Heat conduction 3rd edition

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Heat conduction 3rd edition

  1. 1. http://avibert.blogspot.comHeat Conduction
  2. 2. Latif M. JijiHeat ConductionThird Editionhttp://avibert.blogspot.comABC
  3. 3. Professor Latif M. JijiDepartment of Mechanical EngineeringGrove School of EngineeringThe City College ofThe City University of New YorkNew York, New York 10031USAE-Mail: jiji@ccny.cuny.eduISBN 978-3-642-01266-2 e-ISBN 978-3-642-01267-9DOI 10.1007/978-3-642-01267-9Library of Congress Control Number: Applied forc 2009 Springer-Verlag Berlin HeidelbergThis work is subject to copyright. All rights are reserved, whether the whole or part of the mate-rial is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation,broadcasting, reproduction on microfilm or in any other way, and storage in data banks. Dupli-cation of this publication or parts thereof is permitted only under the provisions of the GermanCopyright Law of September 9, 1965, in its current version, and permission for use must alwaysbe obtained from Springer. Violations are liable to prosecution under the German Copyright Law.The use of general descriptive names, registered names, trademarks, etc. in this publication doesnot imply, even in the absence of a specific statement, that such names are exempt from the relevantprotective laws and regulations and therefore free for general use.Typesetting by the Author.Production: Scientific Publishing Services Pvt. Ltd., Chennai, India.Cover Design: WMX Design GmbH, Heidelberg.Printed in acid-free paper30/3100/as 5 4 3 2 1 0springer.com
  4. 4. This book is dedicated to my wife Vera for openingmany possibilities and providing balance in my life.
  5. 5. PREFACEThis book is designed to: Provide students with the tools to model, analyze and solve a wide range of engineering applications involving conduction heat transfer. Introduce students to three topics not commonly covered in conduction heat transfer textbooks: perturbation methods, heat transfer in living tissue, and microscale conduction. Take advantage of the mathematical simplicity of one- dimensional conduction to present and explore a variety of physical situations that are of practical interest. Present textbook material in an efficient and concise manner to be covered in its entirety in a one semester graduate course. Drill students in a systematic problem solving methodology with emphasis on thought process, logic, reasoning and verification.To accomplish these objectives requires judgment and balance in theselection of topics and the level of details. Mathematical techniquesare presented in simplified fashion to be used as tools in obtainingsolutions. Examples are carefully selected to illustrate the applicationof principles and the construction of solutions. Solutions follow anorderly approach which is used in all examples. To provideconsistency in solutions logic, I have prepared solutions to allproblems included in the first ten chapters myself. Instructors areurged to make them available electronically rather than posting themor presenting them in class in an abridged form.This edition adds a new chapter, “Microscale Conduction.” This is anew and emerging area in heat transfer. Very little is available onthis subject as textbook material at an introductory level. Indeed thepreparation of such a chapter is a challenging task. I am fortunate
  6. 6. viii PREFACEthat Professor Chris Dames of the University of California,Riverside, agreed to take on this responsibility and prepared all thematerial for chapter 11.Now for the originality of the material in this book. Much that ishere was inspired by publications on conduction. I would like toespecially credit Conduction Heat Transfer by my friend Vedat SArpaci. His book contains a wealth of interesting problems andapplications. My original notes on conduction contained manyexamples and problems taken from the literature. Not having beencareful in my early years about recording references, I tried toeliminate those that I knew were not my own. Nevertheless, a fewmay have been inadvertently included.ACKNOWLEDGMENTSFirst I would like to acknowledge the many teachers who directly orindirectly inspired and shaped my career. Among them I wish tosingle out Professors Ascher H. Shapiro of the MassachusettsInstitute of Technology, Milton Van Dyke of Stanford University,D.W. Ver Planck of Carnegie Institute of Technology and Gordon J.Van Wylen and John A. Clark of the University of Michigan. I onlywish that I had recognized their lasting contributions to my educationdecades earlier.Chapter 11 was carefully reviewed by Professor Gang Chen of theMassachusetts Institute of Technology. The chapter author ChrisDames and I are grateful for his technical comments whichstrengthened the chapter.My wife Vera read the entire manuscript and made constructiveobservations. I would like to thank her for being a supportive, patientand understanding partner throughout this project.Latif M. JijiNew York, New YorkMarch, 2009
  7. 7. CONTENTSPreface vii CHAPTER 1: BASIC CONCEPTS 1 1.1 Examples of Conduction Problems 1 1.2 Focal Point in Conduction Heat Transfer 2 1.3 Fourier’s Law of Conduction 2 1.4 Conservation of Energy: Differential Formulation of the Heat Conduction in Rectangular Coordinates 5 1.5 The Heat Conduction Equation in Cylindrical and Spherical Coordinates 9 1.6 Boundary Conditions 10 1.6.1 Surface Convection: Newton’s Law of Cooling 10 1.6.2 Surface Radiation: Stefan-Boltzmann Law 11 1.6.3 Examples of Boundary Conditions 12 1.7 Problem Solving Format 15 1.8 Units 16 REFERENCES 17 PROBLEMS 18 CHAPTER 2: ONE-DIMENSIONAL STEADY-STATE CONDUCTION 24 2.1 Examples of One-dimensional Conduction 24 2.2 Extended Surfaces: Fins 34 2.2.1 The Function of Fins 34 2.2.2 Types of Fins 34 2.2.2 Heat Transfer and Temperature Distribution in Fins 35 2.2.4 The Fin Approximation 36 2.2.5 The Fin Heat Equation: Convection at Surface 37 2.2.6 Determination of dAs / dx 39 2.2.7 Boundary Conditions 40 2.2.8 Determination of Fin Heat Transfer Rate q f 40 2.2.9 Steady State Applications: Constant Area Fins with Surface Convection 41
  8. 8. x CONTENTS 2.2.10 Corrected Length Lc 44 2.2.11 Fin Efficiency f 44 2.2.12 Moving Fins 45 2.2.13 Application of Moving Fins 47 2.2.14 Variable Area Fins 49 2.3 Bessel Differential Equations and Bessel Functions 52 2.3.1 General Form of Bessel Equations 52 2.3.2 Solutions: Bessel Functions 52 2.3.3 Forms of Bessel Functions 54 2.3.4 Special Closed-form Bessel Functions: n = odd integer/2 54 2.3.5 Special Relations for n = 1, 2, 3, … 55 2.3.6 Derivatives and Integrals of Bessel Functions 56 2.3.7 Tabulation and Graphical Representation of Selected Bessel Functions 56 2.4 Equidimensional (Euler) Equation 58 2.5 Graphically Presented Solutions to Fin Heat Transfer Rate q f 59 REFERENCES 60 PROBLEMS 61 CHAPTER 3: TWO-DIMESIONAL STEADY STATE CONDUCTION 72 3.1 The Heat Conduction Equation 72 3.2 Method of Solution and Limitations 72 3.3 Homogeneous Differential Equations and Boundary Conditions 72 3.4 Sturm-Liouville Boundary-Value Problem: Orthogonality 74 3.5 Procedure for the Application of Separation of Variables Method 76 3.6 Cartesian Coordinates: Examples 83 3.7 Cylindrical Coordinates: Examples 97 3.8 Integrals of Bessel Functions 102 3.9 Non-homogeneous Differential Equations 103 3.10 Non-homogeneous Boundary Conditions: The Method of Superposition 109 REFERENCES 111
