall is 
illustrated in Fig. 4-12. When the wall is first exposed to the surrounding 
medium at T m < T t at t = 0, the entire wall is at its initial temperature T t . But 
the wall temperature at and near the surfaces starts to drop as a result of heat 
transfer from the wall to the surrounding medium. This creates a temperature 



cen58933_ch04.qxd 9/10/2002 9:12 AM Page 217 



h 



Initially 
T=T, 



L x 



(a) A large plane wall 



Initially 
T=T : 



(b) A long cylinder 




(c) A sphere 



217 
CHAPTER 4 



FIGURE 4-1 1 

Schematic of the simple 
geometries in which heat 
transfer is one-dimensional. 



gradient in the wall and initiates heat conduction from the inner parts of the 
wall toward its outer surfaces. Note that the temperature at the center of the 
wall remains at T t until t = t 2 , and that the temperature profile within the wall 
remains symmetric at all times about the center plane. The temperature profile 
gets flatter and flatter as time passes as a result of heat transfer, and eventually 
becomes uniform at T = T m . That is, the wall reaches thermal equilibrium 
with its surroundings. At that point, the heat transfer stops since there is no 
longer a temperature difference. Similar discussions can be given for the long 
cylinder or sphere. 

The formulation of the problems for the determination of the one- 
dimensional transient temperature distribution T(x, t) in a wall results in a par- 
tial differential equation, which can be solved using advanced mathematical 
techniques. The solution, however, normally involves infinite series, which 
are inconvenient and time-consuming to evaluate. Therefore, there is clear 
motivation to present the solution in tabular or graphical form. However, the 
solution involves the parameters x, L, t, k, a, h, T t , and T m which are too many 
to make any graphical presentation of the results practical. In order to reduce 
the number of parameters, we nondimensionalize the problem by defining the 
following dimensionless quantities: 



Dimensionless temperature: 
Dimensionless distance from the center: 
Dimensionless heat transfer coefficient: 
Dimensionless time: 



Q(x, t) -- 
x 



T(x, t) 



X 
Bi 



L 

= hh 
k 

at 



(Biot number) 
(Fourier number) 



T 
















\" 






t=\ 


\\ 




T-__ 




t=^> 




„ f _> CO 









< 


> ►■ 

L x 


h 


Initially 




r„ 




T = 


?i 




h 



FIGURE 4-1 2 

Transient temperature profiles in a 

plane wall exposed to convection 

from its surfaces for T t > TL. 



The nondimensionalization enables us to present the temperature in terms of 
three parameters only: X, Bi, and t. This makes it practical to present the 
solution in graphical form. The dimensionless quantities defined above for a 
plane wall can also be used for a cylinder or sphere by replacing the space 
variable x by r and the half-thickness L by the outer radius r . Note that 
the characteristic length in the definition of the Biot number is taken to be the 



cen58933_ch04.qxd 9/10/2002 9:12 AM Page 21E 



218 

HEAT TRANSFER 



half-thickness L for the plane wall, and the radius r for the long cylinder and 
sphere instead of VIA used in lumped system analysis. 

The one-dimensional transient heat conduction problem just described can 
be solved exactly for any of the three geometries, but the solution involves in- 
finite series, which are difficult to deal with. However, the terms in the solu- 
tions converge rapidly with increasing time, and for t > 0.2, keeping the first 
term and neglecting all the remaining terms in the series results in an error 
under 2 percent. We are usually interested in the solution for times with 
t > 0.2, and thus it is very convenient to express the solution using this one- 
term approximation, given as 



Plane 

wall: U(JC ' ' Km " 

Cylinder: 9(r, f) cy] 

Sphere: 6(r, f) sph = 



T(x, t) 


-T„ 


Ti ~ 


T. x 


T(r, t) 


- r„ 


Ti- 


r„ 


T(r, t) - 


- r„ 



T s - 7*. 



: A,e" x i T cos (\,x/L), t > 0.2 



= A,e- x i T 7 (X 1 r/O, t > 0.2 
A,e- X ' T ' ', , t>0.2 



(4-10) 
(4-11) 
(4-12) 



where the constants A, and \ l are functions of the Bi number only, and their 
values are listed in Table 4-1 against the Bi number for all three geometries. 
The function J is the zeroth-order Bessel function of the first kind, whose 
value can be determined from Table 4-2. Noting that cos (0) = J (0) = 1 and 
the limit of (sin x)lx is also 1, these relations simplify to the next ones at the 
center of a plane wall, cylinder, or sphere: 



Center of plane wall (x = 0): 



Center of cylinder (r = 0): 



Center of sphere (r = 0): 



T - T 

■'O, wall r r y 1 



A { e-^ T 



0. 



T - T^ 

= A,e- x ' T 



0, cyl r r 'Y 

T n -T. 



J 0, sph 



A,e" x i T 



(4-13) 
(4-14) 
(4-15) 



Once the Bi number is known, the above relations can be used to determine 
the temperature anywhere in the medium. The determination of the constants 
A, and X, usually requires interpolation. For those who prefer reading charts 
to interpolating, the relations above are plotted and the one-term approxima- 
tion solutions are presented in graphical form, known as the transient temper- 
ature charts. Note that the charts are sometimes difficult to read, and they are 
subject to reading errors. Therefore, the relations above should be preferred to 
the charts. 

The transient temperature charts in Figs. 4-13, 4-14, and 4-15 for a large 
plane wall, long cylinder, and sphere were presented by M. P. Heisler in 1947 
and are called Heisler charts. They were supplemented in 1961 with transient 
heat transfer charts by H. Grober. There are three charts associated with each 
geometry: the first chart is to determine the temperature T at the center of the 
geometry at a given time t. The second chart is to determine the temperature 
at other locations at the same time in terms of T . The third chart is to deter- 
mine the total amount of heat transfer up to the time t. These plots are valid 
for t > 0.2. 



cen58933_ch04.qxd 9/10/2002 9:12 AM Page 219 



TABLE 4- 



1 



Coefficients used in the one-term approximate solution of transient one- 
dimensional heat conduction in plane walls, cylinders, and spheres (Bi = hL/k 
for a plane wall of thickness 2L, and Bi = hr lk\ax a cylinder or sphere of 



radius r a ) 










Plane Wall 


Cylinder 


Sphere 


Bi 


X, A, 


\j A,. 


*•! A 



0.01 

0.02 

0.04 

0.06 

0.08 

0.1 

0.2 

0.3 

0.4 

0.5 

0.6 

0.7 

0.8 

0.9 

1.0 

2.0 

3.0 

4.0 

5.0 

6.0 

7.0 

8.0 

9.0 

10.0 

20.0 

30.0 

40.0 

50.0 

100.0 



0.0998 
0.1410 
0.1987 
0.2425 
0.2791 
0.3111 
0.4328 
0.5218 
0.5932 
0.6533 
0.7051 
0.7506 
0.7910 
0.8274 
0.8603 
1.0769 
1.1925 
1.2646 
1.3138 
1.3496 
1.3766 
1.3978 
1.4149 
1.4289 
1.4961 
1.5202 
1.5325 
1.5400 
1.5552 
1.5708 



1.0017 
1.0033 
1.0066 
1.0098 
1.0130 
1.0161 
1.0311 
1.0450 
1.0580 
1.0701 
1.0814 
1.0918 
1.1016 
1.1107 
1.1191 
1.1785 
1.2102 
1.2287 
1.2403 
1.2479 
1.2532 
1.2570 
1.2598 
1.2620 
1.2699 
1.2717 
1.2723 
1.2727 
1.2731 
1.2732 



0.1412 
0.1995 
0.2814 
0.3438 
0.3960 
0.4417 
0.6170 
0.7465 
0.8516 
0.9408 
1.0184 
1.0873 
1.1490 
1.2048 
1.2558 
1.5995 
1.7887 
1.9081 
1.9898 



0490 
0937 
1286 
1566 
1795 
2880 
3261 
3455 
3572 
3809 



2.4048 



0025 
0050 
0099 
0148 
0197 
0246 
0483 
0712 
0931 
1143 
1345 
1539 
1724 
1902 
2071 
3384 
4191 
4698 
5029 
5253 
5411 
5526 
5611 
5677 
5919 
5973 
5993 
6002 
6015 
6021 



0.1730 
0.2445 
0.3450 
0.4217 
0.4860 
0.5423 
0.7593 
0.9208 



0528 
1656 
2644 
3525 
4320 
5044 
5708 
0288 
2889 
4556 
5704 
6537 
7165 
7654 
8044 
8363 
9857 
0372 
0632 
0788 
1102 
1416 



1.0030 
1.0060 
1.0120 
1.0179 
1.0239 
1.0298 
1.0592 
1.0880 
1.1164 
1.1441 
1.1713 
1.1978 
1.2236 
1.2488 
1.2732 
1.4793 
1.6227 
1.7202 
1.7870 
1.8338 
1.8673 
1.8920 
1.9106 
1.9249 
1.9781 
1.9898 
1.9942 
1.9962 
1.9990 
2.0000 



219 


CHAPTER 4 




TABLE 4- 


2 




The zeroth- and first-order 
functions of the first kind 


Bessel 


6 


JJL& 


M& 



0.0 
0.1 
0.2 
0.3 
0.4 

0.5 
0.6 
0.7 
0.8 
0.9 

1.0 
1.1 
1.2 
1.3 
1.4 

1.5 
1.6 
1.7 
1.8 
1.9 



1.0000 
0.9975 
0.9900 
0.9776 
0.9604 

0.9385 
0.9120 
0.8812 
0.8463 
0.8075 

0.7652 
0.7196 
0.6711 
0.6201 
0.5669 

0.5118 
0.4554 
0.3980 
0.3400 
0.2818 

0.2239 
0.1666 
0.1104 
0.0555 
0.0025 

-0.0968 
-0.1850 
-0.2601 
-0.3202 



0.0000 
0.0499 
0.0995 
0.1483 
0.1960 

0.2423 
0.2867 
0.3290 
0.3688 
0.4059 

0.4400 
0.4709 
0.4983 
0.5220 
0.5419 

0.5579 
0.5699 
0.5778 
0.5815 
0.5812 

0.5767 
0.5683 
0.5560 
0.5399 
0.5202 

-0.4708 
0.4097 
0.3391 

-0.2613 



Note that the case 1/Bi = k/hL = corresponds to h — > °°, which corre- 
sponds to the case of specified surface temperature T w . That is, the case in 
which the surfaces of the body are suddenly brought to the temperature T„, 
at t = and kept at T m at all times can be handled by setting /; to infinity 
(Fig. 4-16). 

The temperature of the body changes from the initial temperature T t to the 
temperature of the surroundings T«, at the end of the transient heat conduction 
process. Thus, the maximum amount of heat that a body can gain (or lose if 
Tj > T„) is simply the change in the energy content of the body. That is, 



Q n 



mCJT m - r,) = pVCJT. - T : ) 



(kJ) 



(4-16) 



cen58933_ch04.qxd 9/10/2002 9:12 AM Page 22C 



220 
HEAT TRANSFER 



Az 
T.- 



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(a) Midplane temperature (from M. P. Heisler) 



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Initially 
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(b) Temperature distribution (from M. P. Heisler) (c) Heat transfer (from H. Grober et al.) 

FIGURE 4-13 

Transient temperature and heat transfer charts for a plane wall of thickness 2L initially at a uniform temperature T ( 
subjected to convection from both sides to an environment at temperature T^ with a convection coefficient of h. 

where m is the mass, V is the volume, p is the density, and C p is the specific 
heat of the body. Thus, Q max represents the amount of heat transfer for f — > °°. 
The amount of heat transfer Q at a finite time f will obviously be less than this 



cen58933_ch04.qxd 9/10/2002 9:12 AM Page 221 



221 
CHAPTER 4 






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x = at// - ? 



100 120 140 150 250 350 



(a) Centerline temperature (from M. P. Heisler) 



T-T a 



r„-r„ 



Initially 
T=Tj 



0*- 



1.0 
0.9 

0.8 

0.7 

0.6 

0.5 

0.4 

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0.01 0.1 1.0 10 100 

1 _ k 
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(£>) Temperature distribution (from M. P. Heisler) 



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LO 3 



10 4 



(c) Heat transfer (from H. Grober et al.) 

FIGURE 4-14 

Transient temperature and heat transfer charts for a long cylinder of radius r initially at a uniform temperature T t 
subjected to convection from all sides to an environment at temperature T x with a convection coefficient of h. 



cen58933_ch04.qxd 9/10/2002 9:12 AM Page 222 



222 
HEAT TRANSFER 



T; ~ T^ 



1.0 

0.7 

0.5 
0.4 
0.3 

0.2 

0.1 
0.07 
0.05 
0.04 
0.03 

0.02 

0.01 

0.007 
0.005 
0.004 
0.003 

0.002 
0.001 



------- 


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0.5 1.0 1.5 



2.5 3 4 5 6 7 8 9 10 20 
x = at/r? 



30 40 50 100 150 200 250 



(a) Midpoint temperature (from M. P. Heisler) 



T-T rr 



T a -T_ 
1.0 

0.9 

0.8 

0.7 

0.6 

0.5 

0.4 

0.3 

0.2 

0.1 


0.01 










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^ 




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— 


1 




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100 




10"' 1 

Bi 2 x = h 2 at/k 2 



(b) Temperature distribution (from M. P. Heisler) (c) Heat transfer (from H. Grober et al.) 

FIGURE 4-15 

Transient temperature and heat transfer charts for a sphere of radius r initially at a uniform temperature T t subjected to 
convection from all sides to an environment at temperature T^ with a convection coefficient of h. 



maximum. The ratio Q/Q max is plotted in Figures 4-13c, 4-14c, and 4-15c 
against the variables Bi and h 2 at/k 2 for the large plane wall, long cylinder, and 



cen58933_ch04.qxd 9/10/2002 9:12 AM Page 223 



sphere, respectively. Note that once the fraction of heat transfer QIQ mzx has 
been determined from these charts for the given t, the actual amount of heat 
transfer by that time can be evaluated by multiplying this fraction by <2 max - 
A negative sign for <2 max indicates that heat is leaving the body (Fig. 4-17). 
The fraction of heat transfer can also be determined from these relations, 
which are based on the one-term approximations already discussed: 



Plane wall: 

Cylinder: 

Sphere: 



Q 



Q 



sin A. , 



max/ wal , 

Q 



■/i(Xi) 

e- 1 20 | 
max/ cy | A l 



xl 111 A 



1 -36, 



O.sph 



sph 



M 



(4-17) 
(4-18) 
(4-19) 



The use of the Heisler/Grober charts and the one-term solutions already dis- 
cussed is limited to the conditions specified at the beginning of this section: 
the body is initially at a uniform temperature, the temperature of the medium 
surrounding the body and the convection heat transfer coefficient are constant 
and uniform, and there is no energy generation in the body. 

We discussed the physical significance of the Biot number earlier and indi- 
cated that it is a measure of the relative magnitudes of the two heat transfer 
mechanisms: convection at the surface and conduction through the solid. 
A small value of Bi indicates that the inner resistance of the body to heat con- 
duction is small relative to the resistance to convection between the surface 
and the fluid. As a result, the temperature distribution within the solid be- 
comes fairly uniform, and lumped system analysis becomes applicable. Recall 
that when Bi < 0.1, the error in assuming the temperature within the body to 
be uniform is negligible. 

To understand the physical significance of the Fourier number t, we ex- 
press it as (Fig. 4-18) 



at _ ^L-(l/L) Ar. 
L 2 pC p LVt AT 



The rate at which heat is conducted 
across L of a body of volume I? 

The rate at which heat is stored 
in a body of volume L 3 



(4-20) 



223 
CHAPTER 4 




(a) Finite convection coefficient 




(b) Infinite convection coefficient 

FIGURE 4-16 

The specified surface 

temperature corresponds to the case 

of convection to an environment at 

T*, with a convection coefficient h 

that is infinite. 



Therefore, the Fourier number is a measure of heat conducted through a body 
relative to heat stored. Thus, a large value of the Fourier number indicates 
faster propagation of heat through a body. 

Perhaps you are wondering about what constitutes an infinitely large plate 
or an infinitely long cylinder. After all, nothing in this world is infinite. A plate 
whose thickness is small relative to the other dimensions can be modeled as 
an infinitely large plate, except very near the outer edges. But the edge effects 
on large bodies are usually negligible, and thus a large plane wall such as the 
wall of a house can be modeled as an infinitely large wall for heat transfer pur- 
poses. Similarly, a long cylinder whose diameter is small relative to its length 
can be analyzed as an infinitely long cylinder. The use of the transient tem- 
perature charts and the one-term solutions is illustrated in the following 
examples. 



cen58933_ch04.qxd 9/10/2002 9:12 AM Page 224 



224 
HEAT TRANSFER 



t = 





t = 



(a) Maximum heat transfer (t — » °°) 

Q 





Bi = .. 

Ifat 

k 2 



Bi 2 x = 



(Grober chart) 



(b) Actual heat transfer for time t 

FIGURE 4-17 

The fraction of total heat transfer 
Q/Qmax U P t° a specified time t is 
determined using the Grober charts. 



i 

y 



Fourier number: x : 



at 
L 2 



*- stored 

FIGURE 4-18 

Fourier number at time / can be 
viewed as the ratio of the rate of heat 
conducted to the rate of heat stored 
at that time. 



EXAMPLE 4-3 Boiling Eggs 

An ordinary egg can be approximated as a 5-cm-diameter sphere (Fig. 4-19). 
The egg is initially at a uniform temperature of 5 C C and is dropped into boil- 
ing water at 95°C. Taking the convection heat transfer coefficient to be 
h = 1200 W/m 2 ■ °C, determine how long it will take for the center of the egg 
to reach 70°C. 



SOLUTION An egg is cooked in boiling water. The cooking time of the egg is to 
be determined. 

Assumptions 1 The egg is spherical in shape with a radius of r = 2.5 cm. 
2 Heat conduction in the egg is one-dimensional because of thermal symmetry 
about the midpoint. 3 The thermal properties of the egg and the heat transfer 
coefficient are constant. 4 The Fourier number is t > 0.2 so that the one-term 
approximate solutions are applicable. 

Properties The water content of eggs is about 74 percent, and thus the ther- 
mal conductivity and diffusivity of eggs can be approximated by those of water 
at the average temperature of (5 + 70)/2 = 37.5°C; k = 0.627 W/m • °C and 
a = k/pC p = 0.151 X 10- 5 m 2 /s (Table A-9). 

Analysis The temperature within the egg varies with radial distance as well as 
time, and the temperature at a specified location at a given time can be deter- 
mined from the Heisler charts or the one-term solutions. Here we will use the 
latter to demonstrate their use. The Biot number for this problem is 



Bi 






(1200 W/m 2 • °C)(0.025m) 



0.627 W/m • °C 



47.8 



which is much greater than 0.1, and thus the lumped system analysis is not 
applicable. The coefficients X 1 and A 1 for a sphere corresponding to this Bi are, 
from Table 4-1, 



3.0753, 



1.9958 



Substituting these and other values into Eq. 4-15 and solving for t gives 
r - 7L 



T; ~ T„ 



A,e~V 



70 — 95 
-> - „: = 1.9958e-' 30753 ^ 



5-95 



-> t = 0.209 



which is greater than 0.2, and thus the one-term solution is applicable with an 
error of less than 2 percent. Then the cooking time is determined from the de- 
finition of the Fourier number to be 



tt 2 _ (0.209)(0.025 m) 2 
~° r ~ 0.151 X 10- 6 m 2 /s 



865 s = 14.4 min 



Therefore, it will take about 15 min for the center of the egg to be heated from 
5°C to 70°C. 

Discussion Note that the Biot number in lumped system analysis was defined 
differently as Bi = hL c /k = h{r/3)/k. However, either definition can be used in 
determining the applicability of the lumped system analysis unless Bi ~ 0.1. 



cen58933_ch04.qxd 9/10/2002 9:12 AM Page 225 



225 
CHAPTER 4 



EXAMPLE 4-4 Heating of Large Brass Plates in an Oven 

In a production facility, large brass plates of 4 cm thickness that are initially at 
a uniform temperature of 20 C C are heated by passing them through an oven 
that is maintained at 500 C C (Fig. 4-20). The plates remain in the oven for a 
period of 7 min. Taking the combined convection and radiation heat transfer 
coefficient to be h = 120 W/m 2 ■ °C, determine the surface temperature of the 
plates when they come out of the oven. 

SOLUTION Large brass plates are heated in an oven. The surface temperature 
of the plates leaving the oven is to be determined. 

Assumptions 1 Heat conduction in the plate is one-dimensional since the plate 
is large relative to its thickness and there is thermal symmetry about the center 
plane. 2 The thermal properties of the plate and the heat transfer coefficient are 
constant. 3 The Fourier number is t > 0.2 so that the one-term approximate so- 
lutions are applicable. 

Properties The properties of brass at room temperature are k = 110 W/m • °C, 
p = 8530 kg/m 3 , C p = 380 J/kg • °C, and a = 33.9 X lO" 6 m 2 /s (Table A-3). 
More accurate results are obtained by using properties at average temperature. 
Analysis The temperature at a specified location at a given time can be de- 
termined from the Heisler charts or one-term solutions. Here we will use the 
charts to demonstrate their use. Noting that the half-thickness of the plate is 
L = 0.02 m, from Fig. 4-13 we have 



0.46 



1 k 100W/m-°C 


T - 

■*■ o 




Bi hL (120 W/m 2 • °C)(0.02 m) 


- T„ 


at (33.9 X 10- 6 m 2 /s)(7 X 60 s) 


T t - 


- 71 


T L 2 (0.02 m) 2 " 5 - e \ 





Also, 



1 


k 


Bi 


hL 


X 


L 


L 


L 



45.8 



0.99 



Therefore, 



T - 71 T- 71 T ~ T„ 



T, - T„ T - 71 T t - 71 



0.46 X 0.99 = 0.455 



and 



T = 71 + 0.455(7, - 71) = 500 + 0.455(20 - 500) = 282°C 



Therefore, the surface temperature of the plates will be 282°C when they leave 
the oven. 

Discussion We notice that the Biot number in this case is Bi = 1/45.8 = 
0.022, which is much less than 0.1. Therefore, we expect the lumped system 
analysis to be applicable. This is also evident from (7~- TJ/(T - 7"J = 0.99, 
which indicates that the temperatures at the center and the surface of the plate 
relative to the surrounding temperature are within 1 percent of each other. 







Egg X 






T. = 5°C j 


h = 


1200W/m 2 -°C 


7. 


= 95°C 





FIGURE 4-1 9 

Schematic for Example 4-3. 



T rr _ = 500°C 




h = 120 W/m 2 -°C 






/ 


2L = 4cm 


Brass 




plate 




J, = 20°C 






FIGURE 4-20 


Schematic for Example 4-4. 



cen58933_ch04.qxd 9/10/2002 9:12 AM Page 226 



226 
HEAT TRANSFER 



Noting that the error involved in reading the Heisler charts is typically at least a 
few percent, the lumped system analysis in this case may yield just as accurate 
results with less effort. 

The heat transfer surface area of the plate is 2/1, where A is the face area of 
the plate (the plate transfers heat through both of its surfaces), and the volume 
of the plate is V = (2L)A, where L is the half-thickness of the plate. The expo- 
nent b used in the lumped system analysis is determined to be 



hA s 
pC p V 



h(2A) 



h 



P C„(2LA) P C„L 
120 W/m 2 • °C 



(8530 kg/m 3 )(380 J/kg • °C)(0.02 m) 



0.00185 s- 



Then the temperature of the plate at t = 7 min = 420 s is determined from 



Tit) 



T(t) - 500 



It yields 



20 - 500 



T(t ) = 279°C 



-(0.00185 s - ')(420s) 



which is practically identical to the result obtained above using the Heisler 
charts. Therefore, we can use lumped system analysis with confidence when the 
Biot number is sufficiently small. 



T x = 200°C 

h = 80W/m 2 -°C 



Stainless steel 
shaft 



T : = 600°C 



D = 20 cm 



FIGURE 4-21 

Schematic for Example 4-5. 



EXAMPLE 4-5 Cooling of a Long 

Stainless Steel Cylindrical Shaft 

A long 20-cm-diameter cylindrical shaft made of stainless steel 304 comes out 
of an oven at a uniform temperature of 600°C (Fig. 4-21). The shaft is then al- 
lowed to cool slowly in an environment chamber at 200°C with an average heat 
transfer coefficient of h = 80 W/m 2 • °C. Determine the temperature at the cen- 
ter of the shaft 45 min after the start of the cooling process. Also, determine 
the heat transfer per unit length of the shaft during this time period. 

SOLUTION A long cylindrical shaft at 600°C is allowed to cool slowly. The cen- 
ter temperature and the heat transfer per unit length are to be determined. 
Assumptions 1 Heat conduction in the shaft is one-dimensional since it is long 
and it has thermal symmetry about the centerline. 2 The thermal properties of 
the shaft and the heat transfer coefficient are constant. 3 The Fourier number 
is t > 0.2 so that the one-term approximate solutions are applicable. 
Properties The properties of stainless steel 304 at room temperature 
are k = 14.9 W/m ■ °C, p = 7900 kg/m 3 , C p = 477 J/kg • °C, and 
a = 3.95 X 10~ 6 m 2 /s (Table A-3). More accurate results can be obtained by 
using properties at average temperature. 

Analysis The temperature within the shaft may vary with the radial distance r 
as well as time, and the temperature at a specified location at a given time can 



cen58933_ch04.qxd 9/10/2002 9:12 AM Page 227 



be determined from the Heisler charts. Noting that the radius of the shaft is 
r = 0.1 m, from Fig. 4-14 we have 



1 



14.9 W/m • °C 



1.86 



Bi hr (80 W/m 2 • °C)(0.1 m) 

_ at _ (3.95 X 10- 6 m 2 /s)(45 X 60 s) 
J ~Ji ~ (0.1m) 2 



1.07 



0.40 



and 



T = r„ + 0.4(7; - 7*.) = 200 + 0.4(600 - 200) = 360°C 



Therefore, the center temperature of the shaft will drop from 600°C to 360°C 
in 45 min. 

To determine the actual heat transfer, we first need to calculate the maximum 
heat that can be transferred from the cylinder, which is the sensible energy of 
the cylinder relative to its environment. Taking L = 1 m, 

m = pV = pirr 2 L = (7900 kg/m 3 )-jr(0.1 m) 2 (l m) = 248.2 kg 
e raax = mC p (T„ - T,) = (248.2 kg)(0477 kJ/kg ■ °C)(600 - 200)°C 
= 47,354 kJ 

The dimensionless heat transfer ratio is determined from Fig. 4-14cfor a long 
cylinder to be 



Bi 



1 



1 



1/Bi 1.86 



0.537 



Q 



h 2 at 



Bi 2 T = (0.537) 2 (1.07) = 0.309 



Q n 



0.62 



Therefore, 



Q = 0.62g„ 



0.62 X (47,354 kJ) = 29,360 kJ 



which is the total heat transfer from the shaft during the first 45 min of 
the cooling. 

ALTERNATIVE SOLUTION We could also solve this problem using the one-term 
solution relation instead of the transient charts. First we find the Biot number 

_ hr _ (80 W/m 2 • °C)(0.1 m) _ 
Bl " T = 14.9 W/m- °C " °' 537 



The coefficients X 1 and A x for a cylinder corresponding to this Bi are deter- 
mined from Table 4-1 to be 



Substituting these values into Eq. 4-14 gives 
T - 71 



e„ 



rp -j-> /\\& 



1.122e-<°- 970 >-< L07) = 0.41 



227 
CHAPTER 4 



cen58933_ch04.qxd 9/10/2002 9:12 AM Page 22E 



228 
HEAT TRANSFER 



and thus 










T 

± o 


= T a 


f o.4i (r, - 


- rj = 


200 + 0.41(600 - 200) = 364°C 


The value 


of J^XJ for \ : = 


0.97C 


is determined from Table 4-2 to be 0.430. 


Then the fractiona 


1 heat tran 


sfer is 


determined from Eq. 4-18 to be 




Q 

*-max 


= l-26 


UK) 


= 1 - 2 X 0.41 jJUJj = 0.636 


and thus 












Q = 


0.636e ma , 


= 0.636 X (47,354 kJ) = 30,120 kj 


Discussion 


Thes 


ight difference between the two results is due to the reading 


error of the charts 









CO \ 


Plane 
s surface 








T * \ 






h 


n / 


X 



FIGURE 4-22 

Schematic of a semi-infinite body. 



4-3 - TRANSIENT HEAT CONDUCTION 
IN SEMI-INFINITE SOLIDS 

A semi-infinite solid is an idealized body that has a single plane surface and 
extends to infinity in all directions, as shown in Fig. 4-22. This idealized body 
is used to indicate that the temperature change in the part of the body in which 
we are interested (the region close to the surface) is due to the thermal condi- 
tions on a single surface. The earth, for example, can be considered to be a 
semi-infinite medium in determining the variation of temperature near its sur- 
face. Also, a thick wall can be modeled as a semi-infinite medium if all we are 
interested in is the variation of temperature in the region near one of the sur- 
faces, and the other surface is too far to have any impact on the region of in- 
terest during the time of observation. 

Consider a semi-infinite solid that is at a uniform temperature T t . At time 
t = 0, the surface of the solid at x = is exposed to convection by a fluid at a 
constant temperature T m , with a heat transfer coefficient h. This problem can 
be formulated as a partial differential equation, which can be solved analyti- 
cally for the transient temperature distribution T(x, t). The solution obtained is 
presented in Fig. 4-23 graphically for the nondimensionalized temperature 
defined as 



1 - 6(x, f) = 1 



T(x, t) 



T(x,t)-T t 



(4-21) 



against the dimensionless variable x/(2\/at) for various values of the param- 
eter h\faXlk. 

Note that the values on the vertical axis correspond to x = 0, and thus rep- 
resent the surface temperature. The curve hA/ai/k = c° corresponds to /; — > °°, 
which corresponds to the case of specified temperature T m at the surface at 
x = 0. That is, the case in which the surface of the semi-infinite body is sud- 
denly brought to temperature T«, at t = and kept at T„ at all times can be han- 
dled by setting h to infinity. The specified surface temperature case is closely 



cen58933_ch04.qxd 9/10/2002 9:12 AM Page 229 



1.0 



0.5 

0.4 

0.3 

0.2 



0.1 



0.05 
-i 0.04 

0.03 
0.02 



0.01 



229 
CHAPTER 4 























































Ambient 


Ax, t)\ 




























































































































- 












































°.J 














































^p . 






















1 ^ 






















^*0 
























0j \ 











































0.25 



0.5 



0.75 



1.0 



1.25 



1.5 



2Var 



FIGURE 4-23 

Variation of temperature with position and time in a semi-infinite solid initially at T t subjected to convection to an 
environment at T„ with a convection heat transfer coefficient of h (from P. J. Schneider, Ref. 10). 



approximated in practice when condensation or boiling takes place on the 
surface. For a finite heat transfer coefficient /;, the surface temperature 
approaches the fluid temperature T m as the time t approaches infinity. 

The exact solution of the transient one-dimensional heat conduction prob- 
lem in a semi-infinite medium that is initially at a uniform temperature of T, 
and is suddenly subjected to convection at time t = has been obtained, and 
is expressed as 



T(x, t) 



erfc 



hx , 

3 iy 



h 2 at 
k 2 



erfc 



h\/ai 



2y/ai 



(4-22) 



where the quantity erfc (£) is the complementary error function, defined as 



erfc(£) = 1 --^= f e-" 2 du 

Vtt Jo 



(4-23) 



Despite its simple appearance, the integral that appears in the above relation 
cannot be performed analytically. Therefore, it is evaluated numerically for 
different values of £, and the results are listed in Table 4-3. For the special 
case of h — > °o, the surface temperature T s becomes equal to the fluid temper- 
ature r„, and Eq. 4-22 reduces to 



T(x, f) 



erfc 



(4-24) 



cen58933_ch04.qxd 9/10/2002 9:12 AM Page 23C 



230 
HEAT TRANSFER 



TABLE 4-3 



The complementary 


error fu 


iction 


















£ 


erfc (£) 


£ 


erfc (Q 


£ 


erfc (g) 


6 


erfc (Q 


6 


erfc (Q 


e 


erfc (Q 


0.00 


1.00000 


0.38 


0.5910 


0.76 


0.2825 


1.14 


0.1069 


1.52 


0.03159 


1.90 


0.00721 


0.02 


0.9774 


0.40 


0.5716 


0.78 


0.2700 


1.16 


0.10090 


1.54 


0.02941 


1.92 


0.00662 


0.04 


0.9549 


0.42 


0.5525 


0.80 


0.2579 


1.18 


0.09516 


1.56 


0.02737 


1.94 


0.00608 


0.06 


0.9324 


0.44 


0.5338 


0.82 


0.2462 


1.20 


0.08969 


1.58 


0.02545 


1.96 


0.00557 


0.08 


0.9099 


0.46 


0.5153 


0.84 


0.2349 


1.22 


0.08447 


1.60 


0.02365 


1.98 


0.00511 


0.10 


0.8875 


0.48 


0.4973 


0.86 


0.2239 


1.24 


0.07950 


1.62 


0.02196 


2.00 


0.00468 


0.12 


0.8652 


0.50 


0.4795 


0.88 


0.2133 


1.26 


0.07476 


1.64 


0.02038 


2.10 


0.00298 


0.14 


0.8431 


0.52 


0.4621 


0.90 


0.2031 


1.28 


0.07027 


1.66 


0.01890 


2.20 


0.00186 


0.16 


0.8210 


0.54 


0.4451 


0.92 


0.1932 


1.30 


0.06599 


1.68 


0.01751 


2.30 


0.00114 


0.18 


0.7991 


0.56 


0.4284 


0.94 


0.1837 


1.32 


0.06194 


1.70 


0.01612 


2.40 


0.00069 


0.20 


0.7773 


0.58 


0.4121 


0.96 


0.1746 


1.34 


0.05809 


1.72 


0.01500 


2.50 


0.00041 


0.22 


0.7557 


0.60 


0.3961 


0.98 


0.1658 


1.36 


0.05444 


1.74 


0.01387 


2.60 


0.00024 


0.24 


0.7343 


0.62 


0.3806 


1.00 


0.1573 


1.38 


0.05098 


1.76 


0.01281 


2.70 


0.00013 


0.26 


0.7131 


0.64 


0.3654 


1.02 


0.1492 


1.40 


0.04772 


1.78 


0.01183 


2.80 


0.00008 


0.28 


0.6921 


0.66 


0.3506 


1.04 


0.1413 


1.42 


0.04462 


1.80 


0.01091 


2.90 


0.00004 


0.30 


0.6714 


0.68 


0.3362 


1.06 


0.1339 


1.44 


0.04170 


1.82 


0.01006 


3.00 


0.00002 


0.32 


0.6509 


0.70 


0.3222 


1.08 


0.1267 


1.46 


0.03895 


1.84 


0.00926 


3.20 


0.00001 


0.34 


0.6306 


0.72 


0.3086 


1.10 


0.1198 


1.48 


0.03635 


1.86 


0.00853 


3.40 


0.00000 


0.36 


0.6107 


0.74 


0.2953 


1.12 


0.1132 


1.50 


0.03390 


1.88 


0.00784 


3.60 


0.00000 



This solution corresponds to the case when the temperature of the exposed 
surface of the medium is suddenly raised (or lowered) to T s at t = and is 
maintained at that value at all times. Although the graphical solution given in 
Fig. 4-23 is a plot of the exact analytical solution given by Eq. 4-23, it is sub- 
ject to reading errors, and thus is of limited accuracy. 



,T. = -10°C 



Soil 



Water pipe 



r ; =i5°c 



FIGURE 4-24 

Schematic for Example 4-6. 



EXAMPLE 4-6 Minimum Burial Depth of Water Pipes to Avoid 
Freezing 

In areas where the air temperature remains below 0°C for prolonged periods of 
time, the freezing of water in underground pipes is a major concern. Fortu- 
nately, the soil remains relatively warm during those periods, and it takes weeks 
for the subfreezing temperatures to reach the water mains in the ground. Thus, 
the soil effectively serves as an insulation to protect the water from subfreezing 
temperatures in winter. 

The ground at a particular location is covered with snow pack at -10°C for a 
continuous period of three months, and the average soil properties at that loca- 
tion are k = 0.4 W/m • °C and a = 0.15 X 10~ 6 m 2 /s (Fig. 4-24). Assuming an 
initial uniform temperature of 15 C C for the ground, determine the minimum 
burial depth to prevent the water pipes from freezing. 

SOLUTION The water pipes are buried in the ground to prevent freezing. The 
minimum burial depth at a particular location is to be determined. 
Assumptions 1 The temperature in the soil is affected by the thermal condi- 
tions at one surface only, and thus the soil can be considered to be a semi- 
infinite medium with a specified surface temperature of - 10°C. 2 The thermal 
properties of the soil are constant. 



cen58933_ch04.qxd 9/10/2002 9:12 AM Page 231 



Properties The properties of the soil are as given in the problem statement. 
Analysis The temperature of the soil surrounding the pipes will be 0°C after 
three months in the case of minimum burial depth. Therefore, from Fig. 4-23, 
we have 



hVoit 



T(x,t) - T 
1 ~ ' ' - = 1 



(since h —> *■) 

0-(-10) 



H 



15 - (-10) 



0.6 



2 Vat 



0.36 



We note that 

t = (90 days)(24 h/day)(3600 s/h) = 7.78 X 10 6 s 

and thus 

x = 2i Vaf = 2 X 0.36V(0.15 X lO" 6 m 2 /s)(7.78 X 10 6 s) = 0.77 m 

Therefore, the water pipes must be buried to a depth of at least 77 cm to avoid 
freezing under the specified harsh winter conditions. 

