Skip to main content
#
Full text of "heat and mass transfer"

##
See other formats

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 cen58 93 3_fm.qxd 9/11/2002 10:56 AM Page viii 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 cen5 8 93 3_fm.qxd 9/11/2002 10:56 AM Page ix 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 cen58933_fm.qxd 9/11/2002 10:56 AM Page x 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 cen5 8 93 3_fm.qxd 9/11/2002 10:56 AM Page xi 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 cen58 933_fm.qxd 9/11/2002 10:56 AM Page xii 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 cen58 93 3_fm.qxd 9/11/2002 10:56 AM Page xiii 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 cen58 933_fm.qxd 9/11/2002 10:56 AM Page xiv 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 cen5 8 93 3_fm.qxd 9/11/2002 10:56 AM Page xv 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 cen58 933_fm.qxd 9/11/2002 10:56 AM Page xvi 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 cen58 93 3_fm.qxd 9/11/2002 10:56 AM Page xvii 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 cen58933_ch01.qxd 9/10/2002 8:29 AM Page 5 CHAPTER 1 inexpensive, but the results obtained are subject to the accuracy of the assumptions and idealizations made in the analysis. In heat transfer studies, often a good compromise is reached by reducing the choices to just a few by analysis, and then verifying the findings experimentally. Modeling in Heat Transfer The descriptions of most scientific problems involve expressions that relate the changes in some key variables to each other. Usually the smaller the increment chosen in the changing variables, the more general and accurate the description. In the limiting case of infinitesimal or differential changes in variables, we obtain differential equations that provide precise mathematical formulations for the physical principles and laws by representing the rates of changes as derivatives. Therefore, differential equations are used to investi- gate a wide variety of problems in sciences and engineering, including heat transfer. However, most heat transfer problems encountered in practice can be solved without resorting to differential equations and the complications asso- ciated with them. The study of physical phenomena involves two important steps. In the first step, all the variables that affect the phenomena are identified, reasonable as- sumptions and approximations are made, and the interdependence of these variables is studied. The relevant physical laws and principles are invoked, and the problem is formulated mathematically. The equation itself is very in- structive as it shows the degree of dependence of some variables on others, and the relative importance of various terms. In the second step, the problem is solved using an appropriate approach, and the results are interpreted. Many processes that seem to occur in nature randomly and without any or- der are, in fact, being governed by some visible or not-so-visible physical laws. Whether we notice them or not, these laws are there, governing consis- tently and predictably what seem to be ordinary events. Most of these laws are well defined and well understood by scientists. This makes it possible to pre- dict the course of an event before it actually occurs, or to study various aspects of an event mathematically without actually running expensive and time- consuming experiments. This is where the power of analysis lies. Very accu- rate results to meaningful practical problems can be obtained with relatively little effort by using a suitable and realistic mathematical model. The prepara- tion of such models requires an adequate knowledge of the natural phenomena involved and the relevant laws, as well as a sound judgment. An unrealistic model will obviously give inaccurate and thus unacceptable results. An analyst working on an engineering problem often finds himself or her- self in a position to make a choice between a very accurate but complex model, and a simple but not-so-accurate model. The right choice depends on the situation at hand. The right choice is usually the simplest model that yields adequate results. For example, the process of baking potatoes or roasting a round chunk of beef in an oven can be studied analytically in a simple way by modeling the potato or the roast as a spherical solid ball that has the properties of water (Fig. 1-5). The model is quite simple, but the results obtained are suf- ficiently accurate for most practical purposes. As another example, when we analyze the heat losses from a building in order to select the right size for a heater, we determine the heat losses under anticipated worst conditions and select a furnace that will provide sufficient heat to make up for those losses. Oven ( Potato ) ■* Actual 175°C Water ■* Ideal FIGURE 1-5 Modeling is a powerful engineering tool that provides great insight and simplicity at the expense of some accuracy. cen58933_ch01.qxd 9/10/2002 8:29 AM Page 6 6 HEAT TRANSFER Often we tend to choose a larger furnace in anticipation of some future ex- pansion, or just to provide a factor of safety. A very simple analysis will be ad- equate in this case. When selecting heat transfer equipment, it is important to consider the ac- tual operating conditions. For example, when purchasing a heat exchanger that will handle hard water, we must consider that some calcium deposits will form on the heat transfer surfaces over time, causing fouling and thus a grad- ual decline in performance. The heat exchanger must be selected on the basis of operation under these adverse conditions instead of under new conditions. Preparing very accurate but complex models is usually not so difficult. But such models are not much use to an analyst if they are very difficult and time- consuming to solve. At the minimum, the model should reflect the essential features of the physical problem it represents. There are many significant real- world problems that can be analyzed with a simple model. But it should al- ways be kept in mind that the results obtained from an analysis are as accurate as the assumptions made in simplifying the problem. Therefore, the solution obtained should not be applied to situations for which the original assump- tions do not hold. A solution that is not quite consistent with the observed nature of the prob- lem indicates that the mathematical model used is too crude. In that case, a more realistic model should be prepared by eliminating one or more of the questionable assumptions. This will result in a more complex problem that, of course, is more difficult to solve. Thus any solution to a problem should be in- terpreted within the context of its formulation. 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. cen58933_ch01.qxd 9/10/2002 8:29 AM Page 8 THERMODYNAMICS Air Air in = 1 kg m = 1 kg 300^301 K f 1000 -> 1001 K f I 0.718 kJ I 0.855 kJ FIGURE 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). cen58933_ch01.qxd 9/10/2002 8:29 AM Page 24 24 HEAT TRANSFER differential thermometer reads the temperature drop AT across this distance along each sample. When steady operating conditions are reached, the total rate of heat transfer through both samples becomes equal to the electric power drawn by the heater, which is determined by multiplying the electric current by the voltage. In a certain experiment, cylindrical samples of diameter 5 cm and length 10 cm are used. The two thermocouples in each sample are placed 3 cm apart. After initial transients, the electric heater is observed to draw 0.4 A at 110 V, and both differential thermometers read a temperature difference of 15°C. De- termine the thermal conductivity of the sample. SOLUTION The thermal conductivity of a material is to be determined by en- suring one-dimensional heat conduction, and by measuring temperatures when steady operating conditions are reached. Assumptions 1 Steady operating conditions exist since the temperature readings do not change with time. 2 Heat losses through the lateral surfaces of the apparatus are negligible since those surfaces are well insulated, and thus the entire heat generated by the heater is conducted through the samples. 3 The apparatus possesses thermal symmetry. Analysis The electrical power consumed by the resistance heater and con- verted to heat is W. VI = (110V)(0.4A) = 44W The rate of heat flow through each sample is Q = \ W e = \ X (44 W) = 22 W since only half of the heat generated will flow through each sample because of symmetry. Reading the same temperature difference across the same distance in each sample also confirms that the apparatus possesses thermal symmetry. The heat transfer area is the area normal to the direction of heat flow, which is the cross-sectional area of the cylinder in this case: A = \ ttD 2 = \ tt(0.05 m) 2 = 0.00196 m 2 Noting that the temperature drops by 15 C C within 3 cm in the direction of heat flow, the thermal conductivity of the sample is determined to be Q =kA AT QL (22 W)(0.03 m) A AT (0.00196 m 2 )(15°C) 22.4 W/m • °C Discussion Perhaps you are wondering if we really need to use two samples in the apparatus, since the measurements on the second sample do not give any additional information. It seems like we can replace the second sample by in- sulation. Indeed, we do not need the second sample; however, it enables us to verify the temperature measurements on the first sample and provides thermal symmetry, which reduces experimental error. EXAMPLE 1-7 Conversion between SI and English Units An engineer who is working on the heat transfer analysis of a brick building in English units needs the thermal conductivity of brick. But the only value he can cen58933_ch01.qxd 9/10/2002 8:29 AM Page 25 find from his handbooks is 0.72 W/m • °C, which is in SI units. To make mat- i ters worse, the engineer does not have a direct conversion factor between the two unit systems for thermal conductivity. Can you help him out? 25 CHAPTER 1 SOLUTION The situation this engineer is facing is not unique, and most engi- neers often find themselves in a similar position. A person must be very careful during unit conversion not to fall into some common pitfalls and to avoid some costly mistakes. Although unit conversion is a simple process, it requires utmost care and careful reasoning. The conversion factors for W and m are straightforward and are given in con- version tables to be 1 W lm 3.41214 Btu/h 3.2808 ft But the conversion of C C into C F is not so simple, and it can be a source of er- ror if one is not careful. Perhaps the first thought that comes to mind is to re- place °C by (°F - 32V1.8 since 7"(°C) = [T(°F) - 32]/1.8. But this will be wrong since the °C in the unit W/m • °C represents per °C change in tempera- ture. Noting that 1°C change in temperature corresponds to 1.8°F, the proper conversion factor to be used is 1°C = 1.8°F Substituting, we get 1 w/m • ° c = S^ = °- 5778 Btu/h ■ ft ■ ° F which is the desired conversion factor. Therefore, the thermal conductivity of the brick in English units is = 0.72 X (0.5778 Btu/h • ft • °F) = 0.42 Btu/h • ft • °F Discussion Note that the thermal conductivity value of a material in English units is about half that in SI units (Fig. 1-30). Also note that we rounded the result to two significant digits (the same number in the original value) since ex- pressing the result in more significant digits (such as 0.4160 instead of 0.42) would falsely imply a more accurate value than the original one. k = 0.72 W/m-°C = 0.42 Btu/h-ft-°F FIGURE 1-30 The thermal conductivity value in English units is obtained by multiplying the value in SI units by 0.5778. 1-7 ■ CONVECTION Convection is the mode of energy transfer between a solid surface and the adjacent liquid or gas that is in motion, and it involves the combined effects of conduction and fluid motion. The faster the fluid motion, the greater the convection heat transfer. In the absence of any bulk fluid motion, heat trans- fer between a solid surface and the adjacent fluid is by pure conduction. The presence of bulk motion of the fluid enhances the heat transfer between the solid surface and the fluid, but it also complicates the determination of heat transfer rates. cen58933_ch01.qxd 9/10/2002 8:29 AM Page 26 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 cen58933_ch01.qxd 9/10/2002 8:29 AM Page 28 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. cen58 933_ch01.qxd 9/10/2002 8:29 AM Page 3C 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. cen58933_ch01.qxd 9/10/2002 8:30 AM Page 31 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. cen58933_ch01.qxd 9/10/2002 8:30 AM Page 32 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 cen58933_ch01.qxd 9/10/2002 8:30 AM Page 33 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). cen58933_ch01.qxd 9/10/2002 8:30 AM Page 34 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. cen58933_ch01.qxd 9/10/2002 8:30 AM Page 36 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. cen58 933_ch01.qxd 9/10/2002 8:30 AM Page 3E 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. cen58933_ch01.qxd 9/10/2002 8:30 AM Page 39 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 cen58933_ch01.qxd 9/10/2002 8:30 AM Page 40 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. cen58933_ch01.qxd 9/10/2002 8:30 AM Page 41 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 cen58933_ch01.qxd 9/10/2002 8:30 AM Page 42 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 cen58933_ch01.qxd 9/10/2002 8:30 AM Page 43 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 cen58933_ch01.qxd 9/10/2002 8:30 AM Page 45 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, cen58933_ch01.qxd 9/10/2002 8:30 AM Page 48 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? cen58933_ch01.qxd 9/10/2002 8:30 AM Page 49 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 cen58933_ch01.qxd 9/10/2002 8:30 AM Page 50 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? cen58933_ch01.qxd 9/10/2002 8:30 AM Page 51 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. cen58933_ch01.qxd 9/10/2002 8:30 AM Page 52 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. cen58933_ch01.qxd 9/10/2002 8:30 AM Page 54 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. cen58933_ch01.qxd 9/10/2002 8:30 AM Page 58 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 cen58933_ch02.qxd 9/10/2002 8:46 AM Page 62 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. cen58933_ch02.qxd 9/10/2002 8:46 AM Page 68 68 HEAT TRANSFER Discussion Note that heat generation is expressed per unit volume in W/cm 3 or Btu/h • ft 3 , whereas heat flux is expressed per unit surface area in W/cm 2 or Btu/h • ft 2 . G - Volume / element V A' + Ai' £ FIGURE 2-13 One-dimensional heat conduction through a volume element in a large plane wall. 2-2 - ONE-DIMENSIONAL HEAT CONDUCTION EQUATION Consider heat conduction through a large plane wall such as the wall of a house, the glass of a single pane window, the metal plate at the bottom of a pressing iron, a cast iron steam pipe, a cylindrical nuclear fuel element, an electrical resistance wire, the wall of a spherical container, or a spherical metal ball that is being quenched or tempered. Heat conduction in these and many other geometries can be approximated as being one-dimensional since heat conduction through these geometries will be dominant in one direction and negligible in other directions. Below we will develop the one- dimensional heat conduction equation in rectangular, cylindrical, and spheri- cal coordinates. Heat Conduction Equation in a Large Plane Wall Consider a thin element of thickness Ax in a large plane wall, as shown in Fig- ure 2-13. Assume the density of the wall is p, the specific heat is C, and the area of the wall normal to the direction of heat transfer is A. An energy bal- ance on this thin element during a small time interval At can be expressed as /Rate of heat\ conduction I atjc / or /Rate of heat\ conduction \ at x + Ax J \lx x!x + Ax / Rate of heat \ generation inside the element / Rate of change \ of the energy content of the element A£„ At (2-6) But the change in the energy content of the element and the rate of heat gen- eration within the element can be expressed as A£eien,e„t = E, + Af - E, = mC(T, + A , - T,) = pCAAxiT, + Al - T,) ^element — § 'element — §Al±X (2-7) (2-8) Substituting into Equation 2-6, we get Q x -Q I + tu + gAAx = pCAAx- Dividing by AAx gives 1 2.V + A.V Qx , . _ T, + \, + g = pC At Ax At (2-9) (2-10) cen58933_ch02.qxd 9/10/2002 8:46 AM Page 69 Taking the limit as Ax — > and At — > yields 1 d Ua dT \ j- ■ r dT (2-11) 69 CHAPTER 2 since, from the definition of the derivative and Fourier's law of heat conduc- tion, e, + A, - e, dQ lim Ax-> Ax dx dx \ dx (2-12) Noting that the area A is constant for a plane wall, the one-dimensional tran- sient heat conduction equation in a plane wall becomes Variable conductivity: d , dT\ , . n dT (2-13) The thermal conductivity k of a material, in general, depends on the tempera- ture T (and therefore x), and thus it cannot be taken out of the derivative. However, the thermal conductivity in most practical applications can be as- sumed to remain constant at some average value. The equation above in that case reduces to Constant conductivity: PT dx 2 ]_dT a dt (2-14) where the property a = k/pC is the thermal diffusivity of the material and represents how fast heat propagates through a material. It reduces to the fol- lowing forms under specified conditions (Fig. 2-14): (1) Steady-state: (d/dt = 0) (2) Transient, no heat generation: (8 = 0) (3) Steady-state, no heat generation: {d/dt = and g = 0) d 2 T S dx 2 k d 2 T 1 dT dx 2 a dt d 2 T = dx 2 (2-15) (2-16) (2-17) Note that we replaced the partial derivatives by ordinary derivatives in the one-dimensional steady heat conduction case since the partial and ordinary derivatives of a function are identical when the function depends on a single variable only [T = T(x) in this case]. Heat Conduction Equation in a Long Cylinder Now consider a thin cylindrical shell element of thickness Ar in a long cylin- der, as shown in Figure 2-15. Assume the density of the cylinder is p, the spe- cific heat is C, and the length is L. The area of the cylinder normal to the direction of heat transfer at any location is A = 2irrL where r is the value of the radius at that location. Note that the heat transfer area A depends on r in this case, and thus it varies with location. An energy balance on this thin cylindrical shell element during a small time interval A; can be expressed as General, one dimensional: No Steady- generation state Steady, one-dimensional: d 2 T dx 2 = FIGURE 2-14 The simplification of the one- dimensional heat conduction equation in a plane wall for the case of constant conductivity for steady conduction with no heat generation. ■ Volume element FIGURE 2-15 One-dimensional heat conduction through a volume element in a long cylinder. cen58933_ch02.qxd 9/10/2002 8:46 AM Page 7C 70 HEAT TRANSFER ' Rate of heat \ conduction \ at r ( Rate of heat \ conduction + 1 at r + Ar j Rate of heat \ generation inside the element / Rate of change \ of the energy content of the element or Qr ~ Qr + Ar + G c | e A£„ Ar (2-18) The change in the energy content of the element and the rate of heat genera- tion within the element can be expressed as A£„ gV A E, = mC(T, + A , - T,) = pCAAr(T, + A , - T,) (2-19) = gAAr (2-20) Substituting into Eq. 2-18, we get Q r -Q r + Ar + gAAr = pCAAr At (2-21) where A = 2ittL. You may be tempted to express the area at the middle of the element using the average radius as A = 2ir(r + Ar/2)L. But there is nothing we can gain from this complication since later in the analysis we will take the limit as Ar — > and thus the term Ar/2 will drop out. Now dividing the equa- tion above by AAr gives 1 8 r+ A, ~ Qr A Ar P C- Ar (2-22) Taking the limit as Ar — > and Ar — > yields Adr( kA Tr +8 = PC V (2-23) since, from the definition of the derivative and Fourier's law of heat conduction, lim Ar->0 g,. + Ar " Qr dQ d Ar dr dr -kA dT dr (2-24) Noting that the heat transfer area in this case is A = 2irrL, the one- dimensional transient heat conduction equation in a cylinder becomes Variable conductivity: 1 d / , dT\, rTr V k Tr ] P C dT dt (2-25) For the case of constant thermal conductivity, the equation above reduces to 1 dT Constant conductivity: ld_ dT\ .__ >' drV dr k a dt (2-26) cen58933_ch02.qxd 9/10/2002 8:46 AM Page 71 71 CHAPTER 2 where again the property a = k/pC is the thermal diffusivity of the material. Equation 2-26 reduces to the following forms under specified conditions (Fig. 2-16): (1) Steady-state: l_d/ dT\ g (d/dt = 0) '" dr \ r dr) k (2) Transient, no heat generation: Id/ dT\ _ 1 dT (g = 0) r dr\ r dr ) ~~ « dt (3) Steady-state, no heat generation: d I dT (d/dt = and g = 0) d~r\ r ~dr (2-27) (2-28) (2-29) Note that we again replaced the partial derivatives by ordinary derivatives in the one-dimensional steady heat conduction case since the partial and ordinary derivatives of a function are identical when the function depends on a single variable only [T = T(r) in this case]. Heat Conduction Equation in a Sphere Now consider a sphere with density p, specific heat C, and outer radius R. The area of the sphere normal to the direction of heat transfer at any location is A = 4ttt 2 , where r is the value of the radius at that location. Note that the heat transfer area A depends on r in this case also, and thus it varies with location. By considering a thin spherical shell element of thickness Ar and repeating the approach described above for the cylinder by using A = 4 tit 2 instead of A = 2irrL, the one-dimensional transient heat conduction equation for a sphere is determined to be (Fig. 2-17) Variable conductivity: 1 d , r l k — dr \ dr P C dT dt (2-30) (a) The form that is ready to integrate my-" (b) The equivalent alternative form d 2 T dT „ FIGURE 2-16 Two equivalent forms of the differential equation for the one- dimensional steady heat conduction in a cylinder with no heat generation. FIGURE 2-17 One-dimensional heat conduction through a volume element in a sphere. which, in the case of constant thermal conductivity, reduces to 1 dT Constant conductivity: L± r 2§T + _. r 2 dr\ dr k a dt (2-31) where again the property a = k/pC is the thermal diffusivity of the material. It reduces to the following forms under specified conditions: (1) Steady-state: (d/dt = 0) (2) Transient, no heat generation: (g = 0) (3) Steady-state, no heat generation: (d/dt = and g = 0) 11 [ r idT r 2 dr\ dr \_d_l 2 dT r 2 dr\ r dr d_ 2 dT dr \ dr ~k = \_dT a dt dr 2 dr (2-32) (2-33) (2-34) where again we replaced the partial derivatives by ordinary derivatives in the one-dimensional steady heat conduction case. cen58933_ch02.qxd 9/10/2002 8:46 AM Page 72 72 HEAT TRANSFER Combined One-Dimensional Heat Conduction Equation An examination of the one-dimensional transient heat conduction equations for the plane wall, cylinder, and sphere reveals that all three equations can be expressed in a compact form as d_ ; dr r" k dT dr P C dT dt (2-35) where n = for a plane wall, n = 1 for a cylinder, and n = 2 for a sphere. In the case of a plane wall, it is customary to replace the variable r by x. This equation can be simplified for steady-state or no heat generation cases as described before. 800 W FIGURE 2-18 Schematic for Example 2-3. EXAMPLE 2-3 Heat Conduction through the Bottom of a Pan Consider a steel pan placed on top of an electric range to cook spaghetti (Fig. 2-18). The bottom section of the pan is L = 0.4 cm thick and has a diameter of D = 18 cm. The electric heating unit on the range top consumes 800 W of power during cooking, and 80 percent of the heat generated in the heating ele- ment is transferred uniformly to the pan. Assuming constant thermal conduc- tivity, obtain the differential equation that describes the variation of the temperature in the bottom section of the pan during steady operation. SOLUTION The bottom section of the pan has a large surface area relative to its thickness and can be approximated as a large plane wall. Heat flux is ap- plied to the bottom surface of the pan uniformly, and the conditions on the inner surface are also uniform. Therefore, we expect the heat transfer through the bottom section of the pan to be from the bottom surface toward the top, and heat transfer in this case can reasonably be approximated as being one- dimensional. Taking the direction normal to the bottom surface of the pan to be the x-axis, we will have T = T(x) during steady operation since the temperature in this case will depend on x only. The thermal conductivity is given to be constant, and there is no heat gener- ation in the medium (within the bottom section of the pan). Therefore, the dif- ferential equation governing the variation of temperature in the bottom section of the pan in this case is simply Eq. 2-17, d 2 T _ dx 2 ' which is the steady one-dimensional heat conduction equation in rectangular coordinates under the conditions of constant thermal conductivity and no heat generation. Note that the conditions at the surface of the medium have no ef- fect on the differential equation. EXAMPLE 2-4 Heat Conduction in a Resistance Heater A 2-kW resistance heater wire with thermal conductivity k = 15 W/m ■ °C, di- ameter D = 0.4 cm, and length L = 50 cm is used to boil water by immersing cen58933_ch02.qxd 9/10/2002 8:46 AM Page 73 it in water (Fig. 2-19). Assuming the variation of the thermal conductivity of the wire with temperature to be negligible, obtain the differential equation that de- scribes the variation of the temperature in the wire during steady operation. SOLUTION The resistance wire can be considered to be a very long cylinder since its length is more than 100 times its diameter. Also, heat is generated uniformly in the wire and the conditions on the outer surface of the wire are uni- form. Therefore, it is reasonable to expect the temperature in the wire to vary in the radial r direction only and thus the heat transfer to be one-dimensional. Then we will have T = T(r) during steady operation since the temperature in this case will depend on ronly. The rate of heat generation in the wire per unit volume can be determined from G G 2000 W v„ (ttD 2 /4)L [tt(0.004 m) 2 /4](0.5 cm) 0.318 X 10"W/m 3 Noting that the thermal conductivity is given to be constant, the differential equation that governs the variation of temperature in the wire is simply Eq. 2-27, 1 d I dT r dr\ dr which is the steady one-dimensional heat conduction equation in cylindrical co- ordinates for the case of constant thermal conductivity. Note again that the con- ditions at the surface of the wire have no effect on the differential equation. 73 CHAPTER 2 FIGURE 2-19 Schematic for Example 2-4. EXAMPLE 2-5 Cooling of a Hot Metal Ball in Air A spherical metal ball of radius R is heated in an oven to a temperature of 600°F throughout and is then taken out of the oven and allowed to cool in am- bient air at 7" x = 75°F by convection and radiation (Fig. 2-20). The thermal conductivity of the ball material is known to vary linearly with temperature. As- suming the ball is cooled uniformly from the entire outer surface, obtain the dif- ferential equation that describes the variation of the temperature in the ball during cooling. SOLUTION The ball is initially at a uniform temperature and is cooled uni- formly from the entire outer surface. Also, the temperature at any point in the ball will change with time during cooling. Therefore, this is a one-dimensional transient heat conduction problem since the temperature within the ball will change with the radial distance rand the time t. That is, T = T(r, t). The thermal conductivity is given to be variable, and there is no heat genera- tion in the ball. Therefore, the differential equation that governs the variation of temperature in the ball in this case is obtained from Eq. 2-30 by setting the heat generation term equal to zero. We obtain r 2 dr r 2 k §T dr PC dT dt 75°F FIGURE 2-20 Schematic for Example 2-5. cen58933_ch02.qxd 9/10/2002 8:46 AM Page 74 74 HEAT TRANSFER which is the one-dimensional transient heat conduction equation in spherical coordinates under the conditions of variable thermal conductivity and no heat generation. Note again that the conditions at the outer surface of the ball have no effect on the differential equation. FIGURE 2-21 Three-dimensional heat conduction through a rectangular volume element. 