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- Charles H. Forsberg

*Table of contents : Heat Transfer Principles and ApplicationsCopyrightUnit conversionsConstantsPrefaceAcknowledgmentsCopyright.pdfUnit Conversions.pdfConstants.pdfPreface.pdfAcknowledgments.pdfChapter 1 - Introduction to heat transfer 1 - Introduction to heat transfer 1.1 Introduction 1.2 Modes of heat transfer 1.3 Conduction 1.3.1 Conduction through a plane wall 1.4 Convection 1.5 Radiation 1.6 The direction of heat flow 1.7 Temperature continuity and heat balances 1.8 Unit systems 1.9 Recommended approach to problem solving Step 1 - Problem definition Step 2 - Problem givens Step 3 - Determine the appropriate equations Step 4 - Obtain the solution Step 5 - Review the solution 1.10 Significant figures 1.11 Chapter summary and final remarks 1.12 Problems ReferencesChapter 2 - Heat conduction equation and boundary conditions 2 - Heat conduction equation and boundary conditions 2.1 Introduction 2.2 Heat conduction equation 2.2.1 Rectangular coordinates 2.2.1.1 Special cases—rectangular coordinates 2.2.2 Cylindrical coordinates 2.2.2.1 Special cases—cylindrical coordinates 2.2.3 Spherical coordinates 2.2.3.1 Special cases—spherical coordinates 2.3 Boundary conditions 2.3.1 Rectangular coordinates 2.3.1.1 Specified temperature 2.3.1.2 Specified heat flux 2.3.1.3 Insulated boundary 2.3.1.4 Convection 2.3.1.5 Radiation 2.3.1.6 Convection and radiation 2.3.1.7 Symmetry conditions 2.3.1.8 Interfacial boundary 2.3.2 Cylindrical and spherical coordinates 2.3.2.1 Symmetry conditions 2.4 Initial conditions 2.5 Chapter summary and final remarks 2.6 Problems Uncited referencesChapter 3 - Steady-state conduction 3 - Steady-state conduction 3.1 Introduction 3.2 One-dimensional conduction 3.2.1 Plane wall 3.2.1.1 Multilayered Walls 3.2.1.2 Electric-heat analogy and the resistance concept 3.2.1.3 Overall heat transfer coefficient and R-Value 3.2.2 Cylindrical shell 3.2.3 Spherical shell 3.3 Critical insulation thickness 3.4 Heat generation in a cylinder 3.5 Temperature-dependent thermal conductivity 3.6 Multi-dimensional conduction 3.7 Conduction shape factors 3.8 Extended surfaces (fins) 3.8.1 Fins of constant cross section 3.8.1.1 The governing differential equation and boundary conditions 3.8.1.2 The solution for temperature distribution and heat flow 3.8.1.3 Very-long-fin approximation 3.8.1.4 Insulated-at-end fin approximation 3.8.2 Fin efficiency 3.8.3 Fin effectiveness 3.8.4 Fins of varying cross section 3.8.4.1 Circumferential fins 3.8.4.2 Straight triangular fins 3.8.4.3 Conical pin fins 3.8.5 Closing comments on fins 3.9 Chapter summary and final remarks 3.10 Problems ReferencesChapter 4 - Unsteady conduction 4 - Unsteady conduction 4.1 Introduction 4.2 Lumped systems (no spatial variation) 4.2.1 Lumped systems analysis 4.2.2 Application criterion 4.2.3 The time constant 4.3 Systems with spatial variation (large plate, long cylinder, sphere) 4.3.1 Overview 4.3.2 Large plane plates 4.3.3 Long cylinders 4.3.4 Spheres 4.4 Multidimensional systems with spatial variation 4.4.1 Overview 4.4.2 Long bar 4.4.3 Short cylinder 4.4.4 Rectangular solid 4.5 Semi-infinite solid 4.5.1 Overview 4.5.2 Temperature boundary condition 4.5.3 Heat flux boundary condition 4.5.4 Convection boundary condition 4.6 Chapter summary and final remarks 4.7 Problems ReferencesChapter 5 - Numerical methods (steady and unsteady) 5 - Numerical methods (steady and unsteady) 5.1 Introduction 5.2 Finite-difference method 5.2.1 Steady state 5.2.2 Unsteady state 5.3 Finite element method 5.4 Chapter summary and final remarks 5.5 Problems ReferencesChapter 6 - Forced convection 6 - Forced convection 6.1 Introduction 6.2 Basic considerations 6.3 External flow 6.3.1 Flow over a flat plate 6.3.1.1 Laminar boundary layer 6.3.1.1.1 Continuity equation 6.3.1.1.2 Momentum equation 6.3.1.1.3 Drag force 6.3.1.1.4 Energy equation 6.3.1.1.5 Thermal boundary layer thickness 6.3.1.1.6 Convective coefficient and Nusselt number 6.3.1.1.7 Reynolds-Colburn analogy 6.3.1.1.8 Constant heat flux 6.3.1.1.9 Unheated starting length 6.3.1.2 Turbulent boundary layer 6.3.2 Flow over cylinders and spheres 6.3.2.1 Cylinders 6.3.2.1.1 Circular cylinders 6.3.2.1.2 Noncircular cylinders 6.3.2.2 Spheres 6.3.3 Flow through tube banks 6.4 Internal flow 6.4.1 Entrance lengths 6.4.2 Mean velocity and mean temperature 6.4.3 Constant heat flux 6.4.4 Constant surface temperature 6.4.5 Equivalent diameter for flow through noncircular tubes 6.4.6 Correlations for the Nusselt number and convective coefficient 6.4.6.1 Laminar flow; entrance region 6.4.6.2 Laminar flow; fully developed 6.4.6.3 Turbulent flow; fully developed 6.4.7 Annular flow 6.4.7.1 Fully developed laminar flow 6.4.7.2 Fully developed turbulent flow 6.5 Chapter summary and final remarks 6.6 Problems ReferencesChapter 7 - Natural (free) convection 7 - Natural (free) convection 7.1 Introduction 7.2 Basic considerations 7.3 Natural convection for flat plates 7.3.1 Vertical plate 7.3.1.1 Constant temperature surface 7.3.1.2 Constant heat flux surface 7.3.2 Horizontal plate 7.3.2.1 Constant temperature surface 7.3.2.2 Constant heat flux surface 7.3.3 Inclined plate 7.4 Natural convection for cylinders 7.4.1 Horizontal cylinder 7.4.2 Vertical cylinder 7.5 Natural convection for spheres 7.6 Natural convection for other objects 7.7 Natural convection for enclosed spaces 7.7.1 Enclosed rectangular space 7.7.1.1 Horizontal rectangular enclosure 7.7.1.2 Vertical rectangular enclosure 7.7.1.3 Inclined rectangular enclosure 7.7.2 Annular space between concentric cylinders 7.7.3 Space between concentric spheres 7.8 Natural convection between vertical fins 7.9 Chapter summary and final remarks 7.10 Problems ReferencesChapter 8 - Heat exchangers 8 - Heat exchangers 8.1 Introduction 8.2 Types of heat exchangers 8.2.1 Temperature distribution in double-pipe heat exchangers 8.3 The overall heat transfer coefficient 8.4 Analysis methods 8.4.1 Log mean temperature difference method 8.4.1.1 Double-pipe heat exchangers 8.4.1.2 Non–double-pipe heat exchangers 8.4.2 Effectiveness–number of transfer unit method 8.5 Chapter summary and final remarks 8.6 Problems ReferencesChapter 9 - Radiation heat transfer 9 - Radiation heat transfer 9.1 Introduction 9.2 Blackbody emission 9.3 Radiation properties 9.4 Radiation shape factors 9.5 Radiative heat transfer between surfaces 9.5.1 Radiation heat transfer for a two-surface enclosure 9.5.1.1 For surface 1 9.5.1.2 For surface 2 9.5.2 Radiation heat transfer for a three-surface enclosure 9.5.2.1 For surface 1 9.5.2.2 For surface 2 9.5.2.3 For surface 3 9.5.2.4 Three-surface enclosure with an insulated surface 9.6 Radiation shields 9.7 Sky radiation and solar collectors 9.8 Chapter summary and final remarks 9.9 Problems References Further readingChapter 10 - Multimode heat transfer 10 - Multimode heat transfer 10.1 Introduction 10.2 Procedure for solution of multimode problems 10.3 Examples 10.4 Chapter summary and final remarks 10.5 ProblemsChapter 11 - Mass transfer 11 - Mass transfer 11.1 Introduction 11.2 Concentrations in a gas mixture 11.3 Fick's law of diffusion 11.3.1 Binary gas diffusion coefficient 11.3.2 Binary gas–liquid diffusion coefficient 11.4 Diffusion in gases 11.4.1 Stefan's law 11.4.2 Equimolar counterdiffusion 11.5 The mass-heat analogy 11.5.1 Mass transfer through walls and membranes 11.5.2 Transient diffusion 11.6 Gas–liquid diffusion 11.7 Mass transfer coefficient 11.7.1 Dimensionless parameters 11.7.2 Wet-bulb and dry-bulb psychrometer 11.8 Chapter summary and final remarks 11.9 Problems ReferencesChapter 12 - Special topics 12 - Special topics 12.1 Introduction 12.2 Internal heat generation 12.2.1 Heat generation in a plane wall 12.2.2 Heat generation in a sphere 12.3 Contact resistance 12.4 Condensation and boiling 12.4.1 Condensation heat transfer 12.4.1.1 Film condensation for vertical and inclined plates 12.4.1.1.1 Vertical plates 12.4.1.1.2 Inclined plates 12.4.1.2 Film condensation for vertical cylinders 12.4.1.3 Film condensation for horizontal cylinders and for spheres 12.4.2 Boiling heat transfer 12.4.2.1 Regions of pool boiling 12.4.2.2 Nucleate pool boiling 12.4.2.3 Film boiling 12.5 Energy usage in buildings 12.6 Chapter summary and final remarks 12.7 Problems ReferencesIndex A B C D E F G H I K L M N O R S T U V WBlank Page*

Heat Transfer Principles and Applications

Charles H. Forsberg

Academic Press is an imprint of Elsevier 125 London Wall, London EC2Y 5AS, United Kingdom 525 B Street, Suite 1650, San Diego, CA 92101, United States 50 Hampshire Street, 5th Floor, Cambridge, MA 02139, United States The Boulevard, Langford Lane, Kidlington, Oxford OX5 1GB, United Kingdom Copyright © 2021 Elsevier Inc. All rights reserved. No part of this publication may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying, recording, or any information storage and retrieval system, without permission in writing from the publisher. Details on how to seek permission, further information about the Publisher’s permissions policies and our arrangements with organizations such as the Copyright Clearance Center and the Copyright Licensing Agency, can be found at our website: www.elsevier.com/permissions. This book and the individual contributions contained in it are protected under copyright by the Publisher (other than as may be noted herein). Notices Knowledge and best practice in this ﬁeld are constantly changing. As new research and experience broaden our understanding, changes in research methods, professional practices, or medical treatment may become necessary. Practitioners and researchers must always rely on their own experience and knowledge in evaluating and using any information, methods, compounds, or experiments described herein. In using such information or methods they should be mindful of their own safety and the safety of others, including parties for whom they have a professional responsibility. To the fullest extent of the law, neither the Publisher nor the authors, contributors, or editors, assume any liability for any injury and/or damage to persons or property as a matter of products liability, negligence or otherwise, or from any use or operation of any methods, products, instructions, or ideas contained in the material herein. MATLABÒ is a trademark of The MathWorks, Inc. and is used with permission. The MathWorks does not warrant the accuracy of the text or exercises in this book. This book’s use or discussion of MATLABÒ software or related products does not constitute endorsement or sponsorship by The MathWorks of a particular pedagogical approach or particular use of the MATLABÒ software. Library of Congress Cataloging-in-Publication Data A catalog record for this book is available from the Library of Congress British Library Cataloguing-in-Publication Data A catalogue record for this book is available from the British Library ISBN: 978-0-12-802296-2

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Publisher: Katey Birtcher Acquisitions Editor: Stephen Merken Editorial Project Manager: Susan Ikeda Production Project Manager: Sujatha Thirugnana Sambandam Cover Designer: Alan Studholme Typeset by TNQ Technologies

Unit conversions Length Area Volume Velocity Density Force Mass

Pressure

Energy (heat)

Heat flow and power

Heat flux Heat generation rate Heat transfer coefficient Specific heat Thermal conductivity Absolute viscosity Kinematic viscosity Thermal diffusivity Temperature level

Temperature size

1 m ¼ 39.370 in ¼ 3.2808 ft 1 mile ¼ 5280 ft ¼ 1.6093 km 1 m2 ¼ 10.7639 ft2 1 m3 ¼ 35.3134 ft3 1 m3 ¼ 264.17 gal (US) 1 m/s ¼ 3.2808 ft/s 1 kg/m3 ¼ 0.06243 lbm/ft3 1 N ¼ 0.2248 lbf 1 kg ¼ 2.20462 lbm 1 slug ¼ 1 lbf s2/ft ¼ 32.174 lbm 1 ton ¼ 2000 lbm 1 metric ton ¼ 1000 kg 1 Pa ¼ 1 N/m2 ¼ 1.45038 104 lbf/in2 1 MPa ¼ 103 kPa ¼ 106 Pa 1 bar ¼ 100 kPa 1 atm ¼ 101.325 kPa ¼ 760 mm Hg @ 0 C 1 atm ¼ 14.696 psia ¼ 29.92 in Hg @ 32 F 1 torr ¼ 133.322 Pa 1 kJ ¼ 0.94783 Btu 1 Btu ¼ 778.169 ft lbf 1 kWh ¼ 3412.14 Btu 1 W ¼ 1 J/s ¼ 3.4121 Btu/h 1 kW ¼ 3412.1 Btu/h ¼ 1.341 hp 1 hp ¼ 550 ft lbf/s 1 W/m2 ¼ 0.3171 Btu/h ft2 1 W/m3 ¼ 0.09662 Btu/h ft3 1 W/m2 C ¼ 0.1761 Btu/h ft2 F 1 kJ/kg C ¼ 0.23885 Btu/lbm F 1 W/m C ¼ 0.5778 Btu/h ft F 1 kg/m s ¼ 2419.1 lbm/ft h 1 m2/s ¼ 10.764 ft2/s 1 m2/s ¼ 10.764 ft2/s K ¼ C + 273.15 R ¼ F + 459.67 F ¼ (9/5) C + 32 C ¼ (5/9) (F e 32) 1C¼1K 1F¼1R 1 C ¼ (9/5) F 1 F ¼ (5/9) C

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Constants StefaneBoltzmann

s ¼ 5:67 108 W=m2 K4 ¼ 0:1714 108 Btu=h ft2 R4

Universal gas constant

R ¼ 8:31446 kJ=kmol K ¼ 1545:35 ft lbf=lbm mol R

Planck’s constant

h ¼ 6:62607 1034 J$s

Boltzmann’s constant

k ¼ 1:38065 1023 J=K

Standard gravity

g ¼ 9:807 m=s2 ¼ 32:174 ft=s2

Speed of light (vacuum)

co ¼ 2:9979 108 m=s

Avogadro’s number

NA ¼ 6:02214 1026 atoms or molecules=kmol

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Preface Over my years of teaching heat transfer, I have watched as the major textbooks in the field have gotten bigger and bigger. They have reached such a massive size that only a small portion of a book can be successfully covered in a semester-length course. Also, the problems in the books have become so complex that many, or most, students have comprehension difficulties. Inordinate amounts of time are needed to just understand the problems, leaving much less time available for their solution. The overall goal of this book is to assist the student in obtaining the greatest possible benefits from a semester-long undergraduate heat transfer course. The book is of reasonable length. Although it contains more material than can be covered in the course, the extra material is not excessive, and it provides the professor with flexibility in topic selection. The problems in the book are of a practical nature. They require the students to think critically, but are not excessively hard to comprehend. The book is designed for a one-semester heat transfer course for mechanical engineering students taken in the junior or senior year. It is also beneficial for practicing mechanical engineers and for practicing engineers in other disciplines (e. g., civil and electrical) who encounter heat transfer aspects in their projects. Finally, with a chapter devoted to mass transfer, the book is useful for chemical engineering students and engineers. The book makes extensive use of Excel and MATLABÒ for the solution of numerical problems. Excel Goal Seek is used for solution of single nonlinear equations, and Solver is used for simultaneous linear and nonlinear equations. The fzero function of MATLAB is used for single nonlinear equations, and MATLAB programs incorporating the GausseSeidel method are used for simultaneous equations. MATLAB is also used for the solution of transient problems. There are 12 chapters in the book, plus an appendix. Brief descriptions are as follows: Chapter 1 introduces the three modes of heat transfer: conduction, convection, and radiation and gives their basic equations. There are also sections regarding the direction of heat flow, temperature continuity and heat balances, unit systems, significant figures, and the recommended approach to problem-solving. Chapter 2 deals with the differential equations for heat conduction in rectangular, cylindrical, and spherical coordinates. It also discusses boundary conditions for the different geometries. Chapter 3 discusses the solution of steady-state, one-dimensional heat conduction problems. The resistance concept and the electriceheat analogy are introduced. Heat generation in current-carrying wires is covered, as is the conduction shape factor. Finally, there is a discussion of fins. Chapter 4 deals with unsteady conduction. It includes the lumped method; the Heisler method for planes, cylinders, spheres, and multidimensional objects; and conduction in semi-infinite solids. Chapter 5 covers numerical methods applied to both steady and unsteady situations. The appropriate finite-difference equations are derived. Excel and MATLAB are used for the solution of the resulting simultaneous linear and nonlinear equations. Chapters 6 and 7 deal with convection. Chapter 6 is forced convection for internal and external flows. Chapter 7 is natural (free) convection. Both chapters present the pertinent dimensionless numbers and correlation equations for the heat transfer coefficient. Chapter 7 includes sections on natural convection for enclosed spaces such as rectangular spaces between parallel plates and annular spaces between concentric cylinders and spheres. Chapter 8 covers heat exchangers. Both the log mean temperature difference and effectivenesse number of transfer unit methods are discussed.

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Preface

Chapter 9 deals with radiation heat transfer. There are sections on blackbody emission, radiation properties, radiation shape factors, and radiant heat transfer between surfaces of an enclosure. Radiation shields are also discussed. Chapter 10 is unique. It is designed to enhance the students’ problem-solving abilities. Most of the previous chapters were based on a single mode of heat transfer. The problems in this chapter are multimode, incorporating more than one mode of heat transfer. The students have to determine what modes are significant, whether the problem is steady state or unsteady, and whether an analytical or numerical solution should be used. They must also search out and provide any necessary information not given in the problem statement. Chapter 11 covers mass transfer. It discusses concentrations and properties of gas mixtures, Fick’s law of diffusion, diffusion coefficients, and Stefan’s law. Evaporation, venting from containers, and mass transfer through walls and membranes are also covered. Finally, it discusses the wet-bulb and dry-bulb psychrometer. Chapter 12 consists of special topics which can be covered in the course if time permits. The Chapter 3 discussion of heat generation in wires is extended to cover generation in plane walls and spheres. Contact resistance and condensation/boiling are also discussed. Finally, there is a discussion of energy usage in buildings, with special emphasis on the degree-day method. The Appendix includes properties of materials, tables of Bessel and error functions, a discussion of the use of Excel and MATLAB, and listings of MATLAB m-files for the examples of Chapter 5. A few comments regarding teaching the course: A semester-length course at Hofstra is usually 13 or 14 weeks long with two 80-minute lectures per week. I have found that Chapters 1 through 9 can be readily covered during a semester. It also may be possible to include a couple of topics from Chapters 11 or 12 if the professor so desires. I also recommend devoting a class session to Chapter 10. I typically switch the order of Chapters 4 and 5. I teach Chapter 5 before Chapter 4 because I require the students to submit two computer solutions: one for steady state and the other for unsteady state. Switching the order allows the students to jump-start their computer work. I give three exams: one after Chapter 3, one after Chapter 8, and the final exam. I typically assign three or four homework problems per week. I feel that class attendance is very worthwhile for the students, so my grading is usually 60% exams, 15% homework, 15% attendance, and 10% computer problems. The following materials are provided to professors adopting the book: A solutions manual for endof-the-chapter problems; Excel spreadsheets and MATLAB m-files for end-of-the-chapter problems; PowerPoint lecture slides; image bank of figures from the book; To obtain access, qualified instructors should log in, or register and create an account, at www.textbooks.elsevier.com.

Acknowledgments There are many people I wish to thank. First, I gratefully thank my parents, Geneva and Harry, who provided me with a wonderful childhood and to whom I owe everything. The concept for this book started several years ago when I had discussions with publisher’s representative Ronde Bradley in my Hofstra University office. I remember discussing the existing heat transfer texts and expressing my interest in possibly writing a text. Ronde passed my ideas to Joe Hayton, currently Publishing Director of Elsevier Science. I had several emails with Joe, but life intervened and the project never got started until a few years ago. Thanks so much, Ronde. And, thanks so much Joe for your insightful comments and your actions to get this project moving. There are others at Elsevier that I wish to thank. As a new author, my knowledge of the publishing industry was essentially nil. Many thanks to Steve Merken, Senior Acquisitions Editor, who helped me along and gave me much encouragement. Also, many thanks to Susan Ikeda, Senior Editorial Project Manager, for her expert assistance and kindness. Finally, thanks to Sujatha Thirugnana Sambandam, Publishing Services Manager, who has led the production activities. Regarding my Hofstra University associates, I would like to thank John P. LeGault, former Engineering Labs Supervisor, who provided assistance with MATLAB. Thanks also to the staff of the ILLiad Interlibrary Loan department who responded swiftly and competently to my many requests for journal articles. Finally, many thanks to my family: Gail, Nancy, Rich, Andrew, Meaghan, and grandchildren Matthew, Abby, Sarah, and Christopher who have given me much joy during this long project and continue to give me much joy. Special thanks to my wonderful wife, Gail, for her encouragement and love. It is to Gail that I dedicate this book.

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CHAPTER

Introduction to heat transfer

1

Chapter outline 1.1 Introduction ...................................................................................................................................1 1.2 Modes of heat transfer....................................................................................................................2 1.3 Conduction.....................................................................................................................................2 1.3.1 Conduction through a plane wall ..................................................................................3 1.4 Convection.....................................................................................................................................5 1.5 Radiation .......................................................................................................................................9 1.6 The direction of heat flow .............................................................................................................11 1.7 Temperature continuity and heat balances .....................................................................................12 1.8 Unit systems ................................................................................................................................14 1.9 Recommended approach to problem solving...................................................................................15 Step 1 - Problem definition ..................................................................................................15 Step 2 - Problem givens.......................................................................................................15 Step 3 - Determine the appropriate equations........................................................................15 Step 4 - Obtain the solution .................................................................................................16 Step 5 - Review the solution ................................................................................................16 1.10 Significant figures ........................................................................................................................16 1.11 Chapter summary and final remarks ...............................................................................................17 1.12 Problems .....................................................................................................................................17 References ............................................................................................................................................21

1.1 Introduction Let us first discuss the relationship between thermodynamics and heat transfer. Thermodynamics, the first thermal science course in most engineering programs, is a study of equilibrium states of systems. It considers the different types of energy (e.g., mechanical, kinetic, and potential, heat, internal) and the amount of energy transferred during a system’s process from one equilibrium state to another. Thermodynamics has two basic laws. The first law deals with conservation of energy and the conversion of energy from one type to another. The second law deals with observed restrictions on system processes. For example, there cannot be a process which solely involves the transfer of heat from a region of lower temperature to a region of higher temperature. Thermodynamics considers the amount of energy being transferred. It does not usually deal with the time needed for this energy transfer.

Heat Transfer Principles and Applications. https://doi.org/10.1016/B978-0-12-802296-2.00001-9 Copyright © 2021 Elsevier Inc. All rights reserved.

1

2

Chapter 1 Introduction to heat transfer

Heat transfer adds the time dimension to energy processes. It deals with the rate of transfer of heat energy as a system moves from one equilibrium state to another. As an example, let us consider a hot metal object at initial temperature T1 being cooled by a liquid at temperature T2. During the process, the heat flows from the object to the liquid and the temperature of the object decreases. The object finally reaches the temperature T2 of the liquid. Thermodynamics can predict the amount of heat transferred to the liquid during the process. Heat transfer provides further details. It can predict how long it will take for the object’s temperature to go from T1 to T2. Heat transfer can also provide the temperature distribution in the object at different times during the process. In this chapter, we introduce the three modes of heat transfer (conduction, convection, radiation) and present their fundamental equations. We also discuss unit systems and give recommendation regarding approaches to problem solving and the number of significant figures to retain in numerical results.

1.2 Modes of heat transfer Mechanical engineering has two major areasdthermal/fluids and mechanics/machine design. Heat transfer, the subject of this book, is part of the first area. Transfer of heat occurs as a result of a temperature difference. This is the driving force which causes the heat to flow. If there is no temperature difference, there is no heat flow. It is also observed that heat flows in the direction from the higher temperature to the lower temperature. There are three primary modes of heat transfer: conduction, convection, and radiation. Heat conduction occurs if there is a temperature difference between two locations in a solid or two locations in essentially nonmoving liquids or gases. Heat convection typically occurs due to a temperature difference between a surface of an object and an adjacent fluid. Change of phase (i.e., evaporation and condensation) is also included in the convection category. Heat radiation occurs between two surfaces of different temperatures. Conduction and convection require physical material (i.e., solids or fluids) for their transport. Radiation can occur through a vacuum. Knowledge of heat transfer fundamentals is essential for design of many devices, systems, and processes. For example, heat transfer knowledge is needed for design of systems for cooling of computers and other electrical devices; heating and cooling equipment for buildings, refrigerators, boilers, and car engines; and heat exchangers for many different industrial applications. Heat transfer is indeed a major subject area for mechanical engineers and also chemical engineers. We will now introduce the three modes of heat transfer. Later we will discuss unit systems, problem-solving techniques, and significant figures.

1.3 Conduction Transfer of heat by conduction occurs in solids and in essentially nonmoving liquids and gases. It has been observed (through experimentation) that the rate of heat transfer per unit area is proportional to the temperature gradient in the material. That is, qn vT f An vn

(1.1)

1.3 Conduction

3

where n designates the direction of the heat flow; e.g., x, y, or z in Cartesian coordinates. In this equation, qn is the rate of heat flow in direction n, An is the cross-sectional area through which the heat flows, and vT/vn is the temperature gradient in direction n. (Note: The area A is the area perpendicular to the heat flow.) If we introduce a proportionality parameter into Eq. (1.1) and introduce a minus sign so that heat flow in the positive n direction will be numerically positive, we get Fourier’s law [1,2]: qn vT (1.2) ¼ k vn An Eq. (1.2), named for French mathematician and physicist Joseph Fourier (1768e1830), gives the heat flow rate per unit area in the n direction at a point in the solid or fluid. Heat flow rate per unit area is called “heat flux.” Heat flow is a vector quantity which has direction and magnitude. If we are using Cartesian coordinates, there can be heat flow and heat flux components in all three coordinate directions. The heat flow components are qx, qy, and qz. The heat flux components are q vT q vT q vT ¼ k ¼ k ¼ k (1.3) A x vx A y vy A z vz where k is the thermal conductivity of the material. If we are using the SI unit system where q is watts, the area is m2, and the temperature gradient is C/m, then k has the units of W/m C. (Note: As the size of a Celsius degree is the same as the size of a Kelvin degree, the numerical value of k in W/m C is the same as its value in W/m K. That is, 1 W/m C ¼ 1 W/m K.) [Note: In this text, we will usually leave out the symbol used for a temperature degree. That is, we will often use W/m C rather than W/m C for the units of thermal conductivity. And, when we are talking about a temperature of 20 degrees Celsius, we will use 20 C rather than 20 C.] Thermal conductivity is highest for pure metals and a bit lower for alloys. The conductivity for liquids is generally lower (except for liquid metals) and for gases even lower. Building materials such as wood, plaster, and insulation have low conductivity. Multilayer evacuated insulation used for insulating cryogenic tanks has very low conductivity. For most materials, thermal conductivity is isotropic; that is, it is the same in all directions. However, for some materials, it varies with direction. For example, although a single average k value is often given for wood, the conductivity of wood is actually different in the across-the-grain and withthe-grain directions. Table 1.1 gives typical thermal conductivity values for some materials at about 20 C. Appendices A through E give detailed information on the properties of common solids, liquids, and gases. In general, thermal conductivity decreases with temperature for metals, decreases slightly with temperature for many liquids, and increases with temperature for gases.

1.3.1 Conduction through a plane wall In this section we develop the equation for the rate of heat flow through a plane wall. The equation is very useful for estimating heat flow through large walls of finite thickness, such as the exterior walls of a building. It is also useful in other situations where the heat flow can be considered to be onedimensional; that is, the heat flow is in a single direction.

4

Chapter 1 Introduction to heat transfer

Table 1.1 Typical thermal conductivity at 20 C. Material

k (W/m C)

Aluminum Copper Gold Silver Carbon steel Stainless steel Plasterboard Brick Cement (hardened) Hardwoods Softwoods Styrofoam Water Engine oil Air (1 atm pressure) He (1 atm pressure)

240 400 315 430 40 15 0.8 0.7 1.0 0.16 0.12 0.03 0.60 0.14 0.025 0.15

Consider the plane wall shown in Fig. 1.1. The wall has a thickness L and is very large in the y and z directions. The left face of the wall is at temperature T1 and is at location x ¼ 0. The right face of the wall is at temperature T2 and is at location x ¼ L. Let us assume that T1 > T2. The heat flow will be one-dimensional and will be in the positive x direction.

y

T1 q

T2

L O z

FIGURE 1.1 Heat conduction through a plane wall.

x

1.4 Convection

5

As all the heat flow is in the x direction, Eq. (1.2) can be modified by changing “n” to “x” (or just simply use “q”) and changing the partial derivative to a total derivative. That is, q dT ¼ k (1.4) A dx Moving dx to the left-hand side of the equation, we can then integrate both sides of the equation: Z Z T2 q L dx ¼ k dT (1.5) A 0 T1 Z T1 q L¼ k dT (1.6) A T2 In general, the conductivity k is a function of temperature. However, if k is constant or if we use an average k value for the problem, then k may be taken outside of the integral sign in Eq. (1.6), and we have q L ¼ kðT1 T2 Þ (1.7) A Rearranging this equation, we reach its final form: kA ðT1 T2 Þ L Details of conduction heat transfer are given in Chapters 2e4. q¼

(1.8)

Example 1.1 Heat flow through a plane wall Problem A large concrete wall is 300 mm thick and has a thermal conductivity of 1.2 W/m C. The heat flux through the wall is 100 W/m2. The higher-temperature surface of the wall is at 42 C. What is the temperature of the other surface of the wall?

Solution Heat flux is heat flow per unit area, or q/A. From Eq. (1.8), we have kA ðT1 T2 Þ L We want to obtain T2. Rearranging the equation, we get q L 0:3 ¼ 42 ð100Þ ¼ 17 C T2 ¼ T1 A k 1:2 q¼

The temperature of the cooler surface of the wall is 17 C.

1.4 Convection Transfer of heat by convection typically occurs between a surface and an adjacent fluid. Convection is a phenomenon involving conduction in the fluid and fluid motion.

6

Chapter 1 Introduction to heat transfer

Velocity Profile u∞

Fluid Flow T∞ q

u Ts

x

FIGURE 1.2 Fluid flow over a surface.

Fig. 1.2 below shows a fluid flowing over a surface. Observations show that the fluid sticks to the surface. As we move away from the surface, the fluid velocity, u, increases until it reaches the free stream velocity uN. In the free stream, the temperature of the fluid is TN. The temperature of the surface is Ts. Let us consider the case where the surface temperature is greater than the fluid temperature; that is, Ts > TN. Heat flows by conduction from the surface into the stagnant fluid layer on the surface. The heat then proceeds into the adjacent fluid which is moving. Convection is this combination of conduction at the surface and heat transport into the moving fluid. There are two main classifications of convection: forced convection and natural (or free) convection. In forced convection, there is a device causing significant fluid motion. This could be a fan or blower for gases or a pump for liquids. Or, it could be the wind blowing on a building on a windy day. In natural convection, there is no device. The fluid motion is gentler and is caused by buoyancy effects in the fluid caused by thermal expansion. It has been observed that the rate of heat transfer by convection depends on the difference in temperature of the surface and the fluid. In addition, there are several other factors which influence the heat transfer. These include the motion of the fluid, the thermal and physical properties of the fluid, and the geometry and orientation in space of the surface. The equation for convection, Newton’s law of cooling, is q ¼ hAðTs TN Þ

(1.9)

In this equation, named for English mathematician, physicist, astronomer, and theologian Sir Isaac Newton (1643e1727), q is the rate of heat flow, A is the surface area in contact with the fluid, Ts is the temperature of the surface, TN is the temperature of the fluid, and h is the convective coefficient. If we are using the SI unit system, then “h” has the units of W/m2 C. In the English system, the units are Btu/ h ft2 F. Eq. (1.9) looks simple. It really isn’t! The convective coefficient h is a complex parameter which depends on the several factors mentioned above. In some cases, it also depends on the temperature difference between the surface and the fluid.

1.4 Convection

7

Table 1.2 Convective coefficient for a heated horizontal cylinder. Situation

h (W/m2 C)

Natural convection in air Natural convection in oil Forced convection with a flow of air at 5 m/s across the cylinder Forced convection with a flow of oil at 5 m/s across the cylinder

9 89 40 1500

The cylinder has a diameter of 3 cm. The temperature of the surface of the cylinder is 200 C and the temperature of the adjacent fluid is 20 C.

In general, h values are greater for liquids than for gases. And, they are greater for forced convection than for natural convection. Table 1.2 illustrates this generality for the situation of convection from a heated horizontal cylinder. We will learn in Chapters 6 and 7 how to determine the convective coefficient for a variety of situations. Until then, h values will be provided for problems involving convection.

Example 1.2 House wall with conduction and convection Problem A house wall is 8 feet high and 20 feet wide. It is 5 inches thick and has a thermal conductivity of 0.017 Btu/h ft F. It is winter and we want to determine the convective coefficient h on the outside surface of the wall. The temperatures of the inside and outside surfaces of the wall are measured and found to be 72 and 28 F, respectively. The outside air temperature is measured, and it is 25 F. What is the convective coefficient on the outside surface of the wall?

Solution As the wall area is large, we will assume one-dimensional heat transfer, with the heat flow from the inside surface of the wall to the outside air. The rate of heat conduction through the wall is given by Eq. (1.8): q¼

kA ðT1 T2 Þ L

Dividing the heat flow q by cross-sectional area A, we get the heat flux through the wall: q k ¼ ðT1 T2 Þ A L For this problem, k ¼ 0:017 Btu = h ft F T1 ¼ 72 F T2 ¼ 28 F L ¼ 5 in

1 ft ¼ 0:41667 ft 12 in

There is convection from the outer surface of the wall to the outside air. The rate of convection is given by Eq. (1.9): q ¼ hAðTs TN Þ

8

Chapter 1 Introduction to heat transfer

For this problem, Ts ¼ 28 F TN ¼ 25 F There is energy conservation at the outer surface of the wall. That is, the heat flux into the surface by conduction equals the heat flux out of the surface by convection. From the above equations, q k ¼ ðT1 T2 Þ ¼ hðTs TN Þ A L Putting values in this equation, we have 0:017 ð72 28Þ ¼ hð28 25Þ 0:41667 Solving for h, we get h [ 0.598 Btu/h ft2 F. (Note: The h’s on the two sides of this result are different. The italic one on the left side is the convective coefficient. The one on the right side is the unit “hours.”) The convective coefficient on the outside surface of the wall is 0.598 Btu/h ft2 F.

Example 1.3 Fins on a surface Problem Fins are often added to a surface to enhance heat transfer. The subject of fins and extended surfaces is discussed in detail in Chapter 3. There are many different geometries of fins. This example deals with a single cylindrical fin, a “pin” fin, as shown in Fig. 1.3. An aluminum pin fin with a 0.3 cm diameter and a length of 10 cm is attached to a wall whose surface is at 180 C. The conductivity of the fin is 230 W/m C. The convective coefficient between the fin and the room air is 50 W/m2 C and the room air is at 22 C. What is the rate at which heat flows into the fin from the wall, and what is the heat flow rate in the fin at a distance of 3 cm from the wall? Note: The temperature distribution in the fin can be considered to be one-dimensional in the direction of the fin’s axis and, as we will show in Chapter 3, is given by: TðxÞ ¼ TN þ ðTs TN Þemx

x

FIGURE 1.3 A pin fin.

(1.10)

1.5 Radiation

9

where x is the distance from the surface to which the fin is attached to the particular location in the fin, Ts is the temperature of the surface to which the fin is attached, and TN is the fluid (i.e., room air) temperature. rﬃﬃﬃﬃﬃﬃ hP m¼ kA where h is the convective coefficient, and P is the perimeter around the fin. k is the thermal conductivity of the fin, and A is the cross-sectional area of the fin.

Solution This fin problem illustrates a situation including both conduction and convection. At the wall, heat flows into the fin by conduction. The heat is conducted axially through the fin from its point of attachment at the wall to its free end. Heat also goes from the surface of the fin, along its whole length, to the room air by convection. P ¼ p D ¼ (p) (0.3 cm) ¼ 0.9425 cm ¼ 0.009425 m A ¼ p R2 ¼ (p) (0.15 cm)2 ¼ 0.07069 cm2 ¼ 7.069 106 m2 rﬃﬃﬃﬃﬃﬃ sﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ hP ð50Þð0:009425Þ ¼ ¼ 17:025 m¼ kA ð230Þð7:069 106 Þ From Eq. (1.4), the heat flux at any x location in the fin is Differentiating the T(x) function of Eq. (1.10), we get

At the wall (x ¼ 0), dT dx

q A

¼ k dT dx .

dT ¼ ðTs TN Þð mÞemx dx ¼ ð180 22Þð 17:025Þð1Þ ¼ 2690 C m and the heat flow is

dT ¼ ð230Þ 7:069 x 106 ð 2690Þ ¼ 4.37 W dx ð17:025Þð0:03Þ ¼ 1614:1 C m and the heat flow is At location x ¼ 3 cm, dT ¼ ð180 dx 22Þð 17:025Þe dT 6 q ¼ kA dx ¼ ð230Þ 7:069 x 10 ð 1614:1Þ ¼ 2.62 W. q ¼ kA

Note: The heat flows are both positive, so they are both in the positive x direction. The heat flow rate into the fin at the wall is 4.37 W. The heat flow rate in the fin at a distance of 3 cm from the wall is 2.62 W. Discussion: From the wall to the location 3 cm from the wall, the conduction has decreased from 4.37 to 2.62 W, a difference of 1.75 W. From energy conservation, this 1.75 W is the rate at which heat is flowing by convection from the surface of the fin to the room air over the first 3 cm length of the fin.

1.5 Radiation All surfaces emit thermal radiation. The rate of emission depends on the absolute temperature and characteristics of the surface. A black surface is an ideal surface which emits the highest possible amount of thermal radiation. For a black surface emitting in accordance with Planck’s law, the rate of thermal radiant emission is qb ¼ sAT 4

(1.11)

where qb is the rate of radiant energy emission, A is the area of the surface, T is the absolute temperature (Kelvin or Rankine) of the surface, and s is the StefaneBoltzmann constant ¼ 5.67 108 W/m2 K4 ¼ 0.1714 108 Btu/h ft2 R4. Real surfaces emit less radiation than ideal black surfaces. The surface property related to the emission of real surfaces is the emissivity, ε, which is the ratio of the power emitted by a real surface to

10

Chapter 1 Introduction to heat transfer

Table 1.3 Emissivities for various surfaces. Surface

Emissivity ε

Aluminum (polished) Aluminum (oxidized) Copper (highly polished) Copper (oxidized) Iron (rusted) Paint (aluminum) Paint (nonaluminum) Brick Wood

0.05 0.3 0.02 0.6 0.6 0.4 0.9 0.94 0.9

the power emitted by an ideal black surface at the same temperature. For a black surface, ε ¼ 1. It is lower for real surfaces. Table 1.3 gives typical emissivity values for different surfaces. Additional values are given in Appendix F. We are usually primarily concerned with the net transfer of radiant energy between surfaces. Like conduction and convection, a temperature difference is needed for heat flow to occur. However, unlike conduction and convection, no physical material is needed to transfer heat via thermal radiation. The radiation is transmitted as an electromagnetic wave. It passes undiminished through a vacuum and is negligibly affected by many gases (e.g., air). Radiant heat transfer between surfaces depends on the absolute temperatures of the surfaces, the emissivities of the surfaces, the geometries of the surfaces, and the orientation of the surfaces with respect to each other. This can be described by the equation q ¼ sFG;ε T14 T24 (1.12) A where FG,ε is a function of the emissivities, geometries, and orientations of the surfaces. Radiant heat transfer is a complex subject treated in more detail in Chapter 9. One special case of practical interest is shown in Fig. 1.4. This is the case in which a heated convex object radiates to the surfaces of a large surrounding enclosure.

A1,ε 1,T1 q T2

FIGURE 1.4 Radiation of a surface to a large enclosure.

1.6 The direction of heat flow

11

If the object has surface area A1, absolute temperature T1, and emissivity ε1, and the enclosure is at absolute temperature T2, then the radiant heat transfer from the object to the enclosure is q ¼ ε1 sA1 T14 T24 (1.13) Details of radiative heat transfer are given in Chapter 9.

Example 1.4 Convection and Radiation from a Cylinder Problem

A heated aluminum cylinder (k ¼ 237 W/m K) is 5 cm diameter and 1 m long. The temperature of its surface is 200 C. The cylinder is in a room where the air temperature is 20 C and the temperature of the room walls is 15 C. The emissivity of the cylinder’s surface is 0.5 and the convective coefficient at the surface is 25 W/m2 C. What is the heat flow from the side of the cylinder by convection and radiation?

Solution The temperatures for the convective heat transfer can remain in Celsius (C), but the temperatures must be in absolute temperature Kelvin (K) for the radiant heat transfer as follows: TN ¼ room air temperature ¼ 20 C. Ts ¼ surface temperature ¼ 200 C ¼ 200 þ 273.15 ¼ 473.15 K. Tsurr ¼ temperature of the room walls ¼ 15 C ¼ 15 þ 273.15 ¼ 288.15 K. The area of the side of the cylinder is A ¼ p D L ¼ p (0.05 m) (1 m) ¼ 0.1571 m2. The convective heat transfer is The radiant heat transfer is

qconv ¼ hAðTs TN Þ ¼ ð25Þð0:1571Þð200 20Þ ¼ 707.0 W

4 ¼ ð0:5Þ 5:67 x 108 ð0:1571Þ 473:154 288:154 ¼ 192.5 W qrad ¼ εsA Ts4 Tsurr The heat flow from the side of the cylinder by convection is 707.0 W. The heat flow by radiation is 192.5 W. Comments: Radiant heat transfer depends on temperature to the fourth power and becomes very significant at high temperatures. As the temperatures in this problem are quite low, the radiation heat transfer is small compared to the convective heat transfer. The radiant heat transfer could be significantly reduced by polishing the cylinder, thereby reducing the emissivity to 0.1 or less. Also, because the convection and radiation occur at the cylinder’s surface, the information given about the material of the cylinder and the conductivity are extraneous and not relevant to the problem. (They would have been relevant if the problem had dealt with conduction within the cylinder.) Engineers often encounter situations in which a surplus of information is available. Judgments must be made as to which items of information are relevant to the specific situation. On the other hand, engineers often have to make educated judgments for required information that has not been provided in the problem statement.

1.6 The direction of heat flow Heat flow is a vector quantity. It has direction and magnitude. In Eq. (1.2) for conduction, we added a minus sign so that heat flow in the positive n direction would be positive and heat flow in the negative n direction would be negative. For example, if heat is flowing in the positive n direction, then vT/vn is negative because heat flows from a higher temperature toward a lower temperature. The minus sign is needed to make qn positive.

12

Chapter 1 Introduction to heat transfer

Likewise, if heat is flowing in the negative n direction, then vT/vn is positive and the minus sign makes qn negative. The direction of heat flow must also be considered for convection and radiation. Consider, for example, Eq. (1.9) for convection: Say that heat is flowing from the surface to the fluid, as shown in Fig. 1.3. In this case, Ts > TN, and Eq. (1.9) is appropriate because it makes the heat flow positive in the direction it is actually flowing. But, if the heat were flowing from the fluid to the surface, then the equation should be revised to make the heat flow positive in the direction it is actually flowing. That is, if TN > Ts we should revise Eq. (1.9) to q ¼ hAðTN Ts Þ

(1.14)

Heat flow direction must also be considered when formulating equations for radiative heat transfer.

1.7 Temperature continuity and heat balances Due to the broadness of the subject of heat transfer, it is usual to separate heat transfer into its three modes and study conduction, convection, and radiation separately. While advantageous for the learning process, one has to keep in mind that many practical problems involve not only one but often two or all three of the modes occurring simultaneously. For example, consider an electric cartridge heater inside a block of metal sitting on a table in a room. This is shown in Fig. 1.5. When the power to the heater is turned on, heat starts to flow into the block, thereby raising its temperature. The temperature of the block will become greater than that of the air in the room, and there will be heat transfer by convection from the exposed surface of the block to the air. There will also be heat transfer by radiation from the exposed surface of the block to the walls and ceiling of the room. And, because the temperature of the block will become greater than the temperature of the table, there will be conduction from the block into the table. There will be continuity of temperature at the exposed surface of the block in contact with the air. That is, the temperature of the exposed surface of the block and the temperature of the air at the surface will be the same. Disregarding any contact resistance between the block and the tabletop, the temperature of the block’s surface in contact with the tabletop will be the same as that of the tabletop.

Air and Surroundings heater

tabletop

FIGURE 1.5 Heater in a metal block.

1.7 Temperature continuity and heat balances

13

There will also be continuity of heat flows. The rate of heat flow by conduction reaching the exposed surface of the block will be equal to the rate of heat flow leaving the block by convection and radiation. Regarding the surface in contact with the tabletop, the rate of heat flow by conduction within the block reaching the tabletop will equal the rate of heat flow by conduction into the table. If the electric power to the heater remains constant, the temperatures and heat flows will eventually stabilize and reach steady state. When this happens, we can do an energy balance on the block. We can state that the electric power to the heater equals the rate of heat transfer from the block’s exposed surface to the air by convection and to the surroundings by radiation plus the rate of heat transfer by conduction from the block to the tabletop. qpower ¼ qconduction to table þ qconvection to air þ qradiation to surroundings

(1.15)

Example 1.5 Heat Balance Problem Consider again Fig. 1.5 which depicts a metal block with an internal cartridge heater. The block rests on a tabletop. It convects to the room air, radiates to the surroundings, and conducts to the tabletop. The block is 6 inches by 6 inches by 6 inches and has a 300 W internal electric heater. The room air and the surroundings are at 72 F. The convective coefficient is 5 Btu/h ft2 F and the emissivity of the block’s exposed surfaces is 0.4. It can be assumed that the heat flow by conduction to the tabletop is negligible. What is the average temperature of the exposed surfaces of the block?

Solution Eq. (1.15) gives the heat balance for the block. At steady state, the power into the heater equals the sum of the convection to the room air, the radiation to the surroundings, and the conduction to the tabletop. The givens of the problem are Tsurr ¼ 72 F ¼ 72 þ 459.67 ¼ 531.67 R ðRankineÞ TN ¼ 72 F ¼ 531.67 R h ¼ 5 Btu = h ft2 F ε ¼ 0:4 qpower ¼ 300 W ð3.412 Btu = hÞ = W ¼ 1023.6 Btu = h The exposed area of the block is A ¼ 5 sides (6 6) ¼ 180 in2 ¼ 1.25 ft2. Let the unknown temperature of the exposed surfaces be Ts. The convective heat transfer is qconv ¼ hAðTs TN Þ ¼ ð5Þð1:25ÞðTs 531:67Þ ¼ 6:25ðTs 531:67Þ The radiant heat transfer is 4 qrad ¼ εsA Ts4 Tsurr ¼ ð0:4Þ 0:1714 x 108 ð1:25Þ Ts4 531:674 ¼ 8.57 1010 Ts4 531:674 Rearranging Eq. (1.15) above, we get qconduction to table ¼ qpower qconv qrad From the problem statement, the conduction to the table is zero. So, we have qpower qconv qrad ¼ 0.

14

Chapter 1 Introduction to heat transfer

Substituting information from above, we have the following equation: 1023:6 6:25ðTs 531:67Þ 8:57 x 1010 Ts4 531:674 ¼ 0 Reducing this equation, we get the following equation to solve for Ts: 8:57 1010 Ts4 þ 6:25Ts 4415 ¼ 0 We can solve this through several techniques. One way is to use trial and error. A guess is made for Ts and the left-hand side of the equation is evaluated. The result is compared to the right-hand side, which is zero. If the left-hand side is not zero, a new guess is made for Ts and the procedure is repeated. Iterations are made until the left-hand side is zero. A more efficient solution of the equation is through the use of the Goal Seek feature of Microsoft Excel. This useful feature is discussed in Appendix J. It is found that the average temperature of the exposed surfaces of the block is Ts ¼ 677.51 R ¼ 677.51 459.67 ¼ 218 F. The average temperature of the exposed surfaces of the block is 218 F.

1.8 Unit systems Problems in this book use either the SI unit system or the English Engineering unit system. Most problems use SI. Conversion factors are located on the inside front cover of the book for easy reference. Unit conversion is really quite simple. One starts with the item to be converted and merely keeps multiplying it by unity factors until the desired units are reached. For example, say we want to convert 10 Watts to BTU/day. We have

10 W

Btu h 24 h ¼ 819 Btu W day day

3:412

One common area of confusion pertains to temperature. The confusion is with respect to temperature levels versus temperature units. Temperature levels are determined by the following equations: Celsius to Kelvin:

K ¼ C þ 273.15

Fahrenheit to Rankine:

R ¼ F þ 459.67

9 C þ 32 5 5 Fahrenheit to Celsius: C ¼ ðF 32Þ 9 Temperature levels in the four temperature scales are usually quite different. One notable exception is that 40 C ¼ 40 F. Things are quite different when we are considering the units of physical and thermal properties. The size of a Fahrenheit degree is the same as the size of a Rankine degree, and the size of a Celsius degree is the same as the size of a Kelvin degree. For example, a thermal conductivity of 150 W/m C is same as a conductivity of 150 W/m K. And a specific heat of 0.2 Btu/lbm F is the same as 0.2 Btu/lbm R. Celsius to Fahrenheit:

F¼

1.9 Recommended approach to problem solving

15

One must remember that there is no need to convert numerical values from Celsius to Kelvin or from Fahrenheit to Rankine when the temperatures are within the units of the various material properties. However, it is of course necessary to convert the numerical values when we are considering temperature levels. The size of a Fahrenheit (or Rankine) degree is smaller than the size of a Celsius (or Kelvin) degree. Let us consider the freezing and boiling points of water. Water at atmospheric pressure freezes at 32 F and boils at 212 F. There are 180 Fahrenheit degrees between the two points. For Celsius, water freezes at 0 C and boils at 100 C. There are 100 Celsius degrees between freezing and boiling. Therefore, a difference of 180 Fahrenheit degrees is equivalent to a difference of 100 Celsius degrees. A Fahrenheit degree is (100/180) ¼ (5/9) the size of a Celsius degree. If we have a temperature difference in Fahrenheit degrees, the equivalent temperature difference in Celsius degrees is 5/9 that of the Fahrenheit degree difference. Finally, we must remember that absolute temperature units (Kelvin or Rankine) must be used for problems involving thermal radiation.

1.9 Recommended approach to problem solving In solving an engineering problem, be it a problem at the end of the chapters in this book or a general problem in engineering, people often have the urge to solve the problem as quickly as possible, and they immediately start writing equations and inserting numbers. This approach should be avoided. Rather, the approach outlined below is recommended. This approach appears to be slower, but it actually is more efficient because it avoids having to go back and redo hastily and ill-performed steps.

Step 1 - Problem definition This is perhaps the most important step. We need to clearly understand the objective of the problem. What are we looking for? If we get this step wrong, then we may have a perfect solution, but it is not relevant to the problem we want to solve! Write the objective of the problem in one or two sentences.

Step 2 - Problem givens List the information given in the problem statement. In textbook problems, there is usually just enough information given to solve the problem, and there is no extraneous information. This is really unlike problems which you will encounter in the “real world.” In those problems, there is often an excess of information and you will have to determine which parts of this information are relevant to your problem. (Indeed, we included some extraneous information in Example 1.4 above and we will do this occasionally in the end-of-chapter problems.)

Step 3 - Determine the appropriate equations Considering the problem definition and the givens, determine the analytical equations appropriate for the problem. For example, what modes of heat transfer are involved? Do we have conduction, convection, or radiation, or a combination of modes? Is the heat transfer one-, two-, or threedimensional? Is the heat transfer steady state or transient? List the equations.

16

Chapter 1 Introduction to heat transfer

Step 4 - Obtain the solution Put values into the equations and obtain numerical solutions. Make sure that you include units with your numerical results! State the result in a sentence or two.

Step 5 - Review the solution This final step is of major importance! Consider your solution. Look at the problem definition and assure yourself that you have solved the correct problem. And, is the numerical result reasonable? Are the magnitudes of the results consistent with the magnitudes of the problem givens? For example, if your problem was to determine the temperature at a point in a metal plate, and your answer was 900 C when the surrounding air was at 20 C, and there are no internal heat sources or heaters in the problem, then perhaps something is wrong. Perhaps you made an error in your calculations. Check the equations and the values you put into them. Also, check that the correct units were used. The above procedure is a methodical way of solving heat transfer problems. Indeed, the procedure is also very applicable to other areas of engineering. It is highly recommended that the procedure be followed as much as possible. Getting into a methodical mindset will serve you well in your engineering career.

1.10 Significant figures There is a great tendency on the part of students (and often engineers) to retain too many significant figures in a final answer. Indeed, calculators provide 10 or more digits and computers have even more. Why not include all of the digits in a final answer? Well, inclusion of all the digits is usually inappropriate because of the inaccuracies and uncertainties inherent in all engineering calculations. First, simplifications are often made in the problem statement and the analysis technique. For example, surfaces are taken as being perfectly insulated; thermal resistances of thin metal walls are neglected; multidimensional heat flows are assumed to be one-dimensional. Inaccuracies also occur with the use of tabulated material property data from reference sources. Perhaps such data were obtained under stringent laboratory conditions which are inconsistent with your application. And perhaps you were unable to find the material properties for your particular material and therefore substituted the properties of a similar material. (Of course, these concerns are minimized if tests are conducted on your actual specimens, but such is usually not the case.) As another example, inaccuracies may be introduced through use of empirical correlations presented in reference sources; e.g., correlations for the convective coefficient “h.” Such correlations may have been from carefully controlled laboratory experiments. Your conditions may differ considerably from the conditions under which the data were obtained. In short, one must be careful to not include an excess of significant figures in a final answer. Too many figures imply that you know the result with more precision than you actually do. For example, an answer of 24.373 C implies that the solution is accurate to one-thousandth of a degree. Such is usually not the case. A good rule of thumb for textbook problems and most engineering problems is the following: If an answer begins with "1", give the answer to four significant figures; if it begins with any other digit, give the answer to three significant figures. For example, 1.78425 would become 1.784 and 5.8957 would become 5.90. Of course, this rule of thumb only pertains to the final answer of a

1.12 Problems

17

problem. Intermediate calculations leading up to the final answer should use considerably more significant figures to not introduce inaccuracies through rounding. Significant figures are discussed in many freshman engineering and data analysis texts. Two such texts are given in Refs. [3,4].

1.11 Chapter summary and final remarks This chapter provided an introduction to the study of heat transfer. The three modes of heat transfer (conduction, convection, and radiation) were discussed, and basic equations for each mode were presented. The concepts of temperature continuity and heat balance were discussed. Finally, there were discussions of unit systems, problem-solving techniques, and significant figures. These last three topics are pertinent not only to heat transfer, but to most other areas of engineering. We end this chapter with a word of caution. A lot of useful information is available on the Internet. Most is accurate, but some is not. Use material only from reputable sources. Some such sources are government offices (e.g., NIST) and standards/professional organizations (e.g., ASTM, ASME, ASHRAE, IEEE, ANSI). Reputable manufacturing companies are also a useful source of information. It is good to compare information from different Internet sources to gain confidence regarding the accuracy of the information.

1.12 Problems 1-1 A concrete wall is 20.3 cm thick and has a thermal conductivity of 0.8 W/m C. One surface of the wall is at 20 C and the other surface is at 5 C. (a) What is the heat flux through the wall (W/m2)? (b) If the wall is 6 m long and 2.44 m high, what is the heat flow (W) through the wall? 1-2 A steel plate is 1.5 m by 1.2 m and 0.5 cm thick. The thermal conductivity of the plate is 60.2 W/m C. If the temperature difference of the plate’s surfaces is 0.2 C, what is the heat flow rate (W) through the plate? 1-3 A building wall has a layer of fiberglass insulation. The conductivity of the fiberglass is 0.022 Btu/h ft F. The fiberglass is in a channel between wall studs and is 14.5 inches wide, 7.5 feet high, and 3.5 inches thick. If the temperature of the inside surface of the fiberglass is 65 F and that of the outside surface is 40 F, what is the heat flow rate (Btu/h) through the fiberglass layer? 1-4 A layer of insulating material is 15.2 cm thick. The temperature difference across the surfaces is 25 C and the heat flux through the layer is 6.6 W/m2. What is the thermal conductivity of the insulating material (W/m C)? 1-5 A cup has a diameter of 3 inches and height of 4.5 inches. The wall thickness of the cup is 1/8 inch. The cup is filled with coffee at 140 F. Assume that the inside surface of the cup is at the coffee temperature (140 F) and the outside surface of the cup is at 90 F. The conductivity of the cup’s material is 0.8 Btu/h ft F. (a) What is the heat flow by conduction (Btu/h) through the cylindrical wall of the cup? (b) If the room air is at 70 F, what is the convective coefficient at the outer surface of the cup (Btu/h ft2 F)?