ALTERNATIVE SOLUTION The solution of this problem could also be deter- 
mined from Eq. 4-24: 



T(x,t) 



erfc 



2Vat, 



0-15 
-10 - 15 



erfc 



iVat 



0.60 



The argument that corresponds to this value of the complementary error func- 
tion is determined from Table 4-3 to be £ = 0.37. Therefore, 



x = 2£ vaf = 2 X 0.37V(0.15 X lO" 6 m 2 /s)(7.78 X 10 6 s) = 0.80 m 
Again, the slight difference is due to the reading error of the chart. 



231 
CHAPTER 4 



h 



T(r,t) 



Heat 
transfer 



(a) Long cylinder 



4-4 - TRANSIENT HEAT CONDUCTION IN 
MULTIDIMENSIONAL SYSTEMS 

The transient temperature charts presented earlier can be used to determine the 
temperature distribution and heat transfer in one-dimensional heat conduction 
problems associated with a large plane wall, a long cylinder, a sphere, and a 
semi-infinite medium. Using a superposition approach called the product 
solution, these charts can also be used to construct solutions for the two- 
dimensional transient heat conduction problems encountered in geometries 
such as a short cylinder, a long rectangular bar, or a semi-infinite cylinder or 
plate, and even three-dimensional problems associated with geometries such 
as a rectangular prism or a semi-infinite rectangular bar, provided that all sur- 
faces of the solid are subjected to convection to the same fluid at temperature 



h 



T(r,x,t) 



Heat 
' transfer 



(b) Short cylinder (two-dimensional) 

FIGURE 4-25 

The temperature in a short 

cylinder exposed to convection from 

all surfaces varies in both the radial 

and axial directions, and thus heat 

is transferred in both directions. 



cen58933_ch04.qxd 9/10/2002 9:12 AM Page 232 



232 

HEAT TRANSFER 



Plane wall 




^1 

Long 

cylinder 

FIGURE 4-26 

A short cylinder of radius r and 
height a is the intersection of a long 
cylinder of radius r and a plane wall 
of thickness a. 



Plane wall 




Plane wall 

hH 
FIGURE 4-27 

A long solid bar of rectangular 
profile a X b is the intersection 
of two plane walls of 
thicknesses a and b. 



T„, with the same heat transfer coefficient h, and the body involves no heat 
generation (Fig. 4-25). The solution in such multidimensional geometries can 
be expressed as the product of the solutions for the one-dimensional geome- 
tries whose intersection is the multidimensional geometry. 

Consider a short cylinder of height a and radius r initially at a uniform tem- 
perature r,. There is no heat generation in the cylinder. At time t = 0, the 
cylinder is subjected to convection from all surfaces to a medium at temper- 
ature r„ with a heat transfer coefficient h. The temperature within the cylin- 
der will change with x as well as r and time t since heat transfer will occur 
from the top and bottom of the cylinder as well as its side surfaces. That is, 
T = T(r, x, t) and thus this is a two-dimensional transient heat conduction 
problem. When the properties are assumed to be constant, it can be shown that 
the solution of this two-dimensional problem can be expressed as 



T(r, x, t) - T a 



short 
cylinder 



T(x, t) 



plane 
w;ill 



T(r, t) 



infinite 
cylinder 



(4-25) 



That is, the solution for the two-dimensional short cylinder of height a and 
radius r is equal to the product of the nondimensionalized solutions for the 
one-dimensional plane wall of thickness a and the long cylinder of radius r , 
which are the two geometries whose intersection is the short cylinder, as 
shown in Fig. 4-26. We generalize this as follows: the solution for a multi- 
dimensional geometry is the product of the solutions of the one-dimensional 
geometries whose intersection is the multidimensional body. 
For convenience, the one-dimensional solutions are denoted by 



0waii(*. r ) 



(r,f) 



J c\ I 



,-i„f(*, t) 



(T(x, t) 


- T a 


[ T,~ 


r„ 


(T(r, t) 


-rj 


\ T,~ 


r- , 


(T(x, t) 


- r. 



plane 

wall 



infinite 
cylinder 



T, - T m 



(4-26) 



For example, the solution for a long solid bar whose cross section is an a X b 
rectangle is the intersection of the two infinite plane walls of thicknesses 
a and b, as shown in Fig. 4-27, and thus the transient temperature distribution 
for this rectangular bar can be expressed as 



T(x, y, t) 



W-*. OQwaiiCy. 



(4-27) 



The proper forms of the product solutions for some other geometries are given 
in Table 4—4. It is important to note that the x-coordinate is measured from the 
surface in a semi-infinite solid, and from the midplane in a plane wall. The ra- 
dial distance r is always measured from the centerline. 

Note that the solution of a two-dimensional problem involves the product of 
two one-dimensional solutions, whereas the solution of a three-dimensional 
problem involves the product of three one-dimensional solutions. 

A modified form of the product solution can also be used to determine 
the total transient heat transfer to or from a multidimensional geometry by 
using the one-dimensional values, as shown by L. S. Langston in 1982. The 



cen58933_ch04.qxd 9/10/2002 9:13 AM Page 233 



233 
CHAPTER 4 



TABLE 4-4 

Multidimensional solutions expressed as products of one-dimensional solutions for bodies that are initially at a 
uniform temperature 7j and exposed to convection from all surfaces to a medium at T« 



9(''.0 = e cyl (n() 

Infinite cylinder 



Hw.') = e cy|(f ,oe s „, inf (x,0 

Semi-infinite cylinder 



Hx,r,t) = e cyl (r,t)() wM (x,f) 
Short cylinder 




Semi-infinite medium 





K.v,v.O = e semWnf (x,f)e semWnf ^o 

Quarter-infinite medium 



B(x,y,z,t) = 

9 semi-inf (' V > f )8 scmi -i„f & e s emi-inf fc f ) 

Corner region of a large medium 



2L 



e(*,o = e wall teO 

Infinite plate (or plane wall) 



2L 




H^y.O = e wall fcOe semi . m ,Cv.O 

Semi-infinite plate 



Hx,y,z,t) = 

^wall^Oe^.^O'.Oe^.infCZ.O 

Quarter-infinite plate 



i j 
;- 

/ 
/ 


if 



ife^ 



/\ 


1 


1 v 







H^y.O = e wall (x,oe wdl (y,f) 

Infinite rectangular bar 



e(jc,y,z,f) = 

e w a ii(^')e wall Cv,oe scmi , nf (z,o 

Semi-infinite rectangular bar 



Q(x,y,z,t) = 

e „an^oe wall (.v,oe wall ( Z ,o 

Rectangular parallelepiped 



cen58933_ch04.qxd 9/10/2002 9:13 AM Page 234 



234 
HEAT TRANSFER 



transient heat transfer for a two-dimensional geometry formed by the inter- 
section of two one-dimensional geometries 1 and 2 is 



_e 

*iraa 



total, 2D 



_Q_ 

Cm. 



_Q_ 



_Q_ 



(4-28) 



Transient heat transfer for a three-dimensional body formed by the inter- 
section of three one-dimensional bodies 1, 2, and 3 is given by 



_Q_ 

Sc m a 






_Q_ 

iima 



_Q_ 

i/ma 



_Q_ 

fcraa 



_Q_ 

*£ma 



_Q_ 

sima 



(4-29) 



The use of the product solution in transient two- and three-dimensional heat 
conduction problems is illustrated in the following examples. 



r„ = 25°C 

h = 60W/m 2 -°C 



T t = 120 D C 



FIGURE 4-28 

Schematic for Example 4-7. 



EXAMPLE 4-7 Cooling of a Short Brass Cylinder 

" A short brass cylinder of diameter D = 10 cm and height H = 12 cm is initially 
I at a uniform temperature 7) = 120°C. The cylinder is now placed in atmo- 
spheric air at 25°C, where heat transfer takes place by convection, with a heat 
transfer coefficient of h = 60 W/m 2 ■ °C. Calculate the temperature at (a) the 
center of the cylinder and (b) the center of the top surface of the cylinder 
15 min after the start of the cooling. 






SOLUTION A short cylinder is allowed to cool in atmospheric air. The temper- 
atures at the centers of the cylinder and the top surface are to be determined. 



Assumptions 1 Heat conduction in the short cylinder is two-dimensional, and 
thus the temperature varies in both the axial x- and the radial r-directions. 2 The 
thermal properties of the cylinder and the heat transfer coefficient are constant. 
3 The Fourier number is t > 0.2 so that the one-term approximate solutions are 
applicable. 



Properties The properties of brass at room temperature are k= 110 W/m • °C 
and a = 33.9 X 10~ 5 m 2 /s (Table A-3). More accurate results can be obtained 
by using properties at average temperature. 



Analysis (a) This short cylinder can physically be formed by the intersection of 
a long cylinder of radius r = 5 cm and a plane wall of thickness 2L = 12 cm, 
as shown in Fig. 4-28. The dimensionless temperature at the center of the 
plane wall is determined from Figure 4-13a to be 



at (3.39 X KT 5 m 2 /s)(900s) 1 

T = ^ = r^ = 8.48 

U (0.06 m) 2 

J_ == _L- 110 W/m • °c 


' 6wall(0,O = 


T(0,t)-T a 


Bi hL (60 w/m 2 . °c)(0.06 m ) . 





cen58933_ch04.qxd 9/10/2002 9:13 AM Page 235 



Similarly, at the center of the cylinder, we have 
at (3.39 X Kr 5 m 2 /s)(900s) 





r 2 (0.05 m) 2 


1 


_ k _ HOW/m • °C 


Bi 


hr o (60 W/m 2 • °C)(0.05 m) 


dte 


ore, 

/r(o,o,o-r„\ 

1 'Y T 1 Ishort "wall' 

cylinder 



12.2 



36.7 



T(0, t)-T a 

■ e cy[ (o,o= T _ T " = o.5 



and 



T(0,0,t) 



(0, t ) X 6 cyl (0, t ) = 0.8 X 0.5 = 0.4 



0.4(7, - r.) = 25 + 0.4(120 - 25) = 63°C 



This is the temperature at the center of the short cylinder, which is also the cen- 
ter of both the long cylinder and the plate. 

(b) The center of the top surface of the cylinder is still at the center of the long 
cylinder (r = 0), but at the outer surface of the plane wall (x = L). Therefore, 
we first need to find the surface temperature of the wall. Noting that x = L = 
0.06 m, 



x _ 0.06 m 
L ~ 0.06 m 



1 



_L = A = 110 W/m ■ °c 

Bi hL (60 w/m 2 ■ °C)(0.06 m) 



30.6 



T -T x . 



0.98 



Then 



9 wa ii(£> 



T{L,t)-T^ (T(L,t)-TA(T -T, 



T t -T, 

Therefore, 

T(L, 0, t)-T = 



T 71 



T- — T 



X 0.8 = 0.784 



Ti-Tn 



shor. = 6 wall (L, f )0 cyl (0, t) = 0.784 X 0.5 = 0.392 

cylinder 



and 



T(L, 0, t) = T a + 0.392(7; - r„) = 25 + 0.392(120 - 25) = 62.2°C 
which is the temperature at the center of the top surface of the cylinder. 



235 
CHAPTER 4 



J EXAMPLE 4-8 Heat Transfer from a Short Cylinder 






Determine the total heat transfer from the short brass cylinder 
, kg/m 3 , C p = 0.380 kJ/kg • °C) discussed in Example 4-7. 


(P = 


= 8530 



cen58933_ch04.qxd 9/10/2002 9:13 AM Page 236 



236 
HEAT TRANSFER 



SOLUTION We first determine the maximum heat that can be transferred from 
the cylinder, which is the sensible energy content of the cylinder relative to its 
environment: 

m = pV = pirr 2 L = (8530 kg/m 3 )Tr(0.05 m) 2 (0.06 m) = 4.02 kg 
g raax = mC p (T, - r„) = (4.02 kg)(0.380 kJ/kg • °C)(120 - 25)°C = 145.1 kJ 

Then we determine the dimensionless heat transfer ratios for both geometries. 
For the plane wall, it is determined from Fig. 4-13c to be 



Bi = IM = 3^6 = °-° 327 



h 2 at 



Bi 2 T = (0.0327) 2 (8.48) = 0.0091 



!cma 



0.23 



plane 



Similarly, for the cylinder, we have 



Bi = W = i = °- 0272 
1/Bi 36.7 



h 2 at 



Bi 2 T = (0.0272) 2 (12.2) = 0.0090 






0.47 



nfinite 
cylinder 



Then the heat transfer ratio for the short cylinder is, from Eq. 4-28, 



^ max ' short cyl V^max/j VtJmax/, 



_G 



= 0.23 + 0.47(1 - 0.23) = 0.592 

Therefore, the total heat transfer from the cylinder during the first 15 min of 
cooling is 

Q = 0.592g raax = 0.592 X (145.1 kJ) = 85.9 kJ 



EXAMPLE 4-9 Cooling of a Long Cylinder by Water 

A semi-infinite aluminum cylinder of diameter D = 20 cm is initially at a uni- 
form temperature 7", = 200 C C. The cylinder is now placed in water at 15°C 
where heat transfer takes place by convection, with a heat transfer coefficient 
of h = 120 W/m 2 • °C. Determine the temperature at the center of the cylinder 
15 cm from the end surface 5 min after the start of the cooling. 

SOLUTION A semi-infinite aluminum cylinder is cooled by water. The tem- 
perature at the center of the cylinder 15 cm from the end surface is to be 
determined. 

Assumptions 1 Heat conduction in the semi-infinite cylinder is two- 
dimensional, and thus the temperature varies in both the axial x- and the radial 
r-directions. 2 The thermal properties of the cylinder and the heat transfer co- 
efficient are constant. 3 The Fourier number is t > 0.2 so that the one-term 
approximate solutions are applicable. 



cen58933_ch04.qxd 9/10/2002 9:13 AM Page 237 



Properties The properties of aluminum at room temperature are k = 237 
W/m ■ °C and a = 9.71 X 10~ 6 m 2 /s (Table A-3). More accurate results can be 
obtained by using properties at average temperature. 

Analysis This semi-infinite cylinder can physically be formed by the inter- 
section of an infinite cylinder of radius r = 10 cm and a semi-infinite medium, 
as shown in Fig. 4-29. 

We will solve this problem using the one-term solution relation for the cylin- 
der and the analytic solution for the semi-infinite medium. First we consider the 
infinitely long cylinder and evaluate the Biot number: 



Bi 



hr B _ (120 W/m 2 • °C)(0.1 m) 
T~ 237 W/m ■ °C 



0.05 



The coefficients k 1 and A 1 for a cylinder corresponding to this Bi are deter- 
mined from Table 4-1 to be \ x = 0.3126 and A 1 = 1.0124. The Fourier num- 
ber in this case is 



at _ (9.71 X 10- 5 m 2 /s)(5 X 60 s) 
~7;~ (0.1 m) 2 



2.91 >0.2 



and thus the one-term approximation is applicable. Substituting these values 
into Eq. 4-14 gives 



' = B_ 1 (0,0 =A x e- 



1.0124e-<°- 3126 > 2 ( 21)1 > = 0.762 



The solution for the semi-infinite solid can be determined from 

hx , h 2 at 



1 - e s emi-inftM) = erfc 



exp iy + > 



erfc 



hVal 



at 



First we determine the various quantities in parentheses 
x 0.15 m 



0.44 



2Vaf 2V(9.71 X 10- 5 m 2 /s)(5 X 60s) 

hVat _ (120 W/m 2 • °C)V(9.71 X lO" 5 m 2 /s)(300 s) 



k 237 W/m 

hx (120 W/m 2 • °C)(0.15m) 



h 2 at (hVai 
k 2 



237 W/m ■ °C 

2 



°c 

0.0759 



0.086 



(0.086) 2 = 0.0074 



Substituting and evaluating the complementary error functions from Table 4-3, 

e semHnf (Jc, t) = 1 - erfc (0.44) + exp (0.0759 + 0.0074) erfc (0.44 + 0.086) 
= 1 - 0.5338 + exp (0.0833) X 0.457 
= 0.963 

Now we apply the product solution to get 
T(jc, 0, t) - T- 



semi-infinite 
cylinder 



6 S cmMnf(*> O9cyi(0, O = 0.963 X 0.762 = 0.734 



237 
CHAPTER 4 



I 
Tf = 200°C 

I 
D = 20cm 



r« = i5°c 

h =120W/m 2 -°C 



x = 15 cm 



FIGURE 4-29 

Schematic for Example 4-9. 



cen58933_ch04.qxd 9/10/2002 9:13 AM Page 23E 



238 
HEAT TRANSFER 



and 




















T(x,0,t) = 


= T„ + 


0.734(7; 


-rj 


= 15 + 0.734(200 


" 15) = 


151 


°C 


whic 


*\ is the tem 


perature at the 


center 


of the cylinder 15 


cm from 


the 


exposed 


botto 


m surface. 

















5 C F 



35°F 



Steak 




1 in 



FIGURE 4-30 

Schematic for Example 4-10. 



EXAMPLE 4-10 Refrigerating Steaks while Avoiding Frostbite 

In a meat processing plant, 1-in. -thick steaks initially at 75°F are to be cooled 
in the racks of a large refrigerator that is maintained at 5°F (Fig. 4-30). The 
steaks are placed close to each other, so that heat transfer from the 1-in. -thick 
edges is negligible. The entire steak is to be cooled below 45°F, but its temper- 
ature is not to drop below 35°F at any point during refrigeration to avoid "frost- 
bite." The convection heat transfer coefficient and thus the rate of heat transfer 
from the steak can be controlled by varying the speed of a circulating fan in- 
side. Determine the heat transfer coefficient h that will enable us to meet both 
temperature constraints while keeping the refrigeration time to a minimum. The 
steak can be treated as a homogeneous layer having the properties p = 74.9 
lbm/ft 3 , C p = 0.98 Btu/lbm • °F, k = 0.26 Btu/h ■ ft • °F, and a = 0.0035 ft 2 /h. 

SOLUTION Steaks are to be cooled in a refrigerator maintained at 5 C F. The 
heat transfer coefficient that will allow cooling the steaks below 45°F while 
avoiding frostbite is to be determined. 

Assumptions 1 Heat conduction through the steaks is one-dimensional since 
the steaks form a large layer relative to their thickness and there is thermal sym- 
metry about the center plane. 2 The thermal properties of the steaks and the 
heat transfer coefficient are constant. 3 The Fourier number is t > 0.2 so that 
the one-term approximate solutions are applicable. 

Properties The properties of the steaks are as given in the problem statement. 
Analysis The lowest temperature in the steak will occur at the surfaces and 
the highest temperature at the center at a given time, since the inner part will 
be the last place to be cooled. In the limiting case, the surface temperature at 
x = L = 0.5 in. from the center will be 35°F, while the midplane temperature 
is 45°F in an environment at 5°F. Then, from Fig. 4-136, we obtain 



1.5 



x _ 0.5 in. _ 
L ~ 0.5 in. ~ 

T(L,t)-T„ 35-5 

' - M - 75 
T -T m 45-5 U -°J 




■ 1 _ k 
Bi hL 


which gives 


1 k _ 0.26 Btu/h • ft • °F _ 
1.5 L 1.5(0.5/12 ft) 


4.16 Btu/h 



ft 2 • °F 



Discussion The convection heat transfer coefficient should be kept below this 
value to satisfy the constraints on the temperature of the steak during refriger- 
ation. We can also meet the constraints by using a lower heat transfer coeffi- 
cient, but doing so would extend the refrigeration time unnecessarily. 



cen58933_ch04.qxd 9/10/2002 9:13 AM Page 239 



The restrictions that are inherent in the use of Heisler charts and the one- 
term solutions (or any other analytical solutions) can be lifted by using the nu- 
merical methods discussed in Chapter 5. 



239 
CHAPTER 4 



TOPIC OF SPECIAL INTEREST 



Refrigeration and Freezing of Foods 

Control of Microorganisms in Foods 

Microorganisms such as bacteria, yeasts, molds, and viruses are widely 
encountered in air, water, soil, living organisms, and unprocessed food 
items, and cause off-flavors and odors, slime production, changes in the 
texture and appearances, and the eventual spoilage of foods. Holding per- 
ishable foods at warm temperatures is the primary cause of spoilage, and 
the prevention of food spoilage and the premature degradation of quality 
due to microorganisms is the largest application area of refrigeration. The 
first step in controlling microorganisms is to understand what they are and 
the factors that affect their transmission, growth, and destruction. 

Of the various kinds of microorganisms, bacteria are the prime cause for 
the spoilage of foods, especially moist foods. Dry and acidic foods create 
an undesirable environment for the growth of bacteria, but not for the 
growth of yeasts and molds. Molds are also encountered on moist surfaces, 
cheese, and spoiled foods. Specific viruses are encountered in certain ani- 
mals and humans, and poor sanitation practices such as keeping processed 
foods in the same area as the uncooked ones and being careless about hand- 
washing can cause the contamination of food products. 

When contamination occurs, the microorganisms start to adapt to the 
new environmental conditions. This initial slow or no-growth period is 
called the lag phase, and the shelf life of a food item is directly propor- 
tional to the length of this phase (Fig. 4-31). The adaptation period is fol- 
lowed by an exponential growth period during which the population of 
microorganisms can double two or more times every hour under favorable 
conditions unless drastic sanitation measures are taken. The depletion of 
nutrients and the accumulation of toxins slow down the growth and start 
the death period. 

The rate of growth of microorganisms in a food item depends on the 
characteristics of the food itself such as the chemical structure, pH level, 
presence of inhibitors and competing microorganisms, and water activity as 
well as the environmental conditions such as the temperature and relative 
humidity of the environment and the air motion (Fig. 4-32). 

Microorganisms need food to grow and multiply, and their nutritional 
needs are readily provided by the carbohydrates, proteins, minerals, and 
vitamins in a food. Different types of microorganisms have different nu- 
tritional needs, and the types of nutrients in a food determine the types of 
microorganisms that may dwell on them. The preservatives added to the 



*This section can be skipped without a loss of continuity. 



Microorganism 
population 




Time 

FIGURE 4-31 

Typical growth curve of 
microorganisms. 



ENVIRONMENT 



Temperature 
Air motion 




100 



Oxygen 
level 



Relative 
humidity 




Water content 

Chemical composition 

Contamination level 

The use of inhibitors 

pH level 



FIGURE 4-32 

The factors that affect the rate of 
growth of microorganisms. 



cen58933_ch04.qxd 9/10/2002 9:13 AM Page 24C 



240 
HEAT TRANSFER 



Rate of 
growth 




Temperature 

FIGURE 4-33 

The rate of growth of microorganisms 
in a food product increases 
exponentially with increasing 
environmental temperature. 



food may also inhibit the growth of certain microorganisms. Different 
kinds of microorganisms that exist compete for the same food supply, and 
thus the composition of microorganisms in a food at any time depends on 
the initial make-up of the microorganisms. 

All living organisms need water to grow, and microorganisms cannot 
grow in foods that are not sufficiently moist. Microbiological growth in 
refrigerated foods such as fresh fruits, vegetables, and meats starts at the 
exposed surfaces where contamination is most likely to occur. Fresh meat 
in a package left in a room will spoil quickly, as you may have noticed. 
A meat carcass hung in a controlled environment, on the other hand, will 
age healthily as a result of dehydration on the outer surface, which inhibits 
microbiological growth there and protects the carcass. 

Microorganism growth in a food item is governed by the combined ef- 
fects of the characteristics of the food and the environmental factors. We 
cannot do much about the characteristics of the food, but we certainly can 
alter the environmental conditions to more desirable levels through heat- 
ing, cooling, ventilating, humidification, dehumidification, and control of 
the oxygen levels. The growth rate of microorganisms in foods is a strong 
function of temperature, and temperature control is the single most effec- 
tive mechanism for controlling the growth rate. 

Microorganisms grow best at "warm" temperatures, usually between 
20 and 60°C. The growth rate declines at high temperatures, and death 
occurs at still higher temperatures, usually above 70°C for most micro- 
organisms. Cooling is an effective and practical way of reducing the 
growth rate of microorganisms and thus extending the shelf life of perish- 
able foods. A temperature of 4°C or lower is considered to be a safe re- 
frigeration temperature. Sometimes a small increase in refrigeration 
temperature may cause a large increase in the growth rate, and thus a 
considerable decrease in shelf life of the food (Fig. 4-33). The growth 
rate of some microorganisms, for example, doubles for each 3°C rise in 
temperature. 

Another factor that affects microbiological growth and transmission is 
the relative humidity of the environment, which is a measure of the water 
content of the air. High humidity in cold rooms should be avoided since 
condensation that forms on the walls and ceiling creates the proper envi- 
ronment for mold growth and buildups. The drip of contaminated conden- 
sate onto food products in the room poses a potential health hazard. 

Different microorganisms react differently to the presence of oxygen in 
the environment. Some microorganisms such as molds require oxygen for 
growth, while some others cannot grow in the presence of oxygen. Some 
grow best in low-oxygen environments, while others grow in environments 
regardless of the amount of oxygen. Therefore, the growth of certain 
microorganisms can be controlled by controlling the amount of oxygen in 
the environment. For example, vacuum packaging inhibits the growth of 
microorganisms that require oxygen. Also, the storage life of some fruits 
can be extended by reducing the oxygen level in the storage room. 

Microorganisms in food products can be controlled by (1) preventing 
contamination by following strict sanitation practices, (2) inhibiting growth 
by altering the environmental conditions, and (3) destroying the organisms 
by heat treatment or chemicals. The best way to minimize contamination 



cen58933_ch04.qxd 9/10/2002 9:13 AM Page 241 



241 
CHAPTER 4 



in food processing areas is to use fine air filters in ventilation systems to 
capture the dust particles that transport the bacteria in the air. Of course, 
the filters must remain dry since microorganisms can grow in wet filters. 
Also, the ventilation system must maintain a positive pressure in the food 
processing areas to prevent any airborne contaminants from entering inside 
by infiltration. The elimination of condensation on the walls and the ceil- 
ing of the facility and the diversion of plumbing condensation drip pans of 
refrigerators to the drain system are two other preventive measures against 
contamination. Drip systems must be cleaned regularly to prevent micro- 
biological growth in them. Also, any contact between raw and cooked food 
products should be minimized, and cooked products must be stored in 
rooms with positive pressures. Frozen foods must be kept at — 18°C or be- 
low, and utmost care should be exercised when food products are packaged 
after they are frozen to avoid contamination during packaging. 

The growth of microorganisms is best controlled by keeping the temper- 
ature and relative humidity of the environment in the desirable range. 
Keeping the relative humidity below 60 percent, for example, prevents the 
growth of all microorganisms on the surfaces. Microorganisms can be de- 
stroyed by heating the food product to high temperatures (usually above 
70°C), by treating them with chemicals, or by exposing them to ultraviolet 
light or solar radiation. 

Distinction should be made between survival and growth of micro- 
organisms. A particular microorganism that may not grow at some low tem- 
perature may be able to survive at that temperature for a very long time 
(Fig. 4-34). Therefore, freezing is not an effective way of killing micro- 
organisms. In fact, some microorganism cultures are preserved by freezing 
them at very low temperatures. The rate of freezing is also an important 
consideration in the refrigeration of foods since some microorganisms 
adapt to low temperatures and grow at those temperatures when the cool- 
ing rate is very low. 

Refrigeration and Freezing of Foods 

The storage life of fresh perishable foods such as meats, fish, vegetables, 
and fruits can be extended by several days by storing them at temperatures 
just above freezing, usually between 1 and 4°C. The storage life of foods 
can be extended by several months by freezing and storing them at sub- 
freezing temperatures, usually between — 18 and — 35°C, depending on the 
particular food (Fig. 4-35). 

Refrigeration slows down the chemical and biological processes in foods, 
and the accompanying deterioration and loss of quality and nutrients. 
Sweet corn, for example, may lose half of its initial sugar content in one 
day at 21°C, but only 5 percent of it at 0°C. Fresh asparagus may lose 
50 percent of its vitamin C content in one day at 20°C, but in 12 days at 
0°C. Refrigeration also extends the shelf life of products. The first appear- 
ance of unsightly yellowing of broccoli, for example, may be delayed by 
three or more days by refrigeration. 

Early attempts to freeze food items resulted in poor-quality products 
because of the large ice crystals that formed. It was determined that the rate 
of freezing has a major effect on the size of ice crystals and the quality, 
texture, and nutritional and sensory properties of many foods. During slow 



Z z 




Frozen 



FIGURE 4-34 

Freezing may stop the growth of 

microorganisms, but it may not 

necessarily kill them. 



Freezer 
-18to-35°C 

o 



Frozen 
foods 



'-' 



Refrigerator 
1 to 4°C 

ft 



y 



Fresh 
foods 



FIGURE 4-35 

Recommended refrigeration and 

freezing temperatures for 

most perishable foods. 



cen58933_ch04.qxd 9/10/2002 9:13 AM Page 242 



242 
HEAT TRANSFER 



Temperature 




End of 
freezing 



Time 

FIGURE 4-36 

Typical freezing curve of a food item. 



TABLE 4-5 


Thermal properties 


of beef 


Quantity 


Typical value 


Average density 


1070 kg/m 3 


Specific heat: 




Above freezing 


3.14 kJ/kg • °C 


Below freezing 


1.70 kJ/kg • °C 


Freezing point 


-2.7°C 


Latent heat of fusion 249 kJ/kg 


Thermal 


0.41 W/m • °C 


conductivity 


(at 6°C) 



freezing, ice crystals can grow to a large size, whereas during fast freezing 
a large number of ice crystals start forming at once and are much smaller in 
size. Large ice crystals are not desirable since they can puncture the walls 
of the cells, causing a degradation of texture and a loss of natural juices 
during thawing. A crust forms rapidly on the outer layer of the product and 
seals in the juices, aromatics, and flavoring agents. The product quality is 
also affected adversely by temperature fluctuations of the storage room. 
The ordinary refrigeration of foods involves cooling only without any 
phase change. The freezing of foods, on the other hand, involves three 
stages: cooling to the freezing point (removing the sensible heat), freezing 
(removing the latent heat), and further cooling to the desired subfreezing 
temperature (removing the sensible heat of frozen food), as shown in Fig- 
ure 4-36. 

Beef Products 

Meat carcasses in slaughterhouses should be cooled as fast as possible to a 
uniform temperature of about 1.7°C to reduce the growth rate of micro- 
organisms that may be present on carcass surfaces, and thus minimize 
spoilage. The right level of temperature, humidity, and air motion should 
be selected to prevent excessive shrinkage, toughening, and discoloration. 

The deep body temperature of an animal is about 39°C, but this temper- 
ature tends to rise a couple of degrees in the midsections after slaughter as 
a result of the heat generated during the biological reactions that occur in 
the cells. The temperature of the exposed surfaces, on the other hand, tends 
to drop as a result of heat losses. The thickest part of the carcass is the 
round, and the center of the round is the last place to cool during chilling. 
Therefore, the cooling of the carcass can best be monitored by inserting a 
thermometer deep into the central part of the round. 

About 70 percent of the beef carcass is water, and the carcass is cooled 
mostly by evaporative cooling as a result of moisture migration toward the 
surface where evaporation occurs. But this shrinking translates into a loss 
of salable mass that can amount to 2 percent of the total mass during an 
overnight chilling. To prevent excessive loss of mass, carcasses are usually 
washed or sprayed with water prior to cooling. With adequate care, spray 
chilling can eliminate carcass cooling shrinkage almost entirely. 

The average total mass of dressed beef, which is normally split into two 
sides, is about 300 kg, and the average specific heat of the carcass is about 
3.14 kJ/kg • °C (Table 4-5). The chilling room must have a capacity equal 
to the daily kill of the slaughterhouse, which may be several hundred. 
A beef carcass is washed before it enters the chilling room and absorbs a 
large amount of water (about 3.6 kg) at its surface during the washing 
process. This does not represent a net mass gain, however, since it is lost by 
dripping or evaporation in the chilling room during cooling. Ideally, the 
carcass does not lose or gain any net weight as it is cooled in the chilling 
room. However, it does lose about 0.5 percent of the total mass in the hold- 
ing room as it continues to cool. The actual product loss is determined by 
first weighing the dry carcass before washing and then weighing it again 
after it is cooled. 

The refrigerated air temperature in the chilling room of beef carcasses 
must be sufficiently high to avoid freezing and discoloration on the outer 



cen58933_ch04.qxd 9/10/2002 9:13 AM Page 243 




20 24 28 32 36 40 44 48 52 56 60 64 
Time from start of chill, hours 



72 



243 
CHAPTER 4 



FIGURE 4-37 

Typical cooling curve of a beef carcass 
in the chilling and holding rooms at an 
average temperature of 0°C (from 
ASHRAE, Handbook: Refrigeration, 
Ref. 3, Chap. 11, Fig. 2). 



surfaces of the carcass. This means a long residence time for the massive 
beef carcasses in the chilling room to cool to the desired temperature. Beef 
carcasses are only partially cooled at the end of an overnight stay in the 
chilling room. The temperature of a beef carcass drops to 1.7 to 7°C at the 
surface and to about 15°C in mid parts of the round in 10 h. It takes another 
day or two in the holding room maintained at 1 to 2°C to complete chilling 
and temperature equalization. But hog carcasses are fully chilled during 
that period because of their smaller size. The air circulation in the holding 
room is kept at minimum levels to avoid excessive moisture loss and dis- 
coloration. The refrigeration load of the holding room is much smaller than 
that of the chilling room, and thus it requires a smaller refrigeration system. 

Beef carcasses intended for distant markets are shipped the day after 
slaughter in refrigerated trucks, where the rest of the cooling is done. This 
practice makes it possible to deliver fresh meat long distances in a timely 
manner. 

The variation in temperature of the beef carcass during cooling is given 
in Figure 4-37. Initially, the cooling process is dominated by sensible heat 
transfer. Note that the average temperature of the carcass is reduced by 
about 28°C (from 36 to 8°C) in 20 h. The cooling rate of the carcass could 
be increased by lowering the refrigerated air temperature and increasing 
the air velocity, but such measures also increase the risk of surface freezing. 

Most meats are judged on their tenderness, and the preservation of ten- 
derness is an important consideration in the refrigeration and freezing of 
meats. Meat consists primarily of bundles of tiny muscle fibers bundled to- 
gether inside long strings of connective tissues that hold it together. The 
tenderness of a certain cut of beef depends on the location of the cut, the 
age, and the activity level of the animal. Cuts from the relatively inactive 
mid-backbone section of the animal such as short loins, sirloin, and prime 
ribs are more tender than the cuts from the active parts such as the legs and 
the neck (Fig. 4-38). The more active the animal, the more the connective 
tissue, and the tougher the meat. The meat of an older animal is more fla- 
vorful, however, and is preferred for stewing since the toughness of the 
meat does not pose a problem for moist-heat cooking such as boiling. The 



Chuck 



Sirloin 




Brisket Flank Round 

Foreshank Short plate 

FIGURE 4-38 

Various cuts of beef (from National 
Livestock and Meat Board). 



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244 
HEAT TRANSFER 




Time in days 

FIGURE 4-39 

Variation of tenderness of meat stored 
at 2°C with time after slaughter. 



Meat freezer 



Air 

-40 to -30°C 

2.5 to 5 m/s 




FIGURE 4-40 

The freezing time of meat can be 
reduced considerably by using low 
temperature air at high velocity. 



TABLE 4-6 

Storage life of frozen meat products 
at different storage temperatures 
(from ASHRAE Handbook: 
Refrigeration, Chap. 10, Table 7) 







Storage Life, Months 




Temperature 


Product 


-12°C 


-18°C-23°C 


Beef 




4-12 


6-18 12-24 


Lamb 




3-8 


6-16 12-18 


Veal 




3-4 


4-14 8 


Pork 




2-6 


4-12 8-15 


Chopped 


beef 


3-4 


4-6 8 


Cooked foods 


2-3 


2-4 



protein collagen, which is the main component of the connective tissue, 
softens and dissolves in hot and moist environments and gradually trans- 
forms into gelatin, and tenderizes the meat. 

The old saying "one should either cook an animal immediately after 
slaughter or wait at least two days" has a lot of truth in it. The biomechan- 
ical reactions in the muscle continue after the slaughter until the energy 
supplied to the muscle to do work diminishes. The muscle then stiffens and 
goes into rigor mortis. This process begins several hours after the animal is 
slaughtered and continues for 12 to 36 h until an enzymatic action sets in 
and tenderizes the connective tissue, as shown in Figure 4-39. It takes 
about seven days to complete tenderization naturally in storage facilities 
maintained at 2°C. Electrical stimulation also causes the meat to be tender. 
To avoid toughness, fresh meat should not be frozen before rigor mortis has 
passed. 

You have probably noticed that steaks are tender and rather tasty when 
they are hot but toughen as they cool. This is because the gelatin that 
formed during cooking thickens as it cools, and meat loses its tenderness. 
So it is no surprise that first-class restaurants serve their steak on hot thick 
plates that keep the steaks warm for a long time. Also, cooking softens the 
connective tissue but toughens the tender muscle fibers. Therefore, barbe- 
cuing on low heat for a long time results in a tough steak. 