  9. 9. CONTENTS xi PROBLEMS 111CHAPTER 4: TRANSIENT CONDUCTION 1194.1 Simplified Model: Lumped-Capacity Method 119 4.1.1 Criterion for Neglecting Spatial Temperature Variation 119 4.1.2 Lumped-Capacity Analysis 1214.2 Transient Conduction in Plates 1244.3 Non-homogeneous Equations and Boundary Conditions 1284.4 Transient Conduction in Cylinders 1324.5 Transient Conduction in Spheres 1384.6 Time Dependent Boundary Conditions: Duhamel’s Superposition Integral 141 4.6.1 Formulation of Duhamel’s Integral 142 4.6.2 Extension to Discontinuous Boundary Conditions 144 4.6.3 Applications 1454.7 Conduction in Semi-infinite Regions: The Similarity Transformation Method 150 REFERENCES 154 PROBLEMS 154CHAPTER 5: POROUS MEDIA 1635.1 Examples of Conduction in Porous Media 1635.2 Simplified Heat Transfer Model 164 5.2.1 Porosity 164 5.2.2 Heat Conduction Equation: Cartesian Coordinates 165 5.2.3 Boundary Conditions 167 5.2.4 Heat Conduction Equation: Cylindrical Coordinates 1685.3 Applications 168 REFERENCES 174 PROBLEMS 175
  10. 10. xii CONTENTS CHAPTER 6: CONDUCTION WITH PHASE CHANGE: MOVING BOUNDARY PROBLEMS 184 6.1 Introduction 184 6.2 The Heat Equation 185 6.3 Moving Interface Boundary Conditions 185 6.4 Non-linearity of the Interface Energy Equation 188 6.5 Non-dimensional Form of the Governing Equations: Governing Parameters 189 6.6 Simplified Model: Quasi-Steady Approximation 190 6.7 Exact Solutions 197 6.7.1 Stefan’s Solution 197 6.7.2 Neumann’s Solution: Solidification of Semi- Infinite Region 200 6.7.3 Neumann’s Solution: Melting of Semi-Infinite Region 203 6.8 Effect of Density Change on the Liquid Phase 204 6.9 Radial Conduction with Phase Change 205 6.10 Phase Change in Finite Regions 209 REFERENCES 210 PROBLEMS 210 CHAPTER 7: NON-LINEAR CONDUCTION PROBLEMS 215 7.1 Introduction 215 7.2 Sources of Non-linearity 215 7.2.1 Non-linear Differential Equations 215 7.2.2 Non-linear Boundary Conditions 216 7.3 Taylor Series Method 216 7.4 Kirchhoff Transformation 220 7.4.1 Transformation of Differential Equations 220 7.4.2 Transformation of Boundary Conditions 221 7.5 Boltzmann Transformation 224 7.6 Combining Boltzmann and Kirchhoff Transformations 226 7.7 Exact Solutions 227 REFERENCES 230 PROBLEMS 230
  11. 11. CONTENTS xiiiCHAPTER 8: APPROXIMATE SOLUTIONS: THE INTEGRAL METHOD 2368.1 Integral Method Approximation: Mathematical Simplification 2368.2 Procedure 2368.3 Accuracy of the Integral Method 2378.4 Application to Cartesian Coordinates 2388.5 Application to Cylindrical Coordinates 2468.6 Non-linear Problems 2518.7 Energy Generation 260 REFERENCES 264 PROBLEMS 264CHAPTER 9: PERTURBATION SOLUTIONS 2699.1 Introduction 2699.2 Solution Procedure 2709.3 Examples of Perturbation Problems in Conduction 2719.4 Perturbation Solutions: Examples 2739.5 Useful Expansions 296 REFERENCES 296 PROBLEMS 297CHAPTER 10: HEAT TRANSFER IN LIVING TISSUE 30210.1 Introduction 30210.2 Vascular Architecture and Blood Flow 30210.3 Blood Temperature Variation 30410.4 Mathematical Modeling of Vessels-Tissue Heat Transfer 305 10.4.1 Pennes Bioheat Equation 305 10.4.2 Chen-Holmes Equation 312 10.4.3 Three-Temperature Model for Peripheral Tissue 313 10.4.4 Weinbaum-Jiji Simplified Bioheat Equation for Peripheral Tissue 315 10.4.5 The s-Vessel Tissue Cylinder Model 323 REFERENCES 332 PROBLEMS 334
  12. 12. xiv CONTENTS CHAPTER 11: MICROSCALE CONDUCTION 347 11.1 Introduction 347 11.1.1 Categories of Microscale Phenomena 348 11.1.2 Purpose and Scope of this Chapter 350 11.2 Understanding the Essential Physics of Thermal Conductivity Using the Kinetic Theory of Gases 351 11.2.1 Determination of Fourier’s Law and Expression for Thermal Conductivity 351 11.3 Energy Carriers 355 11.3.1 Ideal Gas: Heat is Conducted by Gas Molecules 355 11.3.2 Metals: Heat is Conducted by Electrons 359 11.3.3 Electrical Insulators and Semiconductors: Heat is Conducted by Phonons (Sound Waves) 361 11.3.4 Radiation: Heat is Carried by Photons (Light Waves) 372 11.4 Thermal Conductivity Reduction by Boundary Scattering: The Classical Size Effect 376 11.4.1 Accounting for Multiple Scattering Mechanisms: Matthiessen’s Rule 377 11.4.2 Boundary Scattering for Heat Flow Parallel to Boundaries 379 11.4.3 Boundary Scattering for Heat Flow Perpendicular to Boundaries 387 11.5 Closing Thoughts 391 REFERENCES 394 PROBLEMS 397 APPENDIX A: Ordinary Differential Equations 402 (1) Second Order Differential Equations with Constant Coefficients 402 (2) First Order Ordinary Differential Equations with Variable Coefficients 404 APPENDIX B: Integrals of Bessel Functions 405 APPENDIX C: Values of Bessel Functions 406 APPENDIX D: Fundamental Physical Constants and Material Properties 412 D-1 Fundamental Physical Constants 412 D-2 Unit conversions 412
  13. 13. CONTENTS xv D-3 Properties of Helium Gas 412 D-4 Properties of Copper at 300 K 412 D-5 Properties of Fused Silica 412 (Amorphous Silicon Dioxide, SiO2) at 300 K 413 D-6 Properties of Silicon 413 D-7 Measured Thermal Conductivity of a 56 nm Diameter Silicon Nanowire at Selected Temperatures 414 D-8 Calculated Thermal Conductivity of Single-Walled Carbon Nanotubes, Selected Values 414INDEX 416
  14. 14. http://avibert.blogspot.com1BASIC CONCEPTS1.1 Examples of Conduction ProblemsConduction heat transfer problems are encountered in many engineeringapplications. The following examples illustrate the broad range ofconduction problems.(1) Design. A small electronic package is to be cooled by free convection.A heat sink consisting of fins is recommended to maintain the electroniccomponents below a specified temperature. Determine the requirednumber of fins, configuration, size and material.(2) Nuclear Reactor Core. In the event of coolant pump failure, thetemperature of a nuclear element begins to rise. How long does it takebefore meltdown occurs?(3) Glaciology. As a glacier advances slowly due to gravity, its frontrecedes due to melting. Estimate the location of the front in the year 2050.(4) Re-entry Shield. A heat shield is used to protect a space vehicle duringre-entry. The shield ablates as it passes through the atmosphere. Specifythe required shield thickness and material to protect a space vehicle duringre-entry.(5) Cryosurgery. Cryoprobes are used in the treatment of certain skincancers by freezing malignant tissue. However, prolonged contact of acryoprobe with the skin can damage healthy tissue. Determine tissuetemperature history in the vicinity of a cold probe subsequent to contactwith the skin.
