Principles of Heat Transfer -Kreith 7th, Manuais, Projetos, Pesquisas de Engenharia Elétrica

Baixe Principles of Heat Transfer -Kreith 7th e outras Manuais, Projetos, Pesquisas em PDF para Engenharia Elétrica, somente na Docsity! PRINCIPLES. HEAT TRANSFER aja ui] [ca Gi ELA no Mark S. Bohn Conversion Factors for Commonly Used Quantities in Heat Transfer Quantity SI : English English : SI* Area 1 m2  10.764 ft2 1 ft2  0.0929 m2  1550.0 in2 1 in2  6.452  104 m2 Density 1 kg/m3  0.06243 lbm/ft 3 1 lbm/ft 3  16.018 kg/m3 1 slug/ft3  515.38 kg/m3 Energy† 1 J  9.4787  104 Btu 1 Btu  1055.06 J 1 cal  4.1868 J 1 lbf  ft  1.3558 J 1 hp  h  2.685  106 J Energy per unit mass 1 J/kg  4.2995  104 Btu/lbm 1 Btu/lbm  2326 J/kg Force 1 N  0.22481 lbf 1 lbf  4.448 N Heat flux 1 W/m2  0.3171 Btu/(h  ft2) 1 Btu/(h  ft2)  3.1525 W/m2 1 kcal/(h  m2)  1.163 W/m2 Heat generation 1 W/m3  0.09665 Btu/(h  ft3) 1 Btu/(h  ft3)  10.343 W/m3 per unit volume Heat transfer coefficient 1 W/(m2  K)  0.1761 Btu/(h  ft2  °F) 1 Btu/(h  ft2  °F)  5.678 W/(m2  K) Heat transfer rate 1 W  3.412 Btu/h 1 Btu/h  0.2931 W 1 ton  12,000 Btu/h  3517.2 W Length 1m  3.281 ft 1 ft  0.3048 m  39.37 in 1 in  0.0254 m Mass 1 kg  2.2046 lbm 1 lbm  0.4536 kg 1 slug  14.594 kg Mass flow rate 1 kg/s  7936.6 lbm/h 1 lbm/h  0.000126 kg/s  2.2046 lbm/s 1 lbm/s  0.4536 kg/s Power 1 W  3.4123 Btu/h 1 Btu/h  0.2931 W 1 Btu/s  1055.1 W 1 lbf  ft/s  1.3558 W 1 hp  745.7 W Pressure and stress 1 N/m2  0.02089 lbf/ft 2 1 lbf/ft 2  47.88 N/m2 (Note: 1 Pa  1N/m2)  1.4504  104 lbf/in 2 1 psi  1 lbf/in 2  6894.8 N/m2  4.015  103 in water 1 standard atmosphere  1.0133  105 N/m2  2.953  104 in Hg 1 bar  1  105 N/m2 67706_IFC.qxd 5/14/10 7:30 AM Page ii Copyright 2011 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Seventh Edition Principles of HEAT TRANSFER Frank Kreith Professor Emeritus, University of Colorado at Boulder, Boulder, Colorado Raj M. Manglik Professor, University of Cincinnati, Cincinnati, Ohio Mark S. Bohn Former Vice President, Engineering Rentech, Inc., Denver, Colorado Australia • Brazil • Japan • Korea • Mexico • Singapore • Spain • United Kingdom • United States 67706_00_FM_pi-xxiii.qxd 5/14/10 9:32 AM Page iii Copyright 2011 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. 52609_00_fm_pi-pxxvi.indd ii 2/1/10 11:37:43 PM This ia an electronic version of the print textbook. Due to electronic rights restrictions, some third party may be suppressed. Edition review has deemed that any suppres ed content does not materially affect the over all learning experience. The publisher reserves the right to remove the contents from this title at any time if subsequent rights restrictions require it. For valuable information on pricing, previous editions, changes to current editions, and alternate format, please visit www.cengage.com/highered to search by ISBN#, author, title, or keyword for materials in your areas of interest. s Copyright 2011 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Principles of Heat Transfer, Seventh Edition Authors Frank Kreith, Raj M. Manglik, Mark S. Bohn Publisher, Global Engineering: Christopher M. Shortt Senior Developmental Editor: Hilda Gowans Editorial Assistant: Tanya Altieri Team Assistant: Carly Rizzo Marketing Manager: Lauren Betsos Media Editor: Chris Valentine Director, Content and Media Production: Barbara Fuller-Jacobsen Content Project Manager: Cliff Kallemeyn Production Service: RPK Editorial Services, Inc. Copyeditor: Fred Dahl Proofreader: Martha McMaster/Erin Wagner Indexer: Shelly Gerger-Knechtl Compositor: Integra Senior Art Director: Michelle Kunkler Cover Designer: Andrew Adams Cover Image: Abengoa Solar; SkyTrough™ © Shirley Speer/SkyFuel, Inc. 2009 Internal Designer: Jennifer Lambert/jen2design Text and Image Permissions Researcher: Kristiina Paul First Print Buyer: Arethea Thomas © 2011, 2003 Cengage Learning ALL RIGHTS RESERVED. No part of this work covered by the copyright herein may be reproduced, transmitted, stored, or used in any form or by any means graphic, electronic, or mechanical, including but not limited to photocopying, recording, scanning, digitizing, taping, web distribution, information networks, or information storage and retrieval systems, except as permitted under Section 107 or 108 of the 1976 United States Copyright Act, without the prior written permission of the publisher. For product information and technology assistance, contact us at Cengage Learning Customer & Sales Support, 1-800-354-9706. For permission to use material from this text or product, submit all requests online at www.cengage.com/permissions. Further permissions questions can be emailed to permissionrequest@cengage.com Library of Congress Control Number: 2010922630 ISBN-13: 978-0-495-66770-4 ISBN-10: 0-495-66770-6 Cengage Learning 200 First Stamford Place, Suite 400 Stamford, CT 06902 USA Cengage Learning is a leading provider of customized learning solutions with office locations around the globe, including Singapore, the United Kingdom, Australia, Mexico, Brazil, and Japan. Locate your local office at: international.cengage.com/region. Cengage Learning products are represented in Canada by Nelson Education Ltd. For your course and learning solutions, visit www.cengage.com/engineering. Purchase any of our products at your local college store or at our preferred online store www.CengageBrain.com. Printed in the United States of America 1 2 3 4 5 6 7 13 12 11 10 09 67706_00_FM_pi-xxiii.qxd 5/14/10 9:32 AM Page iv Copyright 2011 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. asterisks can be omitted without breaking the continuity of the presentation. If all the sections marked with an asterisk are omitted, the material in the book can be covered in a single quarter. For a full semester course, the instructor can select five or six of these sections and thus emphasize his or her own areas of interest and expertise. The senior author would also like to express his appreciation to Professor Raj M. Manglik, who assisted in the task of updating and refreshing the sixth edition to bring it up to speed for students in the twenty-first century. In turn, Raj Manglik is profoundly grateful for the opportunity to join in the authorship of this revised edition, which should continue to provide students worldwide an engaging learning experience in heat transfer. Although Dr. Mark Bohn decided not to participate in the seventh edition, we wish to express our appreciation for his previous contribu- tion. In addition, the authors would like to acknowledge the contributions by the reviewers of the sixth edition who have provided input and suggestions for the update leading to the new edition of the book: B. Rabi Baliga, McGill University; F.C. Lai, University of Oklahoma; S. Mostafa Ghiaasiaan, Georgia Tech; Michael Pate, Iowa State University; and Forman A. Williams, University of California, San Diego. The authors would also like to thank Hilda Gowans, the Senior Developmental Editor for Engineering at Cengage Learning, who has provided sup- port and encouragement throughout the preparation of the new edition. On a more personal level, Frank Kreith would like to express his appreciation to his assistant, Bev Weiler, who has supported his work in many tangible and intangible ways, and to his wife, Marion Kreith, whose forbearance with the time taken in writing books has been of invaluable help. Raj Manglik would like to thank his graduate students Prashant Patel, Rohit Gupta, and Deepak S. Kalaikadal for the computational solu- tions and algorithms in the book. Also, he would like to express his fond gratitude to his wife, Vandana Manglik, for her patient encouragement during the long hours needed in this endeavor, and to his children, Aditi and Animaesh, for their affection and willingness to forego some of our shared time. viii Preface 67706_00_FM_pi-xxiii.qxd 5/14/10 9:32 AM Page viii Copyright 2011 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. CONTENTS Chapter 1 Basic Modes of Heat Transfer 2 1.1 The Relation of Heat Transfer to Thermodynamics 3 1.2 Dimensions and Units 7 1.3 Heat Conduction 9 1.4 Convection 17 1.5 Radiation 21 1.6 Combined Heat Transfer Systems 23 1.7 Thermal Insulation 45 1.8 Heat Transfer and the Law of Energy Conservation 51 References 58 Problems 58 Design Problems 68 Chapter 2 Heat Conduction 70 2.1 Introduction 71 2.2 The Conduction Equation 71 2.3 Steady Heat Conduction in Simple Geometries 78 2.4 Extended Surfaces 95 2.5* Multidimensional Steady Conduction 105 2.6 Unsteady or Transient Heat Conduction 116 2.7* Charts for Transient Heat Conduction 134 2.8 Closing Remarks 150 References 150 Problems 151 Design Problems 163 Chapter 3 Numerical Analysis of Heat Conduction 166 3.1 Introduction 167 3.2 One-Dimensional Steady Conduction 168 3.3 One-Dimensional Unsteady Conduction 180 ix 67706_00_FM_pi-xxiii.qxd 5/14/10 9:32 AM Page ix Copyright 2011 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. 3.4* Two-Dimensional Steady and Unsteady Conduction 195 3.5* Cylindrical Coordinates 215 3.6* Irregular Boundaries 217 3.7 Closing Remarks 221 References 221 Problems 222 Design Problems 228 Chapter 4 Analysis of Convection Heat Transfer 230 4.1 Introduction 231 4.2 Convection Heat Transfer 231 4.3 Boundary Layer Fundamentals 233 4.4 Conservation Equations of Mass, Momentum, and Energy for Laminar Flow Over a Flat Plate 235 4.5 Dimensionless Boundary Layer Equations and Similarity Parameters 239 4.6 Evaluation of Convection Heat Transfer Coefficients 243 4.7 Dimensional Analysis 245 4.8* Analytic Solution for Laminar Boundary Layer Flow Over a Flat Plate 252 4.9* Approximate Integral Boundary Layer Analysis 261 4.10* Analogy Between Momentum and Heat Transfer in Turbulent Flow Over a Flat Surface 267 4.11 Reynolds Analogy for Turbulent Flow Over Plane Surfaces 273 4.12 Mixed Boundary Layer 274 4.13* Special Boundary Conditions and High-Speed Flow 277 4.14 Closing Remarks 282 References 283 Problems 284 Design Problems 294 Chapter 5 Natural Convection 296 5.1 Introduction 297 5.2 Similarity Parameters for Natural Convection 299 5.3 Empirical Correlation for Various Shapes 308 5.4* Rotating Cylinders, Disks, and Spheres 322 5.5 Combined Forced and Natural Convection 325 5.6* Finned Surfaces 328 x Contents 67706_00_FM_pi-xxiii.qxd 5/14/10 9:32 AM Page x Copyright 2011 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Liquid Metals A24 Thermodynamic Properties of Gases A26 Miscellaneous Properties and Error Function A37 Correlation Equations for Physical Properties A45 Appendix 3 Tridiagonal Matrix Computer Programs A50 Solution of a Tridiagonal System of Equations A50 Appendix 4 Computer Codes for Heat Transfer A56 Appendix 5 The Heat Transfer Literature A57 Index I1 Contents xiii 67706_00_FM_pi-xxiii.qxd 5/14/10 9:32 AM Page xiii Copyright 2011 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. NOMENCLATURE International System of English System Symbol Quantity Units of Units a velocity of sound m/s ft/s a acceleration m/s2 ft/s2 A area; Ac cross-sectional area; Ap, m 2 ft2 projected area of a body normal to the direction of flow; Aq, area through which rate of heat flow is q; As, surface area; Ao, outside surface area; Ai, inside surface area b breadth or width m ft c specific heat; cp, specific heat at J/kg K Btu/lbm °F constant pressure; c, specific heat at constant volume C constant C thermal capacity J/K Btu/°F C hourly heat capacity rate in Chapter 8; W/K Btu/h °F Cc, hourly heat capacity rate of colder fluid in a heat exchanger; Ch, hourly heat capacity rate of warmer fluid in a heat exchanger CD total drag coefficient Cf skin friction coefficient; Cfx, local value of Cf at distance x from leading edge; , average value of Cf defined by Eq. (4.31) d, D diameter; DH, hydraulic diameter; Do, m ft outside diameter; Di, inside diameter e base of natural or Napierian logarithm e internal energy per unit mass J/kg Btu/lbm E internal energy J Btu E emissive power of a radiating body; Eb, W/m 2 Btu/h ft2 emissive power of blackbody Cqf (Continued) xv 67706_00_FM_pi-xxiii.qxd 5/14/10 9:32 AM Page xv Copyright 2011 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. International System of English System Symbol Quantity Units of Units E monochromatic emissive power per W/m 2 m Btu/h ft2 micron micron at wavelength   heat exchanger effectiveness defined by Eq. (8.22) f Darcy friction factor for flow through a pipe or a duct, defined by Eq. (6.13) f friction coefficient for flow over banks of tubes defined by Eq. (7.37) F force N lbf FT temperature factor defined by Eq. (9.119) F1–2 geometric shape factor for radiation from one blackbody to another 1–2 geometric shape and emissivity factor for radiation from one graybody to another g acceleration due to gravity m/s2 ft/s2 gc dimensional conversion factor 1.0 kg m/N s 2 32.2 ft lbm/lbf s 2 G mass flow rate per unit kg/m2 s lbm/h ft 2 area (G  U) G irradiation incident on unit surface W/m2 Btu/h ft2 in unit time h enthalpy per unit mass J/kg Btu/lbm hc local convection heat transfer coefficient W/m 2 K Btu/h ft2°F combined heat transfer coefficient W/m2 K Btu/h ft2°F ; hb, heat transfer coefficient of a boiling liquid, defined by Eq. (10.1); , average convection heat transfer coefficient; , average heat transfer coefficient for radiation hfg latent heat of condensation J/kg Btu/lbm or evaporation i angle between sun direction rad deg and surface normal i electric current amp amp I intensity of radiation W/sr Btu/h sr I intensity per unit wavelength W/sr m Btu/h sr micron J radiosity W/m2 Btu/h ft2 hqr hqc hq = hqc + hqr hq xvi Nomenclature 67706_00_FM_pi-xxiii.qxd 5/14/10 9:32 AM Page xvi Copyright 2011 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. International System of English System Symbol Quantity Units of Units x coordinate m ft x quality y coordinate m ft y distance from a solid boundary measured in direction normal to surface m ft z coordinate m ft Z ratio of hourly