Variety meats intended for long-term storage must be frozen rapidly to 
reduce spoilage and preserve quality. Perhaps the first thought that comes 
to mind to freeze meat is to place the meat packages into the freezer and 
wait. But the freezing time is too long in this case, especially for large 
boxes. For example, the core temperature of a 4-cm-deep box containing 
32 kg of variety meat can be as high as 16°C 24 h after it is placed into a 

— 30°C freezer. The freezing time of large boxes can be shortened consid- 
erably by adding some dry ice into it. 

A more effective method of freezing, called quick chilling, involves the 
use of lower air temperatures, —40 to — 30°C, with higher velocities of 
2.5 m/s to 5 m/s over the product (Fig. 4-40). The internal temperature 
should be lowered to — 4°C for products to be transferred to a storage 
freezer and to — 18°C for products to be shipped immediately. The rate of 
freezing depends on the package material and its insulating properties, the 
thickness of the largest box, the type of meat, and the capacity of the re- 
frigeration system. Note that the air temperature will rise excessively dur- 
ing initial stages of freezing and increase the freezing time if the capacity 
of the system is inadequate. A smaller refrigeration system will be adequate 
if dry ice is to be used in packages. Shrinkage during freezing varies from 
about 0.5 to 1 percent. 

Although the average freezing point of lean meat can be taken to be 

— 2°C with a latent heat of 249 kJ/kg, it should be remembered that freez- 
ing occurs over a temperature range, with most freezing occurring between 

— 1 and — 4°C. Therefore, cooling the meat through this temperature range 
and removing the latent heat takes the most time during freezing. 

Meat can be kept at an internal temperature of —2 to — 1°C for local use 
and storage for under a week. Meat must be frozen and stored at much 
lower temperatures for long-term storage. The lower the storage tempera- 
ture, the longer the storage life of meat products, as shown in Table 4-6. 



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245 
CHAPTER 4 



The internal temperature of carcasses entering the cooling sections 
varies from 38 to 41°C for hogs and from 37 to 39°C for lambs and calves. 
It takes about 15 h to cool the hogs and calves to the recommended tem- 
perature of 3 to 4°C. The cooling-room temperature is maintained at — 1 to 
0°C and the temperature difference between the refrigerant and the cooling 
air is kept at about 6°C. Air is circulated at a rate of about 7 to 12 air 
changes per hour. Lamb carcasses are cooled to an internal temperature of 
1 to 2°C, which takes about 12 to 14 h, and are held at that temperature 
with 85 to 90 percent relative humidity until shipped or processed. The rec- 
ommended rate of air circulation is 50 to 60 air changes per hour during 
the first 4 to 6 h, which is reduced to 10 to 12 changes per hour afterward. 

Freezing does not seem to affect the flavor of meat much, but it affects 
the quality in several ways. The rate and temperature of freezing may in- 
fluence color, tenderness, and drip. Rapid freezing increases tenderness and 
reduces the tissue damage and the amount of drip after thawing. Storage at 
low freezing temperatures causes significant changes in animal fat. Frozen 
pork experiences more undesirable changes during storage because of its 
fat structure, and thus its acceptable storage period is shorter than that of 
beef, veal, or lamb. 

Meat storage facilities usually have a refrigerated shipping dock where 
the orders are assembled and shipped out. Such docks save valuable stor- 
age space from being used for shipping purposes and provide a more ac- 
ceptable working environment for the employees. Packing plants that ship 
whole or half carcasses in bulk quantities may not need a shipping dock; a 
load-out door is often adequate for such cases. 

A refrigerated shipping dock, as shown in Figure 4-41, reduces the re- 
frigeration load of freezers or coolers and prevents temperature fluctua- 
tions in the storage area. It is often adequate to maintain the shipping docks 
at 4 to 7°C for the coolers and about 1.5°C for the freezers. The dew point 
of the dock air should be below the product temperature to avoid conden- 
sation on the surface of the products and loss of quality. The rate of airflow 
through the loading doors and other openings is proportional to the square 
root of the temperature difference, and thus reducing the temperature dif- 
ference at the opening by half by keeping the shipping dock at the average 
temperature reduces the rate of airflow into the dock and thus into the 
freezer by 1 — \/03 = 0.3, or 30 percent. Also, the air that flows into 
the freezer is already cooled to about 1 .5°C by the refrigeration unit of the 
dock, which represents about 50 percent of the cooling load of the in- 
coming air. Thus, the net effect of the refrigerated shipping dock is a 
reduction of the infiltration load of the freezer by about 65 percent since 
1 — 0.7 X 0.5 = 0.65. The net gain is equal to the difference between the 
reduction of the infiltration load of the freezer and the refrigeration load of 
the shipping dock. Note that the dock refrigerators operate at much higher 
temperatures (1.5°C instead of about — 23°C), and thus they consume 
much less power for the same amount of cooling. 



Freezer 
-23°C 



Refrigerated 
dock 

1.5°C 

| Sliding 
door 



Refrigerated 
truck 



FIGURE 4-41 

A refrigerated truck dock for loading 
frozen items to a refrigerated truck. 



Poultry Products 

Poultry products can be preserved by ice-chilling to 1 to 2°C or deep chill- 
ing to about — 2°C for short-term storage, or by freezing them to — 18°C or 



cen58933_ch04.qxd 9/10/2002 9:13 AM Page 246 



246 
HEAT TRANSFER 




FIGURE 4-42 

Air chilling causes dehydration and 
thus weight loss for poultry, whereas 
immersion chilling causes a weight 
gain as a result of water absorption. 



below for long-term storage. Poultry processing plants are completely 
automated, and the small size of the birds makes continuous conveyor line 
operation feasible. 

The birds are first electrically stunned before cutting to prevent strug- 
gling. Following a 90- to 120-s bleeding time, the birds are scalded by 
immersing them into a tank of warm water, usually at 51 to 55°C, for up 
to 120 s to loosen the feathers. Then the feathers are removed by feather- 
picking machines, and the eviscerated carcass is washed thoroughly before 
chilling. The internal temperature of the birds ranges from 24 to 35°C after 
washing, depending on the temperatures of the ambient air and the wash- 
ing water as well as the extent of washing. 

To control the microbial growth, the USDA regulations require that poul- 
try be chilled to 4°C or below in less than 4 h for carcasses of less than 
1.8 kg, in less than 6 h for carcasses of 1.8 to 3.6 kg. and in less than 8 h for 
carcasses more than 3.6 kg. Meeting these requirements today is not diffi- 
cult since the slow air chilling is largely replaced by the rapid immersion 
chilling in tanks of slush ice. Immersion chilling has the added benefit that 
it not only prevents dehydration, but it causes a net absorption of water and 
thus increases the mass of salable product. Cool air chilling of unpacked 
poultry can cause a moisture loss of 1 to 2 percent, while water immersion 
chilling can cause a moisture absorption of 4 to 15 percent (Fig. 4-42). 
Water spray chilling can cause a moisture absorption of up to 4 percent. 
Most water absorbed is held between the flesh and the skin and the 
connective tissues in the skin. In immersion chilling, some soluble solids 
are lost from the carcass to the water, but the loss has no significant effect 
on flavor. 

Many slush ice tank chillers today are replaced by continuous flow-type 
immersion slush ice chillers. Continuous slush ice-chillers can reduce the 
internal temperature of poultry from 32 to 4°C in about 30 minutes at a rate 
up to 10, 000 birds per hour. Ice requirements depend on the inlet and exit 
temperatures of the carcass and the water, but 0.25 kg of ice per kg of car- 
cass is usually adequate. However, bacterial contamination such as salmo- 
nella remains a concern with this method, and it may be necessary to 
chloride the water to control contamination. 

Tenderness is an important consideration for poultry products just as it is 
for red meat, and preserving tenderness is an important consideration in the 
cooling and freezing of poultry. Birds cooked or frozen before passing 
through rigor mortis remain very tough. Natural tenderization begins soon 
after slaughter and is completed within 24 h when birds are held at 4°C. 
Tenderization is rapid during the first three hours and slows down there- 
after. Immersion in hot water and cutting into the muscle adversely affect 
tenderization. Increasing the scalding temperature or the scalding time has 
been observed to increase toughness, and decreasing the scalding time has 
been observed to increase tenderness. The beating action of mechanical 
feather-picking machines causes considerable toughening. Therefore, it is 
recommended that any cutting be done after tenderization. Cutting up the 
bird into pieces before natural tenderization is completed reduces tender- 
ness considerably. Therefore, it is recommended that any cutting be done 
after tenderization. Rapid chilling of poultry can also have a toughening 



cen58933_ch04.qxd 9/10/2002 9:13 AM Page 247 



247 
CHAPTER 4 



effect. It is found that the tenderization process can be speeded up consid- 
erably by a patented electrical stunning process. 

Poultry products are highly perishable, and thus they should be kept at 
the lowest possible temperature to maximize their shelf life. Studies have 
shown that the populations of certain bacteria double every 36 h at — 2°C, 
14 h at 0°C, 7 h at 5°C, and less than 1 h at 25°C (Fig. 4-43). Studies have 
also shown that the total bacterial counts on birds held at 2°C for 14 days 
are equivalent to those held at 10°C for 5 days or 24°C for 1 day. It has also 
been found that birds held at — 1°C had 8 days of additional shelf life over 
those held at 4°C. 

The growth of microorganisms on the surfaces of the poultry causes the 
development of an off-odor and bacterial slime. The higher the initial 
amount of bacterial contamination, the faster the sliming occurs. Therefore, 
good sanitation practices during processing such as cleaning the equipment 
frequently and washing the carcasses are as important as the storage tem- 
perature in extending shelf life. 

Poultry must be frozen rapidly to ensure a light, pleasing appearance. 
Poultry that is frozen slowly appears dark and develops large ice crystals 
that damage the tissue. The ice crystals formed during rapid freezing are 
small. Delaying freezing of poultry causes the ice crystals to become larger. 
Rapid freezing can be accomplished by forced air at temperatures of —23 
to — 40°C and velocities of 1.5 to 5 m/s in air-blast tunnel freezers. Most 
poultry is frozen this way. Also, the packaged birds freeze much faster on 
open shelves than they do in boxes. If poultry packages must be frozen in 
boxes, then it is very desirable to leave the boxes open or to cut holes on 
the boxes in the direction of airflow during freezing. For best results, the 
blast tunnel should be fully loaded across its cross-section with even spac- 
ing between the products to assure uniform airflow around all sides of the 
packages. The freezing time of poultry as a function of refrigerated air tem- 
perature is given in Figure 4-44. Thermal properties of poultry are given in 
Table 4-7. 

Other freezing methods for poultry include sandwiching between cold 
plates, immersion into a refrigerated liquid such as glycol or calcium chlo- 
ride brine, and cryogenic cooling with liquid nitrogen. Poultry can be 
frozen in several hours by cold plates. Very high freezing rates can be ob- 
tained by immersing the packaged birds into a low-temperature brine. The 
freezing time of birds in — 29°C brine can be as low as 20 min, depending 
on the size of the bird (Fig. 4-45). Also, immersion freezing produces a 
very appealing light appearance, and the high rates of heat transfer make 
continuous line operation feasible. It also has lower initial and maintenance 
costs than forced air, but leaks into the packages through some small holes 
or cracks remain a concern. The convection heat transfer coefficient is 17 
W/m 2 • °C for air at -29°C and 2.5 m/s whereas it is 170 W/m 2 • °C for 
sodium chloride brine at — 18°C and a velocity of 0.02 m/s. Sometimes liq- 
uid nitrogen is used to crust freeze the poultry products to — 73°C. The 
freezing is then completed with air in a holding room at — 23°C. 

Properly packaged poultry products can be stored frozen for up to about 
a year at temperatures of — 18°C or lower. The storage life drops consider- 
ably at higher (but still below-freezing) temperatures. Significant changes 



Storage life (days) 




15 20 25 
Storage temperature, °C 

FIGURE 4-43 

The storage life of fresh poultry 

decreases exponentially with 

increasing storage temperature. 



7 

on 

3 6 

o 
v 5 



<D -1 



D 


Giblets 










■ Inside surface 
o 1 3 ram depth 
• Under skin 
























□ / 


























S 11/ 


o 




























9 


o — 














• — 






-• — f 





-84 -73 -62 -51 -40 -29 -18 -7 
Air temperature, degrees Celsius 

Note: Freezing time is the time required for 
temperature to fall from to — 4°C. The values 
are for 2.3 to 3.6 kg chickens with initial 
temperature of to 2°C and with air velocity 
of 2.3 to 2.8 m/s. 

FIGURE 4-44 

The variation of freezing time of 

poultry with air temperature (from 

van der Berg and Lentz, Ref. 11). 



cen58933_ch04.qxd 9/10/2002 9:13 AM Page 24E 



248 
HEAT TRANSFER 



FIGURE 4-45 

The variation of temperature of the 

breast of 6. 8 -kg turkeys initially at 

1°C with depth during immersion 

cooling at — 29°C (from van der Berg 

and Lentz, Ref. 11). 







Giblets 



Inside surface 



38 mm depth 

25 mm depth 
j— 13 mm depth 
\\6.5 mm depth 
\ Under skin 
Skin surface 



100 125 150 
Time, min. 



250 



TABLE 4-7 

Thermal properties of poultry 



Quantity 



Typical value 



Average density: 

Muscle 

Skin 
Specific heat: 

Above freezing 

Below freezing 
Freezing point 
Latent heat of fusion 



1070 kg/m 3 
1030 kg/m 3 

2.94 kJ/kg • °C 
1.55 kJ/kg • °C 
-2.8°C 
247 kJ/kg 
Thermal conductivity: (in W/m • °C) 
Breast muscle 0.502 at 20°C 
1.384 at -20°C 
1.506 at -40°C 
Dark muscle 1.557 at -40°C 



occur in flavor and juiciness when poultry is frozen for too long, and a stale 
rancid odor develops. Frozen poultry may become dehydrated and experi- 
ence freezer burn, which may reduce the eye appeal of the product and 
cause toughening of the affected area. Dehydration and thus freezer burn 
can be controlled by humidification, lowering the storage temperature, and 
packaging the product in essentially impermeable film. The storage life can 
be extended by packing the poultry in an oxygen-free environment. The 
bacterial counts in precooked frozen products must be kept at safe levels 
since bacteria may not be destroyed completely during the reheating 
process at home. 

Frozen poultry can be thawed in ambient air, water, refrigerator, or oven 
without any significant difference in taste. Big birds like turkey should be 
thawed safely by holding it in a refrigerator at 2 to 4°C for two to four days, 
depending on the size of the bird. They can also be thawed by immersing 
them into cool water in a large container for 4 to 6 h, or holding them in a 
paper bag. Care must be exercised to keep the bird's surface cool to mini- 
mize microbiological growth when thawing in air or water. 



H 



EXAMPLE 4-5 Chilling of Beef Carcasses in a Meat Plant 



HThe chilling room of a meat plant is 18 m X 20 m X 5.5 m in size and has a 
capacity of 450 beef carcasses. The power consumed by the fans and the lights 
of the chilling room are 26 and 3 kW, respectively, and the room gains heat 
through its envelope at a rate of 13 kW. The average mass of beef carcasses is 
285 kg. The carcasses enter the chilling room at 36°C after they are washed to 
facilitate evaporative cooling and are cooled to 15°C in 10 h. The water is ex- 
pected to evaporate at a rate of 0.080 kg/s. The air enters the evaporator sec- 
tion of the refrigeration system at 0.7°C and leaves at -2°C. The air side of 
the evaporator is heavily finned, and the overall heat transfer coefficient of the 
evaporator based on the air side is 20 W/m 2 • °C. Also, the average temperature 
■ difference between the air and the refrigerant in the evaporator is 5.5°C. 



cen58933_ch04.qxd 9/10/2002 9:13 AM Page 249 



Determine (a) the refrigeration load of the chilling room, (b) the volume flow 
rate of air, and (c) the heat transfer surface area of the evaporator on the air 
side, assuming all the vapor and the fog in the air freezes in the evaporator. 

SOLUTION The chilling room of a meat plant with a capacity of 450 beef car- 
casses is considered. The cooling load, the airflow rate, and the heat transfer 
area of the evaporator are to be determined. 

Assumptions 1 Water evaporates at a rate of 0.080 kg/s. 2 All the moisture in 
the air freezes in the evaporator. 

Properties The heat of fusion and the heat of vaporization of water at 0°C are 
333.7 kJ/kg and 2501 kJ/kg (Table A-9). The density and specific heat of air at 
0°C are 1.292 kg/m 3 and 1.006 kJ/kg ■ °C (Table A-15). Also, the specific heat 
of beef carcass is determined from the relation in Table A-7b to be 



Analysis (a) A sketch of the chilling room is given in Figure 4-46. The amount 
of beef mass that needs to be cooled per unit time is 

m beef = (Total beef mass cooled)/(Cooling time) 

= (450 carcasses)(285 kg/carcass)/(10 X 3600 s) = 3.56 kg/s 

The product refrigeration load can be viewed as the energy that needs to be 
removed from the beef carcass as it is cooled from 36 to 15°C at a rate of 
3.56 kg/s and is determined to be 



Gb 



(mCAT) h 



(3.56 kg/s)(3.14 kJ/kg • °C)(36 - 15)°C = 235 kW 



Then the total refrigeration load of the chilling room becomes 

G total, chillroom = G beef + G fan + G lights + G heat gain = 235 + 26 + 3 + 13 = 277 kW 

The amount of carcass cooling due to evaporative cooling of water is 

G beef, evaporative = («^)water = (0-080 kg/ S )(2490 kJ/kg) = 199 kW 

which is 199/235 = 85 percent of the total product cooling load. The remain- 
ing 15 percent of the heat is transferred by convection and radiation. 

(£>) Heat is transferred to air at the rate determined above, and the tempera- 
ture of the air rises from -2°C to 0.7°C as a result. Therefore, the mass flow 
rate of air is 



Ga 



277 kW 



(C„A7/ air ) (1.006 kJ/kg • °C)[0.7 - (-2)°C] 



102.0 kg/s 



Then the volume flow rate of air becomes 

h ml 102 kg/s 



V 



Pa.r 1.292 kg/m 3 



78.9 m 3 /s 



(c) Normally the heat transfer load of the evaporator is the same as the refriger- 
ation load. But in this case the water that enters the evaporator as a liquid is 



249 
CHAPTER 4 



L~ 



H 



Lights, 3 kW 



13 kW 



Fans, 26 kW 







Evaporation 
0.080 kg/s 



Refrigerated 

air 



^tlllllllt 



*-*- 0.7°C 



Evaporator 



f 



-2°C 



J 



t-evap 

FIGURE 4-46 

Schematic for Example 4-5. 



cen58933_ch04.qxd 9/10/2002 9:13 AM Page 25C 



250 
HEAT TRANSFER 



frozen as the temperature drops to -2°C, and the evaporator must also remove 
the latent heat of freezing, which is determined from 

G freezing = (* ^ a .e„,)„ater = (0-080 kg/ S )(334 kj/kg) = 27 kW 

Therefore, the total rate of heat removal at the evaporator is 

G evaporator = G total, chill room "•" Gfreezing = 277 + 27 = 304 kW 

Then the heat transfer surface area of the evaporator on the air side is deter- 
mined from Q evaporator = (t//l) airside A7", 



Ge 



304,000 W 



UAT (20 W/m 2 ■ °C)(5.5°C) 



2764 m 2 



Obviously, a finned surface must be used to provide such a large surface area 
on the air side. 



SUMMARY 



In this chapter we considered the variation of temperature with 
time as well as position in one- or multidimensional systems. 
We first considered the lumped systems in which the tempera- 
ture varies with time but remains uniform throughout the sys- 
tem at any time. The temperature of a lumped body of arbitrary 
shape of mass m, volume V, surface area A s , density p, and 
specific heat C p initially at a uniform temperature T t that is 
exposed to convection at time t = in a medium at tempera- 
ture T„ with a heat transfer coefficient h is expressed as 

T(t) ~ T x _ ki 



where 



h 



hA, 



pC p V pC p L c 



(1/s) 



is a positive quantity whose dimension is (time) -1 . This rela- 
tion can be used to determine the temperature T(t) of a body at 
time t or, alternately, the time t required for the temperature to 
reach a specified value T{t). Once the temperature Tit) at time 
t is available, the rate of convection heat transfer between the 
body and its environment at that time can be determined from 
Newton's law of cooling as 



Q(t) = hA s [T(t)-T x ] 



(W) 



The total amount of heat transfer between the body and the sur- 
rounding medium over the time interval t = to / is simply the 
change in the energy content of the body, 



Q = mCJT(t) - Ti\ 



(kJ) 



The amount of heat transfer reaches its upper limit when the 
body reaches the surrounding temperature 7^. Therefore, the 
maximum heat transfer between the body and its surround- 
ings is 



Gma* = mC (T a - T { ) 



(kJ) 



The error involved in lumped system analysis is negligible 
when 



hL c 

Bi = — -<0.1 

k 



where Bi is the Biot number and L c = VIA S is the characteristic 
length. 

When the lumped system analysis is not applicable, the vari- 
ation of temperature with position as well as time can be deter- 
mined using the transient temperature charts given in Figs. 
4-13, 4-14, 4-15, and 4-23 for a large plane wall, a long cylin- 
der, a sphere, and a semi-infinite medium, respectively. These 
charts are applicable for one-dimensional heat transfer in those 
geometries. Therefore, their use is limited to situations in 
which the body is initially at a uniform temperature, all sur- 
faces are subjected to the same thermal conditions, and the 
body does not involve any heat generation. These charts can 
also be used to determine the total heat transfer from the body 
up to a specified time /. 



cen58933_ch04.qxd 9/10/2002 9:13 AM Page 251 



Using a one-term approximation, the solutions of one- 
dimensional transient heat conduction problems are expressed 
analytically as 



Plane wall: 



Cylinder: 



Sphere: 



Q(x, f) wa n 



T(x, t) 



6(r, 0, 



cyl 



8(r, 0, 



sph 



2 




A,e" XlT cos (XjX/L), 


t>0.2 


T(r, t) - T„ 




T - T 

2 




Ate-m JfartrJ, 


t>0.2 


T(r, t) - r« 




r, - r. 




, 2 sin^r/O 
\ x rlr„ 


t>0.2 



where the constants A l and \ { are functions of the Bi number 
only, and their values are listed in Table 4-1 against the Bi 
number for all three geometries. The error involved in one- 
term solutions is less than 2 percent when t > 0.2. 

Using the one-term solutions, the fractional heat transfers in 
different geometries are expressed as 



Plane wall: 

Cylinder: 

Sphere: 



Q 

siniax 

_Q_ 

Q 

tCmax 



cy I 



spli 



1 

1 -26, 
1 -36, 



sin \, 



/i(Xi) 



O.cyl X) 

sin \, — X, cos X, 



0, sph 



K 



The analytic solution for one-dimensional transient heat 
conduction in a semi-infinite solid subjected to convection is 
given by 



T(x, t) 



erfc 
erfc 



hx , h 2 at 

2V^/ exp U + ^ 



2\fat 



+ 



h\/ai 



251 
CHAPTER 4 



where the quantity erfc (£) is the complementary error func- 
tion. For the special case of /i — > =°, the surface temperature T s 
becomes equal to the fluid temperature T m and the above equa- 
tion reduces to 



T(x, t) 



erfc 



(T s = constant) 



Using a clever superposition principle called the product so- 
lution these charts can also be used to construct solutions for 
the two-dimensional transient heat conduction problems en- 
countered in geometries such as a short cylinder, a long rectan- 
gular bar, or a semi-infinite cylinder or plate, and even 
three-dimensional problems associated with geometries such 
as a rectangular prism or a semi-infinite rectangular bar, pro- 
vided that all surfaces of the solid are subjected to convection 
to the same fluid at temperature T m with the same convection 
heat transfer coefficient h, and the body involves no heat 
generation. The solution in such multidimensional geometries 
can be expressed as the product of the solutions for the 
one-dimensional geometries whose intersection is the multi- 
dimensional geometry. 

The total heat transfer to or from a multidimensional geom- 
etry can also be determined by using the one-dimensional val- 
ues. The transient heat transfer for a two-dimensional geometry 
formed by the intersection of two one-dimensional geometries 
1 and 2 is 



_Q_ 






Q_ 

!cma 



_Q_ 



Transient heat transfer for a three-dimensional body formed by 
the intersection of three one -dimensional bodies 1 , 2, and 3 is 
given by 



_Q_ 



kl- m a 



+ 



Jima 



_Q 
_Q 



_Q_ 

*>ma 



_Q_ 

J&ma 



REFERENCES AND SUGGESTED READING 



1. ASHRAE. Handbook of Fundamentals. SI version. 
Atlanta, GA: American Society of Heating, Refrigerating, 
and Air-Conditioning Engineers, Inc., 1993. 

2. ASHRAE. Handbook of Fundamentals. SI version. 
Atlanta, GA: American Society of Heating, Refrigerating, 
and Air-Conditioning Engineers, Inc., 1994. 

3. H. S. Carslaw and J. C. Jaeger. Conduction of Heat in 
Solids. 2nd ed. London: Oxford University Press, 1959. 



4. H. Grober, S. Erk, and U. Grigull. Fundamentals of Heat 
Transfer. New York: McGraw-Hill, 1961. 

5. M. P. Heisler. "Temperature Charts for Induction and 
Constant Temperature Heating." ASME Transactions 69 
(1947), pp. 227-36. 

6. H. Hillman. Kitchen Science. Mount Vernon, NY: 
Consumers Union, 1981. 

7. F. P. Incropera and D. P. DeWitt. Introduction to Heat 
Transfer. 4th ed. New York: John Wiley & Sons, 2002. 



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252 
HEAT TRANSFER 



8. L. S. Langston. "Heat Transfer from Multidimensional 
Objects Using One-Dimensional Solutions for Heat 
Loss." International Journal of Heat and Mass Transfer 
25(1982), pp. 149-50. 

9. M. N. Ozisik, Heat Transfer — A Basic Approach. New 
York: McGraw-Hill, 1985. 



10. P. J. Schneider. Conduction Heat Transfer. Reading, MA: 
Addison -Wesley, 1955. 

11. L. van der Berg and C. P. Lentz. "Factors Affecting 
Freezing Rate and Appearance of Eviscerated Poultry 
Frozen in Air." Food Technology 12 (1958). 



PROBLEMS 



Lumped System Analysis 

4-1 C What is lumped system analysis? When is it 
applicable? 

4-2C Consider heat transfer between two identical hot solid 
bodies and the air surrounding them. The first solid is being 
cooled by a fan while the second one is allowed to cool natu- 
rally. For which solid is the lumped system analysis more 
likely to be applicable? Why? 

4-3C Consider heat transfer between two identical hot solid 
bodies and their environments. The first solid is dropped in a 
large container filled with water, while the second one is al- 
lowed to cool naturally in the air. For which solid is the lumped 
system analysis more likely to be applicable? Why? 

4-4C Consider a hot baked potato on a plate. The tempera- 
ture of the potato is observed to drop by 4°C during the first 
minute. Will the temperature drop during the second minute be 
less than, equal to, or more than 4°C? Why? 

Hot 

baked 
potato 




FIGURE P4-4C 

4-5C Consider a potato being baked in an oven that is main- 
tained at a constant temperature. The temperature of the potato 
is observed to rise by 5°C during the first minute. Will the tem- 
perature rise during the second minute be less than, equal to, or 
more than 5°C? Why? 

4-6C What is the physical significance of the Biot number? 
Is the Biot number more likely to be larger for highly conduct- 
ing solids or poorly conducting ones? 

*Problems designated by a "C" are concept questions, and 
students are encouraged to answer them all. Problems designated 
by an "E" are in English units, and the SI users can ignore them. 
Problems with an EES-CD icon ® are solved using EES, and 
complete solutions together with parametric studies are included 
on the enclosed CD. Problems with a computer-EES icon H are 
comprehensive in nature, and are intended to be solved with a 
computer, preferably using the EES software that accompanies 
this text. 



4-7C Consider two identical 4-kg pieces of roast beef. The 
first piece is baked as a whole, while the second is baked after 
being cut into two equal pieces in the same oven. Will there be 
any difference between the cooking times of the whole and cut 
roasts? Why? 

4-8C Consider a sphere and a cylinder of equal volume 
made of copper. Both the sphere and the cylinder are initially at 
the same temperature and are exposed to convection in the 
same environment. Which do you think will cool faster, the 
cylinder or the sphere? Why? 

4-9C In what medium is the lumped system analysis more 
likely to be applicable: in water or in air? Why? 

4-10C For which solid is the lumped system analysis more 
likely to be applicable: an actual apple or a golden apple of the 
same size? Why? 

4-1 1C For which kind of bodies made of the same material 
is the lumped system analysis more likely to be applicable: 
slender ones or well-rounded ones of the same volume? Why? 

4-12 Obtain relations for the characteristic lengths of a large 
plane wall of thickness 2L, a very long cylinder of radius r , 
and a sphere of radius r . 

4-13 Obtain a relation for the time required for a lumped 
system to reach the average temperature | (T; + TJ), where 
Tj is the initial temperature and T a is the temperature of the 
environment. 

4-14 The temperature of a gas stream is to be measured by 
a thermocouple whose junction can be approximated as a 
1 .2-mm-diameter sphere. The properties of the junction are 
k = 35 W/m • °C, p = 8500 kg/m 3 , and C p = 320 J/kg ■ °C, and 
the heat transfer coefficient between the junction and the gas 
is h = 65 W/m 2 ■ °C. Determine how long it will take for 
the thermocouple to read 99 percent of the initial temperature 
difference. Answer: 38.5 s 

4-15E In a manufacturing facility, 2 -in. -diameter brass balls 
(k = 64.1 Btu/h • ft • °F, p = 532 lbm/ft 3 , and C p = 0.092 
Btu/lbm • °F) initially at 250°F are quenched in a water bath at 
120°F for a period of 2 min at a rate of 120 balls per minute. 
If the convection heat transfer coefficient is 42 Btu/h • ft 2 • °F, 
determine (a) the temperature of the balls after quenching and 
(b) the rate at which heat needs to be removed from the water 
in order to keep its temperature constant at 120°F 



cen58933_ch04.qxd 9/10/2002 9:13 AM Page 253 



250°F 

oo< 


/ 


Brass 
balls 




.OOO 




> 120°F r 

0000000 


V 






Water 
bath 





FIGURE P4-15E 

4-16E Repeat Problem 4-1 5E for aluminum balls. 

4-17 To warm up some milk for a baby, a mother pours milk 
into a thin-walled glass whose diameter is 6 cm. The height of 
the milk in the glass is 7 cm. She then places the glass into a 
large pan filled with hot water at 60°C. The milk is stirred con- 
stantly, so that its temperature is uniform at all times. If the 
heat transfer coefficient between the water and the glass is 
120 W/m 2 • °C, determine how long it will take for the milk to 
warm up from 3°C to 38°C. Take the properties of the milk 
to be the same as those of water. Can the milk in this case be 
treated as a lumped system? Why? Answer: 5.8 min 

4-18 Repeat Problem 4-17 for the case of water also 
being stirred, so that the heat transfer coefficient is doubled to 
240 W/m 2 • °C. 

4-1 9E During a picnic on a hot summer day, all the cold 
drinks disappeared quickly, and the only available drinks were 
those at the ambient temperature of 80°F. In an effort to cool a 
12-fluid-oz drink in a can, which is 5 in. high and has a diame- 
ter of 2.5 in., a person grabs the can and starts shaking it in the 
iced water of the chest at 32°F. The temperature of the drink 
can be assumed to be uniform at all times, and the heat transfer 
coefficient between the iced water and the aluminum can is 
30 Btu/h • ft 2 ■ °F. Using the properties of water for the drink, 
estimate how long it will take for the canned drink to cool 
to 45 °F 




253 
CHAPTER 4 



4-20 Consider a 1000-W iron whose base plate is made of 
0.5-cm-thick aluminum alloy 2024-T6 (p = 2770 kg/m 3 , C p = 
875 J/kg • °C, a = 7.3 X 10~ 5 m 2 /s). The base plate has a sur- 
face area of 0.03 m 2 . Initially, the iron is in thermal equilibrium 
with the ambient air at 22°C. Taking the heat transfer 
coefficient at the surface of the base plate to be 12 W/m 2 ■ °C 
and assuming 85 percent of the heat generated in the resistance 
wires is transferred to the plate, determine how long it will take 
for the plate temperature to reach 140°C. Is it realistic to as- 
sume the plate temperature to be uniform at all times? 




1000 w 
iron 



FIGURE P4-20 



4-21 



Reconsider Problem 4-20. Using EES (or other) 
software, investigate the effects of the heat trans- 
fer coefficient and the final plate temperature on the time it will 
take for the plate to reach this temperature. Let the heat trans- 
fer coefficient vary from 5 W/m 2 ■ °C to 25 W/m 2 ■ °C and the 
temperature from 30°C to 200°C. Plot the time as functions of 
the heat transfer coefficient and the temperature, and discuss 
the results. 

4-22 Stainless steel ball bearings (p = 8085 kg/m 3 , k = 15.1 
W/m ■ °C, C p = 0.480 kJ/kg ■ °C, and a = 3.91 X lO" 6 m 2 /s) 
having a diameter of 1.2 cm are to be quenched in water. The 
balls leave the oven at a uniform temperature of 900°C and are 
exposed to air at 30°C for a while before they are dropped into 
the water. If the temperature of the balls is not to fall below 
850°C prior to quenching and the heat transfer coefficient in 
the air is 125 W/m 2 • °C, determine how long they can stand in 
the air before being dropped into the water. Answer: 3.7 s 

Carbon steel balls (p = 7833 kg/m 3 , k = 54 W/m ■ °C, 
°C, and a = 1.474 X 10~ 6 m 2 /s) 8 mm in 



4-23 



C p = 0.465 kJ/kg 



Furnace 


-,900°C 

o 


Air, 35°C 

, Steel ball 100 o C 

OOO 





FIGURE P4-1 9E 



FIGURE P4-23 



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254 
HEAT TRANSFER 



diameter are annealed by heating them first to 900°C in a fur- 
nace and then allowing them to cool slowly to 100°C in am- 
bient air at 35°C. If the average heat transfer coefficient is 
75 W/m 2 ■ °C, determine how long the annealing process will 
take. If 2500 balls are to be annealed per hour, determine the 
total rate of heat transfer from the balls to the ambient air. 

4-24 rSpM Reconsider Problem 4-23. Using EES (or other) 
b^ti software, investigate the effect of the initial tem- 
perature of the balls on the annealing time and the total rate of 
heat transfer. Let the temperature vary from 500°C to 1000°C. 
Plot the time and the total rate of heat transfer as a function of 
the initial temperature, and discuss the results. 

4-25 An electronic device dissipating 30 W has a mass of 
20 g, a specific heat of 850 J/kg • °C, and a surface area of 
5 cm 2 . The device is lightly used, and it is on for 5 min and 
then off for several hours, during which it cools to the ambient 
temperature of 25°C. Taking the heat transfer coefficient to be 
12 W/m 2 • °C, determine the temperature of the device at the 
end of the 5 -min operating period. What would your answer be 
if the device were attached to an aluminum heat sink having a 
mass of 200 g and a surface area of 80 cm 2 ? Assume the device 
and the heat sink to be nearly isothermal. 



Transient Heat Conduction in Large Plane Walls, 
Long Cylinders, and Spheres with Spatial Effects 

4-26C What is an infinitely long cylinder? When is it proper 
to treat an actual cylinder as being infinitely long, and when is 
it not? For example, is it proper to use this model when finding 
the temperatures near the bottom or top surfaces of a cylinder? 
Explain. 

4-27C Can the transient temperature charts in Fig. 4-13 for 
a plane wall exposed to convection on both sides be used for a 
plane wall with one side exposed to convection while the other 
side is insulated? Explain. 

4-28C Why are the transient temperature charts prepared 
using nondimensionalized quantities such as the Biot and 
Fourier numbers instead of the actual variables such as thermal 
conductivity and time? 

4-29C What is the physical significance of the Fourier num- 
ber? Will the Fourier number for a specified heat transfer prob- 
lem double when the time is doubled? 

4-30C How can we use the transient temperature charts 
when the surface temperature of the geometry is specified in- 
stead of the temperature of the surrounding medium and the 
convection heat transfer coefficient? 

4-31 C A body at an initial temperature of T t is brought into a 
medium at a constant temperature of T„. How can you deter- 
mine the maximum possible amount of heat transfer between 
the body and the surrounding medium? 

4-32C The Biot number during a heat transfer process be- 
tween a sphere and its surroundings is determined to be 0.02. 



Would you use lumped system analysis or the transient tem- 
perature charts when determining the midpoint temperature of 
the sphere? Why? 

4-33 A student calculates that the total heat transfer from 
a spherical copper ball of diameter 15 cm initially at 200°C 
and its environment at a constant temperature of 25°C during 
the first 20 min of cooling is 4520 kJ. Is this result reason- 
able? Why? 