2-3 - GENERAL HEAT CONDUCTION EQUATION In the last section we considered one-dimensional heat conduction and assumed heat conduction in other directions to be negligible. Most heat trans- fer problems encountered in practice can be approximated as being one- dimensional, and we will mostly deal with such problems in this text. However, this is not always the case, and sometimes we need to consider heat transfer in other directions as well. In such cases heat conduction is said to be multidimensional, and in this section we will develop the governing differen- tial equation in such systems in rectangular, cylindrical, and spherical coordi- nate systems. Rectangular Coordinates Consider a small rectangular element of length Ax, width Ay, and height Az, as shown in Figure 2-21. Assume the density of the body is p and the specific heat is C. An energy balance on this element during a small time interval At can be expressed as \ / Rate of heat \ generation inside the / Rate of heat \ conduction at \ x, y, and z J Rate of heat conduction at x + Ax, y + Ay, and z + Az I element Rate of change \ of the energy content of the element or Qt+Qy + Qz-Qx + Ax-Qy Q-. A£„ At (2-36) Noting that the volume of the element is V e i ement = AxAyAz, the change in the energy content of the element and the rate of heat generation within the ele- ment can be expressed as A£„ E-i + At o 'element E, = mC(T, + A , = gAxAyAz T,) = pCAxAyAz{T, + At -T t ) Substituting into Eq. 2-36, we get Qx+Qy + Qz-Qx Qy + . Q-. g AxAyAz = pCAxAyAz T'f + Af Ar Dividing by AxAyAz gives G.Y+A.V _ Qx 1 Gy + Ay 1 AyAz e, Ax AxAz Av 1 Q l + Az- AxAy Az Qz PC T t +At At (2-37) cen58933_ch02.qxd 9/10/2002 8:46 AM Page 75 Noting that the heat transfer areas of the element for heat conduction in the x, y, and z directions are A x = AyAz, A y = AxAz, and A, = AxAy, respectively, and taking the limit as Ax, A_y, Az and At — > yields dx\ dx I dy \ dy + l*fH pC dT dt (2-38) 75 CHAPTER 2 since, from the definition of the derivative and Fourier's law of heat conduction, l Q + Ax Q x Ax-» AyAz Ax 1 Q. + A.v ~ e, Ay-»0 AxAz Ay 1 Q + Az — Qz Az^oAxAy Az 1 dQ x AyAz dx 1 9Qy AxAz dy 1 dQ z AxAy dz AxAy dz 1 d AyAz dx 1 d AxAz dy 1 a -kAyAz -kAxAz -kAxAy i)T dx 8T dy dT dz dx \ dx ^- k- dy\ dy dz r dz Equation 2-38 is the general heat conduction equation in rectangular coordi- nates. In the case of constant thermal conductivity, it reduces to d 2 T d 2 T d 2 T 8 = 1 dT dx 2 dy 2 dz 2 k « dt (2-39) where the property a = k/pC is again the thermal dijfusivity of the material. Equation 2-39 is known as the Fourier-Biot equation, and it reduces to these forms under specified conditions: (1) Steady-state: (called the Poisson equation) (2) Transient, no heat generation: (called the diffusion equation) (3) Steady-state, no heat generation: (called the Laplace equation) Note that in the special case of one-dimensional heat transfer in the x-direction, the derivatives with respect to y and z drop out and the equations above reduce to the ones developed in the previous section for a plane wall (Fig. 2-22). d 2 T | d 2 T | d 2 T | g _ dx 2 dy 2 dz 2 k (2-40) d 2 T d 2 T d 2 T 1 dT dx 2 3y 2 dz 2 ~ a dt (2-41) d_^ + djT + d^T =0 dx 2 9y 2 dz 2 (2-42) Cylindrical Coordinates The general heat conduction equation in cylindrical coordinates can be ob- tained from an energy balance on a volume element in cylindrical coordinates, shown in Figure 2-23, by following the steps just outlined. It can also be ob- tained directly from Eq. 2-38 by coordinate transformation using the follow- ing relations between the coordinates of a point in rectangular and cylindrical coordinate systems: x = r cos <f>, y = r sin 4>, and z = Z FIGURE 2-22 The three-dimensional heat conduction equations reduce to the one-dimensional ones when the temperature varies in one dimension only. FIGURE 2-23 A differential volume element in cylindrical coordinates. cen58933_ch02.qxd 9/10/2002 8:46 AM Page 76 76 HEAT TRANSFER FIGURE 2-24 A differential volume element in spherical coordinates. After lengthy manipulations, we obtain r dr kr dT dr ,11 , dT r 2 acp \ 5cj> d_ dz dT dz + g n dT P c s7 (2-43) Spherical Coordinates The general heat conduction equations in spherical coordinates can be ob- tained from an energy balance on a volume element in spherical coordinates, shown in Figure 2-24, by following the steps outlined above. It can also be obtained directly from Eq. 2-38 by coordinate transformation using the fol- lowing relations between the coordinates of a point in rectangular and spheri- cal coordinate systems: x = r cos 4> sin G, y = r sin 4> sin G, and cos G Again after lengthy manipulations, we obtain r 2 dr \ dr 1 d dT r 2 sin 2 G d<t> V" ^ 6 96 k sin 6 dT ae P c dT dt (2-44) Obtaining analytical solutions to these differential equations requires a knowledge of the solution techniques of partial differential equations, which is beyond the scope of this introductory text. Here we limit our consideration to one-dimensional steady-state cases or lumped systems, since they result in ordinary differential equations. r Metal Heat loss 600°F billet > T„ = 65°F FIGURE 2-25 Schematic for Example 2-6. EXAMPLE 2-6 Heat Conduction in a Short Cylinder A short cylindrical metal billet of radius R and height h is heated in an oven to a temperature of 600 C F throughout and is then taken out of the oven and al- lowed to cool in ambient air at 7" x = 65°F by convection and radiation. Assum- ing the billet is cooled uniformly from all outer surfaces and the variation of the thermal conductivity of the material with temperature is negligible, obtain the differential equation that describes the variation of the temperature in the bil- let during this cooling process. SOLUTION The billet shown in Figure 2-25 is initially at a uniform tempera- ture and is cooled uniformly from the top and bottom surfaces in the z-direction as well as the lateral surface in the radial r-direction. Also, the temperature at any point in the ball will change with time during cooling. Therefore, this is a two-dimensional transient heat conduction problem since the temperature within the billet will change with the radial and axial distances rand zand with time t. That is, T= T(r, z, f). The thermal conductivity is given to be constant, and there is no heat gener- ation in the billet. Therefore, the differential equation that governs the variation of temperature in the billet in this case is obtained from Eq. 2-43 by setting the heat generation term and the derivatives with respect to 4> equal to zero. We obtain IA r dr kr + f dT dz P c dT dt cen58933_ch02.qxd 9/10/2002 8:46 AM Page 77 In the case of constc nt therma cond JCt vity, it reduces to 1A r dr + dz 2 _ 1 dT a dt which is the desired equation. 77 CHAPTER 2 2-A - BOUNDARY AND INITIAL CONDITIONS The heat conduction equations above were developed using an energy balance on a differential element inside the medium, and they remain the same re- gardless of the thermal conditions on the surfaces of the medium. That is, the differential equations do not incorporate any information related to the condi- tions on the surfaces such as the surface temperature or a specified heat flux. Yet we know that the heat flux and the temperature distribution in a medium depend on the conditions at the surfaces, and the description of a heat transfer problem in a medium is not complete without a full description of the thermal conditions at the bounding surfaces of the medium. The mathematical expres- sions of the thermal conditions at the boundaries are called the boundary conditions. From a mathematical point of view, solving a differential equation is essen- tially a process of removing derivatives, or an integration process, and thus the solution of a differential equation typically involves arbitrary constants (Fig. 2-26). It follows that to obtain a unique solution to a problem, we need to specify more than just the governing differential equation. We need to spec- ify some conditions (such as the value of the function or its derivatives at some value of the independent variable) so that forcing the solution to satisfy these conditions at specified points will result in unique values for the arbi- trary constants and thus a unique solution. But since the differential equation has no place for the additional information or conditions, we need to supply them separately in the form of boundary or initial conditions. Consider the variation of temperature along the wall of a brick house in winter. The temperature at any point in the wall depends on, among other things, the conditions at the two surfaces of the wall such as the air tempera- ture of the house, the velocity and direction of the winds, and the solar energy incident on the outer surface. That is, the temperature distribution in a medium depends on the conditions at the boundaries of the medium as well as the heat transfer mechanism inside the medium. To describe a heat transfer problem completely, two boundary conditions must be given for each direction of the coordinate system along which heat transfer is significant (Fig. 2-27). Therefore, we need to specify two boundary conditions for one-dimensional problems, four boundary conditions for two-dimensional problems, and six boundary conditions for three-dimensional problems. In the case of the wall of a house, for example, we need to specify the conditions at two locations (the inner and the outer surfaces) of the wall since heat transfer in this case is one-dimensional. But in the case of a parallelepiped, we need to specify six boundary conditions (one at each face) when heat transfer in all three dimen- sions is significant. The differential equation: d 2 T dx 2 C,JE T c, General solution: T(x) Arbitrary constants Some specific solutions: T(x) = 2x + 5 T(x) = -x + 12 T(x) = -3 T(x) = 6.2* FIGURE 2-26 The general solution of a typical differential equation involves arbitrary constants, and thus an infinite number of solutions. 50°C Some solutions of d 2 T dx 2 ' :0 15°C The only solution that satisfies the conditions 7/(0) = 50°C and T(L) = 15°C. FIGURE 2-27 To describe a heat transfer problem completely, two boundary conditions must be given for each direction along which heat transfer is significant. cen58933_ch02.qxd 9/10/2002 8:46 AM Page 7E 78 HEAT TRANSFER The physical argument presented above is consistent with the mathematical nature of the problem since the heat conduction equation is second order (i.e., involves second derivatives with respect to the space variables) in all direc- tions along which heat conduction is significant, and the general solution of a second-order linear differential equation involves two arbitrary constants for each direction. That is, the number of boundary conditions that needs to be specified in a direction is equal to the order of the differential equation in that direction. Reconsider the brick wall already discussed. The temperature at any point on the wall at a specified time also depends on the condition of the wall at the beginning of the heat conduction process. Such a condition, which is usually specified at time t = 0, is called the initial condition, which is a mathemati- cal expression for the temperature distribution of the medium initially. Note that we need only one initial condition for a heat conduction problem regard- less of the dimension since the conduction equation is first order in time (it in- volves the first derivative of temperature with respect to time). In rectangular coordinates, the initial condition can be specified in the gen- eral form as T(x, y, z, 0) = f(x, y, z) (2-45) 150°C T(x, t) m 70°C 7X0, = 150°C T(L, f) = 70°C FIGURE 2-28 Specified temperature boundary conditions on both surfaces of a plane wall. where the function /(x, y, z) represents the temperature distribution throughout the medium at time t = 0. When the medium is initially at a uniform tem- perature of T h the initial condition of Eq. 2-45 can be expressed as T(x, y, z, 0) = Tj. Note that under steady conditions, the heat conduction equa- tion does not involve any time derivatives, and thus we do not need to specify an initial condition. The heat conduction equation is first order in time, and thus the initial con- dition cannot involve any derivatives (it is limited to a specified temperature). However, the heat conduction equation is second order in space coordinates, and thus a boundary condition may involve first derivatives at the boundaries as well as specified values of temperature. Boundary conditions most com- monly encountered in practice are the specified temperature, specified heat flux, convection, and radiation boundary conditions. 1 Specified Temperature Boundary Condition The temperature of an exposed surface can usually be measured directly and easily. Therefore, one of the easiest ways to specify the thermal conditions on a surface is to specify the temperature. For one-dimensional heat transfer through a plane wall of thickness L, for example, the specified temperature boundary conditions can be expressed as (Fig. 2-28) 7/(0, t) = T x T(L, t) = T 2 (2-46) where T { and T 2 are the specified temperatures at surfaces at x = and x = L, respectively. The specified temperatures can be constant, which is the case for steady heat conduction, or may vary with time. cen58933_ch02.qxd 9/10/2002 8:46 AM Page 79 2 Specified Heat Flux Boundary Condition When there is sufficient information about energy interactions at a surface, it may be possible to determine the rate of heat transfer and thus the heat flux q (heat transfer rate per unit surface area, W/m 2 ) on that surface, and this infor- mation can be used as one of the boundary conditions. The heat flux in the positive x-direction anywhere in the medium, including the boundaries, can be expressed by Fourier's law of heat conduction as dT = Heat flux in the \ dx ^positive ^-direction/ (W/m 2 ) (2-47) Then the boundary condition at a boundary is obtained by setting the specified heat flux equal to —k(dT/dx) at that boundary. The sign of the specified heat flux is determined by inspection: positive if the heat flux is in the positive di- rection of the coordinate axis, and negative if it is in the opposite direction. Note that it is extremely important to have the correct sign for the specified heat flux since the wrong sign will invert the direction of heat transfer and cause the heat gain to be interpreted as heat loss (Fig. 2-29). For a plate of thickness L subjected to heat flux of 50 W/m 2 into the medium from both sides, for example, the specified heat flux boundary conditions can be expressed as 37(0, t) -k—i = 50 dx and dT(L, t) '' dx -50 (2-48) 79 CHAPTER 2 Heat flux %■■ Conduction . 37X0, dx Conduction .dT(L,t) Heat flux 9 L FIGURE 2-29 Specified heat flux boundary conditions on both surfaces of a plane wall. Note that the heat flux at the surface at x = L is in the negative x-direction, and thus it is —50 W/m 2 . Special Case: Insulated Boundary Some surfaces are commonly insulated in practice in order to minimize heat loss (or heat gain) through them. Insulation reduces heat transfer but does not totally eliminate it unless its thickness is infinity. However, heat transfer through a properly insulated surface can be taken to be zero since adequate insulation reduces heat transfer through a surface to negligible levels. There- fore, a well-insulated surface can be modeled as a surface with a specified heat flux of zero. Then the boundary condition on a perfectly insulated surface (at x = 0, for example) can be expressed as (Fig. 2-30) 3T(0, t) dx dT(0, dx (2-49) That is, on an insulated surface, the first derivative of temperature with re- spect to the space variable (the temperature gradient) in the direction normal to the insulated surface is zero. This also means that the temperature function must be perpendicular to an insulated surface since the slope of temperature at the surface must be zero. Another Special Case: Thermal Symmetry Some heat transfer problems possess thermal symmetry as a result of the symmetry in imposed thermal conditions. For example, the two surfaces of a large hot plate of thickness L suspended vertically in air will be subjected to Insulation T(x, t) 97/(0, t) 60°C dx T(L, t) = 60°C FIGURE 2-30 A plane wall with insulation and specified temperature boundary conditions. cen58933_ch02.qxd 9/10/2002 8:46 AM Page 80 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. cen58933_ch02.qxd 9/10/2002 8:46 AM Page 86 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. cen58933_ch02.qxd 9/10/2002 8:46 AM Page 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 cen58933_ch02.qxd 9/10/2002 8:46 AM Page 90 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. cen58933_ch02.qxd 9/10/2002 8:46 AM Page 92 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 cen58933_ch02.qxd 9/10/2002 8:46 AM Page 94 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. cen58933_ch02.qxd 9/10/2002 8:46 AM Page 96 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. cen58933_ch02.qxd 9/10/2002 8:47 AM Page 1C 108 HEAT TRANSFER X x + Ax FIGURE 2-67 The derivative of a function at a point represents the slope of the tangent line of the function at that point. FIGURE 2-68 Graphical representation of partial derivative dzldx. readability of the equations. The value of y at a fixed number a is denoted by y(a). The derivative of a function y(x) at a point is equivalent to the slope of the tangent line to the graph of the function at that point and is defined as (Fig. 2-67) y'W dy(x) dx hm — Ax->0 AX lim y(x + Ax) - y(x) Ax (2-81) Here Ax represents a (small) change in the independent variable x and is called an increment of x. The corresponding change in the function y is called an increment of y and is denoted by Ay. Therefore, the derivative of a function can be viewed as the ratio of the increment Ay of the function to the increment Ax of the independent variable for very small Ax. Note that Ay and thus y'(x) will be zero if the function y does not change with x. Most problems encountered in practice involve quantities that change with time t, and their first derivatives with respect to time represent the rate of change of those quantities with time. For example, if N(t) denotes the population of a bacteria colony at time /, then the first derivative N' = dNIdt represents the rate of change of the population, which is the amount the population increases or decreases per unit time. The derivative of the first derivative of a function y is called the second derivative of y, and is denoted by y" or d 2 y/dx 2 . In general, the derivative of the (n — l)st derivative of y is called the nth derivative of y and is denoted by y (,,) or d n y/dx". Here, n is a positive integer and is called the order of the derivative. The order n should not be confused with the degree of a deriva- tive. For example, y'" is the third-order derivative of y, but (y') 3 is the third degree of the first derivative of y. Note that the first derivative of a function represents the slope or the rate of change of the function with the indepen- dent variable, and the second derivative represents the rate of change of the slope of the function with the independent variable. When a function y depends on two or more independent variables such as x and t, it is sometimes of interest to examine the dependence of the function on one of the variables only. This is done by taking the derivative of the function with respect to that variable while holding the other vari- ables constant. Such derivatives are called partial derivatives. The first partial derivatives of the function y(x, t) with respect to x and t are defined as (Fig. 2-68) dy y(x + Ax, t) - y(x, t) — = hm -. OX Ax -^ o Ax y(x, t + At) - y(x, t) dy T^ = llm at \t^o At (2-82) (2-83) Note that when finding dy/dx we treat (asa constant and differentiate y with respect to x. Likewise, when finding dy/dt we treat x as a constant and differentiate y with respect to /. Integration can be viewed as the inverse process of differentiation. Inte- gration is commonly used in solving differential equations since solving a differential equation is essentially a process of removing the derivatives cen58933_ch02.qxd 9/10/2002 8:47 AM Page 109 from the equation. Differentiation is the process of finding y'(x) when a function y(x) is given, whereas integration is the process of finding the function y(x) when its derivative y'(x) is given. The integral of this deriva- tive is expressed as J y'(x)dx = J dy = y(x) + C (2-84) since y'(x)dx = dy and the integral of the differential of a function is the function itself (plus a constant, of course). In Eq. 2-84, x is the integration variable and C is an arbitrary constant called the integration constant. The derivative of y(x) + C is y'(x) no matter what the value of the con- stant C is. Therefore, two functions that differ by a constant have the same derivative, and we always add a constant C during integration to recover this constant that is lost during differentiation. The integral in Eq. 2-84 is called an indefinite integral since the value of the arbitrary constant C is indefinite. The described procedure can be extended to higher-order deriv- atives (Fig. 2-69). For example, 109 CHAPTER 2 I dy- = y + C .(>' dx = -y + C J" dx = = y' + c \' dx = = y" + C I" dx = --y<"-» + C y"{x)dx = y'(x) + C (2-85) FIGURE 2-69 Some indefinite integrals that involve derivatives. This can be proved by defining a new variable u(x) = y'(x), differentiating it to obtain u'(x) = y"{x), and then applying Eq. 2-84. Therefore, the order of a derivative decreases by one each time it is integrated. Classification of Differential Equations A differential equation that involves only ordinary derivatives is called an ordinary differential equation, and a differential equation that involves partial derivatives is called a partial differential equation. Then it follows that problems that involve a single independent variable result in ordinary differential equations, and problems that involve two or more independent variables result in partial differential equations. A differential equation may involve several derivatives of various orders of an unknown function. The order of the highest derivative in a differential equation is the order of the equation. For example, the order of /" + (y") 4 = 7x 5 is 3 since it contains no fourth or higher order derivatives. You will remember from algebra that the equation 3x — 5 = is much easier to solve than the equation x 4 + 3x — 5 = because the first equation is linear whereas the second one is nonlinear. This is also true for differen- tial equations. Therefore, before we start solving a differential equation, we usually check for linearity. A differential equation is said to be linear if the dependent variable and all of its derivatives are of the first degree and their coefficients depend on the independent variable only. In other words, a dif- ferential equation is linear if it can be written in a form that does not in- volve (1) any powers of the dependent variable or its derivatives such as y 3 or (y') 2 , (2) any products of the dependent variable or its derivatives such as yy' or y'y'", and (3) any other nonlinear functions of the dependent vari- able such as sin y or e>. If any of these conditions apply, it is nonlinear (Fig. 2-70). (a) A nonlinear equation: 30") 2 -4vv' + e 2 Power Product -6x l Other nonlinear functions (b) A linear equation: 3x y" - 4xy' + e y = 6x FIGURE 2-70 A differential equation that is (a) nonlinear and (b) linear. When checking for linearity, we examine the dependent variable only. cen58933_ch02.qxd 9/10/2002 8:47 AM Page 11C 110 HEAT TRANSFER (a) With constant coefficients: y" + 6y' — 2y = xe -2x v V Constant (b) With variable coefficients: y ■> i 2 xe 2 * AT X — 1 V V Variable FIGURE 2-71 A differential equation with (a) constant coefficients and (b) variable coefficients. (a) An algebraic equation: y 2 - 7v - 10 = Solution: y = 2 and y = 5 (b) A differential equation: y' - v.v = o Solution: y = e 7 ' FIGURE 2-72 Unlike those of algebraic equations, the solutions of differential equations are typically functions instead of discrete values. A linear differential equation, however, may contain (1) powers or non- linear functions of the independent variable, such as x 2 and cos x and (2) products of the dependent variable (or its derivatives) and functions of the independent variable, such as x 3 y' , x 2 y, and e^^y". A linear differential equation of order n can be expressed in the most general form as yM +f l (x)yf» -'>+■•■ +f„_ l (x)y' +f„(x)y = R(x) (2-86) A differential equation that cannot be put into this form is nonlinear. A linear differential equation in y is said to be homogeneous as well if R(x) = 0. Otherwise, it is nonhomogeneous. That is, each term in a linear homogeneous equation contains the dependent variable or one of its deriv- atives after the equation is cleared of any common factors. The term R(x) is called the nonhomogeneous term. Differential equations are also classified by the nature of the coefficients of the dependent variable and its derivatives. A differential equation is said to have constant coefficients if the coefficients of all the terms that involve the dependent variable or its derivatives are constants. If, after clearing any common factors, any of the terms with the dependent variable or its deriv- atives involve the independent variable as a coefficient, that equation is said to have variable coefficients (Fig. 2-71). Differential equations with constant coefficients are usually much easier to solve than those with vari- able coefficients. Solutions of Differential Equations Solving a differential equation can be as easy as performing one or more integrations; but such simple differential equations are usually the excep- tion rather than the rule. There is no single general solution method appli- cable to all differential equations. There are different solution techniques, each being applicable to different classes of differential equations. Some- times solving a differential equation requires the use of two or more tech- niques as well as ingenuity and mastery of solution methods. Some differential equations can be solved only by using some very clever tricks. Some cannot be solved analytically at all. In algebra, we usually seek discrete values that satisfy an algebraic equa- tion such as x 1 — Ix —10 = 0. When dealing with differential equations, however, we seek functions that satisfy the equation in a specified interval. For example, the algebraic equation x 2 — Ix — 10 = is satisfied by two numbers only: 2 and 5. But the differential equation y' — ly = is satis- fied by the function e lx for any value of x (Fig. 2-72). Consider the algebraic equation x 3 — 6x 2 + llx — 6 = 0. Obviously, x = 1 satisfies this equation, and thus it is a solution. However, it is not the only solution of this equation. We can easily show by direct substitution that x = 2 and x = 3 also satisfy this equation, and thus they are solutions as well. But there are no other solutions to this equation. Therefore, we say that the set 1, 2, and 3 forms the complete solution to this algebraic equation. The same line of reasoning also applies to differential equations. Typi- cally, differential equations have multiple solutions that contain at least one arbitrary constant. Any function that satisfies the differential equation on an 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. cen58933_ch02.qxd 9/10/2002 8:47 AM Page 112 112 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? cen58933_ch02.qxd 9/10/2002 8:47 AM Page 116 116 