18

Chapter 1 Introduction to heat transfer

1-6 A cup has a diameter of 7.6 cm and height of 11.4 cm. The wall thickness of the cup is 0.32 cm. The cup is filled with coffee at 60 C. Assume that the inside surface of the cup is at the coffee temperature (60 C) and the outside surface of the cup is at 32 C. The conductivity of the cup’s material is 1.4 W/m C. (a) What is the heat flow (W) by conduction through the cylindrical wall of the cup? (b) If the room air is at 21 C, what is the convective coefficient at the outer surface of the cup (W/m2 C)? 1-7 A cylindrical electric heater is used to heat water from 15 to 80 C. The heater has a diameter of 1.5 cm and a length of 25 cm. The surface of the heater is at 95 C during the heating process and the convective coefficient between the heater’s surface and the water is 800 W/m2 C. What is the convective heat flow (W) from the heater’s surface (including the cylindrical side and both ends) when the water temperature is 40 C? 1-8 A 1000 W cylindrical electric heater has a diameter of 1.5 cm and a length of 25 cm. After using the heater to heat water, the heater has unfortunately been taken out of the water and left “on” in the room air. It is convecting and radiating to the room air and the surroundings. Assume that both the room air and the surroundings are at 20 C. Also, assume the emissivity of the heater’s surface is 0.5 and the convective coefficient between the heater and the room air is 10 W/m2 C. What is the temperature of the heater’s surface? 1-9 Cold air at 10 C flows across the outer surface of a 1.5-cm OD tube. The surface of the tube is kept at a constant temperature of 120 C. If the convective coefficient between the outer surface of the tube and the air is 70 W/m2 C, what is the convective heat transfer per meter length of tube (W/m)? 1-10 A solid aluminum cylinder (k ¼ 230 W/m K) has a diameter of 3 cm and a length of 10 cm. It is suspended by a thin wire in the room air. The cylinder has an internal electr-ic heater which keeps the surface of the cylinder at 110 C. The room air is at 25 C and the convective coefficient between the cylinder and the room air is 15 W/m2 C. Assume that the heat flow through the support wire is negligible and that radiation heat transfer is negligible due to the very small emissivity of the surface of the cylinder. What is the heat flow (W) from the cylinder to the room air? 1-11 A solid copper sphere (k ¼ 350 W/m C) has a diameter of 10 cm and is resting on a tabletop. An internal heater keeps the surface of the sphere at 200 C. The surface of the sphere is highly polished and has an emissivity of 0.08. The convective coefficient is 12 W/m2 C. The room air is at 20 C and the surroundings are at 18 C. Assume the heat flow from the sphere to the table by conduction is negligible. (a) What is the heat transfer rate by convection (W)? (b) What is the heat transfer rate by radiation (W)? 1-12 A hot surface at 500 K has an emissivity of 0.7. What is the radiant flux emitted by the surface (W/m2)? 1-13 A sphere of diameter 10 cm is suspended inside a vacuum chamber by a thin wire from the top of the chamber. The sphere’s surface is at 700 C and has an emissivity of 0.3. The vacuum chamber’s walls are at 50 C. What is the heat flow (W) from the sphere? 1-14 Fins are often added to a surface to enhance heat transfer. An aluminum pin fin with a square cross section of 0.5 cm by 0.5 cm and a length of 2 cm is attached to a surface which is at 150 C. The conductivity of the fin is 230 W/m C. The convective coefficient between the

1.12 Problems

19

fin and the room air is 50 W/m2 C and the room air is at 22 C. What is the heat flow rate (W) in the fin at a distance of 1 cm from the surface to which it is attached? Note: The temperature distribution in the fin can be considered to be one-dimensional in the direction of the fin’s axis and is given by TðxÞ ¼ TN þ

ðTs TN Þcosh½mðL xÞ coshðmLÞ

where x is the distance from the surface to which the fin is attached to the particular location in the fin, Ts is the temperature of the surface to which the fin is attached, and TN is the fluid (i.e., room air) temperature.qﬃﬃﬃﬃﬃ m ¼ hP kA where h is the convective coefficient, P is the perimeter around the fin, k is the thermal conductivity of the fin, and A is the cross-sectional area of the fin. For this problem, P ¼ 4 0:5 cm ¼ 2 cm and A ¼ 0:5 cm 0.5 cm ¼ 0:25 cm2 : 1-15 A heated steel plate is 0.5 m by 0.5 m by 2 mm thick and has a thermal conductivity of 45 W/m C. The temperature distribution in the plate (C) is given by Tðx; yÞ ¼ 75eðx0:5Þ $ e2ðy0:5Þ 2

2

Note: The x and y axes are in the plane of the plate. Temperature variation in the z direction is negligible. The bottom left corner of the plate is at x ¼ 0, y ¼ 0 m and the top right corner of the plate is at x ¼ 0.5, y ¼ 0.5 m. (a) What are the x and y components of the heat flux vector in the plate at the center of the plate (x ¼ 0.25, y ¼ 0.25 m) and at the location x ¼ 0.4, y ¼ 0.1 m? (b) For the two points of item (a), combine the x and y components into a vector and determine their magnitudes. Also, determine the direction of the two vectors (i.e., their angle above the positive x axis). 1-16 A heated steel plate is 1 m by 1 m by 2 mm thick and has a thermal conductivity of 45 W/m C. The surface of the plate is in the x-y plane and has the following temperature distribution (C): Tðx; yÞ ¼ 20 þ 100

sinðpxÞ sinhðpyÞ sinhðpÞ

Note: The x and y axes are in the plane of the plate. Temperature variation in the z direction is negligible. The bottom left corner of the plate is at x ¼ 0, y ¼ 0 m and the top right corner of the plate is at x ¼ 1, y ¼ 1 m. (a) What are the x and y components of the heat flux vector in the plate at location x ¼ 0.3, y ¼ 0.6 m? (b) Combine the heat flux components of item (a) into a vector and determine the vector’s magnitude and direction (i.e., angle above the positive x axis). 1-17 A kiln produces 1440 W of heat and has interior dimensions of 10 inches by 9 inches by 6.5 inches. The inside surface of the kiln is at 1200 C and the wall of the kiln has a thermal conductivity of 0.24 W/m C. How thick should the wall be so that the temperature of the exterior surface of the kiln is not greater than 50 C?

20

Chapter 1 Introduction to heat transfer

1-18 A cylindrical electric heater is 0.5 cm diameter and 10 cm long. It has a power output of 50 W and is immersed in a tank of water. What is the convective coefficient (W/m2 C) when the water temperature is 30 C and the surface of the heater is at 95 C? 1-19 It is desired to determine the emissivity of a paint. A sphere of diameter 5 cm is coated with the paint and suspended in a large vacuum chamber. The sphere has a 10 W internal electric heater. At steady-state operation, the surface of the sphere is 200 C. If the walls of the chamber are at 25 C, what is the emissivity of the paint? 1-20 An electronic component is attached to a circuit board. It is 2 cm by 2 cm by 3 mm thick. At steady operation, its power output is 5 W. A fan blows air across the component to keep it cool. The convective coefficient between the component and the air is 80 W/m2 C and the air is at 20 C. If the temperature of the surface of the component cannot be greater than 70 C, what is the rate at which heat must be transferred to the circuit board by conduction (W)? Assume that radiation heat transfer from the component is negligible. 1-21 The thermal conductivity of a material is 63 W/m C. What is its conductivity in the units of W/m K? 1-22 The convective coefficient at a surface is 150 W/m2 C. What is the coefficient in the units of W/m2 K? 1-23 A heater is rated at 1500 W. What is its rating in the units of Btu/h? 1-24 A metal cylindrical container is filled with iceewater and is on the top of a table. The container is 20 cm diameter and 25 cm high. The container gains heat by convection from the room air and radiation from the room surroundings. Heat flow into the container from the table is negligible. Both the air and the surroundings are at 20 C. The emissivity of the outer surface of the container is 0.1 and the convective coefficient at the outer surface is 20 W/m2 C. The container has a thin wall and it can be assumed that the temperature of the wall is at the iceewater temperature of 0 C. (a) What is the heat flow rate into the iceewater mixture (W)? (b) If the heat of fusion of water is 333.5 kJ/kg, how much ice (kg) is melted in 1 hour? 1-25 Consider the problem of Example 1.5 with the following changes: The average temperature of the exposed surfaces is 175 F. The bottom of the block is not perfectly insulated, and the block transfers heat by conduction to the tabletop. The task is to find the rate of this conduction heat transfer (Btu/h). All other givens in the problem remain the same. 1-26 A portable electric heater for heating a room has heating elements in the form of thin metal strips. The total heating surface area of the strips is 0.03 m2. The room air and surroundings are at 20 C. It is desired to find the output of the heater given the surface temperature of the heater’s elements. (a) What are the relevant modes of heat transfer for this problem and what are the directions of these modes? (b) The convective coefficient at the surfaces of the heating elements is 5 W/m2 C and the emissivity of the surfaces is 0.75. If the temperature of the heating surfaces is 950 C, what is the output of the heater (W)? (c) If we want the heat output to be 4000 W, what is the needed surface temperature of the elements?

References

21

1-27 A pipe transfers cold water through a vacuum chamber. The water is at 4 C and the walls of the chamber are at 20 C. It is desired to find the heat transfer from the pipe per meter length of pipe. (a) What are the relevant modes of heat transfer for this problem and what are the directions of these modes? (b) The pipe is 100 Sch 40 galvanized iron. Its outer surface has an emissivity of 0.8 and a temperature of 4.2 C. What is the heat transfer per meter length of pipe and what is its direction? 1-28 A can of soda is 12.2 cm high and 6.3 cm diameter. Its temperature is 2.8 C when it is taken out of the refrigerator and placed on a table. The room is at 22 C. It is desired to determine the change in temperature of the soda after 30 min in the room air. (a) What are the relevant modes of heat transfer for this problem and what are the directions of these modes? (b) If the convective coefficient at the can’s outer surface is 7 W/m2 C and the emissivity of the surface is 0.4, what is the temperature of the soda after 30 min? Assume that soda has the properties of water.

References [1] J. Fourier, Theorie analytique de la chaleur, Paris, 1822. [2] J. Fourier, The Analytical Theory of Heat, Translated by Alexander Freeman, Dover Publications, New York, 1955 (Translation of Reference [1]). [3] G.C. Beakley, H.W. Leach, Engineeringean Introduction to a Creative Profession, Chapter 6, Macmillan, 1982. [4] P.F. Dunn, Measurement and Data Analysis for Engineering and Science, Chapter 2, McGraw-Hill, 2005.

CHAPTER

Heat conduction equation and boundary conditions

2

Chapter Outline 2.1 Introduction ...................................................................................................................................23 2.2 Heat conduction equation ...............................................................................................................24 2.2.1 Rectangular coordinates..............................................................................................24 2.2.1.1 Special casesdrectangular coordinates ................................................................. 25 2.2.2 Cylindrical coordinates................................................................................................27 2.2.2.1 Special casesdcylindrical coordinates................................................................... 32 2.2.3 Spherical coordinates .................................................................................................32 2.2.3.1 Special casesdspherical coordinates .................................................................... 33 2.3 Boundary conditions .......................................................................................................................34 2.3.1 Rectangular coordinates..............................................................................................35 2.3.1.1 Specified temperature ........................................................................................... 35 2.3.1.2 Specified heat flux................................................................................................. 36 2.3.1.3 Insulated boundary ............................................................................................... 37 2.3.1.4 Convection............................................................................................................ 37 2.3.1.5 Radiation .............................................................................................................. 38 2.3.1.6 Convection and radiation ....................................................................................... 39 2.3.1.7 Symmetry conditions............................................................................................. 41 2.3.1.8 Interfacial boundary .............................................................................................. 41 2.3.2 Cylindrical and spherical coordinates ...........................................................................42 2.3.2.1 Symmetry conditions............................................................................................. 43 2.4 Initial conditions ............................................................................................................................45 2.5 Chapter summary and final remarks.................................................................................................45 2.6 Problems .......................................................................................................................................46 Uncited references.................................................................................................................................55

2.1 Introduction Chapter 1 briefly described the three modes of heat transfer: conduction, convection, and radiation. In this chapter, we will look closely at the theoretical aspects of conduction. In particular, we will discuss the governing differential equations and boundary condition equations whose solution will give us the temperature distribution in the body and ultimately the heat flow rates at different points in the body. Heat Transfer Principles and Applications. https://doi.org/10.1016/B978-0-12-802296-2.00002-0 Copyright © 2021 Elsevier Inc. All rights reserved.

23

24

Chapter 2 Heat conduction equation and boundary conditions

We will also discuss the initial condition equations that are needed if the problem is time-dependent, i.e., nonesteady state.

2.2 Heat conduction equation The heat conduction equation is a partial differential equation for conduction of heat through a body. The solution of the equation gives the temperature distribution in the body at different instances of time. As discussed in Chapter 1, differentiation of the temperature distribution will give the heat flows at the various locations in the body. The heat conduction equation is derived for the rectangular (i.e., Cartesian) coordinate system. The equation is then presented for cylindrical and spherical coordinate systems.

2.2.1 Rectangular coordinates In this section, we will derive the differential equation describing the temperature distribution in a body. (By “body” we mean a solid or an essentially nonmoving fluid. In this text, “body” is synonymous with “object” or “material”.) Let us consider an infinitesimal element in the object as shown in Fig. 2.1. The element has dimensions dx, dy, and dz and is located at coordinates x, y, and z, respectively. Fig. 2.1 shows heat flows into the element of qx þ qy þ qz and heat flows out of the element of qxþdx þ qyþdy þ qzþdz . For example, for the x direction: At location x, the incoming heat flow is qx , and at location x þ dx, the outgoing heat flow is qxþdx . In addition to heat flowing in and out of the element, there may be heat generated within the element. For example, an electrical current passing through a wire or busbar will cause resistance (or Joule) heating in the material. Water added to cement will result in hydration, an exothermic heatproducing chemical reaction. Heat is also produced through fission in nuclear fuel elements. Such

y

qy+dy dz

qx

qz qx+dx

dy

qz+dz

dx

qy x x z

FIGURE 2.1 Conductive heat flow through an infinitesimal element.

x + dx

2.2 Heat conduction equation

25

heat-generating phenomena occur within the material, and the rate of heat generation is usually proportional to the volume of the material. The heat flows and generation will, in the most general situation, vary with location in the body and also with time. If there is no variation with time, the heat transfer is “steady” On the other hand, if there is variation with time, the heat transfer is “unsteady” Special cases of unsteady heat transfer are transient and periodic. So, we have heat flowing in and out of the boundaries of the infinitesimal element and possibly also heat input to the material by internal heat generation. If more heat goes into the element than leaves, then the internal energy of the material in the element will increase. If more heat leaves the element than enters, then the internal energy will decrease. Increased internal energy means an increase in temperature, and decreased internal energy means a decrease in temperature. A word equation may be written for the energy balance on the element as follows: Rate of Heat Flow In Rate of Heat Flow Out þ Rate of Heat Generation ¼ Rate of Increase in Internal Energy

(2.1)

The incoming heat flow rate is qx þ qy þ qz . The outgoing heat flow rate is qxþdx þ qyþdy þ qzþdz . The rate of heat generation is qgen dV, where qgen is the generation rate per unit volume and dV is the volume of the element (i.e., dxdy dz). And the rate of increase in internal energy is mass (i.e., density r times volume dxdydz) times specific heat c times change of temperature per unit time. In symbols, Eq. (2.1) is

vT qx þ qy þ qz qxþdx þ qyþdy þ qzþdz þ qgen dx dy dz ¼ rðdx dy dzÞc vt q vT For the x direction, from Eq. (1.3), A ¼ k vx and Ax is dy dz.

(2.2)

x

In addition, qxþdx is qx plus the change in qx over a distance dx. Mathematically, this is v ðqx Þ dx qxþdx ¼ qx þ vx This Taylor expansion can be done similarly for the y and z directions. Substituting these details into Eq. (2.2), canceling dx dy dz, which is in each term, and rearranging the terms, results in the heat conduction equation: v vT v vT v vT vT (2.3) k þ k þ k þ qgen ¼ rc vx vx vy vy vz vz vt Solution of this equation will give the temperature distribution T(x,y,z,t) in the body at different instances of time. To solve the equation, it has to be combined with appropriate boundary conditions that are discussed in Section 2.3 below. And, if the situation is nonesteady state, appropriate initial conditions are needed. These initial conditions are discussed in Section 2.4 below.

2.2.1.1 Special casesdrectangular coordinates The general heat conduction Eq. (2.3) can often be simplified. For example, if the problem is steady state, the vT vt term is zero. If there is no heat generation within the material, then qgen is zero.

26

Chapter 2 Heat conduction equation and boundary conditions

In addition, in many cases, all three dimensions are not needed to accurately model a problem. For example, the wall of a building can often be modeled as one-dimensional, as shown in Example 2.1:

Example 2.1 Heat Flow Through a Wall Problem: To design air conditioning and heating systems for a building, it is necessary to determine the heating and cooling loads for various components of the building’s envelope. Heat flow through walls is often a major contributor to the loads. Let us say we wish to determine the heat flow through the wall shown in Fig. 2.2. We can determine the temperature distribution in the wall by solving a simplified version of Eq. (2.3). The obtained temperature function can then be differentiated to get the heat flow through the wall. The task is to determine the appropriate version of the heat conduction equation for this situation.

Solution: Eq. (2.3) can be simplified as follows: The heat flow is essentially one-dimensional as the wall area is large compared with the wall thickness. In addition, the wall material does not generate heat. Finally, steady-state heat flow is typically used in load calculations. Therefore, in Eq. (2.3), vT vt ¼ 0; qgen ¼ 0; and the y and z directional terms are zero. Eq. (2.3) thus simplifies to d vT k ¼0 (2.4) dx vx Another usual simplification is that thermal conductivity k is constant. This results in the final equation d2 T ¼ 0 dx2

This equation will be solved in Chapter 3.

(2.5)

q

L

x

FIGURE 2.2 A plane wall.

2.2 Heat conduction equation

27

y

qy A

qx

x z

FIGURE 2.3 Heat flow in a thin plate.

Example 2.2 Heat Flow in a Thin Plate As another example, let us consider conduction in a thin plate situated in the x-y plane, as shown in Fig. 2.3.

Problem: Determine the heat conduction equation for a thin, flat plate.

Solution: Like Example 2.1, all three dimensions are not needed to properly describe the temperature distribution in a thin plate. There is temperature variation in the x and y directions, but, as the plate is thin, there is negligible temperature variation in the z direction. Only two dimensions are needed to properly describe the temperature distribution. At a point “A” in the plate, there are heat flow components qx and qy : However, qz is zero. If the heat transfer is steady, Eq. (2.3) simplifies to v vT v vT k þ k þ qgen ¼ 0 (2.6) vx vx vy vy If there is no internal heat generation in the plate, Eq. (2.6) reduces to v vT v vT k þ k ¼0 (2.7) vx vx vy vy And, if the conductivity is constant, there is a further reduction to v2 T v2 T þ 2 ¼0 vx2 vy

(2.8)

2.2.2 Cylindrical coordinates Cylindrical coordinates are used for problems involving cylinders. A point in a body in cylindrical coordinates is located by r, 4, and z, which are radius, azimuthal angle, and axial distance, respectively. These dimensions for a point “Aˮ are shown in Fig. 2.4.

28

Chapter 2 Heat conduction equation and boundary conditions

Z

A z

r

y

φ

x

FIGURE 2.4 Cylindrical coordinates.

The heat conduction equation in cylindrical coordinates can be obtained from the rectangular coordinate equation, Eq. (2.3), through coordinate transformation using the following relations between the parameters of the two coordinate systems. These relations are x ¼ r cos 4 y ¼ r sin 4 z¼z The heat conduction equation in cylindrical coordinates can also be obtained by an energy balance on a cylindrical element, similar to what was done for the rectangular element in Section 2.2.1. A cylindrical element is shown in Fig. 2.5. The heat conduction equation in cylindrical coordinates is 1 v vT 1 v vT v vT vT (2.9) kr þ 2 k þ k þ qgen ¼ rc r vr vr r v4 v4 vz vz vt With the application of appropriate boundary and initial conditions, this equation can be solved, at least in theory, for the temperature distribution in a cylinder at different instances of time.

Example 2.3 Heat Flow in a Wire Problem: Electrical current is flowing through the wire shown in Fig. 2.6 below. This flow causes uniform heat generation (i.e., Joule heating) due to the wire’s resistance. The heat flows and temperatures in the wire are steady, and the conductivity k is constant. As the wire is long, the axial heat flow is negligible.

2.2 Heat conduction equation

29

z

dz

z

y r

x

dφ

dr φ

FIGURE 2.5 A cylindrical element. Electrical Current

r

qr

k, qgen

FIGURE 2.6 Heat flow in a wire. (a) Starting with Eq. (2.3), the heat conduction equation in rectangular coordinates, determine the heat conduction equation for this situation in cylindrical coordinates using the coordinate transformation approach. (b) Determine the heat conduction equation using a heat-balance approach.

Solution: (a) Transforming the equation from rectangular to cylindrical coordinates. The heat conduction equation is v vT v vT v vT vT k þ k þ k þ qgen ¼ rc (2.3) vx vx vy vy vz vz vt As we have steady state, the right side of this equation is zero. As k is a constant, it can be factored out of the first three terms on the left side. Rearrangement of the equation gives qgen v2 T v2 T v2 T ¼0 (2.10) þ 2 þ 2 þ vx2 vy vz k The z direction is along the center axis of the wire. If the wire is long compared with its diameter, the temperature variation in the wire in the z direction will be negligible. Assuming this usual situation, the third term of Eq. (2.10) will be zero, and the equation becomes qgen v2 T v2 T ¼ 0 þ 2 þ 2 vx vy k

(2.11)

30

Chapter 2 Heat conduction equation and boundary conditions

All we have to do now is to convert the first two terms of Eq. (2.11) to cylindrical coordinates. From Fig. 2.4, pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ r ¼ x2 þ y2 Let us first change the first term of Eq. (2.11) to cylindrical coordinates. v2 T v vT ¼ 2 vx vx vx where r is a function of x and y, that is, r ¼ r (x,y). By the chain rule, vT vT vr ¼ vx vr vx Through partial differentiation of Eq. (2.12), it can be shown that

So, Eq. (2.14) becomes

(2.12)

(2.13)

(2.14)

vr x ¼ vx r

(2.15)

vT x vT ¼ vx r vr

(2.16)

From Eqs. (2.13) and (2.16), we have v2 T v x vT ¼ 2 vx vx r vr Again, from the chain rule, the right side of Eq. (2.17) is v x vT v x vT vr v x vT x ¼ ¼ vx r vr vr r vr vx vr r vr r Combining Eqs. (2.17) and (2.18) and performing the differentiation expressed in Eq. (2.18), we get v2 T x x v2 T vT v x ¼ þ 2 2 vx r r vr vr vr r 1 As x ¼ r2 y2 2 , it can be shown that the last term of Eq. (2.19) is

(2.17)

(2.18)

(2.19)

v x r2 x2 ¼ vr r xr2

(2.20)

v2 T x x v2 T vT r2 x2 ¼ þ 2 2 2 vx r r vr vr xr

(2.21)

and Eq. (2.19) becomes

As r 2 ¼ x2 þ y2 , Eq. (2.21) becomes

2 2 2 v2 T x v T vT y ¼ þ (2.22) 2 vx vr2 vr r2 r3 2 A transformation of the vvyT2 term can be done similarly, resulting in 2 2 2 v2 T y v T vT x ¼ þ (2.23) vy2 vr2 vr r2 r3 Adding Eqs. (2.22) and (2.23), we get v2 T v2 T v2 T 1 vT 1 v vT ¼ r (2.24) þ ¼ þ vx2 vy2 vr2 r vr r vr vr Therefore the cylindrical version of Eq. (2.11) for this problem is qgen 1 v vT r þ ¼0 (2.25) r vr vr k (b) Using the heat-balance approach to obtain the heat conduction equation. For this problem, the heat flow is solely in the radial r direction. As shown in Fig. 2.7, the infinitesimal volume element is a ring of wall thickness dr and height dz.

2.2 Heat conduction equation

31

dr r r+dr

dz qr qr+dr

FIGURE 2.7 Infinitesimal volume element for Example 2.3. Heat flows by conduction radially into the volume at r and radially out of the volume at r þ dr. There is also a heat input to the volume by heat generation (i.e., Joule heating). As the problem is steady state, the heat flow rate into the volume equals the heat flow rate out of the volume, that is, the heat balance on the volume element is Heat Flow Rate In By Conduction þ Heat Flow Rate In By Generation ¼ Heat Flow Rate Out By Conduction (2.26) The incoming heat flow rate by conduction is qr and the outgoing heat flow rate by conduction is qrþdr. The area for heat flow at location r is dA ¼ 2 p r dz and the volume of the element is dV ¼ 2pr dr dz. The heat generation rate is qgen per unit volume. Eq. (2.26) thus becomes However, as qrþdr ¼ qr þ

vqr vr

qr þ qgen dV ¼ qrþdr dr, Eq. (2.27) becomes

vqr dr þ qgen dV ¼ 0 dr From Eq. (1.2), conduction heat flow in the r direction is qr ¼ kdA vT vr . Inserting this into Eq. (2.28), we get v vT kdA dr þ qgen dV ¼ 0 vr vr Using the relations for dA and dV, Eq. (2.29) becomes v vT kð2prdzÞ dr þ qgen ð2prdrdzÞ ¼ 0 vr vr As conductivity k is constant, Eq. (2.30) becomes qgen v vT r þ r¼0 vr vr k Dividing Eq. (2.31) by r, we get qgen 1 v vT r þ ¼0 r vr vr k This is identical to Eq. (2.25), which was obtained by coordinate transformation in part (a) of the solution.

(2.27)

(2.28)

(2.29)

(2.30)

(2.31)

(2.32)

32

Chapter 2 Heat conduction equation and boundary conditions

Note that the final heat conduction equation for this problem, Eq. (2.25) or (2.32), could have been obtained directly from the cylindrical heat conduction equation, Eq. (2.9), by eliminating 4 and z variations and assuming steady state and constant conductivity k.

2.2.2.1 Special casesdcylindrical coordinates Depending on the particular problem to be solved, the general Eq. (2.9), repeated here, can often be simplified. 1 v vT 1 v vT v vT vT (2.9) kr þ 2 k þ k þ qgen ¼ rc r vr vr r v4 v4 vz vz vt For example, if the problem is steady state, then the vT vt term is zero. If there is no heat generation by the material, then qgen is zero. And, if the conductivity k is constant, it can be taken outside the parentheses and then each term of the equation can simply be divided by k. Finally, in some problems, the temperature distribution may not depend on all three dimensions. Those terms relating to the nondependent dimensions can then be deleted from Eq. (2.9). For example, in Example 2.3 above, the 4 and z terms could be deleted, the conductivity k was constant, and the problem was steady state.

2.2.3 Spherical coordinates Spherical coordinates are used for problems involving spheres. A point “A” in a body in spherical coordinates is located by radius r, azimuthal angle 4, and polar angle q as shown in Fig. 2.8.

z

A (r,φ ,θ )

θ

r z y φ

x

FIGURE 2.8 Spherical coordinates.

2.2 Heat conduction equation

33

The heat conduction equation in spherical coordinates can be obtained from the rectangular coordinate equation, Eq. (2.3), through coordinate transformation using the following relations between the parameters of the two coordinate systems: x ¼ r cos 4 sin q y ¼ r sin 4 sin q z ¼ r cos q The heat conduction equation can also be obtained by an energy balance on a spherical element, which is shown in Fig. 2.9. The heat conduction equation in spherical coordinates is 1 v 1 v vT 1 v vT vT 2 vT (2.33) kr þ k þ k sin q þ qgen ¼ rc 2 2 2 2 r vr vr v4 r sin q vq vq vt r sin q v4 With the application of appropriate boundary and initial conditions, this equation can be solved, at least in theory, for the temperature distribution in a sphere at different instances of time.

2.2.3.1 Special casesdspherical coordinates The general heat conduction equation, Eq. (2.33), can often be simplified to fit a specific problem. For example, if the problem is steady state, then the vT vt term is zero. If there is no internal heat generation by the material, then qgen is zero. If there is no angular dependence, the terms involving 4 and q can be deleted. Finally, if conductivity k is constant, it can be taken outside the parentheses and then each term of the equation can simply be divided by k. z

dr

θ

r

dθ y φ

x

FIGURE 2.9 Spherical element.

dφ

34

Chapter 2 Heat conduction equation and boundary conditions

Example 2.4 Radioactive Material in a Spherical Cask Problem: A spherical cask is filled with radioactive material. The material is initially at temperature T1 and is generating heat uniformly at a constant rate of qgen per unit volume. The conductivity k of the radioactive material is uniform and constant. The cask is thin-walled and has a radius of r1 . The outer surface of the cask is convecting heat to the surrounding air with a convective coefficient h. The air is at temperature TN . Radiative heat transfer from the cask is negligible. What is the temperature of the material at a distance of x meters from the center of the cask y minutes after the cask is filled? Note: Only determine the appropriate heat conduction equation for the problem. Do not attempt to solve the problem.

Solution: To solve this problem, the first task is to determine the appropriate heat conduction equation. After this is done, boundary and initial conditions can be applied, and the heat conduction equation can be solved for the temperature distribution in the material at various instances of time. In this example, we will just determine the heat conduction equation. Solution of the equation is discussed in later sections of this book. The container is spherical, so the appropriate general heat conduction equation is Eq. (2.33): 1 v vT 1 v vT 1 v vT vT kr2 þ 2 2 k þ 2 k sin q þ qgen ¼ rc (2.33) 2 r vr vr v4 r sin q vq vq vt r sin q v4 Let us tailor this equation for our particular problem, as follows: There is no mention of any angular dependence in the problem description. Hence, the second and third terms of Eq. (2.33) may be deleted, giving 1 v vT 2 vT kr þ qgen ¼ rc (2.34) r2 vr vr vt There is indeed heat generation by the radioactive material and the problem deals with time-dependence. However, the conductivity is constant, so it may be taken outside the parentheses in the first term. All of the terms in the equation can then be divided by k. k. Also, the thermal diffusivity is defined as a ¼ rc Eq. (2.34) thus becomes qgen 1 vT 1 v vT r2 þ (2.35) ¼ 2 r vr vr a vt k which is the heat conduction equation for this particular problem.

2.3 Boundary conditions The heat conduction equations of Section 2.2 apply to the interior parts of the body. In this section, we discuss the conditions at the boundaries of the body. The combination of the heat conduction equation and the boundary condition equations for a body may be solved to determine the temperature distribution in the body. (Note: If the problem is unsteady state, initial conditions are also needed for the solution. These conditions are discussed in Section 2.4 below.) Once the temperature distribution is obtained, it can be differentiated using the equations of Chapter 1 to determine the heat flows at different locations in the body. For example, if the temperature distribution T(x,y,z) is known at a particular instant of time, then, from Eq. (1.3), the heat flow rate per unit area in the x direction at the different points in the body for the particular instant of time is

2.3 Boundary conditions

35

q

vT (1.3) ¼ k A x vx In developing the boundary condition equations, two items of information must be provided. These are (a) the location of the boundary using a particular coordinate system and (b) the condition at the boundary (e.g., specified temperature, heat flux rate, convection, radiation, etc.).

2.3.1 Rectangular coordinates Although most of our discussion deals with boundary condition equations for the rectangular coordinate system, similar equations can easily be developed for cylindrical and spherical geometries. These other geometries are discussed in Section 2.3.2. Let us consider the plane wall of Fig. 2.2 shown earlier and repeated here. The wall has a thickness L in the x direction and large extents in the y and z directions.

q

L

x

FIGURE 2.2 A plane wall.

We will now proceed to obtain boundary equations for several different conditions at the x ¼ 0 and x ¼ L boundaries of the wall.

2.3.1.1 Specified temperature As shown in Fig. 2.10, the left side of the wall is at temperature T1 and the right side is at temperature T2 . The boundary condition equations are At x ¼ 0; T ¼ T1

(2.36a)

At x ¼ L; T ¼ T2

(2.36b)

and

36

Chapter 2 Heat conduction equation and boundary conditions

y

T2

T1 L

x

O

z

FIGURE 2.10 Specified temperature at boundary.

2.3.1.2 Specified heat flux The heat flux

q A

x

at the left side of the wall of Fig. 2.11 has a magnitude of C1 and it is in the þx

direction. Then the boundary condition equation is q At x ¼ 0; ¼ C1 (2.37) A x However, the heat conduction equation deals with the temperature, so this boundary equation should be put in terms of temperature. From Eq. (1.3), q vT (1.3) ¼ k A x vx So, the boundary condition Eq. (2.37) becomes At x ¼ 0; k

vT ¼ C1 vx

(2.38)

y qx = C2 A

qx = C1 A

L O

z

FIGURE 2.11 Specified heat flux at boundary.

x

2.3 Boundary conditions

37

The heat flux at the right side of the wall has a magnitude of C2 and is in the x direction. The appropriate boundary equation is At x ¼ L; k

vT ¼ C2 vx

(2.39)

2.3.1.3 Insulated boundary As shown in Fig. 2.12, the right side of the wall is perfectly insulated and qx ¼ 0. As qx ¼ kA vT vx and both k and A are nonzero, the boundary equation is At x ¼ L;

vT ¼0 vx

(2.40)

2.3.1.4 Convection As shown in Fig. 2.13, let us say there is a fluid on the left side of the wall. The fluid is at temperature TN , and the convective coefficient between the fluid and the wall surface is h. The boundary equation can be developed using the concept of heat flow continuity at the boundary. That is, at the boundary, the convective heat flow is equal to the conductive heat flow, or qconv ¼ qcond. By Eq. (1.3), the conductive heat flow is qcond ¼ kA

vT vx

By Equation (1.9), the convective heat flow is qconv ¼ h A ðT TN Þ However, in our situation the heat flow is from the fluid to the surface. As discussed in Chapter 1, the convective flow at the left surface should be written qconv ¼ h A ðTN TÞ y qx = 0

L O

z

FIGURE 2.12 Insulated boundary.

x

38

Chapter 2 Heat conduction equation and boundary conditions

y A Fluid at T∞

k h qconv

qcond L x

O

z

FIGURE 2.13 Convection at boundary.

Hence, the boundary equation is, after canceling A from both sides of the equation, vT vx

At x ¼ 0; hðTN TÞ ¼ k

(2.41)

2.3.1.5 Radiation Let us say that the wall of Fig. 2.14 is in a vacuum, so there is no convective heat transfer at the wall surfaces. But, let us have radiation heat transfer at both surfaces. The surroundings to the left of the wall are at Tsurr1 and the left surface of the wall has an emissivity ε1. The surroundings to the right of the wall are at Tsurr2 , and the right surface of the wall has an emissivity ε2. Radiation heat transfer is y A Surroundings at Tsurr1

qcond2

ε1 qrad1

qcond1

FIGURE 2.14 Radiation at boundary.

2

ε2

L O

z

qrad2

Surroundings at Tsurr

x

2.3 Boundary conditions

39

discussed in detail in Chapter 9. Until then, we will use Eq. (1.13) for problems involving radiation. For this section, that equation is 4 qrad ¼ εsA Ts4 Tsurr (2.42) where Ts is the temperature of the surface, and Tsurr is the temperature of the surroundings. Note that both temperatures must be in absolute temperature units, i.e., Kelvin or Rankine. Regarding the boundaries at x ¼ 0 and x ¼ L: At the wall surfaces, there is heat flow continuity. That is, the heat flow by conduction equals the heat flow by radiation. Let us assume that all the conduction and radiation heat flows are in the þx direction as shown in Fig. 2.14. The boundary equations at the two surfaces are then At x ¼ 0; qrad1 ¼ qcond1 At x ¼ L; qcond2 ¼ qrad2 This gives, after canceling the area A on both sides of the equations, vT 4 4 (2.43) T At x ¼ 0; ε1 s Tsurr ¼ k 1 vx vT 4 At x ¼ L; k ¼ ε2 s T 4 Tsurr (2.44) 2 vx Like convection, we must be careful in writing the radiation terms. Looking at Fig. 2.14: At x ¼ 0, the assumed radiation flow is from the surroundings to the wall, so the radiation term is

4 4 T 4 and not ε1 s T 4 Tsurr ε1 s Tsurr . At x ¼ L, the assumed radiation flow is from the wall to 1 1 4 the surroundings, so the radiation term is ε2 s T 4 Tsurr . 2

2.3.1.6 Convection and radiation If there is both convection and radiation at a boundary, then the two modes of heat transfer can be added to get the total heat transfer. In Fig. 2.15, there are fluids on both sides of the wall and therefore y A Fluid at T∞1 Surroundings at Tsurr1

qcond2 (qconv+qrad)

2

ε 1,h1

L O

FIGURE 2.15 Convection and radiation at boundary.

ε 2,h2

qcond1

(qconv+qrad)1

z

Fluid at T∞2 Surroundings at Tsurr2

x

40

Chapter 2 Heat conduction equation and boundary conditions

convective heat transfer on both surfaces. There is also radiative heat transfer between both surfaces and their respective surroundings. At the left surface, the conduction heat transfer is equal to the sum of the convective and radiative heat transfers. The same continuity of heat flows concept applies to the right surface. The boundary equations are vT 4 4 (2.45) At x ¼ 0; h1 AðTN1 TÞ þ ε1 sA Tsurr T ¼ kA 1 vx vT 4 At x ¼ L; kA ¼ h2 AðT TN2 Þ þ ε2 sA T 4 Tsurr (2.46) 2 vx The area A can be deleted from the two equations as it is in all the terms. Again, in writing the boundary equations, pay particular attention to the assumed directions of the heat flows so that the terms in the equations will have the correct signs. Let us continue to discuss this boundary condition where convection occurs together with radiation. Ts is the temperature of the surface, TN is the fluid temperature, and Tsurr is the temperature of the surroundings. The total heat flow at the boundary is the sum of the convective heat flow qconv and the radiative heat flow qrad, where qconv ¼ hA ðTs TN Þ 4 and qrad ¼ εsA Ts4 Tsurr

(2.47) (2.48)

It would be nice to convert Eq. (2.48) into an equation that has a difference in temperature rather than a difference in temperature to the fourth power. This can be done by introducing a new coefficient for radiation, hrad , which is defined by the equation 4 ¼ hrad AðTs Tsurr Þ qrad ¼ εsA Ts4 Tsurr (2.49) That is,

4 εs Ts4 Tsurr hrad ¼ Ts Tsurr

Through algebraic manipulation it can be shown that 2 hrad ¼ εs Ts2 þ Tsurr ðTs þ Tsurr Þ

(2.50)

(2.51)

Having definedhrad , we can now write the total heat flow at the surface as qtotal ¼ qconv þ qrad ¼ hAðTs TN Þ þ hrad AðTs Tsurr Þ

(2.52)

In many practical situations, the fluid temperature is about the same as the temperature of the surroundings. If this is the case, we can add convective coefficient h and radiative coefficient hrad to get a total coefficient htotal ¼ h þ hrad , and the total heat transfer at the surface becomes qtotal ¼ htotal AðTs Tavg Þ where Tavg is the average of TN and Tsurr :

(2.53)

2.3 Boundary conditions

41

2.3.1.7 Symmetry conditions There may be situations where the temperature distribution is symmetrical. For example, consider a large plate of thickness 2L, which has uniform internal heat generation. At the two surfaces, there is convection to the adjacent fluids. If the convective coefficients at both surfaces are the same and the temperatures of the adjacent fluids are the same, then the temperature distribution in the plate will be symmetrical about the center plane. And, the temperature of the center plane will be the maximum temperature in the plate. If the center plane is at x ¼ 0 and the surfaces are at x ¼ L and x ¼ þL, then the symmetry boundary condition is At x ¼ 0;

vT ¼0 vx

(2.54)

2.3.1.8 Interfacial boundary Figure 2.16 shows two walls in contact. Wall 1 has a conductivity k1 and a temperature distribution T1 . Wall 2 has a conductivity k2 and a temperature distribution T2 . Both the conductivities and the temperatures may vary with x. At the interface between the two walls, i.e., at x ¼ L1 , there is continuity of temperature if there is no contact resistance between the walls. We will normally assume this condition of zero contact resistance. (Note: Contact resistance is discussed in Section 12.3.) From the temperature continuity, we have the boundary equation At x ¼ L1 ;

T 1 ¼ T2

(2.55)

There is also continuity of heat flows at the interface, i.e., qcond1 ¼ qcond2 . This gives the boundary equation At x ¼ L1 ;

k1 A

vT1 vT2 ¼ k2 A vx vx

y A k1 T1=T2 k2 Wall 2 Wall 1

qcond1 qcond2 O

z

FIGURE 2.16 Walls in contact.

L1

L2

x

(2.56)

42

Chapter 2 Heat conduction equation and boundary conditions

Canceling the area A and the minus signs, the boundary equation for the heat flows becomes At x ¼ L1 ;

k1

vT1 vT2 ¼ k2 vx vx

(2.57)

2.3.2 Cylindrical and spherical coordinates Boundary conditions were developed in Section 2.3.1 for the rectangular coordinate system. Similar equations can be developed for cylindrical and spherical coordinates. For cylindrical objects, the common boundaries are at radial and axial locations. There are also boundaries at angular locations if the objects are asymmetrical. If the object is a hollow cylinder, there are boundaries at the inner and outer surfaces and at the two ends. If it is a solid cylinder, there is only one radial boundary and two axial boundaries. A hollow sphere has boundaries at its inner and outer surfaces. A solid sphere has only one bounding surface. Possible boundary conditions are similar to those for rectangular coordinates, e.g., specified temperature, specified heat flux, convection, and radiation. If a specific temperature or heat flux is not given for a particular boundary, the boundary equation can be obtained by consideration of heat flow continuity at the boundary. For example, we could have, at a boundary, that conduction equals convection, conduction equals radiation, or conduction equals convection plus radiation.

Example 2.5 Boundary Conditions for a Hollow Cylinder Consider Fig. 2.17 that shows a hollow cylinder of inner radius r1 , outer radius r2 , and length L. The boundaries of this cylinder are the inner surface, the outer surface, and the two ends. Depending on the specific problem, many different boundary equations can be developed. The two coordinates of interest are the radial distance r and the axial distance z. We will set z ¼ 0 at the left end of the cylinder. Then the right end of the cylinder is at z ¼ L. Let us say that there is a fluid inside the cylinder at temperature TN1 with a convective coefficient at the inner surface of h1 and a fluid outside the cylinder at temperature TN2 with a convective coefficient at the outer surface of h2 . Let us also say that the two ends of the cylinder are perfectly insulated.

L

r1 r

r2 z

FIGURE 2.17 Finite length hollow cylinder.

2.3 Boundary conditions

43

The conduction heat transfer in the cylinder in the radial direction r is vT (2.58) vr Area A, the area through which heat is flowing, increases linearly with r and is A ¼ 2prL. Thus, from Eq. (2.58), the conduction heat transfer in the þr direction at a distance r from the centerline of the cylinder is qcond ¼ kA

vT (2.59) vr At the inner surface, i.e., at r ¼ r1 , the convective heat transfer equals the conduction heat transfer. Hence, the boundary equation at the inner surface is qcond ¼ kð2prLÞ

kð2pr1 LÞ

vT ¼ h1 ð2pr1 LÞðTN1 TÞ vr

(2.60)

vT ¼ h1 ðTN1 TÞ (2.61) vr For the outer surface of the cylinder, i.e., at r ¼ r2 , the conduction equals the convection. So, the boundary equation, after canceling area A on both sides of the equation, is which simplifies to At r ¼ r1 ;

k

vT ¼ h2 ðT TN2 Þ (2.62) vr Note: Again, be careful with the convection terms in Eqs. (2.61) and (2.62). In writing the equations, we have assumed that both the conduction and the convection heat flows are in the þr direction. Hence, for the convection term at the inner surface, we have ðTN TÞ, and for the convection term at the outer surface, we have ðT TN Þ. This was discussed above in this chapter and in Chapter 1. The two ends of the cylinder are perfectly insulated with no heat flow. Hence, the boundary equations for the ends are At r ¼ r2 ;

k

At z ¼ 0;

vT ¼0 vz

(2.63)

At z ¼ L;

vT ¼0 vz

(2.64)

2.3.2.1 Symmetry conditions If the cylinder or sphere under consideration is solid, then there is only one physical boundary in the radial direction, i.e., at the surface of the cylinder or sphere. However, more than one boundary equation is needed for solution of the heat conduction equation. In many cases, the temperature distribution is symmetrical at the center of the cylinder or sphere. For example, when electricity flows through a wire, there is heating and a symmetrical temperature distribution at the center of the wire. Cooling of a cylinder or sphere can also result in temperature symmetry at the center. For symmetrical heating or cooling, there is a temperature maximum or minimum at the center, and the boundary equation is At r ¼ 0;

vT ¼0 vr

(2.65)

Example 2.6 Cooling of a Heated Sphere Let us consider the heated sphere of radius r1 shown in Fig. 2.18. The sphere has just come out of an oven. The temperature of the sphere is initially uniform throughout the sphere, and it is higher than the room air and the surroundings. The thermal conductivity of the sphere is k. The surface of the sphere has an emissivity ε and a convective coefficient h. The air is at TN and the surroundings are at Tsurr . The temperature distribution in the sphere is T.

44

Chapter 2 Heat conduction equation and boundary conditions

There is heat flow continuity at the surface of the sphere. The heat flow by conduction into the surface equals the heat flow by convection and radiation from the surface. That is;

qcond ¼ qconv þ qrad

vT where qcond ¼ kA vr

(2.66) (2.67)

qconv ¼ hAðT TN Þ 4 qrad ¼ εsA T 4 Tsurr Hence, the boundary equation at the surface of the sphere is

(2.68) (2.69)

vT 4 ¼ hðT TN Þ þ εs T 4 Tsurr (2.70) vr We will assume that k is uniform throughout the sphere. Then the temperature distribution during cooling will be symmetrical with respect to r. Even though r ¼ 0 is not a physical boundary, we can write the following equation for the symmetry: At r ¼ r1 ; k

At r ¼ 0;

vT ¼0 vr

(2.71)

In closing this discussion of boundary equations, let us consider a situation where we have a boundary condition involving an angular coordinate of a cylinder or sphere. In particular, let us look at the segment of a cylinder shown in Fig. 2.19. The segment has inner radius r1 , outer radius r2 , axial length L, and includes the angle from 4 ¼ 0 to 4 ¼ 41 . The 4 ¼ 0 surface is perfectly insulated and the 4 ¼ 41 surface has convection to a fluid at TN with convection coefficient h. For the insulated 4 ¼ 0 surface, the heat flow is zero, so the boundary equation is At 4 ¼ 0;

vT ¼0 v4

(2.72)

For the 4 ¼ 41 surface, the conduction in the þ4 direction equals the convection in that direction. The boundary equation is At 4 ¼ 41 ; kA

vT ¼ hAðT TN Þ v4

(2.73)

The areas cancel as they are the same. Thus, the final boundary equation is At 4 ¼ 41 ; k

vT ¼ hðT TN Þ v4

Fluid at T∞ Surroundings at Tsurr r1 qcond

FIGURE 2.18 Sphere cooled by convection and radiation.

qconv + qrad

(2.74)

2.5 Chapter summary and final remarks

45

Insulated

Convection h, T∞

φ1 φ r1 r2

FIGURE 2.19 Segment of a cylinder.

Eqs. (2.72) and (2.74) are the boundary equations for the 4 coordinate. To solve the heat conduction equation for the temperature distribution in the segment, boundary equations will also be needed for the radial and axial bounding surfaces.

2.4 Initial conditions Eqs. (2.3), (2.9), and (2.33) are the general heat conduction equations for rectangular, cylindrical, and spherical coordinates. If the problem is time-dependent, the vT vt term in the heat conduction equation is nonzero and initial condition(s) are needed for solution of the equation. Two examples of initial conditions are as follows: (1) If all points in the body are at the same temperature Tinit at time zero, then the initial condition equation is At t ¼ 0; Tðx; y; zÞ ¼ Tinit

(2.75)

(2) If the temperature distribution in the body is a known function Tinit ðx; y; zÞ at time zero, then the initial condition equation is At t ¼ 0; Tðx; y; zÞ ¼ Tinit ðx; y; zÞ

(2.76)

2.5 Chapter summary and final remarks In this chapter, we discussed the partial differential equation governing heat conduction within a body. The heat conduction equation was presented for rectangular, cylindrical, and spherical coordinate systems. We also discussed boundary and initial condition equations that are used in conjunction with the heat conduction equation to determine the temperature distribution in the body. Differentiation of the temperature distribution will give the heat flows at the various points in the body. Eq. (1.3) showed this for rectangular coordinates. Chapter 3 will give similar equations for cylindrical and spherical coordinates. In the next chapter we will solve some of the differential equations developed in this chapter for steady-state heat transfer.

46

Chapter 2 Heat conduction equation and boundary conditions

2.6 Problems 2-1 In the derivation of the rectangular heat conduction equation, Eq. (2.3), it was assumed that three flows were incoming to the differential element q ; q ; qz and three were outgoing x y qxþdx ; qyþdy ; qzþdz . Redo the derivation assuming that all six flows are incoming to the element. You should obtain the same heat conduction equation. 2-2 Obtain the cylindrical heat conduction equation, Eq. (2.9), through coordinate transformation of the rectangular heat conduction equation, Eq. (2.3). 2-3 Obtain the cylindrical heat conduction equation, Eq. (2.9), using the heat-balance approach in conjunction with the cylindrical element of Fig. 2.5. For problems 2-4 to 2-8, reduce the rectangular heat conduction equation as much as possible to fit the problem specification. Just give the equation, do not solve it. The general heat conduction equation is v vT v vT v vT vT k þ k þ k þ qgen ¼ rc vx vx vy vy vz vz vt 2-4 A rectangular solid is of dimensions 2 cm by 4 cm by 6 cm and is initially at a uniform temperature of 100 C. Its surfaces are suddenly subjected to a fluid at 50 C with a convective coefficient of 2 W/m2 C. What is the temperature of the center of the solid after 30 s? The conductivity of the solid is 150 W/m C. Its density is 2700 kg/m3 and its specific heat is 880 J/kg C. There is no internal heat generation. 2-5 A long copper bar has a square cross section in the x-y plane. There is negligible heat flow in the z direction. The top and bottom of the bar are held at temperature T1 , while the other two surfaces are held at T2 . Heat transfer in the bar is steady. The material properties, i.e., conductivity, density, and specific heat are uniform throughout the bar and constant. There is no internal heat generation. What is the temperature distribution of the x-z plane that is halfway between the top and bottom surfaces of the bar? (Fig. P2.5).

y

T1

T2

T1 T2

x a

z

FIGURE P2.5

2.6 Problems

47

2-6 A thin rectangular plate is in the x-y plane. There is negligible heat transfer in the z direction. The conductivity, density, and specific heat all vary with temperature. There is constant internal heat generation and the heat flow is unsteady. 2-7 A cube is perfectly insulated from the surroundings. The cube is initially at a uniform temperature and then, at time zero, internal heat generation begins. What is the temperature at a corner of the cube at time t after the heat generation starts? It may be assumed that all material properties are constant. 2-8 There is steady heat conduction in a rectangular solid. The conductivity of the object is constant and there is no heat generation. What is the temperature at the center of one face of the object? 2-9 Consider the equation qgen v2 T 1 vT ¼ þ vx2 a vt k (a) How many dimensions are there (one, two, or three)? (b) Is the heat conduction steady or unsteady? (c) Is there internal heat generation? (d) Is the conductivity constant or variable? 2-10 Consider the equation v vT v vT k þ k ¼0 vy vy vz vz (a) How many dimensions are there (one, two, or three)? (b) Is the heat conduction steady or unsteady? (c) Is there internal heat generation? (d) Is the conductivity constant or variable? 2-11 Consider the equation v2 T v2 T v2 T þ þ 2 ¼0 vx2 vy2 vz (a) How many dimensions are there (one, two, or three)? (b) Is the heat conduction steady or unsteady? (c) Is there internal heat generation? (d) Is the conductivity constant or variable? 2-12 Consider the equation v vT v vT vT k þ k ¼ rc vx vx vy vy vt (a) (b) (c) (d)

How many dimensions are there (one, two, or three)? Is the heat conduction steady or unsteady? Is there internal heat generation? Is the conductivity constant or variable?

48

Chapter 2 Heat conduction equation and boundary conditions

2-13 Consider the equation qgen v2 T v2 T ¼0 þ þ vx2 vy2 k (a) How many dimensions are there (one, two, or three)? (b) Is the heat conduction steady or unsteady? (c) Is there internal heat generation? (d) Is the conductivity constant or variable? For problems 2-14 to 2-17, reduce the cylindrical heat conduction equation as much as possible to fit the problem specification. Just give the equation; don’t solve it. The general heat conduction equation is 1 v vT 1 v vT v vT vT kr þ 2 k þ k þ qgen ¼ rc r vr vr r v4 v4 vz vz vt 2-14 The problem is steady state with no internal heat generation and no angular dependence. The conductivity is constant. 2-15 The problem is transient, with constant and uniform internal heat generation. The heat transfer in the axial direction is negligible. The conductivity varies with temperature. 2-16 The problem is transient, with no internal heat generation. The temperature varies in the radial and axial directions, but there is no angular dependence. The conductivity is constant, but the density and specific heat vary with temperature. 2-17 The problem is steady state with internal heat generation. The temperature variations in the angular and axial directions are negligible. The thermal conductivity is constant. 2-18 Consider the equation in cylindrical coordinates v vT kr ¼ 0 vr vr (a) How many dimensions are there (one, two, or three)? (b) Is the heat conduction steady or unsteady? (c) Is there internal heat generation? (d) Is the conductivity constant or variable? 2-19 Consider the equation in cylindrical coordinates 1 v vT v vT kr þ k þ qgen ¼ 0 r vr vr vz vz (a) (b) (c) (d)

How many dimensions are there (one, two, or three)? Is the heat conduction steady or unsteady? Is there internal heat generation? Is the conductivity constant or variable?

2.6 Problems

49

2-20 Consider the equation in cylindrical coordinates v2 T 1 vT 1 v2 T 1 vT þ þ ¼ vr 2 r vr r 2 v42 a vt (a) How many dimensions are there (one, two, or three)? (b) Is the heat conduction steady or unsteady? (c) Is there internal heat generation? (d) Is the conductivity constant or variable? 2-21 Consider the equation in cylindrical coordinates v2 T 1 vT qgen 1 vT þ ¼ þ vr 2 r vr a vt k (a) How many dimensions are there (one, two, or three)? (b) Is the heat conduction steady or unsteady? (c) Is there internal heat generation? (d) Is the conductivity constant or variable? 2-22 Consider the equation in cylindrical coordinates 1 v vT 1 v vT vT kr þ 2 k ¼ rc r vr vr r v4 v4 vt (a) How many dimensions are there (one, two, or three)? (b) Is the heat conduction steady or unsteady? (c) Is there internal heat generation? (d) Is the conductivity constant or variable? For problems 2-23 to 2-25, reduce the spherical heat conduction equation as much as possible to fit the problem specification. Just give the equation; don’t solve it. The general heat conduction equation is 1 v 1 v vT 1 v vT vT 2 vT kr þ 2 2 k þ 2 k sin q þ qgen ¼ rc r 2 vr vr v4 v4 r vq vq vt sin q r sin q 2-23 The problem is steady state with internal heat generation and constant conductivity. 2-24 The problem is transient, with no internal heat generation and no angular dependence. The conductivity varies with location in the body. 2-25 The problem is steady state with variable conductivity, no internal heat generation, and no angular dependence. 2-26 Consider the equation in spherical coordinates d 2 dT r ¼ 0 dr dr (a) (b) (c) (d)

How many dimensions are there (one, two, or three)? Is the heat conduction steady or unsteady? Is there internal heat generation? Is the conductivity constant or variable?

50

Chapter 2 Heat conduction equation and boundary conditions

2-27 Consider the equation in spherical coordinates 1 v 1 v vT 1 v vT 2 vT k r þ k þ k sin q ¼0 r 2 vr vr v4 r 2 sin q vq vq r 2 sin2 q v4 (a) How many dimensions are there (one, two, or three)? (b) Is the heat conduction steady or unsteady? (c) Is there internal heat generation? (d) Is the conductivity constant or variable? 2-28 Consider the equation in spherical coordinates 1 v 1 v2 T 1 v vT 1 vT 2 vT þ 2 r þ 2 2 sin q ¼ r 2 vr vr r sin q vq vq a vt r sin q v42 (a) How many dimensions are there (one, two, or three)? (b) Is the heat conduction steady or unsteady? (c) Is there internal heat generation? (d) Is the conductivity constant or variable? 2-29 Consider the equation in spherical coordinates v2 T 2 vT qgen 1 vT þ ¼ þ vr 2 r vr a vt k (a) How many dimensions are there (one, two, or three)? (b) Is the heat conduction steady or unsteady? (c) Is there internal heat generation? (d) Is the conductivity constant or variable? 2-30 The thin plate shown in Fig. P2.30 is in the x-y plane. The left and right sides of the plate are perfectly insulated. The top surface of the plate is at 100 C and the bottom is at 20 C. Give the boundary equations for the plate (Fig. P2.30). y 100 C

Insulated 3 cm Insulated

20 C 2 cm

FIGURE P2.30

x

2.6 Problems

51

2-31 For the plate of Problem 2-30, let us say that instead of the left and right sides being perfectly insulated, they both convect to a fluid at TN ¼ 15 C with a convective coefficient of 3 W/m2 C. The conductivity of the plate is 80 W/m C. Give the boundary equations for the left and right sides of the plate. 2-32 The rectangular solid shown in Fig. P2.32 has dimensions 1 cm by 2 cm by 5 cm. The front and back surfaces are perfectly insulated. The left and right surfaces are at a temperature of 50 C, and the top and bottom surfaces have an emissivity of 0.3 and radiate to the surroundings, which are at 10 C. The object is in a vacuum, so there is no convection to a surrounding fluid. The conductivity of the object is 100 W/m C. Give the boundary equations for the six surfaces of the object (Fig. P2.32).

y

1 cm

2 cm

m

5c

x z

FIGURE P2.32

2-33 The left surface of the copper plate (k ¼ 400 W/m C) shown in Fig. P2.33 absorbs a heat flux from the sun of 200 W/m2. The surface also convects to the adjacent air which is at 25 C. The convective coefficient is 5 W/m2 C. Give the boundary equation for this surface of the plate (Fig. P2.33).