4-34 An ordinary egg can be approximated as a 5.5-cm- 
diameter sphere whose properties are roughly k = 0.6 W/m • 
°C and a = 0.14 X 10~ s m 2 /s. The egg is initially at a uniform 
temperature of 8°C and is dropped into boiling water at 97°C. 
Taking the convection heat transfer coefficient to be h = 1400 
W/m 2 ■ °C, determine how long it will take for the center of the 
egg to reach 70°C. 



Boiling 
water y 




97°C 






Egg 






r, = 8°c 





FIGURE P4-34 



4-35 



Reconsider Problem 4-34. Using EES (or other) 
software, investigate the effect of the final center 
temperature of the egg on the time it will take for the center to 
reach this temperature. Let the temperature vary from 50°C 
to 95°C. Plot the time versus the temperature, and discuss the 
results. 

4-36 In a production facility, 3-cm-thick large brass plates 
(Jfc = 110 W/m • °C, p = 8530 kg/m 3 , C p = 380 J/kg • °C, and 
a = 33.9 X 10~ 6 m 2 /s) that are initially at a uniform tempera- 
ture of 25°C are heated by passing them through an oven main- 
tained at 700°C. The plates remain in the oven for a period of 
10 min. Taking the convection heat transfer coefficient to be 
h = 80 W/m 2 • °C, determine the surface temperature of the 
plates when they come out of the oven. 

Furnace, 700°C 




3 cm 



Brass plate 
25°C 

FIGURE P4-36 



cen58933_ch04.qxd 9/10/2002 9:13 AM Page 255 



4-37 ScJU Reconsider Problem 4-36. Using EES (or other) 
|^£^ software, investigate the effects of the tempera- 
ture of the oven and the heating time on the final surface tem- 
perature of the plates. Let the oven temperature vary from 
500°C to 900°C and the time from 2 min to 30 min. Plot the 
surface temperature as the functions of the oven temperature 
and the time, and discuss the results. 

4-38 A long 3 5 -cm-diameter cylindrical shaft made of stain- 
less steel 304 (k = 14.9 W/m • °C, p = 7900 kg/m 3 , C p = All 
J/kg • °C, and a = 3.95 X 10~ 6 m 2 /s) comes out of an oven at 
a uniform temperature of 400°C. The shaft is then allowed to 
cool slowly in a chamber at 150°C with an average convection 
heat transfer coefficient of h = 60 W/m 2 • °C. Determine the 
temperature at the center of the shaft 20 min after the start of 
the cooling process. Also, determine the heat transfer per unit 
length of the shaft during this time period. 
Answers: 390°C, 16,015 kJ/m 

4-39 rfigM Reconsider Problem 4-38. Using EES (or other) 
b^S software, investigate the effect of the cooling 
time on the final center temperature of the shaft and the amount 
of heat transfer. Let the time vary from 5 min to 60 min. Plot 
the center temperature and the heat transfer as a function of the 
time, and discuss the results. 

4-40E Long cylindrical AISI stainless steel rods (k = 7.74 
Btu/h ■ ft • °F and a = 0.135 ft 2 /h) of 4-in. diameter are heat- 
treated by drawing them at a velocity of 10 ft/min through 
a 30-ft-long oven maintained at 1700°F. The heat transfer 
coefficient in the oven is 20 Btu/h ■ ft 2 • °F. If the rods enter 
the oven at 85°F, determine their centerline temperature when 
they leave. 



Oven 



1700°F 



10 ft/min 



. 




/ r 




n 


f 






\ 







Stainless steel 
85°F 

FIGURE P4-40E 



4-41 In a meat processing plant, 2-cm-thick steaks (k = 0.45 
W/m • °C and a = 0.91 X 10~ 7 m 2 /s) that are initially at 25°C 
are to be cooled by passing them through a refrigeration room 
at — 11°C. The heat transfer coefficient on both sides of the 
steaks is 9 W/m 2 • °C. If both surfaces of the steaks are to be 
cooled to 2°C, determine how long the steaks should be kept in 
the refrigeration room. 

4-42 A long cylindrical wood log (k = 0.17 W/m ■ °C and 
a = 1 .28 X 10~ 7 m 2 /s) is 10 cm in diameter and is initially at a 
uniform temperature of 10°C. It is exposed to hot gases at 



255 
CHAPTER 4 



500°C in a fireplace with a heat transfer coefficient of 13.6 
W/m 2 ■ °C on the surface. If the ignition temperature of the 
wood is 420°C, determine how long it will be before the log 
ignites. 

4-43 In Betty Crocker 's Cookbook, it is stated that it takes 2 h 
45 min to roast a 3.2-kg rib initially at 4.5°C "rare" in an oven 
maintained at 163°C. It is recommended that a meat ther- 
mometer be used to monitor the cooking, and the rib is consid- 
ered rare done when the thermometer inserted into the center of 
the thickest part of the meat registers 60°C. The rib can be 
treated as a homogeneous spherical object with the properties 
p = 1200 kg/m 3 , C p = 4.1 kJ/kg ■ °C, k = 0.45 W/m • °C, and 
a = 0.91 X 10~ 7 m 2 /s. Determine (a) the heat transfer coeffi- 
cient at the surface of the rib, (b) the temperature of the outer 
surface of the rib when it is done, and (c) the amount of heat 
transferred to the rib. (d) Using the values obtained, predict 
how long it will take to roast this rib to "medium" level, which 
occurs when the innermost temperature of the rib reaches 
71°C. Compare your result to the listed value of 3 h 20 min. 

If the roast rib is to be set on the counter for about 15 min 
before it is sliced, it is recommended that the rib be taken 
out of the oven when the thermometer registers about 4°C 
below the indicated value because the rib will continue cook- 
ing even after it is taken out of the oven. Do you agree with this 
recommendation? 

Answers: {a) 156.9 W/m 2 • °C, (« 159. 5°C, (c) 1629 kJ, (d) 3.0 h 




Rib 



T t = 4.5°C 

FIGURE P4-43 

4-44 Repeat Problem 4^-3 for a roast rib that is to be "well- 
done" instead of "rare." A rib is considered to be well-done 
when its center temperature reaches 77°C, and the roasting in 
this case takes about 4 h 15 min. 

4-45 For heat transfer purposes, an egg can be considered to 
be a 5. 5-cm -diameter sphere having the properties of water. An 
egg that is initially at 8°C is dropped into the boiling water at 
100°C. The heat transfer coefficient at the surface of the egg is 
estimated to be 800 W/m 2 • °C. If the egg is considered cooked 
when its center temperature reaches 60°C, determine how long 
the egg should be kept in the boiling water. 



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256 
HEAT TRANSFER 



4-46 Repeat Problem 4^5 for a location at 1610-m eleva- 
tion such as Denver, Colorado, where the boiling temperature 
of water is 94.4°C. 

4-47 The author and his 6-year-old son have conducted the 
following experiment to determine the thermal conductivity of 
a hot dog. They first boiled water in a large pan and measured 
the temperature of the boiling water to be 94°C, which is not 
surprising, since they live at an elevation of about 1650 m in 
Reno, Nevada. They then took a hot dog that is 12.5 cm long 
and 2.2 cm in diameter and inserted a thermocouple into the 
midpoint of the hot dog and another thermocouple just under 
the skin. They waited until both thermocouples read 20°C, 
which is the ambient temperature. They then dropped the hot 
dog into boiling water and observed the changes in both tem- 
peratures. Exactly 2 min after the hot dog was dropped into the 
boiling water, they recorded the center and the surface temper- 
atures to be 59°C and 88°C, respectively. The density of the hot 
dog can be taken to be 980 kg/m 3 , which is slightly less than 
the density of water, since the hot dog was observed to be float- 
ing in water while being almost completely immersed. The 
specific heat of a hot dog can be taken to be 3900 J/kg • °C, 
which is slightly less than that of water, since a hot dog is 
mostly water. Using transient temperature charts, determine 
(a) the thermal diffusivity of the hot dog, (b) the thermal con- 
ductivity of the hot dog, and (c) the convection heat transfer 
coefficient. 

Answers: (a) 2.02 x 10" 7 m 2 /s, (b) 0.771 W/m ■ °C, 
(c) 467 W/m 2 ■ °C. 



Refrigerator 
5°F 




FIGURE P4-47 

4-48 Using the data and the answers given in Problem 4^-7, 
determine the center and the surface temperatures of the hot 
dog 4 min after the start of the cooking. Also determine the 
amount of heat transferred to the hot dog. 

4-49E In a chicken processing plant, whole chickens averag- 
ing 5 lb each and initially at 72 °F are to be cooled in the racks 
of a large refrigerator that is maintained at 5°F. The entire 
chicken is to be cooled below 45 °F, but the temperature of the 
chicken is not to drop below 35°F at any point during refrig- 
eration. The convection heat transfer coefficient and thus the 
rate of heat transfer from the chicken can be controlled by 
varying the speed of a circulating fan inside. Determine the 
heat transfer coefficient that will enable us to meet both tem- 
perature constraints while keeping the refrigeration time to a 




FIGURE P4-49E 

minimum. The chicken can be treated as a homogeneous spher- 
ical object having the properties p = 74.9 lbm/ft 3 , C p = 0.98 
Btu/lbm ■ °F, k = 0.26 Btu/h • ft ■ °F, and a = 0.0035 ft 2 /h. 

4-50 A person puts a few apples into the freezer at — 1 5°C to 
cool them quickly for guests who are about to arrive. Initially, 
the apples are at a uniform temperature of 20°C, and the heat 
transfer coefficient on the surfaces is 8 W/m 2 • °C. Treating the 
apples as 9-cm-diameter spheres and taking their properties to 
be p = 840 kg/m 3 , C p = 3.81 kJ/kg • °C, k = 0.418 W/m ■ °C, 
and a = 1.3 X 10~ 7 m 2 /s, determine the center and surface 
temperatures of the apples in 1 h. Also, determine the amount 
of heat transfer from each apple. 

4-51 [ft^S Reconsider Problem 4-50. Using EES (or other) 
1^2 software, investigate the effect of the initial tem- 
perature of the apples on the final center and surface tem- 
peratures and the amount of heat transfer. Let the initial 
temperature vary from 2°C to 30°C. Plot the center tempera- 
ture, the surface temperature, and the amount of heat transfer 
as a function of the initial temperature, and discuss the results. 

4-52 Citrus fruits are very susceptible to cold weather, and 
extended exposure to subfreezing temperatures can destroy 
them. Consider an 8-cm-diameter orange that is initially at 




Orange 
T ; = 15°C 



FIGURE P4-52 



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257 

CHAPTER 4 



15°C. A cold front moves in one night, and the ambient tem- 
perature suddenly drops to — 6°C, with a heat transfer coeffi- 
cient of 15 W/m 2 • °C. Using the properties of water for the 
orange and assuming the ambient conditions to remain con- 
stant for 4 h before the cold front moves out, determine if any 
part of the orange will freeze that night. 

4-53 An 8-cm-diameter potato (p = 1100 kg/m 3 , C p = 3900 
J/kg ■ °C, k = 0.6 W/m • °C, and a = 1.4 X 10" 7 m 2 /s) that is 
initially at a uniform temperature of 25°C is baked in an oven 
at 170°C until a temperature sensor inserted to the center of the 
potato indicates a reading of 70°C. The potato is then taken out 
of the oven and wrapped in thick towels so that almost no heat 
is lost from the baked potato. Assuming the heat transfer coef- 
ficient in the oven to be 25 W/m 2 • °C, determine (a) how long 
the potato is baked in the oven and (b) the final equilibrium 
temperature of the potato after it is wrapped. 

4-54 White potatoes (k = 0.50 W/m ■ °C and a = 0.13 X 
10~ 6 m 2 /s) that are initially at a uniform temperature of 25°C 
and have an average diameter of 6 cm are to be cooled by re- 
frigerated air at 2°C flowing at a velocity of 4 m/s. The average 
heat transfer coefficient between the potatoes and the air is ex- 
perimentally determined to be 19 W/m 2 ■ °C. Determine how 
long it will take for the center temperature of the potatoes to 
drop to 6°C. Also, determine if any part of the potatoes will ex- 
perience chilling injury during this process. 



Air 

2°C 
4 m/s 



FIGURE P4-54 

4-55E Oranges of 2.5-in. diameter (k = 0.26 Btu/h • ft • °F 
and a = 1.4 X 10~ 6 ft 2 /s) initially at a uniform temperature of 
78°F are to be cooled by refrigerated air at 25°F flowing at a 
velocity of 1 ft/s. The average heat transfer coefficient between 
the oranges and the air is experimentally determined to be 4.6 
Btu/h • ft 2 • °F. Determine how long it will take for the center 
temperature of the oranges to drop to 40°F. Also, determine if 
any part of the oranges will freeze during this process. 

4-56 A 65-kg beef carcass {k = 0.47 W/m • °C and a = 
0.13 X 10~ 6 m 2 /s) initially at a uniform temperature of 37°C is 
to be cooled by refrigerated air at — 6°C flowing at a velocity 
of 1.8 m/s. The average heat transfer coefficient between the 
carcass and the air is 22 W/m 2 • °C. Treating the carcass as a 
cylinder of diameter 24 cm and height 1 .4 m and disregarding 
heat transfer from the base and top surfaces, determine how 
long it will take for the center temperature of the carcass to 
drop to 4°C. Also, determine if any part of the carcass will 
freeze during this process. Answer: 14.0 h 










Air - 


Beef 




-6°C - 


37°C 




1.8 m/s ►• 









FIGURE P4-56 

4-57 Layers of 23-cm-thick meat slabs (k = 0.47 W/m ■ °C 
and a = 0.13 X 10~ 6 m 2 /s) initially at a uniform temperature 
of 7°C are to be frozen by refrigerated air at — 30°C flowing at 
a velocity of 1 .4 m/s. The average heat transfer coefficient be- 
tween the meat and the air is 20 W/m 2 • °C. Assuming the size 
of the meat slabs to be large relative to their thickness, deter- 
mine how long it will take for the center temperature of the 
slabs to drop to — 18°C. Also, determine the surface tempera- 
ture of the meat slab at that time. 

4-58E Layers of 6-in. -thick meat slabs (k = 0.26 
Btu/h • ft • °F and a = 1.4 X 10~ 6 ft 2 /s) initially at a uniform 
temperature of 50°F are cooled by refrigerated air at 23°F to a 
temperature of 36°F at their center in 12 h. Estimate the aver- 
age heat transfer coefficient during this cooling process. 
Answer: 1.5 Btu/h • ft 2 • °F 

4-59 Chickens with an average mass of 1.7 kg (k = 0.45 
W/m ■ °C and a = 0.13 X 10~ 6 m 2 /s) initially at a uniform 
temperature of 15°C are to be chilled in agitated brine at 
— 10°C. The average heat transfer coefficient between the 
chicken and the brine is determined experimentally to be 
440 W/m 2 • °C. Taking the average density of the chicken to be 
0.95 g/cm 3 and treating the chicken as a spherical lump, deter- 
mine the center and the surface temperatures of the chicken in 
2 h and 30 min. Also, determine if any part of the chicken will 
freeze during this process. 




FIGURE P4-59 



cen58933_ch04.qxd 9/10/2002 9:13 AM Page 25E 



258 
HEAT TRANSFER 



Transient Heat Conduction in Semi-Infinite Solids 

4-60C What is a semi-infinite medium? Give examples of 
solid bodies that can be treated as semi-infinite mediums for 
heat transfer purposes. 

4-61 C Under what conditions can a plane wall be treated as 
a semi-infinite medium? 

4-62C Consider a hot semi-infinite solid at an initial temper- 
ature of Tj that is exposed to convection to a cooler medium at 
a constant temperature of T m with a heat transfer coefficient of 
h. Explain how you can determine the total amount of heat 
transfer from the solid up to a specified time t B . 

4-63 In areas where the air temperature remains below 0°C 
for prolonged periods of time, the freezing of water in under- 
ground pipes is a major concern. Fortunately, the soil remains 
relatively warm during those periods, and it takes weeks for the 
subfreezing temperatures to reach the water mains in the 
ground. Thus, the soil effectively serves as an insulation to pro- 
tect the water from the freezing atmospheric temperatures in 
winter. 

The ground at a particular location is covered with snow 
pack at — 8°C for a continuous period of 60 days, and the aver- 
age soil properties at that location are k = 0.35 W/m • °C and 
a = 0.15 X 80~ 6 m 2 /s. Assuming an initial uniform tempera- 
ture of 8°C for the ground, determine the minimum burial 
depth to prevent the water pipes from freezing. 

4-64 The soil temperature in the upper layers of the earth 
varies with the variations in the atmospheric conditions. Before 
a cold front moves in, the earth at a location is initially at a uni- 
form temperature of 10°C. Then the area is subjected to a tem- 
perature of — 10°C and high winds that resulted in a convection 
heat transfer coefficient of 40 W/m 2 • °C on the earth's surface 
for a period of 10 h. Taking the properties of the soil at that lo- 
cation to be k = 0.9 W/m ■ °C and a = 1.6 X 10~ 5 m 2 /s, deter- 
mine the soil temperature at distances 0, 10, 20, and 50 cm 
from the earth's surface at the end of this 10-h period. 

► Winds, 

" -10°C 



Soil 



10°C 



FIGURE P4-64 



4-65 



Reconsider Problem 4-64. Using EES (or other) 
software, plot the soil temperature as a function 

of the distance from the earth's surface as the distance varies 

from m to lm, and discuss the results. 

4-66E The walls of a furnace are made of 1.5-ft-thick con- 
crete (k = 0.64 Btu/h ■ ft • °F and a = 0.023 ft 2 /h). Initially, the 



furnace and the surrounding air are in thermal equilibrium at 
70°F. The furnace is then fired, and the inner surfaces of the 
furnace are subjected to hot gases at 1 800°F with a very large 
heat transfer coefficient. Determine how long it will take for 
the temperature of the outer surface of the furnace walls to rise 
to70.1°F Answer: 181 min 

4-67 A thick wood slab (k = 0. 1 7 W/m • °C and a = 1 .28 X 
10~ 7 m 2 /s) that is initially at a uniform temperature of 25°C is 
exposed to hot gases at 550°C for a period of 5 minutes. The 
heat transfer coefficient between the gases and the wood slab is 
35 W/m 2 • °C. If the ignition temperature of the wood is 450°C, 
determine if the wood will ignite. 

4-68 A large cast iron container (k = 52 W/m • °C and a = 
1.70 X 10~ 5 m 2 /s) with 5-cm-thick walls is initially at a uni- 
form temperature of 0°C and is filled with ice at 0°C. Now the 
outer surfaces of the container are exposed to hot water at 60°C 
with a very large heat transfer coefficient. Determine how long 
it will be before the ice inside the container starts melting. 
Also, taking the heat transfer coefficient on the inner surface of 
the container to be 250 W/m 2 • °C, determine the rate of heat 
transfer to the ice through a 1.2-m-wide and 2-m-high section 
of the wall when steady operating conditions are reached. As- 
sume the ice starts melting when its inner surface temperature 
rises to 0.1 °C. 



Hot water 
60°C 



Cast iron 
f chest 









k • 

U o Ice O 

nop O 

O Q o 

Q o 
° o °. Q 



5 cm 



FIGURE P4-68 



Transient Heat Conduction in Multidimensional Systems 

4-69C What is the product solution method? How is it used 
to determine the transient temperature distribution in a two- 
dimensional system? 

4-70C How is the product solution used to determine 
the variation of temperature with time and position in three- 
dimensional systems? 

4-71 C A short cylinder initially at a uniform temperature T : 
is subjected to convection from all of its surfaces to a medium 
at temperature r„. Explain how you can determine the temper- 
ature of the midpoint of the cylinder at a specified time /. 

4-72C Consider a short cylinder whose top and bottom sur- 
faces are insulated. The cylinder is initially at a uniform tem- 
perature Tj and is subjected to convection from its side surface 



cen58933_ch04.qxd 9/10/2002 9:13 AM Page 259 



to a medium at temperature T„ with a heat transfer coefficient 
of h. Is the heat transfer in this short cylinder one- or two- 
dimensional? Explain. 

4-73 A short brass cylinder (p = 8530 kg/m 3 , C p = 0.389 
kJ/kg • °C, k = 110 W/m • °C, and a = 3.39 X 10~ 5 m 2 /s) of 
diameter D = 8 cm and height H = 15 cm is initially at a 
uniform temperature of T; = 150°C. The cylinder is now 
placed in atmospheric air at 20°C, where heat transfer takes 
place by convection with a heat transfer coefficient of h = 
40 W/m 2 • °C. Calculate (a) the center temperature of the cyl- 
inder, {b) the center temperature of the top surface of the cylin- 
der, and (c) the total heat transfer from the cylinder 15 min 
after the start of the cooling. 




15 cm 



FIGURE P4-73 



4-74 



Reconsider Problem 4-73. Using EES (or other) 
software, investigate the effect of the cooling 
time on the center temperature of the cylinder, the center tem- 
perature of the top surface of the cylinder, and the total heat 
transfer. Let the time vary from 5 min to 60 min. Plot the cen- 
ter temperature of the cylinder, the center temperature of the 
top surface, and the total heat transfer as a function of the time, 
and discuss the results. 

4-75 A semi-infinite aluminum cylinder (k = 237 W/m • °C, 
a = 9.71 X 10~ 5 m 2 /s) of diameter D = 15 cm is initially at a 
uniform temperature of T, = 150°C. The cylinder is now 
placed in water at 10°C, where heat transfer takes place 
by convection with a heat transfer coefficient of h = 
140 W/m 2 ■ °C. Determine the temperature at the center of the 
cylinder 5 cm from the end surface 8 min after the start of 
cooling. 

4-76E A hot dog can be considered to be a cylinder 5 in. 
long and 0.8 in. in diameter whose properties are p = 
61.2 lbm/ft 3 , C p = 0.93 Btu/lbm ■ °F, k = 0.44 Btu/h • ft • °F, 
and a = 0.0077 ft 2 /h. A hot dog initially at 40°F is dropped 
into boiling water at 212°F. If the heat transfer coefficient at 
the surface of the hot dog is estimated to be 120 Btu/h • ft 2 • °F, 
determine the center temperature of the hot dog after 5, 10, and 
15 min by treating the hot dog as (a) a finite cylinder and (b) an 
infinitely long cylinder. 



259 
CHAPTER 4 



4-77E Repeat Problem 4-76E for a location at 5300 ft 
elevation such as Denver, Colorado, where the boiling temper- 
ature of water is 202 °F. 

4-78 A 5-cm-high rectangular ice block (k = 2.22 W/m ■ °C 
and a = 0.124 X 10~ 7 m 2 /s) initially at -20°C is placed on a 
table on its square base 4 cm X 4 cm in size in a room at 1 8°C. 
The heat transfer coefficient on the exposed surfaces of the ice 
block is 12 W/m 2 ■ °C. Disregarding any heat transfer from the 
base to the table, determine how long it will be before the ice 
block starts melting. Where on the ice block will the first liquid 
droplets appear? 



Room 




FIGURE P4-78 



4-79 



Reconsider Problem 4-78. Using EES (or other) 
software, investigate the effect of the initial tem- 
perature of the ice block on the time period before the ice block 
starts melting. Let the initial temperature vary from — 26°C to 
— 4°C. Plot the time versus the initial temperature, and discuss 
the results. 

4-80 A 2-cm-high cylindrical ice block (k = 2.22 W/m • °C 
and a = 0.124 X 10~ 7 m 2 /s) is placed on a table on its base of 
diameter 2 cm in a room at 20°C. The heat transfer coefficient 
on the exposed surfaces of the ice block is 13 W/m 2 ■ °C, and 
heat transfer from the base of the ice block to the table is neg- 
ligible. If the ice block is not to start melting at any point for at 
least 2 h, determine what the initial temperature of the ice 
block should be. 

4-81 Consider a cubic block whose sides are 5 cm long and 
a cylindrical block whose height and diameter are also 5 cm. 
Both blocks are initially at 20°C and are made of granite (k = 
2.5 W/m • °C and a = 1.15 X 10~ 6 m 2 /s). Now both blocks are 
exposed to hot gases at 500°C in a furnace on all of their sur- 
faces with a heat transfer coefficient of 40 W/m 2 • °C. Deter- 
mine the center temperature of each geometry after 10, 20, and 
60 min. 

4-82 Repeat Problem 4-81 with the heat transfer coefficient 
at the top and the bottom surfaces of each block being doubled 
to 80 W/m 2 • °C. 



cen58933_ch04.qxd 9/10/2002 9:13 AM Page 26C 



260 
HEAT TRANSFER 



5 cm 



T, = 20°C 




5 cm 



Hot gases, 500°C 

FIGURE P4-81 

4-83 A 20-cm-long cylindrical aluminum block (p = 2702 
kg/m 3 , C p = 0.896 kJ/kg ■ °C, k = 236 W/m ■ °C, and a = 
9.75 X 10~ 5 m 2 /s), 15 cm in diameter, is initially at a uniform 
temperature of 20°C. The block is to be heated in a furnace at 
1200°C until its center temperature rises to 300°C. If the heat 
transfer coefficient on all surfaces of the block is 80 W/m 2 • °C, 
determine how long the block should be kept in the furnace. 
Also, determine the amount of heat transfer from the aluminum 
block if it is allowed to cool in the room until its temperature 
drops to 20°C throughout. 

4-84 Repeat Problem 4-83 for the case where the aluminum 
block is inserted into the furnace on a low-conductivity mater- 
ial so that the heat transfer to or from the bottom surface of the 
block is negligible. 

4-85 rfigM Reconsider Problem 4-83. Using EES (or other) 
b^2 software, investigate the effect of the final center 
temperature of the block on the heating time and the amount of 
heat transfer. Let the final center temperature vary from 50°C 
to 1000°C. Plot the time and the heat transfer as a function of 
the final center temperature, and discuss the results. 

Special Topic: Refrigeration and Freezing of Foods 

4-86C What are the common kinds of microorganisms? 
What undesirable changes do microorganisms cause in foods? 

4-87C How does refrigeration prevent or delay the spoilage 
of foods? Why does freezing extend the storage life of foods 
for months? 

4-88C What are the environmental factors that affect the 
growth rate of microorganisms in foods? 

4-89C What is the effect of cooking on the microorganisms 
in foods? Why is it important that the internal temperature of a 
roast in an oven be raised above 70°C? 

4-90C How can the contamination of foods with micro- 
organisms be prevented or minimized? How can the growth of 
microorganisms in foods be retarded? How can the micro- 
organisms in foods be destroyed? 

4-91 C How does (a) the air motion and (b) the relative hu- 
midity of the environment affect the growth of microorganisms 
in foods? 

4-92C The cooling of a beef carcass from 37°C to 5°C with 
refrigerated air at 0°C in a chilling room takes about 48 h. 



To reduce the cooling time, it is proposed to cool the carcass 
with refrigerated air at -10°C. How would you evaluate this 
proposal? 

4-93C Consider the freezing of packaged meat in boxes with 
refrigerated air. How do (a) the temperature of air, (b) the 
velocity of air, (c) the capacity of the refrigeration system, and 
(d) the size of the meat boxes affect the freezing time? 

4-94C How does the rate of freezing affect the tenderness, 
color, and the drip of meat during thawing? 

4-95C It is claimed that beef can be stored for up to two 
years at -23°C but no more than one year at — 12°C. Is this 
claim reasonable? Explain. 

4-96C What is a refrigerated shipping dock? How does it 
reduce the refrigeration load of the cold storage rooms? 

4-97C How does immersion chilling of poultry compare to 
forced-air chilling with respect to (a) cooling time, (b) mois- 
ture loss of poultry, and (c) microbial growth. 

4-98C What is the proper storage temperature of frozen 
poultry? What are the primary methods of freezing for poultry? 

4-99C What are the factors that affect the quality of 
frozen fish? 

4-100 The chilling room of a meat plant is 15 m X 18 m X 
5.5 m in size and has a capacity of 350 beef carcasses. The 
power consumed by the fans and the lights in the chilling room 
are 22 and 2 kW, respectively, and the room gains heat through 
its envelope at a rate of 1 1 kW. The average mass of beef car- 
casses is 280 kg. The carcasses enter the chilling room at 35°C, 
after they are washed to facilitate evaporative cooling, and are 
cooled to 16°C in 12 h. The air enters the chilling room at 
— 2.2°C and leaves at 0.5°C. Determine (a) the refrigeration 
load of the chilling room and (b) the volume flow rate of air. 
The average specific heats of beef carcasses and air are 3.14 
and 1 .0 kJ/kg • °C, respectively, and the density of air can be 
taken to be 1 .28 kg/m 3 . 

4-101 Turkeys with a water content of 64 percent that are 
initially at 1 °C and have a mass of about 7 kg are to be frozen 
by submerging them into brine at — 29°C. Using Figure 4^-5, 
determine how long it will take to reduce the temperature of 
the turkey breast at a depth of 3.8 cm to — 18°C. If the temper- 
ature at a depth of 3.8 cm in the breast represents the average 



^\ 



Brine -29°C 



II 




FIGURE P4-1 01 



cen58933_ch04.qxd 9/10/2002 9:13 AM Page 261 



temperature of the turkey, determine the amount of heat trans- 
fer per turkey assuming (a) the entire water content of the 
turkey is frozen and (b) only 90 percent of the water content of 
the turkey is frozen at — 18°C. Take the specific heats of turkey 
to be 2.98 and 1.65 kJ/kg ■ °C above and below the freezing 
point of — 2.8°C, respectively, and the latent heat of fusion of 
turkey to be 214 kJ/kg. Answers.- {a) 1753 kJ, (£>) 1617 kJ 

4-102E Chickens with a water content of 74 percent, an 
initial temperature of 32°F, and a mass of about 6 lbm are to be 
frozen by refrigerated air at — 40°F. Using Figure 4^-4, de- 
termine how long it will take to reduce the inner surface 
temperature of chickens to 25 °F. What would your answer be if 
the air temperature were — 80°F? 

4-103 Chickens with an average mass of 2.2 kg and average 
specific heat of 3.54 kJ/kg • °C are to be cooled by chilled wa- 
ter that enters a continuous-flow-type immersion chiller at 
0.5°C. Chickens are dropped into the chiller at a uniform tem- 
perature of 15°C at a rate of 500 chickens per hour and are 
cooled to an average temperature of 3°C before they are taken 
out. The chiller gains heat from the surroundings at a rate of 
210 kJ/min. Determine (a) the rate of heat removal from the 
chicken, in kW, and {b) the mass flow rate of water, in kg/s, if 
the temperature rise of water is not to exceed 2°C. 

4-104 In a meat processing plant, 10-cm-thick beef slabs 
(p = 1090 kg/m 3 , C p = 3.54 kJ/kg • °C, k = 0.47 W/m • °C, 
and a = 0.13 X 10 6 m 2 /s) initially at 15°C are to be cooled in 
the racks of a large freezer that is maintained at — 12°C. The 
meat slabs are placed close to each other so that heat transfer 
from the 10-cm-thick edges is negligible. The entire slab is to 
be cooled below 5°C, but the temperature of the steak is not 
to drop below — 1 °C anywhere during refrigeration to avoid 
"frost bite." The convection heat transfer coefficient and thus 
the rate of heat transfer from the steak can be controlled by 
varying the speed of a circulating fan inside. Determine the 
heat transfer coefficient h that will enable us to meet both tem- 
perature constraints while keeping the refrigeration time to a 
minimum. Answer: 9.9 W/m 2 • °C. 



Aii- 



Meat 




-12°C M0 cm 

FIGURE P4-1 04 

Review Problems 

4-105 Consider two 2-cm-thick large steel plates (k = 43 
W/m ■ °C and a = 1.17 X 10~ 5 m 2 /s) that were put on top of 
each other while wet and left outside during a cold winter night 
at — 15°C. The next day, a worker needs one of the plates, but 
the plates are stuck together because the freezing of the water 



261 
CHAPTER 4 



between the two plates has bonded them together. In an effort 
to melt the ice between the plates and separate them, the 
worker takes a large hairdryer and blows hot air at 50°C all 
over the exposed surface of the plate on the top. The convec- 
tion heat transfer coefficient at the top surface is estimated to 
be 40 W/m 2 • °C. Determine how long the worker must keep 
blowing hot air before the two plates separate. 
Answer: 482 s 

4-106 Consider a curing kiln whose walls are made of 
30-cm-thick concrete whose properties are k = 0.9 W/m • °C 
and a = 0.23 X 10~ 5 m 2 /s. Initially, the kiln and its walls are 
in equilibrium with the surroundings at 2°C. Then all the doors 
are closed and the kiln is heated by steam so that the tempera- 
ture of the inner surface of the walls is raised to 42°C and is 
maintained at that level for 3 h. The curing kiln is then opened 
and exposed to the atmospheric air after the stream flow is 
turned off. If the outer surfaces of the walls of the kiln were in- 
sulated, would it save any energy that day during the period 
the kiln was used for curing for 3 h only, or would it make no 
difference? Base your answer on calculations. 




FIGURE P4-1 06 

4-107 The water main in the cities must be placed at suf- 
ficient depth below the earth's surface to avoid freezing during 
extended periods of subfreezing temperatures. Determine the 
minimum depth at which the water main must be placed at a 
location where the soil is initially at 15°C and the earth's 
surface temperature under the worst conditions is expected 
to remain at — 10°C for a period of 75 days. Take the proper- 
ties of soil at that location to be k = 0.7 W/m • °C and a = 
1.4 X 10~ 5 m 2 /s. Answer: 7.05 m 

4-108 A hot dog can be considered to be a 12-cm-long cyl- 
inder whose diameter is 2 cm and whose properties are 
p = 980 kg/m 3 , C = 3.9 kJ/kg • °C, k = 0.76 W/m ■ °C, and 




FIGURE P4-1 08 



cen58933_ch04.qxd 9/10/2002 9:13 AM Page 262 



262 
HEAT TRANSFER 



a = 2 X 10~ 7 m 2 /s. A hot dog initially at 5°C is dropped into 
boiling water at 100°C. The heat transfer coefficient at the sur- 
face of the hot dog is estimated to be 600 W/m 2 • °C. If the hot 
dog is considered cooked when its center temperature reaches 
80°C, determine how long it will take to cook it in the boiling 
water. 

4-109 A long roll of 2-m-wide and 0.5-cm-thick 1 -Mn man- 
ganese steel plate coming off a furnace at 820°C is to be 
quenched in an oil bath (C p = 2.0 kJ/kg ■ °C) at 45°C. The 
metal sheet is moving at a steady velocity of 10 m/min, and the 
oil bath is 5 m long. Taking the convection heat transfer 
coefficient on both sides of the plate to be 860 W/m 2 • °C, de- 
termine the temperature of the sheet metal when it leaves the 
oil bath. Also, determine the required rate of heat removal from 
the oil to keep its temperature constant at 45°C. 



Furnace 




Oil bath, 45°C 



FIGURE P4-1 09 



4-110E In Betty Crocker 's Cookbook, it is stated that it takes 
5 h to roast a 14-lb stuffed turkey initially at 40°F in an oven 
maintained at 325°F. It is recommended that a meat thermome- 
ter be used to monitor the cooking, and the turkey is considered 
done when the thermometer inserted deep into the thickest part 
of the breast or thigh without touching the bone registers 
185°F. The turkey can be treated as a homogeneous spheri- 
cal object with the properties p = 75 lbm/ft 3 , C p = 0.98 
Btu/lbm ■ °F, k = 0.26 Btu/h • ft ■ °F, and a = 0.0035 ft 2 /h. 
Assuming the tip of the thermometer is at one-third radial 
distance from the center of the turkey, determine (a) the aver- 
age heat transfer coefficient at the surface of the turkey, (b) the 
temperature of the skin of the turkey when it is done, and 
(c) the total amount of heat transferred to the turkey in the 
oven. Will the reading of the thermometer be more or less than 
185°F 5 min after the turkey is taken out of the oven? 




4-111 (~Jb\ During a fire, the trunks of some dry oak trees (k 
W = 0.17 W/m • °C and a = 1.28 X 10" 7 m 2 /s) 
that are initially at a uniform temperature of 30°C are exposed 
to hot gases at 520°C for a period of 5 h, with a heat transfer 
coefficient of 65 W/m 2 • °C on the surface. The ignition tem- 
perature of the trees is 410°C. Treating the trunks of the trees 
as long cylindrical rods of diameter 20 cm, determine if these 
dry trees will ignite as the fire sweeps through them. 




FIGURE P4-1 1 1 

4-112 We often cut a watermelon in half and put it into the 
freezer to cool it quickly. But usually we forget to check on it 
and end up having a watermelon with a frozen layer on the top. 
To avoid this potential problem a person wants to set the timer 
such that it will go off when the temperature of the exposed 
surface of the watermelon drops to 3°C. 

Consider a 30-cm-diameter spherical watermelon that is cut 
into two equal parts and put into a freezer at — 12°C. Initially, 
the entire watermelon is at a uniform temperature of 25°C, and 
the heat transfer coefficient on the surfaces is 30 W/m 2 • °C. 
Assuming the watermelon to have the properties of water, de- 
termine how long it will take for the center of the exposed cut 
surfaces of the watermelon to drop to 3°C. 




FIGURE P4-1 10 



Watermelon, 25°C 

FIGURE P4-1 12 

4-113 The thermal conductivity of a solid whose density and 
specific heat are known can be determined from the relation 
k = a/pC p after evaluating the thermal diffusivity a. 