  15. 15. 2 1 Basic Concepts(6) Rocket Nozzle. One method for protecting the throat of a supersonicrocket nozzle involves inserting a porous ring at the throat. Injection ofhelium through the ring lowers the temperature and protects the nozzle.Determine the amount of helium needed to protect a rocket nozzle during aspecified trajectory.(7) Casting. Heat conduction in casting is accompanied by phase change.Determine the transient temperature distribution and the interface motion ofa solid-liquid front for use in thermal stress analysis.(8) Food Processing. In certain food processing operations conveyor beltsare used to move food products through a refrigerated room. Use transientconduction analysis to determine the required conveyor speed.1.2 Focal Point in Conduction Heat TransferWhat do all these examples have in common? If you guessed temperature,you are on the right track. However, what drives heat is temperaturedifference and not temperature. Refining this observation further, we statethat conduction heat transfer problems involve the determination oftemperature distribution in a region. Once temperature distribution isknown, one can easily determine heat transfer rates. This poses twoquestions: (1) how is the rate of heat transfer related to temperaturedistribution? And (2) what governs temperature distribution in a region?Fourier’s law of conduction provides the answer to the first question whilethe principle of conservation of energy gives the answer to the second.1.3 Fouriers Law of ConductionOur experience shows that if one end of a metal bar is heated, itstemperature at the other end will eventually begin to rise. This transfer ofenergy is due to molecular activity. Molecules at the hot end exchange theirkinetic and vibrational energies with neighboring layers through randommotion and collisions. A temperature gradient, or slope, is established withenergy continuously being transported in the direction of decreasingtemperature. This mode of energy transfer is called conduction.
  16. 16. 1.3 Fourier’s Law of Conduction 3We now turn our attention to Lformulating a law that will help usdetermine the rate of heat transfer by Aconduction. Consider the wall shownin Fig.1.1 . The temperature of one qxsurface (x = 0) is Tsi and of the othersurface (x = L) is Tso . The wallthickness is L and its surface area is Tsi Tso 0 xA. The remaining four surfaces are dxwell insulated and thus heat is Fig. 1.1transferred in the x-direction only.Assume steady state and let q x be the rate of heat transfer in the x-direction. Experiments have shown that q x is directly proportional to Aand (Tsi Tso ) and inversely proportional to L. That is A (Tsi Tso ) qx . LIntroducing a proportionality constant k, we obtain A (Tsi Tso ) qx k , (1.1) Lwhere k is a property of material called thermal conductivity. We mustkeep in mind that eq. (1.1) is valid for: (i) steady state, (ii) constant k and(iii) one-dimensional conduction. These limitations suggest that a re-formulation is in order. Applying eq. (1.1) to the element dx shown in Fig.1.1 and noting that Tsi becomes T (x) , Tso becomes T ( x dx) , andL is replaced by dx, we obtain T ( x) T ( x + dx) T ( x + dx) T ( x) qx = k A = kA . dx dxSince T ( x dx) T ( x) dT , the above gives
  17. 17. 4 1 Basic Concepts dT qx = kA . (1.2) dxIt is useful to introduce the term heat flux q x , which is defined as the heatflow rate per unit surface area normal to x. Thus, qx qx . (1.3) ATherefore, in terms of heat flux, eq. (1.2 ) becomes dT qx k . (1.4) dxAlthough eq. (1.4) is based on one-dimensional conduction, it can begeneralized to three-dimensional and transient conditions by noting thatheat flow is a vector quantity. Thus, the temperature derivativein eq. (1.4) is changed to a partial derivative and adjusted to reflect thedirection of heat flow as follows: T T T qx k , qy k , qz k , (1.5) x y zwhere x, y, and z are the rectangular coordinates. Equation (1.5) is knownas Fouriers law of conduction. Three observations are worth making: (i)The negative sign indicates that when the gradient is negative, heat flow isin the positive direction, i.e., towards the direction of decreasingtemperature, as dictated by the second law of thermodynamics. (ii) Theconductivity k need not be uniform since eq. (1.5) applies at a point in thematerial and not to a finite region. In reality the thermal conductivity varieswith temperature. However, eq. (1.5) is limited to isotropic material, i.e., kis invariant with direction. (iii) Returning to our previous observation thatthe focal point in conduction is the determination of temperaturedistribution, we now recognize that once T ( x, y, z, t ) is known, the heatflux in any direction can be easily determined by simply differentiating thefunction T and using eq. (1.5).
  18. 18. 1.4 Conservation of Energy 51.4 Conservation of Energy: Differential Formulation of the Heat Conduction Equation in Rectangular CoordinatesWhat determines temperature distribution in a region? Is the processgoverned by a fundamental law and therefore predictable? The answer isthat the temperature at each point is adjusted such that the principle ofconservation of energy is satisfied everywhere. Thus the starting point indifferential formulation must be based on an infinitesimal element. Fig.1.2shows a region described by rectangularcoordinates in which heat conduction is ythree-dimensional. To generalize theformulation, the material is assumed tobe moving. In addition, energy isgenerated throughout the material at a xrate q per unit volume. Examples ofvolumetric energy generation include zheat conduction in nuclear elements,metabolic heat production in tissue, andelectrical energy loss in transmission Fig. 1.2lines,We select an infinitesimal element dxdydz and apply the principle ofconservation of energy (first law of thermodynamics)Rate of energy added + Rate of energy generated Rate of energy removed = Rate of energy change within elementDenoting these terms by the symbols Ein , E g , Eout , and E , respectively,we obtain E in + E g E out = E . (1.6)This form of conservation of energy is not helpful in solving conductionproblems. Specifically, temperature, which is the focal point inconduction, does not appear explicitly in the equation. The next step is toexpress eq.(1.6) in terms of the dependent variable T. To simplify theformulation, the following assumptions are made: (1) uniform velocity, (2)
  19. 19. 6 1 Basic Conceptsconstant pressure, (3) constant density and (4) negligible changes inpotential energy. qy (q y dy )dxdz y qx qx dydz ( qx dx )dydz dy x dx q y dxdz (a) ˆ h ˆ V (h dy )dxdz y ˆ h ˆ Uhdydz ˆ U (h dx )dydz dy x dx ˆ Vhdxdz (b) Fig. 1.3Energy is exchanged with the element by conduction and mass motion.These two modes of energy transfer are shown in Fig. 1.3a and Fig. 1.3b,respectively. Not shown in these figures are the z-components, which canbe formulated by analogy with the x and y-components. Energy enters theelement by conduction at fluxes qx , q y and q z , in the x, y and zdirections, respectively. Since each flux represents energy per unit area perunit time, it must be multiplied by the area normal to it. Energy also entersthe element through mass flow. The mass flow rate entering the element in
  20. 20. 1.4 Conservation of Energy 7the x-direction is U dy dz, where is density and U is the velocitycomponent in the x-direction. The rate of energy carried by this mass ˆ ˆis U h dy dz , where h is enthalpy per unit mass. The corresponding ˆ ˆcomponents in the y and z directions are V h dx dz and W h dx dy ,where V and W are the velocity components in the y and z directions,respectively. Thus, E in is given by Ein q x dy dz q y dx dz q z dx dy (a) ˆ U h dy dz ˆ V h dx dz ˆ W h dx dy.Energy generation E g is Eg q dx dy dz . (b)Formulation of energy leaving the element is constructed using Taylorseries expansion qx qy qz Eout (q x dx) dy dz (q y dy ) dx dz (q z dz ) dx dy x y z ˆ h ˆ h ˆ h ˆ U(h dx ) dy dz ˆ V(h dy ) dx dz ˆ W(h dz )dx dy. x y z (c)Note that U, V and W are constant since material motion is assumeduniform. Energy change within the element E is expressed as ˆ u E dx dy dz , (d) t ˆwhere u is internal energy per unit mass and t is time. Substituting (a)-(d)into eq. (1.6) qx qy qz ˆ h ˆ h ˆ h ˆ u U V W q . (e) x y z x y z t ˆEnthalpy h is defined as