heat capacity rates in heat exchangers Greek Letters absorptivity for radiation; , monochromatic absorptivity at wavelength  thermal diffusivity  k/c m2/s ft2/s temperature coefficient 1/K 1/R of volume expansion k temperature coefficient 1/K 1/R of thermal conductivity specific heat ratio, cp/c  body force per unit mass N/kg lbf/lbm c mass rate of flow of condensate per unit breadth for a vertical tube kg/s m lbm/h ft  boundary-layer thickness; h, m ft hydrodynamic boundary-layer thickness; th, thermal boundary-layer thickness  difference between values  packed bed void fraction  emissivity for radiation; , monochromatic emissivity at wavelength ; , emissivity in direction of  H thermal eddy diffusivity m 2/s ft2/s M momentum eddy diffusivity m 2/s ft2/s  ratio of thermal to hydrodynamic boundary-layer thickness, th/h Nomenclature xix (Continued) 67706_00_FM_pi-xxiii.qxd 5/14/10 9:32 AM Page xix Copyright 2011 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. International System of English System Symbol Quantity Units of Units f fin efficiency  time s h or s  wavelength; max, wavelength m micron at which monochromatic emissive power Eb is a maximum  latent heat of vaporization J/kg Btu/lbm  absolute viscosity N s/m2 lbm/ft s  kinematic viscosity, / m2/s ft2/s r frequency of radiation 1/s 1/s  mass density, 1/; l, density kg/m 3 lbm/ft 3 of liquid; , density of vapor  reflectivity for radiation  shearing stress; s, shearing N/m 2 lbf/ft 2 stress at surface; w, shear at wall of a tube or a duct  transmissivity for radiation  Stefan–Boltzmann constant W/m2 K4 Btu/h ft2 R4  surface tension N/m lbf/ft  angle rad rad  angular velocity rad/s rad/s  solid angle sr steradian Dimensionless Numbers Bi Fo Gz Graetz number  (/4)RePr(D/L) Gr Grashof number  gL 3 T/2 Ja Jakob number  (T  Tsat)cpl/hfg M Mach number  U/a Nux local Nusselt number at a distance x from leading edge, hcx/kf average Nusselt number for blot plate, average Nusselt number for cylinder, hqcD/kfNuD hqcL/kfNuL Fourier modulus = au/L2 or au/ro 2 Biot number = hqL/ks or hqro/ks xx Nomenclature 67706_00_FM_pi-xxiii.qxd 5/14/10 9:32 AM Page xx Copyright 2011 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Symbol Quantity Pe Peclet number  RePr Pr Prandtl number  cp/k or / Ra Rayleigh number  GrPr ReL Reynolds number  UL/; Rex  Ux/ Local value of Re at a distance x from leading edge ReD  UD/ Diameter Reynolds number Reb  DbGb/l Bubble Reynolds number  St Miscellaneous a  b a greater than b a  b a smaller than b  proportional sign approximately equal sign  infinity sign  summation sign M Stanton number = hqc/rUqcp or Nu/RePr Boundary Fourier modulus = hq 2au/ks 2 Nomenclature xxi 67706_00_FM_pi-xxiii.qxd 5/14/10 9:32 AM Page xxi Copyright 2011 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. 1.1 The Relation of Heat Transfer to Thermodynamics Whenever a temperature gradient exists within a system, or whenever two systems at different temperatures are brought into contact. energy is transferred. The process by which the energy transport taltes place is known as heat transfer. The thing in transit, called heat, cannot be observed or measured directly. However, its effects can be identified and quantified through measurements and analysis. The flow of heat, like the performance of work, is a process by which the initial energy of a system is changed. The branch of science that deals with the relation between heat and other forms of energy, including mechanical work in particular, is called thermodynamics. Its principles, like all laws of nature, are based on observations and have been gen- eralized into laws that are believed to hold for all processes occurring in nature because no exceptions have ever been found. For example, the first law of thermo- dynamics states that energy can be neither created nor destroyed but only changed from one form to another. It governs all energy transformations quantitatively, but places no restrictions on the direction of the transformation. It is known, however, from experience that no process is possible whose sole result is the net transfer of heat from a region of lower temperature to a region of higher temperature. This state- ment of experimental truth is known as the second law of thermodynamics. All heat transfer processes involve the exchange and/or conversion of energy. They must, therefore, obey the first as well as the second law of thermodynamics. At first glance, one might therefore be tempted to assume that the principles of heat transfer can be derived from the basic laws of thermodynamics. This conclusion, however, would be erroneous, because classical thermodynamics is restricted pri- marily to the study of equilibrium states including mechanical, chemical, and thermal equilibriums, and is therefore, by itself, of little help in determining quantitavely the transformations that occur from a lack of equilibrium in engineering processes. Since heat flow is the result of temperature nonequilibriuin, its quantitative treatment must be based on other branches of science. The same reasoning applies to other types of transport processes such as mass transfer and diffusion. Limitations of Classical Thermodynamics Classical thermodynamics deals with the states of systems from a macroscopic view and makes no hypotheses about the structure of matter. To perform a thermodynamic analysis it is necessary to describe the state of a system in terms of gross characteristics, such as pressure, volume, and temperature, that can be measured directly and involve no special assumptions regarding the structure of matter. These variables (or thermodynamic properties) are of significance for the system as a whole only when they are uniform throughout it, that is, when the system is in equilibrium. Thus, classical thermodynamics is not concerned with the details of a process but rather with equilibrium states and the relations among them. The processes employed in a thermodynamic analysis are idealized processes devised to give information concerning equilibrium states. 3 67706_01_ch01_p002-069.qxd 5/14/10 7:46 AM Page 3 Copyright 2011 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. 4 Chapter 1 Basic Modes of Heat Transfer The schematic example of an automobile engine in Fig. 1.1 is illustrative of the distinctions between thermodynamic and heat transfer analysis. While the basic law of energy conservation is applicable in both, from a thermodynamic viewpoint, the amount of heat transferred during a process simply equals the difference between the energy change of the system and the work done. It is evident that this type of analy- sis considers neither the mechanism of heat flow nor the time required to transfer the heat. It simply prescribes how much heat to supply to or reject from a system dur- ing a process between specified end states without considering whether, or how, this could be accomplished. The question of how long it would take to transfer a speci- fied amount of heat, via different mechanisms or modes of heat transfer and their processes (both in terms of space and time) by which they occur, although of great practical importance, does not usually enter into the thermodynamic analysis. Engineering Heat Transfer From an engineering viewpoint, the key problem is the determination of the rate of heat transfer at a specified temperature difference. Combustion Cylinder-Piston Assembly Automobile Engine Cylinder wall Heat Transfer Model Engine casing Combustion chamber qcond qconv qL qrad qrad qconv= qcond=+ Internal combustion engine Control volume EE WC EAEEqL Exhaust gases Crarks shaft Atr In Work out Fuel In Heat loss Theromodynamic Model =qL WC+ + EF EA EE 0+ −− FIGURE 1.1 A classical thermodynamics model and a heat transfer model of a typical automobile (spark-ignition internal combustion) engine. Source: Photo of automobile engine courtesy of Ajancso/shutterstock. 67706_01_ch01_p002-069.qxd 5/14/10 7:46 AM Page 4 Copyright 2011 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. TABLE 1.1 Significance and diverse practical applications of heat transfer Chemical, petrochemical, and process industry: Heat exchangers, reactors, reboilers, etc. Power generation and distribution: Boilers, condensers, cooling towers, feed heaters, transformer cooling, transmission cable cooling, etc. Aviation and space exploration: Gas turbine blade cooling, vehicle heat shields, rocket engine/nozzle cooling, space suits, space power generation, etc. Electrical machines and electronic equipment: Cooling of motors, generators, computers and microelectronic devices, etc. Manufacturing and material processing: Metal processing, heat treating, composite material processing, crystal growth, micromachining, laser machining, etc. Transportation: Engine cooling, automobile radiators, climate control, mobile food storage, etc. Fire and combustion Health care and biomedical applications: Blood warmers, organ and tissue storage, hypothermia, etc. Comfort heating, ventilation, and air-conditioning: Air conditioners, water heaters, furnaces, chillers, refrigerators, etc. Weather and environmental changes Renewable Energy System: Flat plate collectors, thermal energy storage, PV module cooling, etc. 1.1 The Relation of Heat Transfer to Thermodynamics 5 To estimate the cost, the feasibility, and the size of equipment necessary to transfer a specified amount of heat in a given time, a detailed heat transfer analysis must be made. The dimensions of boilers, heaters, refrigerators, and heat exchangers depends not only on the amount of heat to be transmitted but also on the rate at which the heat is to be transferred under given conditions. The successful operation of equipment components such as turbine blades, or the walls of combustion chambers, depends on the possibility of cooling certain metal parts by continuously removing heat from a surface at a rapid rate. A heat transfer analysis must also be made in the design of electric machines, transformers, and bearings to avoid conditions that will cause overheating and damage the equipment. The listing in Table 1.1, which by no means is comprehensive, gives an indication of the extensive significance of heat transfer and its different practical applications. These examples show that almost every branch of engineering encounters heat transfer problems, which shows that they are not capable of solution by thermodynamic reasoning alone but require an analysis based on the science of heat transfer. In heat transfer, as in other branches of engineering, the successful solution of a problem requires assumptions and idealizations. It is almost impossible to describe physical phenomena exactly, and in order to express a problem in the form of an equation that can be solved, it is necessary to make some approximations. In electri- cal circuit calculations, for example, it is usually assumed that the values of the resistances, capacitances, and inductances are independent of the current flowing through them. This assumption simplifies the analysis but may in certain cases severely limit the accuracy of the results. 67706_01_ch01_p002-069.qxd 5/14/10 7:46 AM Page 5 Copyright 2011 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. 8 Chapter 1 Basic Modes of Heat Transfer where g is the local acceleration due to gravity. Weight has the dimensions of a force and a 1-kgmass will weigh 9.8 N at sea level. It should be noted that g and gc are not similar quantities. The gravitational acceleration g depends on the location and the altitude, whereas gc is a constant whose value depends on the system of units. One of the great conveniences of the SI system is that gc is numerically equal to one and therefore need not be shown specif- ically. In the English system, on the other hand, the omission of gc will affect the numerical answer, and it is therefore imperative that it be included and clearly displayed in analysis, especially in numerical calculations. With the fundamental units of meter, kilogram, second, and kelvin, the units for both force and energy or heat are derived units. For quantifying heat, rate of heat transfer, its flux, and its temperature, the units employed as per the international con- vention are given in Table 1.2. Also listed are their counterparts in English units, along with the respective conversion factors, in cognizance of the fact that such units are still prevalent in practice in the United States. The joule (newton meter) is the only energy unit in the SI system, and the watt (joule per second) is the corresponding unit of power. In the engineering system of units, on the other hand, the Btu (British ther- mal unit) is the unit for heat or energy. It is defined as the energy required to raise the temperature of 1 lb of water by 1°F at 60°F and one atmosphere pressure. The SI unit of temperature is the kelvin, but use of the Celsius temperature scale is widespread and generally considered permissible. The kelvin is based on the ther- modynamic scale, while zero on the Celsius scale (0°C) corresponds to the freezing temperature of water and is equivalent to 273.15 K on the thermodynamic scale. Note, however, that temperature differences are numerically equivalent in K and °C, since 1 K is equal to 1°C. In the English system of units, the temperature is usually expressed in degrees Fahrenheit (°F) or, on the thermodynamic temperature scale, in degrees Rankine (°R). Here, 1 K is equal to 1.8°R and conversions for other temperature scales are given °C = °F - 32 1.8 TABLE 1.2 Dimensions and units of heat and temperature Quantity SI units English units Conversion Q, quantity of heat J Btu 1 J  9.4787  104 Btu q, rate of heat transfer J/s or W Btu/h 1 W  3.4123 Btu/h q”, heat flux W/m2 Btu/h ft2 1 W/m2  0.3171 Btu/h ft2 T, temperature K ˚R or ˚F T ˚C = (T ˚F–32)/1.8 [K] = [˚C] + 273.15 [R] = [˚F] + 459.67 T K = T ˚R/1.8 ## EXAMPLE 1.1 A masonry brick wall of a house has an inside surface temperature of 55°F and an average outside surface temperature of 45°F. The wall is 1.0 ft thick, and because of the temperature difference, the heat loss through the wall per square foot is 3.4 Btu/hft2. Express the heat loss in SI units. Also, calculate the value of this heat 67706_01_ch01_p002-069.qxd 5/14/10 7:46 AM Page 8 Copyright 2011 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. 