Consider a 2-cm-diameter cylindrical rod made of a sample 
material whose density and specific heat are 3700 kg/m 3 and 
920 J/kg ■ °C, respectively. The sample is initially at a uniform 
temperature of 25°C. In order to measure the temperatures of 



cen58933_ch04.qxd 9/10/2002 9:13 AM Page 263 



r 



Thermocouples 



Rod «= 



Boiling water center 

100°C 

FIGURE P4-1 13 



the sample at its surface and its center, a thermocouple is 
inserted to the center of the sample along the centerline, and 
another thermocouple is welded into a small hole drilled on 
the surface. The sample is dropped into boiling water at 100°C. 
After 3 min, the surface and the center temperatures are re- 
corded to be 93°C and 75°C, respectively. Determine the ther- 
mal diffusivity and the thermal conductivity of the material. 

4-114 In desert climates, rainfall is not a common occurrence 
since the rain droplets formed in the upper layer of the atmo- 
sphere often evaporate before they reach the ground. Consider 
a raindrop that is initially at a temperature of 5°C and has a 
diameter of 5 mm. Determine how long it will take for the 
diameter of the raindrop to reduce to 3 mm as it falls through 
ambient air at 18°C with a heat transfer coefficient of 400 
W/m 2 ■ °C. The water temperature can be assumed to remain 
constant and uniform at 5°C at all times. 

4-115E Consider a plate of thickness 1 in., a long cylinder of 
diameter 1 in., and a sphere of diameter 1 in., all initially at 
400°F and all made of bronze (k = 15.0 Btu/h • ft • °F and a = 
0.333 ft 2 /h). Now all three of these geometries are exposed to 
cool air at 75°F on all of their surfaces, with a heat transfer co- 
efficient of 7 Btu/h • ft 2 • °F. Determine the center temperature 
of each geometry after 5, 10, and 30 min. Explain why the cen- 
ter temperature of the sphere is always the lowest. 




Cylinder 



tin. 



J 



Sphere 

<2L, 



FIGURE P4-1 15 

4-116E Repeat Problem 4-1 15E for cast iron geometries 
(k = 29 Btu/h • ft ■ °F and a = 0.61 ft 2 /h). 

4-117E rra| Reconsider Problem 4-1 15E. Using EES (or 
b^2 other) software, plot the center temperature of 
each geometry as a function of the cooling time as the time 
varies fom 5 min to 60 min, and discuss the results. 



263 
CHAPTER 4 



4-118 Engine valves (k = 48 W/m • °C, C p = 440 J/kg ■ °C, 
and p = 7840 kg/m 3 ) are heated to 800°C in the heat treatment 
section of a valve manufacturing facility. The valves are then 
quenched in a large oil bath at an average temperature of 45 °C. 
The heat transfer coefficient in the oil bath is 650 W/m 2 • °C. 
The valves have a cylindrical stem with a diameter of 8 mm 
and a length of 10 cm. The valve head and the stem may be as- 
sumed to be of equal surface area, and the volume of the valve 
head can be taken to be 80 percent of the volume of steam. De- 
termine how long will it take for the valve temperature to drop 
to (a) 400°C, (b) 200°C, and (c) 46°C and (d) the maximum 
heat transfer from a single valve. 

4-119 A watermelon initially at 35°C is to be cooled by 
dropping it into a lake at 1 5°C. After 4 h and 40 min of cooling, 
the center temperature of watermelon is measured to be 20°C. 
Treating the watermelon as a 20-cm -diameter sphere and using 
the properties k = 0.618 W/m • °C, a = 0.15 X 10~ 6 m 2 /s, 
p = 995 kg/m 3 , and C p = 4.18 kJ/kg • °C, determine the aver- 
age heat transfer coefficient and the surface temperature of 
watermelon at the end of the cooling period. 

4-120 10-cm-thick large food slabs tightly wrapped by thin 
paper are to be cooled in a refrigeration room maintained 
at 0°C. The heat transfer coefficient on the box surfaces is 
25 W/m 2 ■ °C and the boxes are to be kept in the refrigeration 
room for a period of 6 h. If the initial temperature of the boxes 
is 30°C determine the center temperature of the boxes if the 
boxes contain (a) margarine (k = 0.233 W/m • °C and a = 
0.11 X 10- 6 m 2 /s), (b) white cake (k = 0.082 W/m • °C and 
a = 0.10 X 10~ 6 m 2 /s), and (c) chocolate cake (k = 0.106 
W/m ■ °C and a = 0.12 X lO" 6 m 2 /s). 

4-121 A 30-cm-diameter, 3.5-m-high cylindrical column of 
a house made of concrete (k = 0.79 W/m • °C, a = 5.94 X 
10- 7 m 2 /s, p = 1600 kg/m 3 , and C p = 0.84 kJ/kg • °C) cooled 
to 16°C during a cold night is heated again during the day by 
being exposed to ambient air at an average temperature of 
28°C with an average heat transfer coefficient of 14 W/m 2 ■ °C. 
Determine (a) how long it will take for the column surface 
temperature to rise to 27°C, (b) the amount of heat transfer 
until the center temperature reaches to 28°C, and (c) the 
amount of heat transfer until the surface temperature reaches 
to 27°C. 

4-122 Long aluminum wires of diameter 3 mm (p = 2702 
kg/m 3 , C p = 0.896 kJ/kg ■ °C, k = 236 W/m • °C, and a = 
9.75 X 10~ 5 m 2 /s) are extruded at a temperature of 350°C and 
exposed to atmospheric air at 30°C with a heat transfer coeffi- 
cient of 35 W/m 2 ■ °C. (a) Determine how long it will take for 
the wire temperature to drop to 50°C. (b) If the wire is extruded 
at a velocity of 10 m/min, determine how far the wire travels 
after extrusion by the time its temperature drops to 50°C. What 
change in the cooling process would you propose to shorten 
this distance? (c) Assuming the aluminum wire leaves the ex- 
trusion room at 50°C, determine the rate of heat transfer from 
the wire to the extrusion room. 

Answers: (a) 144 s, (b) 24 m, (c) 856 W 



cen58933_ch04.qxd 9/10/2002 9:13 AM Page 264 



264 
HEAT TRANSFER 



350°C 




10 m/min 
Aluminum wire 

FIGURE P4-1 22 

4-123 Repeat Problem 4-122 for a copper wire (p = 
8950 kg/m 3 , C p = 0.383 kJ/kg • °C, k = 386 W/m • °C, and 
a = 1.13 X 10- 4 m 2 /s). 

4-124 Consider a brick house (k = 0.72 W/m • °C and a = 
0.45 X 10~ 6 m 2 /s) whose walls are 10 m long, 3 m high, and 
0.3 m thick. The heater of the house broke down one night, and 
the entire house, including its walls, was observed to be 5°C 
throughout in the morning. The outdoors warmed up as the day 
progressed, but no change was felt in the house, which was 
tightly sealed. Assuming the outer surface temperature of the 
house to remain constant at 15°C, determine how long it would 
take for the temperature of the inner surfaces of the walls to 
riseto5.1°C. 




15°C 



~5°C 
FIGURE P4-1 24 

4-125 A 40-cm-thick brick wall (k = 0.72 W/m • °C, and a = 
1.6 X 10~ 7 m 2 /s) is heated to an average temperature of 18°C 
by the heating system and the solar radiation incident on it dur- 
ing the day. During the night, the outer surface of the wall is ex- 
posed to cold air at 2°C with an average heat transfer coefficient 
of 20 W/m 2 ■ °C, determine the wall temperatures at distances 
15,30, and 40 cm from the outer surface for a period of 2 hours. 

4-126 Consider the engine block of a car made of cast iron 
(k = 52 W/m • °C and a = 1 .7 X 10~ 5 m 2 /s). The engine can be 
considered to be a rectangular block whose sides are 80 cm, 
40 cm, and 40 cm. The engine is at a temperature of 150°C 
when it is turned off. The engine is then exposed to atmospheric 
air at 17°C with a heat transfer coefficient of 6 W/m 2 • °C. De- 
termine (a) the center temperature of the top surface whose 
sides are 80 cm and 40 cm and (b) the comer temperature after 
45 min of cooling. 

4-127 A man is found dead in a room at 16°C. The surface 
temperature on his waist is measured to be 23°C and the heat 
transfer coefficient is estimated to be 9 W/m 2 • °C. Modeling the 



body as 28-cm diameter, 1.80-m-long cylinder, estimate how 
long it has been since he died. Take the properties of the body to 
be k = 0.62 W/m • °C and a = 0.15 X 10~ 6 m 2 /s, and assume 
the initial temperature of the body to be 36°C. 

Computer, Design, and Essay Problems 

4-128 Conduct the following experiment at home to deter- 
mine the combined convection and radiation heat transfer co- 
efficient at the surface of an apple exposed to the room air. You 
will need two thermometers and a clock. 

First, weigh the apple and measure its diameter. You may 
measure its volume by placing it in a large measuring cup 
halfway filled with water, and measuring the change in volume 
when it is completely immersed in the water. Refrigerate the 
apple overnight so that it is at a uniform temperature in the 
morning and measure the air temperature in the kitchen. Then 
take the apple out and stick one of the thermometers to its mid- 
dle and the other just under the skin. Record both temperatures 
every 5 min for an hour. Using these two temperatures, calcu- 
late the heat transfer coefficient for each interval and take their 
average. The result is the combined convection and radiation 
heat transfer coefficient for this heat transfer process. Using 
your experimental data, also calculate the thermal conductivity 
and thermal diffusivity of the apple and compare them to the 
values given above. 

4-129 Repeat Problem 4-128 using a banana instead of an 
apple. The thermal properties of bananas are practically the 
same as those of apples. 

4-130 Conduct the following experiment to determine the 
time constant for a can of soda and then predict the temperature 
of the soda at different times. Leave the soda in the refrigerator 
overnight. Measure the air temperature in the kitchen and the 
temperature of the soda while it is still in the refrigerator by 
taping the sensor of the thermometer to the outer surface of the 
can. Then take the soda out and measure its temperature again 
in 5 min. Using these values, calculate the exponent b. Using 
this b-value, predict the temperatures of the soda in 10, 15, 20, 
30, and 60 min and compare the results with the actual temper- 
ature measurements. Do you think the lumped system analysis 
is valid in this case? 

4-131 Citrus trees are very susceptible to cold weather, and 
extended exposure to subfreezing temperatures can destroy the 
crop. In order to protect the trees from occasional cold fronts 
with subfreezing temperatures, tree growers in Florida usually 
install water sprinklers on the trees. When the temperature 
drops below a certain level, the sprinklers spray water on the 
trees and their fruits to protect them against the damage the 
subfreezing temperatures can cause. Explain the basic mecha- 
nism behind this protection measure and write an essay on how 
the system works in practice. 



cen5 8 93 3_ch05.qxd 9/4/2002 11:41 AM Page 265 



NUMERICAL METHODS 
IN HEAT CONDUCTION 



CHAPTER 



So far we have mostly considered relatively simple heat conduction prob- 
lems involving simple geometries with simple boundary conditions be- 
cause only such simple problems can be solved analytically. But many 
problems encountered in practice involve complicated geometries with com- 
plex boundary conditions or variable properties and cannot be solved ana- 
lytically. In such cases, sufficiently accurate approximate solutions can be 
obtained by computers using a numerical method. 

Analytical solution methods such as those presented in Chapter 2 are based 
on solving the governing differential equation together with the boundary con- 
ditions. They result in solution functions for the temperature at every point in 
the medium. Numerical methods, on the other hand, are based on replacing 
the differential equation by a set of n algebraic equations for the unknown 
temperatures at n selected points in the medium, and the simultaneous solu- 
tion of these equations results in the temperature values at those discrete 
points. 

There are several ways of obtaining the numerical formulation of a heat 
conduction problem, such as the finite difference method, the finite element 
method, the boundary element method, and the energy balance (or control 
volume) method. Each method has its own advantages and disadvantages, and 
each is used in practice. In this chapter we will use primarily the energy bal- 
ance approach since it is based on the familiar energy balances on control vol- 
umes instead of heavy mathematical formulations, and thus it gives a better 
physical feel for the problem. Besides, it results in the same set of algebraic 
equations as the finite difference method. In this chapter, the numerical for- 
mulation and solution of heat conduction problems are demonstrated for both 
steady and transient cases in various geometries. 



CONTENTS 



Why Numerical Methods 266 

Finite Difference Formulation of 
Differential Equations 269 

One-Dimensional 

Steady Heat Conduction 272 

Two-Dimensional 

Steady Heat Conduction 282 

Transient Heat Conduction 291 

Topic of Special Interest: 

Controlling the 
Numerical Error 309 



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266 
HEAT TRANSFER 




Solution: 
Q(r) = -kA 



1 ' 6k v 



4ty 3 



dr 3 

FIGURE 5-1 

The analytical solution of a problem 
requires solving the governing 
differential equation and applying 
the boundary conditions. 



5-1 ■ WHY NUMERICAL METHODS? 

The ready availability of high-speed computers and easy-to-use powerful soft- 
ware packages has had a major impact on engineering education and practice 
in recent years. Engineers in the past had to rely on analytical skills to solve 
significant engineering problems, and thus they had to undergo a rigorous 
training in mathematics. Today's engineers, on the other hand, have access to 
a tremendous amount of computation power under their fingertips, and they 
mostly need to understand the physical nature of the problem and interpret the 
results. But they also need to understand how calculations are performed by 
the computers to develop an awareness of the processes involved and the lim- 
itations, while avoiding any possible pitfalls. 

In Chapter 2 we solved various heat conduction problems in various geo- 
metries in a systematic but highly mathematical manner by (1) deriving the 
governing differential equation by performing an energy balance on a differ- 
ential volume element, (2) expressing the boundary conditions in the proper 
mathematical form, and (3) solving the differential equation and applying the 
boundary conditions to determine the integration constants. This resulted in a 
solution function for the temperature distribution in the medium, and the so- 
lution obtained in this manner is called the analytical solution of the problem. 
For example, the mathematical formulation of one-dimensional steady heat 
conduction in a sphere of radius r whose outer surface is maintained at a uni- 
form temperature of T, with uniform heat generation at a rate of g Q was ex- 
pressed as (Fig. 5-1) 



\_d_t 2 dT\ go 
r 2 dr\ dr) k 



dT(0) 
dr 



whose (analytical) solution is 



and 







T(r ) = T x 



T{r) = T l +f k {ri-r^) 



(5-1) 



(5-2) 



This is certainly a very desirable form of solution since the temperature at 
any point within the sphere can be determined simply by substituting the 
r-coordinate of the point into the analytical solution function above. The ana- 
lytical solution of a problem is also referred to as the exact solution since it 
satisfies the differential equation and the boundary conditions. This can be 
verified by substituting the solution function into the differential equation and 
the boundary conditions. Further, the rate of heat flow at any location within 
the sphere or its surface can be determined by taking the derivative of the so- 
lution function T(r) and substituting it into Fourier's law as 



Q(r) 



-kA 



dT 
dr 



-£(4irr 2 ) 



gof_ 
~3k 



4irg Q r' : 



(5-3) 



The analysis above did not require any mathematical sophistication beyond 
the level of simple integration, and you are probably wondering why anyone 
would ask for something else. After all, the solutions obtained are exact and 



cen58 93 3_ch05.qxd 9/4/2002 11:41 AM Page 267 



easy to use. Besides, they are instructive since they show clearly the func- 
tional dependence of temperature and heat transfer on the independent vari- 
able r. Well, there are several reasons for searching for alternative solution 
methods. 



1 Limitations 

Analytical solution methods are limited to highly simplified problems in sim- 
ple geometries (Fig. 5-2). The geometry must be such that its entire surface 
can be described mathematically in a coordinate system by setting the vari- 
ables equal to constants. That is, it must fit into a coordinate system perfectly 
with nothing sticking out or in. In the case of one-dimensional heat conduc- 
tion in a solid sphere of radius r , for example, the entire outer surface can be 
described by r = r . Likewise, the surfaces of a finite solid cylinder of radius 
r and height H can be described by r = r for the side surface and z = and 
z = H for the bottom and top surfaces, respectively. Even minor complica- 
tions in geometry can make an analytical solution impossible. For example, a 
spherical object with an extrusion like a handle at some location is impossible 
to handle analytically since the boundary conditions in this case cannot be ex- 
pressed in any familiar coordinate system. 

Even in simple geometries, heat transfer problems cannot be solved analyt- 
ically if the thermal conditions are not sufficiently simple. For example, the 
consideration of the variation of thermal conductivity with temperature, the 
variation of the heat transfer coefficient over the surface, or the radiation heat 
transfer on the surfaces can make it impossible to obtain an analytical solu- 
tion. Therefore, analytical solutions are limited to problems that are simple or 
can be simplified with reasonable approximations. 



2 Better Modeling 

We mentioned earlier that analytical solutions are exact solutions since they 
do not involve any approximations. But this statement needs some clarifica- 
tion. Distinction should be made between an actual real-world problem and 
the mathematical model that is an idealized representation of it. The solutions 
we get are the solutions of mathematical models, and the degree of applica- 
bility of these solutions to the actual physical problems depends on the accu- 
racy of the model. An "approximate" solution of a realistic model of a 
physical problem is usually more accurate than the "exact" solution of a crude 
mathematical model (Fig. 5-3). 

When attempting to get an analytical solution to a physical problem, there 
is always the tendency to oversimplify the problem to make the mathematical 
model sufficiently simple to warrant an analytical solution. Therefore, it is 
common practice to ignore any effects that cause mathematical complications 
such as nonlinearities in the differential equation or the boundary conditions. 
So it comes as no surprise that nonlinearities such as temperature dependence 
of thermal conductivity and the radiation boundary conditions are seldom con- 
sidered in analytical solutions. A mathematical model intended for a numeri- 
cal solution is likely to represent the actual problem better. Therefore, the 
numerical solution of engineering problems has now become the norm rather 
than the exception even when analytical solutions are available. 



T„h 



267 
CHAPTER 5 



k = constant 



No 
radiation 



T„h 



Long 
cylinder 



h,T„ 



No 
radiation 



h,T„ 



h = constant 
T„ = constant 

FIGURE 5-2 

Analytical solution methods are 

limited to simplified problems 

in simple geometries. 




Exact (analytical) Approximate (numerical) 
solution of model, solution of model, 

but crude solution but accurate solution 

of actual problem of actual problem 

FIGURE 5-3 

The approximate numerical solution 

of a real-world problem may be more 

accurate than the exact (analytical) 

solution of an oversimplified 

model of that problem. 



cen58 93 3_ch05.qxd 9/4/2002 11:41 AM Page 268 



268 
HEAT TRANSFER 




T(r, z) 




Analytical solution: 

T(r,z)-T-^ ~ J (Ar) sinhA„(L-z) 

where A 's are roots of 7 n (A r n ) = 

FIGURE 5-4 

Some analytical solutions are very 
complex and difficult to use. 



3 Flexibility 



Engineering problems often require extensive parametric studies to under- 
stand the influence of some variables on the solution in order to choose the 
right set of variables and to answer some "what-if " questions. This is an iter- 
ative process that is extremely tedious and time-consuming if done by hand. 
Computers and numerical methods are ideally suited for such calculations, 
and a wide range of related problems can be solved by minor modifications in 
the code or input variables. Today it is almost unthinkable to perform any sig- 
nificant optimization studies in engineering without the power and flexibility 
of computers and numerical methods. 



4 Complications 



Some problems can be solved analytically, but the solution procedure is so 
complex and the resulting solution expressions so complicated that it is not 
worth all that effort. With the exception of steady one-dimensional or transient 
lumped system problems, all heat conduction problems result in partial 
differential equations. Solving such equations usually requires mathematical 
sophistication beyond that acquired at the undergraduate level, such as orthog- 
onality, eigenvalues, Fourier and Laplace transforms, Bessel and Legendre 
functions, and infinite series. In such cases, the evaluation of the solution, 
which often involves double or triple summations of infinite series at a speci- 
fied point, is a challenge in itself (Fig. 5-4). Therefore, even when the solu- 
tions are available in some handbooks, they are intimidating enough to scare 
prospective users away. 




FIGURE 5-5 

The ready availability of high- 
powered computers with sophisticated 
software packages has made 
numerical solution the norm 
rather than the exception. 



5 Human Nature 

As human beings, we like to sit back and make wishes, and we like our wishes 
to come true without much effort. The invention of TV remote controls made 
us feel like kings in our homes since the commands we give in our comfort- 
able chairs by pressing buttons are immediately carried out by the obedient 
TV sets. After all, what good is cable TV without a remote control. We cer- 
tainly would love to continue being the king in our little cubicle in the engi- 
neering office by solving problems at the press of a button on a computer 
(until they invent a remote control for the computers, of course). Well, this 
might have been a fantasy yesterday, but it is a reality today. Practically all 
engineering offices today are equipped with high-powered computers with 
sophisticated software packages, with impressive presentation-style colorful 
output in graphical and tabular form (Fig. 5-5). Besides, the results are as 
accurate as the analytical results for all practical purposes. The computers 
have certainly changed the way engineering is practiced. 

The discussions above should not lead you to believe that analytical solu- 
tions are unnecessary and that they should be discarded from the engineering 
curriculum. On the contrary, insight to the physical phenomena and engineer- 
ing wisdom is gained primarily through analysis. The "feel" that engineers 
develop during the analysis of simple but fundamental problems serves as 
an invaluable tool when interpreting a huge pile of results obtained from a 
computer when solving a complex problem. A simple analysis by hand for 
a limiting case can be used to check if the results are in the proper range. Also, 



cen58 93 3_ch05.qxd 9/4/2002 11:41 AM Page 269 



269 
CHAPTER 5 



nothing can take the place of getting "ball park" results on a piece of paper 
during preliminary discussions. The calculators made the basic arithmetic 
operations by hand a thing of the past, but they did not eliminate the need for 
instructing grade school children how to add or multiply. 

In this chapter, you will learn how to formulate and solve heat transfer 
problems numerically using one or more approaches. In your professional life, 
you will probably solve the heat transfer problems you come across using a 
professional software package, and you are highly unlikely to write your own 
programs to solve such problems. (Besides, people will be highly skeptical 
of the results obtained using your own program instead of using a well- 
established commercial software package that has stood the test of time.) The 
insight you will gain in this chapter by formulating and solving some heat 
transfer problems will help you better understand the available software pack- 
ages and be an informed and responsible user. 



5-2 - FINITE DIFFERENCE FORMULATION 
OF DIFFERENTIAL EQUATIONS 

The numerical methods for solving differential equations are based on replac- 
ing the differential equations by algebraic equations. In the case of the popu- 
lar finite difference method, this is done by replacing the derivatives by 
differences. Below we will demonstrate this with both first- and second-order 
derivatives. But first we give a motivational example. 

Consider a man who deposits his money in the amount of A Q = $100 in a 
savings account at an annual interest rate of 18 percent, and let us try to de- 
termine the amount of money he will have after one year if interest is com- 
pounded continuously (or instantaneously). In the case of simple interest, the 
money will earn $18 interest, and the man will have 100 + 100 X 0.18 = 
$118.00 in his account after one year. But in the case of compounding, the 
interest earned during a compounding period will also earn interest for the 
remaining part of the year, and the year-end balance will be greater than $118. 
For example, if the money is compounded twice a year, the balance will be 
100 + 100 X (0.18/2) = $109 after six months, and 109 + 109 X (0.18/2) = 
$118.81 at the end of the year. We could also determine the balance A di- 
rectly from 



A = A (l + 0" = ($100)(1 + 0.09) 2 = $118.81 



(5-4) 



where i is the interest rate for the compounding period and n is the number of 
periods. Using the same formula, the year-end balance is determined for 
monthly, daily, hourly, minutely, and even secondly compounding, and the re- 
sults are given in Table 5—1. 

Note that in the case of daily compounding, the year-end balance will be 
$119.72, which is $1.72 more than the simple interest case. (So it is no wonder 
that the credit card companies usually charge interest compounded daily when 
determining the balance.) Also note that compounding at smaller time inter- 
vals, even at the end of each second, does not change the result, and we sus- 
pect that instantaneous compounding using "differential" time intervals dt will 
give the same result. This suspicion is confirmed by obtaining the differential 



TABLE 5-1 



Year-end balance of a $100 account 
earning interest at an annual rate 
of 18 percent for various 
compounding periods 





Number 




Compounding 


of 


Year-End 


Period 


Periods, n 


Balance 


1 year 


1 


$118.00 


6 months 


2 


118.81 


1 month 


12 


119.56 


1 week 


52 


119.68 


1 day 


365 


119.72 


1 hour 


8760 


119.72 


1 minute 


525,600 


119.72 


1 second 


31,536,000 


119.72 


Instantaneous 


00 


119.72 



cen58933_ch05.qxd 9/4/2002 11:41 AM Page 27C 



270 
HEAT TRANSFER 



equation dAldt 
stitution yields 



iA for the balance A, whose solution is A = A exp(it). Sub- 



A = ($100)exp(0.18 X 1) = $119.72 



m 






f(x + Ax) 














/y\^f 


m 




£ —1 1 




'1 






Ax i 




Tangent line 





x x + Ax x 

FIGURE 5-6 

The derivative of a function at a point 
represents the slope of the function 
at that point. 




FIGURE 5-7 

Schematic of the nodes and the nodal 
temperatures used in the development 
of the finite difference formulation 
of heat transfer in a plane wall. 



which is identical to the result for daily compounding. Therefore, replacing a 
differential time interval dt by a finite time interval of At = 1 day gave the 
same result, which leads us into believing that reasonably accurate results can 
be obtained by replacing differential quantities by sufficiently small differ- 
ences. Next, we develop the finite difference formulation of heat conduction 
problems by replacing the derivatives in the differential equations by differ- 
ences. In the following section we will do it using the energy balance method, 
which does not require any knowledge of differential equations. 

Derivatives are the building blocks of differential equations, and thus we 
first give a brief review of derivatives. Consider a function /that depends on 
x, as shown in Figure 5-6. The first derivative of fix) at a point is equivalent 
to the slope of a line tangent to the curve at that point and is defined as 



df(x) 

dx 



A/ 
lim - — : 
a*->o Ax 



lim 



fix + Ax) -fix) 
Ax 



(5-5) 



which is the ratio of the increment A/of the function to the increment Ax of the 
independent variable as Ax — > 0. If we don't take the indicated limit, we will 
have the following approximate relation for the derivative: 



dfix) fix + Ax) - fix) 



dx 



Ax 



(5-6) 



This approximate expression of the derivative in terms of differences is the 
finite difference form of the first derivative. The equation above can also be 
obtained by writing the Taylor series expansion of the function / about the 
point x, 



fix + Ax) = fix) + Ax 



dfix) 
dx 



1 . J 2 fix) 



(5-7) 



and neglecting all the terms in the expansion except the first two. The first 
term neglected is proportional to Ax 2 , and thus the error involved in each step 
of this approximation is also proportional to Ax 2 . However, the commutative 
error involved after M steps in the direction of length L is proportional to Ax 
since MAx 2 = {LI Ax) Ax 2 = LAx. Therefore, the smaller the Ax, the smaller 
the error, and thus the more accurate the approximation. 

Now consider steady one-dimensional heat transfer in a plane wall of thick- 
ness L with heat generation. The wall is subdivided into M sections of equal 
thickness Ax = LIM in the x-direction, separated by planes passing through 
M + 1 points 0, 1, 2, . . . , m — 1, m, m + 1, . . . , M called nodes or nodal 
points, as shown in Figure 5-7. The x-coordinate of any point m is simply 
x,„ = mAx, and the temperature at that point is simply T(x m ) = T„ r 

The heat conduction equation involves the second derivatives of tempera- 
ture with respect to the space variables, such as d 2 T/dx 2 , and the finite differ- 
ence formulation is based on replacing the second derivatives by appropriate 



cen58 93 3_ch05.qxd 9/4/2002 11:41 AM Page 271 



differences. But we need to start the process with first derivatives. Using 
Eq. 5-6, the first derivative of temperature dTldx at the midpoints m — i and 
m + ~ of the sections surrounding the node m can be expressed as 



271 
CHAPTER 5 



dT 

dx 



Ax 



and 



dT 
dx 



T„, + 1 



Ax 



(5-8) 



Noting that the second derivative is simply the derivative of the first deriva- 
tive, the second derivative of temperature at node m can be expressed as 



d 2 T 
dx 2 



dT 

dx 



dT 

dx 



Ax 
" 27* 



Ax 2 



Ax 



Ax 



(5-9) 



which is the finite difference representation of the second derivative at a gen- 
eral internal node m. Note that the second derivative of temperature at a node 
m is expressed in terms of the temperatures at node m and its two neighboring 
nodes. Then the differential equation 



d 2 T 
dx 2 ' 



8_ 
k 







(5-10) 



which is the governing equation for steady one-dimensional heat transfer in a 
plane wall with heat generation and constant thermal conductivity, can be ex- 
pressed in the finite difference form as (Fig. 5-8) 



2T,„ 



Ax 2 



- m + 1 . 6 »; 



0, 



m = 1,2,3, 



,M - 1 



(5-11) 



where g m is the rate of heat generation per unit volume at node m. If the sur- 
face temperatures T and T M are specified, the application of this equation to 
each of the M — \ interior nodes results in M — 1 equations for the determi- 
nation of M — 1 unknown temperatures at the interior nodes. Solving these 
equations simultaneously gives the temperature values at the nodes. If the 
temperatures at the outer surfaces are not known, then we need to obtain two 
more equations in a similar manner using the specified boundary conditions. 
Then the unknown temperatures at M + 1 nodes are determined by solving 
the resulting system of M + 1 equations in M + 1 unknowns simultaneously. 

Note that the boundary conditions have no effect on the finite difference 
formulation of interior nodes of the medium. This is not surprising since the 
control volume used in the development of the formulation does not involve 
any part of the boundary. You may recall that the boundary conditions had no 
effect on the differential equation of heat conduction in the medium either. 

The finite difference formulation above can easily be extended to two- or 
three-dimensional heat transfer problems by replacing each second derivative 
by a difference equation in that direction. For example, the finite difference 
formulation for steady two-dimensional heat conduction in a region with 



Plane wall 
Differential equation: 

dx 2 k 
Valid at every point 



Finite difference equation: 
T .-27/ +7/ . g 

111 - I III III + I ^jn _ 

Ax 2 k ~ 

Valid at discrete points 



KaH 



FIGURE 5-8 

The differential equation is valid at 

every point of a medium, whereas the 

finite difference equation is valid at 

discrete points (the nodes) only. 



cen58 93 3_ch05.qxd 9/4/2002 11:41 AM Page 272 



272 
HEAT TRANSFER 



n+1 




FIGURE 5-9 

Finite difference mesh for two- 
dimensional conduction in 
rectangular coordinates. 



heat generation and constant thermal conductivity can be expressed in rectan- 
gular coordinates as (Fig. 5-9) 



jj, IT rp 

m,n m—l,n m, n + 1 

~Tx 1 + 



2T 



Ay 2 



- m, n — 1 6m, n 



(5-12) 



for m = 1,2,3, ... ,M —1 and n = 1, 2, 3, . . . , N — 1 at any interior node 
(m, n). Note that a rectangular region that is divided into M equal subregions 
in the x-direction and N equal subregions in the y-direction has a total of 
(M + 1)(N + 1) nodes, and Eq. 5-12 can be used to obtain the finite differ- 
ence equations at (M — 1)(N — 1) of these nodes (i.e., all nodes except those 
at the boundaries). 

The finite difference formulation is given above to demonstrate how differ- 
ence equations are obtained from differential equations. However, we will use 
the energy balance approach in the following sections to obtain the numerical 
formulation because it is more intuitive and can handle boundary conditions 
more easily. Besides, the energy balance approach does not require having the 
differential equation before the analysis. 



Plane wall 




Volume 
* element 


















of node m 








°m 








*^cond, left ^^^1 


x * 


^^^W *~cond, ri; 

A general 
interior node 

/ 


L 




1 2 m-1 


m 


m + 1 


M 


X 


U ►!- -1 






r A.v n Ax 






Ax 



















FIGURE 5-10 

The nodal points and volume 
elements for the finite difference 
formulation of one-dimensional 
conduction in a plane wall. 



5-3 - ONE-DIMENSIONAL 

STEADY HEAT CONDUCTION 

In this section we will develop the finite difference formulation of heat con- 
duction in a plane wall using the energy balance approach and discuss how to 
solve the resulting equations. The energy balance method is based on sub- 
dividing the medium into a sufficient number of volume elements and then 
applying an energy balance on each element. This is done by first selecting 
the nodal points (or nodes) at which the temperatures are to be determined and 
then forming elements (or control volumes) over the nodes by drawing lines 
through the midpoints between the nodes. This way, the interior nodes remain 
at the middle of the elements, and the properties at the node such as the 
temperature and the rate of heat generation represent the average properties of 
the element. Sometimes it is convenient to think of temperature as varying 
linearly between the nodes, especially when expressing heat conduction be- 
tween the elements using Fourier's law. 

To demonstrate the approach, again consider steady one-dimensional heat 
transfer in a plane wall of thickness L with heat generation g(x) and constant 
conductivity k. The wall is now subdivided into M equal regions of thickness 
Ax = LIM in the x-direction, and the divisions between the regions are 
selected as the nodes. Therefore, we have M + 1 nodes labeled 0, 1, 2, ... , 
m — 1, m, m + 1, . . . , M, as shown in Figure 5-10. The x-coordinate of any 
node m is simply x m = mAx, and the temperature at that point is T(x m ) = T m . 
Elements are formed by drawing vertical lines through the midpoints between 
the nodes. Note that all interior elements represented by interior nodes are 
full-size elements (they have a thickness of Ax), whereas the two elements at 
the boundaries are half-sized. 

To obtain a general difference equation for the interior nodes, consider the 
element represented by node m and the two neighboring nodes m — 1 and 
m + 1. Assuming the heat conduction to be into the element on all surfaces, 
an energy balance on the element can be expressed as 



cen58 93 3_ch05.qxd 9/4/2002 11:41 AM Page 273 



/Rate of heat\ 
conduction 
at the left 
surface / 



Rate of heat\ /Rate of heat\ 



conduction 

at the right 

surface 



generation 

inside the 

element 



/Rate of change \ 
of the energy 
content of 
the element / 



273 
CHAPTER 5 



or 



2 cond, left ' 2 



cond, right 



A£„ 



At 



(5-13) 



since the energy content of a medium (or any part of it) does not change under 
steady conditions and thus AE element = 0. The rate of heat generation within 
the element can be expressed as 



tin * element 



om 



AAx 



(5-14) 



where g m is the rate of heat generation per unit volume in W/m 3 evaluated at 
node m and treated as a constant for the entire element, and A is heat transfer 
area, which is simply the inner (or outer) surface area of the wall. 

Recall that when temperature varies linearly, the steady rate of heat con- 
duction across a plane wall of thickness L can be expressed as 



e c 



kA 



AT 



(5-15) 



where AT is the temperature change across the wall and the direction of heat 
transfer is from the high temperature side to the low temperature. In the case 
of a plane wall with heat generation, the variation of temperature is not linear 
and thus the relation above is not applicable. However, the variation of tem- 
perature between the nodes can be approximated as being linear in the deter- 
mination of heat conduction across a thin layer of thickness Ax between two 
nodes (Fig. 5-11). Obviously the smaller the distance Ax between two nodes, 
the more accurate is this approximation. (In fact, such approximations are the 
reason for classifying the numerical methods as approximate solution meth- 
ods. In the limiting case of Ax approaching zero, the formulation becomes ex- 
act and we obtain a differential equation.) Noting that the direction of heat 
transfer on both surfaces of the element is assumed to be toward the node m, 
the rate of heat conduction at the left and right surfaces can be expressed as 



T — T 

M cond, left K/i \ 



and Q 



cond, right 



kA 



Ax 



(5-16) 



Substituting Eqs. 5-14 and 5-16 into Eq. 5-13 gives 



kA 



Ax 



kA- 



Ax 



, AAx = 



(5-17) 



which simplifies to 



TT + T 

Ax 2 



m = 1,2,3, 



,M~ 1 



(5-18) 




Linear 



T ,-T f 

kA-a^ *i 

Ax 



kA- 



A.v 



FIGURE 5-1 1 

In finite difference formulation, the 

temperature is assumed to vary 

linearly between the nodes. 



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274 
HEAT TRANSFER 



kA- 



T x -T 2 
Ax 



g 2 AAx 



-Volume 
element 
of node 2 



T - T, T - T . 

kA—. — - ~kA—. — - + e n AAx = 



v kA- 



T -T 

2 3 

Ax 



Ax 



Ax 



T. - 271 + 71 + g 1 AAx z /k = 



(a) Assuming heat transfer to be out of the 
volume element at the right surface. 



kA- 



T x -T 2 
Ax 



g 2 AAx 



-Volume 
element 
of node 2 



T -71 71 - T . 

kA -± — - + kA -*- — - + g^AAx = 



<kA- 



Ax 



Ax 



Ax 



T l - 2T 2 + T 3 + g 2 AAx l /k = 

(b) Assuming heat transfer to be into the 
volume element at all surfaces. 

FIGURE 5-12 

The assumed direction of heat transfer 
at surfaces of a volume element has 
no effect on the finite difference 
formulation. 



which is identical to the difference equation (Eq. 5-11) obtained earlier. 
Again, this equation is applicable to each of the M — 1 interior nodes, and its 
application gives M — 1 equations for the determination of temperatures at 
M + 1 nodes. The two additional equations needed to solve for the M + 1 un- 
known nodal temperatures are obtained by applying the energy balance on the 
two elements at the boundaries (unless, of course, the boundary temperatures 
are specified). 