Air at 25 C W h=5 m2 C ⎞q ⎞ W = 200 2 ⎠A ⎠ m solar

qconv x

FIGURE P2.33

52

Chapter 2 Heat conduction equation and boundary conditions

2-34 The cube shown in Fig. P2.34 has uniform internal heat generation. All six surfaces of the cube convect to the surrounding fluid that is at TN ¼ 30 C. The convective coefficient for all the surfaces is h ¼ 5 W/m2 C and the conductivity of the cube is 40 W/m C. Give the boundary equations for all six surfaces of the cube (Fig. P2.34). 4 cm

z

4 cm

4 cm x y

FIGURE P2.34

2-35 For the cube of Problem 2-34: Let us say that the six surfaces of the cube are perfectly insulated rather that being convectors to a surrounding fluid. Give the boundary equations for all six surfaces for this situation. 2-36 The solid cylinder (k ¼ 15 W/m C) shown in Fig. P2.36 has a radius ro of 6 cm and is 20 cm long. Let us make z ¼ 0 at the left end of the cylinder and z ¼ 20 cm at the right end. The left end of the cylinder is at 80 C and the right end is at 50 C. The cylindrical surface convects to the surrounding fluid, which is at 30 C with a convective coefficient of 6 W/m2 C. Give the boundary equations for this problem (Fig. P2.36).

r ro = 6 cm 20cm z

FIGURE P2.36

2-37 A long steel pipe of conductivity 60 W/m C has an inner radius of 0.5 cm and an outer radius of 1.2 cm. Water at 90 C is flowing through the pipe, and there is convection from the hot water to the inner surface of the pipe with a convective coefficient of 50 W/m2 C. At the outer surface of the pipe, there is convection to the 20 C room air with a convective coefficient of 2 W/m2 C and radiation to the surrounding walls, which are at 15 C. The emissivity of the outer surface of the pipe is 0.2. Give the boundary equations for the inner and outer surfaces of the pipe. 2-38 An electrical current is flowing through a #12 copper wire. The wire has a diameter of 2.053 mm and a conductivity of 400 W/m C. The heat generated in the wire passes to the

2.6 Problems

53

surrounding 20 C air with a convective coefficient of 1.5 W/m2 C. Give two boundary equations for this problem. 2-39 It is desired to determine the temperature distribution in a sphere of radioactive material buried in the ground. The radioactive material is in a thin-walled container that has negligible thermal resistance. The radius of the sphere is ro . The conductivity of the radioactive material is k1 and the conductivity of the soil is k2 . The temperature distribution in the radioactive material is T1 and in the soil it is T2 : Give the boundary equations at the outer surface of the radioactive material. 2-40 It is desired to determine the temperature distribution for radioactive material (k ¼ 0.23 W/m C) inside a cylindrical container. The thin-walled container is shown in Fig. P2.40. It is on a concrete floor, and it may be assumed that the bottom of the container is perfectly insulated. The top of the container and the curved surface of the container transfer heat to the surrounding air and walls. The air temperature is 20 C and the surrounding walls are also at 20 C. The emissivity of the container’s top and side is 0.45 and the convective coefficient is 1.5 W/m2 C. The container is 3 m high and has a diameter of 1 m. Give the heat conduction equation and boundary equations for this problem. The heat transfer is steady (Fig. P2.40).

1m

z

r

3m

FIGURE P2.40

2-41 A metal cube of dimensions a by a by a and constant conductivity k initially has a uniform temperature Ti . The cube is suddenly placed on a hot plate that keeps the bottom surface of the cube at a temperature Tbox , which is greater than Ti . The top surface of the cube is perfectly insulated, but the remaining four sides of the cube transfer heat to the surrounding fluid, which is at temperature Tfluid . The convective coefficient between the vertical sides of the cube and the fluid is h. It is desired to find the temperature of the center of the cube after tfinal seconds. Sketch the cube on an appropriate set of axes. Provide the appropriate heat conduction equation, boundary equations, and the initial condition equation. 2-42 A large wall of a house has a thickness L. The inside surface of the wall convects to the air that is at temperature Ti The convective coefficient is hi . The outside surface convects to the outside air that is at To with a convective coefficient ho . The sun is beaming on the wall, giving it a solar heat flux input of qsolar =A. The wall has conductivity k, density r, and specific heat c. It is

54

Chapter 2 Heat conduction equation and boundary conditions

desired to determine the temperature distribution of the wall. This is a steady heat transfer problem. Sketch the wall on appropriate axes. Give the heat conduction equation and boundary equations. 2-43 Electric current is flowing through a long wire of radius ro : The flow of current produces internal heat generation of a strength qgen per unit volume. The current is then turned-off and the wire starts to cool due to convection to the surrounding air. The air is at temperature Tair and the convective coefficient is h. It can be assumed that the wire has a uniform temperature Tbegin when the current is turned-off. It is desired to find the temperature distribution in the wire at a time t seconds after the current stops. The radial distance from the centerline of the wire is r and the axial coordinate is z. Give the heat conduction equation, the boundary equations, and the initial condition equation for this problem. 2-44 The square plate shown in Fig. P2.44 has thickness L and cross section a by a. The x ¼ 0 side of the plate has a net heat flux input from a radiant heater of qheater =A. The right side of the plate convects to the adjacent fluid at TN , with convective coefficient h. It also radiates to the surroundings at Tsurr with a surface emissivity ε. The four edges of the plate are perfectly insulated. It is desired to determine the steady-state temperature distribution in the plate. The plate has conductivity k, density r, and specific heat c. Give the heat conduction equation and boundary equations for this problem (Fig. P2.44). 4 edges insulated

y a

Air T∞

qconv

Heat Flux

a Surroundings Tsurr

+ qrad

q heater A

x L z

FIGURE P2.44

2-45 A sphere of radius ro is heated in an oven to a uniform temperature Thot . The sphere is then taken out of the oven and placed in a hemispherical metal holder. See Fig. P2.45A and B. The temperature of the holder is Tholder . The temperature of the holder remains constant, but the sphere cools due to convection from its upper hemispherical area to the room air. The room air

Uncited references

55

is at Tair and the convective coefficient is h. It is desired to determine the temperature in the sphere as it cools. Give the heat conduction equation, boundary equations, and the initial condition equation for this problem (Fig. P2.45A and B).

(A)

ro

(B)

r

Plan View

Top View

FIGURE P2.45A,B

Uncited references [1] J.P. Holman, Heat Transfer, ninth ed., McGraw-Hill, 2002. [2] Y.A. Cengel, A.J. Ghajar, Heat and Mass Transfer Fundamentals & Applications, fourth ed., McGraw-Hill, 2011. [3] F.P. Incropera, D.P. DeWitt, Fundamentals of Heat and Mass Transfer, fourth ed., Wiley, 1996. [4] F. Kreith, M.S. Bohn, Principles of Heat Transfer, sixth ed., Thomson Learning, 2001.

CHAPTER

Steady-state conduction

3

Chapter outline 3.1 Introduction .................................................................................................................................58 3.2 One-dimensional conduction .........................................................................................................58 3.2.1 Plane wall................................................................................................................58 3.2.1.1 Multilayered Walls ............................................................................................... 60 3.2.1.2 Electric-heat analogy and the resistance concept ................................................. 61 3.2.1.3 Overall heat transfer coefficient and R-Value........................................................ 69 3.2.2 Cylindrical shell .......................................................................................................70 3.2.3 Spherical shell .........................................................................................................74 3.3 Critical insulation thickness..........................................................................................................76 3.4 Heat generation in a cylinder........................................................................................................78 3.5 Temperature-dependent thermal conductivity .................................................................................82 3.6 Multi-dimensional conduction .......................................................................................................89 3.7 Conduction shape factors..............................................................................................................89 3.8 Extended surfaces (fins)................................................................................................................94 3.8.1 Fins of constant cross section....................................................................................96 3.8.1.1 The governing differential equation and boundary conditions................................ 96 3.8.1.2 The solution for temperature distribution and heat flow ........................................ 97 3.8.1.3 Very-long-fin approximation ................................................................................. 98 3.8.1.4 Insulated-at-end fin approximation....................................................................... 99 3.8.2 Fin efficiency .........................................................................................................100 3.8.3 Fin effectiveness ....................................................................................................100 3.8.4 Fins of varying cross section ....................................................................................102 3.8.4.1 Circumferential fins ........................................................................................... 102 3.8.4.2 Straight triangular fins ....................................................................................... 106 3.8.4.3 Conical pin fins ................................................................................................. 107 3.8.5 Closing comments on fins........................................................................................109 3.9 Chapter summary and final remarks ............................................................................................. 109 3.10 Problems ...................................................................................................................................110 References ..........................................................................................................................................120

Heat Transfer Principles and Applications. https://doi.org/10.1016/B978-0-12-802296-2.00003-2 Copyright © 2021 Elsevier Inc. All rights reserved.

57

58

Chapter 3 Steady-state conduction

3.1 Introduction Chapter 2 discussed the heat conduction equation and the boundary and initial conditions needed to determine the temperature distribution in a body. In this chapter, we reduce the heat conduction equation to its one-dimensional and steady-state form. We discuss practical problems that can be appropriately modeled using only one dimension. Examples of such problems include heat flow through building walls, fluid and heat flow through pipes, and heat generation in electrical wires. The very useful resistance concept from electric theory is discussed, as is the conduction shape factor that can be applied to many unique, practical problems. Finally, there is a discussion of extended surfaces, i.e., fins. Fins are used in many devices to enhance heat flow.

3.2 One-dimensional conduction This section discusses conduction in plane walls, cylindrical shells, and spherical shells.

3.2.1 Plane wall The plane wall shown in Fig. 3.1 has a finite thickness L in the x direction, is very large in the y and z directions, and has a cross-sectional area A. The left surface of the wall is at temperature T1 and the right surface is at T2 . Our tasks are to determine the temperature distribution in the wall and the heat flow rate through the wall. This can be done as follows: The general heat conduction equation in the rectangular coordinate system is v vT v vT v vT vT (3.1) k þ k þ k þ qgen ¼ rc vx vx vy vy vz vz vt Because the wall is large in the y and z directions, the problem can be reduced to one-dimensional with only the x direction. The problem is also steady state. Eq. (3.1) becomes v vT (3.2) k þ qgen ¼ 0 vx vx y

A

T2

T1 L

O

FIGURE 3.1 Heat flow through a plane wall.

O

x

3.2 One-dimensional conduction

59

We will first consider the situation where the conductivity is constant and there is no heat generation. Conductivity k can be removed from the bracket and qgen is zero. Eq. (3.2) then reduces to d2 T ¼0 dx2 Integrating Eq. (3.3) twice with respect to x, we obtain

(3.3)

TðxÞ ¼ C1 x þ C2

(3.4)

where C1 and C2 are constants. Looking at Eq. (3.4), it is seen that the temperature distribution in the wall is linear, i.e., straightlined, as shown in Fig. 3.2. To obtain the values of C1 and C2 , we have to apply boundary conditions to Eq. (3.4). For our problem, we have the temperature boundary conditions: At x ¼ 0; T ¼ T1

(3.5)

and At x ¼ L; T ¼ T2

(3.6)

T1 ¼ C1 ð0Þ þ C2

(3.7)

Applying these to Eq. (3.4) gives T2 ¼ C1 L þ C2

and

(3.8)

Solving Eqs. (3.7) and (3.8) for C1 and C2, we get C1 ¼ ðT2 T1 Þ=L and C2 ¼ T1 . Substituting this into Eq. (3.4), we get the final temperature distribution in the wall: T2 T1 (3.9) TðxÞ ¼ x þ T1 L To find the heat flow rate through the wall, we use Eq. (1.3), which is q dT ¼ k A x dx

T1 T2

O

FIGURE 3.2 Temperature distribution in a plane wall.

L

x

(1.3)

60

Chapter 3 Steady-state conduction

From Eq. (3.9),

dT dx

is

T2 T1 L

so the heat flow rate is

kA ðT1 T2 Þ (3.10) L From Fig. 3.2, temperature T1 on the left surface of the wall is greater than temperature T2 on the right surface. Hence, qx in Eq. (3.10) is positive. In Chapter 1, we discussed the sign convention for heat flow. Positive qx means that flow is in the positive x direction. This makes sense for our problem. Heat flows from high temperature to low temperature. The left surface of the wall is at a higher temperature than the right surface, and the heat is indeed flowing in the þx direction. qx ¼

3.2.1.1 Multilayered Walls Sometimes walls consist of multiple layers. Consider, for example, the wall of Fig. 3.3, which has four layers. The thicknesses of the layers are L1 ; L2 ; L3 ; and L4 : The respective conductivities are k1 ; k2 ; k3 ; and k4 : As shown in the figure, the temperatures at the interfaces and boundaries are labeled T1 through T5 : Finally, the cross-sectional area of the wall is A. Let us say that there are temperature boundary conditions at the outer surfaces of the wall. The left surface is held at T1 and the right surface is held at T5 . If T1 equals T5 , there is no heat flow. If T1 is greater than T5 , heat flows from left to right. And, if T5 is greater than T1 ; heat flows from right to left. Let us make T1 greater than T5 : Then the heat flow will be from left to right in each layer. Because of continuity of heat flow, the heat flow q through each of the four layers will be the same, and Eq. (3.10) can be written for each layer: q¼

k1 A ðT1 T2 Þ L1

q¼

k2 A ðT2 T3 Þ L2

q¼

k3 A k4 A ðT3 T4 Þ q ¼ ðT4 T5 Þ L3 L4

(3.11)

These four equations can be rearranged so that the temperature differences are on the right-hand sides of each equation, as follows: L1 L2 L3 L4 q ¼ ðT1 T2 Þ q ¼ ðT2 T3 Þ q ¼ ðT3 T4 Þ q ¼ ðT4 T5 Þ (3.12) k1 A k2 A k3 A k4 A

A k1

k2

k3

k4

T1 T2 T3

FIGURE 3.3 Multilayered wall.

T5 L1

L2

L3

L4

T4

3.2 One-dimensional conduction

61

We can now sum all the left-hand sides of the equations and the right-hand sides and make the two resulting sums equal. Doing this, and factoring out q, we get L1 L2 L3 L4 þ þ þ (3.13) q ¼ T1 T 5 k1 A k2 A k3 A k4 A Rearranging this, we finally arrive at the equation for the rate of heat flow through the wall: q¼

T 1 T5 L1 L2 L3 L4 þ þ þ k1 A k2 A k3 A k4 A

(3.14)

This result leads directly to a discussion of the electric-heat analogy, as follows:

3.2.1.2 Electric-heat analogy and the resistance concept There is an analogy between the flow of electric current and the flow of heat. The flow of electric current is caused by a potential difference, and the flow of heat is caused by a temperature difference. Ohm’s law states that the flow of electric current I in a conductor is proportional to the potential difference E across the conductor. That is, I¼

E Ohm’s Law R

(3.15)

where R is the resistance of the conductor. Let us look at Eq. (3.10) that is for the flow of heat through a plane wall. The equation can be rearranged to T 1 T2 q¼ L kA

(3.16)

Let us compare Eq. (3.15) that is for electric flow and Eq. (3.16) that is for heat flow. It is seen that the left-hand sides of the equations are the flow rates and the numerators on the right-hand sides are the potentials causing the flows. Applying the electric-heat analogy, the denominator of the right-hand side of Eq. (3.15) is the electrical resistance, so the denominator of the right-hand side of Eq. (3.16) must be the heat flow resistance. That is, the thermal resistance of a plane wall is the thickness of the wall divided by the product of the conductivity and the area of the wall, or L kA Let us look again at Eq. (3.14) for the heat flow through our four-layer wall. Rwall ¼

q¼

T 1 T5 L1 L2 L3 L4 þ þ þ k1 A k2 A k3 A k4 A

(3.17)

(3.14)

The left-hand side of the equation is the heat flow, and the numerator on the right-hand side is the temperature difference or potential that is causing the flow. The four terms in the denominator of the right-hand side are the resistances of the four layers of the wall. The sum of these four resistances is

62

Chapter 3 Steady-state conduction

the total thermal resistance of the wall, i.e., the sum of the resistances between the two temperatures causing the flow. This can be generalized to the following equation, which will be very useful in solving later problems: DToverall q¼ P R

(3.18)

That is, the heat flow rate is the overall temperature difference divided by the sum of the resistances between the two temperatures.

Example 3.1 Heat Flow Through a Wall Problem: The house wall shown in Fig. 3.4 below consists of three layers. The layers, from outside to inside, are T-111 siding (5/8 inch thick, k ¼ 0.8 Btu in/h ft2 F), fiberglass insulation (3-1/2 inch thick, k ¼ 0.27 Btu in/h ft2 F), and plasterboard (1/2 inch thick, k ¼ 1.2 Btu in/h ft2 F). The wall is 8 ft high by 15 ft wide. The temperature of the outside surface of the wall is 35 F, and the temperature of the inside surface of the wall is 72 F. Determine the rate of heat flow through the wall (Btu/h).

Solution: Let us call the T-111 layer “1", the fiberglass layer “2", and the plasterboard layer “3." This is a problem in English units. The first thing to do is to make sure the units are consistent. Usually the best approach is to change all lengths to feet if the problem is in the English system and all lengths to meters if the problem is SI. For this problem, however, the conductivities are in units of Btu in/h ft2 F and it is not necessary to change the thicknesses of the layers to feet. They can remain in inches. However, if the conductivities were given in the usual units of Btu/h ft F, we would have to change the thicknesses to feet in order for the heat flow to have the units of Btu/h. To get the heat flow through the wall, we use Eq. (3.18) DToverall q¼ P R

(3.18)

T-111

Fiberglass

1

2

Plaster

15 ft.

72 F

8 ft.

3

35 F

Inside

Outside

5" 8

FIGURE 3.4 A house wall.

3

1" 2

1" 2

3.2 One-dimensional conduction

T2

T1

T3

T4

63

T5

L1

L2

L3

L4

k1A

k2A

k3A

k4A

FIGURE 3.5 Circuit for four-layer wall. For a wall, the resistance is L/k A. So, for our problem Eq. (3.18) is q¼

DToverall R1 þ R2 þ R3

(3.19)

where R1 ¼ R2 ¼

L2 3:5 12:963 ¼ ¼ A k2 A 0:27 A

R3 ¼ Applying Eq. (3.19), we have

L1 5=8 0:781 ¼ ¼ A k1 A 0:8 A

L3 0:5 0:417 ¼ ¼ A k3 A 1:2 A

72 35 ¼ 2:613 A q¼ 0:781 12:963 0:417 þ þ A A A The wall is 8 ft high and 15 ft wide, so A ¼ 8 15 ¼ 120 ft2 We therefore arrive at the result q ¼ 2.613 120 ¼ 314 Btu/h. The heat flow through the wall is 314 Btu/h.

(3.20)

In Example 3.1 and also the four-layered wall discussed earlier, the same amount of heat flows through each layer of the wall. The situation is analogous to an electric current flowing through several resistances in series. Let us return to the 4-layer wall of Fig. 3.3. For that wall, the heat flow was T1 T5 q ¼ (3.14) L1 L2 L3 L4 þ þ þ k1 A k2 A k3 A k4 A As discussed above, heat flow is analogous to electric flow. Hence, the 4-layer wall can be modeled as a circuit with four resistances in series between the temperature potentials T1 and T5 (Fig. 3.5). As noted above in Eq. (3.17), the resistance of a wall is L/k A. Heat flow by convection is qconv ¼ h A ðTs TN Þ, or qconv ¼ Ts TN, so the thermal resistance of convection is 1 1 h A Rconv ¼ (3.21) hA In Example 3.1, we had a house wall with temperature boundary conditions at both the inside and outside surfaces. Let us redo Example 3.1 but with convection boundary conditions at the surfaces.

Example 3.2 Heat Flow Through a Wall with Convection Problem: The house wall of Example 3.1 is now modified to have convection boundary conditions rather than temperature boundary conditions, as shown in Fig. 3.6 below. The outside air is at 31 F with ho ¼ 2 Btu = h ft2 F. The inside air is at 74 F with hi ¼ 0.7 Btu/h ft2 F.

64

Chapter 3 Steady-state conduction

A

Inside

Outside T-111

ho = 2 Btu h ft2 F

Fiberglass

Plaster

hi = 0.7

Btu h ft2 F

T∞i = 74 F

T∞o = 31 F

FIGURE 3.6 House wall with convection. (a) What is the rate of heat flow through the wall? (b) What is the temperature at the interface between the T-111 and the fiberglass?

Solution: Rate of Heat Flow We can again use Eq. (3.18), but now we have five resistances in the denominator, three for the wall layers and two for the convections at the wall surfaces. DToverall (3.18) q¼ P R where DToverall is the temperature difference across the whole system, which now includes the convection. Thus, the overall temperature difference is the difference between the temperature of the room air and that of the outside air. Eq. (3.20) is revised to 74 31 ¼ 2:673A (3.22) 1 0:781 12:963 0:417 1 þ þ þ þ 2A A A A 0:7A From Example 3.1, A ¼ 120 ft2, so q ¼ 2.673 120 ¼ 321 Btu/h. Interface Temperature The series electrical circuit for the problem is shown in Fig. 3.7. The heat flow through each layer is the same. Once we have determined q, we can determine the temperature at any location in the system by using Eq. (3.18). We wish to find temperature T2 . We can apply Eq. (3.18) between T2 and the outside air temperature TNo . Between these two temperatures there is one conductive resistance (the T-111 layer) and the outside convective resistance. Hence, Eq. (3.18) becomes q¼

T TNo 2 L 1 þ kA T111 ho A

q¼

T1

T∞o = 31 F

T2

1 ho A

L kA

Outside Convection

T-111

FIGURE 3.7 Electrical circuit for Example 3.2.

T3 L kA

(3.23)

T∞i = 74 F

T4 L kA

Fiberglass Plaster

1 hi A Inside Convection

3.2 One-dimensional conduction

65

Putting in values in Eq. (3.23), we have 321 ¼

T 31 2 0:781 1 þ 120 T111 2$120

(3.24)

Solving for T2 we have T2 [ 34.4 F. We could also have found T2 by applying Eq. (3.18) between T2 and the inside air temperature TNi . Between these two temperatures there are two conductive resistances (the fiberglass and plasterboard layers) and the inside convective resistance. The equations for this approach are q¼

TNi T2 L L 1 þ þ kA fiberglass kA plaster hi A

(3.25)

74 T2 (3.26) 321 ¼ 12:963 0:417 1 þ þ 120 fiberglass 120 plaster 0:7$120 As expected, solving Eq. (3.26) for T2 , we get the same T2 ¼ 34.4 F. We can also get the interface temperature T2 without first obtaining heat flow q, as follows: As the same amount of heat flows through each of the five components of this problem (i.e., the two convective components and the three physical layers), we can apply the resistance technique between any two locations in the system. In particular, we can apply the technique between T2 and the outside air and between the inside air and T2 . That is, DToverall DToverall P P ¼ (3.27) R R T2 to outside air inside air to T2 Looking at the resistances between the particular locations, we get T2 TNo TNi T2 ¼ 1 L L L 1 þ þ þ ho A kA T111 kA fiberglass kA plaster hi A

(3.28)

The only unknown in Eq. (3.28) is T2 . When we put the values for the other variables in the equation and solve for T2 , we get T2 ¼ 34.4 F, which is the same answer as from the other two approaches.

In the above examples we had series heat flow. That is, the same amount of heat flows through all components of the system. In addition to series heat flow, we can also have parallel heat flow. Consider Fig. 3.8 that shows a situation where heat flow is both in series and in parallel.

W

HB T2

qA HA

A

B qB qC C

T1

L1

FIGURE 3.8 Series and parallel heat flow.

L2

HC T2

66

Chapter 3 Steady-state conduction

L2 kB HB W

T1

T2 L1 kA HA W L2 kC HC W

FIGURE 3.9 Circuit diagram for series and parallel heat flow.

There are two layers in the wall shown in Fig. 3.8. The first layer is of material A with conductivity kA, thickness L1 , height HA , and width W. The second layer has two materials: B and C. Material B has conductivity kB, thickness L2 , height HB , and width W. Material C has conductivity kC, thickness L2 , height HC , and width W. The left surface of the wall is at temperature T1 and the right surface is at T2 . The heat flow through the first layer is qA . In the second layer, there are two heat flows, qB and qC , which are parallel to each other. From continuity of heat flow, qA ¼ qB þ qC . The circuit diagram for this situation is shown in Fig. 3.9. In this circuit, there are two resistances in parallel. Let us review circuit theory to get the equivalent resistance for two resistances in parallel. If we have two resistances R1 and R2 in parallel, then, by circuit theory, the equivalent resistance is Requiv ¼

1 1 1 þ R1 R2

(3.29)

Using this information and also the resistance technique of Eq. (3.18), we can easily write the equation for the heat flow through the wall, as follows: q ¼ qA ¼

T 1 T2 L1 1 þk H W k H W B B kA H A W þ C LC L 2

(3.30)

2

Example 3.3 Heat Flow Through a Wall with Studs Unlike the house wall of Example 3.2, a real house wall has studs. This example will determine the effect of the studs on the heat flow through the wall. We will consider the same wall as Example 3.2 except the middle fiberglass layer will now have fiberglass batts with studs. The inclusion of studs causes the heat flow in the middle layer to be one of parallel flow rather than series flow. Some of the heat goes through the fiberglass, and the remainder goes through the studs. A top view of the wall is shown in Fig. 3.10.

3.2 One-dimensional conduction

5" 8

3

67

1" 2

1" 2

14

1" 2

Outside

Inside

T∞o = 31 F 1

1" 2 q

T∞i = 74 F

studs Fiberglass T-111

Plaster

FIGURE 3.10 Top view of house wall with studs. Problem:

The house wall shown in Fig. 3.10 consists of three layers. The first layer is T-111 siding (5/8 inch thick, k ¼ 0.8 Btu in/h ft2 F). The second layer consists of fiberglass batts (3-1/2 inch thick, k ¼ 0.27 Btu in/h ft2 F) separated by studs on 16 inch centers. The studs are 2 4s. (Their actual dimensions are 1-1/2 inch by 3-1/2 inch.) Like the T-111, the studs are softwood, so we will use the same conductivity for them as the T-111. The third layer is plasterboard (1/2 inch thick, k ¼ 1.2 Btu in/h ft2 F). The wall is 8 ft high by 15 ft wide. The outside air is at 31 F and ho ¼ 2 Btu/h ft2 F. The inside air is at 74 F and hi ¼ 0.7 Btu/h ft2 F. Determine the rate of heat flow through the wall.

Solution: Fig. 3.10 shows four fiberglass batts in the middle layer of the wall. For a wall that is 15 feet long, there will actually be 11 such batts, each 14-1/2 inches wide. Also, as shown in the figure, we will assume that there are double studs at both ends of the wall. For a 15 foot wall there will be 14 studs. Let us check that 11 batts and 14 studs will equal 15 feet. We have (11) (14.5) þ (14) (1.5) ¼ 180.5 inches ¼ 15.04 ft. So, we are OK. Let us combine all the studs together and all the fiberglass batts together to result in two parallel heat flows through the center layer of the wall. The wall is 8 feet high and 15 feet wide. For the T-111 and plaster layers, the heat flow area is 8 15 ¼ 120 ft2. For the middle layer, the flow area for the studs is 14 studs x 1.5/12 ft per stud x 8 ft ¼ 14 ft2. The flow area for the fiberglass is then 120e 14 ¼ 106 ft2. The circuit diagram for the wall is shown in Fig. 3.11. Using the circuit diagram, we can write the equation for the heat flow through the wall as follows: q¼

TNi TNo 1 L 1 þ þ ho A kA T111 kA kA þ L L FG

L 1 þ kA plaster hi A

þ studs

(3.31)

68

Chapter 3 Steady-state conduction

L kA fiberglass

T∞o 31 F

1 ho A

L kA

Outside Convection

T 111 T-

L kA studs

L kA

1 hi A

plaster

Inside Convection

T∞i 74 F

FIGURE 3.11 Circuit diagram for Example 3.3.

Putting values into this equation, we get q¼

1 0:781 þ þ 2,120 120 T111

74 31 1 106 14 þ 4:375 12:963 FG

þ

0:417 1 þ 120 plaster 0:7,120

(3.32)

studs

q ¼ 377 Btu/h. We got a result of 321 Btu/h for Example 3.2. Hence, the inclusion of studs in the wall increased the heat transfer by 17%.

We have discussed the thermal resistance of a plane wall, Rwall , and the thermal resistance of convection at a surface, Rconv . They are, from Eqs. (3.17) and (3.21). L (3.17) kA 1 (3.21) Rconv ¼ hA Let us now discuss the thermal resistance of a surface having radiative heat transfer. Similar to the convective coefficient h, we will define the radiative coefficient hrad by Rwall ¼

q ¼ hrad ðTs Tsurr Þ where Ts is the surface temperature and Tsurr is the temperature of the surroundings. From Eq. (1.13), the radiant heat transfer from a surface to a large enclosure is 4 q ¼ ε sA Ts4 Tsurr Equating the right sides of Eqs. (3.33) and (3.34), we get the following for hrad : 4 εs Ts4 Tsurr 2 ¼ εs Ts2 þ Tsurr hrad ¼ ðTs þ Tsurr Þ Ts Tsurr

(3.33)

(3.34)

(3.35)

From Eq. (3.33), the thermal resistance of a surface for radiation is Rrad ¼

1 hrad A

(3.36)

3.2 One-dimensional conduction

69

Defining the radiation coefficient hrad significantly simplifies electrical circuit diagrams. The nonlinearities are in the coefficient. There are no T 4 nodes. In many situations, there is concurrent convection from a surface to an adjacent fluid and radiation from the surface to the surroundings. The total heat transfer from the surface is then qtotal ¼ qconv þ qrad ¼ hAðTs TN Þ þ hrad AðTs Tsurr Þ

(3.37)

If fluid temperature TN and surroundings temperature Tsurr are equal, a combined coefficient hcombined may be defined, which includes both the convective and radiative heat transfers. hcombined ¼ h þ hrad

(3.38)

If this is done, the heat transfer at the surface is qtotal ¼ hcombined AðTs TN Þ

(3.39)

and the resistance for the combined convection and radiation is Rconvþrad ¼

1 hcombined A

(3.40)

3.2.1.3 Overall heat transfer coefficient and R-Value Two terms often used in heat transfer are the Overall Heat Transfer Coefficient, U, and the R-value. The overall heat transfer coefficient is defined by the equation q ¼ UADT

(3.41)

From this equation, U is the heat flow rate per unit area per unit temperature difference. U is also sometimes called the Conductance, C. Let us return to the four-layer wall that we considered earlier. For that wall, the heat flow was given T1 T5 . by Eq. (3.14): q ¼ L1 L2 L3 L4 þ þ þ k1 A k2 A k3 A k4 A Comparing Eqs. (3.14) and (3.41), we can see that, for the four-layer wall, the overall heat transfer coefficient U is U¼

1 L1 L2 L3 L4 þ þ þ k1 k2 k3 k4

(3.42)

The R-value is another term that is frequently used in heat transfer. It is the reciprocal of the Overall Heat Transfer Coefficient U. R-value is commonly used in the specification of thermal insulation. The higher the R-value, the more resistance the insulation has to heat flow. As an example, 3-1/2 inch thick fiberglass batts used between studs in a 2 4 wall typically have an R-value of R-11 or R-13. For a 2 6 wall, the fiberglass insulation would be 5-1/2 inch thick and the R-value would be R-19. R-values are also used by states in their energy codes. For example, in New York, the energy code specifies the minimum Rvalue requirements for walls, floors, and ceilings of buildings. The heat flow through a wall of conductivity k, thickness L, and area A is q ¼ kA L DT. From Eq. (3.41), q ¼ UADT. Hence, for a wall, the overall heat transfer coefficient is U ¼ k/L. The R-value is the reciprocal of U, so the R-value for the wall is L/k.

70

Chapter 3 Steady-state conduction

Let us determine the R-value for the fiberglass insulation used in the above examples. The fiberglass is 3-1/2 inches thick and its conductivity is k ¼ 0.27 Btu in/h ft2 F. The R-value is L/k ¼ 3.5 in/ (0.27 Btu in/h ft2 F) ¼ 13 h ft2 F/Btu. When purchasing this insulation in a store, you will only see R-13 on the insulation. The units will not be included in the labeling.

3.2.2 Cylindrical shell The cylindrical shell (or hollow cylinder) shown in Fig. 3.12 has an inner radius ri , an outer radius ro , and a length L. The thermal conductivity is k. The cylinder has temperature boundary conditions with the inner surface at Ti and the outer surface at To . Heat conduction through the wall is steady and is only in the radial r direction. We want to determine the temperature distribution in the cylindrical wall and the heat flow rate through the wall. From Chapter 2, the heat conduction equation is 1 v vT 1 v vT v vT vT (3.43) kr þ 2 k þ k þ qgen ¼ rc r vr vr r v4 v4 vz vz vt We have steady state and flow only in the r direction, so Eq. (3.43) reduces to 1 d dT kr þ qgen ¼ 0 r dr dr

(3.44)

In the future, we will consider heat generation, but for now let us make qgen zero. Also, let us assume that the conductivity is constant. Then Eq. (3.44) becomes d dT r ¼0 (3.45) dr dr Integrating Eq. (3.45) with respect to r, we get r

dT ¼ C1 dr

(3.46)

To ri Ti

r ro L

FIGURE 3.12 Cylindrical shell.

3.2 One-dimensional conduction

71

Integrating again, we get the solution TðrÞ ¼ C1 ln r þ C2

(3.47)

To get constants C1 and C2 , we apply, to Eq. (3.47), the boundary conditions: At r ¼ ri ;

T ¼ Ti

(3.48)

r ¼ r o ; T ¼ To

(3.49)

This gives

Ti ¼ C1 ln ri þ C2

(3.50)

and

To ¼ C1 ln ro þ C2

(3.51)

and At

Eqs. (3.50) and (3.51) are solved simultaneously to give To Ti C1 ¼ ro ln ri

and

C2 ¼ Ti

To Ti ln ri ro ln ri

(3.52)

Putting these constants into Eq. (3.47) and doing algebraic manipulations, we arrive at the final equation for the temperature distribution in the cylinder: r ln ri TðrÞ ¼ Ti ðTi To Þ (3.53) ro ln ri Heat flow through the cylinder’s wall is in the r direction. Hence, to get the heat flow we can use the relation dT (3.54) dr A is the area through which the heat flows. This area varies with the distance from the centerline of the cylinder. At radial distance r from the centerline, the area is the circumference times the length of the cylinder or A ¼ 2 p r L. dT/dr may be obtained by differentiating Eq. (3.53) or by using Eqs. (3.46) and (3.52). Putting the expressions for A and dT/dr into Eq. (3.54), we arrive at the expression for the heat flow through the cylinder’s wall: q ¼ kA

2pkL q ¼ ðTi To Þ ro ln ri

(3.55)

Inspection of Eq. (3.55) shows that we can define the thermal resistance for a cylindrical shell as ro ln ri Rcyl ¼ (3.56) 2pkL This thermal resistance is very useful in problems involving concentric cylinders. One such problem is an insulated pipe, as illustrated in Example 3.4 below.

72

Chapter 3 Steady-state conduction

Before leaving this discussion of cylindrical shells, let us discuss the overall heat transfer coefficient U: Unlike a plane wall, a cylinder has no single area through which heat is flowing. The area changes with r. At the inner surface of the cylinder, the area is Ai ¼ 2pri L and at the outer surface it is Ao ¼ 2pro L. Because there is no unique area, the value of U must be stated with respect to a specific area. For example, we can give the value of U with respect to the inner area or the value of U with respect to the outer area. From Eq. (3.41) above, U is defined by the equation q ¼ UA DT. For a cylinder, the two physical areas are Ai and Ao , and the corresponding U values are Ui and Uo . Thus, for a cylinder, q ¼ Ui Ai DT ¼ Uo Ao DT

(3.57)

From Eq. (3.57), Ui Ai ¼ Uo Ao and Ui ¼ ðAo =Ai Þ Uo . Putting in the relations for Ai and Ao, we arrive at Ui ¼ ðro = ri Þ Uo

(3.58)

Example 3.4 Heat Transfer from an Insulated Pipe Problem:

Steam at 150 C is flowing through a 100 mm nominal size standard metric pipe. The pipe is carbon steel (k ¼ 60 W/m C) and is located in a factory where the air temperature is 20 C. The convective coefficient between the pipe and the room air is 4 W/m2 C. The pipe is 50 m long and is uninsulated. (a) What is the rate of heat transfer from the bare pipe to the room air? (b) Insulation is added to the pipe to reduce the heat transfer. The insulation is 51 mm thick and has a conductivity of 0.05 W/m C. What is the new rate of heat transfer? You may make the following assumptions: Radiation heat transfer from the pipe is negligible. The insulated pipe has the same convective coefficient at its outer surface as the uninsulated pipe. The temperature of the inside surface of the pipe is essentially the same as the steam temperature.

Solution: The uninsulated pipe is shown in Fig. 3.13. From an Internet search, the outside diameter of the pipe is 114.3 mm and the inside diameter is 102.26 mm. Therefore, r1 ¼ 0.05113 m and r2 ¼ 0.05715 m. There are two resistances between the two known temperatures T1 and TNo : the resistance of the pipe wall and the convective resistance at the outside surface of the pipe. The circuit for the problem is shown in Fig. 3.14. From the circuit drawing and the resistance technique, we have DToverall T1 TNo ¼ q¼ P lnðr2 =r1 Þ 1 R þ 2pkL ho ð2pr2 LÞ

(3.59)

Putting values into Eq. (3.59), we have 150 20 ¼ 9332 W lnð0:05715=0:05113Þ 1 þ 2,p,60,50 4,ð2,p,0:05715,50Þ The insulated pipe is shown in Fig. 3.15. There is an added 51 mm layer of insulation, making r3 ¼ r2 þ 0:051 ¼ 0:10815 m. The area at the outer surface of the insulation is now 2pr3 L rather than 2pr2 L: The added layer of insulation adds a term to the denominator of Eq. (3.59). The new equation is q¼

T1 TNo (3.60) lnðr2 =r1 Þ lnðr3 =r2 Þ 1 þ þ 2pkpipe L 2pkins L ho ð2pr3 LÞ When values are entered into Eq. (3.60), it is seen that the new heat transfer rate is 2710 W. Addition of insulation has decreased the heat transfer from 9332 to 2710 W, a decrease of 71%. q¼

3.2 One-dimensional conduction

ho = 4

73

W m2 C

Air at T∞o = 20 C T1 = 150 C r1 Steam

r2 L = 50 m

FIGURE 3.13 Original pipe of Example 3.4. T1

T∞o r ln 2 r1

150 C

2π kL

1 ho (2π r2L)

20 C

FIGURE 3.14 Circuit for bare pipe of Example 3.4. W m2 C Air at 20 C ho = 4

Insulation Pipe r1 r2 Steam r3 T1 = 150 C

L = 50 m

FIGURE 3.15 Insulated pipe of Example 3.4. As a final observation, we assumed that the temperature of the inside surface of the pipe was essentially the same as the steam temperature. This is often a very good assumption. The convective coefficient, h, for steam on a surface is very large. The convective heat transfer is q ¼ h A ðTs TN Þ: Rearranging this, we have Ts TN ¼ h qA. So, for very large h, the difference between the surface temperature and the fluid temperature is often very small.

74

Chapter 3 Steady-state conduction

3.2.3 Spherical shell The spherical shell (or hollow sphere) shown in Fig. 3.16 has an inner radius ri , and an outer radius ro . The thermal conductivity is k. The sphere has temperature boundary conditions with the inner surface at Ti and the outer surface at To . Heat conduction through the wall is steady and is only in the radial r direction. We want to determine the temperature distribution in the spherical wall and the heat flow rate through the wall. From Chapter 2, the heat conduction equation is 1 v 1 v vT 1 v vT 2 vT k r þ k þ k sin q þ qgen r 2 vr vr v4 r 2 sin q vq vq r 2 sin2 q v4 vT ¼ r c vt (3.61) We have steady state and flow only in the r direction, so Eq. (3.61) reduces to 1 d 2 dT k r þ qgen ¼ 0 r 2 dr dr

(3.62)

Let us assume that there is no internal heat generation, i.e., qgen ¼ 0 and that the thermal conductivity is constant. Then Eq. (3.62) becomes d dT r2 ¼0 (3.63) dr dr Integrating Eq. (3.63) with respect to r, we get r2

dT ¼ C1 dr

(3.64)

Integrating again, we get the solution TðrÞ ¼ C2

C1 r

To Ti

ri r ro

FIGURE 3.16 Spherical shell.

(3.65)

3.2 One-dimensional conduction

75

To get constants C1 and C2 we apply, to Eq. (3.65), the boundary conditions: At r ¼ ri ; and

T ¼ Ti

(3.66)

At r ¼ ro ; T ¼ To

(3.67)

This gives

Ti ¼ C2 C1 =ri

(3.68)

and

To ¼ C2 C1 =ro

(3.69)

Eqs. (3.68) and (3.69) are solved simultaneously to give Ti To C1 ¼ 1 1 ro ri

and

Ti To C2 ¼ Ti ri 1 ro

(3.70)

Putting these constants into Eq. (3.65) and doing algebraic manipulations we arrive at the final equation for the temperature distribution in the sphere: r 3 2 i 1 r 4 5 TðrÞ ¼ Ti ðTi To Þ (3.71) ri 1 ro Heat flow through the sphere’s wall is in the r direction. Hence, to get the heat flow we can use the relation dT (3.72) dr A is the area through which the heat flows. This area varies with the distance from the center of the sphere. At radial distance r from the center, the area is the surface area of a sphere or A ¼ 4 p r 2 . dT/dr may be obtained by differentiating Eq. (3.71) or by using Eqs. (3.64) and (3.70). Putting the expressions for A and dT/dr into Eq. (3.72), we arrive at the expression for the heat flow through the sphere’s wall: q ¼ kA

4pk ðTi To Þ 1 1 ri ro

q¼

(3.73)

From Eq. (3.73), we can see that the thermal resistance for a spherical shell is 1 1 ri ro Rsphere ¼ (3.74) 4pk This thermal resistance is very useful in problems involving concentric spheres as was done earlier for concentric cylinders. As with cylinders, the overall heat transfer coefficient U for spheres must be related to a specific area.

76

Chapter 3 Steady-state conduction

3.3 Critical insulation thickness Contrary to intuition, addition of insulation to an object can sometimes increase the heat transfer from the object rather than decrease it. However, this phenomenon occurs only in limited situations. Fig. 3.17 shows a pipe or wire covered with a layer of insulation. The bare pipe or wire has an outer radius of ri , and the radius to the outer surface of the insulation is ro . As insulation is added to the pipe or wire, ro goes from ri to its final value when addition of insulation is stopped. The length of the pipe or wire is L. Let us look at the heat transfer from the pipe or wire to the surrounding fluid. The temperature at the interface between the pipe or wire and the insulation is T1 . The insulation has conductivity k and the surrounding fluid is at temperature TN . The convective coefficient at the outer surface of the insulation is h. Between T1 and TN are two resistances: the conductive resistance of the insulation and the convective resistance between the outer surface of the insulation and the fluid. The expression for the heat flow through the insulation is q¼

T1 TN

0 @

ln

ro 1

(3.75)

ri 1 A þ hð2pro LÞ 2pkL

Looking at Eq. (3.75), we see that, as insulation is added to the pipe or wire, i.e., as ro increases, the first term in the denominator decreases and the second term increases. Because of this, there is a certain value of ro for which the heat flow is a maximum. To find this value, called the “critical radius roc ”, we can differentiate the expression for q in Eq. (3.75) with respect to ro and set the result to zero. When this is done, it is found that the critical radius is k/h. roc ¼

k h

(3.76)

Outer surface of pipe or wire

Insulation T1 ri ro

FIGURE 3.17 Pipe or wire with insulation.

h Fluid at T∞

3.3 Critical insulation thickness

77

q

A

B

roc

ro

FIGURE 3.18 Heat flow versus outer radius.

So, the maximum heat flow occurs when the outer radius of the insulation equals k/h. Fig. 3.18 shows a typical plot of q versus ro . We have indicated two regions on the plot: Region A where ro is less than roc and Region B where ro is greater than roc . Let us say that a bare pipe or wire has an outer radius ri that is less than roc . This is in Region A. As insulation is added, ro increases and the heat transfer increases until ro reaches roc . Further addition of insulation will cause the heat transfer to decrease. On the other hand, if the bare pipe or wire has a radius ri that is equal to or greater than roc , i.e., ri is in Region B; then any amount of added insulation will decrease the heat transfer.

Example 3.5 Heat Flow from an Insulated Wire Problem: Please refer to the previous Fig. 3.17 that shows an insulated wire. We have a copper #12 AWG wire with PTFE insulation. The bare wire has a diameter of 2.052 mm and the insulation is 15 mils thick. The thermal conductivity of the insulation is k ¼ 0.25 W/m C. The temperature T1 at the interface between the wire and the insulation is kept at 100 C, and the surrounding air is at 20 C. The convective coefficient h at the outer surface of the insulation is 10 W/m2 C. (a) What is the heat transfer rate from the wire per meter length before additional insulation is added? (b) If more insulation is added to the wire, will the heat transfer rate increase or decrease? (c) If the answer to Part (b) is that the heat transfer will increase, then how much insulation has to be added for the heat transfer to reach a maximum? Also, what is the value of this maximum heat transfer rate?

Solution:

(a) For the original wire with the original insulation, ri ¼ 0.001026 m and ro ¼ 0.001407 m. (Note: 1 mil is 1/1000 inch) Putting the values of this problem into Eq. (3.75), we get that the original heat transfer is q/L ¼ 6.95 W/m. (b) Maximum heat transfer occurs when ro ¼ roc ¼ k=h. 0:25 W = m C

¼ 0.025 m 10 W = m2 C Because our initial ro is 0.001407 m, which is less than roc , addition of insulation will increase the heat transfer until ro reaches roc : roc ¼

78

Chapter 3 Steady-state conduction

(c) The thickness of insulation that has to be added to reach roc is 0.025e0.001407 ¼ 0.0236 m ¼ 23.6 mm. Using roc ¼ 0.025 m in Eq. (3.75), it is found that the maximum heat transfer rate is 29.97 W/m. The below graph shows q=L versus ro for this example. It is seen that the maximum heat flow indeed occurs at ro ¼ 0.025 m and the maximum heat flow is 29.97 W/m. Heat flow vs. ro for Example 3.5

30 29.95

q/L

29.9

(W/m) 29.85 29.8 29.75 0.02

0.022

0.024

0.026

0.028

0.03

0.032

ro (m)

One final note on this example: The inner surface of the insulation had a temperature boundary condition. The temperature was held constant at 100 C. As insulation was added to the wire, up to roc , the heat transfer from the wire increased. To keep the temperature constant as the heat transfer increased, the electrical current would have to be increased to increase the heat generation of the wire. If, instead of the constant temperature boundary condition, we had a constant power boundary condition, then, as the insulation was added, the increased heat transfer would have caused a drop in temperature at the inner surface of the insulation.

At the beginning of this section, we mentioned that the critical insulation phenomenon (i.e., an increase in heat flow with increased insulation) is only important in limited situations. Indeed, for common values of k and h, the phenomenon is only relevant when the diameter of the pipe or tube being insulated is very small or, as in Example 3.5, we are insulating a thin wire. To illustrate this, let us consider a practical situation where we are putting fiberglass insulation on a 1/2 inch Schedule 40 pipe. Fiberglass has a k value of about 0.25 Btu in/h ft2 F. And, we will assume that the pipe is in still air where h is about 2 Btu/h ft2 F. For a 1/2 inch Schedule 40 pipe, the OD is 0.84 inch. This makes ri equal to 0.42 inch. The critical radius k=h is 0.25/2 or 0.125 inch. However, ro ¼ ri þ the thickness of the insulation. So, ro is always greater than 0.42 inch, which is greater than roc . Addition of any amount of insulation to the pipe will decrease the heat transfer.

3.4 Heat generation in a cylinder Fig. 3.19 shows a cylinder of radius ro and length L. The cylinder has a uniformly distributed heat source at strength qgen per unit volume. The heat flow is steady and one-dimensional, i.e., T ¼ TðrÞ, and there is a temperature boundary condition at the surface of the cylinder. The heat conduction equation for this situation is Eq. (3.44), repeated here: 1 d dT (3.44) kr þ qgen ¼ 0 r dr dr Let us assume that the conductivity is constant. Then, Eq. (3.44) becomes qgen r d dT ¼0 r þ dr dr k

(3.77)

3.4 Heat generation in a cylinder

79

To ro

k, qgen r

L

FIGURE 3.19 Cylinder with uniform heat generation.

Integrating Eq. (3.77) with respect to r, we get qgen r 2 dT ¼ þ C1 (3.78) dr 2k Dividing by r and integrating again, we get the following general solution for the temperature distribution: r

qgen r 2 þ C1 ln r þ C2 4k We apply boundary conditions to get constants C1 and C2 : These conditions are: TðrÞ ¼

(3.79)

(1) At r ¼ 0; the temperature must be finite. Therefore, C1 must be zero and Eq. (3.79) becomes TðrÞ ¼

qgen r 2 þ C2 4k

(3.80)

(2) At r ¼ ro , T ¼ To . Applying this to Eq. (3.80) and solving for C2, we have qgen ro2 (3.81) 4k Applying C1 and C2 to Eq. (3.79), we get the final expression for the temperature distribution in the cylinder: " 2 # qgen ro2 r TðrÞ ¼ To þ (3.82) 1 ro 4k C2 ¼ To þ

80

Chapter 3 Steady-state conduction

Inspection of Eq. (3.82) shows that the maximum temperature in the cylinder is on the centerline (r ¼ 0) and is qgen ro2 (3.83) 4k The heat flow in the cylinder is in the radial direction and can be determined using the equation Tmax ¼ To þ

dT (3.84) dr At distance r from the centerline, the area through which heat flows is A ¼ 2prL. The derivative dT/dr can easily be obtained from Eq. (3.78) as C1 ¼ 0. Putting this in Eq. (3.84), we get the expression for the heat flow: q ¼ kA

qðrÞ ¼ pr 2 Lqgen

(3.85)

is the volume of the cylinder from r ¼ 0 to r ¼ r. Let Looking at the right side of Eq. (3.85): us call this volume the “control volume” When the volume is multiplied by heat generation rate qgen , we have the rate of heat generation in the control volume. For steady state, the heat inflow to this volume must equal the heat outflow from this volume. The heat generated in the control volume is removed by heat flow through the r ¼ r boundary of the volume. The heat generation for the whole cylinder is pro 2 Lqgen . Again, as we have steady state, this heat generation is removed by heat flow out of the cylinder at the r ¼ ro surface. Let us say that we have convection at the surface of the cylinder with an h and a TN : This is shown in Fig. 3.20 below. The area of the surface of the cylinder is 2pro L: Because of continuity of heat flow, the heat generated in the cylinder equals the heat into the surrounding fluid by convection. If the temperature of the surface is To , then the boundary equation for the surface of the cylinder is pr 2 L

qðro Þ ¼ pro2 Lqgen ¼ hð2pro LÞðTo TN Þ

To ro

k, qgen

Convection h, Fluid at T∞

r L

FIGURE 3.20 Cylinder with heat generation and convection.

(3.86)

3.4 Heat generation in a cylinder

81

Example 3.6 Heat Generation in a Wire Problem: A #22 AWG copper wire shorts out across a 0.15 V source. This causes the wire to heat up significantly. The wire has a length of 0.2 m and is located in 20 C room air. The wire’s thermal conductivity is k ¼ 385 W/m C. Finally, the convective coefficient at the surface of the wire is 15 W/m2 C. For steady-state conditions, (a) What is the power generated in the wire? (b) What is the heat generation rate qgen (W/m3)? (c) What are the temperatures at the center and surface of the wire?

Solution:

(a) The power generated in the wire is P ¼ i2 R, where i is the current through the wire and R is the resistance of the wire. The resistance of the wire can be obtained from the relation

rL (3.87) A where r is the resistivity of the wire, L is the length of the wire, and A is the cross-sectional area of the wire. From the Internet, the resistivity of copper is 1.68 108 U m and the diameter of the wire is 0.644 mm. The cross-sectional area of the wire is pro2 ¼ 3.257 107 m2. The length of the wire was given as 0.2 m. Putting these values into Eq. (3.87) gives a resistance R ¼ 0.010315 U. The current in the wire by Ohm’s law is i ¼ ER. We were given that E ¼ 0.15 V and we determined that R ¼ 0.010315 U. Hence, the current is i ¼ 0.15/0.010315 ¼ 14.54 A. The power generated in the wire is P ¼ i2 R¼(14.54)2 (0.010,315) ¼ 2.181 W. (b) qgen is the heat generation rate per unit volume. The volume of the wire is cross-sectional area times length. Using values above, we have V ¼ (3.257 107) (0.2) ¼ 6.514 108 m3. Therefore, R¼

qgen ¼

P 2:181 ¼ ¼ 3:348 107 W = m3 V 6:514 108

(c) Rearranging Eq. (3.86), we have r o To ¼ TN þ (3.88) qgen 2h Putting in the values, we get, for the surface temperature 0:322 103 3:348 107 ¼ 379:4 C To ¼ 20 þ 2,15 We can get the center temperature from Eq. (3.83) 2 3.348 107 0:322 103 qgen ro2 ¼ 379:4 þ ¼ 379:402 C Tmax ¼ To þ 4k 4,385 It is seen that the difference between the center and surface temperatures is only 0.002 C. This very small difference is due to the high conductivity of copper and the small diameter of the wire.

As a final item for this section, let us say that, instead of a solid cylinder, we have a hollow cylinder that has uniform heat generation, as shown in Fig. 3.21. We will also have temperature boundary conditions at the inner and outer surfaces. For this situation, the heat conduction equation, Eq. (3.77), is the same as that for the solid cylinder, and the general solution, Eq. (3.79), is also the same as for the solid cylinder. When we applied the boundary conditions for the solid cylinder, we made C1 equal to zero due to the existence of material at

82

Chapter 3 Steady-state conduction

k, qgen ri Ti To

ro r

L

FIGURE 3.21 Hollow cylinder with heat generation.

the center of the cylinder and the logarithmic term. For the hollow cylinder, there is no material at r ¼ 0 and we cannot eliminate the C1 ln r term. The boundary conditions for the hollow cylinder situation are applied to Eq. (3.79) in its entirety. Heat generation for other geometries is given in Chapter 12.

3.5 Temperature-dependent thermal conductivity The thermal conductivity k of materials varies with temperature. However, in many cases, this variation is small and the conductivity can be taken as being constant at the average temperature of the problem. Let us consider conduction through the plane wall of Fig. 3.22, which has conductivity varying with temperature. The left face of the wall is at temperature T1 and the right face is at T2 .

A

k(T) T2

q T1

L

FIGURE 3.22 Plane wall with variable conductivity.

x

3.5 Temperature-dependent thermal conductivity

83

The conductive heat transfer equation is dT dx Rearranging and integrating across the wall, we have q ¼ kA

q A

ZL

(3.89)

ZT2 dx ¼

kdT

(3.90)

T1

0

q 1 or; ¼ A L

ZT1 kdT

(3.91)

T2

The functional relation k ðTÞ can be put into the integral, the integral evaluated, and the final result for the heat flux ðq =AÞ obtained. As a special case, let us assume that the kT relation is linear. That is, k ¼ a þ bT

(3.92)

where a and b are constants. Putting this relation in Eq. (3.91), we have q 1 ¼ A L

ZT1 ða þ bTÞdT

(3.93)

T2

Performing the integration, T q 1 bT 2 1 ¼ aT þ A L 2 T2 q 1 b ¼ aðT1 T2 Þ þ T12 T22 A L 2 q 1 b ¼ aðT1 T2 Þ þ ðT1 T2 ÞðT1 þ T2 Þ A L 2

(3.94) (3.95) (3.96)

Factoring out ðT1 T2 Þ, we have

q 1 T1 þ T2 ¼ ðT1 T2 Þ a þ b A L 2

(3.97)

2 is the average, or mean, of the two surface temperatures, so the bracketed term is the But T1 þT 2 conductivity at this mean temperature km . Therefore, the equation for the heat flow q through the wall is

q¼

km A ðT1 T2 Þ L

(3.98)

84

Chapter 3 Steady-state conduction

k(T) q ri Ti To

ro r L

FIGURE 3.23 Hollow cylinder with variable conductivity.

Comparing this with Eq. (3.10), it is seen that we can use the constant conductivity equation if we use the conductivity value km in the equation. Similar results can be obtained for heat flow through cylinders and spheres when the conductivity is linear with temperature. For example, consider heat flow through the hollow cylinder shown in Fig. 3.23. If the conductivity of the cylinder is linear with temperature, it can be shown that the heat flow q through the wall is 2pkm L q ¼ ðTi To Þ ro ln ri Ti þTo where km is the conductivity at the mean temperature . 2

(3.99)

This is the same equation as Eq. (3.55) derived above for the case of constant conductivity. The constant conductivity k has just been replaced with km .

Example 3.7 Plate With Temperature-Dependent Conductivity Problem: A plate of pure copper is 15 cm by 20 and 1 cm thick. One side of the plate is at 200 C and the other side is at 600 C. Determine the heat flow rate through the plate (W). Consider the temperature dependence of the thermal conductivity.

Solution:

The plate is like the wall of Fig. 3.22. The cross-sectional area A is 0.15 0.20 m ¼ 0.03 m2. The thickness L is 0.01 m. From an Internet search, the conductivity of the copper varies with temperature as shown in Table 3.1. Fig. 3.24 is an Excel chart of the data of Table 3.1.

3.5 Temperature-dependent thermal conductivity

85

Table 3.1 Conductivity of pure copper. T (K)

k (W/m K)

400 600 800 1000 1200

392 383 371 357 342

400 390 380

k

(W / m K)

370 360 350 340 330 400

500

600

700

800

900

1000

1100

1200

Temperature (K)

FIGURE 3.24 Conductivity of pure copper.