  21. 21. 8 1 Basic Concepts ˆ ˆ P h u , (f)here P is pressure, assumed constant. Substituting (f) into (e) andrearranging qx qy qz ˆ h ˆ h ˆ h ˆ h q ( U V W ) . (g) x y z t x y zThe next step is to express the heat flux and enthalpy in terms oftemperature. Fourier’s law, eq.(1.5), relates heat flux to temperaturegradient. Enthalpy change for constant pressure is given by ˆ dh c p dT , (h)where c p is specific heat at constant pressure. Substituting eqs. (1.5) and(h) into (g) T T T (k ) (k ) (k ) q x x y y z z T T T T cp( U V W ). (1.7) t x y zAlthough eq. (1.7) is based on uniform velocity, it can be shown that it isalso applicable to incompressible flow with variable velocity, as long asdissipation, which is work done due to viscous forces, is negligible [1]. Forconstant conductivity and stationary material, eq. (1.7) simplifies to 2 2 2 T T T q T ( ) , (1.8) x2 y2 z 2 cp twhere is a material property called thermal diffusivity, defined as k . (1.9) cp
  22. 22. 1.5 The Heat Conduction Equation in Cylindrical and Spherical Coordinates 91.5 The Heat Conduction Equation in Cylindrical and Spherical CoordinatesTo analyze conduction problems in cylindrical and spherical geometriesrequires the formulation of the heat conduction equation in the cylindricaland spherical coordinates shown in Fig. 1.4. For constant properties and nodissipation, the heat equation takes the following forms: z z (r , z , ) (r , , ) r x x r y y (a) cylindrical (b) spherical Fig. 1.4Cylindrical Coordinates ( r, , z ): 2 21 1 T V T T (r T ) 2 T 2 T 2 q T Vr Vz ,r r r r z cp t r r z (1.10)where Vr , V and Vz are the velocity components in the r, and zdirections, respectively. For a stationary material this equation simplifies to 2 2 1 (r T ) 1 T T q T . (1.11) r r r r2 2 z 2 cp tSpherical Coordinates ( r , , ): 2 1 2 T 1 T 1 T q 2 (r ) 2 2 2 2 (sin ) r r r r sin r sin cp T T V T V T Vr , (1.12) t r r r sin
  23. 23. 10 1 Basic Conceptswhere V is the velocity component in the direction. For stationarymaterial this equation reduces to 2 1 T 1 T 1 T q (r 2 ) (sin ) r2 r r r 2 sin 2 2 r 2 sin cp T . (1.13) t1.6 Boundary ConditionsAlthough the heat conduction equation governs temperature behavior in aregion, it does not give temperature distribution. To obtain temperaturedistribution it is necessary to solve the equation. However, to construct acomplete solution, boundary and initial conditions must be specified. Forexample, eq.(1.8) requires six boundary conditions, two in each of thevariables x, y and z, and one initial condition in time t. Boundary conditionsare mathematical equations describing what takes place physically at aboundary. Similarly, an initial condition describes the temperaturedistribution at time t = 0.Since boundary conditions involve thermal interaction with thesurroundings, it is necessary to first describe two common modes ofsurface heat transfer: convection and radiation.1.6.1 Surface Convection: Newtons Law of CoolingIn this mode of heat transfer, energy is exchanged between a surface and afluid moving over it. Based on experimental observations, it is postulatedthat the flux in convection is directly proportional to the difference intemperature between the surface and the streaming fluid. That is qs (Ts T ),where q s is surface flux, Ts is surface temperature and T is the fluidtemperature far away from the surface. Introducing a proportionalityconstant to express this relationship as equality, we obtain qs h (Ts T ). (1.14)
  24. 24. 1.6 Boundary Conditions 11This result is known as Newtons law of cooling. The constant ofproportionality h is called heat transfer coefficient. This coefficientdepends on geometry, fluid properties, motion, and in some casestemperature difference ( Ts T ). Thus, unlike thermal conductivity, h isnot a property of material.1.6.2 Surface Radiation: Stefan-Boltzmann LawWhile conduction and convection require a medium to transport energy,radiation does not. Furthermore, radiation energy is transmitted byelectromagnetic waves, which travel best in a vacuum. The maximumpossible radiation is described by the Stefan-Boltzmann law, which givessurface radiation flux for an ideal body called blackbody as qb Ts4 , (1.15)where qb is blackbody radiation flux, Ts is surface temperature, measuredin absolute degrees, and is the Stefan-Boltzmann constant given by 8 = 5.67 x 10 W/m2- K 4 . (1.16)To determine the radiation flux q r emitted from a real surface, a radiationproperty called emissivity, , is defined as qr . (1.17) qbCombining (1.15) and (1.17) qr Ts4 . (1.18)From the definition of , it follows that its maximum value is unity(blackbody).Radiation energy exchange between two surfaces depends on the geometry,shape, area, orientation and emissivity of the two surfaces. In addition, itdepends on the absorptivity of each surface. Absorptivity is a surfaceproperty defined as the fraction of radiation energy incident on a surface
  25. 25. 12 1 Basic Conceptswhich is absorbed by the surface. Although the determination of the netheat exchange by radiation between two surfaces, q12 , can be complex, theanalysis is simplified for an ideal model for which = . Such an idealsurface is called a gray surface. For the special case of a gray surfacewhich is completely enclosed by a much larger surface, q12 is given by q12 1 A1 (T14 T24 ), (1.19)where 1 is the emissivity of the small surface, A1 its area, T1 its absolutetemperature and T2 is the absolute temperature of the surrounding surface.Note that for this special case neither the area A2 of the large surface norits emissivity 2 affects the result.1.6.3 Examples of Boundary ConditionsThere are several common physical conditions that can take place atboundaries. Fig. 1.5 shows four typical boundary conditions for two-dimensional conduction in a rectangular plate. Fig. 1.6 shows an interfaceof two materials. Two boundary conditions are associated with this case. y insulation k k 1 2 To W qo T T 1 2 0 L 0 x 0 convection, h, T Fig. 1.5 Fig. 1.6Before writing boundary conditions, an origin and coordinate axes must beselected. Boundary conditions for Fig. 1.5 and Fig. 1.6 are expressedmathematically as follows:
  26. 26. 1.6 Boundary Condition 13(1) Specified temperature. Along boundary (0, y ) the temperature is To .This temperature can be uniform or can vary along y as well as with time.Mathematically this condition is expressed as T (0, y ) To . (1.20)(2) Specified flux. The heat flux along boundary ( L, y ) is q o . Accordingto Fourier’s law this condition is expressed as T ( L, y ) qo k . (1.21) x(3) Convection. Neither the temperature nor the flux are known alongboundary (x,0). Instead, heat is exchanged by convection with the ambientfluid. The simplest way to formulate this condition mathematically is topretend that heat flows in the positive coordinate direction. In this examplethe fluid, which is at temperature T , adds energy by convection to theboundary (x,0). Conservation of energy requires that the added energy beconducted to the interior in the positive y-direction. Therefore, equatingNewtons law of cooling with Fouriers law of conduction gives T ( x ,0 ) h [T T ( x,0)] k . (1.22) yNote that surface temperature T (x,0) is not equal to the ambienttemperature T .(4) Insulated boundary. The boundary at ( x, W ) is thermally insulated.Thus, according to Fourier’s law, this condition is expressed as T ( x, W ) 0. (1.23) yNote that this is a special case of eq. (1.21) with qo = 0, or of eq. (1.22)with h = 0.