You are probably thinking that if heat is conducted into the element from 
both sides, as assumed in the formulation, the temperature of the medium will 
have to rise and thus heat conduction cannot be steady. Perhaps a more realis- 
tic approach would be to assume the heat conduction to be into the element on 
the left side and out of the element on the right side. If you repeat the formu- 
lation using this assumption, you will again obtain the same result since the 
heat conduction term on the right side in this case will involve T m — T m + j in- 
stead of T m + j — T m , which is subtracted instead of being added. Therefore, 
the assumed direction of heat conduction at the surfaces of the volume ele- 
ments has no effect on the formulation, as shown in Figure 5-12. (Besides, the 
actual direction of heat transfer is usually not known.) However, it is conve- 
nient to assume heat conduction to be into the element at all surfaces and not 
worry about the sign of the conduction terms. Then all temperature differences 
in conduction relations are expressed as the temperature of the neighboring 
node minus the temperature of the node under consideration, and all conduc- 
tion terms are added. 



Boundary Conditions 



Above we have developed a general relation for obtaining the finite difference 
equation for each interior node of a plane wall. This relation is not applicable 
to the nodes on the boundaries, however, since it requires the presence of 
nodes on both sides of the node under consideration, and a boundary node 
does not have a neighboring node on at least one side. Therefore, we need to 
obtain the finite difference equations of boundary nodes separately. This is 
best done by applying an energy balance on the volume elements of boundary 
nodes. 

Boundary conditions most commonly encountered in practice are the spec- 
ified temperature, specified heat flux, convection, and radiation boundary 
conditions, and here we develop the finite difference formulations for them 
for the case of steady one-dimensional heat conduction in a plane wall of 
thickness L as an example. The node number at the left surface at x = is 0, 
and at the right surface at x = L it is M. Note that the width of the volume el- 
ement for either boundary node is Ax/2. 

The specified temperature boundary condition is the simplest boundary 
condition to deal with. For one-dimensional heat transfer through a plane wall 
of thickness L, the specified temperature boundary conditions on both the left 
and right surfaces can be expressed as (Fig. 5-13) 



7"(0) = T Q = Specified value 
T{L) = T M = Specified value 



(5-19) 



where T and T„, are the specified temperatures at surfaces at x = and x = L, 
respectively. Therefore, the specified temperature boundary conditions are 



cen58 93 3_ch05.qxd 9/4/2002 11:41 AM Page 275 



incorporated by simply assigning the given surface temperatures to the bound- 
ary nodes. We do not need to write an energy balance in this case unless we 
decide to determine the rate of heat transfer into or out of the medium after the 
temperatures at the interior nodes are determined. 

When other boundary conditions such as the specified heat flux, convection, 
radiation, or combined convection and radiation conditions are specified at a 
boundary, the finite difference equation for the node at that boundary is ob- 
tained by writing an energy balance on the volume element at that boundary. 
The energy balance is again expressed as 



2 e 







(5-20) 



275 
CHAPTER 5 



35°C 







Plane wall 




82°C 










L 





1 


2 ••• 




M 



for heat transfer under steady conditions. Again we assume all heat transfer to 
be into the volume element from all surfaces for convenience in formulation, 
except for specified heat flux since its direction is already specified. Specified 
heat flux is taken to be a positive quantity if into the medium and a negative 
quantity if out of the medium. Then the finite difference formulation at the 
node m = (at the left boundary where x = 0) of a plane wall of thickness L 
during steady one-dimensional heat conduction can be expressed as (Fig. 
5-14) 



Qv 



kA 



Ax 



g (AAx/2) = 



(5-21) 



where AAx/2 is the volume of the volume element (note that the boundary ele- 
ment has half thickness), g is the rate of heat generation per unit volume (in 
W/m 3 ) at x = 0, and A is the heat transfer area, which is constant for a plane 
wall. Note that we have Ax in the denominator of the second term instead of 
Ax/2. This is because the ratio in that term involves the temperature difference 
between nodes and 1, and thus we must use the distance between those two 
nodes, which is Ax. 

The finite difference form of various boundary conditions can be obtained 
from Eq. 5-21 by replacing Q left surface by a suitable expression. Next this is 
done for various boundary conditions at the left boundary. 

1. Specified Heat Flux Boundary Condition 



FIGURE 5-13 

Finite difference formulation of 

specified temperature boundary 

conditions on both surfaces 

of a plane wall. 





Ax 

2 




















^- Volume element 










of node 








So 










k i^^B T -T 






surface 




Ax 












L 







1 2 ••• 

x * • Ax— H 




X 



+ kA- 



■-0 



o • , Ax 

FIGURE 5-14 

Schematic for the finite difference 

formulation of the left boundary 

node of a plane wall. 



(joA + kA 



Ax 



+ g (AAx/2) = 



(5-22) 



Special case: Insulated Boundary (q = 0) 

T, -T, 



. -^j^ + £ (AAx/2) = 



(5-23) 



2. Convection Boundary Condition 

Tt-T 



hA{T.„ - T ) + kA 



Ax 



+ g (AAx/2) = 



(5-24) 



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276 
HEAT TRANSFER 




FIGURE 5-15 

Schematic for the finite difference 
formulation of combined convection 
and radiation on the left boundary 
of a plane wall. 



k 


Medium A 

k A 

T , -T 

A 11! - 1 1)1 

» A Ax 


*4,m 


&B,m 


Interface 

Medium B 
k B 

V '" + Ax 


m 




m-\ 




m 


m + 1 

H 


X 












Ax 

2 


Ax 

2 


VI 



T ,-T T ,-T 

k.A-^ m -+k„A-^ ffi 

4 Ajc b Ax 



Ax 



A^ = 
71 2 



FIGURE 5-16 

Schematic for the finite difference 
formulation of the interface boundary 
condition for two mediums A and B 
that are in perfect thermal contact. 



3. Radiation Boundary Condition 



eaA(T* m - T 4 ) + kA ■ 



Ax 



MAAx/2) = 



(5-25) 



4. Combined Convection and Radiation Boundary Condition 

(Fig. 5-15) 

hA(T m - T ) + euA(T* an - T+) + kA -^j^ + g (AAx/2) = 



or 



,A(T x -T a ) + kA 



Ax 



+ g a {AAxl2) = 



(5-26) 



(5-27) 



5. Combined Convection, Radiation, and Heat Flux Boundary 
Condition 

qoA + *A(r« - r ) + eaA(r s 4 urr - T 4 ) + kA -^j^ + g (AA.x/2) = (5-28) 

6. Interface Boundary Condition Two different solid media A and B 
are assumed to be in perfect contact, and thus at the same temperature 
at the interface at node m (Fig. 5-16). Subscripts A and B indicate 
properties of media A and B, respectively. 



T 



Ax 



+ k R A 



T 



Ax 



g A . ,„{AAxl2) + g Bt m (AAx/2) = (5-29) 



In these relations, q is the specified heat flux in W/m 2 , h is the convection 
coefficient, h combmed is the combined convection and radiation coefficient, r m is 
the temperature of the surrounding medium, T san . is the temperature of the 
surrounding surfaces, e is the emissivity of the surface, and cr is the Stefan- 
Boltzman constant. The relations above can also be used for node M on the 
right boundary by replacing the subscript "0" by "M" and the subscript "1" by 
"M - 1". 

Note that absolute temperatures must be used in radiation heat transfer 
calculations, and all temperatures should be expressed in K or R when a 
boundary condition involves radiation to avoid mistakes. We usually try to 
avoid the radiation boundary condition even in numerical solutions since it 
causes the finite difference equations to be nonlinear, which are more difficult 
to solve. 



Treating Insulated Boundary Nodes as Interior Nodes: 
The Mirror Image Concept 

One way of obtaining the finite difference formulation of a node on an insu- 
lated boundary is to treat insulation as "zero" heat flux and to write an energy 
balance, as done in Eq. 5-23. Another and more practical way is to treat the 
node on an insulated boundary as an interior node. Conceptually this is done 



cen58 93 3_ch05.qxd 9/4/2002 11:41 AM Page 277 



by replacing the insulation on the boundary by a mirror and considering the 
reflection of the medium as its extension (Fig. 5-17). This way the node next 
to the boundary node appears on both sides of the boundary node because of 
symmetry, converting it into an interior node. Then using the general formula 
(Eq. 5-18) for an interior node, which involves the sum of the temperatures of 
the adjoining nodes minus twice the node temperature, the finite difference 
formulation of a node m = on an insulated boundary of a plane wall can be 
expressed as 



2T„, + T m 



Sm 



Ax 2 







Mn 



Ax 2 







(5-30) 



which is equivalent to Eq. 5-23 obtained by the energy balance approach. 

The mirror image approach can also be used for problems that possess ther- 
mal symmetry by replacing the plane of symmetry by a mirror. Alternately, we 
can replace the plane of symmetry by insulation and consider only half of the 
medium in the solution. The solution in the other half of the medium is sim- 
ply the mirror image of the solution obtained. 



277 
CHAPTER 5 



Insulation 



Insulated 
boundary 
,- node 



Mirror 



Mirror 
image 



1 



Equivalent 
interior 
, node 



1 



FIGURE 5-17 

A node on an insulated boundary 
can be treated as an interior node by 
replacing the insulation by a mirror. 



EXAMPLE 5-1 Steady Heat Conduction in a Large Uranium Plate 

Consider a large uranium plate of thickness L = 4 cm and thermal conductivity 
k = 28 W/m • °C in which heat is generated uniformly at a constant rate of 
g = 5 X 10 6 W/m 3 . One side of the plate is maintained at 0°C by iced water 
while the other side is subjected to convection to an environment at T x = 30°C 
with a heat transfer coefficient of h = 45 W/m 2 ■ C C, as shown in Figure 5-18. 
Considering a total of three equally spaced nodes in the medium, two at the 
boundaries and one at the middle, estimate the exposed surface temperature of 
the plate under steady conditions using the finite difference approach. 

SOLUTION A uranium plate is subjected to specified temperature on one side 
and convection on the other. The unknown surface temperature of the plate is 
to be determined numerically using three equally spaced nodes. 
Assumptions 1 Heat transfer through the wall is steady since there is no in- 
dication of any change with time. 2 Heat transfer is one-dimensional since 
the plate is large relative to its thickness. 3 Thermal conductivity is constant. 
4 Radiation heat transfer is negligible. 

Properties The thermal conductivity is given to be k = 28 W/m • °C. 
Analysis The number of nodes is specified to be M = 3, and they are chosen 
to be at the two surfaces of the plate and the midpoint, as shown in the figure. 
Then the nodal spacing Ax becomes 



Ax 



M- 1 



0.04 m 
3 - 1 



0.02 m 



We number the nodes 0, 1, and 2. The temperature at node is given to be 
T = C C, and the temperatures at nodes 1 and 2 are to be determined. This 
problem involves only two unknown nodal temperatures, and thus we need to 
have only two equations to determine them uniquely. These equations are ob- 
tained by applying the finite difference method to nodes 1 and 2. 





Uranium 








plate 






o°c 


k = 28 W/m-°C 
g = 5x 10 6 W/m 3 




h 
T„ 


0< 






L 


1 


2 


X 



FIGURE 5-18 

Schematic for Example 5—1. 



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278 
HEAT TRANSFER 



Plate 



It 



Finite difference solution: 



Exact solution: 



T 2 = 136.0°C 

FIGURE 5-19 

Despite being approximate in nature, 
highly accurate results can be 
obtained by numerical methods. 



Node 1 is an interior node, and the finite difference formulation at that node 
is obtained directly from Eq. 5-18 by setting m = 1: 



2T, + T 2 



Ax 2 



- 277] + T 2 
I? 



2T, - T, 



gA* 2 



(1) 



Node 2 is a boundary node subjected to convection, and the finite difference 
formulation at that node is obtained by writing an energy balance on the volume 
element of thickness Ax/2 at that boundary by assuming heat transfer to be into 
the medium at all sides: 



hA(T a -T 2 ) + kA- 



T 2 



Ax 



+ g\(AAxl2) = 



Canceling the heat transfer area A and rearranging give 

r, 



1 + h -f\T 7 



hAx gAx 2 

k °° 2k 



(2) 



Equations (1) and (2) form a system of two equations in two unknowns T x and 
T 2 . Substituting the given quantities and simplifying gives 



2T { - T 2 
- 1.0327; 



71.43 
-36.68 



(in °C) 
(in °C) 



This is a system of two algebraic equations in two unknowns and can be solved 
easily by the elimination method. Solving the first equation for T x and substi- 
tuting into the second equation result in an equation in 7~ 2 whose solution is 

T 2 = 136.1°C 

This is the temperature of the surface exposed to convection, which is the 
desired result. Substitution of this result into the first equation gives 7"! = 
103. 8 C C, which is the temperature at the middle of the plate. 

Discussion The purpose of this example is to demonstrate the use of the finite 
difference method with minimal calculations, and the accuracy of the result 
was not a major concern. But you might still be wondering how accurate the re- 
sult obtained above is. After all, we used a mesh of only three nodes for the 
entire plate, which seems to be rather crude. This problem can be solved ana- 
lytically as described in Chapter 2, and the analytical (exact) solution can be 
shown to be 



T(x) 



0.5ghL 2 /k + gL+ T^h gx 2 
hL+ k X ~2k 



Substituting the given quantities, the temperature of the exposed surface of the 
plate at x = L = 0.04 m is determined to be 136. 0°C, which is almost identi- 
cal to the result obtained here with the approximate finite difference method 
(Fig. 5-19). Therefore, highly accurate results can be obtained with numerical 
methods by using a limited number of nodes. 



cen58 93 3_ch05.qxd 9/4/2002 11:41 AM Page 279 



279 
CHAPTER 5 



EXAMPLE 5-2 Heat Transfer from Triangular Fins 

Consider an aluminum alloy fin (k = 180 W/m • °C) of triangular cross section 
with length L = 5 cm, base thickness b = 1 cm, and very large width w in the 
direction normal to the plane of paper, as shown in Figure 5-20. The base of 
the fin is maintained at a temperature of T = 200°C. The fin is losing heat 
to the surrounding medium at 7" x = 25°C with a heat transfer coefficient of 
ft = 15 W/m 2 • °C. Using the finite difference method with six equally spaced 
nodes along the fin in the x-direction, determine (a) the temperatures at the 
nodes, (£>) the rate of heat transfer from the fin for w = 1 m, and (c) the fin 
efficiency. 



SOLUTION A long triangular fin attached to a surface is considered. The nodal 
temperatures, the rate of heat transfer, and the fin efficiency are to be deter- 
mined numerically using six equally spaced nodes. 

Assumptions 1 Heat transfer is steady since there is no indication of any 
change with time. 2 The temperature along the fin varies in the x direction only. 
3 Thermal conductivity is constant. 4 Radiation heat transfer is negligible. 

Properties The thermal conductivity is given to be k = 180 W/m ■ °C. 

Analysis (a) The number of nodes in the fin is specified to be M = 6, and their 
location is as shown in the figure. Then the nodal spacing Ax becomes 



Ax 



M- 1 



0.05 m 
6 - 1 



0.01 m 



The temperature at node is given to be T = 200°C, and the temperatures at 
the remaining five nodes are to be determined. Therefore, we need to have five 
equations to determine them uniquely. Nodes 1, 2, 3, and 4 are interior nodes, 
and the finite difference formulation for a general interior node m is obtained 
by applying an energy balance on the volume element of this node. Noting that 
heat transfer is steady and there is no heat generation in the fin and assuming 
heat transfer to be into the medium at all sides, the energy balance can be ex- 
pressed as 



2 e = o 



M„ 



Ax 



kA 



right 



Ax 



hA conv (T^ - TJ = 



Note that heat transfer areas are different for each node in this case, and using 
geometrical relations, they can be expressed as 



A lcft = (Height X Width) s 
(Height X Width) 4 



right 



2w[L - (m - l/2)Ax]tan Q 
2w[L - (m + l/2)Ax]tan 6 



2 X Length X Width = 2w(Ax/cos 0) 



Substituting, 



2kw[L - (m - |)Ax]tan ■ 



2kw[L - (m + i)Ax]tan Q 



Ax 



Ax 



l,n | , 2wAx,_ „ . „ 

— + h ^(T„ - TJ = 

cos 




[L-(m — )A.v]tan ( 

FIGURE 5-20 

Schematic for Example 5-2 and the 

volume element of a general 

interior node of the fin. 



cen58933_ch05.qxd 9/4/2002 11:41 AM Page 2i 



280 
HEAT TRANSFER 



Ax/2 




FIGURE 5-21 

Schematic of the volume element of 
node 5 at the tip of a triangular fin. 



Dividing each term by 2kwL tan 0/Ax gives 
(.T m - 1 ~T m ) + 



1 - (m - |) — 



1 / _l_ K A* 

1 - (m + i) y 



+ 



h(Ax) 1 



kL sin 



(r. - rj = o 



Note that 



tan 



fr/2 = 0.5 cm 
L 5 cm 



0.1 



= tan-'O.l = 5.71° 



Also, sin 5.71° = 0.0995. Then the substitution of known quantities gives 



(5.5 - m)T„, 



(10.00838 - 2m)T„, + (4.5 - m)T„, 



-0.209 



Now substituting 1, 2, 3, and 4 for m results in these finite difference equa- 
tions for the interior nodes.- 



m = 1 

m = 2 

m = 3 

m = 4 



-900.209 

3.57*! - 6.008387 2 + 2.5T 3 = -0.209 
2.5T 2 - 4.008387 3 + 1.5T 4 = -0.209 
I.57/3 - 2.008387 4 + 0.57 5 = -0.209 



(1) 
(2) 
(3) 
(4) 



The finite difference equation for the boundary node 5 is obtained by writing an 
energy balance on the volume element of length Ax/2 at that boundary, again by 
assuming heat transfer to be into the medium at all sides (Fig. 5-21): 



kA„ 



7 4 ~7 5 

Ax 



+ M conv (T« - 7 5 ) = 



where 



2w — tan 



and 



A- 



2w 



Ax/2 
cos 



Canceling w in all terms and substituting the known quantities gives 
7 4 - 1.008387 5 = -0.209 



(5) 



Equations (1) through (5) form a linear system of five algebraic equations in five 
unknowns. Solving them simultaneously using an equation solver gives 

7, = 198.6°C, 7 2 = 197.1°C, 7 3 = 195.7°C, 
7 4 = 194.3°C, 7 5 = 192.9°C 

which is the desired solution for the nodal temperatures. 

(b) The total rate of heat transfer from the fin is simply the sum of the heat 
transfer from each volume element to the ambient, and for w= 1 m it is deter- 
mined from 



cen58 93 3_ch05.qxd 9/4/2002 11:41 AM Page 281 



5 5 

x£ fin £j x£ element, m 

in = m = 



£j '^*conv, fflV m ^ «=/ 



Noting that the heat transfer surface area is wAx/cos 6 for the boundary nodes 
and 5, and twice as large for the interior nodes 1, 2, 3, and 4, we have 



G fi „ = h 



h 



wAx 



[(T - t„) + 2(r, - rj + 2(T 2 - r„) + 2(r 3 - r„) 



cos G 

2(r 4 - r„) + (r 5 - r„)] 



wAx 
cos G 



[r + 2(T, + r 2 + r 3 + r 4 ) + r 5 - iotj 



(1 m)(0.01 m) 
(15 W/m 2 • °C) ,- 10 [200 + 2 X 785.7 + 192.9 - 10 X 25] 



cos 5.71° 



258.4 W 



(c) If the entire fin were at the base temperature of T = 200°C, the total rate 
of heat transfer from the fin for w = 1 m would be 



e„ 



hA„ 



(T - r„) = h(2wL/cos Q)(T - T m ) 



(15 W/m 2 • °C)[2(1 m)(0.05 m)/cos5.71°](200 - 25)°C 
263.8 W 



Then the fin efficiency is determined from 

gfin 258.4 W 



%in 



e, 



263.8 W 



0.98 



which is less than 1, as expected. We could also determine the fin efficiency in 
this case from the proper fin efficiency curve in Chapter 3, which is based on 
the analytical solution. We would read 0.98 for the fin efficiency, which is iden- 
tical to the value determined above numerically. 



281 
CHAPTER 5 



The finite difference formulation of steady heat conduction problems usu- 
ally results in a system of N algebraic equations in N unknown nodal temper- 
atures that need to be solved simultaneously. When N is small (such as 2 or 3), 
we can use the elementary elimination method to eliminate all unknowns ex- 
cept one and then solve for that unknown (see Example 5-1). The other un- 
knowns are then determined by back substitution. When N is large, which is 
usually the case, the elimination method is not practical and we need to use a 
more systematic approach that can be adapted to computers. 

There are numerous systematic approaches available in the literature, and 
they are broadly classified as direct and iterative methods. The direct meth- 
ods are based on a fixed number of well-defined steps that result in the solu- 
tion in a systematic manner. The iterative methods, on the other hand, are 
based on an initial guess for the solution that is refined by iteration until a 
specified convergence criterion is satisfied (Fig. 5-22). The direct methods 
usually require a large amount of computer memory and computation time, 



Direct methods: 

Solve in a systematic manner following a 

series of well-defined steps. 
Iterative methods: 

Start with an initial guess for the solution, 

and iterate until solution converges. 



FIGURE 5-22 

Two general categories of solution 

methods for solving systems 

of algebraic equations. 



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) 




n-l 


\y 














































































2 

1 








































Ax 


Ar 





































1 2 



| « 1 • • • M x 

m — 1 m + 1 



FIGURE 5-23 

The nodal network for the finite 
difference formulation of two- 
dimensional conduction in 
rectangular coordinates. 



n + 1 



Ay 



Volume 
element n 



m — 1, n I 



Ay 



n-l 



-Ax- 



m, n + 1 



m, n 



-+- 



I 

m, n-l 



-Ax- 



m + 1 , n 



I 



y t m - 1 



m + 1 



FIGURE 5-24 

The volume element of a general 
interior node (m, n) for two- 
dimensional conduction in 
rectangular coordinates. 



5^ - TWO-DIMENSIONAL 

STEADY HEAT CONDUCTION 

In Section 5-3 we considered one-dimensional heat conduction and assumed 
heat conduction in other directions to be negligible. Many heat transfer prob- 
lems encountered in practice can be approximated as being one-dimensional, 
but this is not always the case. Sometimes we need to consider heat transfer in 
other directions as well when the variation of temperature in other directions 
is significant. In this section we will consider the numerical formulation and 
solution of two-dimensional steady heat conduction in rectangular coordinates 
using the finite difference method. The approach presented below can be ex- 
tended to three-dimensional cases. 

Consider a rectangular region in which heat conduction is significant in the 
x- and y-directions. Now divide the x-y plane of the region into a rectangular 
mesh of nodal points spaced Ax and Ay apart in the x- and y-directions, 
respectively, as shown in Figure 5-23, and consider a unit depth of Az = 1 
in the z-direction. Our goal is to determine the temperatures at the nodes, 
and it is convenient to number the nodes and describe their position by 
the numbers instead of actual coordinates. A logical numbering scheme for 
two-dimensional problems is the double subscript notation (m, n) where 
m = 0,1,2, ... ,M is the node count in the x-direction and n = 0, 1, 2, . . . ,N 
is the node count in the _y-direction. The coordinates of the node (m, n) are 
simply x = mAx and y = nAy, and the temperature at the node (m, n) is 
denoted by T nun . 

Now consider a volume element of size Ax X Ay X 1 centered about a gen- 
eral interior node (m, n) in a region in which heat is generated at a rate of g and 
the thermal conductivity k is constant, as shown in Figure 5-24. Again 
assuming the direction of heat conduction to be toward the node under 
consideration at all surfaces, the energy balance on the volume element can be 
expressed as 



f Rate of heat conduction \ 

at the left, top, right, 
\ and bottom surfaces / 



+ 



I Rate of heat | 
generation inside 
\ the element 



I Rate of change of \ 

the energy content 

\ of the element / 



or 



t^cond, left ~*~ second, top ~*~ \i cond. right "*" ticond, bottom ~*~ ^eL 



A£„ 



At 



(5-31) 



for the steady case. Again assuming the temperatures between the adja- 
cent nodes to vary linearly and noting that the heat transfer area is 
A x = Ay X 1 = Ay in the .^-direction and A y = Ax X 1 = Ax in the y-direction, 
the energy balance relation above becomes 



cen58 93 3_ch05.qxd 9/4/2002 11:41 AM Page 283 



T — T 

m—\,n m, n 



kAy — + kAx — : — r + kAy 



Ax Ay Ax 



kAx — + g nh „ Ax Ay = (5-32) 



Dividing each term by Ax X Ay and simplifying gives 

(5-33) 



* m— I, n ^"* m, n m+ 1, n * m, n — 1 in, n * m, n + 1 am, n 

Ax~ 2 + A^ + ~k~ 

form= 1, 2, 3, ... ,M — 1 and n = 1, 2, 3, . . . , N — 1. This equation is iden- 
tical to Eq. 5-12 obtained earlier by replacing the derivatives in the differen- 
tial equation by differences for an interior node (m, n). Again a rectangular 
region M equally spaced nodes in the x-direction and N equally spaced nodes 
in the y-direction has a total of (M + 1)(7V + 1) nodes, and Eq. 5-33 can be 
used to obtain the finite difference equations at all interior nodes. 

In finite difference analysis, usually a square mesh is used for sim- 
plicity (except when the magnitudes of temperature gradients in the x- and 
y-directions are very different), and thus Ax and Ay are taken to be the same. 
Then Ax = Ay = /, and the relation above simplifies to 

g mn p 

T m - i, „ + T m + (] „ + T m „ + i + T m „ _ j — 4T m ^ „ H - = (5-34) 

That is, the finite difference formulation of an interior node is obtained by 

adding the temperatures of the four nearest neighbors of the node, subtracting 
four times the temperature of the node itself and adding the heat generation 
term. It can also be expressed in this form, which is easy to remember: 

'ka "r" T , op + T nght + i bottom — 4i node H - = (5-35) 

When there is no heat generation in the medium, the finite difference equa- 
tion for an interior node further simplifies to r node = (r left + r top + r right + 
r bottom )/4, which has the interesting interpretation that the temperature of each 
interior node is the arithmetic average of the temperatures of the four neigh- 
boring nodes. This statement is also true for the three-dimensional problems 
except that the interior nodes in that case will have six neighboring nodes in- 
stead of four. 

Boundary Nodes 

The development of finite difference formulation of boundary nodes in two- 
(or three-) dimensional problems is similar to the development in the one- 
dimensional case discussed earlier. Again, the region is partitioned between 
the nodes by forming volume elements around the nodes, and an energy bal- 
ance is written for each boundary node. Various boundary conditions can be 
handled as discussed for a plane wall, except that the volume elements 
in the two-dimensional case involve heat transfer in the y-direction as well as 
the x-direction. Insulated surfaces can still be viewed as "mirrors, " and the 



283 
CHAPTER 5 



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284 
HEAT TRANSFER 



Volume element 
of node 2 



Boundary 

subjected 

/ j to convection 




e,ef, + e top + Q nght + e bollom + — = o 

FIGURE 5-25 

The finite difference formulation of 
a boundary node is obtained by 
writing an energy balance 
on its volume element. 







Convection 






y 


i /', T„ 




1 2 /" 


3 / \ 


t I 

Av 

J 






/ \ Ax = Ay = I 


4 i 


5 ! 


6 7 \ t 8 9 t Vr 


t 
Av 

1 


10] 


— h — 


h — 

12] 


h — 

13] 


h — 

14] 15 










I 
90°C 






A' 




*-Ax^ 


*-Ax^ 


-^A*^ 


--Ajt- 


»-A*- 





FIGURE 5-26 

Schematic for Example 5-3 and 
the nodal network (the boundaries 
of volume elements of the nodes are 
indicated by dashed lines). 



h,T„ 



I 



h,T a 



Ti 



(b) Node 2 



(a) Node 1 

FIGURE 5-27 

Schematics for energy balances on the 
volume elements of nodes 1 and 2. 



mirror image concept can be used to treat nodes on insulated boundaries as in- 
terior nodes. 

For heat transfer under steady conditions, the basic equation to keep in mind 
when writing an energy balance on a volume element is (Fig. 5-25) 



S Q + SKl 







(5-36) 



whether the problem is one-, two-, or three-dimensional. Again we assume, 
for convenience in formulation, all heat transfer to be into the volume ele- 
ment from all surfaces except for specified heat flux, whose direction is al- 
ready specified. This is demonstrated in Example 5-3 for various boundary 
conditions. 



EXAMPLE 5-3 Steady Two-Dimensional Heat Conduction 
in L-Bars 

Consider steady heat transfer in an L-shaped solid body whose cross section is 
given in Figure 5-26. Heat transfer in the direction normal to the plane of the 
paper is negligible, and thus heat transfer in the body is two-dimensional. The 
thermal conductivity of the body is k = 15 W/m • °C, and heat is generated in 
the body at a rate of g = 2 X 10 6 W/m 3 . The left surface of the body is insu- 
lated, and the bottom surface is maintained at a uniform temperature of 90°C. 
The entire top surface is subjected to convection to ambient air at 7" x = 25°C 
with a convection coefficient of h = 80 W/m 2 • °C, and the right surface is sub- 
jected to heat flux at a uniform rate of q R = 5000 W/m 2 . The nodal network of 
the problem consists of 15 equally spaced nodes with Ax = Ay = 1.2 cm, as 
shown in the figure. Five of the nodes are at the bottom surface, and thus their 
temperatures are known. Obtain the finite difference equations at the remain- 
ing nine nodes and determine the nodal temperatures by solving them. 

SOLUTION Heat transfer in a long L-shaped solid bar with specified boundary 
conditions is considered. The nine unknown nodal temperatures are to be de- 
termined with the finite difference method. 

Assumptions 1 Heat transfer is steady and two-dimensional, as stated. 2 Ther- 
mal conductivity is constant. 3 Heat generation is uniform. 4 Radiation heat 
transfer is negligible. 

Properties The thermal conductivity is given to be k = 15 W/m • °C. 
Analysis We observe that all nodes are boundary nodes except node 5, which 
is an interior node. Therefore, we will have to rely on energy balances to obtain 
the finite difference equations. But first we form the volume elements by parti- 
tioning the region among the nodes equitably by drawing dashed lines between 
the nodes. If we consider the volume element represented by an interior node 
to be full size (i.e., Ax X Ay X 1), then the element represented by a regular 
boundary node such as node 2 becomes half size (i.e., Ax X Ay/2 X 1), and 
a corner node such as node 1 is quarter size (i.e., Ax/2 X Ay/2 X 1). Keeping 
Eq. 5-36 in mind for the energy balance, the finite difference equations for 
each of the nine nodes are obtained as follows: 

(a) Node 1. The volume element of this corner node is insulated on the left and 
subjected to convection at the top and to conduction at the right and bottom 
surfaces. An energy balance on this element gives [Fig. 5-27a] 



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. 5-276] 

Ay 7, -T 2 T s - T 2 Ay T, - T 2 Av 

hax(T^ - T 2 ) + k^-^—^ + kax^—^ + k^f^—^ + g 2 Ax— = 
2 Ax Ay 2 ax 2 

Taking Ax = Ay = /, it simplifies to 

/ 2hl\ 2hl R2I 2 

r t - (4 + ±f\ T 2 + T,+ 2T 5 = -^r_ - =p 

(c) Node 3. The volume element of this corner node is subjected to convection 
at the top and right surfaces and to conduction at the bottom and left surfaces. 
An energy balance on this element gives [Fig. 5-28a] 



^♦£W. 



T 3 ) 



axT 6 



Ay 



+ k 



Ay T 2 - T, 



Ax Ay 



Taking Ax = Ay = /, it simplifies to 

, 2hl 



T 3 + T 6 



2hl, 



Ax + ^ 3 "2 2 



2k 



(d) Node 4. This node is on the insulated boundary and can be treated as an 
interior node by replacing the insulation by a mirror. This puts a reflected image 
of node 5 to the left of node 4. Noting that Ax = Ay = /, the general interior 
node relation for the steady two-dimensional case (Eq. 5-35) gives [Fig. 5-286] 

T 5 + F, + T 5 + T l0 -4T 4 + ^ = 



or, noting that T 10 = 90° C, 



T, - 47*. + 27\ 



-90 



&/ 2 



(e) Node 5. This is an interior node, and noting that Ax = Ay = /, the finite 
difference formulation of this node is obtained directly from Eq. 5-35 to be 
[Fig. 5-29a] 

8sl 2 
T 4 + T 2 + T 6 + T u -4T 5 + — = 



h,T m 



Mirror 



(5) 



h,T~ 



EH 

— 1 ♦ 



10 



(a) Node 3 



(b) Node 4 

FIGURE 5-28 

Schematics for energy balances on the 
volume elements of nodes 3 and 4. 



♦ 2 



4 1 

4 1- 



— 4 1 — P, 

5 1 1 



ill 



12 



(a) Node 5 



(6) Node 6 

FIGURE 5-29 

Schematics for energy balances on the 
volume elements of nodes 5 and 6. 



cen58 93 3_ch05.qxd 9/4/2002 11:41 AM Page 286 



286 
HEAT TRANSFER 






tv i 



1r 



15 



13 J 

FIGURE 5-30 

Schematics for energy balances on the 
volume elements of nodes 7 and 9. 



or, noting that T n = 90°C, 

T 2 + T 4 - AT, + T 6 



-90 



g S l 7 



if) Node 6. The volume element of this inner corner node is subjected to con- 
vection at the L-shaped exposed surface and to conduction at other surfaces. 
An energy balance on this element gives [Fig. 5-296] 



# + t)<- 



Ay TV 



kAx- 



T„ 



, a ^5 ~~ T 6 \ x T 3 - T 6 

kAy — ; h k 



2 Ax Ay 

3AxAy 



Ax 



Ay 







Taking Ax = Ay = /and noting that T l2 = 90°C, it simplifies to 



T 3 + 2T S - 6 + 



2hl 



T< + T 7 



2hl 
180 -^r. 
k 



3g 6 / 2 
2fc 



(g) Node 7. The volume element of this boundary node is subjected to convec- 
tion at the top and to conduction at the right, bottom, and left surfaces. An en- 
ergy balance on this element gives [Fig. 5-30a] 



hbx(T a - r 7 ) 

+ k 



AyT s 



2 Ax 
Ay T 6 - TV 



kAx 



T 13 



Ay 



Ax 



Av 
g 7 Ax— = 



Taking Ax = Ay = /and noting that T 13 = 90°C, it simplifies to 



4 + ^]: 



2hl 

■180 -^tv 

k 



k 



{h) Node 8. This node is identical to Node 7, and the finite difference formu- 
lation of this node can be obtained from that of Node 7 by shifting the node 
numbers by 1 (i.e., replacing subscript m by m + 1). It gives 



T 7 



4+fr 



. 180 _2« *; 

k k 



(/') Node 9. The volume element of this corner node is subjected to convection 
at the top surface, to heat flux at the right surface, and to conduction at the 
bottom and left surfaces. An energy balance on this element gives [Fig. 5-306] 



, Ax 
h^(T„ 



T 9 ) + q R 



Ay , , Ax T v . 



AyT % 



Ay 



Ax 



Ax Ay 
2 2 







Taking Ax = Ay = /and noting that 7" 15 = 90°C, it simplifies to 
T K -\2 + ^)T Q = -90 



k k " 2k 



cen58 93 3_ch05.qxd 9/4/2002 11:41 AM Page 287 



This completes the development of finite difference formulation for this prob- 
lem. Substituting the given quantities, the system of nine equations for the 
determination of nine unknown nodal temperatures becomes 



-2.0647*, + T 2 + T 4 = 


-11.2 


4.1287; + r 3 + 27* 5 = 


-22.4 


T 2 - 2.1287* 3 + T 6 = 


-12.8 


T { - 47* 4 + 27* 5 = 


-109.2 


T 2 + T 4 - 4T 5 + T 6 = 


-109.2 


2T 5 - 6.1287* 6 + T 7 = 


-212.0 


T 6 - 4A28T-, + T g = 


-202.4 


T 7 - 4.128r 8 + T 9 = 


-202.4 


7g - 2.0647*, = 


-105.2 



which is a system of nine algebraic equations with nine unknowns. Using an 
equation solver, its solution is determined to be 



T,= 


112.1°C 


T 2 = 


110.8°C 


7 , 3 = 


106.6°C 


T 4 = 


109.4°C 


T s = 


108.1°C 


T 6 = 


103.2°C 


T 7 = 


97.3°C 


T s = 


96.3°C 


T 9 = 


97.6°C 



Note that the temperature is the highest at node 1 and the lowest at node 8. 
This is consistent with our expectations since node 1 is the farthest away from 
the bottom surface, which is maintained at 90°C and has one side insulated, 
and node 8 has the largest exposed area relative to its volume while being close 
to the surface at 90°C. 