At first glance, the figure appears to show that the conductivity varies significantly with temperature. However, the figure is misleading due to the scaling on the vertical axis. If the vertical axis had started at zero, then the figure would have shown a much more gradual variation of conductivity with temperature. This is shown in Fig. 3.25. Indeed, the conductivity varies only 13.6% over the entire 800 K temperature range. And, the temperature relationship is quite linear. Using Excel, the linear trendline fit of Table 3.1 data is k ¼ 419:4 0:063 T with an R2 of 0.991. (Note: A second-degree polynomial fit is even better. It is k ¼ 0:000025 T 2 0:023 T þ 405:4 with anR2 of 0.998.) The units in these equations are k (W/m K) and T (K). If we assume that k varies linearly with T, then we can use Eq. (3.98): km A ðT1 T2 Þ L For this problem, km is k at the mean temperature (200 þ 600)/2 ¼ 400 C ¼ 673 K. Using the Excel linear trendline equation k ¼ 419:4 0:063T, we have km ¼ 419:4 0:063 ð673Þ ¼ 377 W = m K. From Eq. (3.98), the heat ð0:03Þ q ¼ 3770:01 ð600 200Þ ¼ 4:524 x 105 W. q¼

(3.98)

flow

through

the

wall

is

86

Chapter 3 Steady-state conduction

450 400 350 300 250

k

(W / m K) 200 150 100 50 0 400

500

600

700

800

900

1000

1100

1200

Temperature (K)

FIGURE 3.25 Conductivity of pure copper (revised scale).

Alternatively, we could have used the equation from the second-degree polynomial fit of the conductivity and perform the integration indicated in Eq. (3.91), as follows: q¼

A L

ZT1 kdT T2

q¼

0:03 0:01

Z

0:000025T 2 0:023T þ 405:4 dT

600þ273

200þ273

873 0:03 0:000025 3 0:023 2 T T þ 405:4T 0:01 3 2 473 0:03 0:000025 0:023 8733 4733 8732 4732 þ 405:4ð873 473Þ q¼ 0:01 3 2 q ¼ 4.819 105 W. The difference between heat flows using these two data fits is only 6%. An even quicker solution would be to merely look at Fig. 3.24 and use the k value at the mean temperature of 400 C ¼ 673 K. Eyeballing the figure, we get that k is about 377 W/m K. Hence, q¼

kA 377ð0:03Þ ðT1 T2 Þ ¼ ð400Þ ¼ 4:52 105 W. L 0:01 The closeness of these three results for q is due to pure copper’s small variation of conductivity with temperature and the essentially linear characteristic of the keT function. q¼

Let us next consider a problem involving a material having significant temperature-dependent conductivity.

3.5 Temperature-dependent thermal conductivity

87

Example 3.8 Cylinder With Temperature-Dependent Conductivity Problem: A long aluminum oxide cylinder has an inner diameter of 1 cm and an outer diameter of 2 cm. The temperature of the inner surface is 500 C and that of the outer surface is 100 C. Determine the heat flow rate through the cylinder’s wall per meter length of the cylinder. Consider the temperature dependence of the thermal conductivity.

Solution:

Consider Fig. 3.23 above. For this problem, ri ¼ 0.005 m, ro ¼ 0.01 m, Ti ¼ 500 C ¼ 773 K, and To ¼ 100 C ¼ 373 K. From an Internet search, the conductivity of the aluminum oxide varies with temperature as shown in Table 3.2. Fig. 3.26 is an Excel chart of the data of Table 3.2. It is seen that the conductivity of aluminum oxide varies significantly with temperature and is quite nonlinear. Let us look again at Fig. 3.23 and consider conduction through a cylindrical shell having variable thermal conductivity. The heat conduction for our problem is radially outward. The heat flow at location r ¼ r is q ¼ k A dT dr and the area through which heat is flowing at this location is A ¼ 2prL: Hence, at r ¼ r, dT (3.100) dr Rearranging Eq. (3.100) and integrating between the inner and outer surfaces of the cylinder, we have Z ro Z To dr q ¼ 2pL kdT (3.101) r ri Ti Note: q does not vary with r. From heat flow continuity, it is the same at all radial locations. We therefore were able to move it outside the left integral in Eq. (3.101). Continuing the analysis: Z Ti ro q ln kdT (3.102) ¼ 2pL ri To And finally, Z Ti 2pL kdT (3.103) q¼ ro To ln ri To perform the integration, we need to know function kðTÞ: It is often possible to accurately fit thermal conductivity data to a polynomial of second, third, or, at most, fourth degree. Using Excel, it is found that the following third degree polynomial is an excellent fit to the data of Table 3.2: q ¼ kð2prLÞ

k ¼ 4 108 T 3 þ 0:0001 T 2 0:1537 T þ 69:052

Table 3.2 Conductivity of aluminum oxide. T (K)

k (W/m K)

400 600 800 1000 1200 1400

26 15.2 10 8 6.5 5.8

(3.104)

88

Chapter 3 Steady-state conduction

30

25

20

k

(W / m K)

15

10

5

0 400

500

600

700

800

900

1000

Temperature (K)

FIGURE 3.26 Conductivity of aluminum oxide.

Inserting this function and other information into Eq. (3.103), we have Z 773 q 2p ¼ 4 108 T 3 þ 0:0001 T 2 0:1537 T þ 69:052 dT 0:01 L 373 ln 0:005 Performing the integration and calculation, we have the result q ¼ 5:32 104 W=m L It is always good to check a final answer, especially if the calculation involves several steps as in the above integration. So, let us look at Fig. 3.26 and get an average value of k for this problem. The average temperature of the two surfaces is (373 þ 773)/2 ¼ 573 K. Looking at Fig. 3.26, k is about 16 W/m K for this temperature. If we use an average k value, Eq. (3.103) becomes q 2p ¼ kavg ðTi To Þ ro L ln ri

(3.105)

Putting in values for our problem, we have q 2p ¼ ð16Þð773 373Þ ¼ 5.80 104 W=m. L lnð2Þ This approximate solution is only 10% different from our earlier value of 5.32 104 so we are quite confident that no calculation errors were made in our earlier solution.

3.7 Conduction shape factors

89

3.6 Multi-dimensional conduction The above sections discussed one-dimensional conduction. Multi-dimensional problems (i.e., two and three dimensions) can be solved using a variety of methods: classical analytical techniques, conduction shape factor, numerical methods, or packaged software developed specifically for such problems. Classical analytical techniques for solving the partial differential equations of heat transfer have been in existence for many years. Several books have been written on such solutions, including those by Carslaw and Jaeger [1], Schneider [2], Myers [3], and Arpaci [4]. Analytical solutions were very important and necessary before the advent of calculators and computers. They are also important for research and mathematical investigations today. However, analytical solutions are limited in scope and applicability and are often very complex. It is felt that the conduction shape factor and numerical methods are much more useful for students and practicing engineers. Therefore, analytical techniques such as separation of variables are not covered in this book. Shape factors are discussed in the next section, and numerical methods in Chapter 5. Also, several companies have developed software packages for engineering design and heat transfer applications. These include COMSOL [5], Solidworks [6], Ansys [7], and CD-Adapco (acquired by Siemens) [8].

3.7 Conduction shape factors The conduction shape factor is very useful in solving specialized two- and three-dimensional problems involving two temperatures. Examples of applicability include heat transfer from buried pipes and heat transfer through rectangular enclosures such as kilns. The heat transfer equation for this method is q ¼ kSðT1 T2 Þ

(3.106)

where q is the rate of heat transfer between the two surfaces at T1 and T2 k is the thermal conductivity of the medium between T1 and T2 S is the conduction shape factor. 1 . This resistance can be used in solving From Eq. (3.106), the resistance for shape factors is kS problems by means of the resistance concept. It is particularly useful for problems involving multilayered objects or surfaces with convection. Some of the more useful conduction shape factors are given in Table 3.3. More shape factors can be obtained from Refs. [9,10] and [11].

Table 3.3 Conduction shape factors “S.”

90

Shape Factor (1) Cylinder buried in a semi-infinite medium if L[D 2d D S ¼ 2pL if L[D and d > 1.5D 4d ln D S ¼

2pL

T2

d

T1

D

Shape Factor (2) Two Parallel Cylinders in an infinite medium

S ¼ cosh1

2pL

4w2 D21 D22 2D1 D2

if L[w; L[D1 and L[D2

w

T2

T1 D1

D2

Shape Factor (3) Equally-Spaced Parallel Cylinders buried in a semi-infinite medium Sðper cylinderÞ ¼ ln

2pL if L[d; L[D; and w > 1:5D 2w 2pd sinh pD w

T2 T1

d

w

D

Shape Factor (4) Vertical Cylinder buried in a semi-infinite medium S ¼ 2pL if L[D 4L ln D

T2

T1

L

D

Shape Factor (5) Cylinder at center of a square solid bar

Chapter 3 Steady-state conduction

cosh1

S ¼ ln

2pL

1:08w D

T2 T1 D w

Shape Factor (6) Sphere buried in a semi-infinite medium S ¼

T2

2pD

D 4d

1

d

T1

D

Shape Factor (7) Square Flow Channel 2pL

S ¼

0:785 ln

a b

2pL

S ¼

0:93 ln 0:948

T2

if ab < 1:4 a b

if

a b

> 1:4

T1

a

Shape Factor (8) Large Plane Wall S ¼ AL T2 A L

91

T1

3.7 Conduction shape factors

b

Continued

92

Table 3.3 Conduction shape factors “S.”dcont’d Shape Factor (9) Edge of Two Adjoining Walls T2 L T2

T1 L

w

Shape Factor (10) Corner of Three Walls S ¼ 0:15L

L T1 inside

L

L

T2 outside

Notes: For shape factors 1,2,3,5, and 7, L is the length of the object, i.e., the length of the cylinder (s) or flow channel perpendicular to the page. For buried objects, d is the depth into the medium and D is the diameter of the pipe or sphere.

Chapter 3 Steady-state conduction

S ¼ 0:54w

3.7 Conduction shape factors

93

Example 3.9 Heat Flow Through a Kiln Wall Problem: A rectangular electric kiln for pottery has a maximum operating temperature of 1000 C. Its inside dimensions are 60 cm wide by 50 cm deep by 30 cm high. The walls of the kiln are 8 cm thick and of refractory firebrick having a conductivity of 1.2 W/m C. During maximum operation, the inner surface of the kiln is at 700 C and the outer surface is at 375 C. (a) Determine the heat flow through the wall of the kiln (W) for the kiln at its maximum operating level (b) The factory has 240V single-phase electric service. What is the maximum current draw of the kiln (A)? What size wiring would you use for the kiln? Finally, what size circuit breaker would you use in the electric panel?

Solution:

(a) If the kiln were spherical, we could use the earlier derived equation for heat flow through a spherical shell. However, the kiln is rectangular so we have to take into account heat flow at edges and corners of the kiln. The conduction shape factor technique is excellent for this problem. Indeed, there are shape factors in Table 3.3 which cover our situation; namely shape factors no. 8 (wall), 9 (edge), and 10 (corner). The kiln has 6 plane walls, 12 edges, and 8 corners. From Table 3.3, the shape factor calculations are Swalls ¼ 2ð60 50Þ=8 þ 2ð60 30Þ=8 þ 2ð50 30Þ=8 ¼ 1575 cm Sedges ¼ 4 ð0.54 60Þ þ 4ð0.54 50Þ þ 4ð0.54 30Þ ¼ 302.4 cm Scorners ¼ 8 ð0.15Þ ð8Þ ¼ 9.6 cm

Stotal ¼ 1575 þ 302.4 þ 9.6 ¼ 1887 cm ¼ 18.87 m The heat flow rate through the kiln’s wall is q ¼ kStotal DT ¼ 1:2ð18:87Þð700 375Þ ¼ 7360 W Although the shape factor technique is one way to model this problem, there are other possible ways. For example, let us replace the rectangular wall of the kiln with a spherical wall and use the equation for conduction through a spherical wall. The inner surface area of the kiln is 2 (60 50) þ 2 (60 30) ¼ 2 (50 30) ¼ 12,600 cm2. The surface area of a sphere is 4pr2 , so this inner surface area of 1.26 m2 corresponds to a spherical surface with a radius of 0.3167 m. The thickness of the kiln wall is 8 cm. Therefore, let us consider conduction through a spherical wall of inner radius 0.3167 m and outer radius 0.3167 þ 0.08 ¼ 0.3967 m. Conduction through a spherical wall is given by Eq. (3.73). 4pk ðTi To Þ 1 1 ri ro

q¼

(3.73)

Putting in the values for our problem, we have 4pð1:2Þ ð700 375Þ ¼ 7700 W 1 1 0:3167 0:3967 There is only a 4.6% difference from the shape factor result. (b) Power ¼ Voltage Current for single-phase service. For 240 V single-phase service, the 7360 W result from the shape factor technique corresponds to a current of 7360/240 ¼ 30.67 A. From the Internet, AWG #10 copper wire has an ampacity of 30 A, AWG #8 wire has 40 A, and #6 has 55 A. Therefore, #8 AWG wire should be used. Finally, a 40 A circuit breaker would be appropriate for the circuit. q¼

The conduction shape factor is very useful for calculating heat transfer from buried pipes. The following example illustrates this situation.

94

Chapter 3 Steady-state conduction

Example 3.10 Heat Loss from a Buried Pipe Problem: This problem is based on a study made by the author of heat loss from a buried hot water pipe at a dog kennel in New York. Water at 190 F flows at a rate of 5 gpm through a buried one-inch Schedule 40 steel pipe from one building to another. The pipe is horizontal and uninsulated. It is 60 feet long and is buried at a depth of 2 feet. The soil has a conductivity of 0.833 BTU/h ft F, and the ground surface temperature is 35 F. Determine the rate of heat loss from the pipe (BTU/h).

Solution: The outside diameter of 1 inch Sch 40 pipe is 1.315 inch. We will assume that the outer surface of the pipe is at the water temperature of 190 F. This is a reasonable assumption as the pipe is metal, relatively thin-walled, and the convective coefficient at the water/pipe interface is quite large. The appropriate shape factor is #1 from Table 3.3. 2pL S¼ 4d ln D

(3.107)

Putting in the values for this problem, we have 2pð60Þ ¼ 87:87 ft S¼ 4x2 ln 1:315= 12 (Note: We could have used the other equation in Table 3.3 to get the shape factor. The value for S would have been the same.) The rate of heat loss from the pipe is q ¼ kSDT ¼ ð0.833Þ ð87.87Þ ð190 35Þ ¼ 11340 Btu=h (Note: The flow rate of 5 gpm was extraneous information not used in the solution.)

3.8 Extended surfaces (fins) This section gives an introduction to the topic of extended surfaces or fins. Much research has been done on this topic, and references [12e14] may be consulted for more detailed information. As shown in Chapter 1, the heat transfer rate is proportional to heat flow area for all three modes of heat transfer. One way of increasing heat transfer from a surface is to add extended surfaces (or “fins”) to the surface. This increases the heat transfer area and, in turn, the rate of heat transfer. Fins are used on a variety of devices. Some applications include coils for air conditioning and refrigeration; baseboard heaters for residential and commercial buildings; heat sinks for electronic devices, car radiators, and engine and air compressor cylinders. Some examples of these are shown in Fig. 3.27: Let us first consider fins of constant cross-sectional area. We will derive the governing differential equation for the temperature distribution in the fin and discuss the appropriate boundary condition equations. We will see how the addition of fins enhances the heat transfer from a surface. Common fins of varying cross-sectional area will then be discussed. Examples will be given for a single fin on a surface and also for multiple fins on a surface.

3.8 Extended surfaces (fins)

95

FIGURE 3.27 Some devices with fins. (A) Power transformer with straight fins. (B) Reciprocating air compressor with circumferential fins. (C) Baseboard heating tube with circumferential fins. (D) Heat sink with pin fins on a circuit board. (E) Heat sink with fan and straight fins. (F) Heat sink for computer video card. Many thanks to Slant Fin Corporation of Greenvale, NY for providing Figure 3.27 (c).

96

Chapter 3 Steady-state conduction

3.8.1 Fins of constant cross section 3.8.1.1 The governing differential equation and boundary conditions Let us consider the fin of square cross section shown in Fig. 3.28. The cross section is of dimension a x a and the fin has a length L. It is attached to a surface that is at temperature Ts . The surrounding fluid is at temperature TN and the convective coefficient at the side surfaces and end of the fin is h. Finally, the heat transfer is steady state. Our analytical model is one-dimensional. This assumes that the temperature in the fin only depends on x, the distance from the surface on which the fin is attached. This 1D model is usually very acceptable as fins are normally thin and of high thermal conductivity. To derive the equation for the temperature variation TðxÞ of the fin, let us consider the heat flow in and out of a differential vertical slice of the fin. The slice (or “celement”) is from x¼x to x¼xþdx and has a thickness dx. There is conduction qx entering the left side of the element and conduction qxþdx leaving the right side. There is also convection qconv from the element to the adjacent fluid. For steady state, the rate of heat flow into the element equals the rate of heat flow out of the element or qx ¼ qxþdx þ qconv However; qx ¼ kA qxþdx ¼ qx þ and

(3.108) dT dx

(3.109)

dqx dx dx

(3.110)

qconv ¼ hAðT TN Þ

(3.111)

Note that the A’s in Eqs. (3.109) and (3.111) are different. In Eqs. (3.109), A is the area for conduction heat transfer or a2 for our square fin. In Eqs. (3.111), A is the surface area for convection or P dx, where P is the perimeter around the fin. For our square fin of side “a," the perimeter is 4 a.

Fluid at T∞

Wall at Ts

h a

x

FIGURE 3.28 Single fin of square cross section.

L

a

3.8 Extended surfaces (fins)

Putting Eqs. (3.109e3.111) into the main Eq. (3.108) and rearranging, we have d dT kA hPðT TN Þ ¼ 0 dx dx

97

(3.112)

For our fin, the cross-sectional area is constant. We will also assume that conductivity k is constant. Eq. (3.112) then becomes

We will assume that TN

d2 T hP ðT TN Þ ¼ 0 dx2 kA is constant. Then Eq. (3.113) can be changed to

(3.113)

d2 ðT TN Þ hP ðT TN Þ ¼ 0 (3.114) dx2 kA This is the differential equation to be solved for the temperature distribution TðxÞ in the fin. The general solution for Eq. (3.114) is TðxÞ TN ¼ C1 emx þ C2 emx (3.115) qﬃﬃﬃﬃﬃ hP where m ¼ kA. We need two boundary conditions to determine constants C1 and C2 in Eq. (3.115). The boundaries are at the wall (x ¼ 0) and at the end of the fin (x ¼ L). The temperature at the wall is Ts so the boundary condition there is At x ¼ 0;

T TN ¼ Ts TN

(3.116)

There is convection from the fin to the fluid at the x ¼ L end of the fin. At this location, the heat flow by conduction equals the heat flow by convection. The boundary condition is therefore dT ¼ hAðT TN Þ dx The A’s are the same for conduction and convection at the end of the fin, and the fluid temperature TN is constant. Hence, the boundary condition at the end of the fin is At x ¼ L;

At x ¼ L;

kA

dðT TN Þ h ¼ ðT TN Þ dx k

(3.117)

3.8.1.2 The solution for temperature distribution and heat flow It can be shown that the solution of differential Eq. (3.114) with boundary condition Eqs. (3.116) and (3.117) is h cosh½mðL xÞ þ sinh½mðL xÞ mk TðxÞ TN ¼ ðTs TN Þ (3.118) h sinhðmLÞ coshðmLÞ þ mk qﬃﬃﬃﬃﬃ where m ¼ hP kA.

98

Chapter 3 Steady-state conduction

Eq. (3.118) gives the temperature distribution along the total length of the fin, i.e., from x ¼ 0 to x ¼ L. However, the objective in most situations is to determine the heat transfer from the fin to the fluid. This can be done in two different ways. One way involves differentiation of the TðxÞ function of Eq. (3.118). The other way involves integration of the TðxÞ function. For the first way, we consider the heat transfer into the fin at location x ¼ 0. This is conduction heat transfer and is given by dT q ¼ kA (3.119) dx x¼0 The temperature function of Eq. (3.118) is differentiated with respect to x and the result is evaluated at x ¼ 0. The result is then used in Eq. (3.119) to obtain the heat flow into the fin. In the second way, we look at the convection heat transfer from the fin’s surfaces into the fluid. In so doing, we integrate the function of Eq. (3.118) to get the heat transfer from the sides of the fin into the fluid and use the temperature at the x ¼ L end of the fin to get the heat transfer from the end of the fin into the fluid. The total convection heat transfer to the fluid is given by ZL q¼

hPðT TN Þdx þ hAðTx¼L TN Þ

(3.120)

0

As we have steady-state conditions, the heat flow into the fin at x ¼ 0 equals the heat flow into the fluid by convection from the fin’s surfaces. That is, the q’s in Eqs. (3.119) and (3.120) are equal. Regardless of whether the first or second approach is used to get the heat flow, the result is h sinhðmLÞ þ coshðmLÞ pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ mk q ¼ h P k A ðTs TN Þ (3.121) h sinhðmLÞ coshðmLÞ þ mk qﬃﬃﬃﬃﬃ hP where m ¼ kA. The equations for the temperature distribution in the fin (Eq. 3.118) and for the heat flow from the fin (Eq. 3.121) involve hyperbolic trigonometric functions. These functions are readily available on scientific calculators and in software packages. However, such was not always the case. In years past, evaluation of the equations was very tedious and time-consuming as slide rules and tables had to be used to evaluate the exponential and trigonometric functions. To speed up calculations, two frequently used approximations were developed. These are discussed in the next two sections. Indeed, they will speed up hand calculations. However, one must keep in mind that they are approximations and should be used only in situations consistent with the assumptions made in their derivation.

3.8.1.3 Very-long-fin approximation The first approximation assumes that the fin is very long. This means that the temperature at the end of the fin approaches the fluid temperature. Mathematically, the boundary condition at the end of the fin is As

x / N; T/TN

(3.122)

3.8 Extended surfaces (fins)

99

The previous differential equation (Eq. 3.114) and boundary condition equation at the base (Eq. 3.116) remain applicable. The solution for the temperature distribution in the very long fin is rﬃﬃﬃﬃﬃﬃ hP mx TðxÞ TN ¼ ðTs TN Þe where m ¼ (3.123) kA and the solution for the heat transfer to the fluid is pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ q ¼ hPkAðTs TN Þ (3.124)

3.8.1.4 Insulated-at-end fin approximation The other approximation assumes that there is negligible heat transfer from the tip of the fin. This is often an excellent approximation as most fins have a lateral area much larger than the tip area. With no heat flow from the tip, the boundary condition equation at the end of the fin is q ¼ 0 ¼ kA

dT dx

which simplifies to dT ¼ 0 (3.125) dx The differential equation for the temperature distribution in the fin (Eq. 3.114) and boundary condition equation at the base (Eq. 3.116) remain applicable. The solution for the temperature distribution in the insulated-at-end fin is At x ¼ L;

TðxÞ TN ¼ ðTs TN Þ

cosh½mðL xÞ coshðmLÞ

(3.126)

and the solution for the heat transfer to the fluid is pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ q ¼ hPkAðTs TN ÞtanhðmLÞ (3.127) rﬃﬃﬃﬃﬃﬃ hP As before; m ¼ kA As mentioned above, this approximation often gives excellent results. To make the results even better, the lateral area can be increased to compensate for the neglected heat transfer area at the tip. The lateral area is increased by adding an amount equal to A=P to the length of the fin. This “corrected” length Lc is used instead of L in Eqs. (3.126) and (3.127). For example, consider a pin fin (i.e., a cylindrical fin of circular cross section) of length L and diameter D. The corrected length is pD2 =4 D ¼Lþ pD 4 Similarly, a fin of square cross section of side a will have a corrected length Lc ¼ L þ A=P ¼ L þ

Lc ¼ L þ A=P ¼ L þ

a2 a ¼Lþ 4 4a

100

Chapter 3 Steady-state conduction

Finally, a thin fin of rectangular cross section with width w and thickness t with t 0:2 qi Ti TN From Eq. (4.24), the temperature at the center plane of the plate, i.e., at x ¼ 0; is 2 qo Tð0; tÞ TN ¼ ¼ C1 el1 Fo for Fo > 0.2 qi Ti TN

(4.24)

(4.25)

If we are interested in an earlier time of the system’s response, that is, when Fo ¼ Lat2 0:2, we can just use more terms of the infinite series for the temperature.

4.3 Systems with spatial variation (large plate, long cylinder, sphere)

131

Table 4.1 Coefficients for the one-term approximation. Plate Bi 0.01 0.02 0.04 0.06 0.08 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.5 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0 15.0 20.0 30.0 40.0 50.0 100.0 N

l1 0.0998 0.1410 0.1987 0.2425 0.2791 0.3111 0.4328 0.5218 0.5932 0.6533 0.7051 0.7506 0.7910 0.8274 0.8603 0.9883 1.0769 1.1925 1.2646 1.3138 1.3496 1.3766 1.3978 1.4149 1.4289 1.4729 1.4961 1.5202 1.5325 1.5400 1.5552 1.5708

Cylinder C1 1.0017 1.0033 1.0066 1.0098 1.0130 1.0161 1.0311 1.0450 1.0580 1.0701 1.0814 1.0918 1.1016 1.1107 1.1191 1.1537 1.1785 1.2102 1.2287 1.2403 1.2479 1.2532 1.2570 1.2598 1.2620 1.2676 1.2699 1.2717 1.2723 1.2727 1.2731 1.2732

l1 0.1412 0.1995 0.2814 0.3438 0.3960 0.4417 0.6170 0.7465 0.8516 0.9408 1.0184 1.0873 1.1490 1.2048 1.2558 1.4570 1.5995 1.7887 1.9081 1.9898 2.0490 2.0937 2.1286 2.1566 2.1795 2.2509 2.2880 2.3261 2.3455 2.3572 2.3809 2.4048

/

Sphere C1

l1

C1

1.0025 1.0050 1.0099 1.0148 1.0197 1.0246 1.0483 1.0712 1.0931 1.1143 1.1345 1.1539 1.1724 1.1902 1.2071 1.2807 1.3384 1.4191 1.4698 1.5029 1.5253 1.5411 1.5526 1.5611 1.5677 1.5800 1.5919 1.5973 1.5993 1.6002 1.6015 1.6021

0.1730 0.2445 0.3450 0.4217 0.4860 0.5423 0.7593 0.9208 1.0528 1.1656 1.2644 1.3525 1.4320 1.5044 1.5708 1.8364 2.0288 2.2889 2.4556 2.5704 2.6537 2.7165 2.7654 2.8044 2.8363 2.9349 2.9857 3.0372 3.0632 3.0788 3.1102 3.1416

1.0030 1.0060 1.0120 1.0179 1.0239 1.0298 1.0592 1.0880 1.1164 1.1441 1.1713 1.1978 1.2236 1.2488 1.2732 1.3849 1.4793 1.6227 1.7202 1.7870 1.8338 1.8673 1.8920 1.9106 1.9249 1.9630 1.9781 1.9898 1.9942 1.9962 1.9990 2.0000

The l1 and C1 values are given for different Biot numbers in Table 4.1. If more than one term is used in the series solution, then the ln can be obtained through solution of Eq. (4.22) and the Cn can then be obtained from Eq. (4.23). Before continuing, let us look at Table 4.1. It is seen that there are entries for a Biot number of N. As Bi is h L = k for a plate, and Bi is h ro = k for a cylinder or sphere; an infinite Biot number

132

Chapter 4 Unsteady conduction

means that the convective coefficient is infinite. Infinite h means there is zero convective resistance at the surface. So, the N entry in the table corresponds to a temperature boundary condition. At time zero, a constant temperature is imposed on the two surfaces of the plate (or the surface of the cylinder or the surface of the sphere). This temperature is used in place of TN in the above equations. We are often interested in the amount of energy that is transferred from (or to) the surroundings during a time period of the process. If a plate is initially uniform at Ti and the process is fully completed, i.e., the plate is finally uniform at TN , the decrease in internal energy of the plate is Qmax ¼ rcVðTi TN Þ

(4.26)

where r ¼ density of the plate. c ¼ specific heat of the plate. V ¼ volume of the plate. If the process is only partially completed, the decrease in internal energy is Z Q ¼ rc ðTi TÞ dV

(4.27)

V

The decrease in internal energy given in Eq. (4.27) is equal to the energy going into the surrounding medium by convection. For the plane plate, dV ¼ A dx, where A is the cross sectional area of the plate. Therefore, Eq. (4.27) becomes ZL Q ¼ rcA

½Ti Tðx; tÞ dx

(4.28)

L

As the temperature distribution is symmetrical about the center plane, Eq. (4.28) is equivalent to ZL ½Ti Tðx; tÞ dx

Q ¼ 2rcA

(4.29)

0

The temperature in the plate at time t is given by Eq. (4.20). Putting this into Eq. (4.29), and remembering that Fo ¼ at L2 , we have # ZL " N X l2n Fo Q ¼ 2rcA Cn e cosðln x = LÞ dx (4.30) Ti TN ðTi TN Þ 0

n¼1

Rearranging and recognizing that we are integrating only over x and not t, Eq. (4.30) becomes 2 L 3 Z ZL N X 2 Q ¼ 2rcA 4 ðTi TN Þ dx ðTi TN Þ Cn eln Fo cosðln x = LÞ dx5 (4.31) 0

n¼1

0

4.3 Systems with spatial variation (large plate, long cylinder, sphere)

Performing the integrations and factoring out ðTi TN Þ, we get " # N X l2n Fo sin ln Cn e Q ¼ 2rcAL ðTi TN Þ 1 ln n¼1

133

(4.32)

If we divide this result by Qmax and recognize that the volume of the plate is 2LA, we get the final dimensionless result: N X 2 Q ¼ 1 Cn eln Qmax n¼1

Fo

sin ln ln

(4.33)

where Qmax ¼ 2rcLA ðTi TN Þ For the one-term approximation, we have 2 Q sin l1 ¼ 1 C1 el1 Fo Qmax l1

(4.34)

Example 4.2 Cooling of a large plate Problem

The large stainless steel plate (r ¼ 7900 kg/m3, c ¼ 580 J/kg C, k ¼ 23 W/m C) shown below is 30 cm thick. It is initially in an oven and at a uniform temperature of 800 C. It is taken out of the oven, and it starts cooling to the 25 C room air. The convective coefficient is 75 W/m2 C. (a) Can the lumped method be used for this problem? (b) How long does it take for the center of the plate to reach 300 C? (c) What is the surface temperature of the plate at that time? (d) During the cooling, how much heat goes into the room air? If your answer is “no” to item (a), you may use the one-term approximation method for items (b), (c), and (d).

h = 75 W / m2 C (All surfaces) Air at 25 C Ti = 800 C 30 cm

Solution

(a) We first calculate Bilumped to see if we can use the lumped method.

hðV=AÞ ð75Þð0:15Þ ¼ 0:489 ¼ 23 k As Bilumped > 0:1 the lumped method is inappropriate. We will consider the spatial distribution of temperatures in the plate and use the one-term approximation method. Bilumped ¼

134

Chapter 4 Unsteady conduction

(b) For the center temperature, we use Eq. (4.25): 2 qo Tð0; tÞ TN ¼ ¼ C1 el1 Fo for Fo > 0.2 qi Ti TN

(4.25)

hL ð75Þð0:15Þ ¼ ¼ 0:489 k 23 For Bi ¼ 0.489, linear interpolation of entries in Table 4.1 gives l1 ¼ 0:6467 and C1 ¼ 1:0688: Bi ¼

23 ¼ 5:02 106 Therefore, a ¼ ð7900Þð580Þ Putting the values in Eq. (4.25), we have

at k Fo ¼ 2 and a ¼ L rc 6 m2 s and Fo ¼ ð5:02102 Þt ¼ 2:231 104 t. ð0:15Þ

2 4 300 25 ¼ ð1:0688Þe0:6467 ð2:231 10 Þt 800 25

5

or e9:3305 10 t ¼ 0:3320 Taking the natural log of both sides, we get 9:3305 105 t ¼ lnð0:3320Þ ¼ 1:1026 and arrive at the answer t [ 11817 s. Let us just check Fo and make sure that Fo is greater than 0.2, which is the criterion for use of the one-term approximation method: Fo ¼ 2.231 104 t ¼ 2.231 104 ð11817Þ ¼ 2.636 As Fo > 0.2, our use of the one-term approximation was fine. (c) To get the surface temperature, we use Eq. (4.24): 2 q Tðx; tÞ TN ¼ ¼ C1 el1 Fo cosðl1 x = LÞ for Fo > 0:2 qi Ti TN

(4.24)

2 TðL ¼ 0:15m; 11817sÞ 25 ¼ 1:0688 e0:6467 ð2:636Þ cosð0:6467Þ 800 25

T ¼ 244.5 C The surface temperature after 11817 s is 244.5 C. (d) Using Eq. (4.34), we have 2 Q sin l1 ¼ 1 C1 el1 Fo Qmax l1

(4.34)

Q 2 sinð0:6467Þ ¼ 0:6693 ¼ 1 1:0688e0:6467 ð2:636Þ Qmax 0:6467 Qmax ¼ rcVðTi TN Þ ¼ rcð2LAÞðTi TN Þ ¼ ð7900Þð580Þð0:3Þð800 25ÞA ¼ 1:065 109 A where A is the cross-sectional area of the plate. Therefore, Q=A ¼ 0.6693ð1.065 109Þ ¼ 7.13 108 J m2 . During the process, 7.13 3 108 J goes into the surrounding fluid per square meter of cross-sectional area.

4.3.3 Long cylinders From Chapter 2, the general heat conduction equation in cylindrical coordinates is 1 v vT 1 v vT v vT vT kr þ 2 k þ k þ qgen ¼ r c r vr vr r v4 v4 vz vz vt

(4.35)

4.3 Systems with spatial variation (large plate, long cylinder, sphere)

135

Consider the cylinder of Fig. 4.3 above. The cylinder has no internal heat generation and the conductivity is constant. The temperature distribution depends only on r. Eq. (4.35) then reduces to the one-dimensional transient equation. v2 T 1 vT 1 vT ¼ þ vr 2 r vr a vt

(4.36)

k where a ¼ thermal diffusivity ¼ rc All points in the cylinder are initially at temperature Ti . Then, at time zero, the surface of the cylinder at r ¼ ro is suddenly subjected to convection to the environment at TN with a convective coefficient h. The temperature solution is symmetrical about r ¼ 0. Hence, the boundary condition equation at the centerline of the cylinder is

vT ¼0 vr At the surface of the cylinder, the convective boundary condition equation is At r ¼ 0;

At r ¼ ro ;

k

vT ¼ hðT TN Þ vr

The initial condition is: At t ¼ 0; TðrÞ ¼ Ti

(4.37)

(4.38) (4.39)

The solution of Eq. (4.36) with conditions (4.37) through (4.39) is an infinite series: Tðr; tÞ ¼ TN þ ðTi TN Þ

N X

Cn e

l2 n at ro2

J0 ðln r = ro Þ

(4.40)

n¼1

This equation gives the temperature at any point in the cylinder at any time. This solution given by Eq. (4.40) is often written in dimensionless form: N 2 q Tðr; tÞ TN X ¼ ¼ Cn eln Fo J0 ðln r = ro Þ qi Ti TN n¼1

Fo ¼

(4.41)

at ¼ Fourier Number ro2

The eigenvalues ln are the positive roots of the transcendental equation ln

J1 ðln Þ ¼ Bi J0 ðln Þ

(4.42)

where Bi is the Biot number. For cylinders, Bi ¼ hrko . Note: J0 and J1 are Bessel functions of the first kind. The functions are available in software such as Excel and Matlab, and a table of the functions is in Appendix G. Among other methods, Eq. (4.42) may be solved for the ln through use of the Goal Seek or Solver features of Excel or the fzero function of Matlab. See Appendix J for details.

136

Chapter 4 Unsteady conduction

Once the ln values are determined, the Cn values in Eqs. (4.40) and (4.41) may be obtained from Cn ¼

2 J1 ðln Þ ln J20 ðln Þ þ J21 ðln Þ

(4.43)

The series in Eqs. (4.40) and (4.41) converges rapidly with increasing time due to the exponential term. If the Fourier number is greater than 0.2, excellent results are obtained using just the first term of the series. This is called the one-term approximation. Keeping just the first term, Eq. (4.41) becomes 2 q Tðr; tÞ TN ¼ ¼ C1 el1 Fo J0 ðl1 r = ro Þ for Fo > 0.2 qi Ti TN

(4.44)

From Eq. (4.44), the temperature on the centerline of the cylinder, i.e., at r ¼ 0 is 2 qo Tð0; tÞ TN ¼ ¼ C1 el1 Fo for Fo > 0.2 qi Ti TN

(4.45)

If we are interested in an earlier time of the system’s response, that is, when Fo ¼ at ro2 0:2, we can just use more terms of the infinite series for the temperature. The l1 and C1 values are given for different Biot numbers in Table 4.1. If more than one term is used in the series solution, then the ln can be obtained through solution of Eq. (4.42) and the Cn can then be obtained from Eq. (4.43). We discussed above the amount of energy transferred from a plate to the surroundings during a time period of the process. We will now do this for a cylinder: If the cylinder is initially uniform at Ti and the process is fully completed, i.e., the cylinder is finally uniform at TN , the decrease in internal energy of the cylinder is Qmax ¼ rcVðTi TN Þ where r ¼ density of the cylinder. c ¼ specific heat of the cylinder. V ¼ volume of the cylinder ¼ pro2 L, where L is the length of the cylinder. If the process is only partially completed, the decrease in internal energy is Z Q ¼ rc ðTi TÞdV

(4.46)

(4.47)

V

The decrease in internal energy given in Eq. (4.47) is equal to the energy going into the surrounding medium by convection. For the cylinder, dV ¼ 2prL dr, where L is the length of the cylinder. Therefore, Eq. (4.47) becomes Zro Q ¼ 2prcL

½Ti Tðr; tÞrdr 0

(4.48)

4.3 Systems with spatial variation (large plate, long cylinder, sphere)

137

The temperature in the cylinder at time t is given by Eq. (4.40). Putting this into Eq. (4.48), and remembering that Fo ¼ at ro2 , we have # Zro " N X 2 Q ¼ 2prcL Cn eln Fo J0 ðln r = ro Þ rdr (4.49) Ti TN ðTi TN Þ n¼1

0

Rearranging and recognizing that we are integrating only over r and not t, Eq. (4.49) becomes 2 r 3 Zro Zo N X 2 Q ¼ 2prcL4 ðTi TN Þrdr ðTi TN Þ Cn eln Fo J0 ðln r = ro Þ rdr 5 (4.50) n¼1

0

0

Performing the integrations and then factoring out ðTi TN Þ and ro2, we get " # N X 2 l2n Fo J1 ðln Þ Cn e Q ¼ rcLpro ðTi TN Þ 1 2 ln n¼1

(4.51)

If we divide this result by Qmax from Eq. (4.46) and recognize that the volume of the cylinder is pro2 L, we get the final dimensionless result: N X 2 Q J1 ðln Þ ¼1 2 Cn eln Fo Qmax ln n¼1

(4.52)

For the one-term approximation, we have 2 Q J1 ðl1 Þ ¼ 1 2C1 el1 Fo Qmax l1

(4.53)

where Qmax ¼ rcpro2 LðTi TN Þ.

4.3.4 Spheres From Chapter 2, the general heat conduction equation in spherical coordinates is 1 v 1 v vT 1 v vT vT 2 vT kr þ 2 2 k þ 2 k sin q þ qgen ¼ rc r 2 vr vr v4 r sin q vq vq vt r sin q v4

(4.54)

Consider the sphere of Fig. 4.4 above. The sphere has no internal heat generation and the conductivity is constant. The temperature distribution depends only on r. Eq. (4.54) then reduces to the one-dimensional transient equation v2 T 2 vT 1 vT ¼ þ vr 2 r vr a vt

(4.55)

k. where a ¼ thermal diffusivity ¼ rc All points in the sphere are initially at temperature Ti . Then, at time zero, the surface of the sphere at r ¼ ro is suddenly subjected to convection to the environment at TN with a convective coefficient h.

138

Chapter 4 Unsteady conduction

The temperature solution is symmetrical about r ¼ 0. Hence, the boundary condition equation at the center of the sphere is: vT ¼0 vr At the surface of the sphere, the convective boundary condition equation is At r ¼ 0;

vT ¼ hðT TN Þ vr The initial condition is: At t ¼ 0; TðrÞ ¼ Ti At r ¼ ro ; k

(4.56)

(4.57) (4.58)

The solution of Eq. (4.55) with conditions (4.56) through (4.58) is an infinite series: Tðr; tÞ ¼ TN þ ðTi TN Þ

N X

Cn e

n¼1

l2 n at ro2

sinðln r=ro Þ ln r=ro

(4.59)

This equation gives the temperature at any point in the sphere at any time. This solution given by Eq. (4.59) is often written in dimensionless form: N 2 q Tðr; tÞ TN X sinðln r=ro Þ ¼ ¼ Cn eln Fo qi ln r=ro Ti TN n¼1

Fo ¼

(4.60)

at ¼ Fourier Number ro2

The eigenvalues ln are the positive roots of the transcendental equation 1 ln cotln ¼ Bi

(4.61)

where Bi is the Biot number. For spheres, Bi ¼ Among other methods, Eq. (4.61) may be solved for the ln through use of the Goal Seek or Solver features of Excel or the fzero function of Matlab. See Appendix J for details. Once the ln values are determined, the Cn values in Eqs. (4.59) and (4.60) may be obtained from h ro k .

Cn ¼

4ðsin ln ln cos ln Þ 2ln sinð2ln Þ

(4.62)

Note: The arguments of the trig functions are in radians. The series in Eqs. (4.59) and (4.60) converge rapidly with increasing time due to the exponential term. If the Fourier number is greater than 0.2, excellent results are obtained using just the first term of the series. This is called the one-term approximation. Keeping just the first term, Eq. (4.60) becomes 2 q Tðr; tÞ TN sinðl1 r=ro Þ ¼ ¼ C1 el1 Fo for Fo > 0.2 qi l1 r=ro Ti TN

As r goes to 0,

sinðl1 r=ro Þ l1 r=ro goes

(4.63)

to 1. Therefore, the temperature at the center of the sphere is

2 qo Tð0; tÞ TN ¼ ¼ C1 el1 Fo for Fo > 0.2 qi Ti TN

(4.64)

4.3 Systems with spatial variation (large plate, long cylinder, sphere)

139

If we are interested in an earlier time of the system’s response, that is, when Fo ¼ at ro2 0:2, we can just use more terms of the infinite series for the temperature. The l1 and C1 values are given for different Biot numbers in Table 4.1. If more than one term is used in the series solution, then the ln can be obtained through solution of Eq. (4.61) and the Cn can then be obtained from Eq. (4.62). Regarding the amount of energy transferred from a sphere to the surroundings during a time period of the process: If the sphere is initially uniform at Ti and the process is fully completed, i.e., the sphere is finally uniform at TN , the decrease in internal energy of the sphere is Qmax ¼ rcVðTi TN Þ

(4.65)

where r ¼ density of the sphere. c ¼ specific heat of the sphere. V ¼ volume of the sphere ¼ 43 pro3 . If the process is only partially completed, the decrease in internal energy is Z Q ¼ rc ðTi TÞdV

(4.66)

V

The decrease in internal energy given in Eq. (4.66) is equal to the energy going into the surrounding medium by convection. For the sphere, dV ¼ 4pr 2 dr. Therefore, Eq. (4.66) becomes Zro Q ¼ 4prc

½Ti Tðr; tÞ r 2 dr

(4.67)

0

The temperature in the sphere at time t is given by Eq. (4.59). Putting this into Eq. (4.67), and remembering that Fo ¼ at ro2 , we have # Zro " N X l2n Fo sinðln r=ro Þ Q ¼ 4prc Cn e (4.68) Ti TN ðTi TN Þ r 2 dr l r=r n o n¼1 0

Rearranging and recognizing that we are integrating only over r and not t, Eq. (4.68) becomes 2 r 3 Zro Zo N X 2 sinðl r=r Þ n o Q ¼ 4prc4 ðTi TN Þr 2 dr ðTi TN Þ Cn eln Fo r 2 dr 5 (4.69) ln r=ro n¼1 0

0

Performing the integrations and then rearranging the equation, we get " # N X 4 3 l2n Fo sin ln ln cos ln Q ¼ rc pro ðTi TN Þ 1 3 Cn e 3 l3n n¼1

(4.70)

140

Chapter 4 Unsteady conduction

If we divide this result by Qmax from Eq. (4.65) and recognize that the volume of the sphere is 43 pro3 , we get the final dimensionless result: N X 2 Q sin ln ln cos ln ¼1 3 Cn eln Fo Qmax l3n n¼1

(4.71)

For the one-term approximation, we have 2 Q sin l1 l1 cos l1 ¼ 1 3C1 el1 Fo Qmax l31

where Qmax ¼ rc

4 pr 3 o 3

(4.72)

ðTi TN Þ.

Example 4.3 Cooling of a glass sphere Problem

A glass sphere (r ¼ 2800 kg/m3, c ¼ 800 J/kg C, k ¼ 0.8 W/m C) is shown in the figure below. It has a diameter of 2 cm and is initially at a uniform temperature of 200 C. The sphere then begins cooling to the surrounding fluid that is at 20 C. The convective coefficient is 20 W/m2 C. (a) What is the temperature of the center of the sphere after 10 min? (b) What is the temperature of the surface of the sphere after 10 min?

h = 20 W / m2 C

Ti = 200 C

Fluid at 20 C 2 cm

Solution The first thing to do is to calculate Bilumped to see if the lumped method can be used. ! 4 pr 3 o 3 h 4pro2 hðV=AÞ hro ð20Þð0:01Þ ¼ ¼ ¼ 0:0833 ¼ Bilumped ¼ k k 3ð0:8Þ 3k As Bilumped is less than 0.1, the criterion for use of the lumped method is met. However, we are quite close to 0.1, so let us solve the problem in both waysdwith the lumped method and also with the Heisler method. Doing this, we will be able to see the amount of spatial variation of temperature in the sphere. Lumped method. We will use Eq. (4.8): TðtÞ ¼ TN þ ðTi TN Þe 4pro2

hA rcV

t

A 3 3 ¼ ¼ 300 ¼ ¼ V 4 3 ro 0:01 pro ð20Þð300Þð600Þ 3 so, Eq. (4.8) becomes T ¼ 20 þ 180 e ð2800Þð800Þ ¼ 20 þ 180 e1:6071 ¼ 56.1 C. Using the lumped method, the temperature of the sphere after 600 s is 56.1 C. t ¼ 10 minutes ¼ 600s;

(4.8)

4.3 Systems with spatial variation (large plate, long cylinder, sphere)

141

Heisler method (a) We will first check the Fourier number to see if the one-term approximation can be used. at k 0:8 ¼ 3:571 107 m2 =s a¼ ¼ ro2 rc ð2800Þð800Þ ¼ 2:143 > 0:2, so the one-term approximation is OK. Fo ¼

7

Fo ¼ 3:571 10 2 ð600Þ ð0:01Þ

Eq. (4.64) gives the temperature at the center of the sphere: 2 qo Tð0; tÞ TN ¼ ¼ C1 el1 Fo (4.64) qi Ti TN Eigenvalue l1 can be obtained from Table 4.1 or by solving the transcendental equation given in Eq. (4.61). Both methods require the Biot number.

hro ð20Þð0:01Þ ¼ 0:25 ¼ 0:8 k From linear interpolation of Table 4.1 for this Biot number, l1 ¼ 0:8401 and C1 ¼ 1:0736. Putting values in Eq. (4.64), we get Bi ¼

Tð0; 600 sÞ ¼ 20 þ ð200 20Þð1:0736Þe0:8401 The temperature of the center of the sphere after 600 s is 61.9 C. (b) Eq. (4.63) is used to get the surface temperature.

2

ð2:143Þ

¼ 62.6 C.

2 q Tðr; tÞ TN sinðl1 r=ro Þ ¼ ¼ C1 el1 Fo qi l1 r=ro Ti TN At the surface, r ¼ ro : Putting values into Eq. (4.63), we have

Tðro ; 600 sÞ ¼ 20 þ ð200 20Þð1:0736Þe0:8401

2

(4.63)

ð2:143Þ sinð0:8401Þ

¼ 57.8 C. 0:8401 The temperature of the surface of the sphere after 600 s is 57.8 C. For this problem, the value of Bilumped met the 0.1 criterion for use of the lumped method. However, we see that there is still some variation of temperature within the sphere. We conclude that in cases where theBilumped meets the 0.1 criterion for use of the lumped method, but where Bilumped is fairly close to 0.1, it is prudent to also perform a Heisler method solution. We will now do a problem in which the Fourier number is less than 0.2 and multiple terms of the series solutions must be included for accurate results.

Example 4.4 Cooling of a large sphere Problem Note: This example is the same as Example 4.3 except the sphere is considerably larger. Its diameter is 10 cm rather than 2 cm. A glass sphere (r ¼ 2800 kg/m3, c ¼ 800 J/kg C, k ¼ 0.8 W/m C) has a diameter of 10 cm and is initially at a uniform temperature of 200 C. The sphere then begins cooling to the surrounding fluid, which is at 20 C. The convective coefficient is 20 W/m2 C. (a) What is the temperature of the center of the sphere after 10 min? (b) What is the temperature of the surface of the sphere after 10 min?

Solution We first check Bilumped to see if the lumped method can be used. hro ð20Þð0:05Þ ¼ 0:417 ¼ 3ð0:8Þ 3k This is considerably greater than 0.1 so the lumped method is inappropriate. We will use the Heisler method. Bilumped ¼

142

Chapter 4 Unsteady conduction

Let us check the Fourier number to see if the one-term approximation can be used. (From Example 4.3, a ¼ 3:571 x 107 m = s2 :

3:571 107 ð600Þ at ¼ 0:0857 Fo ¼ 2 ¼ 2 ro 0:05 As Fo < 0.2, we are too early in the process to use the one-term approximation. We should use more than one term in the infinite series. (a) The solution for the temperatures in the sphere is given by Eq. (4.59) N X

l2 n at ro2

sinðln r=ro Þ ln r=ro n¼1 n r=ro Þ As r goes to 0, sinðl ln r=ro goes to 1. Therefore, for the center temperature, we have Tðr; tÞ ¼ TN þ ðTi TN Þ

Cn e

Tð0; tÞ ¼ TN þ ðTi TN Þ

N X

Cn e

l2 n at ro2

(4.73)

(4.74)

n¼1

Let us arbitrarily keep four terms of the series. Also, Fo ¼ at ro2 . Then Eq. (4.74) becomes h i 2 2 2 2 Tð0; tÞ ¼ TN þ ðTi TN Þ C1 el1 Fo þ C2 el2 Fo þ C3 el3 Fo þ C4 el4 Fo

(4.75)

The values of ln are from the solution of the transcendental Eq. (4.61). Once we have the ln , we get the Cn from Eq. (4.62). ¼ 1:25. For this Biot number, the ln values are 1.7155, 4.7648, 7.8857, and The Biot number is Bi ¼ hrko ¼ ð20Þð0:05Þ 0:8 11.0183. The corresponding Cn values are 1.3313, 0.5183, 0.3157, and 0.2265. Putting values in Eq. (4.75), we get i h 2 2 2 2 Tð0; tÞ ¼ 20 þ 180 1:3313e1:7155 ð0:0857Þ 0:5183e4:7648 ð0:0857Þ þ 0:3157e7:8857 ð0:0857Þ 0:2265e11:0183 ð0:0857Þ

Tð0; tÞ ¼ 20 þ 180 1:0345 0:0741 þ 0:0015 6:9 106 ¼ 193.1 C. The center temperature of the sphere is 193.1 C after 600 s of cooling. It is seen that the series converges very quickly. If we had only used the first two terms in the series, the result for the center temperature would have been 192.9 C. We showed above that the one-term approximation was inappropriate for this problem as the Fourier number was too small. Indeed, if we had only used the first term of the series, the center temperature after 600 s would have been Tð0; tÞ ¼ 20 þ 180 ð1:0345Þ ¼ 206:2 C. This is obviously wrong as the sphere is being cooled and it starts out with an initial temperature of 200 C. (b) Eq. (4.59) gives the temperature at any location in the sphere. N X

l2 n at ro2

sinðln r=ro Þ (4.76) ln r=ro n¼1 At the surface of the sphere, r ¼ ro . And, like Part (a) above, we will keep the first four terms of the series. Eq. (4.76) then becomes 2 2 2 2 sin l1 sin l2 sin l3 sin l4 Tðro ; tÞ ¼ TN þ ðTi TN Þ C1 el1 Fo þ C2 el2 Fo þ C3 el3 Fo þ C4 el4 Fo l1 l2 l3 l4 Tðr; tÞ ¼ TN þ ðTi TN Þ

Cn e

Putting values into this equation, we get

Tðro ; tÞ ¼ 20 þ 180 0:5967 0:0155 þ 0:0002 6:3 107 ¼ 124.7 C. The surface temperature after 600 s of cooling is 124.7 C. Like Part (a), it is seen that the series converges quickly and only the first two terms are significant.

4.4 Multidimensional systems with spatial variation

143

4.4 Multidimensional systems with spatial variation 4.4.1 Overview Section 4.3 dealt with three geometries: Large (actually infinite) plates of thickness 2L, long (actually infinitely long) cylinders of radius ro, and spheres of radius ro. These geometries are one-dimensional. In all three cases, a location in the body can be described by a single parameter: “x” for the plates and “r” for the cylinders and spheres. The Heisler method of Section 4.3 can be extended to two- and three-dimensional bodies. Let us start by visualizing how we can get multidimensional objects from the one-dimensional plates and cylinders. First, let us consider two large plates. If they intersect each other at right angles, the intersected volume is a long (actually infinitely long) bar. If the two plates have the same thickness 2L, then the cross section of the bar is square. If the two plates have different thicknesses 2L1 and 2L2 , then the cross section of the bar is rectangular. This is shown in Fig. 4.5. A second object is a short cylinder of length 2L and radius ro. As shown in Fig. 4.6, this object is created by intersecting a long cylinder of radius ro with a large plate of thickness 2L. These two examples are two-dimensional objects as a location in the object can be described by two parameters. In the long bar, a location in the cross section is defined by the “x” locations in the two

2 L1 P(x,y) 2 L2

FIGURE 4.5 Long bar created by two large plates.

144

Chapter 4 Unsteady conduction

2L ro

P(x, r)

FIGURE 4.6 Short cylinder created by a long cylinder and a large plate.

intersecting large plates. For the short cylinder, a location in the cylinder is defined by the “x” location in the plate and the “r” location in the long cylinder. An example of a three-dimensional object is the rectangular solid. This object is formed by the intersection of three large plates at right angles to each other. Consider Fig. 4.5 above and envision an additional vertical plate at right angles to the shown vertical plate. If this second vertical plate cuts through both the shown vertical plate and the horizontal plate, the intersected volume is a rectangular solid. Said another way, the rectangular solid is formed by the intersection of three large plates at right angles to each other. If all three plates have the same thickness, then the object is a cube of side 2L. If the plates are not all the same thickness, then we have a rectangular solid of dimensions 2L1 by 2L2 by 2L3 : A location in the solid is defined by the “x” locations for the three intersecting large plates. See Fig. 4.7. The rectangular solid is formed by two vertical plates of thickness 2L1 and 2L2 , respectively, and a horizontal plate of thickness 2L3 : P(x, y, z)

2 L3

2 L2 2 L1

FIGURE 4.7 Rectangular solid created by three large plates.

4.4 Multidimensional systems with spatial variation

145

It turns out that we can use the Heisler method, discussed in Section 4.3 for one-dimensional objects, to solve problems involving two- and three-dimensional objects, which are formed by the intersection of the one-dimensional objects. We simply solve the one-dimensional problems separately and then multiply these results to get the solutions for the two- and three-dimensional objects. That is, the product of the solutions for the one-dimensional objects is the solution for the multidimensional object. The theory behind this technique is discussed in Refs. [3e5]. The following equations outline the solutions for the three multidimensional objects described above: For the long bar, Tðx; y; tÞ TN Tðx; tÞ TN Tðy; tÞ TN ¼ (4.77) Ti TN Ti TN Ti TN long bar plate 1 plate 2 For the one-dimensional plates on the right side of Eq. (4.77): The “x” and “y” are the “x” locations for Plate 1 and Plate 2, respectively. Eq. (4.77) could also be written as q q q ¼ (4.78) qi long bar qi plate 1 qi plate 2 For the short cylinder, Tðx; r; tÞ TN Tðr; tÞ TN Tðx; tÞ TN ¼ Ti TN Ti TN Ti TN short cylinder long cylinder plate Eq. (4.79) could also be written as q q q ¼ qi short cylinder qi long cylinder qi plate

(4.79)

(4.80)

For the rectangular solid, Tðx; y; z; tÞ TN Tðx; tÞ TN Tðy; tÞ TN Tðz; tÞ TN ¼ Ti TN Ti TN Ti TN Ti TN rect. solid plate 1 plate 2 plate 3 (4.81) For the one-dimensional plates on the right side of Eq. (4.81): The “x,” “y,” and “z” are the “x” locations for Plates 1, 2 and 3, respectively. Eq. (4.81) could also be written as q q q q ¼ (4.82) qi rect. solid qi plate 1 qi plate 2 qi plate 3 For the product solution of multidimensional objects, the fluid temperature TN for the object must be the same as that for all the one-dimensional objects which were combined to form the object. The convective coefficients, however, need not be the same on all surfaces of the multidimensional object. However, they must be the same for the different surfaces of each one-dimensional object forming the multidimensional object. To clarify this: If we have a short cylinder, the h value on the ends of the cylinder can be different from the h value on the side of the cylinder. However, both ends of the cylinder must have the same h value. For the rectangular solid, we can have a maximum of

146

Chapter 4 Unsteady conduction

three different h values. The two surfaces of each intersecting plate must have the same h value, but the different plates can have different h values. Regarding the heat transfer from a multidimensional object, Langston [6] showed how the results from the one-dimensional objects could be combined to get the heat transfer results for the two- or three-dimensional object formed from the intersection of the one-dimensional objects, as follows: For a two-dimensional object formed by the intersection of one-dimensional objects “1” and “2”: Q Q Q Q ¼ þ 1 (4.83) Qmax 2D object Qmax 1 Qmax 2 Qmax 1 For a three-dimensional object formed by the intersection of one-dimensional objects “1”, “2”, and “3”: Q Q Q Q Q Q ¼ þ 1 þ 1 Qmax 3D object Qmax 1 Qmax 2 Qmax 1 Qmax 3 Qmax 1 Q 1 (4.84) Qmax 2 The following three sections (Sections 4.4.2e4.4.4) include problems illustrating the use of these equations for the multidimensional objects discussed above, i.e., a long bar, a short cylinder, and a rectangular solid. The application of the Heisler method to additional geometries may be found in Refs. [7e10].