1.3 Heat Conduction 9 loss for a 100-ft2 surface over a 24-h period if the house is heated by an electric resistance heater and the cost of electricity is 10 ¢ kWh. SOLUTION The rate of heat loss per unit surface area in SI units is The total heat loss to the environment over the specified surface area of the house wall in 24 hours is This can be expressed in SI units as And at 10 ¢ kWh, this amounts to  24 ¢ as the cost of heat loss in 24 h. 1.3 Heat Conduction Whenever a temperature gradient exists in a solid medium, heat will flow from the higher-temperature to the lower-temperature region. The rate at which heat is trans- ferred by conduction, qk, is proportional to the temperature gradient times the area A through which heat is transferred: In this relation, T(x) is the local temperature and x is the distance in the direction of the heat flow. The actual rate of heat flow depends on the thermal conductivity k, which is a physical property of the medium. For conduction through a homogeneous medium, the rate of heat transfer is then (1.2) The minus sign is a consequence of the second law of thermodynamics, which requires that heat must flow in the direction from higher to lower temperature. As illustrated in Fig. 1.2 on the next page, the temperature gradient will be negative if the temperature decreases with increasing values of x. Therefore, if heat trans- ferred in the positive x direction is to be a positive quantity, a negative sign must be inserted on the right side of Eq. (1.2). Equation (1.2) defines the thermal conductivity. It is called Fourier’s law of conduction in honor of the French scientist J. B. J. Fourier, who proposed it in 1822. qk = -kA dT dx qk r A dT dx dT>dx > Q = 8160 * 0.2931 * 10-3akWh Btu b = 2.392 [kWh] Q = 3.4aBtu ft2h b * 100(ft2) * 24(h) = 8160 [Btu] q– = 3.4aBtu ft2h b * 0.2931a W Btu/h b * 1 0.0929 a ft2 m2 b = 10.72[W/m2] > 67706_01_ch01_p002-069.qxd 5/14/10 7:46 AM Page 9 Copyright 2011 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. 10 Chapter 1 Basic Modes of Heat Transfer TABLE 1.3 Thermal conductivities of some metals, nonmetallic solids, liquids, and gases Thermal Conductivity at 300 K (540 °R) Material W/m K Btu/h ft °F Copper 399 231 Aluminum 237 137 Carbon steel, 1% C 43 25 Glass 0.81 0.47 Plastics 0.2–0.3 0.12–0.17 Water 0.6 0.35 Ethylene glycol 0.26 0.15 Engine oil 0.15 0.09 Freon (liquid) 0.07 0.04 Hydrogen 0.18 0.10 Air 0.026 0.02 Direction of Heat Flow T T(x) +ΔT +Δx is (+)dTdx Direction of Heat Flow T(x) T x x −ΔT +Δx is (−)dTdx FIGURE 1.2 The sign convention for conduction heat flow. The thermal conductivity in Eq. (1.2) is a material property that indicates the amount of heat that will flow per unit time across a unit area when the temperature gradient is unity. In the SI system, as reviewed in Section 1.2, the area is in square meters (m2), the temperature in kelvins (K), x in meters (m), and the rate of heat flow in watts (W). The thermal conductivity therefore has the units of watts per meter per kelvin (W/m K). In the English system, which is still widely used by engineers in the United States, the area is expressed in square feet (ft2), x in feet (ft), the temperature in degrees Fahrenheit (°F), and the rate of heat flow in Btu/h. Thus, k, has the units Btu/h ft °F. The conversion constant for k between the SI and English systems is Orders of magnitude of the thermal conductivity of various types of materials are presented in Table 1.3. Although, in general, the thermal conductivity varies with temperature, in many engineering problems the variation is sufficiently small to be neglected. 1 W/m K = 0.578 Btu/h ft °F 67706_01_ch01_p002-069.qxd 5/14/10 7:46 AM Page 10 Copyright 2011 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. 1.3 Heat Conduction 13 24°C Glass Window Pane 0.5 cm Glass24.5°C qk T1 Rk T2 FIGURE 1.5 Heat transfer by conduction through a window pane. EXAMPLE 1.2 Calculate the thermal resistance and the rate of heat transfer through a pane of win- dow glass 1 m high, 0.5 m wide, and 0.5 cm thick, if the outer- surface temperature is 24°C and the inner-surface temperature is 24.5°C. SOLUTION A schematic diagram of the system is shown in Fig. 1.5. Assume that steady state exists and that the temperature is uniform over the inner and outer surfaces. The ther- mal resistance to conduction Rk is from Eq. (1.4) Rk = L kA = 0.005 m 0.81 W/m K * 1 m * 0.5 m = 0.0123 K/W (k = 0.81 W/m K) k = 0 k > 0qk T2 k < 0 Physical System T(x) L x β β β FIGURE 1.4 Temperature distribution in conduction through a plane wall with constant and variable thermal conductivity. 67706_01_ch01_p002-069.qxd 5/14/10 7:46 AM Page 13 Copyright 2011 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. The rate of heat loss from the interior to the exterior surface is obtained from Eq. (1.3): Note that a temperature difference of 1°C is equal to a temperature difference of 1 K. Therefore, °C and K can be used interchangeably when temperature differences are indicated. If a temperature level is involved, however, it must be remembered that zero on the Celsius scale (0°C) is equivalent to 273.15 K on the thermodynamic or absolute temperature scale and 1.3.2 Thermal Conductivity According to Fourier’s law, Eq. (1.2), the thermal conductivity is defined as For engineering calculations we generally use experimentally measured values of thermal conductivity, although for gases at moderate temperatures the kinetic theory of gases can be used to predict the experimental values accurately. Theories have also been proposed to calculate thermal conductivities for other materials, but in the case of liquids and solids, theories are not adequate to predict thermal conductivity with satisfactory accuracy [1, 2]. Table 1.3 lists values of thermal conductivity for several materials. Note that the best conductors are pure metals and the poorest ones are gases. In between lie alloys, nonmetallic solids, and liquids. The mechanism of thermal conduction in a gas can be explained on a molecu- lar level from basic concepts of the kinetic theory of gases. The kinetic energy of a molecule is related to its temperature. Molecules in a high-temperature region have higher velocities than those in a lower-temperature region. But molecules are in continuous random motion, and as they collide with one another they exchange energy as well as momentum. When a molecule moves from a higher-temperature region to a lower-temperature region, it transports kinetic energy from the higher- to the lower-temperature part of the system. Upon collision with slower molecules, it gives up some of this energy and increases the energy of molecules with a lower energy content. In this manner, thermal energy is transferred from higher- to lower- temperature regions in a gas by molecular action. In accordance with the above simplified description, the faster molecules move, the faster they will transport energy. Consequently, the transport property that we have called thermal conductivity should depend on the temperature of the gas. A somewhat simplified analytical treatment (for example, see [3]) indicates that the thermal conductivity of a gas is proportional to the square root of the absolute temperature. At moderate pressures the space between molecules is large compared k K qk>A ƒdT>dx ƒ T(K) = T(°C) + 273.15 qk = T1 - T2 Rk = (24.5 - 24.0)°C 0.0123 K/W = 40 W 14 Chapter 1 Basic Modes of Heat Transfer 67706_01_ch01_p002-069.qxd 5/14/10 7:46 AM Page 14 Copyright 2011 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. 1.3 Heat Conduction 15 to the size of a molecule; thermal conductivity of gases is therefore essentially inde- pendent of pressure. The curves in Fig. 1.6(a) show how the thermal conductivities of some typical gases vary with temperature. The basic mechanism of energy conduction in liquids is qualitatively similar to that in gases. However, molecular conditions in liquids are more difficult to describe and the details of the conduction mechanisms in liquids are not as well understood. The curves in Fig. 1.6(b) show the thermal conductivity of some nonmetallic liquids as a function of temperature. For most liquids, the thermal conductivity decreases with increasing temperature, but water is a notable exception. The thermal conduc- tivity of liquids is insensitive to pressure except near the critical point. As a general rule, the thermal conductivity of liquids decreases with increasing molecular weight. For engineering purposes, values of the thermal conductivity of liquids are taken from tables as a function of temperature in the saturated state. Appendix 2 presents such data for several common liquids. Metallic liquids have much higher conductiv- ities than nonmetallic liquids and their properties are listed separately in Tables 25 through 27 in Appendix 2. According to current theories, solid materials consist of free electrons and atoms in a periodic lattice arrangement. Thermal energy can thus be conducted by two mech- anisms: migration of free electrons and lattice vibration. These two effects are addi- tive, but in general, the transport due to electrons is more effective than the transport Hydrogen, H2 Helium, He 1 0.1 0.01 200 300 400 500 Temperature, T (K) (a) 600 700 800 Methane, CH4 T he rm al c on du ct iv ity , k ( W /m K ) Argon, Ar Air CO2 1 0.1 0.01 200 300 Temperature, T (K) (b) 400 500 Engine oil (unused) Ethylene glycol Glycerine (glycerol) Water (@psat) T he rm al c on du ct iv ity , k ( W /m K ) R134a (@psat) FIGURE 1.6. Variation of thermal conductivity with temperature of typical fluids: (a) gases and (b) liquids. Sources for property data: ASHRAE Handbook 2005, Union Carbide (ethylene glycol) and Dow Chemicals (glycerine or glycerol). 67706_01_ch01_p002-069.qxd 5/14/10 7:46 AM Page 15 Copyright 2011 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. The situation is quite similar in natural convection. The principal difference is that in forced convection the velocity far from the surface approaches the free-stream value imposed by an external force, whereas in natural convection the velocity at first increases with increasing distance from the heat transfer surface and then decreases, as shown in Fig. 1.9. The reason for this behavior is that the action of viscosity diminishes rather rapidly with distance from the surface, while the density difference decreases more slowly. Eventually, however, the buoyant force also decreases as the fluid density approaches the value of the unheated surrounding fluid. This interaction of forces will cause the velocity to reach a maximum and then approach zero far from the heated sur- face. The temperature fields in natural and forced convection have similar shapes, and in both cases the heat transfer mechanism at the fluid-solid interface is conduction. 18 Chapter 1 Basic Modes of Heat Transfer Velocity profile y y = 0 u(y) ∂T β ∂y qc g Tsurface Tfluid Temperature profile T(y) FIGURE 1.9 Velocity and temperature distribution for natural convection over a heated flat plate inclined at angle b from the horizontal. The preceding discussion indicates that convection heat transfer depends on the density, viscosity, and velocity of the fluid as well as on its thermal properties (ther- mal conductivity and specific heat). Whereas in forced convection the velocity is usually imposed on the system by a pump or a fan and can be directly specified, in natural convection the velocity depends on the temperature difference between the surface and the fluid, the coefficient of thermal expansion of the fluid (which deter- mines the density change per unit temperature difference), and the body force field, which in systems located on the earth is simply the gravitational force. In later chapters we will develop methods for relating the temperature gradient at the interface to the external flow conditions, but for the time being we shall use a simpler approach to calculate the rate of convection heat transfer, as shown below. Irrespective of the details of the mechanism, the rate of heat transfer by convec- tion between a surface and a fluid can be calculated from the relation (1.10) where qc = rate of heat transfer by convection, W (Btu/h) A = heat transfer area, m2 (ft2) qc = hc A¢T 67706_01_ch01_p002-069.qxd 5/14/10 7:46 AM Page 18 Copyright 2011 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. 1.4 Convection 19 ¢T = difference between the surface temperature Ts and a temperature of the fluid at some specified location (usually far way from the surface), K (°F) = average convection heat transfer coefficient over the area A (often called the surface coefficient of heat transfer or the convection heat transfer coefficient), W/m2 K (Btu/h ft2 °F) The relation expressed by Eq. (1.10) was originally proposed by the British scien- tist Isaac Newton in 1701. Engineers have used this equation for many years, even though it is a definition of rather than a phenomenological law of convection. Evaluation of the convection heat transfer coefficient is difficult because convec- tion is a very complex phenomenon. The methods and techniques available for a quantitative evaluation of will be presented in later chapters. At this point it is sufficient to note that the numerical value of in a system depends on the geom- etry of the surface, on the velocity as well as the physical properties of the fluid, and often even on the temperature difference ¢T. In view of the fact that these quantities are not necessarily constant over a surface, the convection heat transfer coefficient may also vary from point to point. For this reason, we must distinguish between a local and an average convection heat transfer coefficient. The local coefficient hc is defined by (1.11) while the average coefficient can be defined in terms of the local value by (1.12) For most engineering applications, we are interested in average values. Typical val- ues of the order of magnitude of average convection heat transfer coefficients seen in engineering practice are given in Table 1.4. Using Eq. (1.10), we can define the thermal conductance for convection heat transfer Kc as (1.13)Kc = hc A (W/K) hc = 1 ALLA hc dA hc dqc = hc dA(Ts - Tq) hc hc hc hc T q TABLE 1.4 Order of magnitude of convection heat transfer coefficients Convection Heat Transfer Coefficient Fluid W/m2 K Btu/h ft2 °F Air, free convection 6–30 1–5 Superheated steam or air, forced convection 30–300 5–50 Oil, forced convection 60–1,800 10–300 Water, forced convection 300–18,000 50–3,000 Water, boiling 3,000–60,000 500–10,000 Steam, condensing 6,000–120,000 1,000–20,000 hc 67706_01_ch01_p002-069.qxd 5/14/10 7:46 AM Page 19 Copyright 2011 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. 