287 
CHAPTER 5 



Irregular Boundaries 

In problems with simple geometries, we can fill the entire region using simple 
volume elements such as strips for a plane wall and rectangular elements for 
two-dimensional conduction in a rectangular region. We can also use cylin- 
drical or spherical shell elements to cover the cylindrical and spherical bodies 
entirely. However, many geometries encountered in practice such as turbine 
blades or engine blocks do not have simple shapes, and it is difficult to fill 
such geometries having irregular boundaries with simple volume elements. 
A practical way of dealing with such geometries is to replace the irregular 
geometry by a series of simple volume elements, as shown in Figure 5-31. 
This simple approach is often satisfactory for practical purposes, especially 
when the nodes are closely spaced near the boundary. More sophisticated ap- 
proaches are available for handling irregular boundaries, and they are com- 
monly incorporated into the commercial software packages. 



EXAMPLE 5-4 Heat Loss through Chimneys 

Hot combustion gases of a furnace are flowing through a square chimney made 
of concrete (k = 1.4 W/m • °C). The flow section of the chimney is 20 cm X 
20 cm, and the thickness of the wall is 20 cm. The average temperature of the 



Actual boundary 









^- Appro 


ximation 













FIGURE 5-31 

Approximating an irregular 
boundary with a rectangular mesh. 



cen58933_ch05.qxd 9/4/2002 11:41 AM Page 2i 



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 1-m-long section of the chimney. 



Symmetry lines 
(Equivalent to insulation) 




Representative 
7" k section of chimney 

FIGURE 5-32 

Schematic of the chimney discussed in 
Example 5^4 and the nodal network 
for a representative section. 



h,T„ 



h,T w 



1 


\ 


2 




V 


.4 



(a) Node 1 (b) Node 2 

FIGURE 5-33 

Schematics for energy balances on the 
volume elements of nodes 1 and 2. 



SOLUTION Heat transfer through a square chimney is considered. The nodal 
temperatures and the rate of heat loss per unit length are to be determined with 
the finite difference method. 

Assumptions 1 Heat transfer is steady since there is no indication of change 
with time. 2 Heat transfer through the chimney is two-dimensional since the 
height of the chimney is large relative to its cross section, and thus heat con- 
duction through the chimney in the axial direction is negligible. It is tempting 
to simplify the problem further by considering heat transfer in each wall to be 
one-dimensional, which would be the case if the walls were thin and thus the 
corner effects were negligible. This assumption cannot be justified in this case 
since the walls are very thick and the corner sections constitute a considerable 
portion of the chimney structure. 3 Thermal conductivity is constant. 

Properties The properties of chimney are given to be k = 1.4 W/m • °C and 
b= 0.9. 

Analysis The cross section of the chimney is given in Figure 5-32. The most 
striking aspect of this problem is the apparent symmetry about the horizontal 
and vertical lines passing through the midpoint of the chimney as well as the 
diagonal axes, as indicated on the figure. Therefore, we need to consider only 
one-eighth of the geometry in the solution whose nodal network consists of nine 
equally spaced nodes. 

No heat can cross a symmetry line, and thus symmetry lines can be treated 
as insulated surfaces and thus "mirrors" in the finite difference formulation. 
Then the nodes in the middle of the symmetry lines can be treated as interior 
nodes by using mirror images. Six of the nodes are boundary nodes, so we will 
have to write energy balances to obtain their finite difference formulations. First 
we partition the region among the nodes equitably by drawing dashed lines be- 
tween the nodes through the middle. Then the region around a node surrounded 
by the boundary or the dashed lines represents the volume element of the node. 
Considering a unit depth and using the energy balance approach for the bound- 
ary nodes (again assuming all heat transfer into the volume element for conve- 
nience) and the formula for the interior nodes, the finite difference equations 
for the nine nodes are determined as follows: 

(a) Node 1. On the inner boundary, subjected to convection, Figure 5-33a 



Ax AyTj-r, 

+hl -(T i -T0 + k T ^ r 



Ax T 3 - T, 

1 2 Ay 



+ = 



Taking Ax = Ay = /, it simplifies to 

h,l\ 



Ti + T 2 + T, 



h,l 



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289 
CHAPTER 5 



(b) 


Node 2. 


On the inner boundary, subjected to convection, Figure 


5-336 






k Ay 

K 2 


-4 — - + h t == (Ti -T 2 ) + + kAx -4 — - = 

Ax 2 -' Ay 




Tak 


ng Ax = 


Ay = 


= 1, it simplifies to 

T l -[3 + ^JT 2 + 2T 4 =-^T, 




(c) 


Nodes 3 


, 4, and 5. (Interior nodes, Fig. 5-34) 










Node 3: T 4 + T, + T 4 + T 6 - 47/ 3 = 










Node 4: T, + T 2 + T 5 + T 7 - 4T 4 = 










Node 5: T 4 + T 4 + T s + T s - 47/ 5 = 




(d) 


Node 6 


. (On 


the outer boundary, subjected to convection anc 

A x T 3 ~ T 6 Ay T 7 - T 6 
2 Ay 2 Ax 

+ h ^ (T - T 6 ) + eo- ^ (r s 4 ky - 7 6 4 ) = 


radiation) 


Tak 


ng Ax = 


Ay- 


= /, it simplifies to 








1 h„l\ h„l prrl 

T 2 + T 3 -(2 + ^JT 6 =-^T ~^{Ti y - T*) 




(e) 


Node 7 


(On 


the outer boundary, subjected to convection anc 


radiation, 


Fig 


5-35) 


k 


Ay T 6 - T 7 T 4 - T 7 Ay T, - T 7 

_ . + kAx + k _ . 
2 Ax Ay 2 Ax 

+ h o Ax(T - T 7 ) + eo-Ax(r 4 ky - Tf) = 




Tak 


ng Ax = 


Ay- 


= /, it simplifies to 






2T 4 


+ T 6 


/ 2hJ\ 2hJ 2eo-/ , 

- u + -^j t 7 + t s = -^t -^pca - 


r 4 ) 


(f) 


Node 8. 


Same as Node 7, except shift the node numbers up by 


1 (replace 


4 by 5, 6 by 


7, 7 


by 8, and 8 by 9 in the last relation) 






27/ 5 


+ T 7 


/ 2h„l\ 2h„l 2f(tI 


r 8 4 ) 


(g) 


Node 9 


(On 


the outer boundary, subjected to convection anc 


radiation, 


Fig 


5-35) 










Av r 8 - r, a r Ax 






I I 

(4) i Minor i 




Mirror 

FIGURE 5-34 

Converting the boundary 

nodes 3 and 5 on symmetry lines to 

interior nodes by using mirror images. 



Insulation 




h, T„ 



sky 



FIGURE 5-35 

Schematics for energy balances on the 
volume elements of nodes 7 and 9. 



cen58 93 3_ch05.qxd 9/4/2002 11:42 AM Page 29C 



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 5-36 

The variation of temperature 
in the chimney. 



Taking Ax = Ay = /, it simplifies to 



1 + 



h J 



hj 



eo7 



(n 



sky 



This problem involves radiation, which requires the use of absolute tempera- 
ture, and thus all temperatures should be expressed in Kelvin. Alternately, we 
could use C C for all temperatures provided that the four temperatures in the ra- 
diation terms are expressed in the form (7~+ 273) 4 . Substituting the given 
quantities, the system of nine equations for the determination of nine unknown 
nodal temperatures in a form suitable for use with the Gauss-Seidel iteration 
method becomes 



(T 2 + T 3 + 2865)/7 



T 2 = (T x + 2T 4 + 2865)/8 
T } = (7, + 2T 4 + T 6 )/4 
T 4 = (T 2 + T 3 + T 5 + 7V)/4 
T 5 = (27 4 + 27 8 )/4 



T-, = (27/ 4 + T 6 + T s + 912.4 - 0.729 X 10-' r 7 4 )/7 
r 8 = (2T 5 + T 7 + T g + 912.4 - 0.729 X 10-" r 8 4 )/7 
T 9 = (T s + 456.2 - 0.3645 X 1Q-" T 9 4 )/2.5 



which is a system of nonlinear equations. Using an equation solver, its solution 
is determined to be 



r, 


= 545.7 K = 


272.6°C 


T 2 


= 529.2 K = 


256. 1°C 


T 3 


= 425.2 K = 


152.1°C 


T 4 


= 411.2 K = 


138.0°C 


T 5 


= 362.1 K = 


89.0°C 


T 6 


= 332.9 K = 


59.7°C 


Ti 


= 328.1 K = 


54.9°C 


Ts 


= 313.1 K = 


39.9°C 


T 9 


= 296.5 K = 


23.4°C 



The variation of temperature in the chimney is shown in Figure 5-36. 

Note that the temperatures are highest at the inner wall (but less than 
300°C) and lowest at the outer wall (but more that 260 K), as expected. 

The average temperature at the outer surface of the chimney weighed by the 
surface area is 

_ (o.5r 6 + t 7 + r 8 + o.5r 9 ) 

wall, out (Q5 + 1 + 1+ Q5) 

_ 0.5 X 332.9 + 328.1 + 313.1 + 0.5 X 296.5 _ ..„ ,„ 
— ~ — 318.6 K. 



Then the rate of heat loss through the 1-m-long section of the chimney can be 
determined approximately from 



cen58 93 3_ch05.qxd 9/4/2002 11:42 AM Page 291 



Q 



chimney 



"o^o (-' w 



T ) + ectA (r wa n_ oul r sky ) 



= (21 W/m 2 • K)[4 X (0.6 m)(l m)](318.6 - 293)K 

+ 0.9(5.67 X 10- 8 W/m 2 ■ K 4 ) 

[4 X (0.6 m)(l m)](318.6 K) 4 - (260 K) 4 ] 

= 1291 + 702 = 1993 W 

We could also determine the heat transfer by finding the average temperature of 
the inner wall, which is (272.6 + 256.D/2 = 264.4°C, and applying Newton's 
law of cooling at that surface: 

G chimney = "i A (Xi ~ ^ wall, in) 

= (70 W/m 2 • K)[4 X (0.2 m)(l m)](300 - 264.4)°C = 1994 W 

The difference between the two results is due to the approximate nature of the 
numerical analysis. 

Discussion We used a relatively crude numerical model to solve this problem 
to keep the complexities at a manageable level. The accuracy of the solution ob- 
tained can be improved by using a finer mesh and thus a greater number of 
nodes. Also, when radiation is involved, it is more accurate (but more laborious) 
to determine the heat losses for each node and add them up instead of using 
the average temperature. 



291 
CHAPTER 5 



5-5 - TRANSIENT HEAT CONDUCTION 

So far in this chapter we have applied the finite difference method to steady 
heat transfer problems. In this section we extend the method to solve transient 
problems. 

We applied the finite difference method to steady problems by discretizing 
the problem in the space variables and solving for temperatures at discrete 
points called the nodes. The solution obtained is valid for any time since under 
steady conditions the temperatures do not change with time. In transient prob- 
lems, however, the temperatures change with time as well as position, and 
thus the finite difference solution of transient problems requires discretization 
in time in addition to discretization in space, as shown in Figure 5-37. This is 
done by selecting a suitable time step At and solving for the unknown nodal 
temperatures repeatedly for each A; until the solution at the desired time is ob- 
tained. For example, consider a hot metal object that is taken out of the oven 
at an initial temperature of T, at time t = and is allowed to cool in ambient 
air. If a time step of At = 5 min is chosen, the determination of the tempera- 
ture distribution in the metal piece after 3 h requires the determination of the 
temperatures 3 X 60/5 = 36 times, or in 36 time steps. Therefore, the compu- 
tation time of this problem will be 36 times that of a steady problem. Choos- 
ing a smaller Af will increase the accuracy of the solution, but it will also 
increase the computation time. 

In transient problems, the superscript i is used as the index or counter 
of time steps, with i = corresponding to the specified initial condition. 
In the case of the hot metal piece discussed above, i = 1 corresponds to 
t = 1 X Af = 5 min, i = 2 corresponds to t = 2 X At = 10 min, and a general 



r 


, 






















fi+i 

J m-1 


y/+l 
m 


m+1 






h 






m—s 


r 

in 


T' 

m+1 




















1 
















\At 
J Ax 






Ax 


Ax 








' 




. 



1 m — lmm + 1 x 

FIGURE 5-37 

Finite difference formulation of time- 
dependent problems involves discrete 
points in time as well as space. 



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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 



(5-37) 



where the rate of heat transfer Q normally consists of conduction terms for 
interior nodes, but may involve convection, heat flux, and radiation for bound- 
ary nodes. 

Noting that A2i element = mCAT = pV e | ement CAT, where p is density and C is 
the specific heat of the element, dividing the earlier relation by Af gives 



All sides 



A£„ 



Af 



pv« 



AT 

element ^ a * 



(5-38) 



Volume element 
(can be any shape) 




p = density 

V = volume 
pV = mass 

C = specific heat 
A T = temperature change 



AC/ = pVCAT = pVC{T<+ ' - r ) 

FIGURE 5-38 

The change in the energy content of 
the volume element of a node 
during a time interval Af. 



or, for any node m in the medium and its volume element, 

Y i + 1 y i 

' ' X? ' ^element P ^element ^ 

All sides 



Af 



(5-39) 



where T l m and T,' n + ' are the temperatures of node m at times f, = iAt and t i + 1 = 
(i + l)At, respectively, and T[ + l — T,' n represents the temperature change 
of the node during the time interval Af between the time steps i and i + 1 
(Fig. 5-38). 

Note that the ratio (T,;, + ' — T' n ^)IAt is simply the finite difference approxi- 
mation of the partial derivative dT/dt that appears in the differential equations 
of transient problems. Therefore, we would obtain the same result for the 
finite difference formulation if we followed a strict mathematical approach 
instead of the energy balance approach used above. Also note that the finite 
difference formulations of steady and transient problems differ by the single 
term on the right side of the equal sign, and the format of that term remains the 
same in all coordinate systems regardless of whether heat transfer is one-, 
two-, or three-dimensional. For the special case of T]+ ' = T' m (i.e., no change 
in temperature with time), the formulation reduces to that of steady case, as 
expected. 



cen58 93 3_ch05.qxd 9/4/2002 11:42 AM Page 293 



The nodal temperatures in transient problems normally change during each 
time step, and you may be wondering whether to use temperatures at the pre- 
vious time step ;' or the new time step i + 1 for the terms on the left side of Eq. 
5-39. Well, both are reasonable approaches and both are used in practice. The 
finite difference approach is called the explicit method in the first case and 
the implicit method in the second case, and they are expressed in the general 
form as (Fig. 5-39) 



Explicit method: 
Implicit method: 



^j & "*" ^element P "element ^ 



t^ i + I 'r 



m -* ill 



^j & "*" ^element P 'element ^' 



At 



T'/+ 1 'T'i 

1 m in 



At 



(5-40) 



(5-41) 



It appears that the time derivative is expressed in forward difference form in 
the explicit case and backward difference form in the implicit case. Of course, 
it is also possible to mix the two fundamental formulations of Eqs. 5-40 and 
5-41 and come up with more elaborate formulations, but such formulations 
offer little insight and are beyond the scope of this text. Note that both for- 
mulations are simply expressions between the nodal temperatures before and 
after a time interval and are based on determining the new temperatures T'„f ' 
using the previous temperatures T] n . The explicit and implicit formulations 
given here are quite general and can be used in any coordinate system re- 
gardless of the dimension of heat transfer. The volume elements in multi- 
dimensional cases simply have more surfaces and thus involve more terms in 
the summation. 

The explicit and implicit methods have their advantages and disadvantages, 
and one method is not necessarily better than the other one. Next you will see 
that the explicit method is easy to implement but imposes a limit on the al- 
lowable time step to avoid instabilities in the solution, and the implicit method 
requires the nodal temperatures to be solved simultaneously for each time step 
but imposes no limit on the magnitude of the time step. We will limit the dis- 
cussion to one- and two-dimensional cases to keep the complexities at a man- 
ageable level, but the analysis can readily be extended to three-dimensional 
cases and other coordinate systems. 



Transient Heat Conduction in a Plane Wall 

Consider transient one-dimensional heat conduction in a plane wall of thick- 
ness L with heat generation g(x, t) that may vary with time and position and 
constant conductivity k with a mesh size of Ax = L/M and nodes 0, 1, 2, ... , 
M in the x-direction, as shown in Figure 5-40. Noting that the volume ele- 
ment of a general interior node m involves heat conduction from two sides and 
the volume of the element is V e i ement = AAx, the transient finite difference for- 
mulation for an interior node can be expressed on the basis of Eq. 5-39 as 



kA 



Ax 



T T 

*■ m , . , in 

1- kA — 



Ax 



+ g„,AAx = pAAxC : 



At 



(5-42) 



293 
CHAPTER 5 



If expressed at i + 1 : Implicit method 




-py A 



in in_ 

At 



If expressed at i: Explicit method 

FIGURE 5-39 

The formulation of explicit and 

implicit methods differs at the time 

step (previous or new) at which the 

heat transfer and heat generation 

terms are expressed. 



Plane wall 



kA _m^A m 

Ax 



Ax 



1 2 



m-1 



,- Volume 
element 
of node m 
-1 



l kA _m±L 



Ax 



Ax 



m+1 M-1 



M * 



FIGURE 5-40 

The nodal points and volume elements 

for the transient finite difference 

formulation of one-dimensional 

conduction in a plane wall. 



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294 
HEAT TRANSFER 



Canceling the surface area A and multiplying by Ax/k, it simplifies to 

s Ajc 2 a -P- 

7 1 of j_ f _i_ ^HL — ^^ A ( r ri J r\ Ti\ 

'«-] Z - L m ■ ~ L m + 1 ~ K n\t " 



(5-43) 



where a = k/pC is the thermal dijfusivity of the wall material. We now define 
a dimensionless mesh Fourier number as 



ocAf 

Ajc 2 



(5-44) 



Then Eq. 5-43 reduces to 



2T + T 



g„Ax 2 



(5-45) 



Note that the left side of this equation is simply the finite difference formula- 
tion of the problem for the steady case. This is not surprising since the formu- 
lation must reduce to the steady case for T£~ ' = T' m . Also, we are still not 
committed to explicit or implicit formulation since we did not indicate the 
time step on the left side of the equation. We now obtain the explicit finite dif- 
ference formulation by expressing the left side at time step i as 



2T'< 



T' 

1 m+l 



gln^X 2 



(explicit) 



(5-46) 



This equation can be solved explicitly for the new temperature T^ +l (and thus 
the name explicit method) to give 



t(T,U + 7£ +1 ) + (1 -2t)71 + t 



glA* 2 



(5-47) 



A 

M(r.-r^) 


Ax 
2 


pA Ax C ° ° 

y 2 m 

W^kA^ °- 

Ax 

Ax Ax 

















1 2 ••• 


L x 



FIGURE 5-41 

Schematic for the explicit finite 
difference formulation of the 
convection condition at the left 
boundary of a plane wall. 



for all interior nodes m = 1, 2, 3, . . . , M — 1 in a plane wall. Expressing the 
left side of Eq. 5-45 at time step i + 1 instead of i would give the implicit 
finite difference formulation as 



T i + 1 ITi+1 X T/+] _l_ 

1 m-\ LL m ~ 1 m+[ ^ 



Ax 2 Ti 



(implicit) (5-48) 



which can be rearranged as 



x7;;+\-(i + 2T)7r i + T7;;,V 1 + T 



g^Ax 2 



+ Ti = 



(5-49) 



The application of either the explicit or the implicit formulation to each of the 
M — 1 interior nodes gives M — 1 equations. The remaining two equations are 
obtained by applying the same method to the two boundary nodes unless, of 
course, the boundary temperatures are specified as constants (invariant with 
time). For example, the formulation of the convection boundary condition at 
the left boundary (node 0) for the explicit case can be expressed as (Fig. 5^4-1) 



hA(T x - Ti) + kA ■ 



Ax 



+ &A 



Ax 



. Ajc £o 

PA -r- C : — 

v 2 At 



(5-50) 



cen58 93 3_ch05.qxd 9/4/2002 11:42 AM Page 295 



which simplifies to 



1 - 2t - 2t 



hAx 



2t77 



2T nr T - 



gh^x 2 



(5-51) 



295 
CHAPTER 5 



Note that in the case of no heat generation and t = 0.5, the explicit 
finite difference formulation for a general interior node reduces to T,'„ +1 = 
(Tm-i + Tm- 1)/2, which has the interesting interpretation that the temperature 
of an interior node at the new time step is simply the average of the tempera- 
tures of its neighboring nodes at the previous time step. 

Once the formulation (explicit or implicit) is complete and the initial condi- 
tion is specified, the solution of a transient problem is obtained by marching 
in time using a step size of Af as follows: select a suitable time step At and de- 
termine the nodal temperatures from the initial condition. Taking the initial 
temperatures as the previous solution T,'„ at t = 0, obtain the new solution T'„f ' 
at all nodes at time t = At using the transient finite difference relations. Now 
using the solution just obtained at t = At as the previous solution T,' n , obtain 
the new solution T,'„ + ' at t = 2At using the same relations. Repeat the process 
until the solution at the desired time is obtained. 



Stability Criterion for Explicit Method: Limitation on At 

The explicit method is easy to use, but it suffers from an undesirable feature 
that severely restricts its utility: the explicit method is not unconditionally sta- 
ble, and the largest permissible value of the time step Af is limited by the sta- 
bility criterion. If the time step Af is not sufficiently small, the solutions 
obtained by the explicit method may oscillate wildly and diverge from the ac- 
tual solution. To avoid such divergent oscillations in nodal temperatures, the 
value of Af must be maintained below a certain upper limit established by the 
stability criterion. It can be shown mathematically or by a physical argument 
based on the second law of thermodynamics that the stability criterion is sat- 
isfied if the coefficients of all T' ln in the T]„ +[ expressions (called the primary 
coefficients) are greater than or equal to zero for all nodes m (Fig. 5-42). Of 
course, all the terms involving T' m for a particular node must be grouped to- 
gether before this criterion is applied. 

Different equations for different nodes may result in different restrictions on 
the size of the time step Af, and the criterion that is most restrictive should be 
used in the solution of the problem. A practical approach is to identify the 
equation with the smallest primary coefficient since it is the most restrictive 
and to determine the allowable values of Af by applying the stability criterion 
to that equation only. A Af value obtained this way will also satisfy the stabil- 
ity criterion for all other equations in the system. 

For example, in the case of transient one-dimensional heat conduction in a 
plane wall with specified surface temperatures, the explicit finite difference 
equations for all the nodes (which are interior nodes) are obtained from 
Eq. 5-47. The coefficient of T l m in the T^ [ expression is 1 — 2t, which is 
independent of the node number m, and thus the stability criterion for all 
nodes in this case is 1 — 2t ^ or 



Explicit formulation: 






Ti + l = aJi + - 






TS + ' = aJS + - 






t;„ + ' = a,j;„ + ■ 






*m = a M*M + 






Stability criterion: 






a,„>0, m = 0,1,2,.. 


. m, . 


..M 



FIGURE 5-42 

The stability criterion of the 

explicit method requires all primary 

coefficients to be positive or zero. 



ctAf __, J_ /interior nodes, one-dimensional heat 
Ajc 2 ~ 2 \ transfer in rectangular coordinates 



(5-52) 



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296 
HEAT TRANSFER 



When the material of the medium and thus its thermal diffusivity a is known 
and the value of the mesh size Ax is specified, the largest allowable value of 
the time step At can be determined from this relation. For example, in the case 
of a brick wall (a = 0.45 X 10~ 6 m 2 /s) with a mesh size of Ax = 0.01 m, the 
upper limit of the time step is 



At: 



1 Ax 2 

2 « 



(0.01 m) 2 



2(0.45 X 10~ 6 m 2 /s) 



Ills = 1.85 min 



o°c 



50°C 



50°C 



20°C 



m — \ m m + \ 



m—\ m m + 1 
Time step: i + 1 



Time step: i 

FIGURE 5-43 

The violation of the stability criterion 
in the explicit method may result in 
the violation of the second law of 
thermodynamics and thus 
divergence of solution. 







^^~ ~^^^^te- *^^m 










Uranium plate 






o°c 




k = 28 W/m-°C 
g = 5x 10 6 W/m 3 
a = 12.5xl(r 6 m 2 /s 

Ax Ax 




h 
L 


0< 




'' 


■ 





1 


2 


X 






T ... . = 200°C 

initial 







FIGURE 5-44 

Schematic for Example 5-5. 



The boundary nodes involving convection and/or radiation are more re- 
strictive than the interior nodes and thus require smaller time steps. Therefore, 
the most restrictive boundary node should be used in the determination of the 
maximum allowable time step At when a transient problem is solved with 
the explicit method. 

To gain a better understanding of the stability criterion, consider the explicit 
finite difference formulation for an interior node of a plane wall (Eq. 5^47) for 
the case of no heat generation, 

^ + ' = t(7;;,_ 1 + 7;;, +1 ) + (i-2t)7^ 

Assume that at some time step i the temperatures 7^ , and T' m+ , are equal but 
less than T*, (say, T^.j = T;„ +l = 50°C and T l m = 80°C). At the next time 
step, we expect the temperature of node m to be between the two values (say, 
70°C). However, if the value of t exceeds 0.5 (say, t = 1), the temperature of 
node m at the next time step will be less than the temperature of the neighbor- 
ing nodes (it will be 20°C), which is physically impossible and violates the 
second law of thermodynamics (Fig. 5-43). Requiring the new temperature of 
node m to remain above the temperature of the neighboring nodes is equiva- 
lent to requiring the value of t to remain below 0.5. 

The implicit method is unconditionally stable, and thus we can use any time 
step we please with that method (of course, the smaller the time step, the bet- 
ter the accuracy of the solution). The disadvantage of the implicit method is 
that it results in a set of equations that must be solved simultaneously for each 
time step. Both methods are used in practice. 



EXAMPLE 5-5 Transient Heat Conduction in a Large Uranium 
Plate 

Consider a large uranium plate of thickness L = 4 cm, thermal conductivity k = 
28 W/m ■ °C, and thermal diffusivity a = 12.5 X lO" 6 m 2 /s that is initially at 
a uniform temperature of 200°C. Heat is generated uniformly in the plate at a 
constant rate of g = 5 X 10 6 W/m 3 . At time t = 0, one side of the plate is 
brought into contact with iced water and is maintained at 0°C at all times, while 
the other side is subjected to convection to an environment at 7"^ = 30°C with 
a heat transfer coefficient of h = 45 W/m 2 • °C, as shown in Figure 5-44. Con- 
sidering a total of three equally spaced nodes in the medium, two at the bound- 
aries and one at the middle, estimate the exposed surface temperature of the 
plate 2.5 min after the start of cooling using (a) the explicit method and (b) the 
implicit method. 



cen58 93 3_ch05.qxd 9/4/2002 11:42 AM Page 297 



SOLUTION We have solved this problem in Example 5-1 for the steady case, 
and here we repeat it for the transient case to demonstrate the application of 
the transient finite difference methods. Again we assume one-dimensional heat 
transfer in rectangular coordinates and constant thermal conductivity. The num- 
ber of nodes is specified to be M = 3, and they are chosen to be at the two sur- 
faces of the plate and at the middle, as shown in the figure. Then the nodal 
spacing Ax becomes 



Ax 



M- 1 



0.04 m 
3 - 1 



0.02 m 



We number the nodes as 0, 1, and 2. The temperature at node is given to be 
T = 0°C at all times, and the temperatures at nodes 1 and 2 are to be deter- 
mined. This problem involves only two unknown nodal temperatures, and thus 
we need to have only two equations to determine them uniquely. These equa- 
tions are obtained by applying the finite difference method to nodes 1 and 2. 

(a) Node 1 is an interior node, and the explicit finite difference formulation at 
that node is obtained directly from Eq. 5-47 by setting m = 1: 



t(T + Tj) + (1 - 2t) 77 + t 



,?i 



Ax 2 



(1) 



Node 2 is a boundary node subjected to convection, and the finite difference 
formulation at that node is obtained by writing an energy balance on the volume 
element of thickness Ax/2 at that boundary by assuming heat transfer to be into 
the medium at all sides (Fig. 5-45): 



hA{T rj - Tj) + kA 



Ax 



n Ax 



P A 



Ax Tj +l -Tj 



C 



Ax 



Dividing by kA/2Axand using the definitions of thermal diffusivity a = A/pCand 
the dimensionless mesh Fourier number t = aAf/(Ax) 2 gives 

2hAx g^Ax 2 Tj +1 -Tj 
^(T„ - Tj) + 2(77 " Tj) + ^r- = "^f " 



which can be solved for Tj +1 to give 
T i+i 



hAx\ • / • hAx fcAx 2 

! _ 2t _ 2t _ n + J 2T , + 2 — T„ + — 



(2) 



Note that we did not use the superscript /for quantities that do not change with 
time. Next we need to determine the upper limit of the time step Af from the 
stability criterion, which requires the coefficient of T{ in Equation 1 and the co- 
efficient of Tj in the second equation to be greater than or equal to zero. The 
coefficient of Tj is smaller in this case, and thus the stability criterion for this 
problem can be expressed as 



1 - 2t - 2t 



hAx . 







1 



2(1 + hAxlk) 



At: 



Ax 2 



2a(l + hAxlk) 



297 
CHAPTER 5 



Volume element - 
of node 2 



kA- 



Ax 



Si 



T' + 



hA(T- x _ - T') 



Ax 

2 



FIGURE 5-45 

Schematic for the explicit 

finite difference formulation of the 

convection condition at the right 

boundary of a plane wall. 



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298 
HEAT TRANSFER 



TABLE 5-2 

The variation of the nodal 
temperatures in Example 5-5 with 
time obtained by the explicit 
method 







Node 


Time 


Time, 


Temperature, °C 


Step, / 


s 


T[ 


n 








200.0 


200.0 


1 


15 


139.7 


228.4 


2 


30 


149.3 


172.8 


3 


45 


123.8 


179.9 


4 


60 


125.6 


156.3 


5 


75 


114.6 


157.1 


6 


90 


114.3 


146.9 


7 


105 


109.5 


146.3 


8 


120 


108.9 


141.8 


9 


135 


106.7 


141.1 


10 


150 


106.3 


139.0 


20 


300 


103.8 


136.1 


30 


450 


103.7 


136.0 


40 


600 


103.7 


136.0 



since t = aAf/(Ax) 2 . Substituting the given quantities, the maximum allowable 
value of the time step is determined to be 

(0.02 m) 2 

Af < — - ^ = 15.5 s 

2(12.5 X 10- 6 m 2 /s)[l + (45 W/m 2 • °C)(0.02 m)/28 W/m • °C] 

Therefore, any time step less than 15.5 s can be used to solve this problem. For 
convenience, let us choose the time step to be Af = 15 s. Then the mesh 
Fourier number becomes 



ctA? (12-5 X 10- 6 m 2 /s)(15s) 



(A*) 2 



(0.02 m) 2 



0.46875 (forAf = 15 s) 



Substituting this value of t and other given quantities, the explicit finite differ- 
ence equations (1) and (2) developed here reduce to 



77 +l 
7V' +I 



0.06257/ + 0.4687577 + 33.482 
0.93757/ + 0.03236677 + 34.386 



The initial temperature of the medium at t = and / = is given to be 200°C 
throughout, and thus T° = T° = 200°C. Then the nodal temperatures at 7"/ 
and 77 1 at f = Af = 15 s are determined from these equations to be 



0.06257/,° 



0.4687577° + 33.482 
0.0625 X 200 + 0.46875 X 200 + 33.482 = 139.7°C 
0.93757/,° + 0.03236677° + 34.386 
0.9375 X 200 + 0.032366 X 200 + 34.386 = 228.4°C 



Similarly, the nodal temperatures 77/ and 7 2 2 at t = 2Af = 2 X 15 = 30 s are 
determined to be 



Tf = 0.06257/, 1 + 0.4687577 + 33.482 

= 0.0625 X 139.7 + 0.46875 X 228.4 + 33.482 
Ti = 0.93757/,' + 0.03236677 + 34.386 



149.3°C 



= 0.9375 X 139.7 + 0.032366 X 228.4 + 34.386 = 172.8°C 

Continuing in the same manner, the temperatures at nodes 1 and 2 are de- 
termined for / = 1, 2, 3, 4, 5, . . . , 50 and are given in Table 5-2. Therefore, 
the temperature at the exposed boundary surface 2.5 min after the start of 
cooling is 

r L 25rain = 77° = 139.0°C 

(b) Node 1 is an interior node, and the implicit finite difference formulation at 
that node is obtained directly from Eq. 5-49 by setting m = 1: 



J?n A* 2 

t7/ - (1 + 2t) 77+ > + t77 +1 + t—, — + 77 



(3) 



Node 2 is a boundary node subjected to convection, and the implicit finite dif- 
ference formulation at that node can be obtained from this formulation by ex- 
pressing the left side of the equation at time step / + 1 instead of / as 



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 5-3, and the temperature 
at the exposed boundary surface (node 2) 2.5 min after the start of cooling is 
obtained to be 



r 2.5min = j\a = 143.9°C 



which is close to the result obtained by the explicit method. Note that either 
method could be used to obtain satisfactory results to transient problems, ex- 
cept, perhaps, for the first few time steps. The implicit method is preferred 
when it is desirable to use large time steps, and the explicit method is preferred 
when one wishes to avoid the simultaneous solution of a system of algebraic 
equations. 



TABLE 5-3 

The variation of the nodal 
temperatures in Example 5-5 with 
time obtained by the implicit 
method 







Node 


Time 


Time, 


Temperature, °C 


Step, / 


s 


T[ 


n 








200.0 


200.0 


1 


15 


168.8 


199.6 


2 


30 


150.5 


190.6 


3 


45 


138.6 


180.4 


4 


60 


130.3 


171.2 


5 


75 


124.1 


163.6 


6 


90 


119.5 


157.6 


7 


105 


115.9 


152.8 


8 


120 


113.2 


149.0 


9 


135 


111.0 


146.1 


10 


150 


109.4 


143.9 


20 


300 


104.2 


136.7 


30 


450 


103.8 


136.1 


40 


600 


103.8 


136.1 



cen58933_ch05.qxd 9/4/2002 11:42 AM Page 3C 



300 
HEAT TRANSFER 



South 




FIGURE 5-46 

Schematic of a Trombe wall 
(Example 5-6). 



TABLE 5-4 

The hourly variation of monthly 
average ambient temperature and 
solar heat flux incident on a vertical 
surface for January in Reno, Nevada 



Time 


Ambient 


Solar 


of Temperature 


Radiation, 


Day 


°F 


Btu/h ■ ft 2 


7 AM-10 AM 


33 


114 


10 AM-1 PM 


43 


242 


1 PM-4 PM 


45 


178 


4 PM-7 PM 


37 





7 PM-10 PM 


32 





10 PM-1 AM 


27 





1 AM-4 AM 


26 





4 AM-7 AM 


25 






EXAMPLE 5-6 Solar Energy Storage in Trombe Walls 

Dark painted thick masonry walls called Trombe walls are commonly used on 
south sides of passive solar homes to absorb solar energy, store it during the 
day, and release it to the house during the night (Fig. 5-46). The idea was pro- 
posed by E. L. Morse of Massachusetts in 1881 and is named after Professor 
Felix Trombe of France, who used it extensively in his designs in the 1970s. 
Usually a single or double layer of glazing is placed outside the wall and trans- 
mits most of the solar energy while blocking heat losses from the exposed sur- 
face of the wall to the outside. Also, air vents are commonly installed at the 
bottom and top of the Trombe walls so that the house air enters the parallel flow 
channel between the Trombe wall and the glazing, rises as it is heated, and en- 
ters the room through the top vent. 

Consider a house in Reno, Nevada, whose south wall consists of a 1-ft-thick 
Trombe wall whose thermal conductivity is k = 0.40 Btu/h • ft • °F and whose 
thermal diffusivity is a = 4.78 X 10~ s ft 2 /s. The variation of the ambient tem- 
perature 7" out and the solar heat flux <j S0 | ar incident on a south-facing vertical sur- 
face throughout the day for a typical day in January is given in Table 5-4 in 3-h 
intervals. The Trombe wall has single glazing with an absorptivity-transmissivity 
product of k = 0.77 (that is, 77 percent of the solar energy incident is ab- 
sorbed by the exposed surface of the Trombe wall), and the average combined 
heat transfer coefficient for heat loss from the Trombe wall to the ambient is de- 
termined to be h out = 0.7 Btu/h ■ ft 2 • °F. The interior of the house is maintained 
at T m = 70°F at all times, and the heat transfer coefficient at the interior sur- 
face of the Trombe wall is h m =1.8 Btu/h ■ ft 2 ■ °F. Also, the vents on the 
Trombe wall are kept closed, and thus the only heat transfer between the air in 
the house and the Trombe wall is through the interior surface of the wall. As- 
suming the temperature of the Trombe wall to vary linearly between 70°F at the 
interior surface and 30°F at the exterior surface at 7 am and using the explicit 
finite difference method with a uniform nodal spacing of Ax = 0.2 ft, determine 
the temperature distribution along the thickness of the Trombe wall after 12, 
24, 36, and 48 h. Also, determine the net amount of heat transferred to the 
house from the Trombe wall during the first day and the second day. Assume the 
wall is 10 ft high and 25 ft long. 

SOLUTION The passive solar heating of a house through a Trombe wall is con- 
sidered. The temperature distribution in the wall in 12-h intervals and the 
amount of heat transfer during the first and second days are to be determined. 
Assumptions 1 Heat transfer is one-dimensional since the exposed surface of 
the wall is large relative to its thickness. 2 Thermal conductivity is constant. 
3 The heat transfer coefficients are constant. 