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. cen58933_ch03.qxd 9/10/2002 8:59 AM Page 132 132 HEAT TRANSFER FIGURE 3-8 The temperature drop across a layer is proportional to its thermal resistance. ►e= iow conv, 1 I »i* — VWWV^ 2°C/W wall 15°C/W AT=QR Y conv, 2 -^^WVVW — ♦ r »2 3°C/W where "total 7T7 + 7T + 7^7 (° C/W ) /?,A kA h-iA (3-16) Note that the heat transfer area A is constant for a plane wall, and the rate of heat transfer through a wall separating two mediums is equal to the tempera- ture difference divided by the total thermal resistance between the mediums. Also note that the thermal resistances are in series, and the equivalent thermal resistance is determined by simply adding the individual resistances, just like the electrical resistances connected in series. Thus, the electrical analogy still applies. We summarize this as the rate of steady heat transfer between two surfaces is equal to the temperature difference divided by the total thermal re- sistance between those two surfaces. Another observation that can be made from Eq. 3-15 is that the ratio of the temperature drop to the thermal resistance across any layer is constant, and thus the temperature drop across any layer is proportional to the thermal resistance of the layer. The larger the resistance, the larger the temperature drop. In fact, the equation Q = AT/R can be rearranged as AT = QR (°C) (3-17) which indicates that the temperature drop across any layer is equal to the rate of heat transfer times the thermal resistance across that layer (Fig. 3-8). You may recall that this is also true for voltage drop across an electrical resistance when the electric current is constant. It is sometimes convenient to express heat transfer through a medium in an analogous manner to Newton's law of cooling as Q = UA AT (W) (3-18) where U is the overall heat transfer coefficient. A comparison of Eqs. 3-15 and 3-18 reveals that cen58933_ch03.qxd 9/10/2002 8:59 AM Page 133 r »i« WWW WWW R, = L l 1 k t A -A/WWW R 1= L 2 -AA/WW » r «2 conv ' 2 h 2 A 133 CHAPTER 3 FIGURE 3-9 The thermal resistance network for heat transfer through a two-layer plane wall subjected to convection on both sides. UA R„ (3-19) Therefore, for a unit area, the overall heat transfer coefficient is equal to the inverse of the total thermal resistance. Note that we do not need to know the surface temperatures of the wall in or- der to evaluate the rate of steady heat transfer through it. All we need to know is the convection heat transfer coefficients and the fluid temperatures on both sides of the wall. The surface temperature of the wall can be determined as described above using the thermal resistance concept, but by taking the surface at which the temperature is to be determined as one of the terminal surfaces. For example, once Q is evaluated, the surface temperature T x can be determined from Q l//i, A (3-20) Multilayer Plane Walls In practice we often encounter plane walls that consist of several layers of dif- ferent materials. The thermal resistance concept can still be used to determine the rate of steady heat transfer through such composite walls. As you may have already guessed, this is done by simply noting that the conduction resis- tance of each wall is LlkA connected in series, and using the electrical analogy. That is, by dividing the temperature difference between two surfaces at known temperatures by the total thermal resistance between them. Consider a plane wall that consists of two layers (such as a brick wall with a layer of insulation). The rate of steady heat transfer through this two-layer composite wall can be expressed as (Fig. 3-9) Q (3-21) cen58933_ch03.qxd 9/10/2002 8:59 AM Page 134 134 HEAT TRANSFER To find 7,: G=^r L K conv,l To find T 2 : Q=I^zJl To find Ty Q «... FIGURE 3-10 The evaluation of the surface and interface temperatures when T„ { and 7^2 are given and Q is calculated. where i? tota i is the total thermal resistance, expressed as R„- R conv, l + R* "wall, 2 ' R* conv. 2 (3-22) The subscripts 1 and 2 in the R waU relations above indicate the first and the second layers, respectively. We could also obtain this result by following the approach used above for the single-layer case by noting that the rate of steady heat transfer Q through a multilayer medium is constant, and thus it must be the same through each layer. Note from the thermal resistance network that the resistances are in series, and thus the total thermal resistance is simply the arithmetic sum of the individual thermal resistances in the path of heat flow. This result for the two-layer case is analogous to the single-layer case, ex- cept that an additional resistance is added for the additional layer. This result can be extended to plane walls that consist of three or more layers by adding an additional resistance for each additional layer. Once Q is known, an unknown surface temperature Tj at any surface or in- terface j can be determined from Q v total, i—j total, ( — j (3-23) is the total thermal where T t is a known temperature at location i and R resistance between locations i and j. For example, when the fluid temperatures T^ and T m2 for the two-layer case shown in Fig. 3-9 are available and Q is calculated from Eq. 3-21, the interface temperature T 2 between the two walls can be determined from (Fig. 3-10) Q T-, r, -"conv, 1 "T "wall, 1 (3-24) 16°C Wall •2°C 3 m L = 0.3m FIGURE 3-1 1 Schematic for Example 3-1. The temperature drop across a layer is easily determined from Eq. 3-17 by multiplying Q by the thermal resistance of that layer. The thermal resistance concept is widely used in practice because it is intu- itively easy to understand and it has proven to be a powerful tool in the solu- tion of a wide range of heat transfer problems. But its use is limited to systems through which the rate of heat transfer Q remains constant; that is, to systems involving steady heat transfer with no heat generation (such as resistance heating or chemical reactions) within the medium. EXAMPLE 3-1 Heat Loss through a Wall Consider a 3-m-high, 5-m-wide, and 0.3-m-thick wall whose thermal con- ductivity is k = 0.9 W/m • °C (Fig. 3-11). On a certain day, the temperatures of the inner and the outer surfaces of the wall are measured to be 16°C and 2 C C, respectively. Determine the rate of heat loss through the wall on that day. cen58933_ch03.qxd 9/10/2002 8:59 AM Page 135 SOLUTION The two surfaces of a wall are maintained at specified tempera- tures. The rate of heat loss through the wall is to be determined. Assumptions 1 Heat transfer through the wall is steady since the surface temperatures remain constant at the specified values. 2 Heat transfer is one- dimensional since any significant temperature gradients will exist in the direc- tion from the indoors to the outdoors. 3 Thermal conductivity is constant. Properties The thermal conductivity is given to be k = 0.9 W/m ■ C C. Analysis Noting that the heat transfer through the wall is by conduction and the area of the wall is/! = 3mX5m=15 m 2 , the steady rate of heat transfer through the wall can be determined from Eq. 3-3 to be Q=kA- T 2 , (16 — = (0.9 W/m • °C)(15 m 2 ) 2)°C 0.3 m 630 W We could also determine the steady rate of heat transfer through the wall by making use of the thermal resistance concept from Q Ar„ where ft„ 0.3 m lwa " kA (0.9 W/m • °C)(15m 2 ) Substituting, we get Q 0.02222°C/W (16 - 2)°C 0.02222°C/W 630 W Discussion This is the same result obtained earlier. Note that heat conduction through a plane wall with specified surface temperatures can be determined directly and easily without utilizing the thermal resistance concept. However, the thermal resistance concept serves as a valuable tool in more complex heat transfer problems, as you will see in the following examples. EXAMPLE 3-2 Heat Loss through a Single-Pane Window Consider a 0.8-m-high and 1.5-m-wide glass window with a thickness of 8 mm and a thermal conductivity of k = 0.78 W/m • °C. Determine the steady rate of heat transfer through this glass window and the temperature of its inner surface for a day during which the room is maintained at 20 C C while the temperature of the outdoors is — 10°C. Take the heat transfer coefficients on the inner and outer surfaces of the window to be h x = 10 W/m 2 • °C and h 2 = 40 W/m 2 • °C, which includes the effects of radiation. SOLUTION Heat loss through a window glass is considered. The rate of heat transfer through the window and the inner surface temperature are to be determined. 135 CHAPTER 3 cen58933_ch03.qxd 9/10/2002 8:59 AM Page 136 136 HEAT TRANSFER 20°C- h, = 10 W/m 2 -°C glass ♦-WWW-h — WWW-^^WWW— ♦ Glass -10°C h 2 = 40 W/m 2 -°C L = 8 mm FIGURE 3-1 2 Schematic for Example 3-2. Assumptions 1 Heat transfer through the window is steady since the surface temperatures remain constant at the specified values. 2 Heat transfer through the wall is one-dimensional since any significant temperature gradients will ex- ist in the direction from the indoors to the outdoors. 3 Thermal conductivity is constant. Properties The thermal conductivity is given to be k = 0.78 W/m • °C. Analysis This problem involves conduction through the glass window and con- vection at its surfaces, and can best be handled by making use of the thermal resistance concept and drawing the thermal resistance network, as shown in Fig. 3-12. Noting that the area of the window is A = 0.8 m X 1.5 m = 1.2 m 2 , the individual resistances are evaluated from their definitions to be 1 1 R h { A (10W/m 2 • °C)(1.2m 2 ) 0.008 m 0.08333°C/W slass kA (0.78 W/m • °C)( 1.2 m 2 ) 1 1 Rn — R n • 2 h 2 A (40 W/m 2 • °C)(1.2m 2 ) 0.00855°C/W 0.02083°C/W Noting that all three resistances are in series, the total resistance is glass 0.1127°C/W R„ 0.08333 + 0.00855 + 0.02083 Then the steady rate of heat transfer through the window becomes T^-T-^ [20-(-10)]°C Q = -^ " = L TTT77^7^7- = 266 W Knowing the rate of heat transfer, the inner surface temperature of the window glass can be determined from Q T m , - T, *conv, 1 -> Tj - r„] G^conv, 1 = 20°C - (266 W)(0.08333°C/W) = -2.2°C Discussion Note that the inner surface temperature of the window glass will be -2.2°C even though the temperature of the air in the room is maintained at 20°C. Such low surface temperatures are highly undesirable since they cause the formation of fog or even frost on the inner surfaces of the glass when the humidity in the room is high. EXAMPLE 3-3 Heat Loss through Double-Pane Windows Consider a 0.8-m-high and 1.5-m-wide double-pane window consisting of two 4-mm-thick layers of glass (k = 0.78 W/m • °C) separated by a 10-mm-wide stagnant air space (k = 0.026 W/m • °C). Determine the steady rate of heat cen58933_ch03.qxd 9/10/2002 8:59 AM Page 137 transfer through this double-pane window and the temperature of its inner sur- face for a day during which the room is maintained at 20°C while the tempera- ture of the outdoors is - 10°C. Take the convection heat transfer coefficients on the inner and outer surfaces of the window to be /?, = 10 W/m 2 • °C and h z = 40 W/m 2 • °C, which includes the effects of radiation. SOLUTION A double-pane window is considered. The rate of heat transfer through the window and the inner surface temperature are to be determined. Analysis This example problem is identical to the previous one except that the single 8-mm-thick window glass is replaced by two 4-mm-thick glasses that enclose a 10-mm-wide stagnant air space. Therefore, the thermal resistance network of this problem will involve two additional conduction resistances cor- responding to the two additional layers, as shown in Fig. 3-13. Noting that the area of the window is again A = 0.8 m X 1.5 m = 1.2 m 2 , the individual re- sistances are evaluated from their definitions to be 1 1 R ' ' Rconv • 1 h y A (10 W/m 2 • °C)(1.2m 2 ) R t — R 3 — R^ 0.004 m M (0.78 W/m ■ °C)(1.2m 2 ) 0.01 m 0.08333°C/W 0.00427°C/W Rn — Rr. k 2 A (0.026 W/m ■ °C)(1 .2 m 2 ) 1 1 '■ h 2 A (40 W/m 2 • °C)(1.2m 2 ) 0.3205°C/W 0.02083°C/W Noting that all three resistances are in series, the total resistance is R. R,„- + R. Rr, glass, 1 air glass, 2 conv, 2 = 0.08333 + 0.00427 + 0.3205 + 0.00427 + 0.02083 = 0.4332°C/W Then the steady rate of heat transfer through the window becomes Q R„ r„ 2 [2o-(-io)]°c 0.4332°C/W 69.2 W which is about one-fourth of the result obtained in the previous example. This explains the popularity of the double- and even triple-pane windows in cold climates. The drastic reduction in the heat transfer rate in this case is due to the large thermal resistance of the air layer between the glasses. The inner surface temperature of the window in this case will be QRc 20°C - (69.2 W)(0.08333°C/W) = 14.2°C which is considerably higher than the -2.2°C obtained in the previous ex- ample. Therefore, a double-pane window will rarely get fogged. A double-pane window will also reduce the heat gain in summer, and thus reduce the air- conditioning costs. 137 CHAPTER 3 Glass Glass 20°C • '\AAAAA/- 10°C AAAAAV • 3 o ^-2 FIGURE 3-13 Schematic for Example 3-3. cen58933_ch03.qxd 9/10/2002 8:59 AM Page 13E 138 HEAT TRANSFER FIGURE 3-14 Temperature distribution and heat flow lines along two solid plates pressed against each other for the case of perfect and imperfect contact. (a) Ideal (perfect) thermal contact (b) Actual (imperfect) thermal contact Applied load Loading shaft Alignment collar Top plate Steel ball Pencil heaters Heaters block Upper test specimen — Lower test specimen — Lower heat flux meter -SC Thermocouples Interface y= -t Cold fluid ZP* Cold plate Load cell - Steel ball - Bottom plate — H x I Bell jar * base plate FIGURE 3-15 A typical experimental setup for the determination of thermal contact resistance (from Song et al., Ref. 11). 3-2 ■ THERMAL CONTACT RESISTANCE In the analysis of heat conduction through multilayer solids, we assumed "perfect contact" at the interface of two layers, and thus no temperature drop at the interface. This would be the case when the surfaces are perfectly smooth and they produce a perfect contact at each point. In reality, however, even flat surfaces that appear smooth to the eye turn out to be rather rough when ex- amined under a microscope, as shown in Fig. 3-14, with numerous peaks and valleys. That is, a surface is microscopically rough no matter how smooth it appears to be. When two such surfaces are pressed against each other, the peaks will form good material contact but the valleys will form voids filled with air. As a re- sult, an interface will contain numerous air gaps of varying sizes that act as insulation because of the low thermal conductivity of air. Thus, an interface offers some resistance to heat transfer, and this resistance per unit interface area is called the thermal contact resistance, R c . The value of R c is deter- mined experimentally using a setup like the one shown in Fig. 3-15, and as expected, there is considerable scatter of data because of the difficulty in char- acterizing the surfaces. Consider heat transfer through two metal rods of cross-sectional area A that are pressed against each other. Heat transfer through the interface of these two rods is the sum of the heat transfers through the solid contact spots and the gaps in the noncontact areas and can be expressed as q =e t + e E (3-25) It can also be expressed in an analogous manner to Newton's law of cooling as Q =ft f AAr interfacc (3-26) cen58933_ch03.qxd 9/10/2002 8:59 AM Page 139 where A is the apparent interface area (which is the same as the cross-sectional area of the rods) and AT interface is the effective temperature difference at the interface. The quantity h c , which corresponds to the convection heat transfer coefficient, is called the thermal contact conductance and is expressed as QIA K = ^= (W/m 2 • °C) (3-27) ^-* interface It is related to thermal contact resistance by h c QIA That is, thermal contact resistance is the inverse of thermal contact conduc- tance. Usually, thermal contact conductance is reported in the literature, but the concept of thermal contact resistance serves as a better vehicle for ex- plaining the effect of interface on heat transfer. Note that R c represents ther- mal contact resistance per unit area. The thermal resistance for the entire interface is obtained by dividing R c by the apparent interface area A. The thermal contact resistance can be determined from Eq. 3-28 by measuring the temperature drop at the interface and dividing it by the heat flux under steady conditions. The value of thermal contact resistance depends on the surface roughness and the material properties as well as the tem- perature and pressure at the interface and the type of fluid trapped at the interface. The situation becomes more complex when plates are fastened by bolts, screws, or rivets since the interface pressure in this case is nonuniform. The thermal contact resistance in that case also depends on the plate thick- ness, the bolt radius, and the size of the contact zone. Thermal contact resistance is observed to decrease with decreasing surface roughness and increasing interface pressure, as expected. Most experimentally deter- mined values of the thermal contact resistance fall between 0.000005 and 0.0005 m 2 • °C/W (the corresponding range of thermal contact conductance is 2000 to 200,000 W/m 2 • °C). When we analyze heat transfer in a medium consisting of two or more lay- ers, the first thing we need to know is whether the thermal contact resistance is significant or not. We can answer this question by comparing the magni- tudes of the thermal resistances of the layers with typical values of thermal contact resistance. For example, the thermal resistance of a 1-cm-thick layer of an insulating material per unit surface area is r, £j U.U1 111 „ __ 2 Of~< f\\J K c , insulation ~ J ~ 0.04 W/m ■ °C ~ ^ ' whereas for a 1-cm-thick layer of copper, it is ^•»«- = t = 386W/m m °C = °- 000026 m ' ■ ° C/W Comparing the values above with typical values of thermal contact resistance, we conclude that thermal contact resistance is significant and can even domi- nate the heat transfer for good heat conductors such as metals, but can be 139 CHAPTER 3 cen58933_ch03.qxd 9/10/2002 8:59 AM Page 14C 140 HEAT TRANSFER TABLE 3-1 Thermal contact conductance for aluminum plates with different fluids at the interface for a surface roughness of 10 |xm and interface pressure of 1 atm (from Fried, Ref. 5) Contact Fluid at the Conductance, h c , Interface W/m 2 • °C Air 3640 Helium 9520 Hydrogen 13,900 Silicone oil 19,000 Glycerin 37,700 LU- LU" 10-- Contact pressure (psi) 10 2 10 3 Coated with tin/nickel alloy Bronze Coated with nickel alloy Coated aluminum alloy Jji-^T 10 4 10-' io- 10 J 10 J 10' Contact pressure (kN/irr) ■ Uncoated ■ Coated FIGURE 3-16 Effect of metallic coatings on thermal contact conductance (from Peterson, Ref. 10). disregarded for poor heat conductors such as insulations. This is not surpris- ing since insulating materials consist mostly of air space just like the inter- face itself. The thermal contact resistance can be minimized by applying a thermally conducting liquid called a thermal grease such as silicon oil on the surfaces before they are pressed against each other. This is commonly done when at- taching electronic components such as power transistors to heat sinks. The thermal contact resistance can also be reduced by replacing the air at the in- terface by a better conducting gas such as helium or hydrogen, as shown in Table 3-1. Another way to minimize the contact resistance is to insert a soft metallic foil such as tin, silver, copper, nickel, or aluminum between the two surfaces. Experimental studies show that the thermal contact resistance can be reduced by a factor of up to 7 by a metallic foil at the interface. For maximum effec- tiveness, the foils must be very thin. The effect of metallic coatings on thermal contact conductance is shown in Fig. 3-16 for various metal surfaces. There is considerable uncertainty in the contact conductance data reported in the literature, and care should be exercised when using them. In Table 3-2 some experimental results are given for the contact conductance between sim- ilar and dissimilar metal surfaces for use in preliminary design calculations. Note that the thermal contact conductance is highest (and thus the contact re- sistance is lowest) for soft metals with smooth surfaces at high pressure. EXAMPLE 3-4 Equivalent Thickness for Contact Resistance The thermal contact conductance at the interface of two 1-cm-thick aluminum plates is measured to be 11,000 W/m 2 ■ °C. Determine the thickness of the alu- minum plate whose thermal resistance is equal to the thermal resistance of the interface between the plates (Fig. 3-17). SOLUTION The thickness of the aluminum plate whose thermal resistance is equal to the thermal contact resistance is to be determined. Properties The thermal conductivity of aluminum at room temperature is k = 237 W/m • °C (Table A-3). Analysis Noting that thermal contact resistance is the inverse of thermal con- tact conductance, the thermal contact resistance is R, 1 1 h c 1 1,000 W/nr 0.909 X 10- 4 m 2 -°C/W For a unit surface area, the thermal resistance of a flat plate is defined as R L k where L is the thickness of the plate and k is the thermal conductivity. Setting R = R cl the equivalent thickness is determined from the relation above to be kR r (237 W/m • °C)(0.909 X 10~ 4 nr • °C/W) = 0.0215 m = 2.15 cm cen58 93 3_ch03.qxd 9/10/2002 8:59 AM Page 141 141 CHAPTER 3 TABLE 3-2 Thermal contact conductance of some metal surfaces in air (from various sources) Surface Rough- Tempera- Material Condition ness, |xm ture, °C Pressure, MPa W/m 2 °C Identical Metal Pairs 416 Stainless steel Ground 2.54 90-200 0.3-2.5 3800 304 Stainless steel Ground 1.14 20 4-7 1900 Aluminum Ground 2.54 150 1.2-2.5 11,400 Copper Ground 1.27 20 1.2-20 143,000 Copper Milled 3.81 20 1-5 55,500 Copper (vacuum) Milled 0.25 30 0.7-7 11,400 Dissimilar Metal Pairs Stainless steel- 10 2900 Aluminum 20-30 20 20 3600 Stainless steel- Aluminum 1.0-2.0 20 10 20 16,400 20,800 Steel Ct-30- Aluminum Ground 1.4-2.0 20 10 15-35 50,000 59,000 Steel Ct-30- Aluminum Milled 4.5-7.2 20 10 30 4800 8300 Aluminum-Copper Ground 1.3-1.4 20 5 15 42,000 56,000 Aluminum-Copper Milled 4.4-4.5 20 10 20-35 12,000 22,000 *Divide the given values by 5.678 to convert to Btu/h • ft 2 • °F. Discussion Note that the interface between the two plates offers as much re- sistance to heat transfer as a 2.3-cm-thick aluminum plate. It is interesting that the thermal contact resistance in this case is greater than the sum of the thermal resistances of both plates. EXAMPLE 3-5 Contact Resistance of Transistors Four identical power transistors with aluminum casing are attached on one side of a 1-cm-thick 20-cm X 20-cm square copper plate (k = 386 W/m • °C) by screws that exert an average pressure of 6 MPa (Fig. 3-18). The base area of each transistor is 8 cm 2 , and each transistor is placed at the center of a 10-cm X 10-cm quarter section of the plate. The interface roughness is estimated to be about 1.5 |xm. All transistors are covered by a thick Plexiglas layer, which is a poor conductor of heat, and thus all the heat generated at the junction of the transistor must be dissipated to the ambient at 20°C through the back surface of the copper plate. The combined convection/radiation heat transfer coefficient at the back surface can be taken to be 25 W/m 2 • °C. If the case temperature of Plate 1 1 cm Plate 2 1 cm Interface Plate 1 1 cm Equivalent ] Plate aluminum i 2 layer 2.15 cm i 1 cm FIGURE 3-17 Schematic for Example 3-4. cen58933_ch03.qxd 9/10/2002 8:59 AM Page 142 142 HEAT TRANSFER 20°C Copper plate FIGURE 3- Schematic 70°C 18 for Example Plexiglas cover 3-5. the transistor is not to exceed 70°C, determine the maximum power each transistor can dissipate safely, and the temperature jump at the case-plate interface. SOLUTION Four identical power transistors are attached on a copper plate. For a maximum case temperature of 70°C, the maximum power dissipation and the temperature jump at the interface are to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat transfer can be ap- proximated as being one-dimensional, although it is recognized that heat con- duction in some parts of the plate will be two-dimensional since the plate area is much larger than the base area of the transistor. But the large thermal con- ductivity of copper will minimize this effect. 3 All the heat generated at the junction is dissipated through the back surface of the plate since the transistors are covered by a thick Plexiglas layer. 4 Thermal conductivities are constant. Properties The thermal conductivity of copper is given to be k = 386 W/m ■ °C. The contact conductance is obtained from Table 3-2 to be h c = 42,000 W/m 2 • °C, which corresponds to copper-aluminum interface for the case of 1.3-1.4 |xm roughness and 5 MPa pressure, which is sufficiently close to what we have. Analysis The contact area between the case and the plate is given to be 8 cm 2 , and the plate area for each transistor is 100 cm 2 . The thermal resistance net- work of this problem consists of three resistances in series (interface, plate, and convection), which are determined to be R 1 I aterface ^ (42,000 W/m 2 • °C)(8 X 10- 4 m 2 ) L 0.01 m 0.030°C/W plate kA (386 W/m ■ °C)(0.01 m 2 ) 1 1 h B A (25 W/m 2 • °C)(0.01 m 2 ) 0.0026°C/W 4.0°C/W The total thermal resistance is then R„ D _(_ p _i_ p "•interface ' Opiate ' "~a\ 0.030 + 0.0026 + 4.0 = 4.0326°C/W Note that the thermal resistance of a copper plate is very small and can be ignored altogether. Then the rate of heat transfer is determined to be Q AT (70 - 20)°C R,. 4.0326°C/W 12.4 W Therefore, the power transistor should not be operated at power levels greater than 12.4 W if the case temperature is not to exceed 70°C. The temperature jump at the interface is determined from AT QRm (12.4 W)(0.030°C/W) = 0.37°C which is not very large. Therefore, even if we eliminate the thermal contact re- sistance at the interface completely, we will lower the operating temperature of the transistor in this case by less than 0.4°C. cen58933_ch03.qxd 9/10/2002 8:59 AM Page 143 3-3 - GENERALIZED THERMAL RESISTANCE NETWORKS The thermal resistance concept or the electrical analogy can also be used to solve steady heat transfer problems that involve parallel layers or combined series-parallel arrangements. Although such problems are often two- or even three-dimensional, approximate solutions can be obtained by assuming one- dimensional heat transfer and using the thermal resistance network. Consider the composite wall shown in Fig. 3-19, which consists of two par- allel layers. The thermal resistance network, which consists of two parallel re- sistances, can be represented as shown in the figure. Noting that the total heat transfer is the sum of the heat transfers through each layer, we have Q=Q l + Qi Utilizing electrical analogy, we get R 7 (7, - T 2 ) -1 + 1 R, R. where 1 ^total Q R, R, T 2 "> K RiR 2 r7Tr~, (3-29) (3-30) (3-31) since the resistances are in parallel. Now consider the combined series-parallel arrangement shown in Fig. 3-20. The total rate of heat transfer through this composite system can again be expressed as Q (3-32) 143 CHAPTER 3 Insulation -V ,4, CD *, © k 2 « L ► — www R 2 Q = Q\+Q 2 FIGURE 3-1 9 Thermal resistance network for two parallel layers. Insulation <D *i fc. L, = L n h, r„ where and ^12 + ^3 + ^conv R 2 L 2 R t R 2 /?! + R 2 L 3 k 3 A 3 ' R 3 + R D 1 (3-33) (3-34) Once the individual thermal resistances are evaluated, the total resistance and the total rate of heat transfer can easily be determined from the relations above. The result obtained will be somewhat approximate, since the surfaces of the third layer will probably not be isothermal, and heat transfer between the first two layers is likely to occur. Two assumptions commonly used in solving complex multidimensional heat transfer problems by treating them as one-dimensional (say, in the -MAMAA- Qi- ■AWI/VV^ Q w — *3 *c -vww vwvw — ♦ conv FIGURE 3-20 Thermal resistance network for combined series-parallel arrangement. cen58933_ch03.qxd 9/10/2002 8:59 AM Page 144 144 HEAT TRANSFER Foam Plaster ■ Brick 1.5 cm 22 cm 1.5 cm FIGURE 3-21 Schematic for Example 3-6. x-direction) using the thermal resistance network are (1) any plane wall nor- mal to the x-axis is isothermal (i.e., to assume the temperature to vary in the x-direction only) and (2) any plane parallel to the x-axis is adiabatic (i.e., to assume heat transfer to occur in the x-direction only). These two assumptions result in different resistance networks, and thus different (but usually close) values for the total thermal resistance and thus heat transfer. The actual result lies between these two values. In geometries in which heat transfer occurs pre- dominantly in one direction, either approach gives satisfactory results. EXAMPLE 3-6 Heat Loss through a Composite Wall A 3-m-high and 5-m-wide wall consists of long 16-cm X 22-cm cross section horizontal bricks (k = 0.72 W/m • °C) separated by 3-cm-thick plaster layers (k = 0.22 W/m • °C). There are also 2-cm-thick plaster layers on each side of the brick and a 3-cm-thick rigid foam (k = 0.026 W/m ■ °C) on the inner side of the wall, as shown in Fig. 3-21. The indoor and the outdoor temperatures are 20°C and -10°C, and the convection heat transfer coefficients on the inner and the outer sides are h 1 = 10 W/m 2 • °C and h 2 = 25 W/m 2 • °C, respectively. Assuming one-dimensional heat transfer and disregarding radiation, determine the rate of heat transfer through the wall. SOLUTION The composition of a composite wall is given. The rate of heat transfer through the wall is to be determined. Assumptions 1 Heat transfer is steady since there is no indication of change with time. 2 Heat transfer can be approximated as being one-dimensional since it is predominantly in the x-direction. 