20 Chapter 1 Basic Modes of Heat Transfer and the thermal resistance to convection heat transfer Rc, which is equal to the reciprocal of the conductance, as (1.14) EXAMPLE 1.3 Calculate the rate of heat transfer by natural convection between a shed roof of area and ambient air, if the roof surface temperature is 27°C, the air- temperature -3°C, and the average convection heat transfer coefficient 10 W/m2 K (see Fig. 1.10). SOLUTION Assume that steady state exists and the direction of heat flow is from the air to the roof. The rate of heat transfer by convection from the air to the roof is then given by Eq. (1.10): Note that in using Eq. (1.10), we initially assumed that the heat transfer would be from the air to the roof. But since the heat flow under this assumption turns out to be a negative quantity, the direction of heat flow is actually from the roof to the air. We could, of course, have deduced this at the outset by applying the second law of thermodynamics, which tells us that heat will always flow from a higher to a lower temperature if there is no external intervention. But as we shall see in a later section, thermodynamic arguments cannot always he used at the outset in heat trans- fer problems because in many real situations the surface temperature is not known. = -120,000 W = 10 (W/m2 K) * 400 m2(-3 - 27)°C qc = hc Aroof (Tair - Troof) 20 m * 20 m Rc = 1 hc A (K/W) Tair = 3°C Troof = 27°C 20 m20 m FIGURE 1.10 Schematic sketch of shed for analysis of roof temperature in Example 1.3. 67706_01_ch01_p002-069.qxd 5/14/10 7:46 AM Page 20 Copyright 2011 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. 1.6 Combined Heat Transfer Systems 23 2-cm diameter Interior walls of furnace at 800K Heating rod at 1000K FIGURE 1.12 Schematic diagram of vacuum furnace with heating rod for Example 1.4. SOLUTION Assume that steady state has been reached. Moreover, note that since the walls of the furnace completely enclose the heating rod, all the radiant energy emitted by the surface of the rod is intercepted by the furnace walls. Thus, for a black enclosure, Eq. (1.17) applies and the net heat loss from the rod of surface A1 is Note that in order for steady state to exist, the heating rod must dissipate electrical energy at the rate of 1893 W and the rate of heat loss through the furnace walls must equal the rate of electric input to the system, that is, to the rod. From Eq. (1.17), , and therefore the radiation heat transfer coeffi- cient, according to its definition in Eq. (1.21), is Here, we have used T2 as the reference temperature . 1.6 Combined Heat Transfer Systems In the preceding sections the three basic mechanisms of heat transfer have been treated separately. In practice, however, heat is usually transferred by several of the basic mechanisms occurring simultaneously. For example, in the winter, heat is T¿2 hr = e1s(T1 4 - T2 4) T1 - T2 = 151 W/m2 K f1-2 = e1 = 1893 W = p(0.02 m)(1.0 m)(0.9)a5.67 * 10-8 W m2K4 b (10004 - 8004)(K4) qr = Aes(T1 4 - T2 4) = pD1Les(T1 4 - T2 4) 67706_01_ch01_p002-069.qxd 5/14/10 7:46 AM Page 23 Copyright 2011 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. 24 Chapter 1 Basic Modes of Heat Transfer TABLE 1.5 The three modes of heat transfer One dimensional conduction heat transfer through a stationary medium Convection heat transfer from a surface to a moving fluid Net radiation heat transfer from surface 1 to surface 2 Rr = T1 - T2 A1f1-2s(T1 4 - T2 4) qr = A1f1-2s(T1 4 - T2 4) = T2 - T2 Rr Rc = 1 hc A qc = hc A(Ts - Tq) = Ts - Tq Rc Rk = L kA qk = kA L (T1 - T2) = T1 - T2 Rk A T1 T1 > T2 L Ts > T∞ qc T2 Thermal conductivity, k Average convection heat transfer coefficient, hc Solid or stationary fluid Surface 1 at T1 Surface 2 at T2 Moving fluid at T∞ Surface at TsA A1 qr, 1 qr, 2 T1 > T2 qk qr, net transferred from the roof of a house to the colder ambient environment not only by convection but also by radiation, while the heat transfer through the roof from the interior to the exterior surface is by conduction. Heat transfer between the panes of a double-glazed window occurs by convection and radiation acting in parallel, while the transfer through the panes of glass is by conduction with some radiation passing directly through the entire window system. In this section, we will examine com- bined heat transfer problems. We will set up and solve these problems by dividing the heat transfer path into sections that can be connected in series, just like an elec- trical circuit, with heat being transferred in each section by one or more mechanisms acting in parallel. Table 1.5 summarizes the basic relations for the rate equation of each of the three basic heat transfer mechanisms to aid in setting up the thermal cir- cuits for solving combined heat transfer problems. 1.6.1 Plane Walls in Series and Parallel If heat is conducted through several plane walls in good thermal contact, as through a multilayer wall of a building, the rate of heat conduction is the same through all sections. However, as shown in Fig. 1.13 for a three-layer system, the temperature 67706_01_ch01_p002-069.qxd 5/14/10 7:46 AM Page 24 Copyright 2011 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. 1.6 Combined Heat Transfer Systems 25 Physical System Material A kA qk Material B kB Material C kC LCLB qk LA Thermal Circuit LA kAA qk T1 R1 = T2 T3 T4 LB kBA R2 = LC kCA R3 = FIGURE 1.13 Conduction through a three-layer system in series. gradients in the layers are different. The rate of heat conduction through each layer is qk, and from Eq. (1.2) we get (1.23) Eliminating the intermediate temperatures T2 and T3 in Eq. (1.23), qk can be expressed in the form Similarly, for N layers in series we have (1.24) where T1 is the outer-surface temperature of layer 1 and is the outer-surface temperature of layer N. Using the definition of thermal resistance from Eq. (1.4), Eq. (1.24) becomes (1.25) where ¢T is the overall temperature difference, often called the temperature poten- tial. The rate of heat flow is proportional to the temperature potential. qk = T1 - TN+1 a n=N n=1 Rk, n = ¢T a n=N n=1 Rk, n TN+1 qk = T1 - TN+1 a n=N n=1 (L>kA)n qk = T1 - T4 1L>kA2A + 1L>kA2B + 1L>kA2C qk = a kAL bA(T1 - T2) = a kA L b B (T2 - T3) = a kAL bC(T3 - T4) 67706_01_ch01_p002-069.qxd 5/14/10 7:46 AM Page 25 Copyright 2011 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. 28 Chapter 1 Basic Modes of Heat Transfer Physical System Thermal Circuit kAAA T1 A kB kA AB L qk T2 T1 T2 L kBAB R2 = L kAAA R1 = FIGURE 1.16 Heat conduction through a wall section with two paths in parallel. Conduction can occur in a section with two different materials in parallel. For example, Fig. 1.16 shows the cross-section of a slab with two different materials of areas AA and AB in parallel. If the temperatures over the left and right faces are uni- form at T1 and T2, we can analyze the problem in terms of the thermal circuit shown to the right of the physical systems. Since heat is conducted through the two mate- rials along separate paths between the same potential, the total rate of heat flow is the sum of the flows through AA and AB: (1.26) Note that the total heat transfer area is the sum of AA and AB and that the total resist- ance equals the product of the individual resistances divided by their sum, as in any parallel circuit. A more complex application of the thermal network approach is illustrated in Fig. 1.17, where heat is transferred through a composite structure involving thermal resistances in series and in parallel. For this system the resistance of the middle layer, R2 becomes and the rate of heat flow is (1.27) where N = number of layers in series (three) Rn = thermal resistance of nth layer = temperature difference across two outer surfaces¢Toverall qk = ¢Toverall a n=3 n=1 Rn R2 = RBRC RB + RC = T1 - T2 1L>k A2A + T1 - T2 1L>k A2B = T1 - T2 R1R2 >1R1 + R22 qk = q1 + q2 67706_01_ch01_p002-069.qxd 5/14/10 7:46 AM Page 28 Copyright 2011 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. 1.6 Combined Heat Transfer Systems 29 Physical System Material A kAqk T1 Section 1 Section 2 Section 3 Material B kB Material C kC Material D kD LA LB = LC LD T2 LA kAAA qk T1 R1 = Tx Ty T2 qkqk LB kBAB RB = LC kCAC RC = LD kDAD R3 = FIGURE 1.17 Conduction through a wall consisting of series and parallel thermal paths. By analogy to Eqs. (1.4) and (1.5), Eq. (1.27) can also be used to obtain an overall conductance between the two outer surfaces: (1.28) EXAMPLE 1.7 A layer of 2-in.-thick firebrick is placed between two -in.- thick steel plates . The faces of the brick adjacent to the plates are rough, having solid-to-solid contact over only 30 percent of the total area, with the average height of asperities being in. If the surface temperatures of the steel plates are 200° and 800°F, respectively, determine the rate of heat flow per unit area. SOLUTION The real system is first idealized by assuming that the asperities of the surface are dis- tributed as shown in Fig. 1.18 on the next page. We note that the composite wall is symmetrical with respect to the center plane and therefore consider only half of the system. The overall unit conductance for half the composite wall is then, from Eq. (1.28), from an inspection of the thermal circuit. Kk = 1 R1 + [R4R5>(R4 + R5)] + R3 1>32 (ks = 30 Btu/h ft °F) 1>4(kb = 1.0 Btu/h ft °F) Kk = aa n=N n=1 Rnb -1 67706_01_ch01_p002-069.qxd 5/14/10 7:46 AM Page 29 Copyright 2011 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. 30 Chapter 1 Basic Modes of Heat Transfer T1 T2 R1 R4 R2 R3 R5 Firebrick Center Physical System Thermal Circuit (a) (c) (b) Steel plates 2 in. 1/4 in.1/4 in. Air ka b1 + b2 Steel plate ka ks ka b1 + b2 b3L1 L2 b2 b1 kb Firebrick FIGURE 1.18 Thermal circuit for the parallel-series composite wall in Example 1.7. ; ; ; T1 is at the center.L3 = 1>4 in.L2 = 1>32 in.L1 = 1 in. The thermal resistance of the steel plate R3 is, on the basis of a unit wall area, equal to The thermal resistance of the brick asperities R4 is, on the basis of a unit wall area, equal to Since the air is trapped in very small compartments, the effects of convection are small and it will be assumed that heat flows through the air by conduction. At a temperature of 300°F, the conductivity of air ka is about 0.02 Btu/h ft°F. Then R5, the thermal resistance of the air trapped between the asperities, is, on the basis of a unit area, equal to The factors 0.3 and 0.7 in R4 and R5, respectively, represent the percent of the total area of the two separately heat flow paths. R5 = L2 0.7ka = 11>32 in.2 112 in./ft210.02 Btu/h °F ft2 = 186 * 10-3 (Btu/h ft2 °F)-1 R4 = L2 0.3kb = 11>32 in.2 112 in./ft210.3211 Btu/h °F ft2 = 8.68 * 10-3(Btu/h ft2 °F)-1 R3 = L3 ks = 11>4 in.2 112 in./ft2130 Btu/h °F ft2 = 0.694 * 10-3 M 0.69 * 10-3(Btu/h ft2 °F)-1 67706_01_ch01_p002-069.qxd 5/14/10 7:46 AM Page 30 Copyright 2011 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. 1.6 Combined Heat Transfer Systems 33 TABLE 1.7 Thermal contact resistance for aluminum– aluminum interfacea with different interfacial fluids [5] Interfacial Fluid Resistance, Air Helium Hydrogen Silicone oil Glycerin a 10-mm surface roughness under 105 N/m2 contact pressure. 0.265 * 10-4 0.525 * 10-4 0.720 * 10-4 1.05 * 10-4 2.75 * 10-4 Ri(m 2 K/W) found. Each situation must be treated separately. The results of many different conditions and materials have been summarized by Fletcher [6]. In Fig. 1.20 some experimental results for the contact resistance between dissimilar base metal sur- faces at atmospheric pressure are plotted as a function of contact pressure. Efforts have been made to reduce the contact resistance by placing a soft metallic foil, a grease, or a viscous liquid at the interface between the contacting materials. 0 0.001 0.01 C on ta ct r es is ta nc e R i ( m 2 K /k W ) 0.1 1.0 5 10 15 20 Contact pressure (MPa) 25 30 35 m k g p i n o j h, l f c a b q e d FIGURE 1.20 Contact resistances between dissimilar bare metal surfaces. Solid metal blocks in air at 1 atmosphere absolute pressure (see legend, on the next page). 67706_01_ch01_p002-069.qxd 5/14/10 7:46 AM Page 33 Copyright 2011 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. 