4.4.2 Long bar As shown in Fig. 4.5 above, a long bar is the intersection of two large plates. Example 4.5 illustrates the application of the Heisler method to this geometry.

Example 4.5 Extrusion of a PVC bar Problem

A continuous PVC bar r ¼ 1400 kg = m3 ; c ¼ 900 J = kg C; k ¼ 0.19 W = m C is being produced by extrusion. The bar has a rectangular cross section of 2 cm by 4 cm. It leaves the extrusion machine and enters a water bath for cooling. The water bath is at 35 C and the bar is 140 C when it enters the water. The bar is in the bath for 10 min. It may be assumed that the convective coefficient while the bar is in the water bath is 200 W/m2 C.

h = 200 W / m2 C (4 surfaces) Ti = 140 C Water at 35 C

L

2 cm

4 cm

4.4 Multidimensional systems with spatial variation

147

(a) What is the center temperature of the extruded bar as it leaves the water bath? (b) What is the temperature on the surface of the bar, at the center of one of the 2-cm-wide sides as the bar leaves the bath? (c) How much heat is transferred to the water bath per meter length of bar passing through it?

Solution We first check Bilumped to see if the lumped method can be used. ð0:02Þð0:04ÞL ð200Þ hðV=AÞ ½2ð02Þ þ 2ð04ÞL ¼ 7:018 ¼ Bilumped ¼ 0:19 k (Note: In this calculation, “L” is the length of the bar. In the calculations below, “L” is one-half the thickness of the particular intersecting plate.) As Bilumped is greater than 0.1, the lumped method is inappropriate. We will use the Heisler method applied to a twodimensional problem. The bar is the intersection of a 2-cm-thick large plate and a 4-cm-thick large plate. k 0:19 ¼ ¼ 1:508 107 m2 =s rc ð1400Þð900Þ Let us calculate the Fourier numbers to see if the one-term approximation can be used. a¼

Fo ¼

at t ¼ 10 minutes ¼ 600s: L2

Fo ð2 cm plateÞ ¼

1:508 107 ð600Þ ð0:01Þ2

¼ 0:905

1:508 107 ð600Þ

¼ 0:226 ð0:02Þ2 For both plates, the Fourier number is greater than 0.2, so the one-term approximation is appropriate. The product solution gives q q q ¼ qi bar qi 2 cm plate qi 4 cm plate Fo ð4 cm plateÞ ¼

(4.85)

(a) For this part of the problem, the location of interest, the center of the bar, is on the center plane of both plates. For the 2 cm plate (L ¼ 0.01 m), Bi ¼

hL ð200Þð0:01Þ ¼ ¼ 10:526 k 0:19

2 qo Tð0; tÞ TN ¼ ¼ C1 el1 Fo (4.86) qi Ti TN l1 is from Eq. (4.22). Once we have l1 we can get C1 from Eq. (4.23). For the above Biot number, using Table 4.1 or from software mentioned above, we find that l1 ¼ 1:435 and C1 ¼ 1:263: Putting these values into Eq. (4.86), 2 qo ¼ 1:263eð1:435Þ ð0:905Þ ¼ 0:1959 (4.87) qi 2 cm plate For the 4 cm plate (L ¼ 0.02 m),

From Equation ð4.25Þ;

Bi ¼

hL ð200Þð0:02Þ ¼ ¼ 21:053 k 0:19

2 qo Tð0; tÞ TN ¼ ¼ C1 el1 Fo (4.88) qi Ti TN l1 is from Eq. (4.22). Once we have l1 we can get C1 from Eq. (4.23). For the above Biot number, using Table 4.1 or from software mentioned above, we find that l1 ¼ 1:500 and C1 ¼ 1:270:

148

Chapter 4 Unsteady conduction

Putting these values into Eq. (4.88),

For the bar,

2 qo ¼ 1:270eð1:500Þ ð0:226Þ ¼ 0:7638 qi 4 cm plate

(4.89)

qo To TN qo qo ¼ ¼ ¼ ð0:1959Þð0:7638Þ ¼ 0:1496 qi bar Ti TN qi 2 cm plate qi 4 cm plate

(4.90)

To ¼ 35 þ ð140 35Þð0.1496Þ ¼ 50.7 C. The temperature on the centerline of the bar as it leaves the water bath is 50.7 C. (b) The location of interest is at the center of the 2 cm plate and the surface of the 4 cm plate. Using values calculated in Part (a), and from Eq. (4.24), we have at the surface of the 4 cm plate 2 q Tðx ¼ L; tÞ TN ¼ ¼ C1 el1 Fo cos l1 ¼ 0:7638 cosð1:500Þ ¼ 0:05403 qi 4 cm plate Ti TN So, for the bar, we have q T TN q q ¼ ¼ ¼ ð0:1959Þð0:05403Þ ¼ 0:01058 qi bar Ti TN qi 2 cm plate qi 4 cm plate T ¼ TN þ ðTi TN Þð0:01058Þ ¼ 35 þ ð140 35Þð0:01058Þ ¼ 36.1 C. The temperature at the center of one of the 2-cm-wide sides is 36.1 C as the bar leaves the water bath. (c) From Eq. (4.83), we have " # Q Q Q Q ¼ þ 1 Qmax bar Qmax 2 cm plate Qmax 4 cm plate Qmax 2 cm plate For the 2 cm plate using Eq. (4.34), 2 Q sin l1 sinð1:435Þ ¼ 0:8647 ¼ 1 C1 el1 Fo ¼ 1 0:1959 Qmax 1:435 l1 For the 4 cm plate using Eq. (4.34), 2 Q sin l1 sinð1:500Þ ¼ 0:4921 ¼ 1 C1 el1 Fo ¼ 1 0:7638 Qmax 1:500 l1 Q ¼ 0:8647 þ ð0:4921Þð1 0:8647Þ ¼ 0:9313 So; from Equation ð4.93Þ; Qmax bar

ðQmax Þbar ¼ rcVðTi TN Þ ¼ ð1400Þð900Þð0:02Þð0:04ÞLð140 35Þ ¼ 1:0584 105 L

From Eqs. (4.96) and (4.97), QL ¼ ð0:9313Þ 1:0584 105 ¼ 9:86 104 J=m length 4 The heat transfer to the water bath is 9.86 3 10 J per meter length of bar passing through the bath.

(4.91)

(4.92)

(4.93)

(4.94)

(4.95) (4.96) (4.97)

4.4.3 Short cylinder As shown in Fig. 4.6 above, a short cylinder is the intersection of a long cylinder and a large plate. Example 4.6 illustrates the application of the Heisler method to this geometry.

Example 4.6 Cooling of a short cylinder Problem

A stainless steel disk r ¼ 7900 kg = m3 ; c ¼ 480 J = kg C; k ¼ 15 W = m C is 10 cm diameter and 5 cm thick. It goes into an oven where it is heated to a uniform temperature of 1200 C. It is then taken out of the oven and cooled by

4.4 Multidimensional systems with spatial variation

149

convection to a fluid, which is at 150 C. The convective coefficient on the curved surface of the disk is 100 W/m2 C, and the convective coefficient on the two flat surfaces is 250 W/m2 C. (a) How long does it take for the center of the disk to reach 500 C? (b) At the time obtained in Part (a): What is the temperature at a point located 3 cm from the centerline and 1.5 cm from one end?

h = 250 W / m2 C (top)

Ti = 1200 C 10 cm Fluid at 150 C

5 cm

h = 100 W / m2 C (side) h = 250 W / m2 C (bottom)

Solution

(a) Let us first calculate Bilumped . To be conservative, we will use the larger h value that will give the larger Biot number. ! pð0:05Þ2 ð0:05Þ pro2 L ð250Þ ð250Þ 2pð0:05Þð0:05Þ þ 2pð0:05Þ2 2pro L þ 2pro2 hðV=AÞ ¼ 0:208 ¼ ¼ Bilumped ¼ 15 k k As Bilumped is greater than 0.1, we should not use the lumped method. Therefore, we will use the Heisler method for multidimensional objects. The disk is the intersection of a long cylinder of 10 cm diameter and a large plate of 5 cm thickness. We can use the product solution q q q ¼ (4.98) qi disk qi cylinder qi plate The point of interest is the center of the disk, this is, at the center plane of the plate and the centerline of the cylinder. a¼ For the cylinder (ro ¼ 5 cm),

k 15 ¼ ¼ 3:956 106 m2 =s rc ð7900Þð480Þ

at 3:956 106 t ¼ ¼ 0:001582t ro2 ð0:05Þ2 From Eq. (4.45), the one-term approximation solution is Fo ¼

2 qo Tð0; tÞ TN ¼ ¼ C1 el1 Fo qi Ti TN l1 is from Eq. (4.42). Once we have l1 we can get C1 from Eq. (4.43).

(4.99)

hro ð100Þð0:05Þ ¼ 0:3333 ¼ 15 k For this Biot number, using Table 4.1 or from software mentioned above, we find that l1 ¼ 0:7833 and C1 ¼ 1:0785: Putting values into Eq. (4.98), 2 qo 4 ¼ 1:0785eð0:7833Þ ð0:0:001582tÞ ¼ 1:0785e9:707 10 t (4.100) qi cylinder For the plate (L ¼ 2.5 cm), Bi ¼

Fo ¼

at 3:956 106 t ¼ ¼ 0:006330t L2 ð0:025Þ2

150

Chapter 4 Unsteady conduction

From Eq. (4.25), the one-term approximation solution is 2 qo Tð0; tÞ TN ¼ ¼ C1 el1 Fo qi Ti TN l1 is from Eq. (4.22). Once we have l1 , we can get C1 from Eq. (4.23).

(4.101)

hL ð250Þð0:025Þ ¼ ¼ 0:4167 k 15 For this Biot number, using Table 4.1 or from software mentioned above, we find that l1 ¼ 0:6040 and C1 ¼ 1:0601: Putting values into Eq. (4.101), 2 qo 3 ¼ 1:0601eð0:6040Þ ð0:0:00633tÞ ¼ 1:0601e2:309 10 t qi plate Now we can put everything together. From Equation (4. 98) above, q q q ¼ (4.102) qi disk qi cylinder qi plate Bi ¼

For the disk, qo ¼ To TN ¼ 500 150 ¼ 350 qi ¼ Ti TN ¼ 1200 150 ¼ 1050 Continuing with Eq. (4.102), h ih i 4 3 qo 350 ¼ 0:3333 ¼ 1:0785e9:707 10 t 1:0601e2:309 10 t ¼ qi disk 1050 0:3333 ¼ 1:1433e0:003280t Rearranging and taking natural logs, we have e0:003280 t ¼ 0:2915 0.003280t ¼ lnð0:2915Þ ¼ 1:2327 t ¼ 376 s [ 6.26 minutes The center of the disk cools to 500 C after 6.26 min. To complete the solution, we should check the Fourier numbers to make sure they are greater than 0.2. This would confirm that use of the one-term approximation method was appropriate. For the cylinder, Fo ¼ 0:001582 t ¼ ð0:001582Þð376Þ ¼ 0:595 For the plate, Fo ¼ 0:00633 t ¼ ð0:00633Þð376Þ ¼ 2:380 Both numbers are greater than 0.2, so use of the one-term approximation method was fine. (b) The point of interest for this part of the problem is at r ¼ 3 cm for the cylinder and x ¼ 1 cm for the plate. We will use calculations from Part (a) as much as possible. From Eq. (4.98), q q q ¼ (4.103) qi disk qi cylinder qi plate For the cylinder, From Eq. (4.44), 2 q Tð0:03 m; 376 sÞ TN ¼ ¼ C1 el1 Fo J0 ðl1 r = ro Þ qi cylinder Ti TN 2 q 0:03 ¼ 0:7079 ¼ 1:0785eð0:7833Þ ð0:595Þ J0 ð0:7833Þ qi cylinder 0:05 For the plate, From Eq. (4.24), 2 q Tð0:01 m; 376 sÞ TN ¼ ¼ C1 el1 Fo cosðl1 x = LÞ for Fo > 0:2 qi plate Ti TN

4.4 Multidimensional systems with spatial variation

151

2 q 0:01 ¼ 0.4320 ¼ 1:0601eð0:6040Þ ð2:380Þ cos ð0:6040Þ qi plate 0:025 From Eq. (4.103),

q T TN q q ¼ ¼ ¼ ð0:7079Þð0:4320Þ ¼ 0:3058 qi disk Ti TN qi cylinder qi plate

and T ¼ 150 þ ð1200 150Þð0:3058Þ ¼ 471.1 C The temperature in the disk at a location 3 cm from the centerline and 1.5 cm from one end after 376 s of cooling is 471.1 C.

4.4.4 Rectangular solid A rectangular solid is the intersection of three large plates. This is shown in Fig. 4.7 above. Example 4.7 illustrates the application of the Heisler method to a rectangular solid.

Example 4.7 Cooling of a cube Problem

A steel cube r ¼ 7800 kg = m3 ; c ¼ 470 J = kg C; k ¼ 40 W = m C is shown in the figure below. It is 8 cm by 8 cm by 8 cm. The cube is initially at a uniform temperature of 500 C, and it is suddenly immersed in a fluid bath that is at 150 C. The convective coefficient is 200 W/m2 C. The cube stays in the fluid for 2 min. (a) What is the temperature of the center of the cube when it leaves the fluid bath? (b) What is the temperature of the center of one face of the cube when it leaves the bath? (c) What is the temperature of a corner of the cube when it leaves the bath? (d) If 100 cubes are cooled per hour, what size cooling system is needed to keep the bath at 150 C?

Ti = 500 C h = 200 W / m2 C (all 6 sides) Fluid at 150 C 8 cm

8 cm 8 cm

Solution

The cube is the intersection of three large plates of 8 cm thickness (L ¼ 4 cm). We can use the product solution q q q q ¼ (4.104) qi cube qi plate 1 qi plate 2 qi plate 3

152

Chapter 4 Unsteady conduction

(a) The center of the cube is at the center of all three plates. Let us look at Plate 1. The results for Plates 2 and 3 will be the same. a¼

k 40 ¼ ¼ 1:091 105 m2 =s rc ð7800Þð470Þ t ¼ 2 minutes ¼ 120 s

at 1:091 105 ð120Þ ¼ ¼ 0:8183 L2 ð0:04Þ2 We will use the one-term approximation as Fo > 0.2. Using Eq. (4.25), 2 qo Tð0; tÞ TN ¼ ¼ C1 el1 Fo qi plate 1 Ti TN Fo ¼

(4.105)

hL ð200Þð0:04Þ ¼ ¼ 0:2 k 40 From Table 4.1, for Bi ¼ 0.2, l1 ¼ 0:4328 and C1 ¼ 1:0311: We could have alternatively obtained l1 and C1 from Eqs. (4.22) and (4.23). Putting values into Eq. (4.105), we have qo 2 ¼ 1:0311e0:4328 ð0:8183Þ ¼ 0:8846 qi plate 1 As all three plates are the same, from Eq. (4.104), we get qo ¼ ð0:8846Þ3 ¼ 0:6922 qi cube N ¼ 0:6922 and So, TTT i TN Bi ¼

cube

T ¼ TN þ ðTi TN Þð0:6922Þ ¼ 150 þ ð500 150Þð0:6922Þ ¼ 392 C. The temperature at the center of the cube is 392 C after 2 minutes of immersion. (b) For the location at the center of one face of the cube: This location is on the center plane of Plates 1 and 2 and on the surface of Plate 3. At the surface of Plate 3, i.e., at x ¼ L, we have from Eq. (4.24): 2 q ¼ C1 el1 Fo cos l1 (4.106) qi plate 3 From Part (a): Fo ¼ 0.8183, Bi ¼ 0.2, l1 ¼ 0:4328 and C1 ¼ 1:0311: Using the results of calculations in Part (a), we have q ¼ 0:8846 cos l1 ¼ 0:8846 cosð0:4328Þ ¼ 0:8030 qi plate 3 q q So, qqi ¼ qqi ¼ ð0:8846Þð0:8846Þð0:8030Þ ¼ 0:6284. qi qi cube

plate 1

plate 2

plate 3

And T ¼ TN þ ðTi TN Þð0:6284Þ ¼ 150 þ ð500 150Þð0:6284Þ ¼ 370 C. The temperature of the center of one face of the cube after 2 minutes of immersion is 370 C. (c) For the location at a corner of the cube: This location is on the surface of all three plates. From the calculations in ¼ ð0:8030Þ3 ¼ 0:5178 Parts (a) and (b), we have qqi cube

And T ¼ TN þ ðTi TN Þð0:5178Þ ¼ 150 þ ð500 150Þð0:5178Þ ¼ 331 C. The temperature of a corner of the cube after 2 minutes of immersion is 331 C. Looking at the results of Parts (a), (b), and (c): The corner cools quicker than the center of one face, and the center of a face cools quicker than the center of the cube. Makes sense!

4.5 Semi-infinite solid

(d) From Eq. (4.84),

" # Q Q Q Q ¼ þ 1 Qmax cube Qmax plate 1 Qmax plate 2 Qmax plate 1 " # " # Q Q Q þ 1 1 Qmax plate 3 Qmax plate 1 Qmax plate 2

153

(4.107)

All three plates are the same (L ¼ 4 cm). From Eq. (4.34), for each plate, 2 Q sin l1 ¼ 1 C1 el1 Fo Qmax l1 Using the calculations in Parts (a) and (b), we have Q sinð0:4328Þ ¼ 0:1428 ¼ 1 ð0:8846Þ Qmax plates 1; 2; and 3 ð0:4328Þ Therefore, from Eq. (4.107), Q ¼ 0:1428 þ 0:1428ð1 0:1428Þ þ 0:1428ð1 0:1428Þ2 ¼ 0:3701 Qmax cube

3 ¼ 6:569 105 J. For the cube, Qmax ¼ rcVðTi TN Þ ¼ ð7800Þð470Þ 0:08 ð500 150Þ 5 So, Qcube ¼ 0:3701 Qmax ¼ ð0:3701Þ 6:569 10 ¼ 2:431 105 J. Each cube puts 2.431 105 J of heat into the fluid bath. If 100 cubes per hour are processed, the rate of heat into the bath is

ð100 cubes=hourÞ 2.431 105 J=cube ð1 hour=3600 sÞ ¼ 6750 J=s ¼ 6.75 kW The needed size of the cooling system is 6.75 kW. As a final item, let us calculate Bilumped . (Actually, we probably should have done this at the beginning of the problem.) ! ð0:08Þ3 ð200Þ 6ð0:08Þ2 hðV=AÞ ¼ ¼ 0:06667 Bi ¼ k 40 This is less than 0.1, so let us see what results would be obtained from the lumped method. From Eq. (4.8), ! Tð2 minutesÞ ¼ TN þ ðTi TN Þe

hA rcV

t

¼ 150 þ ð500 150Þe

200ð6Þð0:08Þ2 7800ð470Þð0:08Þ3

ð120Þ

¼ 364 C.

Regarding heat flow: From Eq. (4.13),

hA rcV t ¼ ð7800Þð470Þð0:08Þ3 ð500 150Þ 1 e0:4910 Q ¼ rcVðTi TN Þ 1 e

¼ 2:55 105 Joules per cube It was shown above that the temperatures in cube do indeed vary somewhat with location, and it was good that we performed a Heisler analysis even though the Biot number was less than 0.1. However, the lumped method gives results that are very good. The temperature result is well within the range of temperatures found by the Heisler method. And, the heat flow result differs only 4.8% from the result obtained by the Heisler method.

4.5 Semi-infinite solid 4.5.1 Overview In this section, we consider a semi-infinite solid, also called a semi-infinite slab. This geometry has one physical boundary (a plane surface) and extends internally to infinity in all directions. The semiinfinite solid is shown in Fig. 4.8. The distance from the boundary to a location in the solid is “x.”

154

Chapter 4 Unsteady conduction

∞

qo A O

x

∞

∞

FIGURE 4.8 The semi-infinite solid.

A variety of conditions can be imposed at the boundary. We will consider temperature, heat flux, and convection boundary conditions. The semi-infinite geometry is particularly applicable to large, thick bodies and to the earth’s surface and ground. In addition, bodies can often be modeled as semi-infinite solids during short periods of time after a change has been made at a surface of a body. During this time period, the change is felt in the region near the affected surface but has not yet been felt by the other surfaces of the body. In the analysis that follows, the semi-infinite solid has an initial uniform temperature Ti; the imposed boundary condition remains constant during its application; and all thermal and physical properties are constant.

4.5.2 Temperature boundary condition The semi-infinite solid is initially at uniform temperature Ti. At time t ¼ 0, the temperature of the surface is suddenly changed to To . The temperature at a depth x at time t is Tðx; tÞ Ti x ¼ erfc pﬃﬃﬃﬃﬃ (4.108) To Ti 2 at where a ¼ thermal diffusivity ¼ k=rc and erfc ¼ complementary error function. This function is available on the internet and in software, e.g., Excel and Matlab. A table of the function is given in Appendix I. The heat flux qAo at the surface at time t is qo kðTo Ti Þ ðtÞ ¼ pﬃﬃﬃﬃﬃﬃﬃﬃ A pat

(4.109)

4.5 Semi-infinite solid

155

4.5.3 Heat flux boundary condition The semi-infinite solid is initially at uniform temperature Ti. At time t ¼ 0, a constant heat flux qAo is imposed on the surface. The temperature at a depth x at time t is "rﬃﬃﬃﬃﬃﬃﬃ # ðqo =AÞ 4at x2 x e 4at x erfc pﬃﬃﬃﬃﬃ (4.110) Tðx; tÞ ¼ Ti þ k p 2 at

4.5.4 Convection boundary condition The semi-infinite solid is initially at uniform temperature Ti. At time t ¼ 0, convection to a fluid at TN with a convective coefficient h occurs at the surface. The temperature at a depth x from the surface at time t is pﬃﬃﬃﬃﬃ hx h2 at þ k2 k Tðx; tÞ Ti x x h at erfc pﬃﬃﬃﬃﬃ þ ¼ erfc pﬃﬃﬃﬃﬃ e (4.111) k T N Ti 2 at 2 at temperature ﬃ pﬃﬃﬃThe h at ¼ N: Infinite h k

response is also shown in Fig. 4.9. Note that there is a line in the figure for corresponds to the case where the temperature at the surface equals the temperature of the fluid. Therefore, this line actually shows the response due to the sudden imposition of temperature To at the surface at t ¼ 0. On the vertical axis label, just change TN to To for this case. 1

0.1

T ( x, t ) – Ti T∞ – Ti

h αt k ∞ 3 2 1 0.8 0.5 0.4 0.3 0.2 0.15 0.1 0.08 0.05 0.03 0.02 0.01

0.01

0.001

0.0001

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5

x 2 αt

FIGURE 4.9 Temperature response of semi-infinite solid with convection at boundary.

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Chapter 4 Unsteady conduction

Example 4.8 Burial depth of a water pipe Problem

In a town in the Northern United States, the air temperature can be as low as 18 C for as long as 4 weeks. Assume that the ground is initially at a uniform temperature of 10 C. How deep should water pipes be buried to prevent freezing? Assume that the combined convective and radiative coefficient at the ground’s surface is an average 30 W/m2 C. The ground has a thermal conductivity of 2 W/m C, a density of 1800 kg/m3, and a specific heat of 2100 J/kg C.

h = 30 W / m2 C Air at –18 C

Ti = 10 C

x

Water Line

Solution We want to find the depth at which the temperature is the freezing temperature of water (0 C). k 2 ¼ ¼ 5:291 107 m2 =s rc ð1800Þð2100Þ where t ¼ 4 weeks 7 days/week 24 h/day 3600 s/h ¼ 2.42 106 s. We will use Eq. (4.111): pﬃﬃﬃﬃﬃ hx h2 at þ k2 k Tðx; tÞ Ti x x h at erfc pﬃﬃﬃﬃﬃ þ ¼ erfc pﬃﬃﬃﬃﬃ e k TN Ti 2 at 2 at

7 6 at ¼ 5:291 10 2:420 10 ¼ 1:280 Putting values into Eq. (4.111), we have 2 pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 30x 30 ð1:280Þ þ 22 2 0 10 x x 30 1:280 ¼ erfc pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ e erfc pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ þ 18 10 2 2 1:280 2 1:280 a¼

(4.111)

0:3571 ¼ erfcð0:4419xÞ eð15xþ288Þ erfcð0:4419x þ 16:971Þ This equation can be solved by trial and error, that is, guessing a value of x, calculating the right side of the equation, and seeing if it equals 0.3571. If not, choose another x and try again. Of course, it is much more efficient to use software. Solving the equation for x using Excel’s Goal Seek gives x ¼ 1.474 m ¼ 4.84 ft. Being conservative, we will round this answer upwards and conclude that the water pipe should be buried at a depth of 5 feet or greater.

4.7 Problems

157

Tðx;tÞTi Another way pﬃﬃﬃﬃ to solve this problem is to use Fig. 4.9. The left axis is TN Ti ¼ 0:3571 and the lines on the figure are for different h kat ¼ 16:971. For our problem, the left axis is 0.3571 and the line should be the one for 16.971, which is close to the one for N. Reading the figure, we get pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ x h ¼ pﬃﬃﬃﬃﬃ ¼ 0:64 and x ¼ ð0:64Þð2Þ 1:280 ¼ 1:45 m ¼ 4.76 ft 2 at This result agrees with the above result from Eq. (4.111).

4.6 Chapter summary and final remarks In this chapter, we discussed unsteady conduction where temperatures and heat flows in a body vary with time. The first step in an analysis is to calculate the Biot number, Bilumped , to see if the body can be considered to be a lumped system. If so, all locations in the body at a given time have essentially the same temperature and a simplified analysis, the lumped method, can be performed. If the body does not meet the criterion for lumped system analysis, then we have to consider spatial variation of temperature in the body. We discussed the Heisler method that provides solutions for spatial variation in basic geometric shapes such as plates, cylinders, and spheres. We also considered temperature distributions in semi-infinite solids due to different boundary conditions at the exposed surface of the solid. Some bodies have complex shapes that do not fall within the basic geometric categories of this chapter. In addition, some situations do not meet the requirements of the lumped or Heisler methods. For example, thermophysical properties may vary with location or time, and the body may not have a uniform temperature at the beginning of the process. For these cases, problems can often be solved numerically. This is the topic of the next chapter.

4.7 Problems Notes: •

•

•

If needed information is not given in the problem statement, then use the Appendix for material properties and other information. If the Appendix does not have sufficient information, then use the Internet or other reference sources. Some good reference sources are mentioned at the end of Chapter 1. If you use the Internet, double-check the validity of the information by using more than one source. For this chapter, it is often helpful at the beginning of a solution to calculate Bilumped to see if the lumped method can be used. If you use the Heisler method, you should calculate the Fourier number to see if the one-term approximation can be used or if more than one term is needed. Of course, you have to pay close attention to the geometry of the problem to see what sections of the chapter are relevant. Your solutions should include a sketch of the problem. 4-1 A large plate has thickness L and cross-sectional area A. One

4side of4the plate is perfectly insulated. The other side gets a radiation input q ¼ ε sA Tsurr T , where T is the

158

4-2

4-3

4-4

4-5

4-6

4-7

4-8

4-9

Chapter 4 Unsteady conduction

temperature of the plate at time t. Assume that the lumped method can be used. The initial uniform temperature of the plate is Ti . (a) Write the differential equation for the temperature TðtÞ of the plate. (b) Solve the differential equation for TðtÞ. A sphere of radius ro is initially at a uniform temperature Ti . It cools by radiation and convection to the surroundings at Tsurr and the adjacent fluid at TN : The convective coefficient is h and the emissivity of the surface of the sphere is ε. Assume the lumped method can be used. Write the differential equation for the temperature TðtÞ of the sphere. A wire has a radius ro and is initially at the same temperature as the surroundings ðTN Þ Current suddenly starts flowing through the wire, resulting in a uniform and steady volumetric heat generation rate of qgen : The surface of the wire convects to the surrounding fluid at TN with a convective coefficient h. Assuming the lumped method is appropriate. (a) Write the differential equation for the temperature TðtÞ of the wire. (b) Solve the differential equation for TðtÞ.

An aluminum sphere r ¼ 2700 kg = m3 ; c ¼ 900 J = kg C; k ¼ 240 W = m C of 3 cm diameter is initially at a uniform temperature of 300 C. It is suddenly placed in a 75 C air stream. The convective coefficient at the sphere’s surface is 50 W/m2 C. (a) How long does it take for the sphere to cool to 150 C? (b) How much heat

goes into the air during the cooling? A copper cylinder r ¼ 8900 kg m3 ; c ¼ 380 J kg C; k ¼ 400 W m C has a diameter of 5 cm and a length of 3 cm. It is initially at a uniform temperature of 250 C. It is suddenly placed in a fluid whose temperature is 100 C. The convective coefficient at the cylinder’s surface is 150 W/m2 C (a) How long does it take for the center of the cylinder to reach 200 C? (b) How much heat

goes into the fluid during the cooling? A large iron plate r ¼ 7800 kg m3 ; c ¼ 450 J kg C; k ¼ 80 W m C is being heated in an oven. The plate is 1 cm thick and has a cross-sectional area of 3 m2. The plate is initially at a uniform temperature of 20 C. The oven is at 800 C and the convective coefficient at the plate’s surfaces is 150 W/m2 C. The plate stays in the oven for 2 min. What is the temperature

of the plate when it leaves the oven? A large copper plate r ¼ 8900 kg m3 ; c ¼ 380 J kg C; k ¼ 400 W m C is 2 cm thick. The plate is initially at a uniform temperature of 30 C. It is then subjected to convective cooling on both of its sides. The adjacent fluids on both sides of the plate are at 90 C and the convective coefficients are 110 W/m2 C on both sides. How long will it take for the plate to reach 55 C? A large copper plate r ¼ 8900 kg m3 ; c ¼ 380 J kg C; k ¼ 400 W m C is 2 cm thick. The plate is initially at a uniform temperature of 30 C. It is then subjected to convective cooling on both of its sides. On one side of the plate, the adjacent fluid is at 70 C and the convective coefficient is 45 W/m2 C. On the other side of the plate, the adjacent fluid is at 90 C and the convective coefficient is 110 W/m2 C. How long will it take for the plate to reach 55 C? A mercury/glass thermometer initially at 20 C is suddenly immersed into oil that is kept at 120 C. After 4 s, the thermometer reads 90 C. What is the time constant of the thermometer?

4.7 Problems

159

4-10 In a lab experiment, a large horizontal flat plate is heated to a constant temperature by electrical heating tapes under the plate. It is desired to see how uniform the temperature is on the surface of the plate. A handheld surface probe is used to measure 20 different points on the plate. The probe has a time constant of 5 s, and temperature readings are taken at the different points every 5 s. The results are reviewed, and they do not make any sense. (a) What is going wrong? (b) What are two possible ways to

remedy the problem? 4-11 An aluminum cube r ¼ 2700 kg m3 ; c ¼ 900 J kg C; k ¼ 240 W m C is 2 cm by 2 cm by 2 cm. It is initially at a temperature of 200 C. It is suddenly put into an air stream that is at 40 C. In 1 minute, the temperature of the cube drops to 80 C. (a) What is the convective coefficient at the surface of the cube? (b) Using the results of Part (a), how long does it take for the cube to reach the 40 C temperature of the air stream? 4-12 A temperature sensor has a time constant of 1 second and performs as a first-order system. The sensor is initially at 50 C and is then immersed in a fluid that is at 85 C. What temperature does the sensor indicate 10 s after it is immersed? 4-13 A long aluminum cylinder r ¼ 170 lbm ft3 ; c ¼ 0:22 Btu lbm F; k ¼ 140 Btu h ft F has a diameter of 3 inches. It initially has a uniform temperature of 600 F. It is suddenly quenched in a water bath for 2 min. The water bath is at 120 F and the convective coefficient is 100 Btu/h ft2 F. What is the temperature of the cylinder when it leaves the water bath? 4-14 An orange can be considered to be a sphere of 8 cm diameter. The orange is initially at a uniform temperature of 10 C when the air temperature quickly falls to 5 C and remains at that temperature. Assume that the orange has the properties of water and that the convective coefficient at the surface of the orange is 15 W/m2 C. (a) How long will it take for the surface of the orange to freeze, i.e., reach 0 C ? (b) How long will

it take for the center of the orange to freeze, i.e., reach 0C? 4-15 A long hot dog r ¼ 950 kg m3 ; c ¼ 3400 J kg C; k ¼ 0:41 W m C has a diameter of 2.5 cm. It is initially at 5 C and is dropped into boiling water that is at 100 C. Assume that h ¼ 2000 W/m2 C. How long will it take for the center of the hot dog to reach a safe temperature of 65 C? Compare of a cookbook.

your answer to the recommendations 4-16 A ham r ¼ 1030 kg m3 ; c ¼ 3500 J kg C; k ¼ 0:48 W m C is in the shape of a short cylinder: 20 cm diameter and 15 cm long. The ham is initially at 5 C and is placed in a 190 C oven. The convective coefficient for the oven is 30 W/m2 C. (a) How long will it take for the center of the ham to reach 65 C? Compare your answer to the recommendations of a cookbook. (b) If the ham had been modeled as a sphere having the same volume as the cylinder, what would have been the two results.

the answer to Part (a)? Compare 4-17 A 3 kg roast beef r ¼ 950 kg m3 ; c ¼ 3400 J kg C; k ¼ 0:45 W m C is initially at a temperature of 4 C. It is being cooked in a 200 C oven where the convective coefficient is 25 W/m2 C. The roast beef can be considered to be a sphere. (a) How long will it take for the center of the beef to reach 75 C? (b) What is the surface temperature at the time found in Part (a)?

160

Chapter 4 Unsteady conduction

4-18 A furniture company wants to char the surface of some wood rods by passing them through an oven. The rods are 6 cm diameter and 5 m long. Their properties are r ¼ 420 kg m3 ; c ¼ 2700 J kg C; k ¼ 0:15 W m C. The rods are to stay in the oven for 30 min. The rods will char if their surface temperature reaches 300 C. The rods are initially at 20 C before they enter the oven, and the convective coefficient is 40 W/m2 C. What should the oven temperature be? 4-19 Same problem as 4e18 except the company wants to speed up the process and keep the rods in the oven for only 10 min. What should the oven temperature be? 4-20 A large wood sheet (r ¼ 420 kg m3 ; c ¼ 2700 J kg C; k ¼ 0:15 W m C) is 10 cm thick. It is at 20 C when it enters an oven. The convective coefficient on both sides of the wood sheet is 40 W/m2 C. The wood will ignite if its surface temperature reaches 350 C. (a) If the oven is at 360 C, how long will it take for the wood to ignite? (b) If the oven is at 450 C, how long will it take for the wood to ignite? 4-21 A thick slab of wood (r ¼ 400 kg m3 ; c ¼ 2800 J kg C; k ¼ 0:15 W m C) is heated on its surface by hot gases in an oven. The gases are at 600 C, and the convective coefficient at the surface of the wood is 40 W/m2 C. The wood is at 20 C when it enters the oven. How long will it take for the wood is 380 C? to ignite if the ignition temperature 4-22 A large steel plate (r ¼ 7800 kg m3 ; c ¼ 480 J kg C; k ¼ 50 W m C) is 15 cm thick and initially at a uniform temperature of 500 C. The plate is horizontal and on an insulating pad so there is no heat flow through its bottom surface. An air stream at 20 C is blown across the top surface to cool the plate. The convective coefficient at this top surface is 150 W/m2 C. (a) What is the temperature at the top surface of the plate after 10 min of cooling? (b) What is the temperature at the bottom insulated surface after 10 min of cooling? (c) What is the temperature at the midpoint of the plate after 10 min of cooling? (Hint: Take a close look at the geometry and boundary conditions. This can be solved with the Heisler method for a plate.)

4-23 A long stainless steel rod r ¼ 7900 kg m3 ; c ¼ 480 J kg C; k ¼ 15 W m C of 12 cm diameter is initially at 450 C and is being cooled by a 30 C air stream. The convective coefficient at the rod’s surface is 200 W/m2 C. (a) How long does it take for the center of the rod to reach 100 C? (b) How much heat goes into the air during the process? 4-24 Table 4.1 gives eigenvalue l1 and constant C1 for a large plate. Determine the next three eigenvalues and constants for a large plate with a Biot number of 3.5. 4-25 Table 4.1 gives eigenvalue l1 and constant C1 for a long cylinder. Determine the next three eigenvalues and constants for a long cylinder with a Biot number of 3.5. 4-26 Table 4.1 gives eigenvalue l1 and constant C1 for a sphere. Determine the next three eigenvalues and constants of 3.5.

for a sphere with a Biot number 4-27 An aluminum cylinder r ¼ 2700 kg m3 ; c ¼ 900 J kg C; k ¼ 240 W m C has a 40 cm diameter and is 40 cm long. It is heated in an oven until its temperature is uniform at 900 C. It is then taken out of the oven and cooled by convection to the 25 C room air. The convective coefficient between the cylinder and the room air is 20 W/m2 C. What is the temperature of the center of the cylinder 15 min after it is taken out of the oven?

4.7 Problems

161

4-28 A long concrete column r ¼ 2100 kg m3 ; c ¼ 910 J kg C; k ¼ 1:4 W m C of square cross section (10 cm by 10 cm) is initially at a uniform temperature of 220 C. Cooling suddenly starts to the 25 C surroundings with a convective coefficient of 45 W/m2 C. (a) What is the temperature of the center of the column 40 min after the cooling begins? (b) What is the temperature at an edge of the column 40 min after the cooling begins? (c) How much heat leaves the column during the cooling period?

per meter length 4-29 A thick concrete slab r ¼ 2100 kg m3 ; c ¼ 910 J kg C; k ¼ 1:4 W m C is at a uniform initial temperature of 280 C. It suddenly starts cooling from its surface to the room air. The air is at 20 C and the convective coefficient is 25 W/m2 C. (a) What is the temperature at the surface of the slab after 30 min of cooling? (b) What is the temperature at a depth of 5 cm from the surface after 30 min of cooling? 2 4-30 The surface of a thick block of wood is being subjected to a heat flux of 3000 W/m . The 3 properties of the wood are r ¼ 400 kg m ; c ¼ 2500 J kg C; k ¼ 0:12 W m C. The wood is initially at a uniform temperature of 25 C. The wood will ignite if its temperature reaches 400 C. Will the wood ignite if it is subjected to the surface heat flux for 20 min? 2 4-31 The surface of a thick block of wood for 15 min. The 3is subjected to a heat flux of 3500 W/m wood has properties: r ¼ 400 kg m ; c ¼ 2500 J kg C; k ¼ 0:12 W m C. The wood is initially at a uniform temperature of 20 C. What is the temperature at a depth of 1 cm from the surface at the end of the heating? 4-32 The surface of a thick slab of iron r ¼ 8000 kg m3 ; c ¼ 450 J kg C; k ¼ 80 W m C is subjected to a heat flux of 10,000 W/m2 for 1 hour. The slab is initially at a uniform temperature of 30 C. At the end of the heating

process, at what depth is the temperature 50 C? 4-33 A stainless steel cube r ¼ 7900 kg m3 ; c ¼ 500 J kg C; k ¼ 15 W m C is 12 cm by 12 cm by 12 cm. It initially is at 50 C and is heated by hot gases at 900 C with a convective coefficient of 150 W/m2 C on all six sides. (a) What is the temperature at the center of the cube after 30 min of heating? (b) What is the temperature at a corner of the cube after 30 min of heating? (c) How much heat goes into the cube during the process? 30-minute heating 4-34 A stainless steel rectangular solid r ¼ 7900 kg m3 ; c ¼ 500 J kg C; k ¼ 15 W m C is 6 cm by 15 cm by 20 cm. It initially is at 550 C and cools by convection to the room air, which is at 25 C. The convective coefficient on the two 6 cm by 15 cm sides is 20 W/m2C. The convective coefficient on the other four sides is 80 W/m2 C. (a) What is the center temperature of the solid after 40 min of cooling? (b) What is the temperature at the center of a 15 cm

by 20 cm side after 40min of cooling? 4-35 A thick concrete slab r ¼ 2100 kg m3 ; c ¼ 910 J kg C; k ¼ 1:4 W m C is initially at a uniform temperature of 50 C. The temperature of the surface of the slab is suddenly lowered to 10 C. (a) What is the temperature at a depth of 5 cm from the surface 45 min after the surface temperature was lowered? (b) What is the heat flux at the surface 45 min after the surface temperature was lowered?

162

Chapter 4 Unsteady conduction

4-36 A thick steel slab r ¼ 7800 kg m3 ; c ¼ 480 J kg C; k ¼ 45 W m C is initially at 20 C. It is suddenly put into boiling water at 100 C. The convective coefficient at the surface of the slab is 2000 W/m2 C. What is the temperature at a depth 1 cm from the surface 5 min after the slab was put into the water?

4-37 A thick steel slab r ¼ 490 lbm ft3 ; c ¼ 0:1 Btu lbm F; k ¼ 35 Btu h ft F is initially at a uniform temperature of 500 F. Its surface temperature is suddenly lowered to 50 F. (a) How long will it take for the temperature at a depth of 1 inch from the surface to reach 250 F? (b) How much heat per square foot of surface area leaves the slab during this process?

References [1] M.P. Heisler, Temperature charts for induction and constant temperature heating, ASME Trans. 69 (1947) 227e236. [2] H. Gro¨ber, S. Erk, U. Grigull, Fundamentals of Heat Transfer, McGraw-Hill, 1961. [3] H.S. Carslaw, J.C. Jaeger, Conduction of Heat in Solids, Oxford University Press, 1959 (Section 1.15). [4] V.S. Arpaci, Conduction Heat Transfer, Addison-Wesley, 1966 (Section 5.2). [5] J. Sucec, Heat Transfer, W. C. Brown, 1985 (Section 3.2e). [6] L.S. Langston, Heat transfer from multidimensional objects using one-dimensional solutions for heat loss, Int. J. Heat Mass Transf. 25 (1982) 149e150. [7] A.F. Mills, Heat Transfer, second ed., Prentice Hall, 1999 (Section 3.4.4 and Table 3.6). [8] F. Kreith, W.Z. Black, Basic Heat Transfer, Harper and Row, 1980 (Tables 3-2 and 3-3). [9] J.P. Holman, Heat Transfer, tenth ed., McGraw-Hill, 2010 (Section 4-5, Figure 4-18). [10] Y.A. C¸engel, A.J. Ghajar, Heat and Mass Transfer, fourth ed., McGraw-Hill, 2011 (Table 4-5).

CHAPTER

Numerical methods (steady and unsteady)

5

Chapter Outline 5.1 Introduction .................................................................................................................................163 5.2 Finite-difference method ...............................................................................................................164 5.2.1 Steady state.............................................................................................................166 5.2.2 Unsteady state .........................................................................................................183 5.3 Finite element method ..................................................................................................................198 5.4 Chapter summary and final remarks............................................................................................... 199 5.5 Problems .....................................................................................................................................199 References ..........................................................................................................................................209

5.1 Introduction Almost all of the material in previous chapters dealt with bodies having simple geometries, i.e., plates, cylinders, and spheres, and constant and uniform thermophysical properties. For situations involving convection, the convective coefficient h and the fluid temperature TN were constant. They did not vary with time. If radiation was involved, the emissivity ε and the temperature of the surroundings Tsurr were also time invariant. Additional constraints were also present in the Lumped and Heisler Methods of Chapter 4. The bodies were required to have uniform temperatures at the beginning of heat transfer processes. Solutions were for convective, not radiative, boundary conditions. For the flat plate of the Heisler Method, the convective coefficients on both surfaces had to be the same. In this chapter, we discuss numerical solution of heat transfer problems. Through the use of numerical methods, the restrictions mentioned above can be removed. Numerical methods can be applied to complex, as well as basic geometries, and can be used for situations in which properties and boundary conditions are nonuniform and varying with time. Numerical methods for heat transfer problems have been used since at least the 1940s [1e3]. With the widespread availability of digital computers in the 1960s and 1970s, researchers and students became able to solve problems that are very difficult or impossible to be solved analytically. There are several types of numerical methods used in heat transfer, including finite differences, finite elements, boundary elements, and control volumes [4e9]. We discuss the finite-difference method in depth and show its application to both steady and unsteady problems. Numerical solutions to several problems will be presented. Some of these problems were solved earlier using other

Heat Transfer Principles and Applications. https://doi.org/10.1016/B978-0-12-802296-2.00005-6 Copyright © 2021 Elsevier Inc. All rights reserved.

163

164

Chapter 5 Numerical methods (steady and unsteady)

techniques. For those problems, comparison will be made of the numerical results and the previously obtained results. The chapter concludes with a brief discussion of the finite element method. After four chapters of differential equations and relatively complex mathematics, many students will find this chapter to be refreshing and a welcome change, although temporary! The chapter only uses algebra and the three basic equations for the modes of heat transfer: qcond ¼

kA ðT1 T2 Þ L

(5.1)

qconv ¼ hAðTs TN Þ 4 qrad ¼ εsA Ts4 Tsurr

(5.2) (5.3)

In all three equations, A is the area through which heat flows. Eq. (5.1) is for conduction through a wall of thickness L with surface temperatures T1 and T2 : Eq. (5.2) is for convection from a surface at Ts to a fluid at TN with convective coefficient h. And, in Eq. (5.3) for radiation, ε is the emissivity of the surface, s is the StefaneBoltzmann constant, Ts is the surface temperature, and Tsurr is the temperature of the surroundings. We also have to remember that, for radiation, temperatures must be in absolute temperature unitsdRankine or Kelvin.

5.2 Finite-difference method Let us first consider the meaning of “finite difference”. In general, the temperature in a body is a function of location and time. For example, in rectangular Cartesian coordinates, T ¼ Tðx; y; z; tÞ. The parameters x, y, z, and t are continuous variables over their respective ranges. In the finite-difference method, these four parameters are not continuous. They are only defined at specific locations or times: nodes for the spatial parameters and time steps for the time parameter. The first step in a finite-difference analysis is to determine the grid and nodal points for the object. Let us say that we have a large wall of thickness L. The surface temperatures of the wall are, given, and we wish to determine the interior temperatures. We can divide the thickness into any number of parts. For example, if we decide that there should be 10 equally sized parts, the nodal points for the wall will be as shown in Fig. 5.1.

1

Δx 2

FIGURE 5.1 Nodes for wall of thickness L.

2

3

4

5

6

7

Δx L

8

9

10

11

Δx 2

5.2 Finite-difference method

165

As the surface temperatures are of primary importance, the two wall surfaces have been selected as nodal points. The other nodal points are spaced equally across the thickness. The size of increment Dx is given by total length L ¼ (5.4) number of divisions 10 Let us look at Fig. 5.1. Nodes 1 and 11 are on the boundary and hence are called “boundary nodes” The other nodes, nodes 2 through 10, are “interior nodes” The dashed lines in Fig. 5.1 are the boundaries between the different nodes. In our finite-difference analysis, we will be looking at heat flows into and out of the various nodes. The cross-hatched area in Fig. 5.1 is the extent of node 7. The node has node 6 on its left and node 8 on its right. Note that the interior nodes 2e10 have widths of Dx and the boundary nodes 1 and 11 have widths of Dx=2: Fig. 5.2 shows a rectangular plate that has been given a rectangular grid. There are 16 nodes, and the increments Dx and Dy are given by Dx ¼

L1 L2 and Dy ¼ : (5.5) 3 3 There are three cross-hatched areas in Fig. 5.2, which show the extents of corner node 1, interior node 7, and edge node 15. (A hint: To determine the extent of a given node, just go halfway to all adjacent nodes and box-in the area.) For the rectangular plate of Fig. 5.2, L2 ¼ 2L1 . Therefore, we could have used a square grid, as shown in Fig. 5.3. The new grid has 28 nodes instead of the original 16 nodes. Increment Dx is still L1 =3, but Dy is now L2 =6: When the temperatures at these 28 nodes are determined, we will have a better knowledge Dx ¼

1

2

3

4

5

6

7

8

9

10

11 12

L2

Δy 13 Δx

FIGURE 5.2 Nodes for a rectangular plate.

14

L1

15 16

166

Chapter 5 Numerical methods (steady and unsteady)

1

2

3

4

5

6

7

8

9

10

11 12

13

14

15 16 L2

17

18

19 20

21

22

23 24

25

26

27 28

Δy Δx L1

FIGURE 5.3 Square grid for the rectangular plate.

of the temperature distribution in the plate than we would have obtained with the original 16 nodes. Actually, we could have used an even larger number of smaller nodes to get an even more-detailed indication of the temperature distribution in the plate. In the next section, we will be formulating finite difference equations to solve for the nodal temperatures. Such equations can certainly be obtained for a rectangular grid. However, we will see that use of a square grid provides some simplifications. Grids and nodes can also be established for other geometries. For example, consider Fig. 5.4 that shows a circular plate. There are five equally spaced nodes in Fig. 5.4. The distance between nodal points is Dr. If the outer radius of the plate is ro , then the increment Dr ¼ ro =4: The dashed lines in Fig. 5.4 show the boundaries of the various nodes. The center node, node 1, is a circular area. The other four nodes are ring areas. As an example, the cross-hatched area is node 4. Note that the interior nodes 2, 3, and 4 have widths of Dr. Center node 1 is circular with a radius of Dr=2: Boundary node 5 is a ring of width Dr= 2:

5.2.1 Steady state In this section, we will develop finite difference equations for steady-state heat transfer in one and twodimensional objects. Fig. 5.5 shows some nodes in a one-dimensional object. For steady state, temperatures and heat flows are constant; they do not change with time. Let us look at the possible heat flows for node i. There is conduction between node i and its two adjacent nodes i1 and iþ1. And, there may be convection and/or radiation at the upper, lower, front, and/or

5.2 Finite-difference method

1

2

3

4

167

5

Δr

r0

FIGURE 5.4 Nodes for a circular plate.

i–2

i

i–1

i+1

i+2

Δx

FIGURE 5.5 Steady one-dimensional nodes.

back boundaries of node i. (There could also be imposed heat fluxes at these boundaries or the boundaries could be insulated.) Finally, there may also be internal heat generation within node i at a volumetric strength of qgen . For steady state, from conservation of energy, the total rate of heat flow into node i equals the total rate of heat flow out of node i. Fig. 5.6 shows the heat flows for node i. We have assumed that the incoming flows are the conduction from node i1 and the internal heat generation and that the outgoing flows are the convection to the adjacent fluid, the radiation to the surroundings, and the conduction to node iþ1. We actually do not know the directions of the conductive, convective, or radiative flows until we solve the finite-difference equations and determine the temperatures in the object. However, we need to assume directions so we can write the energy

168

Chapter 5 Numerical methods (steady and unsteady)

qconv qrad

A

qi–1 i–2

i–1

qi

i

i

i+1

i+I

i+2

qgen

Δx

FIGURE 5.6 Heat flows for node i.

conservation equation. (Remember that the direction of heat flow is from the higher temperature to the lower temperature.) The energy conservation equation is: heat flow in ¼ heat flow out. For node i we have qði1Þ/i þ qgenerated ¼ qrad þ qconv þ qi/ðiþ1Þ

(5.6)

Using Eqs. (5.1), (5.2), and (5.3) for the various heat flows, Eq. (5.6) becomes kA kA 4 ðTi1 Ti Þ þ qgen A Dx ¼ ðTi Tiþ1 Þ þ hAðTi TN Þ þ εsA Ti4 Tsurr (5.7) Dx Dx The direction of the internal heat generation is, of course, into node i. But, we have assumed the directions for the other heat flows and have written the terms in Eq. (5.7) accordingly. An alternative way to write the energy conservation equation is to assume that all the heat flows are into the node. Then, rather than heat flow in ¼ heat flow out, the energy conservation equation for steady-state would be: total heat flow in ¼ 0. Let us do that. We will assume that all heat flows are into node i. The energy conservation equation is then 4 kA kA ðTi1 Ti Þ þ ðTiþ1 Ti Þ þ hAðTN Ti Þ þ εsA Tsurr Ti4 þ qgen A Dx ¼ 0 (5.8) Dx Dx Comparing Eqs. (5.7) and (5.8), it is seen that the equations are identical. To summarize, we can assume the directions for the different heat flows associated with a node and write an equation using heat flow in ¼ heat flow out. Alternatively, we can assume that all heat flows are into the node and write an equation using total heat flow in ¼ 0. The author prefers the second approach, and that is the way we will determine the finite-difference equations for nodes. Eq. (5.8) is the finite-difference equation for node i. Similar equations can be developed for all the other nodes in the object. The equations may then be solved simultaneously for the nodal temperatures.

5.2 Finite-difference method

169

Continuing our discussion of one-dimensional steady-state numerical analysis, let us now look again at the pin fin of Example 3.11. The heat flow for the fin was determined analytically in Chapter 3. Let us determine the heat flow numerically and compare the two answers.

Example 5.1 Heat transfer from a pin fin Problem

A pin fin (k ¼ 250 W/m C) is attached to a wall. The fin’s diameter is 1.5 cm, and it is 10 cm long. The wall is at 200 C. The surrounding air is at 15 C and the convective coefficient is 9 W/m2 C. Determine, numerically, the rate of heat transfer from the fin to the surrounding air.

Solution As discussed in Chapter 3, fins are attached to surfaces to enhance the heat transfer. This problem was solved analytically as Example 3.11. We now solve it numerically and compare the result with that of the analytical solution. The first step is to determine the grid for the problem. We decided to have the grid shown in Fig. 5.7. There are 11 nodes. Node 1 is at the wall and node 11 is at the end of the fin. These are boundary nodes. The other nodes (nodes 2 through 10) are interior nodes. The spacing between nodes is uniform with Dx ¼ 1 cm ¼ 0.01 m. We wanted nodal points at the wall and at the end of the fin, so nodes 1 and 11 have widths of Dx=2: The other nodes have widths of Dx: The temperature of node 1 is known. It is the wall temperature, i.e., 200 C. The other nodal temperatures are determined through simultaneous solution of the 10 finite-difference equations for nodes 2 through 11. Let us determine the equation for node 2. The node gets heat by conduction from adjacent nodes 1 and 3. It also gets heat by convection from the surrounding fluid. As discussed above, we will assume that all heat flows are into node 2. We have q1/2 þ q3/2 þ qN/2 ¼ 0 Putting in the expressions for the three heat flows, Eq. (5.9) becomes

(5.9)

kA kA ðT1 T2 Þ þ ðT3 T2 Þ þ hPDxðTN T2 Þ ¼ 0 (5.10) Dx Dx In Eq. (5.10), P is the perimeter p D of the fin, and A is the cross section area p D2/4 of the fin. D is the diameter of the fin. Equations similar to Eq. (5.10) can be written for the other eight interior nodes, i.e., nodes 3e10. The only thing that changes is the subscripts on the temperatures. That is. For i ¼ 3 to 10, kA kA ðTi1 Ti Þ þ ðTiþ1 Ti Þ þ hPDxðTN Ti Þ ¼ 0 (5.11) Dx Dx Node 11 is a special node. It is a boundary node and gets conduction from node 10 and convection from the side and end of the fin. Its nodal equation is kA Dx ðT10 T11 Þ þ hP ðTN T11 Þ þ hAðTN T11 Þ ¼ 0 Dx 2

1

2

3 Δx 2

FIGURE 5.7 Grid for example 5.1.