3 Thermal conductivities are constant. 4 Heat transfer by radiation is negligible. Properties The thermal conductivities are given to be k = 0.72 W/m • °C for bricks, k = 0.22 W/m • °C for plaster layers, and k = 0.026 W/m • °C for the rigid foam. Analysis There is a pattern in the construction of this wall that repeats itself every 25-cm distance in the vertical direction. There is no variation in the hori- zontal direction. Therefore, we consider a 1-m-deep and 0.25-m-high portion of the wall, since it is representative of the entire wall. Assuming any cross section of the wall normal to the x-direction to be isothermal, the thermal resistance network for the representative section of the wall becomes as shown in Fig. 3-21. The individual resistances are eval- uated as: 1 1 Ri ' Rconv ' 1 h y A (10 W/m 2 • °C)(0.25 X 1 m 2 ) = L__ 0.03 m 1 - foam - M - (Q Q26 w/m . o C)( q 25 xlm 2) L 0.02 m : 0.4°C/W 4.6°C/W Rl Rb ****«"» kA (0.22 W/m • °C)(0.25 X 1 m 2 ) = 0.36°C/W L 0.16m K } - K 5 - K 5 plaster, center ^ (0.22 W/m ' °C)(0.015 X 1 III 2 ) 48.48°C/W cen58933_ch03.qxd 9/10/2002 8:59 AM Page 145 ^4 — ^h, 0.16m kA (0.72 W/m ■ °C)(0.22 X 1 m 2 1 1 h 2 A (25 W/m 2 • °C)(0.25 X 1 m 2 ) 1.01°C/W 0.16°C/W The three resistances R 3 , ff 4 , and R 5 in the middle are parallel, and their equiv- alent resistance is determined from R mi<l R 3 R A R 5 48.48 1.01 48.48 1.03 W/°C which gives tfmid = 0.97°C/W Now all the resistances are in series, and the total resistance is ^total = Ri + Rl + R 2 + R m U + ^6 + Ro = 0.4 + 4.6 + 0.36 + 0.97 + 0.36 + 0.16 = 6.85°C/W Then the steady rate of heat transfer through the wall becomes Q [20 - (-10)]°C 6.85°C/W 4.38 W (per 0.25 m 2 surface area) or 4.38/0.25 = 17.5 W per m 2 area. The total area of the wall is A = 3 m X 5 m = 15 m 2 . Then the rate of heat transfer through the entire wall becomes fit, (17.5 W/m 2 )(15 m 2 ) = 263 W Of course, this result is approximate, since we assumed the temperature within the wall to vary in one direction only and ignored any temperature change (and thus heat transfer) in the other two directions. Discussion In the above solution, we assumed the temperature at any cross section of the wall normal to the x-direction to be isothermal. We could also solve this problem by going to the other extreme and assuming the surfaces par- allel to the x-direction to be adiabatic. The thermal resistance network in this case will be as shown in Fig. 3-22. By following the approach outlined above, the total thermal resistance in this case is determined to be ff tota | = 6.97°C/W, which is very close to the value 6.85°C/W obtained before. Thus either ap- proach would give roughly the same result in this case. This example demon- strates that either approach can be used in practice to obtain satisfactory results. 145 CHAPTER 3 T V 1, Adiabatic lines R j ■>!♦— VW- -<wv — vw — vw — vw vw— ♦ FIGURE 3-22 Alternative thermal resistance network for Example 3-6 for the case of surfaces parallel to the primary direction of heat transfer being adiabatic. cen58933_ch03.qxd 9/10/2002 8:59 AM Page 146 146 HEAT TRANSFER FIGURE 3-23 Heat is lost from a hot water pipe to the air outside in the radial direction, and thus heat transfer from a long pipe is one-dimensional. FIGURE 3-24 A long cylindrical pipe (or spherical shell) with specified inner and outer surface temperatures T l and T 2 . 3^ - HEAT CONDUCTION IN CYLINDERS AND SPHERES Consider steady heat conduction through a hot water pipe. Heat is continu- ously lost to the outdoors through the wall of the pipe, and we intuitively feel that heat transfer through the pipe is in the normal direction to the pipe surface and no significant heat transfer takes place in the pipe in other directions (Fig. 3-23). The wall of the pipe, whose thickness is rather small, separates two fluids at different temperatures, and thus the temperature gradient in the radial direction will be relatively large. Further, if the fluid temperatures in- side and outside the pipe remain constant, then heat transfer through the pipe is steady. Thus heat transfer through the pipe can be modeled as steady and one-dimensional. The temperature of the pipe in this case will depend on one direction only (the radial r-direction) and can be expressed as T = T(r). The temperature is independent of the azimuthal angle or the axial distance. This situation is approximated in practice in long cylindrical pipes and spherical containers. In steady operation, there is no change in the temperature of the pipe with time at any point. Therefore, the rate of heat transfer into the pipe must be equal to the rate of heat transfer out of it. In other words, heat transfer through the pipe must be constant, Q cond cyl = constant. Consider a long cylindrical layer (such as a circular pipe) of inner radius r x , outer radius r 2 , length L, and average thermal conductivity k (Fig. 3-24). The two surfaces of the cylindrical layer are maintained at constant temperatures Ti and T 2 . There is no heat generation in the layer and the thermal conductiv- ity is constant. For one-dimensional heat conduction through the cylindrical layer, we have T(r). Then Fourier's law of heat conduction for heat transfer through the cylindrical layer can be expressed as e cond, cyl -kA dT dr (W) (3-35) where A = 2ittL is the heat transfer area at location r. Note that A depends on r, and thus it varies in the direction of heat transfer. Separating the variables in the above equation and integrating from r = r u where T(r t ) = 7\, to r = r 2 , where T{r 2 ) = T 2 , gives r 2 Q cond. L ey I dr : kdT (3-36) Substituting A = 2tjtL and performing the integrations give T — T Q coai ^ = 2itLk^j^ (W) (3-37) since Q cond, cyl constant. This equation can be rearranged as xl, con r. d, cyl R. cy I (W) (3-38) cen58933_ch03.qxd 9/10/2002 8:59 AM Page 147 where 147 CHAPTER 3 v cyl ln(r 2 lr x ) ln(Outer radius/Inner radius) 2irLk 2tt X (Length) X (Thermal conductivity) (3-39) is the thermal resistance of the cylindrical layer against heat conduction, or simply the conduction resistance of the cylinder layer. We can repeat the analysis above for a spherical layer by taking A = 4irr 2 and performing the integrations in Eq. 3-36. The result can be expressed as Q cond, sph (3-40) v sp h where Outer radius — Inner radius sph A-ur^k 4Tr(Outer radius)(Inner radius)(Thermal conductivity) (3-41) is the thermal resistance of the spherical layer against heat conduction, or sim- ply the conduction resistance of the spherical layer. Now consider steady one-dimensional heat flow through a cylindrical or spherical layer that is exposed to convection on both sides to fluids at temper- atures r„[ and T m2 with heat transfer coefficients h { and h 2 , respectively, as shown in Fig. 3-25. The thermal resistance network in this case consists of one conduction and two convection resistances in series, just like the one for the plane wall, and the rate of heat transfer under steady conditions can be ex- pressed as Q (3-42) : ^conv,l +S cyI +R conv,: FIGURE 3-25 The thermal resistance network for a cylindrical (or spherical) shell subjected to convection from both the inner and the outer sides. 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 cen58933_ch03.qxd 9/10/2002 8:59 AM Page 154 154 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. cen58933_ch03.qxd 9/10/2002 8:59 AM Page 155 155 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. cen58933_ch03.qxd 9/10/2002 8:59 AM Page 156 156 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 cen58933_ch03.qxd 9/10/2002 8:59 AM Page 157 157 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) cen58933_ch03.qxd 9/10/2002 8:59 AM Page 16C 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 cen58933_ch03.qxd 9/10/2002 8:59 AM Page 161 161 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. cen58933_ch03.qxd 9/10/2002 8:59 AM Page 162 162 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 cen58933_ch03.qxd 9/10/2002 8:59 AM Page 163 163 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) cen58933_ch03.qxd 9/10/2002 8:59 AM Page 164 164 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. cen58933_ch03.qxd 9/10/2002 8:59 AM Page 165 165 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 cen58933_ch03.qxd 9/10/2002 8:59 AM Page 166 166 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). cen58933_ch03.qxd 9/10/2002 8:59 AM Page 167 167 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. cen58933_ch03.qxd 9/10/2002 8:59 AM Page 16E 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. cen58933_ch03.qxd 9/10/2002 8:59 AM Page 17C 170 HEAT TRANSFER An important class of heat transfer problems for which simple solutions are obtained encompasses those involving two surfaces maintained at constant temperatures T [ and T 2 . The steady rate of heat transfer between these two sur- faces is expressed as Q = Sk(T t - T 2 ) (3-79) where S is the conduction shape factor, which has the dimension of length, and k is the thermal conductivity of the medium between the surfaces. The conduction shape factor depends on the geometry of the system only. Conduction shape factors have been determined for a number of configura- tions encountered in practice and are given in Table 3-5 for some common cases. More comprehensive tables are available in the literature. Once the value of the shape factor is known for a specific geometry, the total steady heat transfer rate can be determined from the equation above using the speci- fied two constant temperatures of the two surfaces and the thermal conductiv- ity of the medium between them. Note that conduction shape factors are applicable only when heat transfer between the two surfaces is by conduction. Therefore, they cannot be used when the medium between the surfaces is a liquid or gas, which involves natural or forced convection currents. A comparison of Equations 3-4 and 3-79 reveals that the conduction shape factor S is related to the thermal resistance R by R = 1/kS or S = 1/kR. Thus, these two quantities are the inverse of each other when the thermal conduc- tivity of the medium is unity. The use of the conduction shape factors is illus- trated with examples 3-13 and 3-14. s-T 2 = 10°C Z = 0.5m I_ - ^ | D = 10 cm ) [' ■' ■ ■■' -L = 30 m - — -4 FIGURE 3-49 Schematic for Example 3-13. EXAMPLE 3-13 Heat Loss from Buried Steam Pipes A 30-m-long, 10-cm-diameter hot water pipe of a district heating system is buried in the soil 50 cm below the ground surface, as shown in Figure 3-49. The outer surface temperature of the pipe is 80°C. Taking the surface tempera- ture of the earth to be 10°C and the thermal conductivity of the soil at that lo- cation to be 0.9 W/m • °C, determine the rate of heat loss from the pipe. SOLUTION The hot water pipe of a district heating system is buried in the soil. The rate of heat loss from the pipe is to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat transfer is two- dimensional (no change in the axial direction). 3 Thermal conductivity of the soil is constant. Properties The thermal conductivity of the soil is given to be k = 0.9 W/m • °C. Analysis The shape factor for this configuration is given in Table 3-5 to be S 2ttL ln(4z/D) since z > 1.5D, where z is the distance of the pipe from the ground surface, and D is the diameter of the pipe. Substituting, 2ir X (30 m) ln(4 X 0.5/0.1) 62.9 m cen58933_ch03.qxd 9/10/2002 8:59 AM Page 171 171 CHAPTER 3 TABLE 3-5 Conduction shape factors S for several configurations for use in Q = kS{Ti - T 2 ) to determine the steady rate of heat transfer through a medium of thermal conductivity k between the surfaces at temperatures 7"! and T 2 ( 1 ) Isothermal cylinder of length L buried in a semi-infinite medium (L»D and z>l.5D) rl 2%L " In (4z/D) G m > ) ■■ (2) Vertical isothermal cylinder of length L buried in a semi-infinite medium (L»D) 2%L "ln(4L/D) IX (3) Two parallel isothermal cylinders placed in an infinite medium (L»D l ,D 1 ,z) 2kL cosh Az 2 ~D] ~D\ (4) A row of equally spaced parallel isothermal cylinders buried in a semi-infinite medium (L»D, z andw>1.5D) j£± 2kL ln|^ sinh 2^Z (per cylinder) -j-^-W ; >|» '- w ■ ' »]■■ ; 'Wr»|- (5) Circular isothermal cylinder of length L in the midplane of an infinite wall (Z > 0.5£>) 2kL ln(8ztoD) (6) Circular isothermal cylinder of length L at the center of a square solid bar of the same length 2%L In (1.08 wID) (7) Eccentric circular isothermal cylinder of length L in a cylinder of the same length (L > £>,) j 2kL cosh D]+D\-4z 2 (8) Large plane wall -T, (continued) cen58933_ch03.qxd 9/10/2002 8:59 AM Page 172 172 HEAT TRANSFER TABLE 3-5 (CONCLUDED) (9) A long cylindrical layer 2kL In (D/D.) (10) A square flow passage (a) For alb > 1.4, „ 2%L T 2\/\ 1 J 0.93 In (0.948 alb) (b) For alb < 1.41, /A c 2%L L 0.785 In (alb) \^b^ -> a (1 1) A spherical layer 2%D X D 1 (12) Disk buried parallel to the surface in a semi-infinite medium (z » D) D^-D 1 -u x ^h_ S = 4D (5 = 2£)whenz = 0) & (13) The edge of two adjoining walls of equal thickness 5 = 0.54 w (14) Corner of three walls of equal thickness 5 = 0.15L (15) Isothermal sphere buried in a semi-infinite medium ^ 2kD 1 - 0.25D/Z (16) Isothermal sphere buried in a semi-infinite medium at T 2 whose surface is insulated 2kD Insulated 1 + 0.25D/Z cen58933_ch03.qxd 9/10/2002 8:59 AM Page 173 Then the steady rate of heat transfer from the pipe becomes Q = Sk(T, - T 2 ) = (62.9 m)(0.9 W/m • °C)(80 - 10)°C = 3963 W Discussion Note that this heat is conducted from the pipe surface to the sur- face of the earth through the soil and then transferred to the atmosphere by convection and radiation. 173 CHAPTER 3 EXAMPLE 3-14 Heat Transfer between Hot and Cold Water Pipes A 5-m-long section of hot and cold water pipes run parallel to each other in a thick concrete layer, as shown in Figure 3-50. The diameters of both pipes are 5 cm, and the distance between the centerline of the pipes is 30 cm. The sur- face temperatures of the hot and cold pipes are 70°C and 15°C, respectively. Taking the thermal conductivity of the concrete to be k = 0.75 W/m • °C, de- termine the rate of heat transfer between the pipes. SOLUTION Hot and cold water pipes run parallel to each other in a thick con- crete layer. The rate of heat transfer between the pipes is to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat transfer is two- dimensional (no change in the axial direction). 3 Thermal conductivity of the concrete is constant. Properties The thermal conductivity of concrete is given to be k = 0.75 W/m • °C. Analysis The shape factor for this configuration is given in Table 3-5 to be 2ttL 4z 2 -D]-D} cosh where z is the distance between the centerlines of the pipes and L is their length. Substituting, 2tt X (5 m) 4 X 0.3 2 - 0.05 2 - 0.05 2 cosh 2 X 0.05 X 0.05 6.34 m Then the steady rate of heat transfer between the pipes becomes Q = Sk(T, - T 2 ) = (6.34 m)(0.75 W/m ■ °C)(70 - 15°)C = 262 W Discussion We can reduce this heat loss by placing the hot and cold water pipes further away from each other. 15°C Z = 30 cm FIGURE 3-50 Schematic for Example 3-14. It is well known that insulation reduces heat transfer and saves energy and money. Decisions on the right amount of insulation are based on a heat trans- fer analysis, followed by an economic analysis to determine the "monetary value" of energy loss. This is illustrated with Example 3-15. cen58933_ch03.qxd 9/10/2002 8:59 AM Page 174 174 HEAT TRANSFER 75°F- Wall, R= 13 ■ 45°F 7/cc. ♦-^vwvw^ i — ww- < -^vwwv^-^ 7^ 9 FIGURE 3-51 Schematic for Example 3-15. EXAMPLE 3-15 Cost of Heat Loss through Walls in Winter Consider an electrically heated house whose walls are 9 ft high and have an R-\ia\ue of insulation of 13 (i.e., a thickness-to-thermal conductivity ratio of L/k = 13 h • ft 2 • °F/Btu). Two of the walls of the house are 40 ft long and the others are 30 ft long. The house is maintained at 75°F at all times, while the temperature of the outdoors varies. Determine the amount of heat lost through the walls of the house on a certain day during which the average tem- perature of the outdoors is 45°F. Also, determine the cost of this heat loss to the homeowner if the unit cost of electricity is $0.075/kWh. For combined convec- tion and radiation heat transfer coefficients, use the ASHRAE (American Soci- ety of Heating, Refrigeration, and Air Conditioning Engineers) recommended values of h, = 1.46 Btu/h • ft 2 • °F for the inner surface of the walls and h = 4.0 Btu/h ■ ft 2 • °F for the outer surface of the walls under 15 mph wind condi- tions in winter. SOLUTION An electrically heated house with R-13 insulation is considered. The amount of heat lost through the walls and its cost are to be determined. Assumptions 1 The indoor and outdoor air temperatures have remained at the given values for the entire day so that heat transfer through the walls is steady. 2 Heat transfer through the walls is one-dimensional since any significant temperature gradients in this case will exist in the direction from the indoors to the outdoors. 3 The radiation effects are accounted for in the heat transfer coefficients. Analysis This problem involves conduction through the wall and convection at its surfaces and can best be handled by making use of the thermal resistance concept and drawing the thermal resistance network, as shown in Fig. 3-51. The heat transfer area of the walls is A = Circumference X Height = (2 X 30 ft + 2 X 40 ft)(9 ft) = 1260 ft 2 Then the individual resistances are evaluated from their definitions to be 0.00054 h ■ °F/Btu R=R = J_ = 1 conv '' h,A (1.46 Btu/h ■ ft 2 ■ °F)( 1260 ft 2 ) R» L /{-value 13 h ■ ft 2 ■ °F/Btu kA A 1260 ft 2 1 0.01032 h-°F/Btu 0.00020 h • °F/Btu 1 h c A (4.0 Btu/h • ft 2 ■ °F)(1260 ft 2 ) Noting that all three resistances are in series, the total resistance is fltotai = R i + #waii + R = 0.00054 + 0.01032 + 0.00020 = 0.01106 h • °F/Btu Then the steady rate of heat transfer through the walls of the house becomes Q *Mntnl (75 - 45)°F 0.01 106 h • °F/Btu 2712 Btu/h Finally, the total amount of heat lost through the walls during a 24-h period and its cost to the home owner are Q = Q At = (2712 Btu/h)(24-h/day) = 65,099 Btu/day = 19.1 kWh/day cen58933_ch03.qxd 9/10/2002 8:59 AM Page 175 since 1 kWh = 3412 Btu, and Heating cost = (Energy lost)(Cost of energy) = (19.1 kWh/day)($0.075/kWh) = $1.43/day Discussion The heat losses through the walls of the house that day will cost the home owner $1.43 worth of electricity. 175 CHAPTER 3 TOPIC OF SPECIAL INTEREST Heat Transfer Through Walls and Roofs Under steady conditions, the rate of heat transfer through any section of a building wall or roof can be determined from Q = UA(Ti - T ) MTj ~ T ) R (3-80) where T t and T are the indoor and outdoor air temperatures, A is the heat transfer area, U is the overall heat transfer coefficient (the [/-factor), and R = l/U is the overall unit thermal resistance (the R-value). Walls and roofs of buildings consist of various layers of materials, and the structure and operating conditions of the walls and the roofs may differ significantly from one building to another. Therefore, it is not practical to list the ^-values (or [/-factors) of different kinds of walls or roofs under different conditions. Instead, the overall 7?-value is determined from the thermal resistances of the individual components using the thermal resistance net- work. The overall thermal resistance of a structure can be determined most accurately in a lab by actually assembling the unit and testing it as a whole, but this approach is usually very time consuming and expensive. The ana- lytical approach described here is fast and straightforward, and the results are usually in good agreement with the experimental values. The unit thermal resistance of a plane layer of thickness L and thermal conductivity k can be determined from R = Llk. The thermal conductivity and other properties of common building materials are given in the appen- dix. The unit thermal resistances of various components used in building structures are listed in Table 3-6 for convenience. Heat transfer through a wall or roof section is also affected by the con- vection and radiation heat transfer coefficients at the exposed surfaces. The effects of convection and radiation on the inner and outer surfaces of walls and roofs are usually combined into the combined convection and radiation heat transfer coefficients (also called surface conductances) h t and h a , respectively, whose values are given in Table 3-7 for ordinary surfaces (e = 0.9) and reflective surfaces (e = 0.2 or 0.05). Note that surfaces hav- ing a low emittance also have a low surface conductance due to the reduc- tion in radiation heat transfer. The values in the table are based on a surface TABLE 3-7 Combined convection and radiation heat transfer coefficients at window, wall, or roof surfaces (from ASHRAE Handbook of Fundamentals, Ref. 1, Chap. 22, Table 1). Direc- tion of Posi- Heat h, W/m 2 • c Surface Emittance C* e tion Flow 0.90 0.20 0.05 Still air (both indoors and outdoors) Horiz. UpT 9.26 5.17 4.32 Horiz. Down I 6.13 2.10 1.25 45° slope UpT 9.09 5.00 4.15 45° slope Down I 7.50 3.41 2.56 Vertical Horiz. — > 8.29 4.20 3.35 Moving air (any position, a ny direction) Winter condition (winds at 15 mpr or 24 km/h) 34.0 — — Summer condition (winds at 7.5 mph or 12 km/h) 22.7 — — *This section can be skipped without a loss of continuity. ♦Multiply by 0.176 to convert to Btu/h • ft 2 ■ °F. Surface resistance can be obtained from Ft = llh. cen58933_ch03.qxd 9/10/2002 8:59 AM Page 176 176 HEAT TRANSFER TABLE 3-6 Unit thermal resistance (the ff-value) of common components used in buildings R- value R- value Component -n 2 • °C/W ft 2 • h • °F/Btu Component -n 2 • °C/W ft 2 • h • °F/Btu Outside surface (winter) 0.030 0.17 Wood stud, nominal 2 in. X Outside surface (summer) 0.044 0.25 6 in. (5.5 in. or 140 mm wide) 0.98 5.56 Inside surface, still air 0.12 0.68 Clay tile, 100 mm (4 in.) 0.18 1.01 Plane air space, vertical, orc inary surfaces (e eff = 0.82): Acoustic tile 0.32 1.79 13 mm (1 in.) 0.16 0.90 Asphalt shingle roofing 0.077 0.44 20 mm (| in.) 0.17 0.94 Building paper 0.011 0.06 40 mm (1.5 in.) 0.16 0.90 Concrete block, 100 mm (4 in.): 90 mm (3.5 in.) 0.16 0.91 Lightweight 0.27 1.51 Insulation, 25 mm (1 in ) Heavyweight 0.13 0.71 Glass fiber 0.70 4.00 Plaster or gypsum board, Mineral fiber batt 0.66 3.73 13 mm (i in.) 0.079 0.45 Urethane rigid foam 0.98 5.56 Wood fiberboard, 13 mm (i in.) 0.23 1.31 Stucco, 25 mm (1 in.) 0.037 0.21 Plywood, 13 mm (| in.) 0.11 0.62 Face brick, 100 mm (4 in.) 0.075 0.43 Concrete, 200 mm (8 in.): Common brick, 100 mm (4 n.) 0.12 0.79 Lightweight 1.17 6.67 Steel siding 0.00 0.00 Heavyweight 0.12 0.67 Slag, 13 mm (1 in.) 0.067 0.38 Cement mortar, 13 mm (1/2 in.) 0.018 0.10 Wood, 25 mm (1 in.) 0.22 1.25 Wood bevel lapped siding, Wood stud, nominal 2 in . X 13 mm X 200 mm 4 in. (3.5 in. or 90 mm w de) 0.63 3.58 (1/2 in. X 8 in.) 0.14 0.81 temperature of 21°C (72°F) and a surface-air temperature difference of 5.5°C (10°F). Also, the equivalent surface temperature of the environment is assumed to be equal to the ambient air temperature. Despite the conve- nience it offers, this assumption is not quite accurate because of the addi- tional radiation heat loss from the surface to the clear sky. The effect of sky radiation can be accounted for approximately by taking the outside tem- perature to be the average of the outdoor air and sky temperatures. The inner surface heat transfer coefficient h t remains fairly constant throughout the year, but the value of h varies considerably because of its dependence on the orientation and wind speed, which can vary from less than 1 km/h in calm weather to over 40 km/h during storms. The com- monly used values of ft, and h for peak load calculations are h, = 8.29 W/m 2 • °C = 1 .46 Btu/h • ft 2 ■ °F r 34.0 W/m 2 ■ °C = 6.0 Btu/h ■ ft 2 ■ °F //„ 22.7 W/m 2 ■ °C = 4.0 Btu/h ■ ft 2 (winter and summer) (winter) (summer) which correspond to design wind conditions of 24 km/h (15 mph) for win- ter and 12 km/h (7.5 mph) for summer. The corresponding surface thermal resistances (^-values) are determined from Rj = l/hj and R = l/h a . The surface conductance values under still air conditions can be used for inte- rior surfaces as well as exterior surfaces in calm weather. cen58933_ch03.qxd 9/10/2002 8:59 AM Page 177 177 CHAPTER 3 Building components often involve trapped air spaces between various layers. Thermal resistances of such air spaces depend on the thickness of the layer, the temperature difference across the layer, the mean air temper- ature, the emissivity of each surface, the orientation of the air layer, and the direction of heat transfer. The emissivities of surfaces commonly encoun- tered in buildings are given in Table 3-8. The effective emissivity of a plane-parallel air space is given by Effective e l e 2 1 (3-81) where s { and e 2 are the emissivities of the surfaces of the air space. Table 3-8 also lists the effective emissivities of air spaces for the cases where (1) the emissivity of one surface of the air space is e while the emissivity of the other surface is 0.9 (a building material) and (2) the emissivity of both surfaces is e. Note that the effective emissivity of an air space between building materials is 0.82/0.03 = 27 times that of an air space between sur- faces covered with aluminum foil. For specified surface temperatures, ra- diation heat transfer through an air space is proportional to effective emissivity, and thus the rate of radiation heat transfer in the ordinary sur- face case is 27 times that of the reflective surface case. Table 3-9 lists the thermal resistances of 20-mm-, 40-mm-, and 90-mm- (0.75-in., 1.5-in., and 3.5-in.) thick air spaces under various conditions. The thermal resistance values in the table are applicable to air spaces of uniform thickness bounded by plane, smooth, parallel surfaces with no air leakage. Thermal resistances for other temperatures, emissivities, and air spaces can be obtained by interpolation and moderate extrapolation. Note that the presence of a low-emissivity surface reduces radiation heat transfer across an air space and thus significantly increases the thermal resistance. The thermal effectiveness of a low-emissivity surface will decline, however, if the condition of the surface changes as a result of some effects such as con- densation, surface oxidation, and dust accumulation. The 7?