34 Chapter 1 Basic Modes of Heat Transfer Legend for Fig. 1.20 Curve in Roughness Scatter Fig. 1.20 Material Finish rms (Mm) Temp. (°C) Condition of Data a 416 Stainless Ground 0.76–1.65 93 Heat flow from stainless 7075(75S)T6 Al to aluminum b 7075(75S)T6 Al Ground 1.65–0.76 93–204 Heat flow from to Stainless aluminum to stainless c Stainless 19.94–29.97 20 Clean Aluminum d Stainless 1.02–2.03 20 Clean Aluminum e Bessemer Steel Ground 3.00–3.00 20 Clean Foundry Brass f Steel Ct-30 Milled 7.24–5.13 20 Clean g Steel Ct-30 Ground 1.98–1.52 20 Clean h Steel Ct-30 Milled 7.24–4.47 20 Clean Aluminum i Steel Ct-30 Ground 1.98–1.35 20 Clean Aluminum j Steel Ct-30 Copper Milled 7.24–4.42 20 Clean k Steel Ct-30 Ground 1.98–1.42 20 Clean Copper l Brass Milled 5.13–4.47 20 Clean Aluminum m Brass Ground 1.52–1.35 20 Clean Aluminum n Brass Milled 5.13–4.42 20 Clean Copper o Aluminum Milled 4.47–4.42 20 Clean Copper p Aluminum Ground 1.35–1.42 20 Clean Copper q Uranium Ground 20 Clean Aluminum Source: Abstracted from the Heat Transfer and Fluid Flow Data Books, F. Kreith ed., Genium Pub., Comp., Schenectady, NY, 1991, With permission. ;30% ;26% This procedure can reduce the contact resistance as shown in Table 1.7, but there is no way to predict the effect quantitatively. High-conductivity pastes are often used to mount electronic components to heat sinks. These pastes fill in the interstices and reduce the thermal resistance at the component/heat sink interface. 67706_01_ch01_p002-069.qxd 5/14/10 7:46 AM Page 34 Copyright 2011 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. 1.6 Combined Heat Transfer Systems 35 Instrument package (with insulation removed) Duralumin base plate at 0°C 10 cm 10 cm 1 cm Integrated circuitFastening screws (4) 2 cm FIGURE 1.21 Schematic sketch of instrument for ozone measurement EXAMPLE 1.8 An instrument used to study the ozone depletion near the poles is placed on a large 2-cm-thick duralumin plate. To simplify this analysis the instrument can be thought of as a stainless steel plate 1 cm tall with a 10-cm * 10-cm square base, as shown in Fig. 1.21. The interface roughness of the steel and the duralumin is between 20 and 30 rms (mm). Four screws at the corners provide fastening that exerts an average pressure of 1000 psi. The top and sides of the instruments are thermally insulated. An integrated circuit placed between the insulation and the upper surface of the stainless steel plate generates heat. If this heat is to be transferred to the lower surface of the duralumin, estimated to be at a temperature of 0°C, determine the maximum allowable dissipation rate from the circuit if its temperature is not to exceed 40°C. SOLUTION Since the top and sides of the instrument are insulated, all the heat generated by the cir- cuit must flow downward. The thermal circuit will have three resistances—the stainless steel, the contact, and the duralumin. Using thermal conductivities from Table 10 in Appendix 2, the thermal resistances of the metal plates are calculated from Eq. 1.4: Stainless: Duralumin: The contact resistance is obtained from Fig. 1.20. The contact pressure of 1000 psi equals about or 7 MPa. For that pressure the unit contact resist- ance given by line c in Fig. 1.20 is 0.5 m2 K/kW. Hence, Ri = 0.5 m2K kW * 10-3 kW W * 1 0.01 m2 = 0.05 K W 7 * 106 N/m2 Rk = LA1 AkA1 = 0.02 m 0.01 m2 * 164 W/m K = 0.012 K W Rk = Lss Akss = 0.01 m 0.01 m2 * 144 W/m K = 0.07 K W 67706_01_ch01_p002-069.qxd 5/14/10 7:46 AM Page 35 Copyright 2011 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. 38 Chapter 1 Basic Modes of Heat Transfer Th Ts, 1 Ts, 3 kA LA LB LC kB kC Tc LA kAA 3 2 1 Layer Moving fluid Tc CBA Moving fluid Th R T2 T3 Physical System 1 2 3 LB kBA LC kCA hc, hot A Thot Ts, 1 Ts, 3 TcoldT2 T3 1 hc, cold A 1 kAA LA q Thermal Circuit kBA LB kCA LC FIGURE 1.23 Schematic diagram and thermal circuit for composite three-layer wall with convection over both exterior surfaces. and from Eq. (1.30) the rate of heat transfer per unit area is The approach used in Example 1.9 can also be used for composite walls, and Fig. 1.23 shows the structure, temperature distribution, and equivalent network for a wall with three layers and convection on both surfaces. 1.6.4 Convection and Radiation in Parallel In many engineering problems a surface loses or receives thermal energy by con- vection and radiation simultaneously. For example, the roof of a house heated from the interior is at a higher temperature than the ambient air and thus loses heat by convection as well as radiation. Since both heat flows emanate from the same potential, that is, the roof, they act in parallel. Similarly, the gases in a combustion chamber contain species that emit and absorb radiation. Consequently, the wall of the combustion chamber receives heat by convection as well as radiation. Figure 1.24 illustrates the cocurrent heat transfer from a q A = ¢T R1 + R2 + R3 = 1330 - 2702 K 10.10 + 0.143 + 0.0252 K/W = 223.9 W 67706_01_ch01_p002-069.qxd 5/14/10 7:46 AM Page 38 Copyright 2011 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. 1.6 Combined Heat Transfer Systems 39 Physical System T2 A2 qr = hr A1 (T1 = T2) Surrounding air at T2 qc = hcA1(T1–T2) Surface at T1 T1 T2 Simplified Circuit Rc Rr Rc + Rr R = T1 – T2 R q = = hA(T1 – T2) T1 T2 Thermal Circuit T1 – T2 Rc T1 – T2 Rr 1 hcA1 Rc = q = + 1 hrA1 Rr = FIGURE 1.24 Thermal circuit with convection and radiation acting in parallel. surface to its surroundings by convection and radiation. The total rate of heat transfer is the sum of the rates of heat flow by convection and radiation, or (1.31) where is the average convection heat transfer coefficient between area A1 and the ambient air at T2, and, as shown previously, the radiation heat transfer coefficient between A1 and the surroundings at T2 is (1.32) The analysis of combined heat transfer, especially at boundaries of a complicated geometry or in unsteady-state conduction, can often be simplified by using an effective heat transfer coefficient that combines convection and radiation. The hr = e1s(T1 4 - T2 4) T1 - T2 hc = (hc + hr) A(T1 - T2) = hc A(T1 - T2) + hr A(T1 - T2) q = qc + qr 67706_01_ch01_p002-069.qxd 5/14/10 7:46 AM Page 39 Copyright 2011 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. 40 Chapter 1 Basic Modes of Heat Transfer Pipe Room temperature = 300 K Pipe surface temperature = 500 K = 0.9Steam qr qc FIGURE 1.25 Schematic diagram of steam pipe for Example 1.10. combined heat transfer coefficient (or heat transfer coefficient for short) is defined by (1.33) The combined heat transfer coefficient specifies the average total rate of heat flow between a surface and an adjacent fluid and the surroundings per unit surface area and unit temperature difference between the surface and the fluid. Its units are W/m2 K. EXAMPLE 1.10 A 0.5-m-diameter pipe carrying steam has a surface temperature of 500 K (see Fig. 1.25). The pipe is located in a room at 300 K, and the convection heat trans- fer coefficient between the pipe surface and the air in the room is 20 W/m2 K. Calculate the combined heat transfer coefficient and the rate of heat loss per meter of pipe length. (e = 0.9) h = hc + hr SOLUTION This problem may be idealized as a small object (the pipe) inside a large black enclo- sure (the room). Noting that the radiation heat transfer coefficient is, from Eq. (1.33), The combined heat transfer coefficient is, from Eq. (1.33), and the rate of heat loss per meter is 1.6.4 Overall Heat Transfer Coefficient We noted previously that a common heat transfer problem is to determine the rate of heat flow between two fluids, gaseous or liquid, separated by a wall (see Fig. 1.26.). If the wall is plane and heat is transferred only by convection on both q = pDLh(Tpipe - Tair) = (p)(0.5 m)(1 m)(33.9 W/m 2 K)(200 K) = 10,650 W h = hc + hr = 20 + 13.9 = 33.9 W/m 2 K hr = se(T1 2 + T2 2)(T1 + T2) = 13.9 W/m 2 K T1 4 - T2 4 T1 - T2 = (T1 2 + T2 2)(T1 + T2) 67706_01_ch01_p002-069.qxd 5/14/10 7:46 AM Page 40 Copyright 2011 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. 1.6 Combined Heat Transfer Systems 43 the radiation heat transfer coefficient in the first section (e is assumed unity) convection heat transfer coefficient from gas to wall combined thermal resistance of first section In the steady state, heat is conducted through the shell, the second section of the sys- tem, at the same rate as to the surface and (1.37) where Tsc = surface temperature at wall on coolant side R2 = thermal resistance of second section After passing through the wall, the heat flows through the third section of the sys- tem by convection to the coolant. The rate of heat flow in the third and last step is (1.38) where Tl = temperature of liquid coolant R3 = thermal resistance in third section of system It should be noted that the symbol stands for average convection heat trans- fer coefficient in general, but the numerical values of the convection coefficients in the first, , and third, , sections of the system depend on many factors and will, in general, be different. Also note that the areas of the three-heat-flow sections are not equal, but since the wall is very thin, the change in the heat-flow area is so small that it can be neglected in this system. In practice, often only the temperatures of the hot gas and the coolant are known. If intermediate temperatures are eliminated by algebraic addition of Eqs. (1.36), (1.37), and (1.38), the rate of heat flow is (1.39) where the thermal resistance of the three series-connected sections or heat flow steps in the system are defined in Eqs. (1.36), (1.37), and (1.38). EXAMPLE 1.11 In the design of a heat exchanger for aircraft application (Fig. 1.28 on the next page), the maximum wall temperature in steady state is not to exceed 800 K. For the con- ditions tabulated below, determine the maximum permissible unit thermal resistance per square meter of the metal wall that separates the hot gas from the cold gas. q = Tg - Tl R1 + R2 + R3 = ¢Ttotal R1 + R2 + R3 hc3hc1 hqc = Tsc - Tl R3 q = qc = hc3 A(Tsc - Tl) = Tsg - Tsc R2 q = qk = kA L (Tsg - Tsc) R1 = 1 1hr + hc12A = hc1 = hr1 = s1Tg4 - Tsg4 2 Tg - Tsg = 67706_01_ch01_p002-069.qxd 5/14/10 7:46 AM Page 43 Copyright 2011 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. 44 Chapter 1 Basic Modes of Heat Transfer Hot-gas temperature = Tgh = 1300 K Heat transfer coefficient on hot side Heat transfer coefficient on cold side Coolant temperature SOLUTION In the steady state we can write from hot gas to hot side of wall from hot gas through wall to cold gas Using the nomenclature in Fig. 1.28, we get where Tsg is the hot-surface temperature. Substituting numerical values for the unit thermal resistances and temperatures yields q A = Tgh - Tsg R1 = Tgh - Tgc R1 + R2 + R3 = q A q A = Tgc = 300 K = h3 = hc3 = 400 W/m 2 K = h1 = 200 W/m 2 K Physical System Hot gas Metal wall Coolant (Cold surface)(Hot surface) Tgh Tsg Tsc Tgc k L Detailed Thermal Circuit (a) (b) Tgh Tsg Tsc Tgc R2 R1r = 1 Ahr R1c = 1 Ahc1 R3 = 1 Ahc3 Simplified Circuit Tgh Tsg Tsc Tgc R1 = 1 Ah1 R2 = L kA R3 = 1 Ahc3 Coolant gas Schematic of Aircraft Heat Exchanger Section Hot exhaust gas FIGURE 1.28 Physical system and thermal circuit for Example 1.11. 67706_01_ch01_p002-069.qxd 5/14/10 7:46 AM Page 44 Copyright 2011 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. 1.7 Thermal Insulation 45 Solving for R2 gives Thus, a unit thermal resistance larger than for the wall would raise the inner-wall temperature above 800 K. This value can place an upper limit on the wall thickness. 1.7 Thermal Insulation There are many situations in engineering design when the objective is to reduce the flow of heat. Examples of such cases include the insulation of buildings to minimize heat loss in the winter, a thermos bottle to keep tea or coffee hot, and a ski jacket to prevent excessive heat loss from a skier. All of these examples require the use of thermal insulation. Thermal insulation materials must have a low thermal conductivity. In most cases, this is achieved by trapping air or some other gas inside small cavities in a solid, but sometimes the same effect can be produced by filling the space across which heat flow is to be reduced with small solid particles and trapping air between the particles. These types of thermal insulation materials use the inherently low con- ductivity of a gas to inhibit heat flow. However, since gases are fluids, heat can also be transferred by natural convection inside the gas pockets and by radiation between the solid enclosure walls. The conductivity of insulting materials is therefore not really a material property but rather the result of a combination of heat flow mech- anisms. The thermal conductivity of insulation is an effective value, keff, that changes not only with temperature, but also with pressure and environmental condi- tions, e.g., moisture. The change of keff with temperature can be quite pronounced, especially at elevated temperatures when radiation plays a significant role in the overall heat transport process. The many different types of insulation materials can essentially be classified in the following three broad categories: 1. Fibrous. Fibrous materials consist of small-diameter particles of filaments of low density that can be poured into a gap as “loose fill” or formed into boards, batts, or blankets. Fibrous materials have very high porosity (-90%). Mineral wool is a common fibrous material for applications at temperatures below 700°C, and fiberglass is often used for temperatures below 200°C. For thermal protection at temperatures between 700°C to 1700°C one can use refractory fibers such as alumina or silica .(SiO2)(Al2O3) 06025 m2 K/W R2 = 06025 m 2 K/W 1300 - 800 0.005 = 1300 - 300 R2 + 0.0075 300 - 800 1>200 = 1300 - 300 1>200 + R2 + 1>400 67706_01_ch01_p002-069.qxd 5/14/10 7:46 AM Page 45 Copyright 2011 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. 48 Chapter 1 Basic Modes of Heat Transfer 0.30 Diatomaceous silica (powder) Zirconia powder (980°C) Mineral fiber (~600°C) Silica powder (~1000°C) Perlite (expanded) (980°C) Vermiculite (expanded) (960°C) Alumina-silica (milled) (1260°C) 1 2 3 4 5 6 7 1 6 5 3 4 7 2 0.25 0.20 0.15 E ff ec tiv e th er m al c on du ct iv ity ( W /m K ) Temperature (°C) 0.10 0.05 0 200 400 600 800 FIGURE 1.31 Effective thermal conductivity vs. temperature for some high-temperature insulations. The maximum useful temperature is given in parentheses. or loss of vacuum. Note that except for cellular glass, cellular insulating materials are plastics that are inexpensive and lightweight, i.e., they have densities on the order of 30 kg/m3. All cellular materials are rigid and can be obtained in practically any desired shape. For high-temperature applications refractory materials are used. They come in the form of bricks and can withstand temperatures up to 1700°C. The effective con- ductivities range from 1.5 W/m K for fire clay to about 2.5 W/m K for zirconium. Loose-fill types of insulation have much lower thermal conductivities, as shown in Fig. 1.31, but most of them can only be used below about 900°C. Loose-fill materi- als also trend to “settle,” causing potential problems in places that are difficult to access. 