  27. 27. 14 1 Basic Concepts(5) Interface. Fig.1.6 shows a composite wall of two materials withthermal conductivities k1 and k 2 . For a perfect interface contact, the twotemperatures must be the same at the interface. Thus T1 (0, y ) T2 (0, y ) . (1.24)Conservation of energy at the interface requires that the two fluxes beidentical. Application of Fourier’s law gives T1(0, y ) T2 (0, y ) k1 k2 . (1.25) x x(6) Interface with a heat source. One example of this case is an electricalheating element which is sandwiched between two non-electricallyconducting materials as shown in Fig. 1.7. Another example is frictionalheat, which is generated by relative motion between two surfaces. Energydissipated at an interface can be conducted through both materials orthrough either one of the two. Boundary conditions at the interface areagain based on the continuity of temperatureand conservation of energy. Assuming perfect -contact, continuity results in eq. (1.24). To k1 k2apply conservation of energy at the interface, 1 T1 T2 2it is convenient to pretend that heat isconducted in the positive coordinate direction xthrough both materials. Referring to Fig. 1.7, 0conservation of energy requires that heat flux qiadded to the interface by conduction through Fig. 1.7material 1, plus heat flux generated at theinterface, q i , be equal to the flux removed byconduction through material 2. Thus T1 (0, y ) T2 (0, y ) k1 qi k2 . (1.26) x x(7) Radiation. To illustrate how a radiation boundary condition isformulated, consider Fig. 1.5. Assume that the boundary (x,0) exchangesheat by radiation in addition to convection. Again we pretend that net
  28. 28. 1.7 Problem Solving Format 15radiation energy is added to the surface in the positive y-direction.Conservation of energy at this boundary requires that energy added at thesurface by convection and radiation be equal to energy conducted in thepositive y-direction. Thus qconv qrad qcond .Using Fourier’s law of conduction for qcond , Newton’s law of cooling forqconv and assuming that Stefan-Boltzmann radiation result for q rad , eq.(1.19), is applicable, the above gives T ( x,0) h [T T ( x,0)] [Tsur T 4 ( x,0)] 4 k , (1.27) ywhere Tsur is the surroundings temperature. To formulate the boundarycondition for the case of energy exchange by radiation only, set h = 0 in eq.(1.27).1.7 Problem Solving FormatConduction problems lend themselves to a systematic solution procedure.The following basic format which builds on the work of Ver Plank andTeare [2] is used throughout the book.(1) Observations. Read the problem statement very carefully and noteessential facts and features such as geometry, temperature symmetry, heatflow direction, number of independent variables, etc. Identifycharacteristics that require special attention such as variable properties,composite domain, non-linearity, etc. Note conditions justifying simplifiedsolutions. Show a schematic diagram.(2) Origin and Coordinates. Select an origin and a coordinate systemappropriate to the geometry under consideration.(3) Formulation. In this stage the conduction problem is expressed inmathematical terms. Define all terms. Assign units to all symbols whenevernumerical computations are required. Proceed according to the followingsteps:
  29. 29. 16 1 Basic Concepts (i) Assumptions. Model the problem by making simplifications andapproximations. List all assumptions. (ii) Governing Equations. Based on the coordinate system chosenand the assumptions made, select an appropriate form of the heat equation.Note that composite domains require a heat equation for each layer. (iii) Boundary Conditions. Examine the governing equations anddecide on the number of boundary conditions needed based on the numberof independent variables and the order of the highest derivative. Identifythe physical nature of each boundary condition and express itmathematically. Use the examples of Section 1.6.3 as a guide informulating boundary conditions.(4) Solution. Examine the equation to be solved and select an appropriatemethod of solution. Apply boundary conditions to determine constants ofintegration.(5) Checking. Check each step of the analysis as you proceed. Applydimensional checks and examine limiting cases.(6) Computations. Execute the necessary computations and calculationsto generate the desired numerical results.(7) Comments. Review your solution and comment on such things as therole of assumptions, the form of the solution, the number of governingparameters, etc.1.8 UnitsSI units are used throughout this textbook. The basic units in this systemare Length (L): meter (m) Time (t): second (s) Mass (m): kilogram (kg) Temperature (T): kelvin (K)Temperature on the Celsius scale is related to the kelvin scale by T(oC) = T(K) - 273.15. (1.28)
  30. 30. 1.8 Units 17Note that temperature difference on the two scales is identical. Thus, achange of one kelvin is equal to a change of one Celsius. This means thatquantities that are expressed per unit kelvin, such as thermal conductivity,heat transfer coefficient and specific heat, are numerically the same as perdegree Celsius. That is, W/m2-K = W/ m2-oC.The basic units are used to derive units for other quantities. Force ismeasured in newtons (N). One newton is the force needed to accelerate amass of one kilogram one meter per second per second: Force = mass acceleration N kg.m/s 2 .Energy is measured in joules (J). One joule is the energy associated with aforce of one newton moving a distance of one meter. J N.m kg.m 2 /s 2 .Power is measured in watts (W). One watt is energy rate of one joule persecond. W J/s N.m/s kg.m 2 /s 3 .REFERENCES[1] Bejan, A., Convection Heat Transfer, 2nd Edition, Wiley, New York, 1995.[2] Ver Planck, D.W., and Teare Jr., B.R., Engineering Analysis: An Introduction to Professional Method, Wiley, New York, 1952
  31. 31. 18 1 Basic Concepts PROBLEMS1.1 Write the heat equation for each of the following cases: [a] A wall, steady state, stationary, one-dimensional, incompressible and no energy generation. [b] A wall, transient, stationary, one-dimensional, incompressible, constant k with energy generation. [c] A cylinder, steady state, stationary, two-dimensional (radial and axial), constant k, incompressible, with no energy generation. [d] A wire moving through a furnace with constant velocity, steady state, one-dimensional (axial), incompressible, constant k and no energy generation. [e] A sphere, transient, stationary, one-dimensional (radial), incompressible, constant k with energy generation.1.2 A long electric wire of radius ro generates heat at a rate q . The surface is maintained at uniform temperature To . Write the heat equation and boundary conditions for steady state one-dimensional conduction.1.3 You are interested in analyzing the rate at which a spherical ice ball melts. What heat equation should you use for the ice? List all assumptions.1.4 Consider axial flow of water in a cold tube. Write the heat conduction equation for the ice forming axisymmetrically on the inside surface of the tube. r ice x water qo1.5 Consider two-dimensional conduction in the semi-circular cylinder shown. The cylinder is heated with uniform ro ri flux along its outer surface and is maintained at a variable temperature h, T convection insulation T f( )
  32. 32. Problems 19 along its inner surface. One of the plane surfaces is insulated while the other exchanges heat by convection with an ambient fluid at T . The heat transfer coefficient is h. Write the heat equation and boundary conditions for steady state conduction.1.6 A long hollow cylinder exchanges heat by radiation and convection along its outside surface with an ambient fluid at T . The heat transfer coefficient is h. The surroundings is at Tsur and surface emissivity is . Heat is removed from the inside surface at a uniform flux q i . Use a simplified radiation model and write the heat equation and boundary conditions for one-dimensional steady state conduction. y q o1.7 Write the heat equation and boundary h conditions for steady state two-dimensional To W T conduction in the rectangular plate shown. L 0 x1.8 Heat is generated at a rate q in a spherical insulation shell with inner radius ri and outer radius ro . Heat is added at the outer surface at a uniform flux qo . The inside surface is maintained at uniform temperature Ti . Write the heat equation and boundary conditions for steady state conduction.1.9 A long electric cable generates heat at a rate q . Half the cable is buried underground while the other half exchanges heat by radiation and convection. The ambient and surroundings temperatures are T and Tsur , respectively. The heat transfer coefficient is h. Tsur h, T Neglecting the thickness of the electrical insulation layer and heat loss to the ground, select a model to analyze the temperature distribution q in the cable and write the heat equation and boundary conditions.1.10 A shaft of radius ro rotates inside a sleeve of thickness . Frictional heat is generated at the interface at a rate q i . The outside surface of the sleeve is cooled by convection. The ambient temperature
  33. 33. 20 1 Basic Concepts is T and the heat transfer coefficient is h. Consider steady state one-dimensional conduction in the radial direction. Write the heat equations and boundary conditions for the temperature distribution in the shaft and sleeve.1.11 A rectangular plate of length L and height H slides down an inclined x T surface with a velocity U. Sliding y h U friction results in surface heat flux h 0 g qo . The front and top sides of the T plate exchange heat by convection. The heat transfer coefficient is h and the ambient temperature is T . Neglect heat loss from the back side and assume that no frictional heat is conducted through the inclined surface. Write the two-dimensional steady state heat equation and boundary conditions.1.12 Consider a semi-circular section of a tube with inside radius Ri and outside radius Ro . Heat is exchanged by convection along the inside and outside cylindrical surfaces. The inside temperature and heat transfer ho To coefficient are Ti and hi . The outside Ro temperature and heat transfer r coefficient are To and ho . The two Ri hi Ti plane surfaces are maintained at uniform temperature Tb . Write the Tb 0 Tb two-dimensional steady state heat equation and boundary conditions.1.13 Two large plates of thicknesses L1 and L2 are initially at temperatures T1 and T2 . Their respective conductivities are k1 and k 2 . The two plates are pressed together and insulated along their L2 k2 exposed surfaces. Write the heat x L1 k1 equation and boundary conditions. 0
  34. 34. Problems 211.14 Heat is generated at a r qo volumetric rate q in a rod of radius ro and L/2 h ro length L. Half the z T 0 cylindrical surface is q L insulated while the qo other half is heated at a flux q o . One end is insulated and the other exchanges heat by convection. Write the heat equation and boundary conditions for steady state two-dimensional conduction.1.15 A cable of radius ro , r h, T conductivity k and specific heat c p moves through a U ro 0 z furnace with velocity U and leaves at temperature To . To h, T Electric energy is dissipated furnace in the cable resulting in a volumetric energy generation rate of q . Outside the furnace the cable is cooled by convection and radiation. The heat transfer coefficient is h, emissivity , ambient temperature T and surroundings temperature is Tsur . Use a simplified radiation model and assume that the cable is infinitely long. Write the steady state two-dimensional heat equation and boundary conditions.1.16 Radiation is suddenly directed at the surface of a semi-infinite plate of conductivity k and 0 thermal diffusivity . Due to the thermal To q (x) absorption characteristics of the material the x radiation results in a variable energy generation rate given by bx q ( x) qo e , where qo and b are constants and x is distance along the plate. Surface temperature at x 0 is To . Initially the plate is at uniform temperature Ti . Write the heat equation and boundary conditions.