4

5

6

7

8 Δx

10 cm

9

10 11

(5.12)

1.5 cm D Δx 2

170

Chapter 5 Numerical methods (steady and unsteady)

Eqs. (5.10), (5.11), and (5.12) can be solved simultaneously for the 10 unknown temperatures T2 through T11 : (Remember T1 is a given. It is 200 C) For this problem, k ¼ 250 W/m C, h ¼ 9 W/m2 C, D ¼ 1.5 cm ¼ 0.015 m, Dx ¼ 1 cm ¼ 0.01 m, P ¼ p D ¼ 0.047124 m, A ¼ p D2/4 ¼ 0.00017672 m2, TN ¼ 15 C, and T1 ¼ 200: Putting these values into Eqs. (5.10), (5.11), and (5.12), we have the following 10 equations to solve for the unknown temperatures T2 through T11: 8:840T2 þ 4:418T3 ¼ 883:66

(5.13)

4:418T2 8:840T3 þ 4:418T4 ¼ 0:06362

(5.14)

4:418T3 8:840T4 þ 4:418T5 ¼ 0:06362

(5.15)

4:418T4 8:840T5 þ 4:418T6 ¼ 0:06362

(5.16)

4:418T5 8:840T6 þ 4:418T7 ¼ 0:06362

(5.17)

4:418T6 8:840T7 þ 4:418T8 ¼ 0:06362

(5.18)

4:418T7 8:840T8 þ 4:418T9 ¼ 0:06362

(5.19)

4:418T8 8:840T9 þ 4:418T10 ¼ 0:06362

(5.20)

4:418T9 8:840T10 þ 4:418T11 ¼ 0:06362

(5.21)

4:418T10 4:4217T11 ¼ 0:05567 (5.22) It is seen that these equations are linear equations with constant coefficients. Several methods are available to solve these simultaneous equations. We will discuss their solution by matrix inversion, the GausseSeidel method, and the Excel addin Solver. Matrix Inversion: Equations (5.13) through (5.22) can be written in the matrix form AT ¼ B (5.23) where A is the 10 10 matrix of the coefficients, T is the column vector of the 10 unknown temperatures, and B is the column vector of the constants on the right-hand sides of the equations. That is A[ 8.840 4.418 0 0 0 0 0 0 0 0

4.418 8.840 4.418 0 0 0 0 0 0 0

0 4.418 8.840 4.418 0 0 0 0 0 0

0 0 4.418 8.840 4.418 0 0 0 0 0

0 0 0 4.418 8.840 4.418 0 0 0 0

0 0 0 0 4.418 8.840 4.418 0 0 0

T[

B[

T2 T3 T4 T5 T6 T7 T8 T9 T10 T11

883.66 0.06362 0.06362 0.06362 0.06362 0.06362 0.06362 0.06362 0.06362 0.05567

0 0 0 0 0 4.418 8.840 4.418 0 0

0 0 0 0 0 0 4.418 8.840 4.418 0

0 0 0 0 0 0 0 4.418 8.840 4.418

0 0 0 0 0 0 0 0 4.418 4.4217

5.2 Finite-difference method

171

To obtain the temperatures of vector T, we first determine the inverse AL1 of the coefficients matrix A. The inverse is then multiplied by vector B to get the temperatures. That is (5.24) T ¼ A1 B The following outlines the procedures for solution using Excel and Matlab. Detailed instructions are included for Excel, and the reader is encouraged to run the program to get experience in solving equations by matrix inversion using Excel. Procedures are also given below for Matlab solution. In the Excel spreadsheet, we choose locations for the following four areas: a 10 10 block for matrix A, a 10 10 block for the inverse AL1 of A, a 10-element column for B, and a 10-element column where we want to have the temperatures. For example, let us decide to put matrix A in block A1:J10, matrix AL1 in block A12:J21, vector B in column A23:A32, and vector T in column A34:A43. We enter the values of the elements for A and B and then proceed as follows: To get the inverse matrix AL1, highlight the block A12:J21 and then enter the formula ¼ MINVERSE(A1:J10) in cell A12. Press CtrlþShiftþEnter. You will see the elements of the AL1 matrix appear. As shown in Eq. (5.24), we then have to multiply AL1 by B to get the temperatures. To do this, we highlight the column A34:A43 and then enter the formula ¼ MMULT(A12:J21,A23:A32) in cell A34. Press CtrlþShiftþEnter and the elements of vector T will appear. These elements are the desired values of T2 through T11. For Matlab, the solution may be achieved interactively or through use of an m-file. Matrix A and vector B are defined by assignment statements, and the temperatures T2 through T11 are obtained by the statement T ¼ inv(A)B or T ¼ A\B. The m-file for the solution is given in Appendix K. The results using matrix inversion were the same for both Matlab and Excel. The T2 through T11 temperatures were 198.40, 196.96, 195.69, 194.58, 193.64, 192.85, 192.23, 191.76, 191.45, and 191.31 C. From the problem statement, T1 ¼ 200 C. It should be noted that the matrix inversion method is only appropriate for the solution of linear simultaneous equations. If some equations are nonlinear (e.g., if we have radiative heat transfer and there are T4 terms), then matrix inversion cannot be used. We now discuss the GausseSeidel method and the Excel Solver add-in. These methods can be used for both linear and nonlinear equations. GausseSeidel Method: This is a very useful method that can be used to solve both linear and nonlinear simultaneous equations. Let us return to the 10 equations of this problem, i.e., Eqs. (5.13)e(5.22). Eq. (5.13) was for node 2, Eq. (5.14) was for node 3, etc. Let us rearrange the equations for solution by the GausseSeidel Method: Eq. (5.13), for node two is 8:840T2 þ 4:418T3 ¼ 883:66 Rearrange the equation so that T2 is the only item on the left side of the equal sign. Doing this, we get T2 ¼ 0:4998T3 þ 99:962

(5.13) (5.25)

The equation for node 3 is 4:418T2 8:840T3 þ 4:418T4 ¼ 0:06362 Rearranging this so that T3 is the only item left of the equal sign, we get

(5.14)

T3 ¼ 0:4998ðT2 þ T4 Þ þ 0:007197 Continuing this for the other eight equations, we have

(5.26)

T4 ¼ 0:4998ðT3 þ T5 Þ þ 0:007197

(5.27)

T5 ¼ 0:4998ðT4 þ T6 Þ þ 0:007197

(5.28)

T6 ¼ 0:4998ðT5 þ T7 Þ þ 0:007197

(5.29)

T7 ¼ 0:4998ðT6 þ T8 Þ þ 0:007197

(5.30)

T8 ¼ 0:4998ðT7 þ T9 Þ þ 0:007197

(5.31)

T9 ¼ 0:4998ðT8 þ T10 Þ þ 0:007197

(5.32)

T10 ¼ 0:4998ðT9 þ T11 Þ þ 0:007197

(5.33)

T11 ¼ 0:9992T10 þ 0:01259

(5.34)

172

Chapter 5 Numerical methods (steady and unsteady)

Using Matlab or similar software, the solution of these 10 equations (Eqs. 5.25 through 5.34) is obtained by successive looping through all the equations, one equation after another. Each time an equation is executed, the right side of the equation is calculated and the result becomes a new value for the temperature on the left side of the equation. After each complete loop through all the equations, we have new values for all the temperatures. Before starting the process, we have to give starting values for the unknown temperatures. For this particular problem, the base temperature is 200 C and the fluid temperature is 15 C. Initial guesses in this temperature range would be reasonable. When a loop is performed for all 10 equations, all the temperatures have been given new values. If the GausseSeidel Method is successful, successive looping will result in convergence where the temperatures do not change with additional looping. For our equations, the method is usually successful. As mentioned above, this method can be used for nonlinear, as well as linear, sets of equations. If the equations are nonlinear, convergence may be affected by the initial guesses for the temperatures. If the temperatures do not converge, try other starting temperatures. The Matlab program for this problem is given in Appendix K. The Excel Solver add-in program: Excel Solver, which comes with the Excel software, is a very useful program for solving both linear and nonlinear simultaneous equations. We will use it to solve for the temperatures in this Example 5.1, as follows: We will solve the nodal equations, Eqs. (5.25) through (5.34). Alternatively, we could solve Eqs. (5.13) through (5.22) as they are the same equations as (5.25) through (5.34). Proceeding with Solver: Starting with Eq. (5.25), rearrange the equation so that everything is to the left of the equal sign. T2 ¼ 0:4998T3 þ 99:962

(5.25)

Eq. (5.25) then becomes T2 0:4998T3 99:962 ¼ 0 Do this with the other nine nodal equations. Eqs. (5.26) through (5.34) then become

(5.35)

T3 0:4998ðT2 þ T4 Þ 0:007197 ¼ 0

(5.36)

T4 0:4998ðT3 þ T5 Þ 0:007197 ¼ 0

(5.37)

T5 0:4998ðT4 þ T6 Þ 0:007197 ¼ 0

(5.38)

T6 0:4998ðT5 þ T7 Þ 0:007197 ¼ 0

(5.39)

T7 0:4998ðT6 þ T8 Þ 0:007197 ¼ 0

(5.40)

T8 0:4998ðT7 þ T9 Þ 0:007197 ¼ 0

(5.41)

T9 0:4998ðT8 þ T10 Þ 0:007197 ¼ 0

(5.42)

T10 0:4998ðT9 þ T11 Þ 0:007197 ¼ 0

(5.43)

(5.44) T11 0:9992T10 0:01259 ¼ 0 Looking at these 10 equations, we see that the left sides of the equations are all functions of the unknown temperatures T2 through T11 : Let us call these functions f2 through f11 , respectively. For example, from Eq. (5.35), f2 ¼ T2 0:4998T3 99:962. From Eq. (5.36), f3 ¼ T3 0:4998ðT2 þT4 Þ 0:007197, etc. So, if we find temperatures such that all the functions f2 through f11 are equal to zero, then we have solved the nodal equations. Now comes the main part to get ready for using Solver: The requirement that all 10 functions be zero can be reduced to a single requirement; namely, that the sum of the squares of all the functions be zero. That is, we have the solution to the nodal equations if 11 X i¼2

fi2 ¼ 0

(5.45)

5.2 Finite-difference method

173

Now on to Solver: Open a new Excel spreadsheet. We decided to put the temperatures in the column A2 through A11. We need starting values (i.e., initial guesses) for the temperatures, so we put 200 in these 10 temperature cells. We entered the functions f2 through f11 , one-by-one, into the column C2 through C11. For example, cell C2 for function f2 is ¼ A2 0.4998 2 ; one-by-one, into the column E2 A3 99.962. We then entered the squares of the functions, f22 through f11 through E11. Finally, we entered ¼SUM(E2:E11) into cell G1. Calling up the Solver program, we instructed the program to make G1 zero by changing cells A2 through A11. We pressed Solve. The temperature values changed, but we got a statement that a solution could not be found. This often happens if the program does not get a result close enough to the required value of zero. In our case, the value in G1 was quite small, 0.00124, but not small enough for the program to declare a success. We reran the program asking it to make G1 a minimum rather than zero and got the same answers as before, but now the program declared that it was successful. Even though G1 was not zero, it was quite small, and the values for all the functions were also small. Hence, we decided to accept the temperature values that were, from T2 through T11: 198.38, 196.92, 195.62, 194.48, 193.49, 192.67, 192.01, 191.53, 191.21, and 191.06 C. From the problem statement, T1 ¼ 200 C. To summarize, we solved the nodal equations using the following methods: matrix inversion with Excel and Matlab, GausseSeidel with Matlab, and Excel Solver. The two matrix inversion solutions gave the same results for the temperatures. The GausseSeidel and Solver runs gave results slightly different from each other and slightly different from the matrix inversion results. This was expected, as the GausseSeidel and Solver computational methods are different and are iterative. But the bottom line is that the results from all methods were close to each other and all were deemed acceptable. We will use the temperature results to finally solve this problem, namely, to determine the heat flow from the fin. Recalling the discussion on fins in Chapter 3, the heat flow can be determined by looking at the heat flow into the fin from the base or by considering the heat flow by convection into the fluid. It turns out that looking at the convection to the fluid is the better alternative, for the following reason: The heat flow into the fin at the base is q ¼ kA vT : In our finite-difference model, the partial derivative is vx x¼0

T1 Þ approximated by ðT2Dx : As the grid is quite coarse, this approximation does not accurately represent the partial derivative. If we had taken additional smaller nodes near the base, we would have obtained a better result for ðvT= vxÞ and a more accurate result for heat flow q. So, let us sum up the convective heat flows from the different nodes to the fluid. Looking at Fig. 5.7, we have

q ¼ hPðDx = 2ÞðT1 TN Þ þ

10 X

hPðDxÞðTi TN Þ þ hPðDx = 2ÞðT11 TN Þ þ hAðT11 TN Þ

(5.46)

i¼2

Eq. (5.46) reduces to

" # 10 X 1 ðT1 þ T11 Þ þ Ti 10TN þ hAðT11 TN Þ 2 i¼2

q ¼ hPðDxÞ

(5.47)

Putting, into Eq. (5.47), the above-noted temperatures for the matrix inversion solution and the values of the other parameters in the problem, it is found that the heat flow from the fin to the fluid is 7.89 W. This is only a 0.25% difference from the 7.87 W result obtained for the identical Example 3.11 of Chapter 3 which was done analytically.

After this long, but needed, discussion of solution methods, let us move on to derive nodal equations for two-dimensional objects. Fig. 5.8 shows a flat plate with a rectangular grid. Nodes are of size Dx by Dy. Let us derive the steady-state nodal equation for node (i, j), which is an internal node for the plate, i.e., the node is not on a boundary.

174

Chapter 5 Numerical methods (steady and unsteady)

y

i, j+1 i–1, j

i, j

i+1, j

i, j–1 Δy Δx x

FIGURE 5.8 Two-dimensional plate with rectangular grid.

Node (i, j) is surrounded by nodes (i1, j), (iþ1, j), (i, j1), and (i, jþ1). Let us assume that all heat flows are into node (i, j) and that there could also be internal heat generation at volumetric strength qgen : Node i gets conductive heat flow from the four surrounding nodes and also heat from the internal generation. To better visualize the energy balance, let us give the plate a thickness “b.” This thickness will ultimately cancel out of the equation. Looking at the five heat flows into (i, j), we have kbDy kbDx kbDx kbDy Ti1;j Ti;j þ Tiþ1;j Ti;j þ Ti;j1 Ti;j þ Ti;jþ1 Ti;j þ qgen bDxDy Dx Dx Dy Dy ¼0 (5.48) This can be reduced to kbDx kbDy Ti1;j þ Tiþ1;j 2Ti;j þ Ti;j1 þ Ti;jþ1 2Ti;j þ qgen bDxDy ¼ 0 Dx Dy

(5.49)

If there is no internal heat generation, Eq. (5.49) becomes kbDx kbDy Ti1;j þ Tiþ1;j 2Ti;j þ Ti;j1 þ Ti;jþ1 2Ti;j ¼ 0 Dx Dy

(5.50)

5.2 Finite-difference method

175

Finally, if conductivity k is uniform throughout the body and we have a square grid, i.e., Dx ¼ Dy; Eq. (5.50) becomes Ti1;j þ Tiþ1;j þ Ti;j1 þ Ti;jþ1 4Ti;j ¼ 0

(5.51)

Rearranging this equation, we get Ti1;j þ Tiþ1;j þ Ti;j1 þ Ti;jþ1 (5.52) 4 Eq. (5.52) says that the temperature of a node is the average of the temperatures of the four surrounding nodes. The above nodal equations are for an interior node. Similar equations can be written for the other interior nodes of the object. Eqs. (5.49) through (5.52) were derived for an interior node (i, j) by considering an energy balance on the node. They could have alternatively been obtained through converting the appropriate differential equation to a numerical form, as follows: The general heat conduction equation was presented in Chapter 2: v vT v vT v vT vT (2.3) k þ k þ k þ qgen ¼ rc vx vx vy vy vz vz vt Ti;j ¼

If we have steady-state, constant conductivity, and two-dimensional conduction, then Eq. (2.3) reduces to v2 T v2 T qgen ¼0 þ þ vx2 vy2 k

(a)

Consider the five nodes shown in Fig. 5.8. We will convert the partial derivatives in Eq. (a) to their finite-difference form: Tiþ1;j Ti;j vT (b) z vx iþ1;j Dx 2 Ti;j Ti1;j vT (c) z vx i1;j Dx 2 Ti;jþ1 Ti;j vT (d) z vy i;jþ1 Dy 2 Ti;j Ti;j1 vT (e) z vy i;j1 Dy 2 vT vT vx iþ1;j vx i1;j v2 T 2 2 (f) z vx2 i;j Dx

176

Chapter 5 Numerical methods (steady and unsteady)

v2 T z vx2 i;j

Tiþ1;j Ti;j ðDxÞ2

!

Ti;j Ti1;j

! (g)

ðDxÞ2

Tiþ1;j 2Ti;j þ Ti1;j v2 T z 2 vx i;j ðDxÞ2 Ti;jþ1 2Ti;j þ Ti;j1 v2 T Similarly; for y; 2 z vy i;j ðDyÞ2

(h)

(i)

Putting Eqs. (h) and (i) into Eq. (a), we have the finite-difference form of the differential equation: Tiþ1;j þ Ti1;j 2Ti;j 2

ðDxÞ

þ

Ti;jþ1 þ Ti;j1 2Ti;j ðDyÞ

2

þ

qgen ¼0 k

(j)

If there is no internal heat generation, Eq. (j) becomes Tiþ1;j þ Ti1;j 2Ti;j ðDxÞ

2

þ

Ti;jþ1 þ Ti;j1 2Ti;j ðDyÞ2

¼0

(k)

Finally, if the grid is square with Dx ¼ Dy; Eq. (k) becomes Tiþ1;j þ Ti1;j þ T1;jþ1 þ Ti;j1 4Ti;j ¼ 0

(l)

Eq. (l), obtained from the differential equation, is the same as Eq. (5.51), which was obtained from an energy-balance approach. Let us now consider nodes on the boundary of an object. Fig. 5.9 shows node (i, j) and its surrounding nodes.

i, j+2 Δy i, j+1

qin

i, j

i+1, j

i, j–1 Boundary i, j –2 Δx

FIGURE 5.9 Node (i, j) on a boundary.

i+2, j

5.2 Finite-difference method

177

The boundary node (i, j) gets heat by conduction from its three surrounding nodes, possible heat input qin at the boundary, and possible heat input by internal generation. We will sum all the heat flows into (i, j) and set the result equal to zero as follows: qin þ

kbðDx=2Þ kbDy kbðDx=2Þ Ti;jþ1 Ti;j þ Ti;j1 Ti;j þ Tiþ1;j Ti;j Dy Dy Dx

þ qgen bðDx = 2ÞðDyÞ ¼ 0

(5.53)

If we have convection at the boundary, then

qin ¼ hbDy TN Ti;j

(5.54)

If we have radiation at the boundary, then

4 4 Ti;j qin ¼ εsbDy Tsurr

(5.55)

If the boundary is perfectly insulated, then qin ¼ 0

(5.56)

For example, if we have a square grid ðDx ¼ DyÞ; convection at the boundary, and no internal heat generation, then Eq. (5.53) reduces to the following equation for node (i, j): hDx 1 TN Ti;jþ1 þ Ti;j1 þ 2 k Ti;j ¼ (5.57) hDx 2þ k Another boundary situation is shown in Fig. 5.10, where we have node (i, j) as an exterior corner. The exterior corner node (i, j) gets heat by conduction from its two adjacent nodes, possible heat input qin at the boundary, and possible heat input by internal generation. Tiþ1:j þ

i, j+2

i, j+1 Δy i+1, j

i, j qin

FIGURE 5.10 Node (i, j) at exterior corner.

Δx

i+2, j

178

Chapter 5 Numerical methods (steady and unsteady)

Summing all the heat flows into (i, j) and setting the result equal to zero, we have qin þ

kbðDy=2Þ kbðDx=2Þ Ti;jþ1 Ti;j þ Tiþ1;j Ti;j þ qgen bðDx = 2ÞðDy = 2Þ ¼ 0 Dy Dx

If we have convection at the boundary, then Dx Dy þ qin ¼ hb TN Ti;j 2 2

(5.58)

(5.59)

If we have radiation at the boundary, then Dx Dy 4 4 þ Tsurr Ti;j qin ¼ εsb 2 2

(5.60)

If the boundary is perfectly insulated, then qin ¼ 0

(5.61)

For example, if we have a square grid ðDx ¼ DyÞ; convection at the boundary, and no internal heat generation, then Eq. (5.58) reduces to the following equation for node (i, j): hDx 1 TN Ti;jþ1 þ Tiþ1;j þ k Ti;j ¼ 2 (5.62) hDx 1þ k As a final example of a boundary node, let us look at Fig. 5.11, where we have a node on an interior corner.

i, j+2

i, j+1

i–1, j

i, j qin i, j–1 Δy i, j–2 Δx

FIGURE 5.11 Node (i, j) at interior corner.

i+1, j

i+2, j

5.2 Finite-difference method

179

The interior corner node (i, j) gets heat by conduction from its four adjacent nodes, possible heat input qin at the boundary, and possible heat input by internal generation. Summing all the heat flows into node (i, j) and setting the result to zero, we have kbðDy=2Þ kbDx kbDy kbðDx=2Þ Ti;j1 Ti;j þ Ti1;j Ti;j þ Ti;jþ1 Ti;j þ Tiþ1;j Ti;j Dy Dx Dy Dx 3 þ qgen b DxDy 4

qin þ

¼0 (5.63) If we have convection at the boundary, then Dx Dy qin ¼ hb þ TN Ti;j 2 2 If we have radiation at the boundary, then Dx Dy 4 4 þ qin ¼ εsb Tsurr Ti;j 2 2

(5.64)

(5.65)

If the boundary is perfectly insulated, then qin ¼ 0

(5.66)

For example, if we have a square grid ðDx ¼ DyÞ; convection at the boundary, and no internal heat generation, then Eq. (5.63) reduces to the following equation for node (i, j): Ti;j ¼

Ti;jþ1 þ Tiþ1:j þ

hDx 1 Ti;j1 þ Ti1;j þ TN 2 k hDx 3þ k

(5.67)

Example 5.2 Rectangular plate with temperature boundary conditions Problem The rectangular plate shown in Fig. 5.12 has temperature boundary conditions. Determine, numerically, the temperatures at the four internal nodes for steady conditions.

Solution

The nodal spacing is Dx ¼ 20 cm and Dy ¼ 10 cm. There is no internal heat generation. The four nodes are interior nodes, so we can use Eq. (5.50). kbDx kbDy Ti1;j þ Tiþ1;j 2Ti;j þ Ti;j1 þ Ti;jþ1 2Ti;j ¼ 0 Dx Dy Applying this equation to node 1, we have

(5.50)

0:5ð300 þ T2 2T1 Þ þ 2ðT3 þ 100 2T1 Þ ¼ 0 Rearranging Eq. (5.68), the equation for temperature T1 becomes

(5.68)

180

Chapter 5 Numerical methods (steady and unsteady)

100 C

300 C

1

2

3

4

400 C 30 cm

500 C 60 cm

FIGURE 5.12 Plate for Example 5.2.

T1 ¼ Applying Eq. (5.50) to node 2, we have

0:5T2 þ 2T3 þ 350 5

(5.69)

0:5ðT1 þ 400 2T2 Þ þ 2ðT4 þ 100 2T2 Þ ¼ 0

(5.70)

0:5T1 þ 2T4 þ 400 T2 ¼ 5 Continuing the same procedure for nodes 3 and 4, we get

(5.71)

0:5T4 þ 2T1 þ 1150 5

(5.72)

T3 ¼

0:5T3 þ 2T2 þ 1200 5 Solving Eqs. (5.69), (5.71), (5.72), and (5.73) by GausseSeidel, we got the results and T4 ¼

(5.73)

T 1 ¼ 241.3; T 2 ¼ 255.6; T 3 ¼ 364.4; T 4 ¼ 378.7 C An alternative way of solving this problem is to change the grid from a rectangular grid to a square grid, as shown in Fig. 5.13. There are more nodes and equations (10 vs. 4), but it is easier to write the equations as the ratio Dy=Dx becomes 1. Eq. (5.52) shows that, for an interior node and a square grid, a nodal temperature is simply the average of the temperatures of the four surrounding nodes. The equations for the 10 nodes of this revised grid can be quickly written by inspection. For example, the temperature of node 1 is (300 þ 100 þ T2 þ T6)/4; the temperature of node 2 is (T1 þ 100 þ T3 þ T7)/4, etc. Solving the 10 equations by GausseSeidel, we got the below results: T 1 ¼ 248.0;

T 2 ¼ 238.4;

T 3 ¼ 240.1;

T 4 ¼ 250.9;

T 5 ¼ 285.5;

T 6 ¼ 353.4;

T 7 ¼ 365.7;

T 8 ¼ 371.0;

T 9 ¼ 378.2;

T 10 ¼ 390.9 C

100 C

300 C

1

2

3

4

5

6

7

8

9

10

500 C 60 cm

FIGURE 5.13 Revised grid for Example 5.2.

400 C 30 cm

5.2 Finite-difference method

181

In this revised, finer grid, node 2 is the same as the original node 1; node 4 is the same as the original node 2; node 7 is the same as the original node 3, and node 9 is the same as the original node 4. Using the labeling of the original grid, the result is T 1 ¼ 238.4;

T 2 ¼ 250.9;

T 3 ¼ 365.7;

T 4 ¼ 378.2 C

ð10 node gridÞ

T 1 ¼ 241.3; T 2 ¼ 255.6; T 3 ¼ 364.4; T 4 ¼ 378.7 C ð4 node gridÞ It is seen that there is a small difference between the results of the two models. In general, a finer grid gives more accurate results than a coarser grid unless the grid is so fine that round-off error becomes significant. The Matlab programs for both the 4-node and 10-node models are in Appendix K. (Note: Although this problem was for a plate, it could have been for a long column of rectangular cross section having different temperatures on its four sides.)

Example 5.2 only had interior nodes. The next example has boundary, as well as interior, nodes. It also has internal heat generation.

Example 5.3 Plate with a variety of boundary conditions and heat generation Problem The plate shown in Fig. 5.14 has a variety of boundary conditions. The left surface is isothermal at 500 C. The bottom surface is isothermal at 100 C. The top and right surfaces are insulated, and the inside corner on the left has convection to a fluid. The convective coefficient is 100 W/m2 C, and the fluid temperature is 30 C. The plate is steel and has a conductivity of 43 W/m C. The grid is square with Dx ¼ Dy ¼ 8 cm. (a) If there is no internal heat generation, determine the nodal temperatures for steady state. (b) If the plate has internal heat generation at a strength of 106 W/m3, determine the nodal temperatures for steady state.

24 cm insulated

Δx

500 C

1

2

3

4

5

6

7

8

9

10

h = 100

W

24 cm

2

m C Δy

T∞ = 30 C 11 100 C

FIGURE 5.14 Plate for Example 5.3.

12

182

Chapter 5 Numerical methods (steady and unsteady)

Solution Let us start the solution by deriving the nodal equations. For a particular node, we assume that all heat flows go into the node and we sum up the heat flows to zero. Again, we will use “b" as the thickness of the plate. As Part (b) has heat generation, we will include the heat generation term in the nodal equations. Insulated nodes (2, 3, 4, 8, 10): Each node gets heat flow by conduction from its adjacent nodes. For node 2, we have conduction from nodes 1, 3, and 6. The nodal equation is kbðDy=2Þ kbðDy=2Þ kbDx Dy ðT1 T2 Þ þ ðT3 T2 Þ þ ðT6 T2 Þ þ qgen b Dx ¼ 0 Dx Dx Dy 2 For node 3, we have conduction from nodes 2, 4, and 7, and the nodal equation is kbðDy=2Þ kbðDy=2Þ kbDx Dy ðT2 T3 Þ þ ðT4 T3 Þ þ ðT7 T3 Þ þ qgen b Dx ¼ 0 Dx Dx Dy 2 Similarly for nodes 4, 8, and 10: Node 4

Node 8

Node 10

(5.74)

(5.75)

kbðDy=2Þ kbðDx=2Þ Dy Dx ðT3 T4 Þ þ ðT8 T4 Þ þ qgen b ¼0 Dx Dy 2 2

(5.76)

kbðDx=2Þ kbðDx=2Þ kbDy Dx ðT4 T8 Þ þ ðT10 T8 Þ þ ðT7 T8 Þ þ qgen b Dy ¼ 0 Dy Dy Dx 2

(5.77)

kbðDx=2Þ kbðDx=2Þ kbDy Dx ðT8 T10 Þ þ ðT12 T10 Þ þ ðT9 T10 Þ þ qgen b Dy ¼ 0 Dy Dy Dx 2 Convection nodes (6, 7, 9): These nodes get conduction from adjacent nodes and convection from the adjacent fluid. For node 6, we have conduction from nodes 2, 5, and 7 and convection from the fluid. The nodal equation is kbðDy=2Þ kbðDy=2Þ kbDx Dy ðT5 T6 Þ þ ðT7 T6 Þ þ ðT2 T6 Þ þ hbDxðTN T6 Þ þ qgen b Dx ¼ 0 Dx Dx Dy 2 Similarly for nodes 7 and 9: Node 7 kbðDy=2Þ kbðDx=2Þ kbDy kbDx ðT6 T7 Þ þ ðT9 T7 Þ þ ðT8 T7 Þ þ ðT3 T7 Þ Dx Dy Dx Dy 3 þhbððDx=2Þ þ ðDy=2ÞÞðTN T7 Þ þ qgen b DxDy ¼ 0 4

(5.78)

(5.79)

(5.80)

Node 9 kbðDx=2Þ kbðDx=2Þ kbDy Dx ðT7 T9 Þ þ ðT11 T9 Þ þ ðT10 T9 Þ þ hbDyðTN T9 Þ þ qgen b Dy ¼ 0 Dy Dy Dx 2 Isothermal nodes (1, 5, 11, 12):

(5.81)

T1 ¼ T5 ¼ 500 C and T11 ¼ T12 ¼ 100 C (5.82) We will solve the nodal equations by matrix inversion. We have a square grid ðDx ¼ DyÞ, which simplifies the equations considerably. To use matrix inversion, we need to rearrange the terms in Eqs. (5.74)e(5.82) to facilitate creation of a coefficients matrix and a constants vector. Doing this, and incorporating the square matrix feature, the equations become For Node 1: T1 ¼ 500

(5.83) 2

qgen ðDxÞ 1 1 For Node 2: T1 þ 2T2 T3 T6 ¼ 2 2 2k

(5.84)

qgen ðDxÞ2 1 1 For Node 3: T2 þ 2T3 T4 T7 ¼ 2 2 2k

(5.85)

5.2 Finite-difference method

qgen ðDxÞ2 1 1 For Node 4: T3 þ T4 T8 ¼ 2 2 4k

183

(5.86)

For Node 5: T5 ¼ 500 qgen ðDxÞ2 hDx 1 hDx 1 T6 T7 ¼ TN þ For Node 6: T2 T5 þ 2 þ 2 k 2 k 2k 1 hDx 1 3 qgen ðDxÞ2 hDx T7 T8 T9 ¼ TN þ For Node 7: T3 T6 þ 3 þ 2 k 2 4 k k qgen ðDxÞ2 1 1 For Node 8: T4 T7 þ 2T8 T10 ¼ 2 2 2k qgen ðDxÞ2 hDx 1 hDx 1 T9 T10 T11 ¼ TN þ For Node 9: T7 þ 2 þ 2 k 2 k 2k

(5.87) (5.88)

(5.89) (5.90) (5.91)

qgen ðDxÞ2 1 1 For Node 10: T8 T9 þ 2T10 T12 ¼ 2 2 2k

(5.92)

For Node 11: T11 ¼ 100

(5.93)

For Node 12: T12 ¼ 100 (5.94) For our problem, k ¼ 43 W=mC, h ¼ 100 W/m2 C, TN ¼ 30 C, and Dx ¼ 8 cm ¼ 0.08 m. We put these values into Eqs. (5.83)e(5.94) and created the coefficients matrix and the constants vector. For Part (a) of the problem, qgen ¼ 0: For Part (b) of the problem, qgen ¼ 106 W/m3. The problem was solved using matrix inversion and Matlab, with the following results: (a) qgen [ 0 T1 T2 T3 T4 T5 T6 T7 T8 T9 T10 T11 T12

500 C 353.9 256.7 231.0 500 329.3 221.0 205.3 143.8 148.2 100 100

(b) qgen [ 106 W=m3 500 C 589.8 617.0 628.5 500 546.6 550.4 565.6 361.1 384.1 100 100

The Matlab program for this solution is in Appendix K. (Note: Although this problem was for a plate, it could have been for a long L-shaped structural support having different conditions on its various surfaces.)

Let us now move on to a discussion of numerical solution of unsteady problems.

5.2.2 Unsteady state If we have unsteady heat flow in an object, we have to consider time, as well as location. For steady state, we had spatial increments such as Dx and Dy. For unsteady state, we have a time increment, or “step,” Dt in addition to the spatial increments. As shown in Fig. 5.15, the time scale progresses from the origin at time zero, to one time step (Dt), two time steps (2Dt), three time steps (3Dt), etc.

184

Chapter 5 Numerical methods (steady and unsteady)

0

Δt

2Δt

3Δt

4Δt

5Δt

6Δt

t

Δt

FIGURE 5.15 The time scale.

In general, temperatures in an object are known at the beginning of a process, i.e., at time zero. Problem solution proceeds from one time step to the next, with temperatures being determined for each successive time step. Let us now determine an energy conservation equation for unsteady state situations. Fig. 5.16 shows an interior node i in a one-dimensional object. Nodes i1 and iþ1 are the two nodes adjacent to node i. There is conductive heat flow between these nodes and node i. There may also be internal heat generation. Like steady state, we will consider all heat flows to be into a node. A positive net heat flow into node i will increase the node’s internal energy. A negative net heat flow into node i will decrease the node’s internal energy. Let us look at the change in internal energy of node i during a time increment Dt: In words, we have The net heat flow into Node i ¼ Change in internal energy of material during time Dt

in node i during time Dt

Mathematically, this is kA b kA b Ti1 Tib Dt þ Tiþ1 Tib Dt þ qgen ADx Dt ¼ rcADx Tie Tib Dx Dx where k ¼ thermal conductivity of the material r ¼ density of the material c ¼ specific heat of the material

i–2

i–1

i

Δx

FIGURE 5.16 Interior nodes for one-dimensional object.

i+1

i+2

(5.95)

(5.96)

5.2 Finite-difference method

185

Dx ¼ spatial increment Dt ¼ time increment A ¼ cross-sectional area qgen ¼ volumetric heat generation rate Eq. (5.96) describes the change in internal energy during one time step Dt. The terms on the left are the heat flows into node i during the time step. We have the heat flow rates times Dt to give the heat quantity into node i during the time step. The change in internal energy of the material in node i during the time step is on the right side of Eq. (5.96). This internal energy change causes a change in temperature of the node. At the beginning of the time step, the temperature of node i is Tib : At the end of the time step, the temperature is Tie : Let us talk a little more about these temperatures. For numerical finite differences, time is not continuous. Rather, it is a series of successive, discrete time steps. Temperatures are defined at the beginning and end of these time steps. In Eq. (5.96), temperature superscript “b” designates temperatures at the beginning of the time step, and superscript “e” designates temperatures at the end of the time step. Looking at the left side of Eq. (5.96), it is seen that the heat flows into the node are based on temperatures at the beginning of the time step. We need temperatures at both the beginning and the end of the time step on the right side of Eq. (5.96) as we need the temperature change for the step. Equations like Eq. (5.96) may be written for all interior nodes in an object. Let us rearrange the equation so that Tie is on the left side of the equation. Doing this, we have ! q Dt aDt aDt b gen e b b (5.97) Ti ¼ Ti 1 2 þ Ti1 þ Tiþ1 þ 2 2 rc ðDxÞ ðDxÞ where a ¼ thermal diffusivity ¼ k/r c. A two-dimensional plate is shown in Fig. 5.17.

y

i, j+1 i–1, j

i, j

i+1, j

i, j–1 Δy Δx x

FIGURE 5.17 Interior nodes for two-dimensional plate.

186

Chapter 5 Numerical methods (steady and unsteady)

Interior node (i, j) gets conduction from surrounding nodes (i1, j), (iþ1, j), (i, j1), and (i, jþ1). It also may get heat from internal heat generation. Like the one-dimensional case, a positive net heat flow into node (i, j) will raise the internal energy of the node. The energy conservation equation for node (i, j) is kbDy b kbDy b kbDx b kbDx b b b b b Ti1;j Ti;j Tiþ1;j Ti;j Ti;j1 Ti;j Ti;jþ1 Ti;j Dt þ Dt þ Dt þ Dt Dx Dx Dy Dy e b Ti;j þqgen bDxDyDt ¼ rcbDxDy Ti;j (5.98) e Ti;j

Rearranging this equation so that is on the left side of the equation, we get ! q Dt aDt aDt aDt b aDt b gen e b b b ¼ Ti;j 2 þ T þ T Ti;j 12 þ T þ T i1;j iþ1;j i;j1 i;jþ1 þ rc ðDxÞ2 ðDyÞ2 ðDxÞ2 ðDyÞ2 (5.99) If we have a square grid (Dx ¼ Dy) and no heat generation, Eq. (5.99) reduces to ! aDt aDt b e b b b b ¼ Ti;j þ T þ T þ T Ti;j 14 þ T i1;j iþ1;j i;j1 i;jþ1 ðDxÞ2 ðDxÞ2

(5.100)

Furthermore, if we also pick the time step Dt such that 1 4 aDt2 ¼ 0. That is, if ðDxÞ ðDxÞ2 Dt ¼ 4a , then Eq. (5.100) becomes 1 b e b b b Ti;j ¼ Ti1;j þ Tiþ1;j þ Ti;j1 þ Ti;jþ1 (5.101) 4 Eq. (5.101) states that the temperature of node (i, j) at the end of the time step is equal to the average of the temperatures of the four surrounding nodes at the beginning of the time step. Energy conservation equations may also be written for boundary nodes. Consider, for example, the corner in Fig. 5.18, which has convection from the adjacent fluid and conduction from nodes (i, jþ1) and (iþ1, j). The energy equation for corner node (i, j) is kbðDx=2Þ b kbðDy=2Þ b Dx Dy b b b þ Ti;jþ1 Ti;j Dt þ Tiþ1;j Ti;j Dt þ hb TN Ti;j Dt Dy Dx 2 2 e b Ti;j þqgen bðDx=2Þ ðDy=2ÞDt ¼ rcbðDx=2Þ ðDy=2Þ Ti;j (5.102) Rearranging this equation so that e b Ti;j ¼ Ti;j

e Ti;j

is on the left side, we get ! aDt aDt hDtðDx þ DyÞ aDt b aDt b þ2 12 2 2 T þ2 Ti;jþ1 2 2 2 iþ1;j rcðDxÞðDyÞ ðDxÞ ðDyÞ ðDxÞ ðDyÞ2

qgen Dt hDtðDx þ DyÞ TN þ þ2 rcðDxÞðDyÞ rc

(5.103)

5.2 Finite-difference method

187

i, j+2

i, j+1 Δy i+1, j

i, j convection h, T∞

i+2, j

Δx

FIGURE 5.18 Exterior corner with convection.

If the grid is square (Dx ¼ Dy) and there is no heat generation, Eq. (5.103) reduces to ! aDt hDt aDt b hDt e b b TN 4 þ T þ 2 T Ti;j ¼ Ti;j 1 4 iþ1;j i;jþ1 þ 4 2 2 rcðDxÞ rcðDxÞ ðDxÞ ðDxÞ

(5.104)

We can write nodal equations similar to Eqs. (5.97), (5.99), (5.100), (5.101), (5.103), and (5.104) for all the interior and boundary nodes of a problem. If we look at these equations, it is seen from the superscripts that the temperature on the left side of the equation is the temperature at the end of a time step and all the temperatures on the right side are temperatures at the beginning of a time step. In solving the equations, we work forward in time. For each time step, we know the temperatures at the beginning of the time step. We calculate the right sides of the equations, and the result becomes the temperatures on the left side of the equation. Then we continue with the next time step. The temperatures we got on the left side of the equations now become the beginning temperatures for the next time step. We use these temperatures on the right side of the equations to calculate end temperatures for the time step. And, the process continues for as many time steps as we desire. For steady-state problems we had to simultaneously solve the nodal equations. For unsteady problems, we simply list the equations, one after another, and go through the equations successively. We calculate the righthand sides of the equations using temperatures that we already know and we get new temperatures on the left sides of the equations that are temperatures at the end of that time step. These temperatures are now used on the right sides of the equations to determine the temperatures at the end of the next time step. This procedure can be easily done using loops in Matlab or iterations in Excel. We will give a few examples of the procedure, but let us first discuss the determination of time increment Dt. When we solve a problem, we first decide on the spatial increments such as Dx, Dy, Dz, Dr, etc. We then decide on what time increment Dt to use. The above nodal equations were developed using the forward-difference or explicit approach. As described above, we know the temperatures at the beginning of the time step and we get the temperatures at the end of the time step using the nodal

188

Chapter 5 Numerical methods (steady and unsteady)

equations. We are working forward in time. It turns out that selection of the spatial and time increments are not independent with this forward-difference approach. If we decide on too large a Dt, the procedure will be unstable and not result in the correct solution. What is “too large” for Dt ? Let us look at Eq. (5.97), which is the equation for an interior node (i) in a one-dimensional object, and Eq. (5.99), which is the equation for an interior node (i, j) in a twodimensional object. The coefficients for nodes (i) and (i, j) in these equations must be zero or positive for a stable solution. If Dt is too large, the coefficients become negative and the solution is unstable. That is, for the forward-difference approach, 12 12

aDt ðDxÞ

2

2

aDt ðDxÞ2

aDt ðDyÞ

2

0 and Dt

0 and Dt

ðDxÞ2 ðOne DimensionalÞ 2a 1

2a ðDxÞ2

þ

2a

! ðTwo DimensionalÞ

(5.105) (5.106)

ðDyÞ2

These restrictions on the size of Dt are based on interior points in a body. If there are boundary nodes, Dt may have to be a bit smaller. See, for example, Eq. (5.103), which is for a two-dimensional boundary node with convection. The coefficient for node (i, j) needs to be zero or positive, and the convection term makes the acceptable Dt smaller than that given by Eq. (5.106). As a practical matter, these restrictions on Dt usually have no impact. Most of the time we are able to use a Dt meeting the criteria. However, if we really need to use a larger Dt, then we can use the backward-difference or implicit approach. With this approach, the spatial and time increments are independent, and there is no restriction on the size of Dt. However, as we will see, much more computation is needed for the backward-difference approach than for the forward-difference approach. Let us look at this backward-difference approach. The previous equation we had for an interior node in a two-dimensional object was kbDy b kbDy b kbDx b kbDx b b b b b Ti1;j Ti;j Tiþ1;j Ti;j Ti;j1 Ti;j Ti;jþ1 Ti;j Dt þ Dt þ Dt þ Dt Dx Dx Dy Dy þ qgen bDxDyDt e b ¼ rcbDxDy Ti;j Ti;j (5.98) The temperature differences on the left side of the equation have superscript “b," which means they are the differences at the beginning of the time step. Let us instead make these differences the differences at the end of the time step. That is, we will give them the superscript “e." kbDy e kbDy e kbDx e kbDx e e e e e Ti1;j Ti;j Tiþ1;j Ti;j Ti;j1 Ti;j Ti;jþ1 Ti;j Dt þ Dt þ Dt þ Dt Dx Dx Dy Dy e b Ti;j þqgen bDxDyDt ¼ rcbDxDy Ti;j (5.107)

5.2 Finite-difference method

189

Rearranging this equation, we get ! q Dt aDt aDt aDt e aDt e gen b e e e þ 2 þ T þ T Ti;j ¼ Ti;j 1 þ 2 T T i1;j iþ1;j i;j1 i;jþ1 rc ðDxÞ2 ðDyÞ2 ðDxÞ2 ðDyÞ2 (5.108) All items in the coefficient for Ti;j in Eq. (5.108) are now positive, and there is no restriction on the size of Dt. Similar equations may be written for all nodes in an object, giving us a set of equations to solve for the temperatures at the end of a time step. Unfortunately, there are many unknowns in each equation: namely, the temperatures of the various nodes at the end of the time step. We have to solve the set of equations simultaneously to get the temperatures at the end of each time step before we can move forward to the next time step. The computational effort is much more than we had with the forward-difference approach. In using the backward-difference approach, we have eliminated the size restriction on Dt, but we have increased the computational effort greatly. With the forward-difference approach, we merely looped through the nodal equations, getting new values for the temperatures and working forward one time step per loop. With the backward-difference approach, we have to solve simultaneous equations for the temperatures for each time step as we work forward in time. We will now give some examples illustrating solution of unsteady state problems.

Example 5.4 Rectangular plate with unsteady conduction Problem

The stainless steel (k ¼ 15 W/m C, r ¼ 7900 kg/m3, c ¼ 477 J/kg C) plate shown in Fig. 5.19 has an initial uniform temperature of 30 C. At time zero, the boundary temperatures shown in the figure are imposed on the plate. (a) How long does it take for node 1 to reach 150 C and what are the temperatures of the other nodes at that time? (b) What are the steady-state temperatures for the nodes? (c) How long does it take for the plate to achieve steady state?

Solution The four nodes of the plate are interior nodes. Each node is surrounded by four other nodes. If we call the node we are writing the equation for “node (i, j),” then the surrounding nodes are nodes (i1, j), (iþ1, j), (i, j1), and (i, jþ1). There is no internal heat generation. Therefore, Eq. (5.99), with qgen ¼ 0, can be used to obtain the nodal equations. This equation is

100 C

300 C

1

2

3

4

500 C 60 cm

FIGURE 5.19 Plate for Example 5.4.

400 C 30 cm

190

Chapter 5 Numerical methods (steady and unsteady)

! aDt aDt b aDt b b b 2 þ Ti1;j þ Tiþ1;j þ Ti;j1 þ Ti;jþ1 (5.109) 2 2 2 2 ðDxÞ ðDyÞ ðDxÞ ðDyÞ Applying Eq. (5.109) to the four nodes, we get the following nodal equations: ! aDt aDt aDt aDt e b 2 þ 300 þ T2b þ 100 þ T3b (5.110) T1 ¼ T1 1 2 ðDxÞ2 ðDyÞ2 ðDxÞ2 ðDyÞ2 ! aDt aDt aDt b aDt b T1 þ 400 þ T4 þ 100 2 (5.111) þ T2e ¼ T2b 1 2 2 2 2 2 ðDxÞ ðDyÞ ðDxÞ ðDyÞ ! aDt aDt aDt aDt e b 300 þ T4b þ 500 þ T1b (5.112) 2 T3 ¼ T3 1 2 þ 2 2 2 2 ðDxÞ ðDyÞ ðDxÞ ðDyÞ ! aDt aDt aDt b aDt b þ T3 þ 400 þ T2 þ 500 T4e ¼ T4b 1 2 2 (5.113) 2 2 2 2 ðDxÞ ðDyÞ ðDxÞ ðDyÞ One of the first things to do is decide on a time step. For the forward-difference approach, the bracketed expression ! e b Ti;j ¼ Ti;j 12

1 2

aDt ðDxÞ2

2

aDt ðDyÞ2

aDt

cannot be negative. This limits the size of Dt. For this problem, Dx ¼ 0:2 m and Dy ¼ 0:1 m

k ¼ 15 6 m2 =s. and a ¼ rc ð7900Þð477Þ ¼ 3:981 10

Putting these values into the bracketed expression, it is found that Dt must be less than or equal to 1004 s. We used a Dt of 1 s, certainly well within the limit. A small Dt will give more-accurate results. A time step of 10 s gave very good results and a time step of 100 s gave good results. Above that time step, the results were not accurate. (a) The Matlab program for this problem is in Appendix K. The four nodes are given initial values of 30 C and then the program loops through Eq. (5.110)e(5.113) successively. Each loop (or pass) moves the time ahead one Dt. When node 1 reaches 150 C, the program stops and prints out all the nodal temperatures. The program took 2134 loops for node 1 to reach 150 C. With Dt ¼ 1 s, this means node 1 reached 150 C in 2134 s [ 35.6 min. At this time, T2 [ 161.1 C, T3 [ 265.3 C, and T4 [ 276.4 C. (b) We ran the program without the T1 ¼ 150 breakpoint to see what final temperatures would be reached. For 19,300 loops and greater, the temperatures did not change. Steady state had been reached. Final temperatures were T1 [ 241.3 C, T2 [ 255.6 C, T3 [ 364.4 C, and T4 [ 378.7 C. These temperatures are the same as we got for the GausseSeidel solution to Example 5.2, which was the steady-state problem for this geometry. (c) As mentioned in Part (b): By running the program to steady-state conditions, we found that temperatures stopped changing and steady state was reached in 19,300 loops, or 19,300 s ¼ 5.4 h. We may wish to define steady state a bit less stringently, as follows: Earlier in this text, we mentioned that, for a first-order system, a process is often considered to be complete after five time constants. At this point, the process is actually 99.33% complete. Let us consider node 1. For its process, the node goes from an initial 30 C to a final 241.3 C. The process is therefore a temperature change of 241.3e30 or 211.3 C. The 99.33% change for the process is 211.3 0.9933 ¼ 209.9 C, which gives a final temperature of 30 þ 209.9 ¼ 239.9 C. Running the Matlab program again, it is found that node 1 reaches 239.9 C in 10,610 s ¼ 2.95 h. Similar results are found for the other three nodes. So, if we use this “five time constant” criterion, we could say that steady state is essentially reached after 3 h. (Note: Although this problem was for a plate, it could have been for a long column of rectangular cross section having different temperatures on its four sides.)

We now look at an example in cylindrical coordinates, which have both convection and radiation boundary conditions.

Example 5.5 Disk with convection and radiation Problem

The stainless steel disk (k ¼ 15 W/m C, r ¼ 7900 kg/m3, c ¼ 477 J/kg C) shown in Fig. 5.20 has a diameter of 8 cm and a thickness of 1 cm. It is perfectly insulated on the bottom and has convection and radiation from the top. The top surface has

5.2 Finite-difference method

191

qconv, qrad

4 cm 1 cm

qin (heater tape)

Insulated

FIGURE 5.20 Disk for Example 5.5. an emissivity of 0.5 and a convective coefficient of 100 W/m2 C. Radiation is to the surroundings that are at 20 C, and convection is to a fluid also at 20 C. There is an electric heater tape on the side of the disk that inputs 300 W into the disk. The disk is initially uniform at 20 C. The heater tape is then turned on. Assume that the temperature in the disk varies only radially. There is no axial variation. (a) Assume there are five nodes as shown in the top view of the disk in Fig. 5.21. What are the nodal temperatures 4 minutes after the heater is turned on? (b) What are the nodal temperatures of the disk at steady state?

Solution Let us first get the areas and volumes for the five nodes. Looking at Fig. 5.21, the areas on the top surface in contact with the surroundings are i i h h A3 ¼ p ð5Dr=2Þ2 ð3Dr=2Þ2 A1 ¼ pðDr=2Þ2 A2 ¼ p ð3Dr=2Þ2 ðDr=2Þ2 i i h h A5 ¼ p ð4DrÞ2 ð7Dr=2Þ2 A4 ¼ p ð7Dr=2Þ2 ð5Dr=2Þ2

qin

1

2

3

4 5

Δr = 1 cm Δr

FIGURE 5.21 Top view of disk showing nodes.

192

Chapter 5 Numerical methods (steady and unsteady)

Each node gets conduction from adjacent nodes. The thickness of the disk is “b." The areas are the cross-sectional areas for the radial conduction. The subscripts on the areas denote the two nodes sharing conduction. For example, A12 is for conduction between nodes 1 and 2. A12 ¼ b½2pðDr = 2Þ The volumes of the nodes are

A23 ¼ b½2pð3Dr = 2Þ

A34 ¼ b½2pð5Dr = 2Þ

A45 ¼ b½2pð7Dr = 2Þ

V1 ¼ bA1 V2 ¼ bA2 V3 ¼ bA3 V4 ¼ bA4 V5 ¼ bA5 In deriving the nodal equations, we use the basic equations for conduction, convection, and radiation. 4 T 4 That is, qcond ¼ kA ðDTÞ q ¼ hAðT T Þ q ¼ εsA T conv N rad surr L For the conduction, A is the cross-sectional area through which the heat flows and L is the distance between nodes. For convection and radiation, A is the surface area of the particular node. As we have done previously, we assume that all heat flows into the node we are considering. Node 1 gets heat by conduction from node 2, convection from the adjacent fluid, and radiation from the surroundings. The rate of heat flow into node 1 is 4 kA12 ðT2 T1 Þ þ hA1 ðTN T1 Þ þ εsA1 Tsurr T14 Dr The rate of heat flows into the other nodes 2, 3, and 4 are q1 ¼

(5.114)

4 kA12 kA23 ðT1 T2 Þ þ ðT3 T2 Þ þ hA2 ðTN T2 Þ þ εsA2 Tsurr T24 (5.115) Dr Dr 4 kA23 kA34 ðT2 T3 Þ þ ðT4 T3 Þ þ hA3 ðTN T3 Þ þ εsA3 Tsurr T34 (5.116) q3 ¼ Dr Dr 4 kA34 kA45 ðT3 T4 Þ þ ðT5 T4 Þ þ hA4 ðTN T4 Þ þ εsA4 Tsurr T44 (5.117) q4 ¼ Dr Dr Node 5 gets conduction from node 4, convection from the adjacent fluid, radiation from the surroundings, and heat input qin from the heater tape. q2 ¼

4 kA45 ðT4 T5 Þ þ hA5 ðTN T5 Þ þ εsA5 Tsurr T54 þ qin (5.118) Dr If we multiply the heat flow rates in Equations (5.114) through (5.118) by Dt, we get the heat going into a node during the time step Dt. This heat input is equal to the increase in the internal energy of the node during the time step. For example, for node 1, from Eq. (5.114), we have 4 kA12 b 4 (5.119) T2 T1b þ hA1 TN T1b þ εsA1 Tsurr T1b Dt ¼ rcV1 T1e T1b q1 Dt ¼ Dr Similarly, for the other four nodes we have kA 4 kA12 b 23 4 q2 Dt ¼ T1 T2b þ T3b T2b þ hA2 TN T2b þ εsA2 Tsurr T2b Dt ¼ rcV2 T2e T2b (5.120) Dr Dr 4 kA kA23 b 34 4 q3 Dt ¼ T3b Dt ¼ rcV3 T3e T3b (5.121) T2 T3b þ T4b T3b þ hA3 TN T3b þ εsA3 Tsurr Dr Dr kA 4 kA34 b 45 4 T3 T4b þ T5b T4b þ hA4 TN T4b þ εsA4 Tsurr T4b Dt ¼ rcV4 T4e T4b (5.122) q4 Dt ¼ Dr Dr 4

kA45 b 4 (5.123) T4 T5b þ hA5 TN T5b þ εsA4 Tsurr q5 Dt ¼ T5b þ qin Dt ¼ rcV5 T5e T5b Dr Rearranging Equations (5.119) through (5.123) so that the temperatures at the end of the time step are on the left side of the equations, we get A12 aDt hA1 Dt εsA1 Dt b 3 A12 aDt b hA1 Dt εsA1 Dt 4 T1e ¼ T1b 1 T1 T þ TN þ T (5.124) þ ðDrÞV1 rcV1 rcV1 ðDrÞV1 2 rcV1 rcV1 surr q5 ¼

5.2 Finite-difference method

A12 aDt A23 aDt hA2 Dt εsA2 Dt b 3 A12 aDt b A23 aDt b T2 T2e ¼ T2b 1 T þ T þ ðDrÞV2 ðDrÞV2 rcV2 rcV2 ðDrÞV2 1 ðDrÞV2 3 hA2 Dt εsA2 Dt 4 þ TN þ T rcV2 rcV2 surr A23 aDt A34 aDt hA3 Dt εsA3 Dt b 3 A23 aDt b A34 aDt b T3 T þ T þ T3e ¼ T3b 1 ðDrÞV3 ðDrÞV3 rcV3 rcV3 ðDrÞV3 2 ðDrÞV3 4 hA3 Dt εsA3 Dt 4 þ TN þ T rcV3 rcV3 surr A34 aDt A45 aDt hA4 Dt εsA4 Dt b 3 A34 aDt b A45 aDt b T4 T þ T þ T4e ¼ T4b 1 ðDrÞV4 ðDrÞV4 rcV4 rcV4 ðDrÞV4 3 ðDrÞV4 5 hA4 Dt εsA4 Dt 4 þ TN þ T rcV4 rcV4 surr A45 aDt hA5 Dt εsA5 Dt b 3 A45 aDt b hA5 Dt T5 T þ TN þ T5e ¼ T5b 1 ðDrÞV5 rcV5 rcV5 ðDrÞV5 4 rcV5 εsA5 Dt 4 qin Dt þ T þ rcV5 surr rcV5

193

(5.125)

(5.126)

(5.127)

(5.128)

(a) A Matlab program was written to solve the problem by the forward-difference approach. The program is in Appendix K. As radiation is present, we needed to use the absolute temperature scale, i.e., Kelvin. The five nodes were given the initial temperature of 20 C ¼ 293.15 K. Eqs. (5.124) through (5.128) were arranged successively in a loop. We picked a time increment Dt of 1 second. As we wanted the nodal temperatures after 4 min ¼ 240 s, the program performed 240 loops and then output the nodal temperatures. The temperatures after 4 min were found to be T1 [ 494.5 K [ 221.4 C T2 [ 503.0 K [ 229.9 C T3 [ 529.0 K [ 255.9 C T4 [ 573.9 K [ 300.8 C T5 [ 640.4 K [ 367.3 C (b) To determine the steady-state temperatures: Instead of only doing 240 loops (4 min), we continued the looping of Eqs. (5.124) through (5.128) until the temperatures stopped changing with successive loops. The steady-state temperatures were found to be T1 [ 711.7 K [ 438.6 C T2 [ 719.9 K [ 446.7 C T3 [ 744.9 K [ 471.7 C T4 [ 788.5 K [ 515.3 C T5 [ 854.1 K [ 580.9 C We also determined the steady-state temperatures using Excel Solver. If we set Eqs. (5.114)e(5.118) all equal to zero, we have the energy equations for steady state. And, the expressions in the equations are the functions for use by Solver. There was excellent agreement between the Solver and finite-difference results; only a maximum of 0.4 C difference in temperatures for all five nodes.

To close this section, we will use numerical forward-difference to solve two problems that were previously solved analytically. The first problem, Example 5.6, is the same as Example 4.8, which determined the acceptable burial depth of a water line using semiinfinite slab theory.

194

Chapter 5 Numerical methods (steady and unsteady)

Example 5.6 Burial depth of a water pipe Problem

In a town in the Northern United States, the air temperature can be as low as 18 C for as long as 4 weeks. Assume that the ground is initially at a uniform temperature of 10 C. How deep should water pipes be buried to prevent freezing? Assume that the combined convective and radiative coefficient at the ground’s surface is an average 30 W/m2 C. The ground has a thermal conductivity of 2 W/m C, a density of 1800 kg/m3, and a specific heat of 2100 J/kg C. Fig. 5.22.

Solution This is a one-dimensional problem. The first step is to determine the grid. We decided to make node 1 at the ground surface. This is a boundary node with convection. We need another boundary at some depth within the ground. This node, node “n", has to be at a depth sufficiently greater than that of the water pipe (More about this later). We decided to make this node an insulated boundary node. The grid is shown in Fig. 5.23. Let us now develop the energy equations. Like we have done previously, we will assume that all heat flows into a node. And, the heat input during time step Dt is equal to the increase in the internal energy of the node during Dt. We have two boundary nodes (nodes 1 and n) and (n2) interior nodes (nodes 2 through n1). Let us consider a crosssectional area A for the heat flow. (This area A eventually cancels out of the equations.) The two boundary nodes have a thickness of Dx/2. The interior nodes have a thickness of Dx. Therefore, the volume of a boundary node is (A Dx/2), and the volume of an interior node is (A Dx). Node 1 has heat input by convection from the adjacent fluid and conduction from node 2. The energy equation is kA

hA TN T1b þ T2b T1b Dt ¼ rcAðDx = 2Þ T1e T1b (5.129) Dx Node n only has conduction from node n1. Its equation is

kA b Tn1 Tnb Dt ¼ rcAðDx = 2Þ Tne Tnb (5.130) Dx Each interior node has conduction from its two adjacent nodes. The nodal equations for the interior nodes are as follow: For nodes i ¼ 2 to i ¼ n1, kA

kA b b Ti1 Tib þ Tiþ1 (5.131) Tib Dt ¼ rcAðDxÞ Tie Tib Dx Dx Rearranging these equations so that the temperatures at the end of a time step are on the left side of the equations and the temperatures at the beginning of a time step are on the right side, we have

h = 30

W m2C

T∞ = –18C

x

Water line

FIGURE 5.22 Buried water line.

ground surface

5.2 Finite-difference method

195

Δx 2

h, T∞ 1 2 3 4

Δx

5 6 7

n–2 n–1 n Insulated

Δx 2

FIGURE 5.23 Nodes for Example 5.6. For node 1: T1e

¼ T1b

! 2hDt 2aDt b 2hDt TN T þ 1 þ 2 2 2 rcðDxÞ rcðDxÞ ðDxÞ ðDxÞ 2aDt

For node n: Tne

¼ Tnb

For nodes i ¼ 2 to i ¼ n1:

1

2aDt

! þ

ðDxÞ2 !