-value of a wall or roof structure that involves layers of uniform thickness is determined easily by simply adding up the unit thermal re- sistances of the layers that are in series. But when a structure involves components such as wood studs and metal connectors, then the ther- mal resistance network involves parallel connections and possible two- dimensional effects. The overall R-yalue in this case can be determined by assuming (1) parallel heat flow paths through areas of different construc- tion or (2) isothermal planes normal to the direction of heat transfer. The first approach usually overpredicts the overall thermal resistance, whereas the second approach usually underpredicts it. The parallel heat flow path approach is more suitable for wood frame walls and roofs, whereas the isothermal planes approach is more suitable for masonry or metal frame walls. The thermal contact resistance between different components of building structures ranges between 0.01 and 0.1 m 2 ■ °C/W, which is negligible in most cases. However, it may be significant for metal building components such as steel framing members. TABLE 3-8 Emissivities s of various surfaces and the effective emissivity of air spaces (from ASHRAE Handbook of Fundamentals, Ref. 1, Chap. 22, Table 3). Effective EmissK ity of ? Air Space 1 = s e 1 = e Surface E £ 2 = 0.9 £ 2 = E Aluminum foil, bright 0.05* 0.05 0.03 Aluminum sheet 0.12 0.12 0.06 Aluminum- coated paper, polished 0.20 0.20 0.11 Steel, galvan ized, bright 0.25 0.24 0.15 Aluminum paint 0.50 0.47 0.35 Building materials: Wood, paper, masonry, nonmetallic paints 0.90 0.82 0.82 Ordinary glass 0.84 0.77 0.72 *Surface emissivity of aluminum foil increases to 0.30 with barely visible condensation, and to 0.70 with clearly visible condensation. cen58933_ch03.qxd 9/10/2002 8:59 AM Page 17E TABLE 3-9 Unit thermal resistances (ff-values) of well-sealed plane air spaces (from ASHRAE Handbook of Fundamentals, Ref. 1, Chap. 22, Table 2) (a) SI units (in m 2 • °C/W) Position Direction of Air of Heat Space Flow Mean Temp., °C Temp. Diff., °C 20-mm Air Space 40-mm Air Space 90-mm Air Spa Effective Emissivity, e e ce Effective Emissivity, e ef Effective Emissivity, e ef t 0.03 0.05 0.5 0.82 0.03 0.05 0.5 0.82 0.03 0.05 0.5 0.82 32.2 5.6 0.41 0.39 0.18 0.13 0.45 0.42 0.19 0.14 0.50 0.47 0.20 0.14 10.0 16.7 0.30 0.29 0.17 0.14 0.33 0.32 0.18 0.14 0.27 0.35 0.19 0.15 Horizontal Up T 10.0 5.6 0.40 0.39 0.20 0.15 0.44 0.42 0.21 0.16 0.49 0.47 0.23 0.16 -17.8 11.1 0.32 0.32 0.20 0.16 0.35 0.34 0.22 0.17 0.40 0.38 0.23 0.18 32.2 5.6 0.52 0.49 0.20 0.14 0.51 0.48 0.20 0.14 0.56 0.52 0.21 0.14 10.0 16.7 0.35 0.34 0.19 0.14 0.38 0.36 0.20 0.15 0.40 0.38 0.20 0.15 45° slope UpT 10.0 5.6 0.51 0.48 0.23 0.17 0.51 0.48 0.23 0.17 0.55 0.52 0.24 0.17 -17.8 11.1 0.37 0.36 0.23 0.18 0.40 0.39 0.24 0.18 0.43 0.41 0.24 0.19 32.2 5.6 0.62 0.57 0.21 0.15 0.70 0.64 0.22 0.15 0.65 0.60 0.22 0.15 10.0 16.7 0.51 0.49 0.23 0.17 0.45 0.43 0.22 0.16 0.47 0.45 0.22 0.16 Vertical Horizontal -> 10.0 5.6 0.65 0.61 0.25 0.18 0.67 0.62 0.26 0.18 0.64 0.60 0.25 0.18 -17.8 11.1 0.55 0.53 0.28 0.21 0.49 0.47 0.26 0.20 0.51 0.49 0.27 0.20 32.2 5.6 0.62 0.58 0.21 0.15 0.89 0.80 0.24 0.16 0.85 0.76 0.24 0.16 10.0 16.7 0.60 0.57 0.24 0.17 0.63 0.59 0.25 0.18 0.62 0.58 0.25 0.18 45° slope Down i 10.0 5.6 0.67 0.63 0.26 0.18 0.90 0.82 0.28 0.19 0.83 0.77 0.28 0.19 -17.8 11.1 0.66 0.63 0.30 0.22 0.68 0.64 0.31 0.22 0.67 0.64 0.31 0.22 32.2 5.6 0.62 0.58 0.21 0.15 1.07 0.94 0.25 0.17 1.77 1.44 0.28 0.18 10.0 16.7 0.66 0.62 0.25 0.18 1.10 0.99 0.30 0.20 1.69 1.44 0.33 0.21 Horizontal Down i 10.0 5.6 0.68 0.63 0.26 0.18 1.16 1.04 0.30 0.20 1.96 1.63 0.34 0.22 V J -17.8 11.1 0.74 0.70 0.32 0.23 1.24 1.13 0.39 0.26 1.92 1.68 0.43 0.29 (b) English units (in h ■ ft 2 ■ °F/Btu) Position Direction of Air of Heat Space Flow Mean Temp., °F Temp. Diff., °F 0.75-in. Air Space 1.5-in. A ir Space 3.5-in. Air Space Effective Emissivity, e ef Effective Emissivity, e ef Effective Emissivity, e eff 0.03 0.05 0.5 0.82 0.03 0.05 0.5 0.82 0.03 0.05 0.5 0.82 90 10 2.34 2.22 1.04 0.75 2.55 2.41 1.08 0.77 2.84 2.66 1.13 0.80 50 30 1.71 1.66 0.99 0.77 1.87 1.81 1.04 0.80 2.09 2.01 1.10 0.84 Horizontal Up T 50 10 2.30 2.21 1.16 0.87 2.50 2.40 1.21 0.89 2.80 2.66 1.28 0.93 20 1.83 1.79 1.16 0.93 2.01 1.95 1.23 0.97 2.25 2.18 1.32 1.03 90 10 2.96 2.78 1.15 0.81 2.92 2.73 1.14 0.80 3.18 2.96 1.18 0.82 50 30 1.99 1.92 1.08 0.82 2.14 2.06 1.12 0.84 2.26 2.17 1.15 0.86 45° slope UpT 50 10 2.90 2.75 1.29 0.94 2.88 2.74 1.29 0.94 3.12 2.95 1.34 0.96 20 2.13 2.07 1.28 1.00 2.30 2.23 1.34 1.04 2.42 2.35 1.38 1.06 90 10 3.50 3.24 1.22 0.84 3.99 3.66 1.27 0.87 3.69 3.40 1.24 0.85 50 30 2.91 2.77 1.30 0.94 2.58 2.46 1.23 0.90 2.67 2.55 1.25 0.91 Vertical Horizontal -* 50 10 3.70 3.46 1.43 1.01 3.79 3.55 1.45 1.02 3.63 3.40 1.42 1.01 20 3.14 3.02 1.58 1.18 2.76 2.66 1.48 1.12 2.88 2.78 1.51 1.14 90 10 3.53 3.27 1.22 0.84 5.07 4.55 1.36 0.91 4.81 4.33 1.34 0.90 50 30 3.43 3.23 1.39 0.99 3.58 3.36 1.42 1.00 3.51 3.30 1.40 1.00 45° slope Down i 50 10 3.81 3.57 1.45 1.02 5.10 4.66 1.60 1.09 4.74 4.36 1.57 1.08 20 3.75 3.57 1.72 1.26 3.85 3.66 1.74 1.27 3.81 3.63 1.74 1.27 90 10 3.55 3.29 1.22 0.85 6.09 5.35 1.43 0.94 10.07 8.19 1.57 1.00 50 30 3.77 3.52 1.44 1.02 6.27 5.63 1.70 1.14 9.60 8.17 1.88 1.22 Horizontal Down i 50 10 3.84 3.59 1.45 1.02 6.61 5.90 1.73 1.15 11.15 9.27 1.93 1.24 20 4.18 3.96 1.81 1.30 7.03 6.43 2.19 1.49 10.90 9.52 2.47 1.62 178 cen58933_ch03.qxd 9/10/2002 8:59 AM Page 179 EXAMPLE 3-16 The /?- Value of a Wood Frame Wall Determine the overall unit thermal resistance (the ff-value) and the overall heat transfer coefficient (the L/-factor) of a wood frame wall that is built around 38-mm X 90-mm (2x4 nominal) wood studs with a center-to-center distance of 400 mm. The 90-mm-wide cavity between the studs is filled with glass fiber insulation. The inside is finished with 13-mm gypsum wallboard and the out- side with 13-mm wood fiberboard and 13-mm X 200-mm wood bevel lapped siding. The insulated cavity constitutes 75 percent of the heat transmission area while the studs, plates, and sills constitute 21 percent. The headers con- stitute 4 percent of the area, and they can be treated as studs. Also, determine the rate of heat loss through the walls of a house whose perimeter is 50 m and wall height is 2.5 m in Las Vegas, Nevada, whose winter design temperature is -2°C. Take the indoor design temperature to be 22°C and assume 20 percent of the wall area is occupied by glazing. 179 CHAPTER 3 SOLUTION The ff-value and the ^/-factor of a wood frame wall as well as the rate of heat loss through such a wall in Las Vegas are to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat transfer through the wall is one-dimensional. 3 Thermal properties of the wall and the heat transfer coefficients are constant. Properties The ff-values of different materials are given in Table 3-6. Analysis The schematic of the wall as well as the different elements used in its construction are shown here. Heat transfer through the insulation and through the studs will meet different resistances, and thus we need to analyze the ther- mal resistance for each path separately. Once the unit thermal resistances and the L/-factors for the insulation and stud sections are available, the overall av- erage thermal resistance for the entire wall can be determined from ''nv era 1 1 1 / L/ rivers 1 1 where overall V ./area/insi (U X/ arca ) stud and the value of the area fraction f area is 0.75 for the insulation section and 0.25 for the stud section since the headers that constitute a small part of the wall are to be treated as studs. Using the available ff-values from Table 3-6 and calculating others, the total ff-values for each section can be determined in a systematic manner in the table in this sample. We conclude that the overall unit thermal resistance of the wall is 2.23 m 2 • °C/W, and this value accounts for the effects of the studs and headers. It corresponds to an ff-value of 2.23 X 5.68 = 12.7 (or nearly ff-13) in English units. Note that if there were no wood studs and headers in the wall, the over- all thermal resistance would be 3.05 m 2 ■ °C/W, which is 37 percent greater than 2.23 m 2 • °C/W. Therefore, the wood studs and headers in this case serve as thermal bridges in wood frame walls, and their effect must be considered in the thermal analysis of buildings. cen58933_ch03.qxd 9/10/2002 8:59 AM Page IE 180 HEAT TRANSFER Schematic ft- value, m 2 • °C/W Construction Between At Studs Studs 1. Outside surface, 24 km/h wind 0.030 0.030 2. Wood bevel lapped siding 0.14 0.14 3. Wood fiberboard sheeting, 13 mm 0.23 0.23 4a. Glass fiber insulation, 90 mm 2.45 — 4b. Wood stud, 38 mm X 6 90 mm — 0.63 5. Gypsum wall board, 13 mm 0.079 0.079 6. Inside surface, still air 0.12 0.12 Total unit thermal resistance of each section, R (in m 2 • °C/W) 3.05 1.23 The U-factor of each section, U = 1/ff, in W/m 2 • °C 0.328 0.813 Area fraction of each section, 4rea 0.75 0.25 Overall U-factor: U= 2f area , U, = 0.75 X 0.328 + 0.25 X 0.813 = 0.449 W/m 2 • °C Overall unit thermal resistance: R = l/U= 2.23 m 2 • °C/W The perimeter of the building is 50 m and the height of the walls is 2.5 m. Noting that glazing constitutes 20 percent of the walls, the total wall area is A wa „ = 0.80(Perimeter)(Height) = 0.80(50 m)(2.5 m) = 100 m 2 Then the rate of heat loss through the walls under design conditions becomes Swan = (HA) wall {T i - T ) = (0.449 W/m 2 • °C)(100 m 2 )[22 - (-2)°C] = 1078 W Discussion Note that a 1-kW resistance heater in this house will make up al- most all the heat lost through the walls, except through the doors and windows, when the outdoor air temperature drops to -2°C. EXAMPLE 3-17 The ff-Value of a Wall with Rigid Foam The 13-mm-thick wood fiberboard sheathing of the wood stud wall discussed in the previous example is replaced by a 25-mm-thick rigid foam insulation. De- termine the percent increase in the R -value of the wall as a result. cen58933_ch03.qxd 9/10/2002 8:59 AM Page 181 SOLUTION The overall ff-value of the existing wall was determined in Example 3-16 to be 2.23 m 2 • °C/W. Noting that the ff-values of the fiberboard and the foam insulation are 0.23 m z • °C/W and 0.98 m 2 • °C/W, respectively, and the added and removed thermal resistances are in series, the overall ff-value of the wall after modification becomes Rr, R„ old removed added 2.23 - 0.23 + 0.98 2.98 m 2 • °C/W This represents an increase of (2.98 - 2.23)/2.23 = 0.34 or 34 percent in the ff-value of the wall. This example demonstrated how to evaluate the new ff-value of a structure when some structural members are added or removed. EXAMPLE 3-18 The ff-Value of a Masonry Wall Determine the overall unit thermal resistance (the ff-value) and the overall heat transfer coefficient (the ^/-factor) of a masonry cavity wall that is built around 6-in. -thick concrete blocks made of lightweight aggregate with 3 cores filled with perlite (ff = 4.2 h • ft 2 • °F/Btu). The outside is finished with 4-in. face brick with i-in. cement mortar between the bricks and concrete blocks. The in- side finish consists of | in. gypsum wallboard separated from the concrete block by f-in.-thick (1-in. X 3-in. nominal) vertical furring (ff = 4.2 h • ft 2 • °F/Btu) whose center-to-center distance is 16 in. Both sides of the |-in. -thick air space between the concrete block and the gypsum board are coated with reflective aluminum foil (e = 0.05) so that the effective emissivity of the air space is 0.03. For a mean temperature of 50°F and a temperature difference of 30°F, the ff-value of the air space is 2.91 h • ft 2 ■ °F/Btu. The reflective air space con- stitutes 80 percent of the heat transmission area, while the vertical furring con- stitutes 20 percent. SOLUTION The ff-value and the U-factor of a masonry cavity wall are to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat transfer through the wall is one-dimensional. 3 Thermal properties of the wall and the heat transfer coefficients are constant. Properties The ff-values of different materials are given in Table 3-6. Analysis The schematic of the wall as well as the different elements used in its construction are shown below. Following the approach described here and using the available ff-values from Table 3-6, the overall ff-value of the wall is deter- mined in this table. 181 CHAPTER 3 cen58933_ch03.qxd 9/10/2002 8:59 AM Page 182 182 HEAT TRANSFER Schematic l fl-va ue, h • ft 2 • °F/Btu Between At Construction Furring Furring 1. Outside surface, 15 mph wind 0.17 0.17 2. Face brick, 4 in. 0.43 0.43 3. Cement mortar, 0.5 in. 0.10 0.10 4. Concrete block, 6 in . 4.20 4.20 5a. Reflective air space, fin- 2.91 — ^ 5b. Nominal 1x3 <\ vertical furring — 0.94 Gypsum wall board, 0.5 in. 0.45 0.45 7. Inside surface, still air 0.68 0.68 Total unit thermal resistance of each section, R 8.94 6.97 The U-factor of each section, U = 1/ff, in Btu/h ■ ft 2 • °F 0.112 0.143 Area fraction of each section, 4rea 0.80 0.20 Overall £/-f actor: U = 2f area ,U, = 0.80 X 0.112 + 0.20 X 0.143 = 0.1 18 Btu/h- ft 2 -°F Overall unit thermal resistance: R= IIU= 8.46 h • ft 2 • °F/Btu Therefore, the overall unit thermal resistance of the wall is 8.46 h • ft 2 • C F/Btu and the overall l/-factor is 0.118 Btu/h • ft 2 • °F. These values account for the effects of the vertical furring. EXAMPLE 3-19 The ff-Value of a Pitched Roof Determine the overall unit thermal resistance (the ff -value) and the overall heat transfer coefficient (the ^/-factor) of a 45° pitched roof built around nominal 2-in. X 4-in. wood studs with a center-to-center distance of 16 in. The 3.5-in.- wide air space between the studs does not have any reflective surface and thus its effective emissivity is 0.84. For a mean temperature of 90°F and a temper- ature difference of 30°F, the ff-value of the air space is 0.86 h • ft 2 • °F/Btu. The lower part of the roof is finished with |-in. gypsum wallboard and the upper part with |-in. plywood, building paper, and asphalt shingle roofing. The air space constitutes 75 percent of the heat transmission area, while the studs and headers constitute 25 percent. SOLUTION The ff-value and the ^/-factor of a 45° pitched roof are to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat transfer through the roof is one-dimensional. 3 Thermal properties of the roof and the heat transfer coefficients are constant. Properties The ff-values of different materials are given in Table 3-6. cen58933_ch03.qxd 9/10/2002 9:00 AM Page 183 183 CHAPTER 3 Analysis The schematic of the pitched roof as well as the different elements used in its construction are shown below. Following the a oproach described above and using the avai able ff-values from Table 3-6, the overall R- /alue of the roof can be determined in the table here. Schematic fl-va h • ft 2 • Between ue, °F/Btu At . . Construction Studs Studs 1. Outside surface, / " V \ X \ X \^ 15 mph wind 0.17 0.17 / "^Cv^O^N^/ 2. Asphalt shingle / /Q/^f^^w/ roofing 0.44 0.44 / /r^j^^^A 4 / 3. Building paper 0.10 0.10 TrCyA 4. Plywood deck, | in. 0.78 0.78 I I \ \ \ \ C ) 5a Nonreflective air 1 2 3 4 5a 5b 6 7 space, 3.5 in. 0.86 — 5b Wood stud, 2 in. by 4 in. — 3.58 6. Gypsum wallboard, 0.5 in. 0.45 0.45 7. Inside surface, 45° slope, still air 0.63 0.63 Total unit thermal resista ice of each section, R 3.43 6.15 The Ofactor of each sect ion, U= 1/fl, in Btu/h -ft 2 - °F 0.292 0.163 Area fraction of each seel ion, 'area 0.75 0.25 Overall U-i actor: U= 2f area iUr- = 0.75 X 0.292 + 0.25 X 0.163 = 0.260 Btu/h • ft 2 • °F Overall unit thermal resistance: R = 1/U = 3.85 h • fl : • °F/Btu Therefore, the overall unit thermal resistance of this pitched roof is 3.85 h • ft 2 • °F/Btu and the overall tZ-factor is 0.260 Btu/h • ft 2 • °F. Note that the wood studs offer much larger thermal resistance to heat flow than the air space between the studs. The construction of wood frame flat ceilings typically involve 2-in. X 6-in. joists on 400-mm (16-in.) or 600-mm (24-in.) centers. The fraction of framing is usually taken to be 0.10 for joists on 400-mm centers and 0.07 for joists on 600-mm centers. Most buildings have a combination of a ceiling and a roof with an attic space in between, and the determination of the 7?-value of the roof-attic- ceiling combination depends on whether the attic is vented or not. For ad- equately ventilated attics, the attic air temperature is practically the same as the outdoor air temperature, and thus heat transfer through the roof is gov- erned by the /lvalue of the ceiling only. However, heat is also transferred between the roof and the ceiling by radiation, and it needs to be considered (Fig. 3-52). The major function of the roof in this case is to serve as a ra- diation shield by blocking off solar radiation. Effectively ventilating the at- tic in summer should not lead one to believe that heat gain to the building through the attic is greatly reduced. This is because most of the heat trans- fer through the attic is by radiation. Aii- exhaust .^ppRadiant s^' barrier K/\/\/w\/w\/\/\/w\/\/\/y K n Ail- intake Air intake FIGURE 3-52 Ventilation paths for a naturally ventilated attic and the appropriate size of the flow areas around the radiant barrier for proper air circulation (from DOE/CE-0335P, U.S. Dept. of Energy). cen58933_ch03.qxd 9/10/2002 9:00 AM Page 184 184 HEAT TRANSFER - Roof decking Air space Roof decking - Roof decking Joist - Insulation Joist — Insulation - (b) At the bottom of rafters (a) Under the roof deck FIGURE 3-53 Three possible locations for an attic radiant barrier (from DOE/CE-0335P, U.S. Dept. of Energy). Joist — Insulation - (c) On top of attic floor insulation Shingles Deck ^Rafter/^ %\ y^ Attic < " A r ceiling attic \ Ceiling joist > ' ^ceiling ?i FIGURE 3-54 Thermal resistance network for a pitched roof-attic-ceiling combination for the case of an unvented attic. Radiation heat transfer between the ceiling and the roof can be mini- mized by covering at least one side of the attic (the roof or the ceiling side) by a reflective material, called radiant barrier, such as aluminum foil or aluminum-coated paper. Tests on houses with R- 19 attic floor insulation have shown that radiant barriers can reduce summer ceiling heat gains by 16 to 42 percent compared to an attic with the same insulation level and no radiant barrier. Considering that the ceiling heat gain represents about 15 to 25 percent of the total cooling load of a house, radiant barriers will reduce the air conditioning costs by 2 to 10 percent. Radiant barriers also reduce the heat loss in winter through the ceiling, but tests have shown that the percentage reduction in heat losses is less. As a result, the percentage reduction in heating costs will be less than the reduction in the air- conditioning costs. Also, the values given are for new and undusted radiant barrier installations, and percentages will be lower for aged or dusty radi- ant barriers. Some possible locations for attic radiant barriers are given in Figure 3-53. In whole house tests on houses with R-\9 attic floor insulation, radi- ant barriers have reduced the ceiling heat gain by an average of 35 percent when the radiant barrier is installed on the attic floor, and by 24 percent when it is attached to the bottom of roof rafters. Test cell tests also demon- strated that the best location for radiant barriers is the attic floor, provided that the attic is not used as a storage area and is kept clean. For unvented attics, any heat transfer must occur through (1) the ceiling, (2) the attic space, and (3) the roof (Fig. 3-54). Therefore, the overall K-value of the roof-ceiling combination with an unvented attic depends on the combined effects of the 7?-value of the ceiling and the K-value of the roof as well as the thermal resistance of the attic space. The attic space can be treated as an air layer in the analysis. But a more practical way of ac- counting for its effect is to consider surface resistances on the roof and ceil- ing surfaces facing each other. In this case, the ^-values of the ceiling and the roof are first determined separately (by using convection resistances for the still-air case for the attic surfaces). Then it can be shown that the over- all i?-value of the ceiling-roof combination per unit area of the ceiling can be expressed as cen58933_ch03.qxd 9/10/2002 9:00 AM Page 185 /? = /? H- /? "■ "ceiling "roof ^ceiling ^roof (3-82) 185 CHAPTER 3 where i4 cei | in g and A mof are the ceiling and roof areas, respectively. The area ratio is equal to 1 for flat roofs and is less than 1 for pitched roofs. For a 45° pitched roof, the area ratio is ^ ce iiin g /^roof = l/\/2 = 0.707. Note that the pitched roof has a greater area for heat transfer than the flat ceiling, and the area ratio accounts for the reduction in the unit lvalue of the roof when expressed per unit area of the ceiling. Also, the direction of heat flow is up in winter (heat loss through the roof) and down in summer (heat gain through the roof). The 7?-value of a structure determined by analysis assumes that the ma- terials used and the quality of workmanship meet the standards. Poor work- manship and substandard materials used during construction may result in ^-values that deviate from predicted values. Therefore, some engineers use a safety factor in their designs based on experience in critical applications. SUMMARY One-dimensional heat transfer through a simple or composite body exposed to convection from both sides to mediums at temperatures T^ t and T rjJ1 can be expressed as Q (W) mediums. For a plane wall exposed to convection on both sides, the total resistance is expressed as R„ R,, -"wiiii *~ Rr, 1 1 This relation can be extended to plane walls that consist of two or more layers by adding an additional resistance for each ad- ditional layer. The elementary thermal resistance relations can be expressed as follows: where h c is the thermal contact conductance, R c is the thermal contact resistance, and the radiation heat transfer coefficient is defined as />,, s(j(T; + Tl Tr )(T s + T sulr ) Once the rate of heat transfer is available, the temperature drop across any layer can be determined from AT= QR The thermal resistance concept can also be used to solve steady heat transfer problems involving parallel layers or combined series-parallel arrangements. Adding insulation to a cylindrical pipe or a spherical shell will increase the rate of heat transfer if the outer radius of the insulation is less than the critical radius of insulation, defined as Interface resistance: Radiation resistance: v cyl Conduction resistance (plane wall): R wa[[ Conduction resistance (cylinder): R Conduction resistance (sphere): Convection resistance: kA Info//-)) 2nLk R sph R R ■"rad 4ir r x r 2 k J_ : onv hA j_ = Rc nterface ^ £ 1 "rad "■ cr, cylinder h 2k, * cr, sphere h The effectiveness of an insulation is often given in terms of its R-value, the thermal resistance of the material per unit sur- face area, expressed as lvalue L (flat insulation) where L is the thickness and k is the thermal conductivity of the material. cen58933_ch03.qxd 9/10/2002 9:00 AM Page 186 186 HEAT TRANSFER Finned surfaces are commonly used in practice to enhance heat transfer. Fins enhance heat transfer from a surface by ex- posing a larger surface area to convection. The temperature distribution along the fin for very long fins and for fins with negligible heat transfer at the fin are given by Very long fin: Adiabatic fin tip: T(x) - T a T(x) - T a p—x\ hplkA cosh a(L — x) cosh ah where a = \/hplkA c , p is the perimeter, and A c is the cross sectional area of the fin. The rates of heat transfer for both cases are given to be Very long ei ongfin = ~kA c fin: Adiabatic Jm ^c insulated tip — ~~ ^c tip: VhpkAc(T b -TJ \/kpkA c (T b - r„) tanh ah Fins exposed to convection at their tips can be treated as fins with insulated tips by using the corrected length L c = L + AJp instead of the actual fin length. The temperature of a fin drops along the fin, and thus the heat transfer from the fin will be less because of the decreasing temperature difference toward the fin tip. To account for the ef- fect of this decrease in temperature on heat transfer, we define fin efficiency as %i„ Q< Q fin, max Actual heat transfer rate from the fin Ideal heat transfer rate from the fin if the entire fin were at base temperature The performance of the fins is judged on the basis of the en- hancement in heat transfer relative to the no-fin case and is ex- pressed in terms of the fin effectiveness s fln , defined as e f ,» g f ,„ e» of ,„ hA t {T t -TJ Heat transfer rate from the fin of base area A b Heat transfer rate from the surface of area A h Here, A h is the cross-sectional area of the fin at the base and Q no fin represents the rate of heat transfer from this area if no fins are attached to the surface. The overall effectiveness for a finned surface is defined as the ratio of the total heat transfer from the finned surface to the heat transfer from the same sur- face if there were no fins, Q total, fin h(A unfm + TlflnAaJfTj, - TJ) Q total, no fin "A no fin (T b T„) Fin efficiency and fin effectiveness are related to each other by Afl„ A, %,„ Certain multidimensional heat transfer problems involve two surfaces maintained at constant temperatures T { and T 2 . The steady rate of heat transfer between these two surfaces is ex- pressed as Q = Sk(T { - T 2 ) where S is the conduction shape factor that has the dimen- sion of length and k is the thermal conductivity of the medium between the surfaces. When the fin efficiency is available, the rate of heat transfer from a fin can be determined from Gfi„ = infi„2f ■nfin^ fin (T b - T„) REFERENCES AND SUGGESTED READING 1. American Society of Heating, Refrigeration, and Air Conditioning Engineers. Handbook of Fundamentals. Atlanta: ASHRAE, 1993. 2. R. V. Andrews. "Solving Conductive Heat Transfer Problems with Electrical-Analogue Shape Factors." Chemical Engineering Progress 5 (1955), p. 67. 3. R. Barron. Cryogenic Systems. New York: McGraw-Hill, 1967. 4. L. S. Fletcher. "Recent Developments in Contact Conductance Heat Transfer." Journal of Heat Transfer 110, no. 4B (1988), pp. 1059-79. 5. E. Fried. "Thermal Conduction Contribution to Heat Transfer at Contacts." Thermal Conductivity, vol. 2, ed. R. R Tye. London: Academic Press, 1969. 6. K. A. Gardner. "Efficiency of Extended Surfaces." Trans. ASME 67 (1945), pp. 621-31. 7. F. P. Incropera and D. P. DeWitt. Introduction to Heat Transfer. 4th ed. New York: John Wiley & Sons, 2002. 8. D. Q. Kern and A. D. Kraus. Extended Surface Heat Transfer. New York: McGraw-Hill, 1972. cen58933_ch03.qxd 9/10/2002 9:00 AM Page 187 9. M. N. Ozisik. Heat Transfer — A Basic Approach. New York: McGraw-Hill, 1985. 10. G. P. Peterson. "Thermal Contact Resistance in Waste Heat Recovery Systems." Proceedings of the 18th ASME/ETCE Hydrocarbon Processing Symposium. Dallas, TX, 1987, pp. 45-51. 11. S. Song, M. M. Yovanovich, and F. O. Goodman. "Thermal Gap Conductance of Conforming Surfaces in Contact." Journal of Heat Transfer 115 (1993), p. 533. PROBLEMS 187 CHAPTER 3 12. J. E. Sunderland and K. R. Johnson. "Shape Factors for Heat Conduction through Bodies with Isothermal or Convective Boundary Conditions," Trans. ASME 10 (1964), pp. 237-41. 13. N. V. Suryanarayana. Engineering Heat Transfer. St. Paul, MN: West Publishing, 1995. Steady Heat Conduction in Plane Walls 3-1 C Consider one-dimensional heat conduction through a cylindrical rod of diameter D and length L. What is the heat transfer area of the rod if (a) the lateral surfaces of the rod are insulated and (b) the top and bottom surfaces of the rod are insulated? 3-2C Consider heat conduction through a plane wall. Does the energy content of the wall change during steady heat con- duction? How about during transient conduction? Explain. 