  35. 35. 22 1 Basic Concepts1.17 A section of a long rotating shaft of radius ro is buried in a material of very low thermal conductivity. The length of the buried section is h L. Frictional heat generated in the buried T section can be modeled as surface heat flux. z Along the buried surface the radial heat flux, q r , is assumed uniform. However, at the qr flat end the axial heat flux, q z , varies with radius according to 0 r qz qz r, where is constant. The exposed surface exchanges heat by convection with the ambient. The heat transfer coefficient is h and the ambient temperature is T . Treat the shaft as semi-infinite and assume that all frictional heat is conducted through the shaft. Write the steady state heat equation and boundary conditions.1.18 A plate of thickness 2L moves through a furnace with velocity U and leaves at temperature To . Outside the furnace it is cooled by convection and radiation. The heat transfer coefficient is h, emissivity , ambient temperature T and surroundings temperatures is Tsur . y h, T Write the two-dimensional steady L U state heat equation and boundary 0 x L conditions. Use a simplified radiation model and assume that the plate is To h, T infinitely long. furnace1.19 A cable of radius ro moves with velocity U through a furnace of length r L L where it is heated at uniform flux qo qo . Far away from the inlet of the furnace the cable is at temperature Ti . 0 z U Assume that no heat is exchanged qo with the cable before it enters and after it leaves the furnace. Write the
  36. 36. Problems 23 two-dimensional steady state heat equation and boundary conditions.1.20 A hollow cylinder of inner radius Ri and outer radius Ro is heated with uniform flux qi at its inner surface. The lower half of the cylinder is insulated and the upper half exchanges heat by convection and radiation. The heat transfer coefficient is h, ambient temperature T , surroundings temperature Tsur and surface emissivity is . Neglecting axial conduction and using a simplified radiation model, write the steady state heat equations and boundary conditions. h, T Tsur Ri Ro qi qi
  37. 37. 2 http://avibert.blogspot.comONE-DIMENSIONAL STEADY STATECONDUCTIONOne-dimensional conduction, with its simple mathematical level, is used inthis chapter to present the essential steps in the analysis of conduction heattransfer problems. The objective is to learn to model problems, set up theapplicable governing equations and boundary conditions and constructsolutions. The simplicity of the mathematical treatment of one-dimensionalconduction enables us to explore a variety of practical applications.2.1 Examples of One-dimensional ConductionExample 2.1: Plate with Energy Generation and Variable ConductivityConsider a plate with internal energy generation q and a variablethermal conductivity k given by k k o (1 T) ,where k o and are constant and T istemperature. Both surfaces in Fig. 2.1 aremaintained at 0 oC . Determine the temperature q 0o C 0o Cdistribution in the plate. 0 x(1) Observations. (i) Thermal conductivity isvariable. (ii) Temperature distribution is Lsymmetrical about the center plane. (iii) Thegeometry can be described by a rectangular Fig. 2.1coordinate system. (iv) The temperature isspecified at both surfaces.
  38. 38. 2.1 Examples of One-dimensional Conduction 25(2) Origin and Coordinates. A rectangular coordinate system is used withthe origin at one of the two surfaces. The coordinate x is oriented as shownin Fig. 2.1.(3) Formulation. (i) Assumptions. (1) One-dimensional, (2) steady state, (3) stationaryand (4) uniform energy generation. (ii) Governing Equations. Introducing the above assumptions into eq.(1.7) gives d dT (k ) q 0, (2.1) dx dxwhere k = thermal conductivity q = rate of energy generation per unit volume T = temperature x = independent variableThe thermal conductivity k varies with temperature according to k k o (1 T) , (a)where k o and are constant. Substituting (a) into eq. ( 2.1) gives d dT [(1 T) ] q 0. (b) dx dx ko (iii) Boundary Conditions. Since (b) is second order with a singleindependent variable x, two boundary conditions are needed. Thetemperature is specified at both boundaries, x = 0 and x = L. Thus T ( 0) 0 , (c)and T ( L) 0 . (d)(4) Solution. Integrating (b) twice q T T2 x2 C1 x C 2 , (e) 2 2k o
  39. 39. 26 2 One-dimensional Steady State Conductionwhere C1 and C2 are constants of integration. Application of boundaryconditions (c) and (d) gives C1 and C2 q L C1 , C2 0. (f) 2k oSubstituting (f) into (e) 2 q Lx T2 T (1 x ) 0. (g) ko LRecognizing that (g) is a quadratic equation, its solution is 1 1 q Lx T 2 (1 x ) . (h) ko LTo satisfy boundary conditions (c) and (d), the minus sign in (h) must beconsidered. The solution becomes 1 1 q Lx T 2 (1 x ) . (i) ko L(5) Checking. Dimensional check: Assigning units to the quantities in (i)shows that each term has units of temperature.Boundary conditions check: Setting x = 0 or x = L in (i) gives T = 0.Thus, boundary conditions (c) and (d) are satisfied.Limiting check: For the special case of q 0, the temperature distributionshould be uniform given by T = 0. Setting q 0 in (i) gives 1 1 T 2 0.Symmetry check: Symmetry requires that the temperature gradient be zeroat the center, x = L/2. Differentiating (i) with respect to x 1 1 1 ( q L )( 2 x 1) . dT q Lx 2 (1 x ) (j) dx 2 2 ko L ko L
  40. 40. 2.1 Examples of One-dimensional Conduction 27 L dTSetting x in (j) gives 0. 2 dxQuantitative check: Conservation of energy and symmetry require that halfthe total energy generated in the plate leaves in the negative x-direction at x 0 and half in the positive x-direction at x = L. That is, for a wall ofsurface area A the heat transfer rate q is q AL q ( 0) , (k) 2and q AL q( L ) . (l) 2The heat transfer rate at x 0 and x L is determined by applyingFourier’s law dT (0) q(0) A k o [1 T (0)] , (m) dxand dT ( L) q ( L) A k o [1 T ( L)] . (n) dxEvaluating the gradient (j) at x = 0, noting that T (0) 0 and substitutinginto (m) gives the same result as (k). Similarly, results in (n) agree with (l).(6) Comments. The solution to the special case of constant thermalconductivity corresponds to 0. Since can not be set equal to zero in(i), the solution to this case must be obtained from (g) by first multiplyingthrough by and then setting 0.Example 2.2: Radial Conduction in a Composite Cylinder with Interface FrictionA shaft of radius Rs rotates inside a sleeve of inner radius R s and outerradius Ro . Frictional heat is generated at the interface at a flux q i . Theoutside surface of the sleeve is cooled by convection with an ambient fluidat T . The heat transfer coefficient is h. Consider one-dimensional steadystate conduction in the radial direction, determine the temperaturedistribution in the shaft and sleeve.