2aDt ðDxÞ2

b Tn1

(5.132)

(5.133)

aDt b b Ti1 þ Tiþ1 (5.134) 2 ðDxÞ ðDxÞ Water pipes in the northern states of the USA are typically buried at depths of four feet or greater to prevent freezing. We decided to use a nodal increment Dx of 1 inch ¼ 0.0254 m. As we are solving this problem using forward differences, we have to make sure that the chosen time step meets the stability criterion. For our problem, the limiting criterion is that the bracketed expression in Eq. (5.132) may not be negative. That is, for stability, ! 2aDt 2hDt 1 0 (5.135) ðDxÞ2 rcðDxÞ Putting in the values of the parameters in Eq. (5.135), it is found that Dt must be equal to or less than 441 s. We decided to make Dt ¼ 5 min ¼ 300 s. From the problem statement, we want nodal temperatures after 4 weeks. This is equivalent to 2.4192 106 s. With a time step of 300 s, we need 2.4192 106 s/300 s ¼ 8064 time steps. Tie ¼ Tib 1

2aDt

2

þ

196

Chapter 5 Numerical methods (steady and unsteady)

The Matlab program used for the solution is in Appendix K. All nodes were given initial temperatures of 10 C and then we had 8064 successive loops of Equations (5.132) through (5.134). Each loop updated the nodal temperatures for a time step. At the end of the 8064 loops, we had the nodal temperatures after 4 weeks. We then had the program determine the first node at which the temperature was above the freezing point of water, 0 C. The location of that node was the required burial depth. Finally, as mentioned above, node n has to be sufficiently deeper than the water pipe. We ran the program for several depths, i.e., several different numbers of nodes. We found that if node n was 10 feet or more from the surface, the results of the program did not change. As we had chosen Dx to be 1 inch, 10 feet is equivalent to 120 Dx. Hence, we used 121 nodes in the program. It was found that the water pipe had to be buried at a depth of at least 4.75 feet to prevent freezing. This result by numerical forward differences is in excellent agreement with the 4.84 feet result of Example 4.8, which was solved using the semiinfinite slab equations.

Our final example of numerical solution is the same as Example 4.6, which used the Multidimensional Heisler Method to determine the required cooling time for a short cylinder and the temperature of a point in the cylinder at that time.

Example 5.7 Cooling of a short cylinder Problem

The stainless steel disk r ¼ 7900 kg m3 ; c ¼ 480 J kg C; k ¼ 15 W mC shown in Fig. 5.24 is 10 cm in diameter and 5 cm thick. It goes into an oven where it is heated to a uniform temperature of 1200 C. It is then taken out of the oven and cooled by convection to a fluid that is at 150 C. The convective coefficient on the curved surface of the disk is 100 W/ m2 C and the convective coefficient on the two flat surfaces is 250 W/m2 C. (a) How long does it take for the center of the disk to reach 500 C? (b) At the time obtained in Part (a): What is the temperature at a point located 3 cm from the centerline and 1.5 cm from one end?

Solution The nodal labeling is shown in Figs. 5.25 and 5.26. Fig. 5.25 shows the radial numbering, and Fig. 5.26A and B show the axial numbering. Because of symmetry, there is no need to write nodal equations for axial nodes 7 through 11 of

h = 250

W m2C

5 cm

5 cm

h = 100

FIGURE 5.24 Cylinder for Example 5.7.

W 2

m C

h = 250

W 2 m C

5.2 Finite-difference method

1

2

3

4

5

6

7

8

9

10

197

11

Δr

FIGURE 5.25 Radial numbering of nodes.

(A)

(B) 1

1

2

2

3

3

4

4

5 6

5 6

7

5

8

4

9

3

10 11

2 1

FIGURE 5.26 (A) Original axial numbering of nodes. (B) Revised axial numbering of nodes. Fig. 5.26A. The temperature at node 7 is the same as that of node 5; the temperature of node 8 is the same as that of node 4, etc. Hence, we can use the revised axial grid of Fig. 5.26B and have fewer nodal equations. Nodal identification is by two subscripts (i, j). For example, the temperature Ti;j denotes the node at radial location “i” and axial location “j.” Nodes on the axis of the cylinder (that is, i ¼ 1) are cylinders of radius Dr/2 with thickness Dx for interior nodes and Dx/2 for the boundary node (j ¼ 1). The other nodes are rings of width Dr except for the boundary node (i ¼ 11), which has a width of Dr/2. The ring nodes have a thickness Dx except for the boundary nodes (j ¼ 1), which have a thickness Dx/2. From the dimensions of the cylinder and the nodal grid, we have Dr ¼ 0.5 cm ¼ 0.005 m and Dx ¼ 0.5 cm ¼ 0.005 m. Writing the nodal equations is straightforward but very tedious. There are many nodes, and one must be very careful in the correct assignment of subscripts and heat flows for a node. We will derive the equations for two nodesdone on the axis of the cylinder and the other on the side of the cylinder. The other equations are in the Matlab solution, which is in Appendix K.

198

Chapter 5 Numerical methods (steady and unsteady)

Node (1, 4) is an interior node on the axis of the cylinder. It is a cylinder of radius Dr/2 and thickness Dx. The node gets conduction from nodes (1, 3), (1, 5), and (2, 4). The heat input during a time step raises the internal energy of the material in the node. The energy equation for the node is kA1 b kA1 b kA2 b b b b e b Dt þ Dt þ Dt ¼ rcV1;4 T1;4 (5.136) T1;3 T1;4 T1;5 T1;4 T2;4 T1;4 T1;4 Dx Dx Dr where A1 ¼ pðDr=2Þ2 A2 ¼ 2pðDr =2ÞðDxÞ and V1;4 ¼ pðDr=2Þ2 ðDxÞ. Rearranging Eq. (5.136) for forward-difference solution, we have

A1 aDt A2 aDt A1 aDt b A2 aDt b e b b þ þ T þ T1;5 T T1;4 ¼ T1;4 12 (5.137) V1;4 ðDxÞ V1;4 ðDrÞ V1;4 ðDxÞ 1;3 V1;4 ðDrÞ 2;4 Node (11, 3) is on the side of the cylinder. It is a ring of width Dr/2 and thickness Dx. Its inner radius is 19 Dr and its 2 outer radius is ð10 DrÞ. The node gets convection on its outer surface and conduction from nodes (11, 2), (11, 4), and (10, 3). The energy equation for the node is kA3 b kA3 b kA4 b b b b b Dt þ Dt þ Dt þ hside A5 TN T11;3 Dt T11;2 T11;3 T11;4 T11;3 T10;3 T11;3 Dx Dx Dr e b ¼ rcV11;3 T11;3 T11;3 (5.138) where " A3 ¼ p ð10 DrÞ2

2 # 19 Dr 2

19 Dr ðDxÞ A4 ¼ 2p 2 A5 ¼ 2pð10 DrÞðDxÞ V11;3 ¼ A3 ðDxÞ Rearranging Eq. (5.138) for forward-difference solution, we have

A3 aDt A4 aDt hside A5 Dt A3 aDt b A4 aDt b hside A5 Dt e b b T11;3 þ T11;2 þ T11;4 T ¼ T11;3 þ TN 12 þ V11;3 ðDxÞ V11;3 ðDrÞ rcV11;3 V11;3 ðDxÞ V11;3 ðDrÞ 10;3 rcV11;3 (5.139) In the Matlab program, all the nodes are given the initial temperature of 1200 C. The program then loops through these two equations and the other nodal equations. With each pass through the equations, the time advances another time step. (a) We used a time step of Dt ¼ 1 s, and it took 376 loops for the center node (node (1,6)) to reach 500 C. So, the center node reached 500 C in 376 s. (b) Node (7,4) is the node that is 3 cm from the centerline of the cylinder and 1.5 cm from an end. After 376 s, the temperature of this node was 470.6 C. These results are in excellent agreement with the results of the Multidimensional Heisler solution of Example 4.6, which were 376 s and 471.1 C. The Matlab program also determines the amount of heat flow into the adjacent fluid during the 376 s of the process. The result was Q [ 1.108 3 106 J. The Heisler solution for the heat flow was 1.110 106 J. Again, there was excellent agreement between the finite-difference and Heisler solutions.

5.3 Finite element method We have discussed the finite-difference method in detail. There are also other methods in general use today. These include the finite element method, the boundary element method, and the control volume approach.

5.5 Problems

199

In the finite element method, an object is divided into spatial regions, i.e., finite elements. Triangles are commonly used for two-dimensional objects and tetrahedrons for three-dimensional objects. The elements are typically larger than the nodal volumes for finite differences, leading to a lesser number of volumes. And, the finite element method gives a better approximation of irregularly and complexshaped geometries and complicated boundary conditions. However, the finite element method is harder to implement than the finite-difference method. References at the end of this chapter discuss the finite element method in detail. There are also many other books available on the subject. The finite element method has historically been the predominant method used for structural mechanics and stress calculations. Determination of the temperature distribution in an object is a necessary precursor to determination of thermal stresses. Hence, several companies have developed finite element software for determining both temperatures and stresses. Some of these companies provide low-cost software packages for colleges and students. Major software providers include ANSYS [11], COMSOL [12], SolidWorks [13], Autodesk [14], and Siemens (formerly, CD-adapco) [15]. This list is by no means inclusive. There are several other providers, both open source and commercial, of software applicable to the heat transfer field.

5.4 Chapter summary and final remarks In this chapter, we discussed the finite-difference method for solving both steady and unsteady heat transfer problems. For steady state, we learned how to divide an object into nodal volumes and develop energy equations for both interior and boundary nodes. We then gave much attention to different ways of solving these simultaneous equations for the nodal temperatures. For unsteady state, we discussed the nodal grid and the development of the nodal equations. We discussed the forward-difference approach in detail and mentioned the stability criteria associated with it. We also outlined the backward-difference approach that eliminates stability concerns. Finally, we briefly discussed the finite element method, which is one alternative to the finite-difference method. Several examples were included in the chapter to demonstrate the application of numerical methods to both steady-state and unsteady-state problems. Some of these examples had previously been solved using the advanced mathematics of differential equations, semi-infinite slab theory, and the Heisler Method. It was found that the numerical methods, using only algebra, gave results in excellent agreement with the results obtained through the higher-level mathematics. Moving on to the next chapter: Up to now, if a problem involved convection, we provided the value of the convective coefficient h. This is very unrealistic. For a problem in the “real world”, h will not normally be a given. One has to either make a good assumption based on experience or determine h using information from heat transfer researchers. A limited amount of this information comes from analytical solutions. A major part of the information is in the form of correlations obtained by experimentation. The next two chapters discuss convection and the determination of convective coefficients for various geometries and situations.

5.5 Problems Notes: •

If needed information is not given in the problem statement, then use the Appendix for material properties and other information. If the Appendix does not have sufficient information, then use the internet, but double-check the validity of the information by using more than one source.

200

• •

•

Chapter 5 Numerical methods (steady and unsteady)

Some problems specify the nodal spacing and time increment (if an unsteady problem). If these are not given, you should choose the values yourself and clearly note them in your solution. Your solutions should include a sketch of the problem. Node points should be numbered and clearly indicated on the sketch. Solutions should include the nodal equations and indicate the method used for solving them. Caution: Make sure that you use absolute temperatures (Kelvin or Rankine) if the problem involves radiation. Problems 5-1 through 5-15 are steady; problems 5-16 through 5-30 are unsteady. 5-1 A long square bar (k ¼ 10 W/m C) has a cross section of 20 cm by 20 cm and has boundary conditions as shown in Fig. P 5.1. T = 200 C

1

2

3 T = 100 C

h = 15

W 2

4

5

6

7

8

9

m C

T∞ = 20 C

h = 15

W m2 C

T∞ = 20 C

FIGURE P 5.1

(a) What are the steady-state temperatures for the nodes? (b) What is the rate of heat flow to the fluid per meter length of the bar? 5-2 A square steel plate (k ¼ 45 W/m C) has temperatures on its four sides as shown in Fig. P 5.2. What are the steady-state nodal temperatures? 100 C

1

2 200 C

300 C

3

150 C

FIGURE P 5.2

4

5.5 Problems

201

5-3 An aluminum conical fin (k ¼ 240 W/m C) is attached to a wall that is at 250 C. The diameter of the fin at the point of attachment is 0.5 cm and the fin is 20 cm long. The surrounding fluid is at 20 C, and the convective coefficient is 15 W/m2 C. (a) Using a nodal spacing of Dx ¼ 1 cm, determine the rate of heat transfer from the fin to the surrounding fluid. (b) Do the problem using Section 3.8.4.3 and compare the result with that of Part (a). 5-4 For the pin fin of Example 5.1: Replace the convection boundary condition with a radiation condition. The surroundings are at 15 C, and the emissivity of the fin surface is 0.8. Determine the rate of heat transfer from the fin to the surroundings. 5-5 The straight triangular fin (k ¼ 150 W/m C) shown in Fig. P 5.5 has a length of 8 cm and a height at the base of 0.5 cm. The temperature of the base is 250 C, and there is convection from the two sides of the fin to the surrounding 50 C fluid with a convective coefficient of 40 W/m2 C. h = 40

W 2

m C

T∞ = 50 C 0.5 cm

1

2

3

4

5

6

7

8

9

8 cm 250 C

h = 40

W 2 m C

T∞ = 50 C

FIGURE P 5.5

(a) What are the steady-state nodal temperatures? (b) What is the rate of heat transfer to the fluid? (c) Determine the rate of heat transfer to the fluid using the information in Section 3.8.4.2 and compare the result with that of Part (b). 5-6 The cross section of a long carbon steel I-beam (k ¼ 62 W/m C) is shown in Fig. P 5.6. The top surface of the beam is at 50 C, and the bottom surface is at 10 C. All other surfaces are perfectly insulated. Determine the rate of heat flow through the beam (top to bottom) per meter length. Use a nodal spacing of Dx ¼ Dy ¼ 0.5 cm. 5-7 The square plate (k ¼ 1.2 W/m C) of Fig. P 5.7 has temperature boundary conditions on the top and right side, a convective boundary condition on the bottom, and is insulated on the left side. The plate is 6 cm by 6 cm. Determine the steady-state nodal temperatures. 5-8 The rectangular cross section of a long bar (k ¼ 5 W/m C) is shown in Fig. P 5.8. The bar has convection on the top surface, a temperature boundary condition on the right side, radiation on the bottom, and is insulated on the left side. (a) What are the steady-state nodal temperatures? (b) What is the rate of heat flow to the fluid from the top surface per meter length of the bar?

202

Chapter 5 Numerical methods (steady and unsteady)

50 C

1 cm

20 cm

1 cm

1 cm

10 cm 10 C

FIGURE P 5.6 T = 50 C

1

2

3

4

5

6

7

8

9

T = 300 C

Insulated h = 80

W 2 m C

T∞ = 10 C

FIGURE P 5.7

(c) What is the rate of heat flow to the surroundings from the bottom surface per meter length of the bar? 5-9 The plate (k ¼ 3 W/m C) of Fig. P 5.9 has temperature, convection, and insulated boundary conditions as shown. The thickness of the plate is 0.5 cm. (a) What are the steady-state nodal temperatures?

5.5 Problems

h = 25

203

W

m2C T∞ = 20 C

4 cm

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

T = 150 C

6 cm Insulated

ε = 0.5 Tsurr = 20 C

FIGURE P 5.8 100 C h = 50

16 cm

W

m2C T∞ = 25 C

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

Insulated

500 C

16 cm Insulated

FIGURE P 5.9

(b) What is the rate of heat flow to the fluid from the sides of the plate that have convection? 5-10 The nichrome wire (k ¼ 11.3 W/m C) shown in Fig. P 5.10 has a diameter of 1.626 mm and a length of 20 cm. An electrical current is flowing through the wire, causing a uniform heat generation of 2 108 W/m3. The two ends of the wire are thermally anchored to sinks at 20 C. The surface of the wire radiates with an emissivity of 0.7 to the surroundings at 20 C. As shown in the figure, use a uniform nodal spacing of Dx ¼ 1 cm. Node 1 is at one end of the wire, and node 21 is at the other end. What is the steady-state temperature of node 11, which is at the center of the wire?

204

Chapter 5 Numerical methods (steady and unsteady)

20 C 1

qgen = 2x10

Δx = 1 cm 11

W

8

m3

20 cm

ε = 0.7 Tsurr = 20 C

21 20 C D = 1.626 mm

FIGURE P 5.10

5-11 A copper circumferential fin (k ¼ 250 W/m C) is attached to a tube that has an outer surface temperature of 80 C. The fin has an inside radius of 1.5 cm, an outside radius of 6 cm, and a thickness of 0.05 cm. Let there be 11 equally spaced nodes between the inner and outer surfaces of the fin (i.e., Dr ¼ 0.45 cm). Node 1 is at the inner surface of the fin, and node 11 is at the outer surface. There is convection from the top, bottom, and edge of the fin to the 18 C surrounding air with a convective coefficient of 7.5 W/m2 C (Fig. P 5.11). (a) Determine the steady-state temperatures of nodes 2 through 11 (node 1 is 80 C) (b) What is the rate of heat flow from the fin to the surrounding air? (c) What is the fin efficiency? (d) Determine the fin efficiency using Fig. 3.30 and compare the result with the numerical result from Part (c). 5-12 Do Problem 5-7, but let the plate also have uniform internal heat generation of 2 106 W/m3. 5-13 A chimney has a cross section as shown in Fig. P 5.13. If the inner surface is at 800 C and the outer surface is at 30 C, what are the steady-state nodal temperatures? (Note: You can use the symmetry of the problem to reduce the number of nodes from 20 to 6.) 5-14 A chimney (k ¼ 0.7 W/m C) is 8 m high and has a cross section as shown in Fig. P 5.14. The inner surface is at 800 C, and the outer surface convects to the 10 C ambient air with a convective coefficient of 25 W/m2 C. (a) What are the steady-state nodal temperatures? (b) What is the rate of heat flow to the ambient air? 5-15 As shown in Fig. P 5-15, two layers of material are attached to a surface. Both layers have the same thickness. Layer A is firmly attached to the surface with a very high melting point adhesive. However, the glue between layers A and B will melt if its temperature is greater than

5.5 Problems

r1 = 1.5 cm

80 C

1 2 3 4 5 6 7 8 9 10 11

r11 = 6 cm

TOP VIEW

FIGURE P 5.11

1

2

3

4 30 C

5 800 C

4m

FIGURE P 5.13

6

3m

205

206

Chapter 5 Numerical methods (steady and unsteady)

h = 25

W 2

m C T∞ = 10 C

1

2

6

3

7

4

8

800 C

5

9

10

11

12

13

14

3m

4m

FIGURE P 5.14 W

h = 200

2 m C T∞ = 30 C

B

1

2

3

4

5

6

7

8

9

10

11 12

Insulated

A Tsurf = ?

FIGURE P 5.15

80 C and the layers will separate from each other. There is convection on the left and top surfaces with the fluid being 30 C and the convective coefficient being 200 W/m2 C. The right surface is perfectly insulated. Layer A has a conductivity of 30 W/m C and layer B has a conductivity of 5 W/m C. The grid is square, with Dx ¼ Dy ¼ 1:5 cm. How high can the temperature of the bottom surface, Tsurf , be without layers A and B separating? 5-16 The nichrome wire (k ¼ 11.3 W/m C, r ¼ 8400 kg/m3, c ¼ 450 J/kg C) of Problem 5e10 initially has a uniform temperature of 20 C. The electric current is then turned-on. How long does it take for the center of the wire (node 11) to reach 700 C? 5-17 A concrete wall (k ¼ 0.8 Btu/h ft F, r ¼ 140 lbm/ft3, c ¼ 0.18 Btu/lbm F) is 8 inches thick and is initially at a uniform temperature of 65 F. The outside surface of the wall is suddenly

5.5 Problems

5-18

5-19

5-20

5-21

5-22

5-23

207

subjected to sunlight that inputs a heat flux of 120 Btu/h ft2 to the wall. The inside surface of the wall has convection to the 70 F room air with a convective coefficient of 2.5 Btu/h ft2 F. (a) How long does it take for the inside surface of the wall to reach 75 F? (b) What is the steady-state temperature of the inside surface? A thick steel plate (k ¼ 45 W/m C, r ¼ 7800 kg/m3, c ¼ 480 J/kg C) is initially at a uniform temperature of 50 C. A surface of the plate is suddenly lowered to 10 C. (a) What is the temperature 5 cm from the surface after 2 min? Use a nodal spacing of 0.1 cm and assume that the furthest node from the surface is an insulated node at a distance of 20 cm from the surface. (b) Do the problem using the semiinfinite slab theory of Section 4.5 and compare the result with the Part (a) result. A large steel plate (k ¼ 45 W/m C, r ¼ 7800 kg/m3, c ¼ 480 J/kg C) is 30 cm thick and initially at a uniform temperature of 500 C. The plate is horizontal. Air at 20 C is blown across the top and bottom surfaces of the plate to cool it. The convective coefficient is 150 W/m2 C for both surfaces. (a) How long does it take for the center of the plate to reach 350 C? (b) What is the temperature of the top surface of the plate at the time found in Part (a)? (c) Do this problem using the Heisler Method of Chapter 4 and compare the results with the numerical results of Parts (a) and (b). A large steel plate (k ¼ 45 W/m C, r ¼ 7800 kg/m3, c ¼ 480 J/kg C) is 30 cm thick and initially at a uniform temperature of 500 C. The plate is horizontal. Air at 20 C is blown across the top and bottom surfaces of the plate to cool it. The convective coefficient on the top surface is 200 W/m2 C and the coefficient on the bottom surface is 50 W/m2 C. (a) How long does it take for the center of the plate to reach 350 C? (b) What are the temperatures of the two surfaces of the plate at the time found in Part (a)? A long aluminum cylinder (k ¼ 240 W/m C, r ¼ 2700 kg/m3, c ¼ 900 J/kg C) has a diameter of 40 cm. It is heated in an oven until its temperature is uniform at 400 C. It is then taken out of the oven and cooled by convection and radiation to the 25 C room air and surroundings. The convective coefficient at the cylinder’s surface is 500 W/m2 C and the emissivity of the cylinder’s surface is 0.3. What is the temperature of the centerline of the cylinder 5 min after it is taken out of the oven? Use Dr ¼ 2 cm. Node 1 is at the centerline of the cylinder and node 11 is at the surface. A long stainless steel cylinder (k ¼ 15 W/m C, r ¼ 7900 kg/m3, c ¼ 480 J/kg C) has a diameter of 12 cm. It is initially at a uniform temperature of 450 C. The surface of the cylinder is suddenly subjected to a 30 C air stream with a convective coefficient of 200 W/m2 C. (a) How long does it take for the center of the cylinder to reach 100 C? (b) How much heat goes into the air during the process? An apple is modeled as a sphere of 10 cm diameter. It is initially at a uniform temperature of 10 C when the air temperature suddenly falls to 5 C and remains at that temperature. Assume that the apple has the properties of water and that the convective coefficient at the apple’s surface is 15 W/m2 C. (a) How long will it take for the surface of the apple to freeze, i.e., reach 0 C? (b) What is the temperature of the center of the apple at the time found in Part (a)?

208

Chapter 5 Numerical methods (steady and unsteady)

5-24 A large iron plate (k ¼ 80 W/m C, r ¼ 7800 kg/m3, c ¼ 450 J/kg C) is being heated in an oven. The plate is 1 cm thick and is initially at a uniform temperature of 20 C. The oven is at 800 C and the convective coefficient at both of the plate’s surfaces is 150 W/m2 C. The plate is in the oven for 2 min. (a) What is the temperature of the center of the plate when it leaves the oven? (b) What is the temperature of the surface of the plate when it leaves the oven? Use a nodal spacing of Dx ¼ 0.1 cm. 5-25 A stainless steel sphere (k ¼ 15 W/m C, r ¼ 7900 kg/m3, c ¼ 480 J/kg C) has a diameter of 16 cm and is initially at a uniform temperature of 750 C. It is suddenly placed in a large vacuum chamber where it radiates to the walls of the chamber. The walls are at 20 C and the emissivity of the surface of the sphere is 0.9. How long does it take for the center of the sphere to reach 500 C? (Use a nodal spacing of Dr ¼ 2 cm. Node 1 is at the center of the sphere and node 5 is at the surface.). 5-26 A thick piece of wood (k ¼ 0.15 W/m C, r ¼ 700 kg/m3, c ¼ 1300 J/kg C) is initially at a uniform temperature of 20 C. The surface of the wood is suddenly exposed to 500 C combustion gases. The convective coefficient at the surface is 35 W/m2 C. (a) How long will it take for the surface to reach 250 C? (b) Do the problem using the semiinfinite solid theory of Section 4.5 and compare the result with the numerical result of Part (a). 5-27 As shown below, a man is soldering an elbow fitting on the end of a long 1/2-inch Type L copper tube. He is using a propane torch that heats the fitting at a rate of about 200 W. The solder melts at 240 C. The man is holding the torch but does not have a third hand to hold the tube. Therefore, he asks his wife to hold the tube about 25 cm from the end that is being heated. He assures her that she will not be burned. The wife is greatly concerned but helps out and holds the tube. The copper elbow has a mass of 0.018 kg. Before the torch is applied, the tube and the fitting are at 20 C. There is convection from the outer surface of the tube to the room air. The air is at 20 C, and the convective coefficient is 15 W/m2 C. Neglect any heat transfer at the inner surface of the tube or in the air inside the tube (Fig. P 5.27).

FIGURE P 5.27

References

209

FIGURE P 5.28

(a) How long does it take for the fitting to reach 240 C and melt the solder? (b) What is the temperature at the wife’s hand at the time found in Part (a)? (c) Was the man correct, or did the wife have to drop the tube to avoid getting burned? 5-28 A man is soldering an elbow fitting on the end of two long 1/2-inch Type L copper tubes as shown below. He is using a propane torch that heats the fitting at a rate of about 200 W. The solder melts at 240 C. The copper elbow has a mass of 0.018 kg. Before the torch is applied, the fitting and the tubes are at 20 C. There is convection from the outer surface of the tubes to the room air. The air is at 20 C and the convective coefficient is 15 W/m2 C. Neglect any heat transfer at the inner surfaces of the tubes or in the air inside the tubes (Fig. P 5.28). (a) How long does it take for the fitting to reach 240 C and melt the solder? (b) What is the temperature of one of the tubes at a location 10 cm from the elbow at the time found in Part (a)? 5-29 The plate of Problem 5e7 (k ¼ 1.2 W/m C, r ¼ 2500 kg/m3, c ¼ 850 J/kg C) has an initial uniform temperature of 10 C. The boundary conditions are suddenly applied at time zero. Plot the temperature of node 3 versus time until the node reaches steady state. 5-30 A long steel cylinder (k ¼ 40 W/m C, r ¼ 7800 kg/m3, c ¼ 470 J/kg C) has a diameter of 12 cm. It has an initial uniform temperature of 300 C. The cylinder is suddenly quenched in an oil bath that is at 40 C. The center of the cylinder cools to 75 C in 300 s. What is the average convective coefficient during the cooling?

References [1] [2] [3] [4]

R.V. Southwell, Relaxation Methods in Engineering Science, Oxford University Press, 1940. H.W. Emmons, The numerical solution of heat-conduction problems, Trans. ASME 65 (1943) 607e612. G.M. Dusinberre, Numerical Analysis of Heat Flow, McGraw-Hill, 1949. E.L. Wilson, R.E. Nickell, Application of the finite element method to heat conduction analysis, Nucl. Eng. Des. 4 (1966) 276e286. [5] Y. Jaluria, Computer Methods for Engineering, Allyn and Bacon, 1988.

210

Chapter 5 Numerical methods (steady and unsteady)

[6] G.E. Myers, Analytical Methods in Conduction Heat Transfer, McGraw-Hill, 1971 (Chapter 8, Finite Differences; Chapter 9, Finite Elements). [7] W.J. Minkowycz, E.M. Sparrow, G.E. Schneider, R.H. Pletcher (Eds.), Handbook of Numerical Heat Transfer, John Wiley & Sons, 1988. [8] C.S. Desai, Elementary Finite Element Method, Prentice-Hall, 1979. [9] L.C. Thomas, Heat Transfer e Professional Version, Capstone Publishing, 1999 (Appendix F - Finite Element Method). [10] W.M. Rohsenow, J.P. Hartnett (Eds.), Handbook of Heat Transfer, McGraw- Hill, 1973 (Section 4 by P. Razelos). [11] www.ansys.com. [12] www.comsol.com. [13] www.solidworks.com. [14] www.autodesk.com. [15] www.plm.automation.siemens.com.

CHAPTER

Forced convection

6

Chapter outline 6.1 Introduction .................................................................................................................................211 6.2 Basic considerations ....................................................................................................................212 6.3 External flow ................................................................................................................................213 6.3.1 Flow over a flat plate.................................................................................................213 6.3.1.1 Laminar boundary layer....................................................................................... 215 6.3.1.2 Turbulent boundary layer .................................................................................... 234 6.3.2 Flow over cylinders and spheres.................................................................................237 6.3.2.1 Cylinders ............................................................................................................ 238 6.3.2.2 Spheres .............................................................................................................. 241 6.3.3 Flow through tube banks ...........................................................................................242 6.4 Internal flow.................................................................................................................................246 6.4.1 Entrance lengths ......................................................................................................247 6.4.2 Mean velocity and mean temperature .........................................................................247 6.4.3 Constant heat flux ....................................................................................................248 6.4.4 Constant surface temperature....................................................................................249 6.4.5 Equivalent diameter for flow through noncircular tubes ................................................250 6.4.6 Correlations for the Nusselt number and convective coefficient.....................................250 6.4.6.1 Laminar flow; entrance region ............................................................................. 250 6.4.6.2 Laminar flow; fully developed .............................................................................. 251 6.4.6.3 Turbulent flow; fully developed ............................................................................ 251 6.4.7 Annular flow.............................................................................................................255 6.4.7.1 Fully developed laminar flow................................................................................ 255 6.4.7.2 Fully developed turbulent flow ............................................................................. 256 6.5 Chapter summary and final remarks............................................................................................... 258 6.6 Problems .....................................................................................................................................258 References ..........................................................................................................................................265

6.1 Introduction Up to now, the convective coefficient, h, has been a given in all problems involving convective heat transfer. This is very unrealistic. Indeed, the coefficient is usually not a given. Rather, one has to Heat Transfer Principles and Applications. https://doi.org/10.1016/B978-0-12-802296-2.00006-8 Copyright © 2021 Elsevier Inc. All rights reserved.

211

212

Chapter 6 Forced convection

determine the coefficient either by making an informed estimate based on previous experience or by using available analytical or experimental results. The extent of analytical solutions is limited, and h-values are usually obtained from correlations of experimental data. There are three major categories of convection: forced convection; natural (or free) convection; and boiling, condensation, and evaporation. Forced convection is covered in this chapter. Natural convection is in Chapter 7. Evaporation is in Chapter 11, and boiling and condensation are in Chapter 12.

6.2 Basic considerations Let us consider convective heat transfer between a surface and a fluid. In forced convection, the fluid is moving over the surface at a significant velocity. This velocity is often created by a pump (for liquids) or a fan or blower (for gases). Other examples of forced convection are wind blowing across the roof or exterior walls of a building and a liquid flowing by gravity over an inclined plate. Experimental data for forced convection are correlated through the use of three dimensionless parameters: the Reynolds number, the Prandtl number, and the Nusselt number. These parameters are as follows: rVL m

Reynolds number ¼ Re ¼

(6.1)

where r ¼ density of the fluid. V ¼ velocity of the fluid L ¼ a characteristic length of the geometry. ðFor example; L may be the length of a plate or the inside diameter of a pipe.Þ m ¼ absolute viscosity of the fluid The kinematic viscosity, y, is related to the absolute viscosity by y ¼ m=r. Hence, Eq. (6.1) may be written as VL y cp m Prandtl number ¼ Pr ¼ k

Reynolds number ¼ Re ¼

(6.2) (6.3)

where cp ¼ specific heat at constant pressure of the fluid m ¼ absolute viscosity of the fluid k ¼ thermal conductivity of the fluid Nusselt number ¼ Nu ¼ where h ¼ convective coefficient L ¼ a characteristic length of the geometry k ¼ thermal conductivity of the fluid

hL k

(6.4)

6.3 External flow

213

Correlation of experimental data gives the Nusselt number as a function of the two other numbers. That is Nu ¼ f ðRe; PrÞ

(6.5)

The functional relationship is often of the form f ¼ C Re where C, a, and b are constants. After the Nusselt number has been determined for a problem, the convective coefficient, h, can be obtained from the definition of Nu given in Eq. (6.4). Fluid motion and velocities for natural (or free) convection are much smaller than those of forced convection. There typically is no equipment moving the fluid over the surface, and the fluid motion occurs through density gradients in the fluid created by the temperature difference between the surface and the fluid. Because velocities are small, the Reynolds number, which has the fluid velocity in it, is no longer significant and relevant. It is replaced by the dimensionless Grashof number, which has the coefficient of thermal expansion, b, and the acceleration of gravity, g, in its definition. a

Grashof Number ¼ Gr ¼

Prb ,

gbðTs TN ÞL3 y2

(6.6)

where g ¼ acceleration of gravity b ¼ coefficient of thermal expansion of the fluid L ¼ a characteristic length of the geometry y ¼ kinematic viscosity of the fluid Ts and TN are the temperatures of the surface and fluid; respectively For natural convection, experimental data are correlated using the Grashof, Prandtl, and Nusselt numbers: hL ¼ f ðGr; PrÞ (6.7) k The functional relationship is often of the form f ¼ C ðGr PrÞa , where C and a are constants. The product of the Grashof and Prandtl numbers is the Rayleigh number, Ra. Like forced convection, once the Nusselt number is determined for a problem, the convective coefficient, h, is obtained from the definition of Nu, i.e., from Eq. (6.4). We will now continue with our discussion of forced convection. We first look at flow over flat plates. Following that, we consider flow over cylinders and spheres and flow through tube banks. Finally, we look at flow through tubes, pipes, and ducts. Nu ¼

6.3 External flow This section considers external flow over surfaces. We first look at flow over a flat plate; then flow over cylinders and spheres; and finally, flow through tube banks.

6.3.1 Flow over a flat plate Consider two-dimensional fluid flow over a flat plate, as shown in Fig. 6.1.

214

Chapter 6 Forced convection

y

u∞ Free Stream Flow u∞ laminar

transition

turbulent

x

L

leading edge

FIGURE 6.1 Flow over a flat plate.

Flow velocity in the x-direction is u and flow velocity in the y direction is v. There is no flow in the z-direction. The incoming flow is solely in the x-direction at a velocity uN. This is called the “freestream” velocity. The fluid encounters the plate at x ¼ 0, which is called the “leading edge” of the plate. Flow of the fluid over the plate causes a “boundary layer” to form on the surface of the plate. At the plate surface (y ¼ 0), the x and y components of the fluid velocity are zero. That is, the fluid sticks to the plate. Within the boundary layer, the fluid has both u and v components. However, u is much larger than v. Outside the boundary layer, the fluid is moving in the x-direction at the free-stream velocity uN. If the plate is short enough in the direction of flow, there is only a laminar boundary layer region. Longer plates will have transition and turbulent boundary layer regions. The transitional region is short, and the flow over a plate is often modeled as a laminar boundary layer immediately followed by a turbulent boundary layer, as shown in Fig. 6.2. y

u∞

laminar

δ

xc

FIGURE 6.2 Laminar and turbulent boundary layers on a flat plate.

turbulent

x

6.3 External flow

215

The Reynolds number was defined in Eqs. (6.1) and (6.2). For flow over a flat plate, the flow velocity in the Reynolds number is the free-stream velocity uN and the characteristic length is the distance x downstream of the leading edge. As shown in Fig. 6.2, the flow changes from laminar to turbulent at a distance xc downstream of the leading edge. This distance, the “critical distance” is determined from the critical Reynolds number. ruN xc (6.8) Critical Reynolds number ¼ Rexc ¼ m The critical Reynolds number varies, depending on the quality of the flow as it passes the leading edge. If the flow is quiet and very smooth at the leading edge, the Critical Reynolds number is higher than if the flow is noisy and rough. The value typically used is 5 105. With this value, the critical distance xc , from Eq. (6.8), is 5 105 m 5 105 y ¼ (6.9) xc ¼ ruN uN The thickness of the boundary layer, d, varies with x. It is usually defined as the distance from the plate at which the x-component of velocity is 99% of the free-stream velocity. That is, u At y ¼ d; ¼ 0:99 (6.10) uN The boundary layer thickness, d, does not increase as rapidly with downstream distance x as shown in Figs. 6.1 and 6.2. For example, let us consider air flowing across a plate at a velocity uN ¼ 2 m=s. If the plate temperature is 100 C and the free-stream air temperature is 20 C, then the boundary layer thickness at a location 1 m downstream of the leading edge is only 1.5 cm.

6.3.1.1 Laminar boundary layer In this section, we discuss the fluid mechanics and heat transfer relevant to the laminar boundary layer. Let us first obtain the continuity and momentum equations for an infinitesimal control volume in the boundary layer. This elemental control volume is shown in Fig. 6.3. Also shown in the figure are the y

u∞

v+

Fluid Flow

∂v dy ∂y

dz

dy

u dx

Control volume for laminar boundary layer.

∂u dx ∂x x

v

FIGURE 6.3

u+

216

Chapter 6 Forced convection

flow velocities entering and leaving the control volume. The flow is two-dimensional. There is no flow in the z-direction. Solution of the continuity and momentum equations will give us the velocity components uðx; yÞ and vðx; yÞ in the boundary layer.

6.3.1.1.1 Continuity equation

The incoming mass flows are ru dydz and rv dxdz; the outgoing flows are r u þvu vx dx dydz and r v þvv vy dy dxdz. We will assume incompressible, steady flow. Then, from mass conservation, we have Mass flow rate in ¼ Mass flow rate out

(6.11)

vu vv ru dydz þ rv dxdz ¼ r u þ dx dydz þ r v þ dy dxdz vx vy

(6.12)

This simplifies to

vu vv þ ¼0 vx vy

(6.13)

6.3.1.1.2 Momentum equation We now look at the momentum in the x-direction. From Newton’s Second Law, X (6.14) Fx ¼ max P where Fx ¼ sum of forces on the element in the x-direction. m ¼ mass inside the element ¼ rdxdydz ax ¼ acceleration of the element in the x-direction ¼ du dt Let us look at Fig. 6.3 and consider the forces acting on the fluid element. There are shear forces on the top and bottom of the element and pressure forces on the left and right faces of the element. The shear forces are due to the velocity gradient in the y direction, vu/vy, and the fluid viscosity m: The shear stress in the x-direction in the fluid is s ¼ m vu vy , and the shear force is the shear stress times the area it acts on. Because of the fluid motion, the fluid at the bottom of the element is moving faster in the x-direction than the fluid below the element. Hence, the fluid below the element is exerting a shear force on the element in the x direction. At the top face of the element, the shear force on the element is in the þx direction as the fluid above the element is moving faster than the fluid at the top face of the element. The pressure forces on the left and right faces of the element act inward to the element. The pressure on the left face is P and that on the right face is P þ vP dx dx: Summing the forces on the element, we have X vu v vu vu vP Fx ¼ m þ dy dxdz m dxdz þ Pdydz P þ dx dydz vy vy vy vy vx Simplifying, we get X

Fx ¼

m

v2 u vP dxdydz vy2 vx

(6.15)

6.3 External flow

217

From Eqs. (6.14) and (6.15), we get m

v2 u vP du ¼r vy2 vx dt

vu dx vu As u ¼ uðx; yÞ; we have; by the chain rule; du dt ¼ vx dt þ vy dy dx And, as u ¼ dt and v ¼ dt ; Eq. (6.16) becomes v2 u vP vu vu ¼ r u þv m 2 vy dx vx vy

(6.16) dy dt

(6.17)

For a flat plate, vP=dx ¼ 0; so we have the x momentum equation u

vu vu v2 u þv ¼ y 2 vx vy vy

(6.18)

From before, we had for continuity vu vv þ ¼0 vx vy

(6.19)

Eqs. (6.18) and (6.19) can be solved simultaneously for the velocity components uðx; yÞ and vðx; yÞ within the laminar boundary layer. This was first done by Blasius [1] in 1908. He changed the two partial differential equations into the ordinary differential equation f

d2 f d3 f þ 2 ¼0 dh2 dh3

(6.20)

This was done by defining a similarity variable h, which is a function of both x and y, and the streamline function jðx; yÞ: For the streamline function j, u¼

vj vj and v ¼ vy vx

(6.21)

If we apply Eq. (6.21) to Eq. (6.19), the continuity equation is satisfied identically, so we only have to consider the momentum equation, Eq. (6.18). The variable hðx; yÞ is defined as rﬃﬃﬃﬃﬃﬃﬃ uN hðx; yÞ ¼ y (6.22) yx The streamline function j is related to the function f ðhÞ by pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ jðx; yÞ ¼ uN yxf ðhÞ

(6.23)

Eq. (6.20) was obtained by applying the definitions of Eqs. (6.22) and (6.23) to the terms of the momentum equation, Eq. (6.18). Using Eqs. (6.21)e(6.23), it can be shown that the velocity components u and v are rﬃﬃﬃﬃﬃﬃﬃﬃﬃ df 1 uN y df and v ¼ (6.24) u ¼ uN h f dh 2 x dh

218

Chapter 6 Forced convection

Boundary conditions are needed for the solution of Eq. (6.20). These are as follows: At the plate surface ðy ¼ 0Þ; u ¼ 0 and v ¼ 0: From Eq. (6.22) and Eq. (6.24), these convert to

•

At h ¼ 0;

•

As y/N;

df ¼ 0 and f ¼ 0 dh

u/uN converts to As h / N;

df /1 dh

Solution of Eq. (6.20) is difficult due to its nonlinearity. Blasius solved it using a power series. More-recent solutions have been obtained through numerical methods. Table 6.1 gives the function f ðhÞ and its first two derivatives. The velocity components u and v may be obtained from this table by using Eq. (6.24). The velocity components are also available from Fig. 6.4: component u from the left axis and component v from the right axis. Note the straight line on the figure. This is the tangent line to the u=uN versus h curve at h ¼ 0. The slope of the line is 0.332, which is consistent with the 0.332 entry in Table 6.1 for d2 f dh2 at h ¼ 0: The “0.332” factor will be discussed later in more detail.

Table 6.1 Blasius function and its derivatives for laminar boundary layer. h

f

df dh

d2 f dh2

0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 2.2 2.4 2.6 2.8 3.0 3.2 3.4 3.6

0.0000 0.0066 0.0266 0.0597 0.1061 0.1656 0.2379 0.3230 0.4203 0.5295 0.6500 0.7812 0.9223 1.0725 1.2310 1.3968 1.5691 1.7470 1.9300

0.0000 0.0664 0.1328 0.1989 0.2647 0.3298 0.3938 0.4563 0.5168 0.5748 0.6298 0.6813 0.7290 0.7725 0.8115 0.8460 0.8761 0.9018 0.9233

0.3321 0.3320 0.3315 0.3301 0.3274 0.3230 0.3166 0.3079 0.2967 0.2829 0.2668 0.2484 0.2281 0.2065 0.1840 0.1614 0.1391 0.1179 0.0981

6.3 External flow

219

Table 6.1 Blasius function and its derivatives for laminar boundary layer.dcont’d h

f

df dh

d2 f dh2

3.8 4.0 4.2 4.4 4.6 4.8 5.0 5.2 5.4 5.6 5.8 6.0

2.1160 2.3057 2.4980 2.6924 2.8882 3.0853 3.2833 3.4819 3.6809 3.8803 4.0800 4.2796

0.9411 0.9555 0.9670 0.9759 0.9827 0.9878 0.9915 0.9942 0.9962 0.9975 0.9984 0.9990

0.0801 0.0642 0.0505 0.0390 0.0295 0.0219 0.0159 0.0113 0.0079 0.0054 0.0036 0.0024

FIGURE 6.4 Velocity components for laminar boundary layer.

We discussed the boundary layer thickness, d, earlier. It is the distance y from the plate surface at which the x-component of velocity is 99% of the free-stream velocity uN. Looking at Table 6.1, we see df that uuN ¼ dh ¼ 0:99 at about h ¼ 5 (actually h ¼ 4:92). From the definition of h, the boundary layer thickness at location x from the leading edge is rﬃﬃﬃﬃﬃﬃﬃ yx d ¼ 4:92 (6.25) uN

220

Chapter 6 Forced convection

In terms of the local Reynolds number Rex ¼ uNy x; the boundary layer thickness is 4:92x d ¼ pﬃﬃﬃﬃﬃﬃﬃﬃ Rex

(6.26)

For external flows such as flow over a plate, fluid properties are evaluated at the so-called film temperature, which is the average of the surface temperature Ts and the fluid temperature TN:

Example 6.1 Flow over a flat plate Problem Air at 20 C flows at a velocity of 2 m/s across a rectangular flat plate. The plate is 25 cm long in the direction of flow and 5 cm wide. It is at a temperature of 160 C. (a) What types of boundary layers are on the plate, laminar or turbulent? (b) What is the boundary layer thickness at a distance 15 cm downstream of the leading edge? (c) What are the velocity components u and v at a location 15 cm downstream of the leading edge and at the middle of the boundary layer?

Solution The first thing to do in this, and similar, problems is to the determine the Reynolds number at the far end of the plate, i.e., at x ¼ L: For this problem, L ¼ 25 cm. N The fluid properties are taken at the film temperature Tf ¼ Ts þT ¼ 160þ20 ¼ 90 C. 2 2 For this problem, we only need the kinematic viscosity of air. At 90 C, y ¼ 2:20 105 m2 s. 2ð0:25Þ 4 5 (a) At the end of the plate, ReL ¼ uNy L ¼ 2:2010 5 ¼ 2:27 10 . As ReL < 5 10 , the boundary layer is laminar at the end of the plate and has a laminar boundary layer on it. Hence, we can use the equations of this section for determining d; u; and v: 4:92x ﬃﬃﬃﬃﬃﬃ. (b) From Eq. (6.26), d ¼ p Re x

Rex ¼ uN x=y ¼ ð2Þð0:15Þ=2:20 105 ¼ 1:364 104 4:92ð0:15Þ d ¼ pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ¼ 0:00632 m ¼ 6.32 mm. 1:364 104 (c) We want u and v at x ¼ 0.15 m and y ¼ 0.00316 m. For these values, sﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ rﬃﬃﬃﬃﬃﬃﬃ uN 2 ¼ 0:00316 ¼ 2:46 h¼y yx ð2:20 105 Þð0:15Þ df dh

2

d f ¼ 0:7421; dh 2 ¼ 0:2216. p ﬃﬃﬃﬃﬃﬃ ﬃ df df and v ¼ 12 uNx y h dh f Using Eq. (6.24), u ¼ uN dh rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 2ð2:20 105 Þ ð2:46ð0:7421Þ 0:9674Þ ¼ 0:00735 m=s u ¼ 2ð0:7421Þ ¼ 1:484 m=s and v ¼ 0:5 0:15 As expected, it is seen that u[v: qﬃﬃﬃﬃﬃﬃﬃ Alternatively, Fig. 6.4 could have been used to solve this part. From the figure, for h ¼ 2:46; uuN z0:74 and v yuxN z0:43. The u and v results are the same as the above results obtained from Table 6.1.

From Table 6.1, for this h, f ¼ 0:9674;

6.3 External flow

221

6.3.1.1.3 Drag force Flow of a fluid across the flat plate exerts a drag force on the plate. The shear stress at the plate surface is vu (6.27) s¼m vy y¼0ðplate surfaceÞ pﬃﬃﬃﬃﬃvu Using the definition of h, vu ¼ uyxN vh and Eq. (6.27) may be written as vy h¼0

y¼0

rﬃﬃﬃﬃﬃﬃﬃ uN vu sx ¼ m yx vh h¼0

(6.28)

Note that the subscript x has been added to s: The shear stress at the surface of the plate varies with x, and sx is the local shear stress at location x downstream of the leading edge. d2 f df vu As u ¼ uN dh , vh ¼ uN dh and Eq. (6.28) becomes 2 h¼0 h¼0 rﬃﬃﬃﬃﬃﬃﬃ 2 uN d f (6.29) sx ¼ muN yx dh2 h¼0 d2 f ¼ 0:332 so Eq. (6.29) becomes From Table 6.1, for h ¼ 0, dh 2 rﬃﬃﬃﬃﬃﬃﬃ uN sx ¼ 0:332 muN (6.30) yx as

The local Reynolds number is Rex ¼ rumN x ¼ uNy x. Using this definition, Eq. (6.30) may be written pﬃﬃﬃﬃﬃﬃﬃﬃ muN Rex x The local drag or friction coefficient Cfx is defined by sx ¼ 0:332

(6.31)

ru2N 2

(6.32)

0:664 Cfx ¼ pﬃﬃﬃﬃﬃﬃﬃﬃ Rex

(6.33)

sx ¼ Cfx From Eqs. (6.31) and (6.32),

By integrating the local friction coefficient over the length L of the plate, we get the average drag coefficient Cf for the plate. 1 Cf ¼ L

ZL 0

1:328 Cfx dx ¼ pﬃﬃﬃﬃﬃﬃﬃﬃ ReL

(6.34)

ReL is the local Reynolds number at the end of the plate, that is, at x ¼ L: Therefore, the average friction coefficient for the whole plate is twice the value of the local friction coefficient at x ¼ L:

222

Chapter 6 Forced convection

The average shear stress on the plate’s surface is s ¼ Cf the fluid is A, then the drag force DF on the plate is DF ¼ sA ¼ Cf A

ru2N 2 .

If the area of the plate in contact with

ru2N 2

(6.35)

Example 6.2 Drag force on a flat plate Problem Air at 20 C flows at a velocity of 2 m/s across a rectangular flat plate. The plate is 25 cm long in the direction of flow and 5 cm wide. It is at a temperature of 160 C. What is the drag force exerted by the fluid on the plate?

Solution

We must first determine the Reynolds number at the far end of the plate, i.e., at x ¼ L, to determine which types of boundary layers are on the plate, i.e., laminar or turbulent, or both. For this problem, L ¼ 25 cm. Ts þTN 160þ20 The fluid properties are taken at the film temperature T f ¼ 2 ¼ 2 ¼ 90 C. At 90 C, y ¼ 2:20 105 m2 s, m ¼ 2:14 105 kg ms, and r ¼ 0:972 kg m3 : 2ð0:25Þ 4 5 At the end of the plate, ReL ¼ uNy L ¼ 2:2010 5 ¼ 2:27 10 . Hence, as ReL < 5 10 , the equations for laminar flow on a flat plate may be used. 1:328 1:328 ﬃ ¼ 0:008814. ﬃﬃﬃﬃﬃﬃ ¼ pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ From Eq. (6.34), Cf ¼ p ReL 2:27104 From Eq. (6.35), ru2 DF ¼ sA ¼ Cf A N ¼ 0:008814ð0:25 0:05Þð0.972Þ 22 =2 ¼ 2:14 104 N 2 L4 The drag force on the plate is 2.14 3 10 N.

6.3.1.1.4 Energy equation Let us now look at the thermal aspects of the laminar boundary layer. As we saw above, the flow of fluid across a plate creates a hydrodynamic boundary layer. If the plate is at a different temperature than the fluid, a thermal boundary layer is also created. Let us say that the plate temperature is Ts and the free-stream fluid temperature is TN : The temperature of the thermal boundary layer at the plate’s surface is Ts . At the outer surface of the boundary layer, the fluid temperature is essentially at TN : Being more specific, the thermal boundary layer thickness is normally defined as the distance from the plate at which the difference between the fluid temperature T and the plate temperature Ts is 99% of the difference between the free-stream fluid temperature TN and the plate temperature Ts : Fig. 6.5 shows an infinitesimal control volume within the laminar thermal boundary layer. Let us look at the energy transfers into and out of the control volume. We assume that the fluid is incompressible and the flow is steady. There is also no internal heat generation in the fluid due to viscous shearing and no changes in potential and kinetic energies. The flow in the boundary layer is two-dimensional. There is no flow in the z-direction. Finally, fluid properties are constant. Energy transfers to and from the control volume are by conduction and flow of enthalpy. Enthalpy is given the symbol hen to distinguish it from the convective coefficient h. The energy flows into and out of the control volume are shown in Fig. 6.5. They are ðEnergy inÞleft ¼ kdydz

vT þ ruhen dydz vx

(6.36)

6.3 External flow

223

y

u∞

Fluid Flow

(Energy Out) top

dz (Energy In) left

dy

dx

(Energy Out) right

x

(Energy In) bottom

FIGURE 6.5 Control volume in the laminar thermal boundary layer.

vT þ rvhen dxdz vy vT v vT vu vhen þ ðEnergy outÞright ¼ kdydz dx þ r u þ dx hen þ dx dydz vx vx vx vx vx vT v vT vv vhen þ dy þ r v þ dy hen þ ðEnergy outÞtop ¼ kdxdz dy dxdz vy vy vy vy vy ðEnergy inÞbottom ¼ kdxdz

(6.37) (6.38) (6.39)

For an incompressible fluid, dhen ¼ cp dT

(6.40)

where cp ¼ specific heat at constant pressure. And, as there is no internal heat generation and conditions are steady, the energy conservation equation for the control volume is (Energy In)left þ (Energy In)bottom ¼ (Energy Out)right þ (Energy Out)top

(6.41)

Using Eqs. (6.36) through (6.40) and the continuity equation, Eq. (6.19), and also neglecting second-order terms, Eq. (6.41) becomes 2 vT vT v T v2 T ¼a u þv þ (6.42) vx vy vx2 vy2 where a ¼ rck p ¼ thermal diffusivity. In a boundary layer,

v2 T vx2

vvyT2 , so the final form of the energy equation is 2

u

vT vT v2 T þv ¼a 2 vx vy vy

(6.43)

224

vT vy

Chapter 6 Forced convection

vq1 s Let us define a dimensionless temperature parameter q1 ¼ TTT . Then, vT vx ¼ ðTN Ts Þ vx , N Ts 2 vq1 v2 q1 v T ¼ ðTN Ts Þ vy and vy2 ¼ ðTN Ts Þ vy2 . Using these expressions in Eq. (6.43), we get

u

vq1 vq1 v2 q1 þv ¼a 2 vx vy vy

(6.44)

Let us now look at the momentum equation obtained earlier. The momentum equation is u vu vy2 2

vu vu v2 u þv ¼ y 2 vx vy vy

Defining a dimensionless velocity parameter as q2 ¼ uuN , we get ¼ uN vvyq22 . Using these expressions in Eq. (6.18), we get 2

u

vq2 vq2 v2 q 2 þv ¼y 2 vx vy vy

(6.18) vu vx

2 ¼ uN vq vx ,

vu vy

2 ¼ uN vq vy , and

(6.45)

Let us look at the boundary conditions for Eqs. (6.44) and (6.45). For Eq. (6.44), the boundary conditions are • • •

At the surface (y ¼ 0), T ¼ Ts and q1 ¼ 0: As y/N; T/TN and q1 /1: vq1 As y/N; vT vy /0 and vy /0: For Eq. (6.45), the boundary conditions are

• • •

At the surface (y ¼ 0), u ¼ 0 and q2 ¼ 0: As y/N; u/uN and q2 /1: vq2 As y/N; vu vy /0 and vy /0:

Summarizing, it is seen that the boundary conditions for Eqs. (6.44) and (6.45) are identical. Hence, the solutions for q1 in Eq. (6.44) and q2 in Eq. (6.45) will be the same if the kinematic viscosity y of the fluid is equal to the thermal diffusivity a of the fluid. However, the ratio ay is the Prandtl number c m s Pr ¼ pk : So, the solutions for q1 ¼ TTT and q2 ¼ uuN will be the same if the Prandtl number of the N Ts fluid is 1. Pohlhausen [2] provided the first solution to the energy equation in 1921. The dimensionless temperature distribution in the laminar boundary layer is shown in Fig. 6.6 for several different Prandtl numbers. As discussed to the u=uN curve of Fig. 6.4. pﬃﬃﬃﬃﬃabove, the curve for Pr ¼ 1pisﬃﬃﬃﬃidentical ﬃ If the abscissa y uy Nx of Fig. 6.6 is changed to y uyxN Pr1=3 , then the curves of Fig. 6.6 converge to a single curve. This is shown in Fig. 6.7.

6.3.1.1.5 Thermal boundary layer thickness Both the hydrodynamic and thermal boundary layers begin at the leading edge for flow over a heated plate. Eq. (6.26) gave the thickness d of the hydrodynamic boundary layer: 4:92x d ¼ pﬃﬃﬃﬃﬃﬃﬃﬃ Rex

(6.26)

6.3 External flow

FIGURE 6.6 Temperature in laminar boundary layer.

FIGURE 6.7 Temperature in laminar boundary layer (consolidated curve).

225

226

Chapter 6 Forced convection

y

u∞ T∞ Flow

δ

Thermal BL

δt

Hydrodynamic BL

x L

FIGURE 6.8 Boundary layers for laminar flow on flat plate.

For laminar flow over a flat plate, the thickness dt of the thermal boundary layer is dt ¼ dPr 1=3 ¼

4:92x pﬃﬃﬃﬃﬃﬃﬃﬃ Rex

Pr1=3

(6.46)

From this equation, if the Prandtl number of the fluid is 1, then the hydrodynamic and thermal boundary layers are identical. If Pr > 1, then the thermal boundary layer thickness is less than the hydrodynamic boundary layer thickness at a given location. If Pr < 1, the thermal boundary layer thickness is greater than the hydrodynamic boundary layer thickness at a given location. Fig. 6.8 shows the hydrodynamic and thermal boundary layers for laminar flow over a flat plate. Considering the relative thicknesses of the layers, the figure is for a fluid having Pr > 1.