67706_01_ch01_p002-069.qxd 5/14/10 7:46 AM Page 48 Copyright 2011 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. 1.7 Thermal Insulation 49 EXAMPLE 1.12 The door for an industrial gas furnace is in surface area and is to be insulated to reduce heat loss to no more than . The door is shown schemat- ically in Fig. 1.32. The interior surface is a -in.-thick Inconel 600 sheet, and the outer surface is -in.-thick sheet of stainless steel 316. Between these metal sheets a suitable thickness of insulation material is to be placed. The effective gas tem- perature inside the furnace is 1200°C, and the overall heat transfer coefficient between the gas and the door is . The heat transfer coefficient between the outer surface of the door and the surroundings at 20°C is . Select a suitable insulation material and size its thickness. SOLUTION From Fig. 1.7 we estimate the thermal conductivity of the metal sheets to be approx- imately 43 W/m K. The thermal resistances of the two metal sheets are approximately These resistances are negligible compared to the other three resistances shown in the simplified thermal circuit below: The temperature drop between the gas and the interior surface of the door at the specified heat flux is: Hence, the temperature of the Inconel will be about 1140°C. This is acceptable since no appreciable structural load is applied. ¢T = q>A U = 1200 W/m2 20 W/m2 K = 60 K Air 1 h Ra = Rins Insulation Gas 1200°C20°C 1 Ui Rg = R = L>k ' 0.625 in. 43 W/m K * 1 m 39.4 in. ' 3.7 * 10-4 m2 K/W h = 5 W/m2 K Ui = 20 W/m 2 K 1>4 3 >81200 W/m 2 2 m * 4 m Insulation 3/8 in. Inconel 6001/4 in. stainless steel 316 Furnance Door Cross Section FIGURE 1.32 Cross section of composite wall of gas furnace door for Example 1.12. 67706_01_ch01_p002-069.qxd 5/14/10 7:46 AM Page 49 Copyright 2011 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. 50 Chapter 1 Basic Modes of Heat Transfer From Fig. 1.31 we see that only milled alumina-silica chips can withstand the maximum temperature in the door. Thermal conductivity data are available only between 300 and 650°C. The trend of the data suggests that at higher temperatures when radiation becomes the dominant mechanism, the increase of keff with T will become more pronounced. We shall select the value at 650°C (0.27 W/mK) and then apply a safety factor to the insulation thickness. The temperature drop at the outer surface is Hence, ¢T across the insulation is . The insula- tion thickness for is: In view of the uncertainty, in the value of keff, and the possibility that the insulation may become more compact with use, a prudent design would double the value of insulation thickness. Additional insulation would also reduce the temperature of the outer surface of the door for safety, comfort, and ease of operation. In engineering practice, especially for building materials, insulation is often characterized by a term called R-value. The temperature difference divided by the R-value gives the rate of heat transfer per unit area. For a large sheet or slab of material: The R-value is generally given in the English units of h ft2 °F/Btu. For example, the R-value of a 3.5-in.-thick sheet of fiberglass ( from Table 11 in Appendix 2) equals R-values can also be assigned to composite structures such as double-glazed win- dows or walls constructed of wood with insulation between the struts. In some cases the R-value is given on a “per inch” basis. Then its units are h ft2 °F/Btu in. In the above example, the R-value per inch of the fiberglass is in. Note that the R-value per inch is equal to when the thermal conductivity is given in units of Btu/h ft °F. Care should be exercised when using manufacturers’ literature for R-values because the per-inch value may be given even though the property may be called simply the R-value. By examining the units given for the property it should be clear which R-value is given. 1>12k8.3>3.5 ' 2.4 h ft2 °F/Btu 13.5 in.2 h ft °F 0.035 Btu * ft 12 in. = 8.3 h °F ft2 Btu keff = 0.035 Btu/h ft °F R-value = thickness effective average thermal conductivity L = k ¢T q>A = 0.27 W/m K * 880 K 1200 W/m2 = 0.2 m k = 0.27 W/m K 1180°C - (240 + 60)°C = 880 K ¢T = 1200 W/m2 5 W/m2 K = 240°C 67706_01_ch01_p002-069.qxd 5/14/10 7:46 AM Page 50 Copyright 2011 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. 1.8 Heat Transfer and the Law of Energy Conservation 53 qr, sun→1 qr, 1→skyqc, air→1 qr, sun→1 qr, sun→1 + qc, air→1 = qr, 1↔2 qr, 1→skyqc, air→1 Surface 2 (sky at 50 K) Heat balance: Surface 1 Control surface (b)(a) Roof FIGURE 1.34 Heat transfer by convection and radiation for roof in Example 1.13. EXAMPLE 1.13 A house has a black tar, flat, horizontal roof. The lower surface of the roof is well insulated, while the upper surface is exposed to ambient air at 300 K through a con- vective heat transfer coefficient of 10 W/m2 K. Calculate the roof equilibrium tem- perature for the following conditions: (a) a clear sunny day with an incident solar radiation flux of 500 W/m2 and the ambient sky at an effective temperature of 50 K and (b) a clear night with an ambient sky temperature of 50 K. SOLUTION A schematic sketch of the system is shown in Fig. 1.34. The control volume is the roof. Assume that there are no obstructions between the roof, called surface 1, and the sky, called surface 2, and that both surfaces are black. The sky behaves as a blackbody because it absorbs all the radiation emitted by the roof and reflects none. Heat is transferred by convection between the ambient air and the roof and by radiation between the sun and roof and between the roof and the sky. This is a closed system in thermal equilibrium. Since there is no generation, storage, or work output, we can express the energy conservation requirement by the conceptual relation Analytically, this relation can be cast in the form Canceling the roof area A1 and substituting the Stefan-Boltzmann relation [Eq. (1.17)] for the net radiation from the roof to the ambient sky gives qr,sun:1 + hc(300 - Troof) = s(T roof 4 - T sky 4 ) A1qr,sun:roof + hc A1(Tair - Troof) = A1qr,roof:ambient sky rate of solar radiation heat transfer to roof + rate of convection heat transfer to roof = net rate of radiation heat transfer from roof to ambient sky 67706_01_ch01_p002-069.qxd 5/14/10 7:46 AM Page 53 Copyright 2011 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. 54 Chapter 1 Basic Modes of Heat Transfer (a) When the solar radiation to the roof, , is 500 W/m2 and Tsky is 50 K, we get Solving by trial and error for the roof temperature, we get Note that the convection term is negative because the sun heats the roof to a temper- ature above the ambient air, so that the roof is not heated but is cooled by convec- tion to the air. (b) At night the term and we get, upon substituting the numerical data in the conservation of energy relation, or Solving this equation for Troof gives At night the roof is cooler than the ambient air and convection occurs from the air to the roof, which is heated in the process. Observe also that the conditions at night and during the day are assumed to be steady and that the change from one steady condition to the other requires a period of transition in which the energy stored in the roof changes and the roof temperature also changes. The energy stored in the roof increases during the morning hours and decreases during the evening after the sun has set, but these periods are not considered in this example. EXAMPLE 1.14 A long, thin copper wire of diameter D and length L has an electrical resistance of per unit length. The wire is initially at steady state in a room at temperature Tair. At time , an electric current i is passed through the wire. The wire tempera- ture begins to increase due to internal electrical heat generation, but at the same time heat is lost from the wire by convection through a convection coefficient to the ambient air. Set up an equation to determine the change in temperature with time in the wire, assuming that the wire temperature is uniform. This is a good assumption because the thermal conductivity of copper is very large and the wire is thin. We will learn in Chapter 2 how to calculate the transient radial temperature distribution if the conductivity is small. hc t = 0 re Troof = 270 K = -3°C (10 W/m2 K)(300 - Troof )(K) = (5.67 * 10 -8 W/m2 K4)(Troof 4 - 504)(K4) hc(Tair - Troof) = s(Troof 4 - Tsky 4 ) Qr,sun:1 = 0 Troof = 303 K = 30°C = (5.67 * 10-8 W/m2 K4)(Troof 4 - 504)(K4) 500 W/m2 + (10 W/m2 K)(300 - Troof)(K) qr,sun:1 67706_01_ch01_p002-069.qxd 5/14/10 7:46 AM Page 54 Copyright 2011 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. 1.8 Heat Transfer and the Law of Energy Conservation 55 Power supply (a) (b) Copper wire L i POWER Control surface Twire qc = hcπDL(Twire – Tair) qG Ammeter ON OFF POWER ON OFF L D FIGURE 1.35 Schematic diagram for electric heat generation system of Example 1.14. SOLUTION The sketch in Fig. 1.35 shows the wire and the control volume. We shall assume that radiation losses are negligible so that the net rate of convection heat flow qc is equal to the rate of heat loss from the wire, qout: The rate of energy generation (or electrical dissipation) in the wire control volume is where , the electrical resistance. The rate of internal energy storage in the control volume is where c is the specific heat and r is the density of the wire material. Applying the conservation of energy relation for a closed system [Eq. (1.39)] to the problem at hand gives since there is no work output and qin is zero. Substituting the appropriate relations for the three energy terms in the conser- vation of energy law gives the different equation i2reL - (hcpDL)(Twire - Tair) = apD 2 4 Lcrb dTwire(t) dt q # G - qout = 0E 0t 0E 0t = d[(pD2>4)LcrTwire(t)] dt Re = reL q # G = i 2Re = i 2reL qout = hc Asurf (Twire - Tair) = hcpDL(Twire - Tair) 67706_01_ch01_p002-069.qxd 5/14/10 7:46 AM Page 55 Copyright 2011 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. 58 Chapter 1 Basic Modes of Heat Transfer –5ºC 20ºC Concrete q = ? 0.2 m References 1. P. G. Klemens, “Theory of the Thermal Conductivity of Solids,” in Thermal Conductivity, R. P. Tye, ed., vol. 1, Academic Press, London, 1969. 2. E. McLaughlin, “Theory of the Thermal Conductivity of Fluids,” in Thermal Conductivity, R. P. Tye, ed., vol. 2. Academic Press, London, 1969. 3. W. G. Vincenti and C. H. Kruger Jr., Introduction to Physical Gas Dynamics, Wiley, New York. 1965. 4. J. F. Mallory, Thermal Insulation, Reinhold, New York. 1969. 5. E. Fried, “Thermal Conduction Contribution to Heat Transfer at Contacts,” Thermal Conductivity, R. P. Tye, ed., vol. 2, Academic press, London, 1969. 6. L. S. Fletcher, “Imperfect Metal-to-Metal Contact,” sec. 502.5 in Heat Transfer and Fluid Flow Data Books, F. Kreith, ed., Genium, Schenectady, NY, 1991. Problems 1.1 The outer surface of a 0.2-m-thick concrete wall is kept at a temperature of -5°C, while the inner surface is kept at 20°C. The thermal conductivity of the concrete is 1.2 W/m K. The problems for this chapter are organized by subject matter as shown below. Topic Problem Number Conduction 1.1–1.11 Convection 1.12–1.21 Radiation 1.22–1.29 Conduction in series and parallel 1.30–1.35 Convection and conduction in series and parallel 1.36–1.43 Convection and radiation in parallel 1.44–1.53 Conduction, convection, and radiation combinations 1.54–1.56 Heat transfer and energy conservation 1.57–1.58 Dimensions and units 1.59–1.65 Heat transfer modes 1.66–1.72 Determine the heat loss through a wall 10 m long and 3 m high. 1.2 The weight of the insulation in a spacecraft may be more important than the space required. Show analytically that the lightest insulation for a plane wall with a specified thermal resistance is the insulation that has the smallest product of density times thermal conductivity. 1.3 A furnace wall is to be constructed of brick having standard dimensions of 9 by 4.5 in. * 3 in. Two kinds of material are available. One has a maximum usable temperature of 1900°F and a thermal conductivity of 1 Btu/h ft °F, and the other has a maximum temperature limit of 1600°F and a thermal conductivity of 0.5 Btu/h ft °F. The bricks have the same cost and can be laid in any manner, but we wish to design the most economical wall for a furnace with a tem- perature of 1900°F on the hot side and 400°F on the cold side. If the maximum amount of heat transfer permissible is 300 Btu/h for each square foot of area, determine the most economical arrangement using the available bricks. 67706_01_ch01_p002-069.qxd 5/14/10 7:46 AM Page 58 Copyright 2011 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Problems 59 Similar specimens Guard ring and insulation Heater Wattmeter Power supply Silicon chip Substrate Synthetic liquid 1.4 To measure thermal conductivity, two similar 1-cm-thick specimens are placed in the apparatus shown in the accompanying sketch. Electric current is supplied to the 6-cm * 6-cm guard heater, and a wattmeter shows that the power dissipation is 10 W. Thermocouples attached to the warmer and to the cooler surfaces show temperatures of 322 and 300 K, respectively. Calculate the thermal conductivity of the material at the mean temperature in Btu/h ft °F and W/m K. 1.5 To determine the thermal conductivity of a structural mate- rial, a large 6-in.