  41. 41. 28 2 One-dimensional Steady State Conduction(1) Observations. (i) The shaft and sleeve h, Tform a composite cylindrical wall. (ii) Thegeometry can be described by a cylindricalcoordinate system. (iii) Heat conduction is Ro Rs rassumed to be in the radial direction only. T2 T1(iv) Steady state requires that all energy shaft 0generated at the interface be conductedthrough the sleeve. Thus, no heat isconducted through the shaft. It follows thatthe shaft must be at a uniform temperature. qi sleeve(v) The heat flux is specified at the inner Fig. 2.2radius of the sleeve and convection takesplace at the outer radius.(2) Origin and Coordinates. A cylindrical coordinate system is used withthe origin at the center of the shaft as shown in Fig. 2.2.(3) Formulation. (i) Assumptions. (1) One-dimensional radial conduction, (2) steadystate, (3) constant thermal conductivities, (4) no energy generation, (5)uniform frictional energy flux at the interface and (6) stationary material(no motion in the radial direction). (ii) Governing Equations. Physical consideration requires that theshaft temperature be uniform. Thus, although this is a composite domainproblem, only one equation is needed to determine the temperaturedistribution. Introducing the above assumptions into eq. (1.11) gives d (r dT1 ) 0, (2.2) dr drwhere r = radial coordinate T1 = temperature distribution in the sleeve (iii) Boundary Conditions. Equation ( 2.2) is second order with asingle independent variable. Thus two boundary conditions are needed.Since all frictional energy is conducted through the sleeve, it follows thatthe heat flux added to the sleeve is specified at the inside radius R s .
  42. 42. 2.1 Examples of One-dimensional Conduction 29Conservation of energy at the interface and Fourier’s law of conductiongive dT1 ( Rs ) qi k1 , (a) drwhere k1 = thermal conductivity of the sleeve qi = interface frictional heat fluxConvection at the outer surface gives the second boundary condition dT1 ( Ro ) k1 h [T1 ( Ro ) T ] . (b) dr(4) Solution. Equation (2.2) is solved by integrating it twice with respectto r to obtain T1 C1 ln r C 2 , (c)where C1 and C 2 are constants of integration. Application of boundaryconditions (a) and (b) gives C1 and C 2 qi Rs C1 , (d) k1and qi Rs k1 C2 T (ln Ro ). (e) k1 hRoSubstituting (d) and (e) into (c) and rearranging gives the dimensionlesstemperature distribution in the sleeve T1 (r ) T Ro k1 ln . (f) qi Rs / k1 r hRoThe dimensionless ratio, hRo / k1 , appearing in the solution is called theBiot number. It is associated with convection boundary conditions.The temperature of the shaft, T2 , is obtained from the interface boundarycondition T2 (r ) T2 ( Rs ) T1 ( Rs ) . (g)
  43. 43. 30 2 One-dimensional Steady State ConductionEvaluating (f) at r Rs and substituting into (g) gives the dimensionlessshaft temperature T2 (r ) T Ro k1 ln . (h) qi Rs / k1 Rs hRo(5) Checking. Dimensional check: Each term in solution (f) isdimensionlessBoundary conditions check: Substituting (f) into boundary conditions (a)and (b) shows that they are satisfied.Limiting check: If frictional heat vanishes, both shaft and sleeve will be atthe ambient temperature. Substituting qi 0 into (f) and (h) givesT1 (r ) T2 (r ) T .(6) Comments. (i) Shaft conductivity plays no role in the solution. (ii) Thesolution is characterized by two dimensionless parameters: the Biotnumber hRo / k1 and the geometric ratio Ro / Rs . (iii) This problem canalso be treated formally as a composite cylindrical wall. This requireswriting two heat equations and four boundary conditions. The heat equationfor the sleeve is given by eq. (2.2). A second similar equation applies to theshaft d (r dT2 ) 0. (i) dr drBoundary condition (b) is still applicable. The remaining three conditionsare: Symmetry of temperature at the center dT2 (0) 0. (j) drConservation of energy and equality of temperature at the interface givetwo conditions dT2 ( Rs ) dT1 ( Rs ) k2 qi k1 , (k) dr drand T1 ( Rs ) T2 ( Rs ) . (l)
  44. 44. 2.1 Examples of One-dimensional Conduction 31where k 2 is the thermal conductivity of the shaft. This approach yields thesolutions to the temperature distribution in the sleeve and shaft obtainedabove.Example 2.3: Composite Wall with Energy GenerationA plate of thickness L1 and conductivity k1 generates heat at a volumetricrate of q . The plate is sandwiched between two plates of conductivity k 2and thickness L2 each. The exposed surfaces of the two plates aremaintained at constant temperature To . Determine the temperaturedistribution in the three plates.(1) Observations. (i) The three x Toplates form a composite wall. (ii) k2 L2The geometry can be described bya rectangular coordinate system. L1 q 0 k1(iii) Due to symmetry, no heat is k2 L2conducted through the centerplane of the heat generating plate. To Fig. 2.3(iv) Heat conduction is assumed tobe in the direction normal to the three plates. (v) Steady state and symmetryrequire that the total energy generated be conducted equally through eachof the outer plates.(2) Origin and Coordinates. To take advantage of symmetry, the originis selected at the center of the heat generating plate. The composite wall isshown in Fig. 2.3.(3) Formulation. (i) Assumptions. (1) One-dimensional, (2) steady state, (3) constantthermal conductivities and (4) perfect interface contact. (ii) Governing Equations.Since temperature is symmetrical x Toabout the origin, only half thecomposite system is considered T2 ( x ) k2 L2(see Fig. 2.4). Thus two heat T1 ( x ) q 0 k1 L1 / 2equations are required. Letsubscripts 1 and 2 refer to the Fig. 2.4center plate and outer plate,
  45. 45. 32 2 One-dimensional Steady State Conductionrespectively. Based on the above assumptions, eq. (1.8) gives d 2T1 q 2 0, (a) dx k1and d 2T2 0. (b) dx 2where k1 = thermal conductivity of the heat generating plate q = rate of energy generation per unit volume T = temperature x = coordinate (iii) Boundary Conditions. Since equations (a) and (b) are secondorder with a single independent variable x, four boundary conditions areneeded. They are:Symmetry of temperature at the center dT1 (0) 0. (c) dxConservation of energy and equality of temperature at the interface dT1 ( L1 / 2) dT2 ( L1 / 2) k1 k2 , (d) dx dxand T1 ( L1 / 2) T2 ( L1 / 2) , (e)where L1 = thickness of the center plate L2 = thickness of the outer plateSpecified temperature at the outer surface T2 ( L1 / 2 L2 ) To . (f)(4) Solution. Integrating (a) twice gives q 2 T1 ( x) x Ax B. (g) 2k1Integration of (b) gives
  46. 46. 2.1 Examples of One-dimensional Conduction 33 T2 ( x) Cx D, (h)where A, B, C and D are constants of integration. Application of the fourboundary conditions gives the constants of integration. Solutions (g) and(h), expressed in dimensionless form, become T1 ( x) To 1 1 k1 L2 1 x2 2 2 , (i) q L1 / k1 8 2 k 2 L1 2 L1and T2 ( x) To 1 k1 1 k1 L2 1 k1 x 2 . (j) q L1 / k1 4 k2 2 k 2 L1 2 k 2 L1(5) Checking. Dimensional check: Each term in solutions (i) and (j) isdimensionless.Boundary conditions check: Substitution of solutions (i) and (j) into (c), (d),(e) and (f) shows that they satisfy the four boundary conditions.Quantitative check: Conservation of energy and symmetry require that halfthe energy generated in the center plate must leave at the interfacex L1 / 2 L1 dT1 ( L1 / 2) q k1 . (k) 2 dxSubstituting (i) into (k) shows that this condition is satisfied. Similarly, halfthe heat generated in the center plate must be conducted through the outerplate L1 dT2 ( L1 / 2 L2 ) q k2 . (l) 2 dxSubstituting (j) into (l) shows that this condition is also satisfied.Limiting check: (i) If energy generation vanishes, the composite wallshould be at a uniform temperature To . Setting q 0 in (i) and (j) givesT1 ( x) T2 ( x) To .(ii) If the thickness of the center plate vanishes, no heat will be generatedand consequently the outer plate will be at uniform temperature To .Setting L1 0 in eq. (j) gives T2 ( x) To .