3-3C Consider heat conduction through a wall of thickness L and area A. Under what conditions will the temperature distri- butions in the wall be a straight line? 3-4C What does the thermal resistance of a medium represent? 3-5C How is the combined heat transfer coefficient defined? What convenience does it offer in heat transfer calculations? 3-6C Can we define the convection resistance per unit surface area as the inverse of the convection heat transfer coefficient? 3-7C Why are the convection and the radiation resistances at a surface in parallel instead of being in series? 3-8C Consider a surface of area A at which the convection and radiation heat transfer coefficients are h com and /z rad , re- spectively. Explain how you would determine (a) the single equivalent heat transfer coefficient, and (b) the equivalent ther- mal resistance. Assume the medium and the surrounding sur- faces are at the same temperature. 3-9C How does the thermal resistance network associated with a single-layer plane wall differ from the one associated with a five-layer composite wall? *Problems designated by a "C" are concept questions, and students are encouraged to answer them all. Problems designated by an "E" are in English units, and the SI users can ignore them. Problems with an EES-CD icon ® are solved using EES, and complete solutions together with parametric studies are included on the enclosed CD. Problems with a computer-EES icon H are comprehensive in nature, and are intended to be solved with a computer, preferably using the EES software that accompanies this text. 3-10C Consider steady one-dimensional heat transfer through a multilayer medium. If the rate of heat transfer Q is known, explain how you would determine the temperature drop across each layer. 3-11C Consider steady one-dimensional heat transfer through a plane wall exposed to convection from both sides to environments at known temperatures T rA and T^ 2 with known heat transfer coefficients h { and h 2 . Once the rate of heat trans- fer Q has been evaluated, explain how you would determine the temperature of each surface. 3-12C Someone comments that a microwave oven can be viewed as a conventional oven with zero convection resistance at the surface of the food. Is this an accurate statement? 3-13C Consider a window glass consisting of two 4-mm- thick glass sheets pressed tightly against each other. Compare the heat transfer rate through this window with that of one con- sisting of a single 8-mm-thick glass sheet under identical con- ditions. 3-14C Consider steady heat transfer through the wall of a room in winter. The convection heat transfer coefficient at the outer surface of the wall is three times that of the inner surface as a result of the winds. On which surface of the wall do you think the temperature will be closer to the surrounding air tem- perature? Explain. 3-1 5C The bottom of a pan is made of a 4-mm-thick alu- minum layer. In order to increase the rate of heat transfer through the bottom of the pan, someone proposes a design for the bottom that consists of a 3-mm-thick copper layer sand- wiched between two 2-mm-thick aluminum layers. Will the new design conduct heat better? Explain. Assume perfect con- tact between the layers. 3-16C Consider two cold canned drinks, one wrapped in a blanket and the other placed on a table in the same room. Which drink will warm up faster? 3-17 Consider a 4-m-high, 6-m-wide, and 0.3-m-thick brick wall whose thermal conductivity is k = 0.8 W/m ■ °C . On a certain day, the temperatures of the inner and the outer surfaces of the wall are measured to be 14°C and 6°C, respectively. De- termine the rate of heat loss through the wall on that day. 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 cen58933_ch03.qxd 9/10/2002 9:00 AM Page 189 189 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. cen58933_ch03.qxd 9/10/2002 9:00 AM Page 19C 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? cen58933_ch03.qxd 9/10/2002 9:00 AM Page 191 191 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 cen58933_ch03.qxd 9/10/2002 9:00 AM Page 192 192 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 cen58933_ch03.qxd 9/10/2002 9:00 AM Page 193 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 cen58933_ch03.qxd 9/10/2002 9:00 AM Page 194 194 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 cen58933_ch03.qxd 9/10/2002 9:00 AM Page 195 195 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 cen58933_ch03.qxd 9/10/2002 9:00 AM Page 196 196 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? cen58933_ch03.qxd 9/10/2002 9:00 AM Page 197 3-88 A 2-mm-diameter and 1 0-m-long electric wire is tightly wrapped with a 1-mm-thick plastic cover whose thermal con- ductivity is k = 0.15 W/m ■ °C. Electrical measurements indi- cate that a current of 10 A passes through the wire and there is a voltage drop of 8 V along the wire. If the insulated wire is ex- posed to a medium at T„ = 30°C with a heat transfer coeffi- cient of h = 24 W/m 2 ■ °C, determine the temperature at the interface of the wire and the plastic cover in steady operation. Also determine if doubling the thickness of the plastic cover will increase or decrease this interface temperature. Electrical wire T. = 30°C - Insulation 2> ■10m FIGURE P3-88 3-89E A 0.083-in. -diameter electrical wire at 115°F is covered by 0.02-in. -thick plastic insulation (k = 0.075 Btu/h • ft • °F). The wire is exposed to a medium at 50°F, with a combined convection and radiation heat transfer coefficient of 2.5 Btu/h • ft 2 • °F. Determine if the plastic insulation on the wire will increase or decrease heat transfer from the wire. Answer: It helps 3-90E Repeat Problem 3-89E, assuming a thermal contact resistance of 0.001 h • ft 2 • °F/Btu at the interface of the wire and the insulation. 3-91 A 5-mm-diameter spherical ball at 50°C is covered by a 1-mm-thick plastic insulation (k = 0.13 W/m • °C). The ball is exposed to a medium at 15°C, with a combined convection and radiation heat transfer coefficient of 20 W/m 2 ■ °C. Determine if the plastic insulation on the ball will help or hurt heat trans- fer from the ball. FIGURE P3-91 3-92 Reconsider Problem 3-91. Using EES (or other) software, plot the rate of heat transfer from the ball as a function of the plastic insulation thickness in the range of 0.5 mm to 20 mm. Discuss the results. Heat Transfer from Finned Surfaces 3-93C What is the reason for the widespread use of fins on surfaces? 197 CHAPTER 3 3-94C What is the difference between the fin effectiveness and the fin efficiency? 3-95C The fins attached to a surface are determined to have an effectiveness of 0.9. Do you think the rate of heat transfer from the surface has increased or decreased as a result of the addition of these fins? 3-96C Explain how the fins enhance heat transfer from a surface. Also, explain how the addition of fins may actually decrease heat transfer from a surface. 3-97C How does the overall effectiveness of a finned sur- face differ from the effectiveness of a single fin? 3-98C Hot water is to be cooled as it flows through the tubes exposed to atmospheric air. Fins are to be attached in order to enhance heat transfer. Would you recommend attaching the fins inside or outside the tubes? Why? 3-99C Hot air is to be cooled as it is forced to flow through the tubes exposed to atmospheric air. Fins are to be added in order to enhance heat transfer. Would you recommend attach- ing the fins inside or outside the tubes? Why? When would you recommend attaching fins both inside and outside the tubes? 3-100C Consider two finned surfaces that are identical except that the fins on the first surface are formed by casting or extrusion, whereas they are attached to the second surface afterwards by welding or tight fitting. For which case do you think the fins will provide greater enhancement in heat trans- fer? Explain. 3-101C The heat transfer surface area of a fin is equal to the sum of all surfaces of the fin exposed to the surrounding medium, including the surface area of the fin tip. Under what conditions can we neglect heat transfer from the fin tip? 3-102C Does the (a) efficiency and (b) effectiveness of a fin increase or decrease as the fin length is increased? 3-103C Two pin fins are identical, except that the diameter of one of them is twice the diameter of the other. For which fin will the (a) fin effectiveness and (b) fin efficiency be higher? Explain. 3-104C Two plate fins of constant rectangular cross section are identical, except that the thickness of one of them is twice the thickness of the other. For which fin will the (a) fin effec- tiveness and (b) fin efficiency be higher? Explain. 3-105C Two finned surfaces are identical, except that the convection heat transfer coefficient of one of them is twice that of the other. For which finned surface will the (a) fin effective- ness and (b) fin efficiency be higher? Explain. 3-106 Obtain a relation for the fin efficiency for a fin of con- stant cross-sectional area A c , perimeter p, length L, and thermal conductivity k exposed to convection to a medium at r„ with a heat transfer coefficient h. Assume the fins are sufficiently long so that the temperature of the fin at the tip is nearly T x . Take the temperature of the fin at the base to be T b and neglect heat cen58933_ch03.qxd 9/10/2002 9:00 AM Page 19E 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 cen58933_ch03.qxd 9/10/2002 9:00 AM Page 199 199 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 cen58933_ch03.qxd 9/10/2002 9:00 AM Page 2C 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 cen58933_ch03.qxd 9/10/2002 9:00 AM Page 201 201 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 cen58933_ch03.qxd 9/10/2002 9:00 AM Page 202 202 HEAT TRANSFER 12 3 4 5 FIGURE P3-1 42 sides of the ceiling are exposed to still air. The air space con- stitutes 82 percent of the heat transmission area, while the studs and headers constitute 1 8 percent. Determine the winter R- value and the {/-factor of the ceiling assuming the 90-mm- wide air space between the studs (a) does not have any reflec- tive surface, (b) has a reflective surface with s = 0.05 on one side, and (c) has reflective surfaces with e = 0.05 on both sides. Assume a mean temperature of 10°C and a temperature difference of 5.6°C for the air space. 3-143 Determine the winter lvalue and the {/-factor of a masonry cavity wall that consists of 100-mm common bricks, a 90-mm air space, 100-mm concrete blocks made of light- weight aggregate, 20-mm air space, and 13-mm gypsum wall- board separated from the concrete block by 20-mm-thick (1-in. X 3 -in. nominal) vertical furring whose center-to-center distance is 400 mm. Neither side of the two air spaces is coated with any reflective films. When determining the ^-value of the air spaces, the temperature difference across them can be taken to be 16.7°C with a mean air temperature of 10°C. The air space constitutes 84 percent of the heat transmission area, while the vertical furring and similar structures constitute 16 percent. Answers: 1.02 m 2 • °C/W, 0.978 W/m 2 • °C 3-144 Repeat Problem 3-143 assuming one side of both air spaces is coated with a reflective film of e = 0.05. 3-145 Determine the winter R-vahie and the [/-factor of a masonry wall that consists of the following layers: 100-mm face bricks, 100-mm common bricks, 25-mm urethane rigid foam insulation, and 13-mm gypsum wallboard. Answers: 1.404 m 2 • °C/W, 0.712 W/m 2 • °C 3-146 The overall heat transfer coefficient (the [/-value) of a wall under winter design conditions is U = 1.55 W/m 2 ■ °C. Determine the [/-value of the wall under summer design conditions. 3-147 The overall heat transfer coefficient (the [/-value) of a wall under winter design conditions is U = 2.25 W/m 2 ■ °C. Now a layer of 100-mm face brick is added to the outside, leaving a 20-mm air space between the wall and the bricks. De- termine the new [/-value of the wall. Also, determine the rate of heat transfer through a 3-m-high, 7-m-long section of the wall after modification when the indoor and outdoor tempera- tures are 22°C and — 5°C, respectively. FIGURE P3-1 47 FIGURE P3-1 43 3-148 Determine the summer and winter ^-values, in m 2 • °C/W, of a masonry wall that consists of 100-mm face bricks, 13-mm of cement mortar, 100-mm lightweight concrete block, 40-mm air space, and 20-mm plasterboard. Answers: 0.809 and 0.795 m 2 ■ °C/W 3-149E The overall heat transfer coefficient of a wall is determined to be U = 0.09 Btu/h • ft 2 ■ °F under the conditions of still air inside and winds of 7.5 mph outside. What will the [/-factor be when the wind velocity outside is doubled? Answer: 0.0907 Btu/h ■ ft 2 • °F 3-150 Two homes are identical, except that the walls of one house consist of 200-mm lightweight concrete blocks, 20-mm air space, and 20-mm plasterboard, while the walls of the other house involve the standard R-2.4 m 2 • °C/W frame wall con- struction. Which house do you think is more energy efficient? cen58933_ch03.qxd 9/10/2002 9:00 AM Page 203 3-151 Determine the /J-value of a ceiling that consists of a layer of 19-mm acoustical tiles whose top surface is covered with a highly reflective aluminum foil for winter conditions. Assume still air below and above the tiles. Highly reflective foil FIGURE P3-1 51 Review Problems 3-152E Steam is produced in the copper tubes (k = 223 Btu/h • ft • °F) of a heat exchanger at a temperature of 250°F by another fluid condensing on the outside surfaces of the tubes at 350°F. The inner and outer diameters of the tube are 1 in. and 1.3 in., respectively. When the heat exchanger was new, the rate of heat transfer per foot length of the tube was 2 X 10 4 Btu/h. Determine the rate of heat transfer per foot length of the tube when a 0.01 -in. -thick layer of limestone (k = 1 .7 Btu/h ■ ft ■ °F) has formed on the inner surface of the tube after extended use. 3-153E Repeat Problem 3-152E, assuming that a 0.01-in.- thick limestone layer has formed on both the inner and outer surfaces of the tube. 3-154 A 1.2-m-diameter and 6-m-long cylindrical propane tank is initially filled with liquid propane whose density is 581 kg/m 3 . The tank is exposed to the ambient air at 30°C, with a heat transfer coefficient of 25 W/m 2 • °C. Now a crack devel- ops at the top of the tank and the pressure inside drops to 1 atm while the temperature drops to — 42°C, which is the boiling temperature of propane at 1 atm. The heat of vaporization of Propane T f ir =30°C vapor "\ PROPANE TANK 1.2 m T = -42°C P = 1 atm / 4 6 m ► 203 CHAPTER 3 propane at 1 atm is 425 kJ/kg. The propane is slowly vaporized as a result of the heat transfer from the ambient air into the tank, and the propane vapor escapes the tank at — 42°C through the crack. Assuming the propane tank to be at about the same temperature as the propane inside at all times, determine how long it will take for the propane tank to empty if the tank is (a) not insulated and (b) insulated with 7.5-cm-thick glass wool insulation (k = 0.038 W/m • °C). 3-155 Hot water is flowing at an average velocity of 1 .5 m/s through a cast iron pipe (k = 52 W/m • °C) whose inner and outer diameters are 3 cm and 3.5 cm, respectively. The pipe passes through a 15-m-long section of a basement whose temperature is 15°C. If the temperature of the water drops from 70°C to 67°C as it passes through the basement and the heat transfer coefficient on the inner surface of the pipe is 400 W/m 2 ■ °C, determine the combined convection and radiation heat transfer coefficient at the outer surface of the pipe. Answer: 272.5 W/m 2 • °C 3-156 Newly formed concrete pipes are usually cured first overnight by steam in a curing kiln maintained at a temperature of 45°C before the pipes are cured for several days outside. The heat and moisture to the kiln is provided by steam flowing in a pipe whose outer diameter is 12 cm. During a plant inspection, it was noticed that the pipe passes through a 10-m section that is completely exposed to the ambient air before it reaches the kiln. The temperature measurements indicate that the average temperature of the outer surface of the steam pipe is 82°C when the ambient temperature is 8°C. The combined convec- tion and radiation heat transfer coefficient at the outer surface of the pipe is estimated to be 25 W/m 2 • °C. Determine the amount of heat lost from the steam during a 10-h curing process that night. Steam is supplied by a gas-fired steam generator that has an efficiency of 80 percent, and the plant pays $0.60/therm of natural gas (1 therm = 105,500 kJ). If the pipe is insulated and 90 percent of the heat loss is saved as a result, determine the amount of money this facility will save a year as a result of insulating the steam pipes. Assume that the concrete pipes are cured 110 nights a year. State your assumptions. 8 "C Furnace FIGURE P3-1 54 FIGURE P3-1 56 3-157 Consider an 18-cm X 18-cm multilayer circuit board dissipating 27 W of heat. The board consists of four layers of 0.2-mm-thick copper (k = 386 W/m • °C) and three layers of cen58933_ch03.qxd 9/10/2002 9:00 AM Page 204 204 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 cen58933_ch03.qxd 9/10/2002 9:00 AM Page 205 205 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 cen58933_ch03.qxd 9/10/2002 9:00 AM Page 206 206 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. cen58933_ch03 . qxd 9/10/2002 9:00 AM Page 208 cen58933_ch04.qxd 9/10/2002 9:12 AM Page 209 TRANSIENT HEAT CONDUCTION CHAPTER The temperature of a body, in general, varies with time as well as position. In rectangular coordinates, this variation is expressed as T(x, y, z, t), where (x, y, z) indicates variation in the x, y, and z directions, respectively, and t indicates variation with time. In the preceding chapter, we considered heat conduction under steady conditions, for which the tempera- ture of a body at any point does not change with time. This certainly simpli- fied the analysis, especially when the temperature varied in one direction only, and we were able to obtain analytical solutions. In this chapter, we consider the variation of temperature with time as well as position in one- and multi- dimensional systems. We start this chapter with the analysis of lumped systems in which the tem- perature of a solid varies with time but remains uniform throughout the solid at any time. Then we consider the variation of temperature with time as well as position for one-dimensional heat conduction problems such as those asso- ciated with a large plane wall, a long cylinder, a sphere, and a semi-infinite medium using transient temperature charts and analytical solutions. Finally, we consider transient heat conduction in multidimensional systems by uti- lizing the product solution. CONTENTS 4-1 Lumped Systems Analysis 210 4-2 Transient Heat Conduction in Large Plane Walls, Long Cylinders, and Spheres with Spatial Effects 216 4-3 Transient Heat Conduction in Semi-Infinite Solids 228 4-4 Transient Heat Conduction in Multidimensional Systems 231 Topic of Special Interest: Refrigeration and Freezing of Foods 239 cen58933_ch04.qxd 9/10/2002 9:12 AM Page 21C 210 HEAT TRANSFER (a) Copper ball (b) Roast beef FIGURE 4-1 A small copper ball can be modeled as a lumped system, but a roast beef cannot. SOLID BODY m = mass V = volume : density = initial temperature T = T(t) Q = hA s [T„-T(t)] FIGURE 4-2 The geometry and parameters involved in the lumped system analysis. 4-1 ■ LUMPED SYSTEM ANALYSIS In heat transfer analysis, some bodies are observed to behave like a "lump" whose interior temperature remains essentially uniform at all times during a heat transfer process. The temperature of such bodies can be taken to be a function of time only, T(t). Heat transfer analysis that utilizes this idealization is known as lumped system analysis, which provides great simplification in certain classes of heat transfer problems without much sacrifice from accuracy. Consider a small hot copper ball coming out of an oven (Fig. 4-1). Mea- surements indicate that the temperature of the copper ball changes with time, but it does not change much with position at any given time. Thus the tem- perature of the ball remains uniform at all times, and we can talk about the temperature of the ball with no reference to a specific location. Now let us go to the other extreme and consider a large roast in an oven. If you have done any roasting, you must have noticed that the temperature dis- tribution within the roast is not even close to being uniform. You can easily verify this by taking the roast out before it is completely done and cutting it in half. You will see that the outer parts of the roast are well done while the cen- ter part is barely warm. Thus, lumped system analysis is not applicable in this case. Before presenting a criterion about applicability of lumped system analysis, we develop the formulation associated with it. Consider a body of arbitrary shape of mass m, volume V, surface area A s , density p, and specific heat C p initially at a uniform temperature T { (Fig. 4-2). At time t = 0, the body is placed into a medium at temperature r„, and heat transfer takes place between the body and its environment, with a heat trans- fer coefficient h. For the sake of discussion, we will assume that T m > T t , but the analysis is equally valid for the opposite case. We assume lumped system analysis to be applicable, so that the temperature remains uniform within the body at all times and changes with time only, T = T(t). During a differential time interval dt, the temperature of the body rises by a differential amount dT. An energy balance of the solid for the time interval dt can be expressed as /Heat transfer into the body\ during dt ( The increase in the | energy of the body \ during dt j or hA s {T a -T)dt = mC„ dT (4-1) Noting that m = pV and dT = d(T — r„) since T„ = constant, Eq. 4-1 can be rearranged as d(T- TJ) T-T a hA s pvc„ dt (4-2) Integrating from t = 0, at which T = T t , to any time t, at which T = T(t), gives In Tit) - T„ T; ~ fZ hA s pvc p ' (4-3) cen58933_ch04.qxd 9/10/2002 9:12 AM Page 211 Taking the exponential of both sides and rearranging, we obtain T(t) - T m _„, where pvc. (1/s) (4-4) (4-5) is a positive quantity whose dimension is (time) -1 . The reciprocal of b has time unit (usually s), and is called the time constant. Equation 4-4 is plotted in Fig. 4-3 for different values of b. There are two observations that can be made from this figure and the relation above: 1. Equation 4-4 enables us to determine the temperature T(t) of a body at time t, or alternatively, the time t required for the temperature to reach a specified value T(t). 2. The temperature of a body approaches the ambient temperature T^ exponentially. The temperature of the body changes rapidly at the beginning, but rather slowly later on. A large value of b indicates that the body will approach the environment temperature in a short time. The larger the value of the exponent b, the higher the rate of decay in temperature. Note that b is proportional to the surface area, but inversely proportional to the mass and the specific heat of the body. This is not surprising since it takes longer to heat or cool a larger mass, especially when it has a large specific heat. Once the temperature T(t) at time t is available from Eq. A-A, the rate of con- vection heat transfer between the body and its environment at that time can be determined from Newton's law of cooling as Q(t) = hA s [T(t) - rj (W) (4-6) The total amount of heat transfer between the body and the surrounding medium over the time interval t = to t is simply the change in the energy content of the body: Q = mC.[T(!) ~ n (kJ) (4-7) The amount of heat transfer reaches its upper limit when the body reaches the surrounding temperature T„. Therefore, the maximum heat transfer between the body and its surroundings is (Fig. 4-4) e„ mC p {T a - r,) (kJ) (4-8) We could also obtain this equation by substituting the T(t) relation from Eq. 4-4 into the Q(t) relation in Eq. 4-6 and integrating it from t = to t — > °°. Criteria for Lumped System Analysis The lumped system analysis certainly provides great convenience in heat transfer analysis, and naturally we would like to know when it is appropriate 211 CHAPTER 4 Tin FIGURE 4-3 The temperature of a lumped system approaches the environment temperature as time gets larger. 1 = h t — > 00 T t T, T a T„ T> r«, T T rJ -_ r, < T a r» Q-- tsraax — ">C p (T r -TJ FIGURE 4-4 Heat transfer to or from a body reaches its maximum value when the body reaches the environment temperature. cen58933_ch04.qxd 9/10/2002 9:12 AM Page 212 212 HEAT TRANSFER Convection » Conduction . SOLID BODY to use it. The first step in establishing a criterion for the applicability of the lumped system analysis is to define a characteristic length as and a Biot number Bi as Bi hL c (4-9) Bi heat convection heat conduction FIGURE 4-5 The Biot number can be viewed as the ratio of the convection at the surface to conduction within the body. It can also be expressed as (Fig. 4-5) Bi h AT Convection at the surface of the body ~klL,.