-thick slab of the material was subjected to a uniform heat flux of 800 Btu/h ft2, while thermocou- ples embedded in the wall at 2-in. intervals were read over a period of time. After the system had reached equilibrium, an operator recorded the thermocouple readings shown below for two different environmental conditions: Distance from the Surface (in.) Temperature (°F) Test 1 0 100 2 150 4 206 6 270 Test 2 0 200 2 265 4 335 6 406 From these data, determine an approximate expression for the thermal conductivity as a function of temperature between 100 and 400°F. 1.6 A square silicone chip 7 mm * 7 mm in size and 0.5 mm thick is mounted on a plastic substrate as shown in the sketch 1.7 A warehouse is to be designed for keeping perishable foods cool prior to transportation to grocery stores. The warehouse has an effective surface area of 20,000 ft2 exposed to an ambient air temperature of 90°F. The ware- house wall insulation (k is 3 in. thick. Determine the rate at which heat must be removed (Btu/h) from the warehouse to maintain the food at 40°F. 1.8 With increasing emphasis on energy conservation, the heat loss from buildings has become a major concern. The typ- ical exterior surface areas and R-factors (area * thermal resistance) for a small tract house are listed below: Element Area (m2) R-Factors (m2 K/W) Walls 150 2.0 Ceiling 120 2.8 Floor 120 2.0 Windows 20 0.1 Doors 5 0.5 (a) Calculate the rate of heat loss from the house when the interior temperature is 22°C and the exterior is -5°C. (b) Suggest ways and means to reduce the heat loss, and show quantitatively the effect of doubling the wall insu- lation and substituting double-glazed windows (thermal resistance = 0.2 m2 K/W) for the single-glazed type in the table above. 1.9 Heat is transferred at a rate of 0.1 kW through glass wool insulation (density = 100 kg/m3) of 5-cm thickness and 2-m2 area. If the hot surface is at 70°C, determine the temperature of the cooler surface. 1.10 A heat flux meter at the outer (cold) wall of a concrete building indicates that the heat loss through a wall of 10 cm thickness is 20 W/m2. If a thermocouple at the = 0.1 Btu/h ft °F) below. The top surface of the chip is cooled by a synthetic liquid flowing over it. Electronic circuits on the bottom of the chip generate heat at a rate of 5 W that must be transferred through the chip. Estimate the steady-state temperature dif- ference between the front and back surfaces of the chip. The thermal conductivity of silicone is 150 W/m K. 67706_01_ch01_p002-069.qxd 5/14/10 7:46 AM Page 59 Copyright 2011 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. 60 Chapter 1 Basic Modes of Heat Transfer 3 mTG = 30°C 0.3 m x Gas inner surface of the wall indicates a temperature of 22°C while another at the outer surface shows 6°C, calculate the thermal conductivity of the concrete and compare your result with the value in Appendix 2, Table 11. 1.11 Calculate the heat loss through a 1 m * 3 m glass window 7 mm thick if the inner surface temperature is 20°C and the outer surface temperature is 17°C. Comment on the possible effect of radiation on your answer. 1.12 If the outer air temperature in Problem 1.11 is -2°C, cal- culate the convection heat transfer coefficient between the outer surface of the window and the air, assuming radiation is negligible. 1.13 Using Table 1.4 as a guide, prepare a similar table show- ing the orders of magnitude of the thermal resistances of a unit area for convection between a surface and various fluids. 1.14 A thermocouple (0.8-mm-diameter wire) used to measure the temperature of the quiescent gas in a furnace gives a reading of 165°C. It is known, however, that the rate of radiant heat flow per meter length from the hotter furnace walls to the thermocouple wire is 1.1 W/m and the con- vection heat transfer coefficient between the wire and the gas is 6.8 W/m2 K. With this information, estimate the true gas temperature. State your assumptions and indicate the equations used. 1.15 Water at a temperature of 77°C is to be evaporated slowly in a vessel. The water is in a low-pressure con- tainer surrounded by steam as shown in the sketch below. The steam is condensing at 107°C. The overall heat transfer coefficient between the water and the steam is 1100 W/m2 K. Calculate the surface area of the con- tainer that would be required to evaporate water at a rate of 0.01 kg/s. 1.18 A cryogenic fluid is stored in a 0.3-m-diameter spherical container in still air. If the convection heat transfer coef- ficient between the outer surface of the container and the air is 6.8 W/m2 K, the temperature of the air is 27°C, and the temperature of the surface of the sphere is -183°C, determine the rate of heat transfer by convection. 1.19 A high-speed computer is located in a temperature-con- trolled room at 26°C. When the machine is operating, its internal heat generation rate is estimated to be 800 W. The external surface temperature of the computer is to be maintained below 85°C. The heat transfer coefficient for Furnace Thermocouple Condensate Water Water vapor Steam 1.16 The heat transfer rate from hot air by convection at 100°C flowing over one side of a flat plate with dimensions 0.1 m by 0.5 m is determined to be 125 W when the surface of the plate is kept at 30°C. What is the average convection heat transfer coefficient between the plate and the air? 1.17 The heat transfer coefficient for a gas flowing over a thin flat plate 3 m long and 0.3 m wide varies with distance from the leading edge according to If the plate temperature is 170°C and the gas temperature is 30°C, calculate (a) the average heat transfer coeffi- cient, (b) the rate of heat transfer between the plate and the gas, and (c) the local heat flux 2 m from the lead- ing edge. hc(x) = 10x -1/4 W m2 K 67706_01_ch01_p002-069.qxd 5/14/10 7:46 AM Page 60 Copyright 2011 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Problems 63 200 ft 30 ft Glass Solar water heater Insulation Water estimated to vary from about 10 Btu/h ft2 °F (not mov- ing) to 150 Btu/h ft2 °F (top speed). For the following wall constructions, determine the minimum size (in kilowatts) of the heating unit required if the seawater temperature varies from 34°F to 55°F during operation. The walls of the submarine are (a) -in. aluminum, (b) -in. stainless steel with a 1-in.-thick layer of fiberglass insulation on the inside, and (c) of sandwich construction with a -in.-thick layer of stainless steel, a 1-in.-thick layer of fiberglass insulation, and a -in.-thick layer of aluminum on the inside. What conclusions can you draw? 1>4 3>4 3>4 1>2 1.40 A simple solar heater consists of a flat plate of glass below which is located a shallow pan filled with water, so that the water is in contact with the glass plate above it. Solar radiation passes through the glass at the rate of 156 Btu/h ft2. The water is at 200°F and the surrounding air is at 80°F. If the heat transfer coefficients between the water and the glass, and between the glass and the air are 5 Btu/h ft2 °F and 1.2 Btu/h ft2 °F, respectively, deter- mine the time required to transfer 100 Btu per square foot of surface to the water in the pan. The lower surface of the pan can be assumed to be insulated. 1.41 A composite refrigerator wall is composed of 2 in. of corkboard sandwiched between a -in.-thick layer of oak and a -in.-thick layer of aluminum lining on the inner surface. The average convection heat transfer coefficients at the interior and exterior wall are 2 and 1.5 Btu/h ft2 °F, respectively. (a) Draw the thermal cir- cuit. (b) Calculate the individual resistances of the com- ponents of this composite wall and the resistances at the surfaces. (c) Calculate the overall heat transfer coefficient through the wall. (d) For an air temperature of 30°F inside the refrigerator and 90°F outside, calcu- late the rate of heat transfer per unit area through the wall. 1.42 An electronic device that internally generates 600 mW of heat has a maximum permissible operating tempera- ture of 70°C. It is to be cooled in 25°C air by attaching aluminum fins with a total surface area of 12 cm2. The convection heat transfer coefficient between the fins and the air is 20 W/m2 K. Estimate the operating temperature when the fins are attached in such a way that (a) there is a contact resistance of approximately 50 K/W between the surface of the device and the fin array and (b) there is no contact resistance (in this case, the construction of the device is more expensive). Comment on the design options. 1>32 1>2 1.43 To reduce home heating requirements, modern building codes in many parts of the country require the use of double- glazed or double-pane windows, i.e., windows with two panes of glass. Some of these so-called thermopane win- dows have an evacuated space between the two glass panes while others trap stagnant air in the space. (a) Consider a double-pane window with the dimensions shown in the following sketch. If this window has Insulation Electronic device Fins 67706_01_ch01_p002-069.qxd 5/14/10 7:46 AM Page 63 Copyright 2011 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. 64 Chapter 1 Basic Modes of Heat Transfer Rocket Motor Combustion Chamber T = 1000°F Gas T = 5000°F stagnant air trapped between the two panes and the convection heat transfer coefficients on the inside and outside surfaces are 4 W/m2 K and 15 W/m2 K, respec- tively, calculate the overall heat transfer coefficient for the system. (b) If the inside air temperature is 22°C and the outside air temperature is -5°C, compare the heat loss through a 4-m2 double-pane window with the heat loss through a single-pane window. Comment on the effect of the window frame on this result. (c) If the total window area of a home heated by electric resistance heaters at a cost of is 80 m2. How much more cost can you justify for the double-pane windows if the average temperature difference during the six winter months when heating is required is about 15°C? $0.10/k Wh 1.44 A flat roof can be modeled as a flat plate insulated on the bottom and placed in the sunlight. If the radiant heat that the roof receives from the sun is 600 W/m2, the convection heat transfer coefficient between the roof and the air is 12 W/m2 K, and the air temperature is 27°C, determine the roof temperature for the following two cases: (a) Radiative heat loss to space is negligible. (b) The roof is black and radiates to space, which is assumed to be a blackbody at 0 K. (e = 1.0) 1.45 A horizontal, 3-mm-thick flat-copper plate, 1 m long and 0.5 m wide, is exposed in air at 27°C to radiation from the sun. If the total rate of solar radiation absorbed is 300 W and the combined radiation and convection heat transfer coefficients on the upper and lower surfaces are 20 and 15 W/m2 K, respectively, determine the equilibrium tem- perature of the plate. 1.46 A small oven with a surface area of 3 ft2 is located in a room in which the walls and the air are at a temperature of 80°F. The exterior surface of the oven is at 300°F, and the next heat transfer by radiation between the oven’s surface and the surroundings is 2000 Btu/h. If the average convection heat transfer coefficient between the oven and the surrounding air is 2.0 Btu/h ft2 °F, cal- culate (a) the net heat transfer between the oven and the surroundings in Btu/h, (b) the thermal resistance at the surface for radiation and convection, respectively, in h °F/Btu, and (c) the combined heat transfer coefficient in Btu/h ft2 °F. 1.47 A steam pipe 200 mm in diameter passes through a large basement room. The temperature of the pipe wall is 500°C, while that of the ambient air in the room is 20°C. Determine the heat transfer rate by convection and radiation per unit length of steam pipe if the emis- sivity of the pipe surface is 0.8 and the natural convec- tion heat transfer coefficient has been determined to be 10 W/m2 K. 1.48 The inner wall of a rocket motor combustion chamber receives 50,000 Btu/h ft2 by radiation from a gas at 5000°F. The convection heat transfer coefficient between the gas and the wall is 20 Btu/h ft2 °F. If the inner wall Flat Roof Insulation Sunlight To = −5°C Ti = 22°C 2 cm Frame A = 4 m2 67706_01_ch01_p002-069.qxd 5/14/10 7:46 AM Page 64 Copyright 2011 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Problems 65 0.003 m 0.01 m of the combustion chamber is at a temperature of 1000°F, determine (a) the total thermal resistance of a unit area of the wall in h ft2 °F/Btu and (b) the heat flux. Also draw the thermal circuit. 1.49 A flat roof of a house absorbs a solar radiation flux of 600 W/m2. The backside of the roof is well insulated, while the outside loses heat by radiation and convec- tion to ambient air at 20°C. If the emittance of the roof is 0.80 and the convection heat transfer coefficient between the roof and the air is 12 W/m2 K, calculate (a) the equilibrium surface temperature of the roof and (b) the ratio of convection to radiation heat loss. Can one or the other of these be neglected? Explain your answer. 1.50 Determine the power requirement of a soldering iron in which the tip is maintained at 400°C. The tip is a cylin- der 3 mm in diameter and 10 mm long. The surround- ing air temperature is 20°C, and the average convection heat transfer coefficient over the tip is 20 W/m2 K. The tip is highly polished initially, giving it a very low emittance. work output divided by the total heat input) is 0.33. If the engine block is aluminum with a graybody emissivity of 0.9, the engine compartment operates at 150°C, and the convection heat transfer coefficient is 30 W/m2 K, determine the average surface temperature of the engine block. Comment on the practicality of the concept. 1.53 A pipe carrying superheated steam in a basement at 10°C has a surface temperature of 150°C. Heat loss from the pipe occurs by radiation and natural convection . Determine the percentage of the total heat loss by these two mechanisms. 