  47. 47. 34 2 One-dimensional Steady State Conduction(6) Comments. (i) The solution is characterized by two dimensionlessparameters: a geometric parameter L2 / L1 and a conductivity parameter k1 / k 2 . (ii) An alternate approach to solving this problem is to considerthe outer plate first and note that the heat flux at the interface is equal tohalf the energy generated in the center plate. Thus the outer plate has twoknown boundary conditions: a specified flux at x L1 / 2 and a specifiedtemperature at x ( L1 / 2) L2 . The solution gives the temperaturedistribution in the outer plate. Equality of temperatures at the interfacegives the temperature of the center plate at x L1 / 2. Thus the center platehas two known boundary conditions: an insulated condition at x 0 and aspecified temperature at x L1 / 2.2.2 Extended Surfaces: Fins2.2.1 The Function of FinsWe begin with Newtons law of cooling for surface heat transfer byconvection q s hAs (Ts T ) . (2.3)Eq. (2.3) provides an insight as to the options available for increasingsurface heat transfer rate q s . One option is to increase the heat transfercoefficient h by changing the fluid and/or manipulating its motion. Asecond option is to lower the ambient temperature T . A third option is toincrease surface area As . This option is exercised in many engineeringapplications in which the heat transfer surface is "extended" by adding fins.Inspect the back side of your refrigerator where the condenser is usuallyplaced and note the many thin rods attached to the condenser’s tube. Therods are added to increase the rate of heat transfer from the tube to thesurrounding air and thus avoid using a fan. Other examples include thehoneycomb surface of a car radiator, the corrugated surface of a motorcycleengine, and the disks attached to a baseboard radiator.2.2.2 Types of FinsVarious geometries and configurations are used to construct fins. Examplesare shown in Fig. 2.5. Each fin is shown attached to a wall or surface. Theend of the fin which is in contact with the surface is called the base whilethe free end is called the tip. The term straight is used to indicate that thebase extends along the wall in a straight fashion as shown in (a) and (b). If
  48. 48. 2.2 Extended Surfaces: Fins 35the cross-sectional area of the fin changes as one moves from the basetowards the tip, the fin is characterized as having a variable cross-sectionalarea. Examples are the fins shown in (b), (c) and (d). A spine or a pin finis distinguished by a circular cross section as in (c). A variation of the pinfin is a bar with a square or other cross-sectional geometry. An annular orcylindrical fin is a disk which is mounted on a tube as shown in (d). Sucha disk can be either of uniform or variable thickness. (a ) constant area (b) variable area fin straight fin straight fin (c) pin fin (d) annular fin Fig. 2.52.2.3 Heat Transfer and Temperature Distribution in FinsIn the pin fin shown in Fig. 2.6,heat is removed from the wall at T rthe base and is carried through hthe fin by conduction in both theaxial and radial directions. At Tthe fin surface, heat is xexchanged with the surroundingfluid by convection. Thus the hdirection of heat flow is two-dimensional. Examining the Fig. 2.6
  49. 49. 36 2 One-dimensional Steady State Conductiontemperature profile at any axial location x we note that temperaturevariation in the lateral or radial direction is barely noticeable near thecenter of the fin. However, it becomes more pronounced near its surface.This profile changes as one proceeds towards the tip. Thus temperaturedistribution is also two-dimensional.2.2.4 The Fin ApproximationAn important simplification made in the analysis of fins is based on theassumption that temperature variation in the lateral direction is negligible.That is, the temperature at any cross section is uniform. This assumptionvastly simplifies the mathematical treatment of fins since it not onlytransforms the governing equation for steady state from partial to ordinary,but also makes it possible to analytically treat fins having irregular cross-sectional areas. The question is, under what conditions can thisapproximation be made? Let us try to develop a criterion for justifying thisassumption. First, the higher the thermal conductivity is the more uniformthe temperature will be at a given cross section. Second, a low heattransfer coefficient tends to act as an insulation layer and thus forcing amore uniform temperature in the interior of the cross section. Third, thesmaller the half thickness is the smaller the temperature drop will bethrough the cross section. Assembling these three factors together gives adimensionless ratio, h / k , which is called the Biot number. Therefore,based on the above reasoning, the criterion for assuming uniformtemperature at a given cross section is a Biot number which is smallcompared to unity. That is h Biot number = Bi = << 1. (2.4) kComparisons between exact and approximate solutions have shown thatthis simplification is justified when the Biot number is less than 0.1. Notethat / k represents the internal conduction resistance and 1/h is theexternal convection resistance. Rewriting the Biot number in eq. ( 2.4) asBi = ( / k ) /(1 / h) shows that it represents the ratio of the internal andexternal resistances.
  50. 50. 2.2 Extended Surfaces: Fins 37 2.2.5 The Fin Heat Equation: Convection at Surface To determine the rate of heat transfer from fins it is first necessary to obtain the temperature distribution. As with other conduction problems, temperature distribution is determined by solving an appropriate heat equation based on the principle of conservation of energy. Since conduction in fins is two-dimensional which is modeled mathematically as one-dimensional, it is necessary to formulate the principle of conservation of energy specifically for fins. dAs dxy C ys dx h, T qx qx qx dx x0 x q dx q dy dqc ds (a ) (b) (c ) Fig. 2.7 We consider a general case of a variable area fin with volumetric energy generation q under transient conditions. The fin exchanges heat with an ambient fluid by convection. The heat transfer coefficient is h and the ambient temperature is T . We select an origin at the base and a coordinate axis x as shown in Fig. 2.7. We limit the analysis to stationary material and assume that the Biot number is small compared to unity and thus invoke the fin approximation that the temperature does not vary within a cross section. Since the temperature depends on a single spatial variable x, our starting point should be the selection of an appropriate element. We select a fin slice at location x of infinitesimal thickness dx which encompasses the entire cross section. This element is enlarged in Fig. 2.7 b. Conservation of energy for the element requires that Ein E g - E out = E , (a) where E = rate of energy change within the element Ein = rate of energy added to the element E g = rate of energy generation per unit volume

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