~AT = ~ or Bi LJk Conduction within the body Conduction resistance within the body \lh Convection resistance at the surface of the body When a solid body is being heated by the hotter fluid surrounding it (such as a potato being baked in an oven), heat is first convected to the body and subsequently conducted within the body. The Biot number is the ratio of the internal resistance of a body to heat conduction to its external resistance to heat convection. Therefore, a small Biot number represents small resistance to heat conduction, and thus small temperature gradients within the body. Lumped system analysis assumes a uniform temperature distribution throughout the body, which will be the case only when the thermal resistance of the body to heat conduction (the conduction resistance) is zero. Thus, lumped system analysis is exact when Bi = and approximate when Bi > 0. Of course, the smaller the Bi number, the more accurate the lumped system analysis. Then the question we must answer is, How much accuracy are we willing to sacrifice for the convenience of the lumped system analysis? Before answering this question, we should mention that a 20 percent uncertainty in the convection heat transfer coefficient h in most cases is con- sidered "normal" and "expected." Assuming h to be constant and uniform is also an approximation of questionable validity, especially for irregular geome- tries. Therefore, in the absence of sufficient experimental data for the specific geometry under consideration, we cannot claim our results to be better than ±20 percent, even when Bi = 0. This being the case, introducing another source of uncertainty in the problem will hardly have any effect on the over- all uncertainty, provided that it is minor. It is generally accepted that lumped system analysis is applicable if Bi<0.1 When this criterion is satisfied, the temperatures within the body relative to the surroundings (i.e., T — T m ) remain within 5 percent of each other even for well-rounded geometries such as a spherical ball. Thus, when Bi < 0.1, the variation of temperature with location within the body will be slight and can reasonably be approximated as being uniform. cen58933_ch04.qxd 9/10/2002 9:12 AM Page 213 213 CHAPTER 4 The first step in the application of lumped system analysis is the calculation of the Biot number, and the assessment of the applicability of this approach. One may still wish to use lumped system analysis even when the criterion Bi < 0.1 is not satisfied, if high accuracy is not a major concern. Note that the Biot number is the ratio of the convection at the surface to con- duction within the body, and this number should be as small as possible for lumped system analysis to be applicable. Therefore, small bodies with high thermal conductivity are good candidates for lumped system analysis, es- pecially when they are in a medium that is a poor conductor of heat (such as air or another gas) and motionless. Thus, the hot small copper ball placed in quiescent air, discussed earlier, is most likely to satisfy the criterion for lumped system analysis (Fig. 4-6). Some Remarks on Heat Transfer in Lumped Systems To understand the heat transfer mechanism during the heating or cooling of a solid by the fluid surrounding it, and the criterion for lumped system analysis, consider this analogy (Fig. 4-7). People from the mainland are to go by boat to an island whose entire shore is a harbor, and from the harbor to their desti- nations on the island by bus. The overcrowding of people at the harbor de- pends on the boat traffic to the island and the ground transportation system on the island. If there is an excellent ground transportation system with plenty of buses, there will be no overcrowding at the harbor, especially when the boat traffic is light. But when the opposite is true, there will be a huge overcrowd- ing at the harbor, creating a large difference between the populations at the harbor and inland. The chance of overcrowding is much lower in a small is- land with plenty of fast buses. In heat transfer, a poor ground transportation system corresponds to poor heat conduction in a body, and overcrowding at the harbor to the accumulation of heat and the subsequent rise in temperature near the surface of the body relative to its inner parts. Lumped system analysis is obviously not applicable when there is overcrowding at the surface. Of course, we have disregarded radiation in this analogy and thus the air traffic to the island. Like passengers at the harbor, heat changes vehicles at the surface from convection to conduc- tion. Noting that a surface has zero thickness and thus cannot store any energy, heat reaching the surface of a body by convection must continue its journey within the body by conduction. Consider heat transfer from a hot body to its cooler surroundings. Heat will be transferred from the body to the surrounding fluid as a result of a tempera- ture difference. But this energy will come from the region near the surface, and thus the temperature of the body near the surface will drop. This creates a temperature gradient between the inner and outer regions of the body and ini- tiates heat flow by conduction from the interior of the body toward the outer surface. When the convection heat transfer coefficient h and thus convection heat transfer from the body are high, the temperature of the body near the surface will drop quickly (Fig. 4-8). This will create a larger temperature difference between the inner and outer regions unless the body is able to transfer heat from the inner to the outer regions just as fast. Thus, the magnitude of the maximum temperature difference within the body depends strongly on the ability of a body to conduct heat toward its surface relative to the ability of 15W/m 2 -°C V k n ° 3 , — -^ 4-£> = 0.02 m 1 f A s kD 2 6 Bi = jfc = j5xap2 = o,ooo 75 < p.] * 401 FIGURE 4-6 Small bodies with high thermal conductivities and low convection coefficients are most likely to satisfy the criterion for lumped system analysis. FIGURE 4-7 Analogy between heat transfer to a solid and passenger traffic to an island. Convection /? = 2000W/m 2 -°C FIGURE 4-8 When the convection coefficient h is high and k is low, large temperature differences occur between the inner and outer regions of a large solid. cen58933_ch04.qxd 9/10/2002 9:12 AM Page 214 214 HEAT TRANSFER the surrounding medium to convect this heat away from the surface. The Biot number is a measure of the relative magnitudes of these two competing effects. Recall that heat conduction in a specified direction n per unit surface area is expressed as q = —k dT/dn, where dT/dn is the temperature gradient and k is the thermal conductivity of the solid. Thus, the temperature distribution in the body will be uniform only when its thermal conductivity is infinite, and no such material is known to exist. Therefore, temperature gradients and thus temperature differences must exist within the body, no matter how small, in order for heat conduction to take place. Of course, the temperature gradient and the thermal conductivity are inversely proportional for a given heat flux. Therefore, the larger the thermal conductivity, the smaller the temperature gradient. Thermocouple Gas T ,h *" ^> Junction ~D = 1 mm T(t) FIGURE 4-9 Schematic for Example 4-1. EXAMPLE 4-1 Temperature Measurement by Thermocouples The temperature of a gas stream is to be measured by a thermocouple whose junction can be approximated as a 1-mm-diameter sphere, as shown in Fig. 4-9. The properties of the junction are k = 35 W/m • °C, p = 8500 kg/m 3 , and C p = 320 J/kg • C C, and the convection heat transfer coefficient between the junction and the gas is h = 210 W/m 2 • °C. Determine how long it will take for the thermocouple to read 99 percent of the initial temperature difference. SOLUTION The temperature of a gas stream is to be measured by a thermo- couple. The time it takes to register 99 percent of the initial A T is to be determined. Assumptions 1 The junction is spherical in shape with a diameter of D = 0.001 m. 2 The thermal properties of the junction and the heat transfer coeffi- cient are constant. 3 Radiation effects are negligible. Properties The properties of the junction are given in the problem statement. Analysis The characteristic length of the junction is V_ A, ttD 2 I D (0.001 m) = 1.67 X 10- Then the Biot number becomes hL c (210 W/m 2 ■ °C)(1.67 X 10- 4 m) Bi = X = 35W/m.°C = 0.001 <0.1 Therefore, lumped system analysis is applicable, and the error involved in this approximation is negligible. In order to read 99 percent of the initial temperature difference 7" - T, M between the junction and the gas, we must have T(t) - 7V„ 0.01 For example, when 7" = C C and T„ = 100 C C, a thermocouple is considered to have read 99 percent of this applied temperature difference when its reading indicates T(t) = 99°C. cen58933_ch04.qxd 9/10/2002 9:12 AM Page 215 The value of the exponent b is hA, h 210 W/m 2 • °C pC p V pC p L c (8500 kg/m 3 )(320 J/kg • °C)( 1.67 X l(T 4 m) We now substitute these values into Eq. 4-4 and obtain T(t) ~ T a _ kt 0.462 s" T, -> 0.01 (0.462 s~> which yields t = 10 s Therefore, we must wait at least 10 s for the temperature of the thermocouple junction to approach within 1 percent of the initial junction-gas temperature difference. Discussion Note that conduction through the wires and radiation exchange with the surrounding surfaces will affect the result, and should be considered in a more refined analysis. 215 CHAPTER 4 EXAMPLE 4-2 Predicting the Time of Death A person is found dead at 5 pm in a room whose temperature is 20°C. The tem- perature of the body is measured to be 25°C when found, and the heat trans- fer coefficient is estimated to be ft = 8 W/m 2 • °C. Modeling the body as a 30-cm-diameter, 1.70-m-long cylinder, estimate the time of death of that per- son (Fig. 4-10). SOLUTION A body is found while still warm. The time of death is to be estimated. Assumptions 1 The body can be modeled as a 30-cm-diameter, 1.70-m-long cylinder. 2 The thermal properties of the body and the heat transfer coefficient are constant. 3 The radiation effects are negligible. 4 The person was healthy(l) when he or she died with a body temperature of 37°C. Properties The average human body is 72 percent water by mass, and thus we can assume the body to have the properties of water at the average temperature of (37 + 25)/2 = 31°C; k J/kg • °C (Table A-9). 0.617 W/m • °C, p = 996 kg/m 3 , and C. = 4178 Analysis The characteristic length of the body is 9 T TTK- L ir(0.15m) 2 (1.7m) L =^ = c A s 2irr D L + 2-nrj 2ir(0.15 m)(1.7 m) + 2ir(0.15 m) 2 Then the Biot number becomes hL c (8 W/m 2 • °C)(0.0689 m) 0.0689 m Bi 0.617 W/m 0.89 > 0.1 FIGURE 4-10 Schematic for Example 4-2. 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.- 1.0 0.7 0.5 0.4 0.3 0.2 0.1 0.07 0.05 0.04 0.03 0.02 0.01 0.007 0.005 0.004 0.003 0.002 0.001 - 1 ' " i i i axil ' ■ fii '4iJ ' ' ' LL ' : : [ V[j~] — k ' '-H- ■ - ~"^0$^^^^^^^^^^^^S : ^^^^^^^^^^^^^^&t=S==b=i==:^^- ' " ~^ - «/ * ~i -Si^c'^ss>C""'^ i S--J , . fi/ * , ■ ■ | i , | : , | | . , | , , | I ■,..: 1 | ■ ^T j 1 ! ! ! | 1 ! ! 1 i 1 1 ! | j ii 1 ! 1 i l | ! 1 II!! Mill! ,,, I i ! i i i ' i 1 i i 1 : i i ! | i 1 ill Sbw>^i ; -iP^i i i ! ! i i i | ! i | i i i i i ] i i ! 1 1 ! mm M r-«^9n-» , 1 1 1 11 1! im :^!ti ;|t n tta^T ifeip ! tit PsN-JS^^""^ ^^t^T-^vV .^^"I^zlt^U^^ 5§§§fffipfc====; i \-— -i ^yh- ri — - -i a -r,- h— -i-i- JuTtlm-H^ mt-^--- +_T -E!-?^ >c + +- L L T3"ZdiJ - S \ \ ^JTOH — ^-_.^ +- t-*'-tg*4-i-I-—t , -n-^>T-+4- + W\ \±t:5jz S V - l- -n-^ m^ \ t^ T " + " N ^ ^rr- — AV \ ^T^ L -+T- \, \\ ^>„ -?o , u V VX sL^5t^5st mypl\V v \i\ s s vo ' A V^ AX As A Y \ >ft VVJV as SA \ v \ \\\\\ S\ V 2 * s \ A\V^ V^\ %N tf& i \ v\ \ i\ \ W\ N aNSF \ \ YSAVvVV n l ? ~" ITT 1 A~ ^ \ s i^ ' ' i ' y^ V* X "" ~ L " ~ ~- N s v u Si ? «\^ ^t s v ^ is? fe\ _S, Vitt^WjV ^ V VAt\ \ f S, S V V '^ _L \\S \ \m i^^\i\ \ X \ i m \ t i \ V Y V^^Cti J i X ^ V \aY Y uViJt.«Ai\^V V V \ Uv A a v\ S \ s irtr i \\ \ ^ v\a 5 ffiiltuVfcA $ V \ U U V \ $ V S V VU YYf 5 \ ^ \ S v \ \W\\ \vfn \ ^ ^ nv AAA Xa V - IV^Vr-lX^AV \\\ ™\\\\Wx\ \ \ M\\ lAjAaaAJaEaAAaAa 4 6 10 14 IS 22 26 30 50 T = at/t 2 70 100 120 1 50 300 400 500 600 700 (a) Midplane temperature (from M. P. Heisler) T-T„ Initially T=Tj 0+ — 2L- 1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.01 IT L = t iffi j _|_i| 1 1 1 — 0. 6 J — - JT — — 1 [If PL II Q 1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 BA^y •lYf, ; 57 " 'fl^y 04- / t' / / ' tt-t <tty 1 -t / / / tj-t 4 it a % t / / 14- t f^^ ll—t f / / tT4 k ^ t^ 1 ' 1 I / / tirt- -4T -J- 4T - / / ill ' / 'iiA i%4- 7 IT- -, / t]S o y Oil 1 - , r\ v ; ill! / MJ^J- -tt-jl 1111/ SV O'j- O'J-J o t Mi \i OlO'l c id <-v__ Iff / ■ _ r - cj r * / ki/ij it t A^/ ir -W -t- ty - /J ,r t i / / / J A v T / K A-t i w / B 1 v^ i / 4a7 1 i+ 7 I / J^ \/ '17 t'Z Jl / .< itz / / -Jz:^ Jffffi/ > ]>' / y Jt --'' iz -'M> '- 7ft' . S — IT+P — '- — " lit"^" - Plate il=S£ a^£ ■:.W^^: £;^l ;;#=•-" 0.1 1.0 1 k 10 1 00 10- io- i ci- io- 2 10-' 1 Bi 2 z = h 2 at/k 2 10 10 2 10 J 10 4 Bi hL (b) Temperature distribution (from M. P. Heisler) (c) Heat transfer (from H. Grober et al.) FIGURE 4-13 Transient temperature and heat transfer charts for a plane wall of thickness 2L initially at a uniform temperature T ( subjected to convection from both sides to an environment at temperature T^ with a convection coefficient of h. where m is the mass, V is the volume, p is the density, and C p is the specific heat of the body. Thus, Q max represents the amount of heat transfer for f — > °°. The amount of heat transfer Q at a finite time f will obviously be less than this cen58933_ch04.qxd 9/10/2002 9:12 AM Page 221 221 CHAPTER 4 1.0 0.7 0.5 0.4 0.3 0.2 0.1 0.07 0.05 0.04 0.03 0.02 0.01 0.007 0.005 0.004 0.003 0.002 0.001 -1 - H -'- - 1 1 1 1 1 1 ijSfcJ 1 ! i ' ! ! i ! I i i ! ! i I ! i Z-LL-l -f-4- J 1 _ ^^- : -i-i-i-i ! L_|_|.J j l_|_l i jI^J __; ~~ ^== ^£p§ >x <t - Ttmt VW 5 ^ t ! i m h~|~ —. --, ■— 5=fc £>*. i 1' c ^ T ' 35^S^ w^xloslSJ "S- - -+- sv ^» ™\lTO>sk ^ P \J > 4> -, [ I V- «/ ^ lu M\M\vtv \ s 4 \ <>s s \ \ \ N x x \ w s S v <* y X s 4-1 VKv v \ ~tfo. \ \ ^ < \ \ \ \ V S s — — — — \ / . ±r- — — — 't- -'/ft — — !*n 'o n ~T \\- u s s > t" i ' \ \ ' : ," \ v n ^ , P ■ \ \ i? V rv 17 s ■ \ \ ■■ Q \ s L ,' V o %, r \ \- s { <o * \ \ <y >j 3 > + 1 ?\illc t? % 5\ »- ■ <• I ' o » \ 5 \ " c . L V i i \ t \ 1 \ \ T \ \ \ T \ \ V .'■ \ \ J i \ ± \ i \ \ u\ \ \ (\ \ , — 1 \ \ ^~ \ ► — 4 6 8 10 14 18 22 26 30 50 70 x = at// - ? 100 120 140 150 250 350 (a) Centerline temperature (from M. P. Heisler) T-T a r„-r„ Initially T=Tj 0*- 1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.01 0.1 1.0 10 100 1 _ k Bi " hr (£>) Temperature distribution (from M. P. Heisler) rir _ 'in 'iltl y*' ' Mf\At 4 'WM 7 -W/~— -$£— — 0. 6-4 -*4 tt I[ jt — 0. 1 S 1 — 0. 1 L^ylind 1 ( er 1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 44 '/' l| [4 / / 1 ' / f /' TjI / / l4 / , _i / / -m / / / / -M 1 / -CSfQ — o -\ " l/fll / 1/ 3 fO — O 4 tiWii i / ii ' Tl 1 ii a ^iW / I ±M i Q ? 1 / / t / / I / / y jfl / / -4- - / / -1 / / / >1 1 /, f J / / / 1// tt / w + / / t 4l / / # Iff / -t / -J / / / , / 1 'W/ ' 1 i |ll bi ^~ W\ ^ 4 III ^4 " i tt Cylinder dm ^^ 10- 5 10" 4 10" 3 10" 2 10" 1 1 Bi 2 T = /7 2 ar/A: 2 10 10 2 LO 3 10 4 (c) Heat transfer (from H. Grober et al.) FIGURE 4-14 Transient temperature and heat transfer charts for a long cylinder of radius r initially at a uniform temperature T t subjected to convection from all sides to an environment at temperature T x with a convection coefficient of h. cen58933_ch04.qxd 9/10/2002 9:12 AM Page 222 222 HEAT TRANSFER T; ~ T^ 1.0 0.7 0.5 0.4 0.3 0.2 0.1 0.07 0.05 0.04 0.03 0.02 0.01 0.007 0.005 0.004 0.003 0.002 0.001 ------- — nt^t— — — — ^T x L-| Sphere X- 35^ -+- o M -+- v.-T - #/ * Y- - -S "4-4= 4^<V ] a\\ O^^x \ "X^V^ sKv^vvS ^rtvjsos -WrWS N ^yA4A\V i\ , r^ \ A^r \\?^p< s v ~rn\rr"W'0\ %^c^ — r^i ltd v o§^*&ra WsJv* SAXW I5$t+ -XS, JlV^ v\v\ V^K^K^a^A V \\x*S i c ^ \\\ M \K \ * ^ 5\f<^- ^ $ \\ > ~ W^\ K $ SV^Sfc I'Z' \ \ n\ u Txt^ \ \ 's \n rv\^s WW* \ \ V m K°v ■ >\ \ \ \ lu u&v? \ \ S A S v\ iv %M%?± \ ^ \ \ ffiM V\ V A ff^J^MsKSuw^ BVlvKSV \ OHvu\ ys ====lH^AA= S ^-E======^=^= »B±3 ^tS5 — r vffrr^r^ HSSt*ri=^ -^ wte^ ™^1TOF= \W\ *\\SSv > \ N JhAV P\^ \ li-O A Vl\^ ■ifo\-rt \ N \ \ v- vV 4 HP V V \ Mlfe V A^ ^ a^ \ v \ \ M \\ pU \ V v v V \ \\ LpjII l-J t V\ i_i_i_L-i_L- L L ^ L V BR V S S V VWAUWA \ \Vff3 \V vv y\w \ > v vV lllA___\__AlIKffiKA_ jMlmLiSffiiAiLSA... 0.5 1.0 1.5 2.5 3 4 5 6 7 8 9 10 20 x = at/r? 30 40 50 100 150 200 250 (a) Midpoint temperature (from M. P. Heisler) T-T rr T a -T_ 1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.01 !i ^ Mil i 1 t — 1 tt [ T [j j . LI n h J_] — 8- H o 1 1+1 (l Sphere II 0.1 1.0 -L= A. Bi fcr 10 100 10"' 1 Bi 2 x = h 2 at/k 2 (b) Temperature distribution (from M. P. Heisler) (c) Heat transfer (from H. Grober et al.) FIGURE 4-15 Transient temperature and heat transfer charts for a sphere of radius r initially at a uniform temperature T t subjected to convection from all sides to an environment at temperature T^ with a convection coefficient of h. maximum. The ratio Q/Q max is plotted in Figures 4-13c, 4-14c, and 4-15c against the variables Bi and h 2 at/k 2 for the large plane wall, long cylinder, and cen58933_ch04.qxd 9/10/2002 9:12 AM Page 223 sphere, respectively. Note that once the fraction of heat transfer QIQ mzx has been determined from these charts for the given t, the actual amount of heat transfer by that time can be evaluated by multiplying this fraction by <2 max - A negative sign for <2 max indicates that heat is leaving the body (Fig. 4-17). The fraction of heat transfer can also be determined from these relations, which are based on the one-term approximations already discussed: Plane wall: Cylinder: Sphere: Q Q sin A. , max/ wal , Q ■/i(Xi) e- 1 20 | max/ cy | A l xl 111 A 1 -36, O.sph sph M (4-17) (4-18) (4-19) The use of the Heisler/Grober charts and the one-term solutions already dis- cussed is limited to the conditions specified at the beginning of this section: the body is initially at a uniform temperature, the temperature of the medium surrounding the body and the convection heat transfer coefficient are constant and uniform, and there is no energy generation in the body. We discussed the physical significance of the Biot number earlier and indi- cated that it is a measure of the relative magnitudes of the two heat transfer mechanisms: convection at the surface and conduction through the solid. A small value of Bi indicates that the inner resistance of the body to heat con- duction is small relative to the resistance to convection between the surface and the fluid. As a result, the temperature distribution within the solid be- comes fairly uniform, and lumped system analysis becomes applicable. Recall that when Bi < 0.1, the error in assuming the temperature within the body to be uniform is negligible. To understand the physical significance of the Fourier number t, we ex- press it as (Fig. 4-18) at _ ^L-(l/L) Ar. L 2 pC p LVt AT The rate at which heat is conducted across L of a body of volume I? The rate at which heat is stored in a body of volume L 3 (4-20) 223 CHAPTER 4 (a) Finite convection coefficient (b) Infinite convection coefficient FIGURE 4-16 The specified surface temperature corresponds to the case of convection to an environment at T*, with a convection coefficient h that is infinite. Therefore, the Fourier number is a measure of heat conducted through a body relative to heat stored. Thus, a large value of the Fourier number indicates faster propagation of heat through a body. Perhaps you are wondering about what constitutes an infinitely large plate or an infinitely long cylinder. After all, nothing in this world is infinite. A plate whose thickness is small relative to the other dimensions can be modeled as an infinitely large plate, except very near the outer edges. But the edge effects on large bodies are usually negligible, and thus a large plane wall such as the wall of a house can be modeled as an infinitely large wall for heat transfer pur- poses. Similarly, a long cylinder whose diameter is small relative to its length can be analyzed as an infinitely long cylinder. The use of the transient tem- perature charts and the one-term solutions is illustrated in the following examples. cen58933_ch04.qxd 9/10/2002 9:12 AM Page 224 224 HEAT TRANSFER t = t = (a) Maximum heat transfer (t — » °°) Q Bi = .. Ifat k 2 Bi 2 x = (Grober chart) (b) Actual heat transfer for time t FIGURE 4-17 The fraction of total heat transfer Q/Qmax U P t° a specified time t is determined using the Grober charts. i y Fourier number: x : at L 2 *- stored FIGURE 4-18 Fourier number at time / can be viewed as the ratio of the rate of heat conducted to the rate of heat stored at that time. EXAMPLE 4-3 Boiling Eggs An ordinary egg can be approximated as a 5-cm-diameter sphere (Fig. 4-19). The egg is initially at a uniform temperature of 5 C C and is dropped into boil- ing water at 95°C. Taking the convection heat transfer coefficient to be h = 1200 W/m 2 ■ °C, determine how long it will take for the center of the egg to reach 70°C. SOLUTION An egg is cooked in boiling water. The cooking time of the egg is to be determined. Assumptions 1 The egg is spherical in shape with a radius of r = 2.5 cm. 2 Heat conduction in the egg is one-dimensional because of thermal symmetry about the midpoint. 3 The thermal properties of the egg and the heat transfer coefficient are constant. 4 The Fourier number is t > 0.2 so that the one-term approximate solutions are applicable. Properties The water content of eggs is about 74 percent, and thus the ther- mal conductivity and diffusivity of eggs can be approximated by those of water at the average temperature of (5 + 70)/2 = 37.5°C; k = 0.627 W/m • °C and a = k/pC p = 0.151 X 10- 5 m 2 /s (Table A-9). Analysis The temperature within the egg varies with radial distance as well as time, and the temperature at a specified location at a given time can be deter- mined from the Heisler charts or the one-term solutions. Here we will use the latter to demonstrate their use. The Biot number for this problem is Bi (1200 W/m 2 • °C)(0.025m) 0.627 W/m • °C 47.8 which is much greater than 0.1, and thus the lumped system analysis is not applicable. The coefficients X 1 and A 1 for a sphere corresponding to this Bi are, from Table 4-1, 3.0753, 1.9958 Substituting these and other values into Eq. 4-15 and solving for t gives r - 7L T; ~ T„ A,e~V 70 — 95 -> - „: = 1.9958e-' 30753 ^ 5-95 -> t = 0.209 which is greater than 0.2, and thus the one-term solution is applicable with an error of less than 2 percent. Then the cooking time is determined from the de- finition of the Fourier number to be tt 2 _ (0.209)(0.025 m) 2 ~° r ~ 0.151 X 10- 6 m 2 /s 865 s = 14.4 min Therefore, it will take about 15 min for the center of the egg to be heated from 5°C to 70°C. Discussion Note that the Biot number in lumped system analysis was defined differently as Bi = hL c /k = h{r/3)/k. However, either definition can be used in determining the applicability of the lumped system analysis unless Bi ~ 0.1. cen58933_ch04.qxd 9/10/2002 9:12 AM Page 225 225 CHAPTER 4 EXAMPLE 4-4 Heating of Large Brass Plates in an Oven In a production facility, large brass plates of 4 cm thickness that are initially at a uniform temperature of 20 C C are heated by passing them through an oven that is maintained at 500 C C (Fig. 4-20). The plates remain in the oven for a period of 7 min. Taking the combined convection and radiation heat transfer coefficient to be h = 120 W/m 2 ■ °C, determine the surface temperature of the plates when they come out of the oven. SOLUTION Large brass plates are heated in an oven. The surface temperature of the plates leaving the oven is to be determined. Assumptions 1 Heat conduction in the plate is one-dimensional since the plate is large relative to its thickness and there is thermal symmetry about the center plane. 2 The thermal properties of the plate and the heat transfer coefficient are constant. 3 The Fourier number is t > 0.2 so that the one-term approximate so- lutions are applicable. Properties The properties of brass at room temperature are k = 110 W/m • °C, p = 8530 kg/m 3 , C p = 380 J/kg • °C, and a = 33.9 X lO" 6 m 2 /s (Table A-3). More accurate results are obtained by using properties at average temperature. Analysis The temperature at a specified location at a given time can be de- termined from the Heisler charts or one-term solutions. Here we will use the charts to demonstrate their use. Noting that the half-thickness of the plate is L = 0.02 m, from Fig. 4-13 we have 0.46 1 k 100W/m-°C T - ■*■ o Bi hL (120 W/m 2 • °C)(0.02 m) - T„ at (33.9 X 10- 6 m 2 /s)(7 X 60 s) T t - - 71 T L 2 (0.02 m) 2 " 5 - e \ Also, 1 k Bi hL X L L L 45.8 0.99 Therefore, T - 71 T- 71 T ~ T„ T, - T„ T - 71 T t - 71 0.46 X 0.99 = 0.455 and T = 71 + 0.455(7, - 71) = 500 + 0.455(20 - 500) = 282°C Therefore, the surface temperature of the plates will be 282°C when they leave the oven. Discussion We notice that the Biot number in this case is Bi = 1/45.8 = 0.022, which is much less than 0.1. Therefore, we expect the lumped system analysis to be applicable. This is also evident from (7~- TJ/(T - 7"J = 0.99, which indicates that the temperatures at the center and the surface of the plate relative to the surrounding temperature are within 1 percent of each other. Egg X T. = 5°C j h = 1200W/m 2 -°C 7. = 95°C FIGURE 4-1 9 Schematic for Example 4-3. T rr _ = 500°C h = 120 W/m 2 -°C / 2L = 4cm Brass plate J, = 20°C FIGURE 4-20 Schematic for Example 4-4. cen58933_ch04.qxd 9/10/2002 9:12 AM Page 226 226 HEAT TRANSFER Noting that the error involved in reading the Heisler charts is typically at least a few percent, the lumped system analysis in this case may yield just as accurate results with less effort. The heat transfer surface area of the plate is 2/1, where A is the face area of the plate (the plate transfers heat through both of its surfaces), and the volume of the plate is V = (2L)A, where L is the half-thickness of the plate. The expo- nent b used in the lumped system analysis is determined to be hA s pC p V h(2A) h P C„(2LA) P C„L 120 W/m 2 • °C (8530 kg/m 3 )(380 J/kg • °C)(0.02 m) 0.00185 s- Then the temperature of the plate at t = 7 min = 420 s is determined from Tit) T(t) - 500 It yields 20 - 500 T(t ) = 279°C -(0.00185 s - ')(420s) which is practically identical to the result obtained above using the Heisler charts. Therefore, we can use lumped system analysis with confidence when the Biot number is sufficiently small. T x = 200°C h = 80W/m 2 -°C Stainless steel shaft T : = 600°C D = 20 cm FIGURE 4-21 Schematic for Example 4-5. EXAMPLE 4-5 Cooling of a Long Stainless Steel Cylindrical Shaft A long 20-cm-diameter cylindrical shaft made of stainless steel 304 comes out of an oven at a uniform temperature of 600°C (Fig. 4-21). The shaft is then al- lowed to cool slowly in an environment chamber at 200°C with an average heat transfer coefficient of h = 80 W/m 2 • °C. Determine the temperature at the cen- ter of the shaft 45 min after the start of the cooling process. Also, determine the heat transfer per unit length of the shaft during this time period. SOLUTION A long cylindrical shaft at 600°C is allowed to cool slowly. The cen- ter temperature and the heat transfer per unit length are to be determined. Assumptions 1 Heat conduction in the shaft is one-dimensional since it is long and it has thermal symmetry about the centerline. 2 The thermal properties of the shaft and the heat transfer coefficient are constant. 3 The Fourier number is t > 0.2 so that the one-term approximate solutions are applicable. Properties The properties of stainless steel 304 at room temperature are k = 14.9 W/m ■ °C, p = 7900 kg/m 3 , C p = 477 J/kg • °C, and a = 3.95 X 10~ 6 m 2 /s (Table A-3). More accurate results can be obtained by using properties at average temperature. Analysis The temperature within the shaft may vary with the radial distance r as well as time, and the temperature at a specified location at a given time can cen58933_ch04.qxd 9/10/2002 9:12 AM Page 227 be determined from the Heisler charts. Noting that the radius of the shaft is r = 0.1 m, from Fig. 4-14 we have 1 14.9 W/m • °C 1.86 Bi hr (80 W/m 2 • °C)(0.1 m) _ at _ (3.95 X 10- 6 m 2 /s)(45 X 60 s) J ~Ji ~ (0.1m) 2 1.07 0.40 and T = r„ + 0.4(7; - 7*.) = 200 + 0.4(600 - 200) = 360°C Therefore, the center temperature of the shaft will drop from 600°C to 360°C in 45 min. To determine the actual heat transfer, we first need to calculate the maximum heat that can be transferred from the cylinder, which is the sensible energy of the cylinder relative to its environment. Taking L = 1 m, m = pV = pirr 2 L = (7900 kg/m 3 )-jr(0.1 m) 2 (l m) = 248.2 kg e raax = mC p (T„ - T,) = (248.2 kg)(0477 kJ/kg ■ °C)(600 - 200)°C = 47,354 kJ The dimensionless heat transfer ratio is determined from Fig. 4-14cfor a long cylinder to be Bi 1 1 1/Bi 1.86 0.537 Q h 2 at Bi 2 T = (0.537) 2 (1.07) = 0.309 Q n 0.62 Therefore, Q = 0.62g„ 0.62 X (47,354 kJ) = 29,360 kJ which is the total heat transfer from the shaft during the first 45 min of the cooling. ALTERNATIVE SOLUTION We could also solve this problem using the one-term solution relation instead of the transient charts. First we find the Biot number _ hr _ (80 W/m 2 • °C)(0.1 m) _ Bl " T = 14.9 W/m- °C " °' 537 The coefficients X 1 and A x for a cylinder corresponding to this Bi are deter- mined from Table 4-1 to be Substituting these values into Eq. 4-14 gives T - 71 e„ rp -j-> /\\& 1.122e-<°- 970 >-< L07) = 0.41 227 CHAPTER 4 cen58933_ch04.qxd 9/10/2002 9:12 AM Page 22E 228 HEAT TRANSFER and thus T ± o = T a f o.4i (r, - - rj = 200 + 0.41(600 - 200) = 364°C The value of J^XJ for \ : = 0.97C is determined from Table 4-2 to be 0.430. Then the fractiona 1 heat tran sfer is determined from Eq. 4-18 to be Q *-max = l-26 UK) = 1 - 2 X 0.41 jJUJj = 0.636 and thus Q = 0.636e ma , = 0.636 X (47,354 kJ) = 30,120 kj Discussion Thes ight difference between the two results is due to the reading error of the charts CO \ Plane s surface T * \ h n / X FIGURE 4-22 Schematic of a semi-infinite body. 4-3 - TRANSIENT HEAT CONDUCTION IN SEMI-INFINITE SOLIDS A semi-infinite solid is an idealized body that has a single plane surface and extends to infinity in all directions, as shown in Fig. 4-22. This idealized body is used to indicate that the temperature change in the part of the body in which we are interested (the region close to the surface) is due to the thermal condi- tions on a single surface. The earth, for example, can be considered to be a semi-infinite medium in determining the variation of temperature near its sur- face. Also, a thick wall can be modeled as a semi-infinite medium if all we are interested in is the variation of temperature in the region near one of the sur- faces, and the other surface is too far to have any impact on the region of in- terest during the time of observation. Consider a semi-infinite solid that is at a uniform temperature T t . At time t = 0, the surface of the solid at x = is exposed to convection by a fluid at a constant temperature T m , with a heat transfer coefficient h. This problem can be formulated as a partial differential equation, which can be solved analyti- cally for the transient temperature distribution T(x, t). The solution obtained is presented in Fig. 4-23 graphically for the nondimensionalized temperature defined as 1 - 6(x, f) = 1 T(x, t) T(x,t)-T t (4-21) against the dimensionless variable x/(2\/at) for various values of the param- eter h\faXlk. Note that the values on the vertical axis correspond to x = 0, and thus rep- resent the surface temperature. The curve hA/ai/k = c° corresponds to /; — > °°, which corresponds to the case of specified temperature T m at the surface at x = 0. That is, the case in which the surface of the semi-infinite body is sud- denly brought to temperature T«, at t = and kept at T„ at all times can be han- dled by setting h to infinity. The specified surface temperature case is closely cen58933_ch04.qxd 9/10/2002 9:12 AM Page 229 1.0 0.5 0.4 0.3 0.2 0.1 0.05 -i 0.04 0.03 0.02 0.01 229 CHAPTER 4 Ambient Ax, t)\ - °.J ^p . 