1.54 For a furnace wall, draw the thermal circuit, determine the rate of heat flow per unit area, and estimate the exterior surface temperature if (a) the convection heat transfer coefficient at the interior surface is 15 W/m2 K (b) the rate of heat flow by radiation from hot gases and soot particles at 2000°C to the interior wall surface is 45,000 W/m2 (c) the unit thermal conductance of the wall (interior surface temperature is about 850°C) is 250 W/m2 K and (d) there is convection from the outer surface. 1.55 Draw the thermal circuit for heat transfer through a dou- ble-glazed window. Identify each of the circuit ele- ments. Include solar radiation to the window and interior space. 1.56 The ceiling of a tract house is constructed of wooden studs with fiberglass insulation between them. On the interior of the ceiling is plaster and on the exterior is a thin layer of sheet metal. A cross section of the ceiling with dimensions is shown below. (hc = 25 W/m 2 K) (e = 0.6) (a) The R-factor describes the thermal resistance of insu- lation and is defined by R - factor = L>keff = ¢T>(q>A) Plaster 16 in. Sheet metal 31/2 in. 1/2 in. 11/2 in. Ti = 22°C To = −5°C Fiberglass Wood stud Wood stud 1.51 The soldering iron tip in Problem 1.50 becomes oxidized with age and its gray-body emittance increases to 0.8. Assuming that the surroundings are at 20°C, determine the power requirement for the soldering iron. 1.52 Some automobile manufacturers are currently working on a ceramic engine block that could operate without a cooling system. Idealize such an engine as a rectangu- lar solid, 45 cm * 30 cm * 30 cm. Suppose that under maximum power output the engine consumes 5.7 L of fuel per hour, the heat released by the fuel is 9.29/k Whr per liter, and the net engine efficiency (useful 67706_01_ch01_p002-069.qxd 5/14/10 7:46 AM Page 65 Copyright 2011 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. 68 Chapter 1 Basic Modes of Heat Transfer Thermocouple Circular cross section Air flow 15 m/s Leads 1 m Design Problems 1.1 Optimum Boiler Insulation Package (Chapter 1) To insulate high-temperature surfaces it is economical to use two layers of insulation. The first layer is placed next to the hot surface and is suitable for high temperature. It is costly and is usually a relatively poor insulator. The second layer is placed outside the first layer and is cheaper and a good insulator, but will not withstand high temperatures. Essentially, the first layer protects the second layer by providing just enough insulating capability so that the second layer is only exposed to moderate temperatures. Given commercially available insulating materials, design the optimum combination of two such materials to insulate a flat 1000°C surface from ambient air at 20°C. Your goal is to reduce the rate of heat transfer to 0.1% of that with- out any insulation, to achieve an outer surface temperature that is safe to personnel, and to minimize cost of the insu- lating package. 1.2 Thermocouple Radiator Error (Chapters 1 and 9) Design a thermocouple installation to measure the temper- ature of air flowing at a velocity of 15 m/s in a 1-m-diameter duct. The air is at approximately 1000°C and the duct walls are at 200°C. Select a type of thermocouple that could be used, and then determine how accurately the thermocouple will measure the air temperature. Prepare a plot of the measurement error vs. air temperature and discuss the result. Use Table 1.4 to estimate the convection heat trans- fer coefficients. This is a multistep problem; after you have studied convection and radiation, you will improve this design to reduce the measurement error by orienting the thermocouple and its leads differently and using a radiation shield. 1.3 Heating Load on Factory (Chapters 1, 4 and 5) Design a heating system for a small factory in Denver, Colorado. This is a multistep problem that will be contin- ued in subsequent chapters. In the first step, you are to determine the heating load on the building, i.e., the rate at which the building loses heat in the winter, if the inside temperature is to be maintained at 20°C. In order to com- pensate for this heat loss, you will subsequently be asked to design a suitable heater that can provide a rate of heat transfer equal to the heat load from the building. A schematic diagram of the building and construction details for the walls and ceilings are shown in the figure. Additional information may be obtained from the ASHRAE Handbook of Fundamentals. For the purpose of this analysis, it may be assumed that the ambient temperature in Denver is equal to or greater than -10°C 97% of the time. Furthermore, air infil- tration through windows and doors may be assumed to be approximately 0.2 times the volume of the building per hour. For the initial estimate of the heat load, you may use average values for the convective heat transfer coefficients over the inside and outside surfaces from Table 1.4. Note that for this design, the outside temperature assumes the worst possible conditions and, if the heater is able to main- tain the temperature under these conditions, it will be able to meet less-severe conditions as well. After you have completed the initial design, examine the results and prepare a report for the architect and the owner of the building, pointing out how the thermal design could be improved. Note especially any areas where exces- sive heat losses may occur. After you have studied Chapters 4 and 5 you will be asked to repeat the heat loss calculations, but calculate the heat transfer coefficient from information presented in these chapters. 67706_01_ch01_p002-069.qxd 5/14/10 7:46 AM Page 68 Copyright 2011 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Design Problems 69 1.5-cm-thick plywood 4-cm-thick pine stud 2-cm gypsum plaster Corrugated sheet metal 1.2-cm hardboard siding 4-cm-thick pine stud Corrugated sheet metal Ceiling Cross Section Wall Cross Section Fiberglass insulation 40 cm 40 cm 14 cm 14 cm Fiberglass insulation 3.0 m Sloping roof 10 m 4 m 25 m 0.75 m Windows (4) 2.5 m 2-cm gypsum plaster Design Problem 1.3 67706_01_ch01_p002-069.qxd 5/14/10 7:46 AM Page 69 Copyright 2011 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. CHAPTER 2 Concepts and Analyses to Be Learned Heat transfer by conduction is a diffusion process, whereby thermal energy is transferred from a hot end of a medium (usually solid) to its colder end via an intermolecular energy exchange. Modeling the heat conduction process requires you to apply thermodynamics of energy conservation along with Fourier’s law of heat conduction. The consequent mathematical descriptions are usually in the form of ordinary as well as partial differen- tial equations. By considering different engineering applications that rep- resent situations for steady as well as time-dependent (or transient) conduction, a study of this chapter will teach you: • How to derive the conduction equation in different coordinate sys- tems for both steady-state and transient conditions. • How to obtain steady-state temperature distributions in simple con- ducting geometries without and with heat generation. • How to develop the mathematical formulation of boundary conditions with insulation, constant heat flux, surface convection, and specified changes in surface temperature. • How to apply the concept of lumped capacitance (conditions under which internal resistance in a conducting body can be neglected) in transient heat transfer. • How to use charts for transient heat conduction to obtain tempera- ture distribution as a function of time in simple geometries. • How to obtain temperature distribution and rate of heat loss or gain from extended surfaces, also called fins, and use them in typical applications. Heat Conduction A typical arrangement of rectangular pin-fin heat sinks mounted on a computer/ microprocessor hardware for electronic cooling. Source: Courtesy of Hardware Canucks. 67706_02_ch02_p070-165.qxd 5/14/10 12:36 PM Page 70 Copyright 2011 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. 2.2 The Conduction Equation 73 FIGURE 2.2 Control volume for one-dimensional conduction in rectangular coordinates. unsteady or transient. If the temperature is independent of time, the problem is called a steady-state problem. If the temperature is a function of a single space coor- dinate, the problem is said to be one-dimensional. If it is a function of two or three coordinate dimensions, the problem is two- or three-dimensional, respectively. If the temperature is a function of time and only one space coordinate, the problem is classified as one-dimensional and transient. 2.2.1 Rectangular Coordinates To illustrate the analytic approach, we will first derive the conduction equation for a one-dimensional, rectangular coordinate system as shown in Fig. 2.2. We will assume that the temperature in the material is a function only of the x coordinate and time; that is, T  T(x,t), and the conductivity k, density , and specific heat c of the solid are all constant. The principle of conservation of energy for the control volume, surface area A, and thickness of Fig. 2.2 can be stated as follows: rate of heat conduction rate of heat conduction into control volume out of control volume + = + (2.1) rate of heat generation rate of energy storage inside control volume inside control volume We will use Fourier’s law to express the two conduction terms and define the symbol as the rate of energy generation per unit volume inside the control volume. Then the word equation (Eq. 2.1) can be expressed in mathematical form: (2.2)- kA 0T 0x ` x + q # GA ¢x = -kA 0T 0x ` x+¢x + rA ¢xc 0T(x + ¢x/2, t) 0t q # G ¢x, T = T(x, t) q(x) qG q(x + Δx) Δxx 67706_02_ch02_p070-165.qxd 5/14/10 12:37 PM Page 73 Copyright 2011 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. 74 Chapter 2 Heat Conduction Dividing Eq. (2.2) by the control volume A x and rearranging, we obtain (2.3) In the limit as x : 0, the first term on the left side of Eq. (2.3) can be expressed in the form (2.4) The right side of Eq. (2.3) can be expanded in a Taylor series as Equation (2.2) then becomes, to the order of x, (2.5) Physically, the first term on the left side represents the net rate of heat conduction into the control volume per unit volume. The second term on the left side is the rate of energy generation per unit volume inside the control volume. The right side rep- resents the rate of increase in internal energy inside the control volume per unit vol- ume. Each term has dimensions of energy per unit time and volume with the units (W/m3) in the SI system and (Btu/h ft3) in the English system. Equation (2.5) applies only to unidimensional heat flow because it was derived on the assumption that the temperature distribution is one-dimensional. If this restriction is now removed and the temperature is assumed to be a function of all three coordinates as well as time, that is, T  T(x, y, z, t), terms similar to the first one in Eq. (2.5) but representing the net rate of conduction per unit volume in the y and z directions will appear. The three-dimensional form of the conduction equation then becomes (see Fig. 2.3) (2.6) where  is the thermal diffusivity, a group of material properties defined as (2.7) The thermal diffusivity has units of (m2/s) in the SI system and (ft2/s) in the English system. Numerical values of the thermal conductivity, density, specific heat, and thermal diffusivity for several engineeering materials are listed in Appendix 2. Solutions to the general conduction equation in the form of Eq. (2.6) can be obtained only for simple geometric shapes and easily specified boundary conditions. However, as shown in the next chapter, solutions by numerical methods can be obtained a = k rc 0 2T 0x2 + 0 2T 0y2 + 0 2T 0z2 + q # G k = 1 a 0T 0t k 0 2T 0x2 + q# G = rc 0T 0t 0T 0t c ax + ¢x 2 b , t d = 0T 0t ` x + 0 2T 0x 0T ` x ¢x 2 + Á 0T 0x ` x+dx = 0T 0x ` x + 0 0x a 0T 0x ` x bdx = 0T 0x ` x + 0 2 T 0x2 ` x dx k 10T/0x2x+¢x - (0T/0x)x ¢x + #qG = rc0T(x + ¢x/2, t) 0t 67706_02_ch02_p070-165.qxd 5/14/10 12:37 PM Page 74 Copyright 2011 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. dx x x z y x + dx dy dxqx qx + ∂qx ∂x dz 2.2 The Conduction Equation 75 FIGURE 2.3 Differential control volume for three-dimensional conduction in rectangular coordinates. quite easily for complex shapes and realistic boundary conditions; this procedure is used in engineering practice today for the majority of conduction problems. Nevertheless, a basic understanding of analytic solutions is important in writing com- puter programs, and in the rest of this chapter we will examine problems for which sim- plifying assumptions can eliminate some terms from Eq. (2.6) and reduce the complexity of the solution. If the temperature of a material is not a function of time, the system is in the steady state and does not store any energy. The steady-state form of a three-dimen- sional conduction equation in rectangular coordinates is (2.8) If the system is in the steady state and no heat is generated internally, the conduc- tion equation further simplifies to (2.9) Equation (2.9) is known as the Laplace equation, in honor of the French mathemati- cian Pierre Laplace. It occurs in a number of areas in addition to heat transfer, for instance, in diffusion of mass or in electromagnetic fields. The operation of taking the second derivatives of the potential in a field has therefore been given a shorthand symbol, 2, called the Laplacian operator. For the rectangular coordinate system Eq. (2.9) becomes (2.10) Since the operator 2 is independent of the coordinate system, the above form will be particularly useful when we want to study conduction in cylindrical and spheri- cal coordinates. 0 2T 0x2 + 0 2T 0y2 + 0 2T 0z2 = § 2T = 0 0 2T 0x2 + 0 2T 0y2 + 0 2T 0z2 = 0 0 2T 0x2 + 0 2T 0y2 + 0 2T 0z2 + q# G k = 0 67706_02_ch02_p070-165.qxd 5/14/10 12:37 PM Page 75 Copyright 2011 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.