Properties The wall properties are given to be k = 0.40 Btu/h ■ ft • °F, a = 
4.78 X 10- 6 ft 2 /s, and k = 0.77. 

Analysis The nodal spacing is given to be Ax = 0.2 ft, and thus the total num- 
ber of nodes along the Trombe wall is 



M 



A +1= _LIL 

Ax 0.2 ft 



1 



We number the nodes as 0, 1, 2, 3, 4, and 5, with node on the interior sur- 
face of the Trombe wall and node 5 on the exterior surface, as shown in Figure 
5-47. Nodes 1 through 4 are interior nodes, and the explicit finite difference 
formulations of these nodes are obtained directly from Eq. 5-47 to be 



cen58 93 3_ch05.qxd 9/4/2002 11:42 AM Page 301 



Node 1 (m = 1) 
Node 2 (m = 2) 
Node 3 (m = 3) 
Node 4 (m = 4) 



77+ ' = T(r j + Ti) + (i - 2T)r/ 

rj +I = T(r/' + r 3 + (l - 2t)t{ 

Ti +[ = T(Ti + Ti) + (1 - 2t)7^ 

rj +1 = T(Ti + r 5 + (i - 2j)Ti 



(i) 

(2) 
(3) 
(4) 



The interior surface is subjected to convection, and thus the explicit formula- 
tion of node can be obtained directly from Eq. 5-51 to be 



1 -2t 



h m Ax\ . 
2x^^)7-,; 



2t77 + 2t—, — T in 



Substituting the quantities h m , Ax, k, and T m , which do not change with time, 
into this equation gives 



(1 - 3.80t) Ti + t(277 + 126.0) 



(5) 



The exterior surface of the Trombe wall is subjected to convection as well as to 
heat flux. The explicit finite difference formulation at that boundary is obtained 
by writing an energy balance on the volume element represented by node 5, 



h out A(Ti ut - Ti) + KAq^ 



kA 



Ti 



Ax 



pA — C 



T i +i 



Af 



(5-53) 



which simplifies to 



j.j+1 



1 - 2t - 2t ■ 



h m „ Ax 



Ti + 2tTI + 2t - 



h„„, Ax 



K qj olm . Ax 
2t : (5-54) 



where t = aAf/Ax 2 is the dimensionless mesh Fourier number. Note that we 
kept the superscript /for quantities that vary with time. Substituting the quan- 
tities h out , Ax, k, and k, which do not change with time, into this equation gives 



(1 - 2.70t) Ti + t(2T\ + 0.7071, + Q.770?^) 



(6) 



where the unit of q* | ar is Btu/h ■ ft 2 . 

Next we need to determine the upper limit of the time step Af from the sta- 
bility criterion since we are using the explicit method. This requires the iden- 
tification of the smallest primary coefficient in the system. We know that the 
boundary nodes are more restrictive than the interior nodes, and thus we exam- 
ine the formulations of the boundary nodes and 5 only. The smallest and thus 
the most restrictive primary coefficient in this case is the coefficient of Ti in the 
formulation of node since 1 - 3.8t < 1 - 2.7t, and thus the stability cri- 
terion for this problem can be expressed as 



1 - 3.80t>0 



aA;t 

A^" : 



1 

3.80 



Substituting the given quantities, the maximum allowable value of the time step 
is determined to be 



At: 



(0.2 ft) 2 



Ax 2 

3.80a 3.80 X (4.78 X lO" 6 ft 2 /s) 



2202 s 



301 
CHAPTER 5 





A": 


Trombe wall 
= 0.40 Btu/h-ft-°F 




V 




a 


= 4.78x 10- 6 ft 2 /s 






70 ; F 




Initial temperature 
/ distribution at 




,T. 

in in 


if 7 AM (t = 0) 


/; „ T 

1 out c 




AX = 


= 0.2 ft 




•30°F 


0' 


* 




1 


2 3 4 


5 


L x 



FIGURE 5-47 

The nodal network for the Trombe 
wall discussed in Example 5-6. 



cen58 93 3_ch05.qxd 9/4/2002 11:42 AM Page 302 



302 
HEAT TRANSFER 



Temperature 



C F 












170 


















— 1 st day 










— 2nd day 














7 pm -s 




110 

90 
















/ /l AM 










1 PM 


70 
















50 


























{ Initial 




-"^7 AM 






temperature 







0.2 0.4 0.6 0.8 1 ft 

Distance along the Trombe wall 

FIGURE 5-48 

The variation of temperatures in 
the Trombe wall discussed 
in Example 5-6. 



Therefore, any time step less than 2202 s can be used to solve this problem. 
For convenience, let us choose the time step to be Af = 900 s = 15 min. Then 
the mesh Fourier number becomes 



ocAf (4-78 X 10- 6 ft 2 /s)(900 s) 



(Ax) 2 



(0.2 ft) 2 



0.10755 (for A? = 15 min) 



Initially (at 7 am or f = 0), the temperature of the wall is said to vary linearly be- 
tween 70 C F at node and 30°F at node 5. Noting that there are five nodal 
spacings of equal length, the temperature change between two neighboring 
nodes is (70 - 30)°F/5 = 8°F. Therefore, the initial nodal temperatures are 



r 3 ° 



70°F, 
46°F, 



T o 

TO 



62°F, 
38°R 



54°F, 
30°F 



Then the nodal temperatures at t = Af = 15 min (at 7:15 am) are determined 
from these equations to be 

Ti = (1 - 3.80t) r ° + T(2r,° 



— v,! j.oui/iq i ni^i < 126.0) 

(1 - 3.80 X 0.10755) 70 + 0.10755(2 X 62 + 126.0) = 68.3° F 



77 



T(r ° + r 2 °) + (i - 2t) r,° 

= 0.10755(70 + 54) + (1 - 2 X 0.10755)62 = 62°F 

7V = t(t? + r 3 °) + (l - 2t) r 2 ° 

= 0.10755(62 + 46) + (1 - 2 X 0.10755)54 = 54°F 

Tl = T(r 2 ° + r 4 °) + (i - 2t) r 3 ° 

= 0.10755(54 + 38) + (1 - 2 X 0.10755)46 = 46°F 

t\ = T(r 3 ° + r 5 °) + (i - 2t) r 4 ° 

= 0.10755(46 + 30) + (1 - 2 X 0.10755)38 = 38°F 
Ti = (1 - 2.70t) T 5 ° + t(2T 4 ° + 0.70r o ° u , + 0.770</° olar ) 

= (1 - 2.70 X 0.10755)30 + 0.10755(2 X 38 + 0.70 X 33 + 0.770 X 114) 
= 41.4°F 

Note that the inner surface temperature of the Trombe wall dropped by 1.7°F 
and the outer surface temperature rose by 11.4°F during the first time step 
while the temperatures at the interior nodes remained the same. This is typical 
of transient problems in mediums that involve no heat generation. The nodal 
temperatures at the following time steps are determined similarly with the help 
of a computer. Note that the data for ambient temperature and the incident 
solar radiation change every 3 hours, which corresponds to 12 time steps, 
and this must be reflected in the computer program. For example, the value of 
q' salar must be taken to be q' solar = 75 for / = 1-12, q' solar = 242 for / = 13-24, 
<7^ ar = 178 for / = 25-36, and q' solar = for / = 37-96. 

The results after 6, 12, 18, 24, 30, 36, 42, and 48 h are given in Table 5-5 
and are plotted in Figure 5-48 for the first day. Note that the interior tempera- 
ture of the Trombe wall drops in early morning hours, but then rises as the solar 
energy absorbed by the exterior surface diffuses through the wall. The exterior 
surface temperature of the Trombe wall rises from 30 to 142°F in just 6 h be- 
cause of the solar energy absorbed, but then drops to 53°F by next morning as 
a result of heat loss at night. Therefore, it may be worthwhile to cover the outer 
surface at night to minimize the heat losses. 



cen58 93 3_ch05.qxd 9/4/2002 11:42 AM Page 303 



TABLE 5-5 

The temperatures at the nodes of a Trombe wall at various times 





Time 
Step, / 






Nodal Temperatures, 


°F 




Time 


To 


T, 


h 


h 


T, 


T 5 


h (7 am) 





70.0 


62.0 


54.0 


46.0 


38.0 


30.0 


6 h (1 pm) 


24 


65.3 


61.7 


61.5 


69.7 


94.1 


142.0 


12 h (7 pm) 


48 


71.6 


74.2 


80.4 


88.4 


91.7 


82.4 


18 h (1 AM) 


72 


73.3 


75.9 


77.4 


76.3 


71.2 


61.2 


24 h (7 am) 


96 


71.2 


71.9 


70.9 


67.7 


61.7 


53.0 


30 h (1 pm) 


120 


70.3 


71.1 


74.3 


84.2 


108.3 


153.2 


36 h (7 pm) 


144 


75.4 


81.1 


89.4 


98.2 


101.0 


89.7 


42 h (1 am) 


168 


75.8 


80.7 


83.5 


83.0 


77.4 


66.2 


48 h (7 am) 


192 


73.0 


75.1 


72.2 


66.0 


66.0 


56.3 



303 
CHAPTER 5 



The rate of heat transfer from the Trombe wall to the interior of the house dur- 
ing each time step is determined from Newton's law using the average temper- 
ature at the inner surface of the wall (node 0) as 



Gt, 



Gxrombewall Af = h iB A(T& - T m ) At = h m A[(Ti + 7T >)/2 - TJAt 



Therefore, the amount of heat transfer during the first time step (/' = 1) or 
during the first 15-min period is 

GTron.be wall = h m A[(Tj + T °)/2 ~ TJ At 

= (1.8 Btu/h • ft 2 ■ °F)(10 X 25 ft 2 )[(68.3 + 70)/2 - 70°F](0.25 h) 
= -95.6 Btu 

The negative sign indicates that heat is transferred to the Trombe wall from the 
air in the house, which represents a heat loss. Then the total heat transfer dur- 
ing a specified time period is determined by adding the heat transfer amounts 
for each time step as 



Q 



Trombe wall 



2e 



Trombe wall 



2 h m A[(T- + Tt l )/2 - TJ At (5-55) 



where / is the total number of time intervals in the specified time period. In this 
case / = 48 for 12 h, 96 for 24 h, and so on. Following the approach described 
here using a computer, the amount of heat transfer between the Trombe wall 
and the interior of the house is determined to be 



Gt 

G 
Q 
Q 



Trombe wall 



Trombe wall 



Trombe wall 



-17, 048 Btu after 12 h 
-2483 Btu after 24 h 
5610 Btu after 36 h 
34, 400 Btu after 48 h 



(-17, 078 Btu during the first 12 h) 
(14, 565 Btu during the second 12 h) 
(8093 Btu during the third 12 h) 
(28, 790 Btu during the fourth 12 h) 



Therefore, the house loses 2483 Btu through the Trombe wall the first day as a 
result of the low start-up temperature but delivers a total of 36,883 Btu of heat 
to the house the second day. It can be shown that the Trombe wall will deliver 
even more heat to the house during the third day since it will start the day at a 
higher average temperature. 



cen58 93 3_ch05.qxd 9/4/2002 11:42 AM Page 304 



n + 1 



Ay 



304 
HEAT TRANSFER 



in, n + 1 



Volume 
element -, 

1 k 
m-l,n I "••" 



Ay 



«-l 



-+- 



-Ax 



m, n - 1 



Ax 



Two-Dimensional Transient Heat Conduction 



m + 1, n 



y t m - 1 



m + 1 



FIGURE 5-49 

The volume element of a 
general interior node (m, «) for two- 
dimensional transient conduction 
in rectangular coordinates. 



Consider a rectangular region in which heat conduction is significant in the 
x- and y-directions, and consider a unit depth of Az = 1 in the z-direction. 
Heat may be generated in the medium at a rate of g(x, y, t), which may vary 
with time and position, with the thermal conductivity k of the medium as- 
sumed to be constant. Now divide the x-y-plane of the region into a rectangu- 
lar mesh of nodal points spaced Ax and A_y apart in the x- and y-directions, 
respectively, and consider a general interior node (m, ri) whose coordinates are 
x = mAx and y = nAy, as shown in Figure 5-49. Noting that the volume ele- 
ment centered about the general interior node (m, n) involves heat conduction 
from four sides (right, left, top, and bottom) and the volume of the element is 
Element = Ax X Ay X 1 = AxAy, the transient finite difference formulation for 
a general interior node can be expressed on the basis of Eq. 5-39 as 



kAy 



T — T 

m— \,n in, 11 

Ax 



T — T T — T 

, , ■* m, n+ 1 7H, h , . -* m+ 1. n m, n 

+ kAx - — — — + kAy - 



kAx 



T — T 

m, n— 1 m, n 



Ay 



Av 
g,„,„AxAy = pAxAyC 



Ax 

'T'i+l r r\ 

1 in 1 m 

At 



(5-56) 



Taking a square mesh {Ax = Ay = I ) and dividing each term by k gives after 
simplifying, 



f m — 1, n m + 1, n 



r + t 

1 m, n + 1 -* m, n — I 



AT 



J : 



(5-57) 



where again a = k/pCis the thermal diffusivity of the material and t = a.At/1 2 
is the dimensionless mesh Fourier number. It can also be expressed in terms of 
the temperatures at the neighboring nodes in the following easy-to-remember 
form: 



, J, _|_ rp 

1 1 top ' J right 



+ T h , 



4r„, 



,/ : 



(5-58) 



Again the left side of this equation is simply the finite difference formulation 
of the problem for the steady case, as expected. Also, we are still not com- 
mitted to explicit or implicit formulation since we did not indicate the time 
step on the left side of the equation. We now obtain the explicit finite differ- 
ence formulation by expressing the left side at time step i as 



+ T' + T 1 + T' 

1 1 top ' J right ' 1 bottom 



ATI, 



cSnode' 



(5-59) 



Expressing the left side at time step i + 1 instead of ;' would give the implicit 
formulation. This equation can be solved explicitly for the new temperature 
Tlodi to give 



TLVc = T<T,' Bft + ri p + T' ght + Ti onom ) + (1 - 4t) 7l de + t ^t^ (5-60) 



for all interior nodes (m, n) where m = 1, 2, 3, . . . , M — 1 and n = 1,2, 
3, . . . , N — 1 in the medium. In the case of no heat generation and t = \, the 



cen58 93 3_ch05.qxd 9/4/2002 11:42 AM Page 305 



explicit finite difference formulation for a general interior node reduces to 
^node = (Tift + T/op + Trfght + T{ onom )/4, which has the interpretation that the 
temperature of an interior node at the new time step is simply the average 
of the temperatures of its neighboring nodes at the previous time step 
(Fig. 5-50). 

The stability criterion that requires the coefficient of T' m in the T f ' n + ' expres- 
sion to be greater than or equal to zero for all nodes is equally valid for two- 
or three-dimensional cases and severely limits the size of the time step At that 
can be used with the explicit method. In the case of transient two-dimensional 
heat transfer in rectangular coordinates, the coefficient of T' m in the r„', + ' ex- 
pression is 1 — 4t, and thus the stability criterion for all interior nodes in this 
case is 1 — 4t > 0, or 



aA? j_ (interior nodes, two-dimensional heat 
I 2 ~4 transfer in rectangular coordinates) 



(5-61) 



where Ax = Ay = /. When the material of the medium and thus its thermal 
diffusivity a are known and the value of the mesh size / is specified, the 
largest allowable value of the time step At can be determined from the relation 
above. Again the boundary nodes involving convection and/or radiation are 
more restrictive than the interior nodes and thus require smaller time steps. 
Therefore, the most restrictive boundary node should be used in the determi- 
nation of the maximum allowable time step At when a transient problem is 
solved with the explicit method. 

The application of Eq. 5-60 to each of the (M — 1) X (N — 1) interior nodes 
gives (M — 1) X (N — 1) equations. The remaining equations are obtained by 
applying the method to the boundary nodes unless, of course, the boundary 
temperatures are specified as being constant. The development of the transient 
finite difference formulation of boundary nodes in two- (or three-) dimen- 
sional problems is similar to the development in the one-dimensional case dis- 
cussed earlier. Again the region is partitioned between the nodes by forming 
volume elements around the nodes, and an energy balance is written for each 
boundary node on the basis of Eq. 5-39. This is illustrated in Example 5-7. 



305 
CHAPTER 5 



Time step i: 



30°C 



20°C 



Node 



40°C 



10°C 



Time step i + 1 : 











yi + 1 

m 


25°C 






Node 


m 











FIGURE 5-50 

In the case of no heat generation 

and t = i the temperature of an 

interior node at the new time step is 

the average of the temperatures of 

its neighboring nodes at the 

previous time step. 



EXAMPLE 5-7 Transient Two-Dimensional Heat Conduction 
in L-Bars 

Consider two-dimensional transient heat transfer in an L-shaped solid body that 
is initially at a uniform temperature of 90°C and whose cross section is given 
in Figure 5-51. The thermal conductivity and diffusivity of the body are k = 
15 W/m • °C and a = 3.2 X 10~ 6 m 2 /s, respectively, and heat is generated in 
the body at a rate of g = 2 X 10 6 W/m 3 . The left surface of the body is insu- 
lated, and the bottom surface is maintained at a uniform temperature of 90°C 
at all times. At time f = 0, the entire top surface is subjected to convection to 
ambient air at T„ = 25°C with a convection coefficient of h = 80 W/m 2 • °C, 
and the right surface is subjected to heat flux at a uniform rate of q R = 5000 
W/m 2 . The nodal network of the problem consists of 15 equally spaced nodes 
with Ax = Ay = 1.2 cm, as shown in the figure. Five of the nodes are at the bot- 
tom surface, and thus their temperatures are known. Using the explicit method, 
determine the temperature at the top corner (node 3) of the body after 1, 3, 5, 
10, and 60 min. 



Convection 



V 






h, T„ 










1 2 /" 


. 3 /\ 


t 






J \ Ax = Ay = I 


Av 


h — 


h — 


6 7 \ 8 9 Qr 


T 

Av 

1 


+ 


+ 

ii! 


h — 

12! 


h — 

13! 


h — 

14| 15 










) 
90°C 






X 




-Ax^ 


-Ax^ 


-A*— 


-A*— 


-Ax^ 





FIGURE 5-51 

Schematic and nodal network for 
Example 5-7. 



cen58 93 3_ch05.qxd 9/4/2002 11:42 AM Page 306 



306 
HEAT TRANSFER 



h,T„ 



h,T m 



] 



X 



(b) Node 2 



(a) Node 1 

FIGURE 5-52 

Schematics for energy balances on the 
volume elements of nodes 1 and 2. 



SOLUTION This is a transient two-dimensional heat transfer problem in rec- 
tangular coordinates, and it was solved in Example 5-3 for the steady case. 
Therefore, the solution of this transient problem should approach the solution 
for the steady case when the time is sufficiently large. The thermal conductiv- 
ity and heat generation rate are given to be constants. We observe that all nodes 
are boundary nodes except node 5, which is an interior node. Therefore, we will 
have to rely on energy balances to obtain the finite difference equations. The re- 
gion is partitioned among the nodes equitably as shown in the figure, and the 
explicit finite difference equations are determined on the basis of the energy 
balance for the transient case expressed as 



2 Q' + GL 



PKle 



c- 



At 



The quantities h, T„, g, and q R do not change with time, and thus we do not 
need to use the superscript /for them. Also, the energy balance expressions are 
simplified using the definitions of thermal diffusivity a = k/pC and the dimen- 
sionless mesh Fourier number t = aAt/P, where Ax = Ay = /. 

(a) Node 1. (Boundary node subjected to convection and insulation, Fig. 
5-52a) 



i^X 



77) 



Ay Ti 



Ax 



+ k 



AxU 



Ay 

Ax Ay 
2 2 



Ax Ay T{ 

H 2 2 



At 



Dividing by fe/4 and simplifying, 



2hl 



8,l 2 Tl +i -T< 



t (T m - 77) + 2(T{ - T{) + 2{T\ - T[) , 



which can be solved for T{ +1 to give 



1 - 4t - 2tj J 77 + 2t\T{ + Ti +^T» + |r 



(b) Node 2. (Boundary node subjected to convection, Fig. 5-526) 



Ay Ti - Ti Ti - Ti 

hAx(T^ -T{) + k-±- — k — - + kAx - 



Ax 



Ay 



Ay T{ - T{ Ay Ay T{ +1 - Ti 

+ k T^ x - + ^-^T = ^T c ^T^ 



Dividing by kl2, simplifying, and solving for T^ +1 gives 



1 - 4t - 2t j J Ti + t (t{ + Ti + 2Ti, + ^j- T x + ^- 



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(c) Node 3. (Boundary node subjected to convection on two sides, Fig. 5-53a) 

AyTl-Ti . AxAy Ax Ay T^ 1 - 7^ 
Dividing by klA, simplifying, and solving for 7"3 +1 gives 



At 



T- + [ = I 1 - 4t - 4t^ ] Ti + 2t( Ti + T£ + 2jT^ + 



hi, 



2k 



(d) Node 4. (On the insulated boundary, and can be treated as an interior node, 
Fig. 5-536). Noting that T 10 = 90°C, Eq. 5-60 gives 



U 1 



(1 - 4t) T\ + t T[ + 2Tj, + 90 + 



(e) Node 5. (Interior node, Fig. 5-54a). Noting that T n = 90°C, Eq. 5-60 
gives 



Ti, + ' = (1 - 4t) Ti + t I H + r^ + Ti + 90 + 



&/ 2 



(f) Node 6. (Boundary node subjected to convection on two sides, Fig. 5-546) 

(Ax Ay\ Av Tj - Ti T' n - Ti k T{ - TI 

Ax Ti - Tl . 3 AxAy 3 AxAy Tf - Ti 
+ T^y~ + S6 ^^ = P ^^ C At 

Dividing by 3k/A, simplifying, and solving for 7"g +1 gives 



tw'+I 
J 6 



l-4 T -4x|]7 



hi, 



2Tj + 4Ti + 2Tj + 4 X 90 + 4 -f T w + 3 ^- 



[g) Node 7. (Boundary node subjected to convection, Fig. 5-55) 



Ay Ti - Tj T( 3 - Tj 

hAx(T^ -Tj) + k-^- -=-; + kAx - 



2 Ax 

Ay U ~ Tj 



Ay 



Ay Ay Tj +I - Tj 



2 Ax +^Ax T = pAx T C- A[ 



Dividing by kl2, simplifying, and solving for Tj +1 gives 



1 -4T-2Ty ] 



2hi , g 7 r- 



Tl + Ti + 2 X 90 + —r- T„ + 



307 
CHAPTER 5 



h, r« 

r 



Mirror 



(5) 



*,r~ 



t 



EH 

— I ♦ 



10 



(a) Node 3 



(b) Node 4 

FIGURE 5-53 

Schematics for energy balances on the 
volume elements of nodes 3 and 4. 



m 



ii 



12 



(a) Node 5 



(7?) Node 6 

FIGURE 5-54 

Schematics for energy balances on the 
volume elements of nodes 5 and 6. 



h, r„ 






1\ 1 

13 1 ^ ^ 



1r 



• 15 

FIGURE 5-55 

Schematics for energy balances on the 
volume elements of nodes 7 and 9. 



cen5 8 93 3_ch05.qxd 9/4/2002 11:42 AM Page 3C 



308 
HEAT TRANSFER 



(h) Node 8. This node is identical to node 7, and the finite difference formula- 
tion of this node can be obtained from that of node 7 by shifting the node num- 
bers by 1 (i.e., replacing subscript m by subscript m + 1). It gives 



1 - 4t - 2t 



hi 



n + T 



2hl, 



Tj + Tj + 2 X 90 + -T- 7^ + 



(/) Node 9. (Boundary node subjected to convection on two sides, Fig. 5-55) 

,Ax,_ _,, , . Ay A.t Tj 5 - Tj 
h T (T«,-T$) + q R — k-—^- 



kAyTi 



Ti , . Ax^J AxAy T<' 



+ 



2 Ax 6J 2 2 

Dividing by klA, simplifying, and solving for 7"g +1 gives 



p TT c " 



At 



T' + l 
*9 



hl\ I 4rI hi 

i - 4t - 2t j r 9 " + 2t m + 90 + ^-+ ^r„ 



2fe 



This completes the finite difference formulation of the problem. Next we need 
to determine the upper limit of the time step Af from the stability criterion, 
which requires the coefficient of T m in the 7~^ +1 expression (the primary coeffi- 
cient) to be greater than or equal to zero for all nodes. The smallest primary co- 
efficient in the nine equations here is the coefficient of Ti in the expression, 
and thus the stability criterion for this problem can be expressed as 



1 - 4t - 4t 



///. 







1 



4(1 + hllk) 



A?: 



/ : 



4a(l + hllk) 



since t = aAf// 2 . Substituting the given quantities, the maximum allowable 
value of the time step is determined to be 



At: 



(0.012 m) 2 



4(3.2 X 10- 6 m 2 /s)[l + (80 W/m 2 • °C)(0.012 m)/(15 W/m ■ °C)] 



10.6 s 



Therefore, any time step less than 10.6 s can be used to solve this problem. For 
convenience, let us choose the time step to be At = 10 s. Then the mesh 
Fourier number becomes 



aAt _ (3-2 X 10- 6 m 2 /s)(10s) 
l r ~ (0.012 m) 2 



0.222 (for At = 10 s) 



Substituting this value of t and other given quantities, the developed transient 
finite difference equations simplify to 



T i+ 



0.08367,' + 0.444(7^ + 7] + 11.2) 
0.0S36Ti + 0.222(7/ + Tj + ITi, + 22.4; 
0.05527J + 0.444(7; + T b + 12.8) 
0.1127] + 0.222(7/ + 27J + 109.2) 
0.1127/ + 0.222(7] + 7j + 7 6 ' + 109.2) 



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0.093 17^ + 0.074(2^ + 47^' + 2Tj + 424) 
0.08367V + 0.222(7,; + TJ + 202.4) 
0.08367V + 0.222(Tj + T$ + 202.4) 
0.08367V + 0.444(7^ + 105.2) 



Using the specified initial condition as the solution at time t = (for / = 0), 
sweeping through these nine equations will give the solution at intervals of 
10 s. The solution at the upper corner node (node 3) is determined to be 
100.2, 105.9, 106.5, 106.6, and 106. 6°C at 1, 3, 5, 10, and 60 min, re- 
spectively. Note that the last three solutions are practically identical to the 
solution for the steady case obtained in Example 5-3. This indicates that steady 
conditions are reached in the medium after about 5 min. 



309 
CHAPTER 5 



TOPIC OF SPECIAL INTEREST 



Controlling the Numerical Error 

A comparison of the numerical results with the exact results for tempera- 
ture distribution in a cylinder would show that the results obtained by a nu- 
merical method are approximate, and they may or may not be sufficiently 
close to the exact (true) solution values. The difference between a numeri- 
cal solution and the exact solution is the error involved in the numerical 
solution, and it is primarily due to two sources: 

• The discretization error (also called the truncation ox formulation 
error), which is caused by the approximations used in the formulation 
of the numerical method. 

• The round-off error, which is caused by the computer's use of a 
limited number of significant digits and continuously rounding (or 
chopping) off the digits it cannot retain. 

Below we discuss both types of errors. 

Discretization Error 

The discretization error involved in numerical methods is due to replacing 
the derivatives by differences in each step, or the actual temperature distri- 
bution between two adjacent nodes by a straight line segment. 

Consider the variation of the solution of a transient heat transfer problem 
with time at a specified nodal point. Both the numerical and actual (exact) 
solutions coincide at the beginning of the first time step, as expected, but 
the numerical solution deviates from the exact solution as the time t in- 
creases. The difference between the two solutions at t = Af is due to the ap- 
proximation at the first time step only and is called the local discretization 
error. One would expect the situation to get worse with each step since the 
second step uses the erroneous result of the first step as its starting point 
and adds a second local discretization error on top of it, as shown in Figure 
5-56. The accumulation of the local discretization errors continues with the 
increasing number of time steps, and the total discretization error at any 



Global 
error 




'3 Time 

FIGURE 5-56 

The local and global discretization 

errors of the finite difference 

method at the third time step 

at a specified nodal point. 



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310 
HEAT TRANSFER 



step is called the global or accumulated discretization error. Note that the 
local and global discretization errors are identical for the first time step. 
The global discretization error usually increases with the increasing num- 
ber of steps, but the opposite may occur when the solution function 
changes direction frequently, giving rise to local discretization errors of op- 
posite signs, which tend to cancel each other. 

To have an idea about the magnitude of the local discretization error, 
consider the Taylor series expansion of the temperature at a specified nodal 
point m about time t h 

dT(x m , t) 1 , d 2 T(x m , t) 
T(x„„ t t + At) = T(x m , f ; ) + M \ m t + i At - "J +■■■ (5-62) 

The finite difference formulation of the time derivative at the same nodal 
point is expressed as 



dT(x„„ t t ) T(x„„ t t + At) - T(x m , t t ) n + ' - T;„ 



dt Ar At 



(5-63) 



or 



dT(x m , tf) 
T(x m , t, + At) = T(x m , t,) + At (5-64) 

which resembles the Taylor series expansion terminated after the first two 
terms. Therefore, the third and later terms in the Taylor series expansion 
represent the error involved in the finite difference approximation. For a 
sufficiently small time step, these terms decay rapidly as the order of de- 
rivative increases, and their contributions become smaller and smaller. The 
first term neglected in the Taylor series expansion is proportional to At 2 , 
and thus the local discretization error of this approximation, which is the 
error involved in each step, is also proportional to At 2 . 

The local discretization error is the formulation error associated with a 
single step and gives an idea about the accuracy of the method used. How- 
ever, the solution results obtained at every step except the first one involve 
the accumulated error up to that point, and the local error alone does not 
have much significance. What we really need to know is the global dis- 
cretization error. At the worst case, the accumulated discretization error 
after / time steps during a time period t is i(At) 2 = (f /A0(Af) 2 = t Q At, 
which is proportional to At Thus, we conclude that the local discretization 
error is proportional to the square of the step size At 2 while the global dis- 
cretization error is proportional to the step size At itself. Therefore, the 
smaller the mesh size (or the size of the time step in transient problems), 
the smaller the error, and thus the more accurate is the approximation. For 
example, halving the step size will reduce the global discretization error by 
half. It should be clear from the discussions above that the discretization er- 
ror can be minimized by decreasing the step size in space or time as much 
as possible. The discretization error approaches zero as the difference 
quantities such as Ax and At approach the differential quantities such as dx 
and dt. 



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311 
CHAPTER 5 



Round-off Error 

If we had a computer that could retain an infinite number of digits for all 
numbers, the difference between the exact solution and the approximate 
(numerical) solution at any point would entirely be due to discretization er- 
ror. But we know that every computer (or calculator) represents numbers 
using a finite number of significant digits. The default value of the number 
of significant digits for many computers is 7, which is referred to as single 
precision. But the user may perform the calculations using 15 significant 
digits for the numbers, if he or she wishes, which is referred to as double 
precision. Of course, performing calculations in double precision will re- 
quire more computer memory and a longer execution time. 

In single precision mode with seven significant digits, a computer will 
register the number 44444.666666 as 44444.67 or 44444.66, depending on 
the method of rounding the computer uses. In the first case, the excess dig- 
its are said to be rounded to the closest integer, whereas in the second case 
they are said to be chopped off. Therefore, the numbers a = AAAAA.123A5 
and b = AAAAA.12032 are equivalent for a computer that performs calcula- 
tions using seven significant digits. Such a computer would give a — b = 
instead of the true value 0.00313. 

The error due to retaining a limited number of digits during calculations 
is called the round-off error. This error is random in nature and there is no 
easy and systematic way of predicting it. It depends on the number of cal- 
culations, the method of rounding off, the type of computer, and even the 
sequence of calculations. 

In algebra you learned that a + b + c = a + c + b, which seems quite 
reasonable. But this is not necessarily true for calculations performed with 
a computer, as demonstrated in Figure 5-57. Note that changing the se- 
quence of calculations results in an error of 30.8 percent in just two opera- 
tions. Considering that any significant problem involves thousands or even 
millions of such operations performed in sequence, we realize that the ac- 
cumulated round-off error has the potential to cause serious error without 
giving any warning signs. Experienced programmers are very much aware 
of this danger, and they structure their programs to prevent any buildup of 
the round-off error. For example, it is much safer to multiply a number by 
10 than to add it 10 times. Also, it is much safer to start any addition 
process with the smallest numbers and continue with larger numbers. This 
rule is particularly important when evaluating series with a large number of 
terms with alternating signs. 

The round-off error is proportional to the number of computations per- 
formed during the solution. In the finite difference method, the number of 
calculations increases as the mesh size or the time step size decreases. 
Halving the mesh or time step size, for example, will double the number of 
calculations and thus the accumulated round-off error. 

Controlling the Error in Numerical Methods 

The total error in any result obtained by a numerical method is the sum of 
the discretization error, which decreases with decreasing step size, and the 
round-off error, which increases with decreasing step size, as shown in Fig- 
ure 5-58. Therefore, decreasing the step size too much in order to get more 



Given 








a 


= 1111111 




b 


= -1111116 




c 


- 0.4444432 


Find: 


D 


= a + b + c 




E 


= a + c + b 


Solution: 




D = 


7777777 


- 1111116 + 0.4444432 




1 + 0.4444432 




1.444443 


(Correct result) 


E = 


7777777 + 0.4444432 - 7777776 




7777777 


- 7777776 




1.000000 


(In eiTor by 30.8%) 



FIGURE 5-57 

A simple arithmetic operation 

performed with a computer 

in single precision using seven 

significant digits, which results in 

30.8 percent error when the order 

of operation is reversed. 



i Error 




Optimum Step size 

step size 

FIGURE 5-58 

As the mesh or time step size 

decreases, the discretization error 

decreases but the round-off 

error increases. 



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312 
HEAT TRANSFER 



accurate results may actually backfire and give less accurate results be- 
cause of a faster increase in the round-off error. We should be careful not to 
let round-off error get out of control by avoiding a large number of compu- 
tations with very small numbers. 

In practice, we will not know the exact solution of the problem, and thus 
we will not be able to determine the magnitude of the error involved in the 
numerical method. Knowing that the global discretization error is propor- 
tional to the step size is not much help either since there is no easy way of 
determining the value of the proportionality constant. Besides, the global 
discretization error alone is meaningless without a true estimate of the 
round-off error. Therefore, we recommend the following practical proce- 
dures to assess the accuracy of the results obtained by a numerical method. 

• Start the calculations with a reasonable mesh size Ax (and time step 
size At for transient problems) based on experience. Then repeat the 
calculations using a mesh size of Ax/2. If the results obtained by 
halving the mesh size do not differ significantly from the results 
obtained with the full mesh size, we conclude that the discretization 
error is at an acceptable level. But if the difference is larger than we 
can accept, then we have to repeat the calculations using a mesh size 
Ax/4 or even a smaller one at regions of high temperature gradients. 
We continue in this manner until halving the mesh size does not cause 
any significant change in the results, which indicates that the 
discretization error is reduced to an acceptable level. 

• Repeat the calculations using double precision holding the mesh size 
(and the size of the time step in transient problems) constant. If the 
changes are not significant, we conclude that the round-off error is not 
a problem. But if the changes are too large to accept, then we may try 
reducing the total number of calculations by increasing the mesh size 
or changing the order of computations. But if the increased mesh size 
gives unacceptable discretization errors, then we may have to find a 
reasonable compromise. 

It should always be kept in mind that the results obtained by any numer- 
ical method may not reflect any trouble spots in certain problems that re- 
quire special consideration such as hot spots or areas of high temperature 
gradients. The results that seem quite reasonable overall may be in consid- 
erable error at certain locations. This is another reason for always repeating 
the calculations at least twice with different mesh sizes before accepting 
them as the solution of the problem. Most commercial software packages 
have built-in routines that vary the mesh size as necessary to obtain highly 
accurate solutions. But it is a good engineering practice to be aware of any 
potential pitfalls of numerical methods and to examine the results obtained 
with a critical eye. 



SUMMARY 



Analytical solution methods are limited to highly simplified 
problems in simple geometries, and it is often necessary to use 



a numerical method to solve real world problems with com- 
plicated geometries or nonuniform thermal conditions. The 



cen58 93 3_ch05.qxd 9/4/2002 11:42 AM Page 313 



numerical finite difference method is based on replacing deriv- 
atives by differences, and the finite difference formulation of a 
heat transfer problem is obtained by selecting a sufficient num- 
ber of points in the region, called the nodal points or nodes, 
and writing energy balances on the volume elements centered 
about the nodes. 

For steady heat transfer, the energy balance on a volume el- 
ement can be expressed in general as 



2 Q + gv d 







313 
CHAPTER 5 



also be used to solve a system of equations simultaneously at 
the press of a button. 

The finite difference formulation of transient heat conduc- 
tion problems is based on an energy balance that also accounts 
for the variation of the energy content of the volume element 
during a time interval At. The heat transfer and heat generation 
terms are expressed at the previous time step i in the explicit 
method, and at the new time step ;' + 1 in the implicit method. 
For a general node m, the finite difference formulations are 
expressed as 



whether the problem is one-, two-, or three-dimensional. For 
convenience in formulation, we always assume all heat trans- 
fer to be into the volume element from all surfaces toward the 
node under