1 ^ ^*0 0j \ 0.25 0.5 0.75 1.0 1.25 1.5 2Var FIGURE 4-23 Variation of temperature with position and time in a semi-infinite solid initially at T t subjected to convection to an environment at T„ with a convection heat transfer coefficient of h (from P. J. Schneider, Ref. 10). approximated in practice when condensation or boiling takes place on the surface. For a finite heat transfer coefficient /;, the surface temperature approaches the fluid temperature T m as the time t approaches infinity. The exact solution of the transient one-dimensional heat conduction prob- lem in a semi-infinite medium that is initially at a uniform temperature of T, and is suddenly subjected to convection at time t = has been obtained, and is expressed as T(x, t) erfc hx , 3 iy h 2 at k 2 erfc h\/ai 2y/ai (4-22) where the quantity erfc (£) is the complementary error function, defined as erfc(£) = 1 --^= f e-" 2 du Vtt Jo (4-23) Despite its simple appearance, the integral that appears in the above relation cannot be performed analytically. Therefore, it is evaluated numerically for different values of £, and the results are listed in Table 4-3. For the special case of h — > °o, the surface temperature T s becomes equal to the fluid temper- ature r„, and Eq. 4-22 reduces to T(x, f) erfc (4-24) cen58933_ch04.qxd 9/10/2002 9:12 AM Page 23C 230 HEAT TRANSFER TABLE 4-3 The complementary error fu iction £ erfc (£) £ erfc (Q £ erfc (g) 6 erfc (Q 6 erfc (Q e erfc (Q 0.00 1.00000 0.38 0.5910 0.76 0.2825 1.14 0.1069 1.52 0.03159 1.90 0.00721 0.02 0.9774 0.40 0.5716 0.78 0.2700 1.16 0.10090 1.54 0.02941 1.92 0.00662 0.04 0.9549 0.42 0.5525 0.80 0.2579 1.18 0.09516 1.56 0.02737 1.94 0.00608 0.06 0.9324 0.44 0.5338 0.82 0.2462 1.20 0.08969 1.58 0.02545 1.96 0.00557 0.08 0.9099 0.46 0.5153 0.84 0.2349 1.22 0.08447 1.60 0.02365 1.98 0.00511 0.10 0.8875 0.48 0.4973 0.86 0.2239 1.24 0.07950 1.62 0.02196 2.00 0.00468 0.12 0.8652 0.50 0.4795 0.88 0.2133 1.26 0.07476 1.64 0.02038 2.10 0.00298 0.14 0.8431 0.52 0.4621 0.90 0.2031 1.28 0.07027 1.66 0.01890 2.20 0.00186 0.16 0.8210 0.54 0.4451 0.92 0.1932 1.30 0.06599 1.68 0.01751 2.30 0.00114 0.18 0.7991 0.56 0.4284 0.94 0.1837 1.32 0.06194 1.70 0.01612 2.40 0.00069 0.20 0.7773 0.58 0.4121 0.96 0.1746 1.34 0.05809 1.72 0.01500 2.50 0.00041 0.22 0.7557 0.60 0.3961 0.98 0.1658 1.36 0.05444 1.74 0.01387 2.60 0.00024 0.24 0.7343 0.62 0.3806 1.00 0.1573 1.38 0.05098 1.76 0.01281 2.70 0.00013 0.26 0.7131 0.64 0.3654 1.02 0.1492 1.40 0.04772 1.78 0.01183 2.80 0.00008 0.28 0.6921 0.66 0.3506 1.04 0.1413 1.42 0.04462 1.80 0.01091 2.90 0.00004 0.30 0.6714 0.68 0.3362 1.06 0.1339 1.44 0.04170 1.82 0.01006 3.00 0.00002 0.32 0.6509 0.70 0.3222 1.08 0.1267 1.46 0.03895 1.84 0.00926 3.20 0.00001 0.34 0.6306 0.72 0.3086 1.10 0.1198 1.48 0.03635 1.86 0.00853 3.40 0.00000 0.36 0.6107 0.74 0.2953 1.12 0.1132 1.50 0.03390 1.88 0.00784 3.60 0.00000 This solution corresponds to the case when the temperature of the exposed surface of the medium is suddenly raised (or lowered) to T s at t = and is maintained at that value at all times. Although the graphical solution given in Fig. 4-23 is a plot of the exact analytical solution given by Eq. 4-23, it is sub- ject to reading errors, and thus is of limited accuracy. ,T. = -10°C Soil Water pipe r ; =i5°c FIGURE 4-24 Schematic for Example 4-6. EXAMPLE 4-6 Minimum Burial Depth of Water Pipes to Avoid Freezing In areas where the air temperature remains below 0°C for prolonged periods of time, the freezing of water in underground pipes is a major concern. Fortu- nately, the soil remains relatively warm during those periods, and it takes weeks for the subfreezing temperatures to reach the water mains in the ground. Thus, the soil effectively serves as an insulation to protect the water from subfreezing temperatures in winter. The ground at a particular location is covered with snow pack at -10°C for a continuous period of three months, and the average soil properties at that loca- tion are k = 0.4 W/m • °C and a = 0.15 X 10~ 6 m 2 /s (Fig. 4-24). Assuming an initial uniform temperature of 15 C C for the ground, determine the minimum burial depth to prevent the water pipes from freezing. SOLUTION The water pipes are buried in the ground to prevent freezing. The minimum burial depth at a particular location is to be determined. Assumptions 1 The temperature in the soil is affected by the thermal condi- tions at one surface only, and thus the soil can be considered to be a semi- infinite medium with a specified surface temperature of - 10°C. 2 The thermal properties of the soil are constant. cen58933_ch04.qxd 9/10/2002 9:12 AM Page 231 Properties The properties of the soil are as given in the problem statement. Analysis The temperature of the soil surrounding the pipes will be 0°C after three months in the case of minimum burial depth. Therefore, from Fig. 4-23, we have hVoit T(x,t) - T 1 ~ ' ' - = 1 (since h —> *■) 0-(-10) H 15 - (-10) 0.6 2 Vat 0.36 We note that t = (90 days)(24 h/day)(3600 s/h) = 7.78 X 10 6 s and thus x = 2i Vaf = 2 X 0.36V(0.15 X lO" 6 m 2 /s)(7.78 X 10 6 s) = 0.77 m Therefore, the water pipes must be buried to a depth of at least 77 cm to avoid freezing under the specified harsh winter conditions. ALTERNATIVE SOLUTION The solution of this problem could also be deter- mined from Eq. 4-24: T(x,t) erfc 2Vat, 0-15 -10 - 15 erfc iVat 0.60 The argument that corresponds to this value of the complementary error func- tion is determined from Table 4-3 to be £ = 0.37. Therefore, x = 2£ vaf = 2 X 0.37V(0.15 X lO" 6 m 2 /s)(7.78 X 10 6 s) = 0.80 m Again, the slight difference is due to the reading error of the chart. 231 CHAPTER 4 h T(r,t) Heat transfer (a) Long cylinder 4-4 - TRANSIENT HEAT CONDUCTION IN MULTIDIMENSIONAL SYSTEMS The transient temperature charts presented earlier can be used to determine the temperature distribution and heat transfer in one-dimensional heat conduction problems associated with a large plane wall, a long cylinder, a sphere, and a semi-infinite medium. Using a superposition approach called the product solution, these charts can also be used to construct solutions for the two- dimensional transient heat conduction problems encountered in geometries such as a short cylinder, a long rectangular bar, or a semi-infinite cylinder or plate, and even three-dimensional problems associated with geometries such as a rectangular prism or a semi-infinite rectangular bar, provided that all sur- faces of the solid are subjected to convection to the same fluid at temperature h T(r,x,t) Heat ' transfer (b) Short cylinder (two-dimensional) FIGURE 4-25 The temperature in a short cylinder exposed to convection from all surfaces varies in both the radial and axial directions, and thus heat is transferred in both directions. cen58933_ch04.qxd 9/10/2002 9:12 AM Page 232 232 HEAT TRANSFER Plane wall ^1 Long cylinder FIGURE 4-26 A short cylinder of radius r and height a is the intersection of a long cylinder of radius r and a plane wall of thickness a. Plane wall Plane wall hH FIGURE 4-27 A long solid bar of rectangular profile a X b is the intersection of two plane walls of thicknesses a and b. T„, with the same heat transfer coefficient h, and the body involves no heat generation (Fig. 4-25). The solution in such multidimensional geometries can be expressed as the product of the solutions for the one-dimensional geome- tries whose intersection is the multidimensional geometry. Consider a short cylinder of height a and radius r initially at a uniform tem- perature r,. There is no heat generation in the cylinder. At time t = 0, the cylinder is subjected to convection from all surfaces to a medium at temper- ature r„ with a heat transfer coefficient h. The temperature within the cylin- der will change with x as well as r and time t since heat transfer will occur from the top and bottom of the cylinder as well as its side surfaces. That is, T = T(r, x, t) and thus this is a two-dimensional transient heat conduction problem. When the properties are assumed to be constant, it can be shown that the solution of this two-dimensional problem can be expressed as T(r, x, t) - T a short cylinder T(x, t) plane w;ill T(r, t) infinite cylinder (4-25) That is, the solution for the two-dimensional short cylinder of height a and radius r is equal to the product of the nondimensionalized solutions for the one-dimensional plane wall of thickness a and the long cylinder of radius r , which are the two geometries whose intersection is the short cylinder, as shown in Fig. 4-26. We generalize this as follows: the solution for a multi- dimensional geometry is the product of the solutions of the one-dimensional geometries whose intersection is the multidimensional body. For convenience, the one-dimensional solutions are denoted by 0waii(*. r ) (r,f) J c\ I ,-i„f(*, t) (T(x, t) - T a [ T,~ r„ (T(r, t) -rj \ T,~ r- , (T(x, t) - r. plane wall infinite cylinder T, - T m (4-26) For example, the solution for a long solid bar whose cross section is an a X b rectangle is the intersection of the two infinite plane walls of thicknesses a and b, as shown in Fig. 4-27, and thus the transient temperature distribution for this rectangular bar can be expressed as T(x, y, t) W-*. OQwaiiCy. (4-27) The proper forms of the product solutions for some other geometries are given in Table 4—4. It is important to note that the x-coordinate is measured from the surface in a semi-infinite solid, and from the midplane in a plane wall. The ra- dial distance r is always measured from the centerline. Note that the solution of a two-dimensional problem involves the product of two one-dimensional solutions, whereas the solution of a three-dimensional problem involves the product of three one-dimensional solutions. A modified form of the product solution can also be used to determine the total transient heat transfer to or from a multidimensional geometry by using the one-dimensional values, as shown by L. S. Langston in 1982. The cen58933_ch04.qxd 9/10/2002 9:13 AM Page 233 233 CHAPTER 4 TABLE 4-4 Multidimensional solutions expressed as products of one-dimensional solutions for bodies that are initially at a uniform temperature 7j and exposed to convection from all surfaces to a medium at T« 9(''.0 = e cyl (n() Infinite cylinder Hw.') = e cy|(f ,oe s „, inf (x,0 Semi-infinite cylinder Hx,r,t) = e cyl (r,t)() wM (x,f) Short cylinder Semi-infinite medium K.v,v.O = e semWnf (x,f)e semWnf ^o Quarter-infinite medium B(x,y,z,t) = 9 semi-inf (' V > f )8 scmi -i„f & e s emi-inf fc f ) Corner region of a large medium 2L e(*,o = e wall teO Infinite plate (or plane wall) 2L H^y.O = e wall fcOe semi . m ,Cv.O Semi-infinite plate Hx,y,z,t) = ^wall^Oe^.^O'.Oe^.infCZ.O Quarter-infinite plate i j ;- / / if ife^ /\ 1 1 v H^y.O = e wall (x,oe wdl (y,f) Infinite rectangular bar e(jc,y,z,f) = e w a ii(^')e wall Cv,oe scmi , nf (z,o Semi-infinite rectangular bar Q(x,y,z,t) = e „an^oe wall (.v,oe wall ( Z ,o Rectangular parallelepiped cen58933_ch04.qxd 9/10/2002 9:13 AM Page 234 234 HEAT TRANSFER transient heat transfer for a two-dimensional geometry formed by the inter- section of two one-dimensional geometries 1 and 2 is _e *iraa total, 2D _Q_ Cm. _Q_ _Q_ (4-28) Transient heat transfer for a three-dimensional body formed by the inter- section of three one-dimensional bodies 1, 2, and 3 is given by _Q_ Sc m a _Q_ iima _Q_ i/ma _Q_ fcraa _Q_ *£ma _Q_ sima (4-29) The use of the product solution in transient two- and three-dimensional heat conduction problems is illustrated in the following examples. r„ = 25°C h = 60W/m 2 -°C T t = 120 D C FIGURE 4-28 Schematic for Example 4-7. EXAMPLE 4-7 Cooling of a Short Brass Cylinder " A short brass cylinder of diameter D = 10 cm and height H = 12 cm is initially I at a uniform temperature 7) = 120°C. The cylinder is now placed in atmo- spheric air at 25°C, where heat transfer takes place by convection, with a heat transfer coefficient of h = 60 W/m 2 ■ °C. Calculate the temperature at (a) the center of the cylinder and (b) the center of the top surface of the cylinder 15 min after the start of the cooling. SOLUTION A short cylinder is allowed to cool in atmospheric air. The temper- atures at the centers of the cylinder and the top surface are to be determined. Assumptions 1 Heat conduction in the short cylinder is two-dimensional, and thus the temperature varies in both the axial x- and the radial r-directions. 2 The thermal properties of the cylinder and the heat transfer coefficient are constant. 3 The Fourier number is t > 0.2 so that the one-term approximate solutions are applicable. Properties The properties of brass at room temperature are k= 110 W/m • °C and a = 33.9 X 10~ 5 m 2 /s (Table A-3). More accurate results can be obtained by using properties at average temperature. Analysis (a) This short cylinder can physically be formed by the intersection of a long cylinder of radius r = 5 cm and a plane wall of thickness 2L = 12 cm, as shown in Fig. 4-28. The dimensionless temperature at the center of the plane wall is determined from Figure 4-13a to be at (3.39 X KT 5 m 2 /s)(900s) 1 T = ^ = r^ = 8.48 U (0.06 m) 2 J_ == _L- 110 W/m • °c ' 6wall(0,O = T(0,t)-T a Bi hL (60 w/m 2 . °c)(0.06 m ) . cen58933_ch04.qxd 9/10/2002 9:13 AM Page 235 Similarly, at the center of the cylinder, we have at (3.39 X Kr 5 m 2 /s)(900s) r 2 (0.05 m) 2 1 _ k _ HOW/m • °C Bi hr o (60 W/m 2 • °C)(0.05 m) dte ore, /r(o,o,o-r„\ 1 'Y T 1 Ishort "wall' cylinder 12.2 36.7 T(0, t)-T a ■ e cy[ (o,o= T _ T " = o.5 and T(0,0,t) (0, t ) X 6 cyl (0, t ) = 0.8 X 0.5 = 0.4 0.4(7, - r.) = 25 + 0.4(120 - 25) = 63°C This is the temperature at the center of the short cylinder, which is also the cen- ter of both the long cylinder and the plate. (b) The center of the top surface of the cylinder is still at the center of the long cylinder (r = 0), but at the outer surface of the plane wall (x = L). Therefore, we first need to find the surface temperature of the wall. Noting that x = L = 0.06 m, x _ 0.06 m L ~ 0.06 m 1 _L = A = 110 W/m ■ °c Bi hL (60 w/m 2 ■ °C)(0.06 m) 30.6 T -T x . 0.98 Then 9 wa ii(£> T{L,t)-T^ (T(L,t)-TA(T -T, T t -T, Therefore, T(L, 0, t)-T = T 71 T- — T X 0.8 = 0.784 Ti-Tn shor. = 6 wall (L, f )0 cyl (0, t) = 0.784 X 0.5 = 0.392 cylinder and T(L, 0, t) = T a + 0.392(7; - r„) = 25 + 0.392(120 - 25) = 62.2°C which is the temperature at the center of the top surface of the cylinder. 235 CHAPTER 4 J EXAMPLE 4-8 Heat Transfer from a Short Cylinder Determine the total heat transfer from the short brass cylinder , kg/m 3 , C p = 0.380 kJ/kg • °C) discussed in Example 4-7. (P = = 8530 cen58933_ch04.qxd 9/10/2002 9:13 AM Page 236 236 HEAT TRANSFER SOLUTION We first determine the maximum heat that can be transferred from the cylinder, which is the sensible energy content of the cylinder relative to its environment: m = pV = pirr 2 L = (8530 kg/m 3 )Tr(0.05 m) 2 (0.06 m) = 4.02 kg g raax = mC p (T, - r„) = (4.02 kg)(0.380 kJ/kg • °C)(120 - 25)°C = 145.1 kJ Then we determine the dimensionless heat transfer ratios for both geometries. For the plane wall, it is determined from Fig. 4-13c to be Bi = IM = 3^6 = °-° 327 h 2 at Bi 2 T = (0.0327) 2 (8.48) = 0.0091 !cma 0.23 plane Similarly, for the cylinder, we have Bi = W = i = °- 0272 1/Bi 36.7 h 2 at Bi 2 T = (0.0272) 2 (12.2) = 0.0090 0.47 nfinite cylinder Then the heat transfer ratio for the short cylinder is, from Eq. 4-28, ^ max ' short cyl V^max/j VtJmax/, _G = 0.23 + 0.47(1 - 0.23) = 0.592 Therefore, the total heat transfer from the cylinder during the first 15 min of cooling is Q = 0.592g raax = 0.592 X (145.1 kJ) = 85.9 kJ EXAMPLE 4-9 Cooling of a Long Cylinder by Water A semi-infinite aluminum cylinder of diameter D = 20 cm is initially at a uni- form temperature 7", = 200 C C. The cylinder is now placed in water at 15°C where heat transfer takes place by convection, with a heat transfer coefficient of h = 120 W/m 2 • °C. Determine the temperature at the center of the cylinder 15 cm from the end surface 5 min after the start of the cooling. SOLUTION A semi-infinite aluminum cylinder is cooled by water. The tem- perature at the center of the cylinder 15 cm from the end surface is to be determined. Assumptions 1 Heat conduction in the semi-infinite cylinder is two- dimensional, and thus the temperature varies in both the axial x- and the radial r-directions. 2 The thermal properties of the cylinder and the heat transfer co- efficient are constant. 3 The Fourier number is t > 0.2 so that the one-term approximate solutions are applicable. cen58933_ch04.qxd 9/10/2002 9:13 AM Page 237 Properties The properties of aluminum at room temperature are k = 237 W/m ■ °C and a = 9.71 X 10~ 6 m 2 /s (Table A-3). More accurate results can be obtained by using properties at average temperature. Analysis This semi-infinite cylinder can physically be formed by the inter- section of an infinite cylinder of radius r = 10 cm and a semi-infinite medium, as shown in Fig. 4-29. We will solve this problem using the one-term solution relation for the cylin- der and the analytic solution for the semi-infinite medium. First we consider the infinitely long cylinder and evaluate the Biot number: Bi hr B _ (120 W/m 2 • °C)(0.1 m) T~ 237 W/m ■ °C 0.05 The coefficients k 1 and A 1 for a cylinder corresponding to this Bi are deter- mined from Table 4-1 to be \ x = 0.3126 and A 1 = 1.0124. The Fourier num- ber in this case is at _ (9.71 X 10- 5 m 2 /s)(5 X 60 s) ~7;~ (0.1 m) 2 2.91 >0.2 and thus the one-term approximation is applicable. Substituting these values into Eq. 4-14 gives ' = B_ 1 (0,0 =A x e- 1.0124e-<°- 3126 > 2 ( 21)1 > = 0.762 The solution for the semi-infinite solid can be determined from hx , h 2 at 1 - e s emi-inftM) = erfc exp iy + > erfc hVal at First we determine the various quantities in parentheses x 0.15 m 0.44 2Vaf 2V(9.71 X 10- 5 m 2 /s)(5 X 60s) hVat _ (120 W/m 2 • °C)V(9.71 X lO" 5 m 2 /s)(300 s) k 237 W/m hx (120 W/m 2 • °C)(0.15m) h 2 at (hVai k 2 237 W/m ■ °C 2 °c 0.0759 0.086 (0.086) 2 = 0.0074 Substituting and evaluating the complementary error functions from Table 4-3, e semHnf (Jc, t) = 1 - erfc (0.44) + exp (0.0759 + 0.0074) erfc (0.44 + 0.086) = 1 - 0.5338 + exp (0.0833) X 0.457 = 0.963 Now we apply the product solution to get T(jc, 0, t) - T- semi-infinite cylinder 6 S cmMnf(*> O9cyi(0, O = 0.963 X 0.762 = 0.734 237 CHAPTER 4 I Tf = 200°C I D = 20cm r« = i5°c h =120W/m 2 -°C x = 15 cm FIGURE 4-29 Schematic for Example 4-9. 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). cen58933_ch04.qxd 9/10/2002 9:13 AM Page 244 244 HEAT TRANSFER Time in days FIGURE 4-39 Variation of tenderness of meat stored at 2°C with time after slaughter. Meat freezer Air -40 to -30°C 2.5 to 5 m/s FIGURE 4-40 The freezing time of meat can be reduced considerably by using low temperature air at high velocity. TABLE 4-6 Storage life of frozen meat products at different storage temperatures (from ASHRAE Handbook: Refrigeration, Chap. 10, Table 7) Storage Life, Months Temperature Product -12°C -18°C-23°C Beef 4-12 6-18 12-24 Lamb 3-8 6-16 12-18 Veal 3-4 4-14 8 Pork 2-6 4-12 8-15 Chopped beef 3-4 4-6 8 Cooked foods 2-3 2-4 protein collagen, which is the main component of the connective tissue, softens and dissolves in hot and moist environments and gradually trans- forms into gelatin, and tenderizes the meat. The old saying "one should either cook an animal immediately after slaughter or wait at least two days" has a lot of truth in it. The biomechan- ical reactions in the muscle continue after the slaughter until the energy supplied to the muscle to do work diminishes. The muscle then stiffens and goes into rigor mortis. This process begins several hours after the animal is slaughtered and continues for 12 to 36 h until an enzymatic action sets in and tenderizes the connective tissue, as shown in Figure 4-39. It takes about seven days to complete tenderization naturally in storage facilities maintained at 2°C. Electrical stimulation also causes the meat to be tender. To avoid toughness, fresh meat should not be frozen before rigor mortis has passed. You have probably noticed that steaks are tender and rather tasty when they are hot but toughen as they cool. This is because the gelatin that formed during cooking thickens as it cools, and meat loses its tenderness. So it is no surprise that first-class restaurants serve their steak on hot thick plates that keep the steaks warm for a long time. Also, cooking softens the connective tissue but toughens the tender muscle fibers. Therefore, barbe- cuing on low heat for a long time results in a tough steak. Variety meats intended for long-term storage must be frozen rapidly to reduce spoilage and preserve quality. Perhaps the first thought that comes to mind to freeze meat is to place the meat packages into the freezer and wait. But the freezing time is too long in this case, especially for large boxes. For example, the core temperature of a 4-cm-deep box containing 32 kg of variety meat can be as high as 16°C 24 h after it is placed into a — 30°C freezer. The freezing time of large boxes can be shortened consid- erably by adding some dry ice into it. A more effective method of freezing, called quick chilling, involves the use of lower air temperatures, —40 to — 30°C, with higher velocities of 2.5 m/s to 5 m/s over the product (Fig. 4-40). The internal temperature should be lowered to — 4°C for products to be transferred to a storage freezer and to — 18°C for products to be shipped immediately. The rate of freezing depends on the package material and its insulating properties, the thickness of the largest box, the type of meat, and the capacity of the re- frigeration system. Note that the air temperature will rise excessively dur- ing initial stages of freezing and increase the freezing time if the capacity of the system is inadequate. A smaller refrigeration system will be adequate if dry ice is to be used in packages. Shrinkage during freezing varies from about 0.5 to 1 percent. Although the average freezing point of lean meat can be taken to be — 2°C with a latent heat of 249 kJ/kg, it should be remembered that freez- ing occurs over a temperature range, with most freezing occurring between — 1 and — 4°C. Therefore, cooling the meat through this temperature range and removing the latent heat takes the most time during freezing. Meat can be kept at an internal temperature of —2 to — 1°C for local use and storage for under a week. Meat must be frozen and stored at much lower temperatures for long-term storage. The lower the storage tempera- ture, the longer the storage life of meat products, as shown in Table 4-6. cen58933_ch04.qxd 9/10/2002 9:13 AM Page 245 245 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. cen58933_ch04.qxd 9/10/2002 9:13 AM Page 252 252 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 cen58933_ch04.qxd 9/10/2002 9:13 AM Page 254 254 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. cen58933_ch04.qxd 9/10/2002 9:13 AM Page 256 256 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 cen58933_ch04.qxd 9/10/2002 9:13 AM Page 257 257 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 cen58 93 3_ch05.qxd 9/4/2002 11:41 AM Page 266 266 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. cen58 93 3_ch05.qxd 9/4/2002 11:41 AM Page 274 274 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) cen58 93 3_ch05.qxd 9/4/2002 11:41 AM Page 276 276 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. cen58 93 3_ch05.qxd 9/4/2002 11:41 AM Page 27E 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 cen58 93 3_ch05.qxd 9/4/2002 11:41 AM Page 284 284 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 cen58 93 3_ch05.qxd 9/4/2002 11:42 AM Page 289 289 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. cen58 93 3_ch05.qxd 9/4/2002 11:42 AM Page 292 292 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. cen58 93 3_ch05.qxd 9/4/2002 11:42 AM Page 294 294 HEAT TRANSFER Canceling the surface area A and multiplying by Ax/k, it simplifies to s Ajc 2 a -P- 7 1 of j_ f _i_ ^HL — ^^ A ( r ri J r\ Ti\ '«-] Z - L m ■ ~ L m + 1 ~ K n\t " (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) cen58 93 3_ch05.qxd 9/4/2002 11:42 AM Page 296 296 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. cen58 93 3_ch05.qxd 9/4/2002 11:42 AM Page 298 298 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 + ^- cen58 93 3_ch05.qxd 9/4/2002 11:42 AM Page 307 (c) Node 3. (Boundary node subjected to convection on two sides, Fig. 5-53a) AyTl-Ti . AxAy Ax Ay T^ 1 - 7^ Dividing by klA, simplifying, and solving for 7"3 +1 gives At T- + [ = I 1 - 4t - 4t^ ] Ti + 2t( Ti + T£ + 2jT^ + hi, 2k (d) Node 4. (On the insulated boundary, and can be treated as an interior node, Fig. 5-536). Noting that T 10 = 90°C, Eq. 5-60 gives U 1 (1 - 4t) T\ + t T[ + 2Tj, + 90 + (e) Node 5. (Interior node, Fig. 5-54a). Noting that T n = 90°C, Eq. 5-60 gives Ti, + ' = (1 - 4t) Ti + t I H + r^ + Ti + 90 + &/ 2 (f) Node 6. (Boundary node subjected to convection on two sides, Fig. 5-546) (Ax Ay\ Av Tj - Ti T' n - Ti k T{ - TI Ax Ti - Tl . 3 AxAy 3 AxAy Tf - Ti + T^y~ + S6 ^^ = P ^^ C At Dividing by 3k/A, simplifying, and solving for 7"g +1 gives tw'+I J 6 l-4 T -4x|]7 hi, 2Tj + 4Ti + 2Tj + 4 X 90 + 4 -f T w + 3 ^- [g) Node 7. (Boundary node subjected to convection, Fig. 5-55) Ay Ti - Tj T( 3 - Tj hAx(T^ -Tj) + k-^- -=-; + kAx - 2 Ax Ay U ~ Tj Ay Ay Ay Tj +I - Tj 2 Ax +^Ax T = pAx T C- A[ Dividing by kl2, simplifying, and solving for Tj +1 gives 1 -4T-2Ty ] 2hi , g 7 r- Tl + Ti + 2 X 90 + —r- T„ + 307 CHAPTER 5 h, r« r Mirror (5) *,r~ t EH — I ♦ 10 (a) Node 3 (b) Node 4 FIGURE 5-53 Schematics for energy balances on the volume elements of nodes 3 and 4. m ii 12 (a) Node 5 (7?) Node 6 FIGURE 5-54 Schematics for energy balances on the volume elements of nodes 5 and 6. h, r„ 1\ 1 13 1 ^ ^ 1r • 15 FIGURE 5-55 Schematics for energy balances on the volume elements of nodes 7 and 9. cen5 8 93 3_ch05.qxd 9/4/2002 11:42 AM Page 3C 308 HEAT TRANSFER (h) Node 8. This node is identical to node 7, and the finite difference formula- tion of this node can be obtained from that of node 7 by shifting the node num- bers by 1 (i.e., replacing subscript m by subscript m + 1). It gives 1 - 4t - 2t hi n + T 2hl, Tj + Tj + 2 X 90 + -T- 7^ + (/) Node 9. (Boundary node subjected to convection on two sides, Fig. 5-55) ,Ax,_ _,, , . Ay A.t Tj 5 - Tj h T (T«,-T$) + q R — k-—^- kAyTi Ti , . Ax^J AxAy T<' + 2 Ax 6J 2 2 Dividing by klA, simplifying, and solving for 7"g +1 gives p TT c " At T' + l *9 hl\ I 4rI hi i - 4t - 2t j r 9 " + 2t m + 90 + ^-+ ^r„ 2fe This completes the finite difference formulation of the problem. Next we need to determine the upper limit of the time step Af from the stability criterion, which requires the coefficient of T m in the 7~^ +1 expression (the primary coeffi- cient) to be greater than or equal to zero for all nodes. The smallest primary co- efficient in the nine equations here is the coefficient of Ti in the expression, and thus the stabil