6.3.1.1.6 Convective coefficient and Nusselt number Let us first determine a relation for the convective coefficient at a location x downstream of the leading edge. This is the local coefficient hx . The coefficient we have previously been using, h, is the average coefficient for an entire surface. As mentioned before, the fluid velocity is zero at the plate’s surface, i.e., at y ¼ 0. Hence, heat transfer from the plate’s surface to the fluid is by conduction. If the area of the plate surface at location x is dA; then the heat transfer to the fluid from area dA is vT qy¼0 ¼ kdA (6.47) vy y¼0 From the definition of the convective coefficient, we have qy¼0 ¼ hx dAðTs TN Þ Equating the right sides of Eqs. (6.47) and (6.48) and rearranging, we get k vT hx ¼ Ts TN vy y¼0

(6.48)

(6.49)

6.3 External flow

We can get

vT

vy y¼0

227

from Fig. 6.7. The straight line on the figure is the tangent to the curve at

y ¼ 0. The line has a slope of 0.332. That is, T Ts v T Ts ﬃﬃﬃﬃﬃﬃﬃ At y ¼ 0; rN ¼ 0:332 uN 1=3 Pr v y yx Everything in Eq. (6.50) is constant except T and y, so we have rﬃﬃﬃﬃﬃﬃﬃ vT uN 1=3 ¼ 0:332ðTN Ts Þ Pr vy y¼0 yx Using this in Eq. (6.49), the local convective coefficient is rﬃﬃﬃﬃﬃﬃﬃ uN 1=3 hx ¼ 0:332k Pr yx

(6.50)

(6.51)

(6.52)

And, as the local Reynolds number is Rex ¼ uNy x, Eq. (6.52) may also be written as 0:332k pﬃﬃﬃﬃﬃﬃﬃﬃ 1=3 (6.53) Rex Pr x Using the definition of the Nusselt number given in Eq. (6.4), the local Nusselt number is hx ¼

pﬃﬃﬃﬃﬃﬃﬃﬃ hx x ¼ 0:332 Rex Pr 1=3 (6.54) k To get the average convective coefficient and average Nusselt number for the entire plate, hx and Nux from Eqs. (6.53) and (6.54) are integrated over the length of the plate. If the length of the plate in the direction of flow is L, the average values are Nux ¼

1 h¼ L

ZL hx dx ¼

0:664k pﬃﬃﬃﬃﬃﬃﬃﬃ 1=3 ReL Pr L

(6.55)

0

1 Nu ¼ L

ZL

pﬃﬃﬃﬃﬃﬃﬃﬃ Nux dx ¼ 0:664 ReL Pr 1=3

(6.56)

0

Looking at Eq. (6.53) through (6.56), it is seen that the average values of the convective coefficient and Nusselt number for the entire plate are twice the local values at the trailing edge, i.e., at x ¼ L: The above equations for laminar flow on a flat plate are valid for fluids with Prandtl numbers between about 0.6 and 50. These fluids include common gases (e.g., air, oxygen, nitrogen, hydrogen, carbon dioxide) and low viscosity liquids such as water, refrigerants, and propane. The equations are not valid for high viscosity fluids such as heavy oils and glycerin that have high Pr or for liquid metals that have low Pr. Churchill and Ozoe [3] have correlated experimental data to arrive at Eq. (6.57), which is valid for all Prandtl numbers.

228

Chapter 6 Forced convection

1=2

Nux ¼

hx x 0:3387 Rex Pr1=3 ¼h k 0:0468 2=3 i1=4 1þ Pr

for

Rex Pr > 100

(6.57)

We saw above that the average Nusselt number and convective coefficient for the plate were twice the local values at the trailing edge. Churchill [4] recommends that this can be done for Eq. (6.57). That is, the average Nusselt number for the plate is Nu ¼ 2 NuL . Using Eq. (6.57), we have 1=2

Nu ¼

0:6774ReL Pr1=3 hL ¼h k 0:0468 2=3 i1=4 1þ Pr

(6.58)

Eqs. (6.57) and (6.58) are valid for all values of Pr.

Example 6.3 Convection from an isothermal flat plate Problem Air at 15 C flows across a rectangular flat plate at a velocity of 3 m/s. The plate is 1.5 m wide and 2 m in the direction of flow. The plate is at a uniform temperature of 85 C. What is the rate of heat transfer from the plate to the air?

Solution The first step is to determine the Reynolds number at the trailing edge to see if the boundary layer on the plate is entirely laminar or laminar followed by turbulent. 85þ15 N The film temperature is Tf ¼ Ts þT 2 ¼ 2 ¼ 50 C. For air at 50 C, k ¼ 0:02735 W m C; y ¼ 1.798 105 m2 s; Pr ¼ 0.7228 uN L 3ð2Þ ¼ ¼ 3.34 105 y 1:798 105 5 As ReL < 5 10 ; there is a laminar boundary layer on the entire plate. Using Eq. (6.55), ReL ¼

0:664k pﬃﬃﬃﬃﬃﬃﬃﬃ 1=3 0:664ð0:02735Þ pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ5ﬃ ReL Pr 3:34 10 ð0:7228Þ1=3 ¼ 4:71 W=m2 C ¼ L 2 The heat transfer is q ¼ hAðTs TN Þ ¼ ð4:71Þ½ð1:5Þð2Þð85 15Þ ¼ 989 W. It is interesting to compare this result with the results obtained from Eq. (6.58). If we had used that equation, Nu ¼ 338.5, h ¼ k LNu ¼ 4:63 W m2 C; and q ¼ 972 W. This is less than a 2% difference from the Pohlhausen result. h¼

6.3.1.1.7 Reynolds-Colburn analogy The electric-heat analogy was discussed in Chapter 3. The analogy used electrical circuits to model heat transfer systems. The circuits were very useful in the visualization of the heat transfer processes and in the development of equations for solution of heat transfer problems. In this section, we discuss another very useful analogy. The analogy relates the drag force for laminar flow over the surface of a flat plate to the convective coefficient for the surface. We will first develop the analogy and then discuss its applications. The local drag coefficient and local Nusselt number for laminar flow over a flat plate are given by Eqs. (6.33) and (6.54): Cfx ¼ 0:664Re1=2 x

(6.33)

6.3 External flow

hx x ¼ 0:332Re1=2 Pr 1=3 k The local Stanton number, Stx , is defined as Nux ¼

Stx ¼

hx rcp uN

229

(6.54)

(6.59)

The Stanton number is dimensionless. It is a combination of three other dimensionless parameters: Stx ¼

Nux Rex Pr

(6.60)

From Eq. (6.33), we have 1 Cfx ¼ 0:332Re1=2 x 2 From Eqs. (6.60) and (6.54), we have

(6.61)

1=2

Stx ¼

0.332 Rex Pr1=3 ¼ 0.332 Re1=2 Pr2=3 x Rex Pr

(6.62)

which can be changed to Stx Pr 2=3 ¼ 0.332 Re1=2 x

(6.63)

The right-hand sides of Eqs. (6.61) and (6.63) are the same. Therefore, 1 Stx Pr 2=3 ¼ Cfx (6.64) 2 Eq. (6.64) is the Reynolds-Colburn analogy. It relates the local Stanton number to the local drag coefficient. Eq. (6.64) may also be written for the average Stanton number and the average drag coefficient: 1 St Pr 2=3 ¼ Cf (6.65) 2 The convective coefficient is in the definition of the Stanton number. Hence, Eqs. (6.64) and (6.65) are relationships between the convective coefficient ðhx or hÞ and the drag coefficient ðCfx or Cf Þ The Reynolds-Colburn analogy is very useful. Measurement of drag force for laminar flow on a flat plate can be used to predict the convective coefficient. That is, convective coefficients can be estimated from mechanical measurements without the performance of heat transfer experiments, which can often be very complex and time-consuming. Conversely, if the convective heat transfer for the plate is known, the drag force on the plate can be estimated without the performance of drag force experiments. Example 6.4 illustrates application of the Reynolds-Colburn analogy.

Example 6.4 Reynolds-Colburn analogy Problem Nitrogen at 20 C flows at a speed of 5 m/s across a flat plate that has a uniform surface temperature of 80 C. The plate is 0.6 m long in the direction of flow and has a width of 0.3 m. The nitrogen exerts a drag force of 0.002 N on the plate. Estimate the average convective coefficient h for the plate’s surface.

230

Chapter 6 Forced convection

Solution Ts þ TN 80 þ 20 ¼ 50 C ¼ Tfilm ¼ 2 2 3 For nitrogen at 50 C, r ¼ 1:049 kg m , cp ¼ 1041 J=kg C, y ¼ 17:97 106 m2 s, and Pr ¼ 0.714. uN L ð5Þð0:6Þ ¼ ¼ 1:669 105 laminar since < 5 105 y 17:97 106 There is a laminar boundary layer2 over the entire surface. ru Eq. (6.35): Drag Force ¼ Cf A 2N C ð0:6Þð0:3Þð1:049Þð5Þ2 Putting in values, we have 0.002 ¼ f 2 At x ¼ L ¼ 0.6 m; ReL ¼

and Cf ¼ 8:474 104 The Reynolds-Colburn analogy is Eq. (6.65): 1 St Pr2=3 ¼ Cf 2

Rearranging and putting in values, we have

1 1 Cf 8:474 104 2 2 St ¼ 2=3 ¼ ¼ 5:304 104 Pr ð0:714Þ2=3 From Eq. (6.59) modified for average coefficients rather than local coefficients: St ¼

h rcp uN

Rearranging and putting in values, we have h ¼ rcp uN St ¼ ð1.049Þð1041Þð5Þ 5.304 104 ¼ 2.90 W=m2 C 2 The estimated convective coefficient for the surface is 2.90 W/m C.

6.3.1.1.8 Constant heat flux

The above discussion is for situations in which the plate has a uniform surface temperature Ts : However, some situations, e.g., electrically heated plates, have a constant heat flux boundary condition at the plate’s surface rather than a constant temperature boundary condition. Let us consider heat flow q to the fluid from an area A on the plate’s surface. The heat flux q/A is constant over the plate’s surface. Although the heat flux is constant, the convective coefficient h and the temperature difference Ts TN vary with the x-location on the plate. For the convective heat transfer, we have

q (6.66) ¼ hx ðTs TN Þx A The local convective coefficient is hx ¼

ðq=AÞ ðTs TN Þx

(6.67)

From this equation, the plate temperature Ts at location x is Ts ðxÞ ¼ TN þ

ðq=AÞ hx

(6.68)

The local Nusselt number for constant heat flux is Nux ¼

hx x 1=3 ¼ 0:453Re1=2 x Pr k

(6.69)

6.3 External flow

231

As Nux ¼ hkx x; the local convective coefficient is 0:453k 1=2 1=3 Rex Pr (6.70) x The average temperature difference ðTs TN Þavg for the entire plate of length L in the direction of flow is hx ¼

1 ðTs TN Þavg ¼ L

ZL ðTs TN Þxdx

(6.71)

0

Using Eqs. (6.68) and (6.70), and performing the integration of Eq. (6.71), we get

q L=k ðTs TN Þavg ¼ A 0:6795Re1=2 Pr1=3 L

(6.72)

where ReL is the local Reynolds number at x ¼ L: The average temperature difference is also ðTs TN Þavg ¼

2 q 3hL A

(6.73)

where hL is the local convective coefficient at x ¼ L: The material properties in Eqs. (6.69), (6.70), and (6.72) are evaluated at the “film” temperature, which is the average of the free-stream and surface temperatures. However, the surface temperature varies along the plate and is not known at the beginning of the solution. An iterative procedure must be used. This procedure is illustrated in Example 6.5.

Example 6.5 Heated plate with uniform heat flux Problem Air at 30 C flows at a speed of 2.4 m/s across the top of a thin, square electrically heated plate. The bottom of the plate is perfectly insulated. The plate is 25 cm on a side and inputs 150 W to the air. (a) What is the average temperature of the plate? (b) What is the maximum temperature of the plate?

Solution Because of the small dimensions of the plate, we will assume that the plate only has a laminar boundary layer. We can check this assumption after plate temperatures are determined. As mentioned above, we need an iterative technique to determine the surface temperature of the plate. One technique is as follows: We first rearrange Eq. (6.73) to get hL on the left side:

q 2 i (6.74) hL ¼ h 3 ðTs Þavg TN A

232

Chapter 6 Forced convection

From Eq. (6.70), the convective coefficient at x ¼ L is 0:453k 1=2 1=3 ReL Pr L Equating the right sides of Eqs. (6.74) and (6.75), we get

q 0:453k 2 1=2 i h ¼ ReL Pr1=3 L 3 ðTs Þ TN A hL ¼

(6.75)

(6.76)

avg

Rearranging Eq. (6.73) for the average temperature of the surface ðTs Þavg, we get ﬃ

q 1 rﬃﬃﬃﬃﬃﬃ Ly ðTs Þavg ¼ TN þ 1:4717 (6.77) A kPr1=3 uN

2 q 150 For our problem, TN ¼ 30 C, A ¼ 2 ¼ 2400 W m , L ¼ 0:25 m, and uN ¼ 2:4 m=s, so Eq. (6.77) becomes ð0:25Þ

y1=2 (6.78) kPr1=3 We can now proceed with the iterative process: guess a value for the average surface temperature ðTs Þavg; determine the ðTs Þavg ¼ 30 þ 1140

ðTs Þ þTN

avg ; and calculate ðTs Þavg using Eq. (6.78) and compare this result with fluid properties at the film temperature Tf ¼ 2 the guessed surface temperature. If they agree, we have a solution. If not, pick another guess for ðTs Þavg and do the process again and again until the guessed and calculated values of the surface temperature agree. Following this procedure, let us pick an initial value for ðTs Þavg of 200 C. Then, Tf ¼ ð200 þ30Þ=2 ¼ 115 C. At 115 C, k ¼ 0:0320 W m C; y ¼ 2.468 105 m2 s; and Pr ¼ 0.7082. Calculating ðTs Þavg using Eq. (6.78) gives ðTs Þavg ¼ 228.6 C, which does not agree with our guess of 200C. Picking 230 C as our new guess, we have Tf ¼ ð230 þ30Þ=2 ¼ 130 C. At 130 C, k ¼ 0:03305 W m C; y ¼ 2.633 105 m2 s; and Pr ¼ 0.7057. And, using Eq. (6.78), we get ðTs Þavg ¼ 228.8 C, which agrees closely with our guess of 230 C. Iterating a final time with 229 C, we have agreement between our guess and the calculated value. So, the solution to Part (a) is: The average surface temperature of the plate is 229 C. From Eqs. (6.68) and (6.70), it can be seen that the maximum surface temperature is at the trailing edge x ¼ L: From Eq. (6.70), we have 1=2 0:453k 1=2 1=3 0:453ð0:033Þ 2:4ð0:25Þ ReL Pr ¼ ð0:7058Þ1=3 ¼ 8:05 W=m2 C hL ¼ 5 L 0:25 2:6275 10 From Eq. (6.68), we have

ðq=AÞ 150=ð0:25Þ2 ¼ 328 C ¼ 30 þ hL 8:05 So, the solution to Part (b) is: The maximum surface temperature of the plate is 328 C. At the beginning, we assumed that the plate had only a laminar boundary layer. Let us check this by calculating the Reynolds number at x ¼ L: Ts ðLÞ ¼ TN þ

ð2:4Þð0:25Þ uN L ¼ ¼ 2:284 104 y 2:6275 105 As ReL < 5 105 ; the boundary layer is laminar at x ¼ L and the plate only has a laminar boundary layer on it. ReL ¼

6.3.1.1.9 Unheated starting length The above sections considered laminar flow over a flat plate where the entire plate was at a uniform surface temperature Ts : Let us now consider flow over a plate where the plate is insulated or unheated for a portion of the plate immediately downstream of the leading edge. This is shown in Fig. 6.9. The plate is unheated or insulated for 0 x xo : For this portion, the plate surface is at the freestream temperature TN : For x > xo ; the plate surface is at uniform temperature Ts : The hydrodynamic boundary layer starts at the leading edge, but the thermal boundary layer starts at x ¼ xo : The boundary layers are laminar over the entire plate.

6.3 External flow

233

y

u∞ T∞ Flow

δ

Ts

T∞ unheated xo

δt

x

heated L

FIGURE 6.9 Plate with unheated or insulated starting length.

The local Nusselt number is [5] 1=2

Nux ¼

hx x 0:332Rex Pr1=3 ¼

x 3=4 1=3 k o 1 x

(6.79)

The average convective coefficient for the heated portion of the plate ðx xo Þ is found by integrating hx over the heated part of the plate. It is

x 3=4 2=3 0:664k o 1=2 h¼ ReL Pr1=3 (6.80) 1 ðL xo Þ L Comparing Eqs. (6.79) and (6.80), it can be seen that the relationship between the average coefficient h for the heated section and the local coefficient hL at the trailing edge x ¼ L is

x 3=4 o 1 L

x i hL h¼2 h (6.81) o 1 L Eqs. (6.79) and (6.80) are for fluids with Prandtl numbers in the range of 0.6e50. For fluids outside this range, Eq. (6.57) modified for an unheated starting length may be used. This is 1=2

Nux ¼

hx x 0:3387Rex Pr1=3 ¼

x 3=4 1=3 h k 0:0468 2=3 i1=4 o 1þ 1 Pr x

(6.82)

234

Chapter 6 Forced convection

Eq. (6.81) may be used to determine the average convective coefficient h for the heated portion of the plate. From Eq. (6.82), the convective coefficient at the trailing edge of the plate is 1=2 0:3387ReL Pr1=3 k hL ¼ (6.83) 2=3 i

x 3=4 1=3 h L 1=4 0:0468 o 1þ 1 Pr L Using this hL in Eq. (6.81), we get the average convective coefficient:

x 3=4 2=3 o 1=2 0:6774 1 ReL Pr1=3 k L h¼ h 2=3 i

x ih L 1=4 0:0468 o 1 1þ Pr L

(6.84)

Summarizing: For fluids with Prandtl numbers in the 0.6e50 range, Eq. (6.80) should be used to determine the average convective coefficient. Eq. (6.84) should be used for fluids with Prandtl numbers outside this range.

6.3.1.2 Turbulent boundary layer If a plate is long enough, there will generally be a laminar boundary layer followed by a turbulent boundary layer. If the laminar boundary layer portion is small or if the flow upstream of the plate is rough and agitated, then it is often reasonable to assume that the plate has a turbulent boundary layer on its entire length. This section presents equations for plates having a turbulent boundary layer region. Eqs. (6.33) and (6.34) gave equations for the local friction coefficientCfx and average drag coefficient Cf for a laminar boundary layer. If the flow is turbulent over the entire plate, the corresponding relations are 1=5

and Cf ¼ 0:074 ReL Cfx ¼ 0:059 Re1=5 x

(6.85)

Eqs. (6.53) and (6.55) gave equations for the local convective coefficient hx and the average convective coefficient h for a laminar boundary layer. If the flow is turbulent over the entire plate, the corresponding relations are 1=3 1=3 hx ¼ 0:0296 Re0:8 ðk = xÞ and h ¼ 0:037 Re0:8 ðk = LÞ x Pr L Pr

(6.86)

Eqs. (6.85) and (6.86) are for Reynolds numbers in the range 5 10 5 105 ; the plate has both laminar and turbulent boundary layers, and Eq. (6.92) should be used for h. h i 1=3 0:8 ðk=LÞ ¼ ð0:037Þ 1:111 106 871 ð0:723Þ1=3 ð0:0274=1Þ h ¼ 0:037Re0:8 L 871 Pr h ¼ 41:0 W=m2 C The heat flow by convection from the plate to the air is q ¼ hAðTs TN Þ ¼ ð41:0Þ½ð0:2Þð1Þð80 20Þ ¼ 492 W (b) The radiative heat transfer from the plate to the surroundings is i h 4 q ¼ εsA Ts4 Tsurr ¼ ð0:8Þ 5:67 108 ½ð0:2Þð1Þ ð80 þ 273:15Þ4 ð20 þ 273:15Þ4 q ¼ 74 W Note that we had to use absolute Kelvin temperatures for the radiative heat transfer and that the radiative heat transfer is small compared with the convective heat transfer.

As a final topic in the category of flow over a flat plate, let us look at how the local convective coefficient hx changes with distance x from the leading edge. It will be seen that hx increases significantly as the flow changes from laminar to turbulent.

Example 6.7 Variation of local convective coefficient with location Problem Air at 30 C flows across a flat plate at a velocity of 8 m/s. The plate is at a uniform temperature of 90 C and is 2.5 m long in the direction of air flow. Determine how the local convective coefficient hx varies with distance x, the downstream distance from the leading edge. Plot hx versus x from the leading edge (x ¼ 0) to the downstream end of the plate (x ¼ L ¼ 2.5 m).

Solution

Let us first calculate the Reynolds number at x ¼ L and confirm that the plate has both laminar and turbulent boundary layers. Properties of air at the film temperature of (90 þ 30)/2 ¼ 60 C are, from Appendix D: k ¼ 0.0281 W=m C; y ¼ 18:97 106 m2 =s; Pr ¼ 0.72. ð8Þð2:5Þ 6 The Reynolds number at x ¼ L ¼ 2.5 m is ReL ¼ uNy L ¼ 18:9710 6 ¼ 1:054 10 . As this is greater than the value of the critical Reynolds number (5 105), the flow at the downstream end of the plate is turbulent, and there are both laminar and turbulent regions on the plate. The flow changes from laminar to turbulent at the critical distance xc where the Reynolds number is 5 105. From the definition of the Reynolds number, we have uN xc 8xc ¼ ¼ 5 105 ; and xc ¼ 1:19 m y 18:97 106 From x ¼ 0 to x ¼ 1.19 m, the flow is laminar. From x ¼ 1.19 m to x ¼ 2.5 m, the flow is turbulent. In the laminar region, the local convective coefficient is given by Eq. (6.54): Rexc ¼

1=3 ðk = xÞ hx ¼ 0:332Re1=2 x Pr When values of the various parameters are put into this equation, we have hx ¼ 5:430x0:5 W = m2 C In the turbulent region, the local convective coefficient is given by Eq. (6.86): 1=3 ðk = xÞ hx ¼ 0:0296Re0:8 x Pr

(6.54) (6.93) (6.86)

6.3 External flow

237

When values of the various parameters are put into this equation, we have (6.94) hx ¼ 23:58x0:2 W = m2 C The following figure shows the variation of hx with x. Eq. (6.93) was used for the laminar region and Eq. (6.94) for the turbulent region. 50 45 40 35

hx

30

(W/m2C) 25 20 15 10 5 0 0

0.5

1

1.5

2

2.5

X (m)

In the laminar region from x ¼ 0 to x ¼ 1.19 m, it is seen that hx decreases from a very large value near the leading edge (x ¼ 0) to a value of about 5 at the end of the laminar region. Then,hx has a jump to about 22.5 at the transition location between laminar and turbulent. As x continues to increase, hx gradually decreases to a value of about 20 at the end of the plate. It is seen that the rate of hx decrease in the laminar region is higher than that in the turbulent region. This is because, from Eqs. (6.93) and (6.94), hx is proportional to x0:5 in the laminar region and proportional to x0:2 in the turbulent region. (Note: The figure shows an abrupt jump in hx where the flow changes from laminar to turbulent. In reality, there will be a transitional region between the laminar and turbulent regions and the change in hx , although still very significant, will not be as abrupt.)

6.3.2 Flow over cylinders and spheres Fig. 6.10 shows crossflow over cylinders and spheres. The free-stream velocity is uN and the freestream temperature is TN . The cylinder’s and sphere’s diameter is D and its surface temperature is Ts . Ts

Flow u∞ T∞

FIGURE 6.10 Cross flow over cylinders and spheres.

D

238

Chapter 6 Forced convection

6.3.2.1 Cylinders 6.3.2.1.1 Circular cylinders Many experimental studies have been performed to determine the convective heat transfer for isothermal circular cylinders in crossflow. Some correlation equations from these studies are given below. For these equations, the characteristic length in the Reynolds number is the diameter D of the cylinder. That is ReD ¼ uNy D. Zukauskas [6] performed experiments with air, water, and transformer oil and recommended the equation 1=4 hD 0:37 Pr ¼ CRem Nu ¼ Pr (6.95) D k Prs All parameters in Eq. (6.95) are at TN except for Prs, which is at Ts . The constants C and m are given in Table 6.2. Churchill and Bernstein [7] recommended the equation " 5=8 #4=5 1=2 0:62ReD Pr1=3 hD ReD ¼ 0:3 þ h Nu ¼ (6.96) 1þ 2=3 i k 2:82 105 1=4 0:4 1þ Pr for ReD Pr > 0:2 The fluid parameters in Eq. (6.96) are evaluated at the film temperature Tf ¼ ðTs þTN Þ=2. Sparrow, Abraham, and Tong [8] recommended the equation 1=4

hD m 1=2 2=3 ¼ 0:25 þ 0:4ReD þ 0:06ReD Pr0:37 Nu ¼ (6.97) k ms for 1 ReD 105 The fluid parameters in Eq. (6.97) are at TN except for ms, which is at Ts .

6.3.2.1.2 Noncircular cylinders Sparrow, Abraham, and Tong [8] reviewed the existing experimental data and correlations for air flow normal to blunt objects such as flat plates and noncircular cylinders. Their recommendations for flow normal to square and rectangular cylinders and normal to flat plates are given in Eq. (6.98)e(6.107) for the objects shown in Fig. 6.11.

Table 6.2 Coefficients for Eq. (6.95). ReD

C

m

1e40 40e1000 1000e2 105 2 105e1 106

0.75 0.51 0.26 0.076

0.4 0.5 0.6 0.7

6.3 External flow

Shape

Correlations

Square Cylinder

Rotated Square Cylinder

239

u∞

u∞

u∞

Flat Plate

D

D

Equation (6.98)

D

Equation (6.99)

Front Surface Equation (6.100) Rear Surface Equation (6.101) Both Surfaces Equation (6.102)

Rectangular Cylinder

Aspect Ratio b

a

a = 0.2 b 0.33 0.67 1.33 1.5

Equation (6.103) Equation (6.104) Equation (6.105) Equation (6.106) Equation (6.107)

FIGURE 6.11 Air flow over noncircular cylinders and flat plates.

Correlations for objects in Fig. 6.11 hD ¼ 0:14Re0:66 for 5; 000 ReD 60; 000 D k hD ¼ 0:27Re0:59 for 6; 000 ReD 60; 000 Rotated square cylinder NuD ¼ D k hD 1=2 ¼ 0:592ReD for 10; 000 ReD 50; 000 Plate ðfront surfaceÞ NuD ¼ k hD 2=3 ¼ 0:17ReD for 7; 000 ReD 80; 000 Plate ðrear surfaceÞ NuD ¼ k

Square cylinder NuD ¼

(6.98) (6.99) (6.100) (6.101)

240

Chapter 6 Forced convection

Plate ðboth surfacesÞ NuD ¼

hD ¼ 0:25Re0:61 D k

for

10; 000 ReD 50; 000

(6.102)

Rectangular cylinder hb ¼ 0:26Re0:60 for 13; 000 Reb 77; 000 b k hb ¼ 0:25Re0:62 for 7; 500 Reb 37; 500 a = b ¼ 0.33 Nub ¼ b k hb ¼ 0:163Re0:667 for 7; 500 Reb 37; 500 a = b ¼ 0.67 Nub ¼ b k h b ¼ 0:127 Re0:667 for 7; 500 Reb 37; 500 a = b ¼ 1.33 Nub ¼ b k hb ¼ 0:116Re0:667 for 7; 500 Reb 37; 500 a = b ¼ 1.5 Nub ¼ b k a = b ¼ 0.2

Nub ¼

(6.103) (6.104) (6.105) (6.106) (6.107)

Example 6.8 Flow over a square exhaust duct Problem A worker in a factory is cold. There is a sheet metal exhaust duct carrying hot air near his station, and he thinks he can get some warmth from the duct by directing a fan over its hot surface. Using an infrared thermometer, he finds that the temperature of the outer surface of the duct is 120 F. The room air is at 55 F. The duct has a square cross section 10 inches by 10 inches. Air from the fan moves at 8 ft/s. (a) To get the greatest heating from the duct, should the fan be directed on a flat side of the duct or on a corner of the duct? (See Fig. 6.12 below) (b) What is the expected convective heat flow from the duct surface to the room air per foot length of duct? Will this have a significant impact on the person’s comfort? (c) Estimate the duct’s radiative heat flow to the room and compare it with the convective heat flow you found in Part (B).

(A) 10"

Fan 10"

(B) Fan 10"

FIGURE 6.12 For Example 6.8: Which is preferred? A or B?

10"

6.3 External flow

241

Solution

N (a) The film temperature is Tf ¼ Ts þT ¼ 120þ55 ¼ 87:5 F. At this temperature, for air, we have 2 2 k ¼ 0:015 Btu h ft F and y ¼ 1:74 104 ft2 s. For Orientation (A): Looking at Fig. 6.11, D ¼ 10" ¼ 0.8333 ft.

uN D ð8Þð0:8333Þ ¼ ¼ 38300 y 1:74 104 For this orientation, the appropriate correlation is Eq. (6.98) Then ReD ¼

hD ¼ 0:14Re0:66 ¼ 0.14ð38300Þ0:66 ¼ 148.3 D k k 0:015 ð148:3Þ ¼ 2:67 Btu=h ft2 F h ¼ NuD ¼ D 0:8333 For orientation (B): pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ Looking at Fig. 6.11, D ¼ 102 þ 102 ¼ 14:14 inch ¼ 1.178 ft. NuD ¼

uN D ð8Þð1:178Þ ¼ ¼ 54160 y 1:74 104 For this orientation, the appropriate correlation is Eq. (6.99): Then ReD ¼

NuD ¼

hD ¼ 0:27Re0:59 ¼ 0.27ð54160Þ0:59 ¼ 167.6 D k

k 0:015 ð167:6Þ ¼ 2:13 Btu=h ft2 F h ¼ NuD ¼ D 1:178 Conclusion: As Orientation (A) gives a larger convective coefficient, the air heating will be larger for that orientation. (b) The surface area for a one-foot length of duct is 4 0.8333 1 ¼ 3.333 ft2. qconv ¼ hADT ¼ ð2:67Þð3:333Þð120 55Þ ¼ 578 Btu=h per foot length of duct If there is, say, about 10 feet of duct near the person, then about 5800 Btu/h will be added to the air near the person. This should result in a positive impact on the person’s comfort but perhaps not a significant impact. (c) We estimate that the sheet metal duct has 4an emissivity of about 0.3. For radiative heat transfer, qrad ¼ εsA Ts4 Tsurr . In English units, s ¼ 0:1714 108 Btu=h ft2 R4 Ts and Tsurr must be in absolute temperature units ðRankineÞ

So; qrad ¼ ð0:3Þ 0:1714 108 ð3:333Þ ð120 þ 459:67Þ4 ð55 þ 459:67Þ4 qrad ¼ 73 Btu=h per foot length of duct. The radiative heat transfer is small compared with the convective heat transfer. (Note: Painting the duct black would increase the emissivity to about 0.9, which would triple the radiative heat transfer.)

6.3.2.2 Spheres Many experimental studies have been performed to determine convective coefficients for spheres in crossflow. Although most of these studies are for heat transfer in air or water, there have been some studies for other fluids, such as oils. We mention a few forced convection correlations for spheres. There are many more in the literature. Vliet and Leppert [9] recommended the following correlation for forced convection for spheres in water: 0:5 m 1=4 hD 0:66 Nu ¼ ¼ 2:7 þ 0:12ReD Pr (6.108) k ms for 50 ReD 300; 000

242

Chapter 6 Forced convection

They also recommended the following correlation for forced convection for spheres in fluids of a wide range of Prandtl numbers: 0:3 m 1=4 hD 0:54 ¼ 1:2 þ 0:53ReD Pr Nu ¼ (6.109) k ms for 1 ReD 300; 000; 2 Pr 380 In Eqs. (6.108) and (6.109), all fluid properties are at TN except for ms, which is at Ts . Will, Kruyt, and Venner [10] studied forced convection for spheres in air and recommended the correlation Nu ¼

hD 1=2 ¼ AReD þ BReD k

(6.110)

where A ¼ 0:493 0:015 B ¼ 0:0011 ð1 0:035Þ 7.8 103 ReD 2:9 105 N In Eq. (6.110), the fluid properties are at the film temperature Tf ¼ Ts þT 2 . The most often used correlation for forced convection for spheres is that by Whitaker [11]: 1=4

hD m 1=2 2=3 ¼ 2 þ 0:4ReD þ 0:06ReD Pr0:4 Nu ¼ (6.111) k ms

for 3:5 ReD 7:6 104 ; 0.71 Pr 380; and 1 ðm = ms Þ 3:2 In Eq. (6.111), all fluid properties are at TN except for ms, which is at Ts .

6.3.3 Flow through tube banks Heat exchangers are used to transfer heat between two fluids. The exchangers often have many parallel tubes arranged in a bundle or “tube bank” One fluid flows through the tubes and the other fluid flows through the tube bank, i.e., across the outer surfaces of the tubes. Convective coefficients for flow over a single cylinder (or tube) were discussed in Section 6.3.2.1. Flow over a tube in a tube bank is more complicated due to the impact on the flow by adjacent tubes in the tube bank. Such altering of the flow has a significant effect on the convective coefficient for the tube in question. This section discusses the convective coefficient for the outer surface of a tube in a tube bank. Section 6.4 covers the convective coefficient for the inner surface of the tube. Two common arrangements of tubes in a tube bank are “inline” and “staggered” as shown in Fig. 6.13. The tube spacing in the flow direction is SL and the spacing in the transverse direction is ST . A diagonal spacing SD is also shown for the staggered arrangement. In terms of SL and ST ; sﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 2ﬃ ST SD ¼ S2L þ (6.112) 2

6.3 External flow

243

Inline SL ST

Vin Tin

Staggered SL

SD

ST

Vin Tin

FIGURE 6.13 Tube arrangements for a tube bank.

Many researchers have determined correlations for the Nusselt number for flow through a tube bank. These correlations contain the Reynolds number, which is based on the maximum fluid velocity in the tube bank, and the outside diameter of the tubes. That is, ReD ¼

rVmax D Vmax D ¼ m y

(6.113)

where D is the outside diameter of the tube. Let us look at the maximum velocity for the two arrangements: For both arrangements, fluid enters the tube bank from the left at uniform velocity Vin and inlet temperature Tin : As the fluid passes the first row of tubes, it speeds up due to the narrowing of the flow area. From continuity, the velocity as it goes T through the opening between two tubes in the first row is STSD Vin : For the inline arrangement, this is the maximum velocity in the tube bank. It is also the maximum velocity for the staggered arrangement unless the spacing of the rows is small. Looking at the geometry and mass flow conservation, it can be shown that if SD < ST þD 2 ; then the maximum velocity is through the diagonal passage of width SD D: To summarize: For the inline arrangement: Vmax ¼ For the staggered arrangement : Vmax ¼

ST Vin ST D

ST ST þ D Vin if SD > 2 ST D

(6.114) (6.115)

244

Chapter 6 Forced convection

Table 6.3 Coefficients for Eq. (6.117).

For inline

For staggered

ReD

C

m

n

1e100 100e1000 1000e2 105 2 105e2 106 1e500 500e1000 1000e2 105 2 105e2 106

0.9 0.52 0.27 0.033 1.04 0.71 0:35ðST =SL Þ0:2 0:031ðST =SL Þ0:2

0.4 0.5 0.63 0.8 0.4 0.5 0.6 0.8

0.36 0.36 0.36 0.4 0.36 0.36 0.36 0.36

Vmax ¼

ST ST þ D Vin if SD < 2 2ðSD DÞ

(6.116)

Several researchers have performed experiments on tube banks and developed correlations for the Nusselt number. The recommendation of Zukauskas [12] is 1=4 hD m n Pr Nu ¼ ¼ CReD Pr F (6.117) k Prs Fluid parameters in Eq. (6.117) are at TN except for Prs, which is at Ts . The constants C, m, and n are given in Table 6.3. Factor F depends on the number of rows in the flow direction. If ReD > 1000 and there are 16 or more rows, then F is 1. For less than 16 rows, F is less than 1. For 13, 10, 7, 5, and 4 rows, it was found that F is, respectively, 0.99, 0.98, 0.96, 0.93, and 0.9. This data can be interpolated for other numbers of rows. These data fit very well to the fourth degree polynomial F ¼ 1:4894 105 R4 þ 7:2245 104 R3 1:3009 102 R2 þ 1:0674 101 R þ 0:63929 where R is the number of rows in the flow direction.

Example 6.9 Flow through a tube bank Problem An inline tube bank consists of 30 tubes. There are five rows of tubes in the flow direction and six rows in the transverse direction. The tube spacing is shown in Fig. 6.14. The tubes are 1/2-inch Type M copper tubes, 0.5 m long. Air enters the tube bank at a velocity of 3 m/s and a temperature of 20 C. The outer surfaces of the tubes are at 110 C. What is the rate of heat transfer by forced convection to the air?

Solution

From an Internet search, the tubes have an outside diameter of 0.625 inch ¼ 0.01588 m. Looking at Fig. 6.14, we have SL ¼ D þ 5 mm ¼ 0.02088 m and ST ¼ D þ 1 cm ¼ 0.02588 m 0:02588 T From Eq. (6.114), Vmax ¼ STSD Vin ¼ 0:025880:01588 ð3Þ ¼ 7:764 m s.

6.3 External flow

245

SL 5 mm

Vin Tin

1.588 cm D ST

1 cm

FIGURE 6.14 Tube spacing for example 6.9.

The Reynolds number is based on the maximum velocity and the outside diameter of the tubes. Fluid properties are to be taken at the average temperature of the incoming and outgoing flows from the tube bank. The air enters at 20 C. Let us assume that the air exits at 40 C properties at this average temperature 30 C are (to be checked later). The k ¼ 0:0259 W=m C; r ¼ 1.164 kg m3 ; y ¼ 1.608 105 m2 s; Pr ¼ 0.728; and cp ¼ 1007 J kg C. We also need Pr at the tube surface temperature of 110 C, that is 0.709. rVmax D Vmax D ð7:764Þð0:01588Þ ¼ ¼ ¼ 7670 m y 1:608 105 From Eq. (6.117) and Table 6.3, the appropriate equation for this Reynolds number is 1=4 hD 0:36 Pr ¼ 0:27Re0:63 F Nu ¼ D Pr k Prs Putting values into this equation and recognizing that F ¼ 0.93 as we only have five rows in the direction of flow, we get hð0:01588Þ 0:728 1=4 ¼ 0:27ð7670Þ0:63 ð0:728Þ0:36 Nu ¼ ð0:93Þ 0:0259 0:709 2 Solving this for the convective coefficient, we get h ¼ 103.0 W/m C. The heat transfer to the air is q ¼ hNtubes Atube ðTs TN Þ; where Ntubes is the number of tubes; Atube is the surface area per tube and TN is the average fluid temperature. We assumed TN was 30 C. So, we have q ¼ ð103:0Þð30Þ½pð0:01588Þð0:5Þð110 30Þ ¼ 6166 W. The rate of heat transfer by forced convection to the air is 6166 W. We assumed that the exit temperature of the air from the tube bank was 40 C. This gave us an average air temperature of 30 C for determination of the fluid properties. Now that we have the heat transfer amount, we can check whether our assumption of 40 C exit temperature was appropriate. The heat gain of the air is q ¼ ðrin Vin Ain Þcp ðTout Tin Þ. The first parenthesis is the mass flow rate of the incoming air. Then we have the specific heat, and finally the change of temperature of the air as it goes through the tube bank. Putting values into this equation and using the calculated 6166 W for q, we get q ¼ 6166 ¼ ð1:172Þð3Þ½ð6Þð0:02588Þð0:5Þð1007ÞðTout 20Þ Solving for Tout , we get Tout ¼ 42:4 C. This gives an average air temperature of (20 þ 42.4)/2 ¼ 31.2 C. Our assumption was very close, 30 C, and there is no need for iteration on the exit temperature and fluid properties to get a better value for q. Our calculated value is fine. The Reynolds number is ReD ¼

246

Chapter 6 Forced convection

6.4 Internal flow The previous section dealt with flow over objects. For example, we looked at flow over a cylinder and discussed correlations for the convective coefficient at the outer surface of the cylinder. This section deals with flow through cylinders, i.e., pipes and tubes, and ducts. We will discuss correlations for the convective coefficient at the inner surface of the cylinders and ducts. Consider Fig. 6.15, which shows a fluid flowing through a circular tube. The fluid enters the tube with a uniform velocity Vin. As the fluid goes through the tube, it is acted on by shear stresses at the wall which impact the velocity profile uðx; rÞ: The fluid sticks to the wall (zero velocity) and a boundary layer forms. If the flow is laminar and the tube is long enough, the velocity profile ultimately becomes parabolic. At this point, the flow is said to be “fully developed hydrodynamically” and the velocity profile stays the same as the fluid progresses further down the tube. If the flow is turbulent, rather than laminar, the ultimate velocity profile is not parabolic. Rather, it is flatter near the center of the tube, as shown in Fig. 6.16. The distance from the inlet to the fully developed region is Lh ; called the “hydrodynamic entrance length” For x > Lh ; vu vx ¼ 0: We have used the terms “laminar” and “turbulent” In laminar flow, fluid particles glide through the tube. There is little cross mixing of the particles. They tend to stay in the same relative layer as they travel down the tube. In turbulent flow, the flow is rougher, with cross mixing of fluid particles. There are random variations of velocity and other fluid properties such as pressure and temperature. Whether y u (x, r)

u (r)

Vin r x

Lh

parabolic

FIGURE 6.15 Laminar flow through a tube. y u (r) Vin r x

Lh

FIGURE 6.16 Fully developed velocity profile for turbulent flow.

6.4 Internal flow

247

a flow is laminar or turbulent depends a lot on the fluid velocity and viscosity. Indeed, the Reynolds number is used to determine whether the flow is laminar or turbulent. The Reynolds number for internal flow is ReD ¼

rVD VD ¼ m y

(6.118)

where D is the inside diameter of the tube. V is the flow velocity r; m; and y are the fluid density; absolute viscosity; and kinematic viscosity From experiments, flow is usually laminar if ReD < 2300 and turbulent if ReD > 4000: (These are rough, not firm, values.) Between these values, the flow is “transitional.” Looking at the definition of the Reynolds number, it is seen that laminar flow generally occurs for fluids of high viscosity flowing at low velocity. Most flows encountered in engineering practice are of the turbulent category. If we consider the thermal aspects of the flow, there is a thermal entrance length Lt analogous to the hydrodynamic entrance length Lh discussed above. For thermally fully developed flow, i.e., for x >

v Ts T ¼ 0: T is the surface temperature, T is the fluid temperature (which can vary with y), and Lt ; vx s Ts Tm Tm is the mean or bulk fluid temperature, all at location x.

6.4.1 Entrance lengths Experimentation has provided the following estimates of entrance lengths Lh and Lt: For laminar flows; Lh z 0:05ReD D

(6.119)

Lt z 0:05ReD DPr

(6.120)

For turbulent flows, the entrance lengths Lh and Lt are about the same and they are essentially independent of Prandtl number Pr. It is often assumed that turbulent flows are fully developed, both hydrodynamically and thermally, about 10 diameters downstream of the entrance.

6.4.2 Mean velocity and mean temperature The mass flow rate of the fluid is m_ ¼ rAc Vavg , where Ac is the cross-sectional area of the flow and Vavg is the average flow velocity at a given location x. The fluid velocity varies over the cross section, and the average (or mean) velocity is obtained by integrating the velocity over the cross section: Z 1 Vavg ¼ udAc (6.121) Ac Ac

_ p Tm ¼ The energy flowing in the fluid at a given location x is E_ ¼ mc

R

rucp TdAc . For constant

Ac

density and specific heat at location x, the average or mean temperature at location x is Z 1 Tm ¼ uTdAc Vavg Ac Ac

Tm is also called the “bulk” fluid temperature.

(6.122)

248

Chapter 6 Forced convection

Later in this chapter, we will be providing correlation equations for the convective coefficient for internal flows. These equations often depend on the boundary conditions at the wall surface. The two common boundary conditions are constant heat flux and constant temperature, which we consider now.

6.4.3 Constant heat flux Fig. 6.17 shows a differential control volume in a fluid flowing through a channel. There is a constant heat flux on the surface of the channel. Such a constant heat flux could have been applied by heating tapes wrapped around the tube or perhaps a radiant heat source beaming on the outer surface of the tube. We will do an energy balance on the control volume. The heat flux q0s is constant, and the rate of heat flow into the control volume is q0s dAs , where dAs is the differential area at the surface of the tube. The heat flow into the volume raises the energy of the _ p dTm : The energy balance is fluid in the volume a net amount mc Net rate of heat in ¼ net rate of energy increase of fluid. _ p ðTm þ dTm Þ mc _ p Tm dqs ¼ q0s dAs ¼ mc

(6.123)

The differential surface area dAs is the perimeter P times dx. Eq. (6.123) then becomes: dTm q0s P ¼ ¼ constant _ p dx mc

(6.124)

It is seen that the mean fluid temperature varies linearly with x. Solving Eq. (6.124) for Tm ðxÞ;between locations x1 and x2 , we get Tm ðx2 Þ ¼ Tm ðx1 Þ þ

q0s P ðx2 x1 Þ _ p mc

(6.125)

In Eqs. (6.124) and (6.125), the specific heat cp should be evaluated at the average of the mean temperatures at x1 and x2 : dqs = q's dAs

Flow ṁ cp (Tm + dTm)

ṁ cp Tm dx

x

FIGURE 6.17 Constant heat flux.

x

x + dx

6.4 Internal flow

249

6.4.4 Constant surface temperature Fig. 6.18 shows a differential control volume in a fluid flowing through a channel. The channel has a constant surface temperature. Perhaps such a constant surface temperature occurred from a fluid condensing or vaporizing on the outer surface of the tube. We will do an energy balance on the control volume. The surface temperature is Ts . The center of the volume is at location x þ dx=2. At this location, the mean fluid temperature is Tm þ dTm =2: The rate of convective heat flow into the control volume is dqs ¼ hdAs ½Ts ðTm þdTm =2Þ, where h is the average convective coefficient for the tube section being considered. Area dAs is the differential area at the surface of the tube. The heat flow into the _ p dTm : The energy balance is volume raises the energy of the fluid in the volume a net amount mc Net Rate of Heat in ¼ Net Rate of Energy Increase of Fluid _ p ðTm þ dTm Þ mc _ p Tm dqs ¼ hdAs ½Ts ðTm þ dTm = 2Þ ¼ mc

(6.126)

The differential surface area dAs is the perimeter P times dx. And, we can eliminate the secondorder term dx dTm : Eq. (6.126) then becomes _ p dTm dqs ¼ hPdxðTs Tm Þ ¼ mc

(6.127)

As Ts is constant, dTm ¼ dðTs Tm Þ: Eq. (6.127) can then be rearranged to dðTs Tm Þ hP ¼ dx _ p Ts T m mc

(6.128)

Integrating Eq. (6.128) between locations x1 and x2 ; we get ½lnðTs Tm Þxx21 ¼ "

hP ðx2 x1 Þ _ p mc

(6.129)

# ðTs Tm Þx2 hP ln ðx2 x1 Þ ¼ _ p mc ðTs Tm Þx1

(6.130)

dTm ⎛ dqs = h dAs Ts – ⎛Tm + ⎝ 2 ⎝

Flow ṁ cp Tm

ṁ cp (Tm + dTm) dx

x x

FIGURE 6.18 Constant surface temperature.

x + dx

250

Chapter 6 Forced convection

And, finally ðTs Tm Þx2 ¼ ðTs Tm Þx1 e

hP mc ðx2 x1 Þ _ p

(6.131)

Fluid properties in h, and the specific heat cp , should be evaluated at the average of the mean temperatures at x1 and x2 :

6.4.5 Equivalent diameter for flow through noncircular tubes Many parameters, including the Reynolds number, Nusselt number, and friction factor, contain the inside diameter D of the tube. If the tube has a noncircular area, then an equivalent diameter, the hydraulic diameter, should be used for the inside diameter. This diameter is defined by 4Ac (6.132) Dh ¼ P where Ac is the flow area and P is the wetted perimeter, i.e., the perimeter of the surface seeing fluid. Some examples are the following: If the flow area is circular with diameter D, then the hydraulic diameter is the same as the actual =4Þ diameter D of the tube. That is, Dh ¼ 4ðpD ¼ D. pD 2 If the flow area is square, with a side a, then Dh ¼ 4a 4a ¼ a. 4ab ¼ 2ab . If the flow area is rectangular, with sides of a and b, then Dh ¼ 2aþ2b aþb pﬃﬃﬃ 32 4 a pﬃﬃ 4 3 If the flow area is an equilateral triangle of side a, then Dh ¼ ¼ 3 a. 3a If the flow area is an open channel of rectangular cross section, with two vertical sides a and bottom 4ab . b, then Dh ¼ 2aþb 2

6.4.6 Correlations for the Nusselt number and convective coefficient Internal flow has been extensively investigated and there are many correlations for the Nusselt number and convective coefficient. We will present some of them here. The correlations are grouped according to type of flowdlaminar or turbulent.

6.4.6.1 Laminar flow; entrance region •

Hausen [13] (for constant surface temperature) NuD ¼

hD 0:0668ðD=LÞReD Pr ¼ 3:66 þ k 1 þ 0:04½ðD=LÞReD Pr2=3

(6.133)

for ReD PrðD = LÞ < 100 Fluid properties are at the mean bulk temperature. •

Sieder and Tate [14] (for constant surface temperature) NuD ¼

0:14 hD m ¼ 1:86ðReD PrÞ1=3 ðD=LÞ1=3 k ms

(6.134)

for 0.5 < Pr < 16; 700 and ReD PrðD = LÞ > 10 All properties are at mean bulk temperature except ms , which is at the surface temperature.

6.4 Internal flow

251

6.4.6.2 Laminar flow; fully developed Nusselt numbers for circular and noncircular cylinders are given in Table 6.4. There are entries for two boundary conditions: constant heat flux and constant surface temperature.

6.4.6.3 Turbulent flow; fully developed In turbulent flow, the flow is usually fully developed within 10 or 20 diameters downstream of the entrance. As the entrance region is so short, we will only include correlation equations for the fully developed portion of the tube and will use these equations also for the entrance region as needed. In addition, the type of boundary condition (e.g., constant flux or constant temperature) has less impact on the convective coefficients for turbulent flow than for laminar flow. Therefore, the following equations may be used for either boundary condition. •

Dittus and Boelter [15] NuD ¼

hD n ¼ 0:023Re0:8 D Pr k

(6.135)

where n ¼ 0.4 for heating of the fluid and 0.3 for cooling of the fluid and 0.6 < Pr < 100. Properties are at the mean bulk temperature. If there is substantial temperature difference between the fluid and the surface of the tube, then the following equation may give better results: •

Sieder and Tate [14] 0:14 hD 0:8 1=3 m ¼ 0:027ReD Pr NuD ¼ k ms

(6.136)

for 0.7 < Pr < 16; 700 Table 6.4 Nusselt numbers for fully developed laminar flow. h Nu [ hD k

Shape

Constant heat flux

Constant surface temperature

Circular cylinder

4.36

3.66

3.61 4.12 5.33 6.49 8.24

2.98 3.39 4.44 5.60 7.54

3.11

2.47

D

Rectangular channel

b a

Triangular channel

3 2

a a

a/b 1 2 4 8 N

252

Chapter 6 Forced convection

All properties are at mean bulk temperature except ms , which is at the surface temperature. Gnielinski performed an extensive review of correlation equations and recommended the following equation which includes friction factor f: •

Gnielinski [16].

" 2=3 # 0:11 hD ðf =8ÞðReD 1000ÞPr D Pr ¼ NuD ¼ 1þ 1=2 2=3 k L Pr s Pr 1 1 þ 12:7ðf =8Þ

(6.137)

for 0.6 < Pr < 105 2300 < Re < 106 All properties are at mean bulk temperature except Prs , which is at the surface temperature. The friction factor f for turbulent flow is a function of the Reynolds number and the relative roughness of the pipe surface. It may be obtained from the Moody Diagram [17] or the following Colebrook Equation [18]. 1 ε=D 2:51 pﬃﬃﬃ ¼ 2 log10 pﬃﬃﬃ þ (6.138) 3:7 ReD f f The values for the pipe roughness ε can vary greatly with manufacturing process and manufacturer. Typical values are Drawn tubing, PVC Commercial steel Galvanized iron Cast iron

0.0015 mm 0.045 mm 0.15 mm 0.25 mm

It is seen that Eq. (6.138) is not explicit for the friction factor f. It is on both sides of the equation. The equation can be solved for f by manual trial-and-error or by using equation solving software. For example, it can be solved using Goal Seek or Solver of Excel or using the fzero function of Matlab.

Example 6.10 Flow through a circular tube Problem We want to cool engine oil from 100 to 80 C. To do this, we plan to pass the oil through a tube whose surface is at a uniform temperature of 40 C. The oil flows through the tube at a velocity of 0.03 m/s, and the tube has an inside diameter of 2 cm. How long does the tube have to be to achieve the desired cooling of the oil?

Solution

The mean bulk temperature of the oil is (100 þ 80)/2 ¼ 90 C. Properties of engine oil at 90 C are r ¼ 846 kg=m3 ; cp ¼ 2176 J=kg C; k ¼ 0:138 W=m C y ¼ 2:81 105 m2 =s; Pr ¼ 360 The Reynolds number isReD ¼

VD ð0:03Þð0:02Þ ¼ ¼ 21:35 Laminar since < 2300 y 2:81 105

6.4 Internal flow

253

Let us check if the flow is fully developed: From Eqs. (6.119) and (6.120), Lh z 0:05ReD D ¼ 0:05ð21:35Þð0:02Þ ¼ 0:021 m Lt z 0:05ReD D Pr ¼ 0:021ð360Þ ¼ 7:6 m The flow is fully hydrodynamically developed, but probably thermally developing. Hence, we cannot say that the flow is fully developed, and we should use either Eq. (6.133) or (6.134) for the Nusselt number. We will use Eq. (6.133): NuD ¼

hD 0:0668ðD=LÞReD Pr ¼ 3:66 þ k 1 þ 0:04½ðD=LÞReD Pr2=3

(6.139)

There are two unknowns in this equation, h and L. We need a second equation with h and L. As the temperature of the tube surface is uniform, we can use the energy equation for constant surface temperature, Eq. (6.130): " # ðTs Tm Þx2 hP ðx2 x1 Þ (6.140) ln ¼ ðTs Tm Þx1 _ p mc Note: L ¼ x2 x1 and P ¼ p D.

p m_ ¼ rAV ¼ r D2 V ¼ ð846Þðp = 4Þð0:02Þ2 ð0:03Þ ¼ 0:00797 kg=s 4 Rearranging Eq. (6.139) for h, we have " # k 0:0668ðD=LÞReD Pr 3:66 þ (6.141) h¼ D 1 þ 0:04½ðD=LÞReD Pr2=3 Rearranging Eq. (6.140) for h, we have _ p mc ðTs Tm Þx¼L h¼ ln (6.142) pDL ðTs Tm Þx¼0 Equating the right-hand sides of Eqs. (6.141) and (6.142), we have # " _ p mc k 0:0668ðD=LÞReD Pr ðTs Tm Þx¼L ln (6.143) ¼ 3:66 þ 2=3 D pDL ðTs Tm Þx¼0 1 þ 0:04½ðD=LÞReD Pr The only unknown in Eq. (6.143) is the desired tube length L. So, if we solve Eq. (6.143) for L, we solve the problem. One way to do this is to use Excel. Moving the right side of Eq. (6.143) to the left of the equal sign, we have " # _ p mc k 0:0668ðD=LÞReD Pr ðTs Tm Þx¼L 3:66 þ þ ln ¼0 (6.144) D pDL ðTs Tm Þx¼0 1 þ 0:04½ðD=LÞReD Pr2=3 Putting values into Eq. (6.144), we have " # 0:138 0:0668ð0:02=LÞð21:35Þð360Þ ð0:00797Þð2176Þ ð40 80Þ ln ¼0 3:66 þ þ 2=3 0:02 pð0:02ÞL ð40 100Þ 1 þ 0:04½ð0:02=LÞð21:35Þð360Þ Reducing this, we have " 3:66 þ

ð10:2685=LÞ

1 þ 0:04ð153:72=LÞ2=3 Using Excel’s Goal Seek, we got L ¼ 2.67 m. The tube has to be 2.67 m long for the desired cooling of the oil.

16:220 ¼0 L

# (6.145)

Example 6.11 Flow through a noncircular tube Problem Water at 20 C having a flow rate of 5 kg/s enters a tube of rectangular cross section. The tube cross section is 3 cm by 6 cm, and it is 15 m long. The inner surface of the tube is maintained at a constant temperature of 75 C. What is the temperature of the water leaving the tube?

254

Chapter 6 Forced convection

Solution As the tube is noncircular, we need the hydraulic diameter. For a rectangular tube with sides of a and b, the hydraulic diameter is Dh ¼

2ab 2ð0:03Þð0:06Þ ¼ ¼ 0:04 m aþb 0:03 þ 0:06

For water at 20 C; r ¼ 998 kg=m3 cp ¼ 4182 J=kg C k ¼ 0:600 W=m C m ¼ 1:002 103 kg=ms Pr ¼ 7:01 We need the flow velocity to get the Reynolds number. The mass flow rate is m_ ¼ rVA, where A is the flow area. The flow 2 m_ . Therefore, area is A ¼ (0.03) (0.06) ¼ 0.0018 m , and V ¼ rA V¼

m_ 5 ¼ ¼ 2:78 m=s rA ð998Þð0:0018Þ

The Reynolds number is rVDh ð998Þð2:78Þð0:04Þ ¼ ¼ 1:108 105 1:002 103 m The flow is turbulent. Using Eq. (6.135) with n ¼ 0.4 as the fluid is being heated, we have Re ¼

NuD ¼

hD n 5 0:8 ¼ 0:023Re0:8 ð7:01Þ0:4 ¼ 544:1 D Pr ¼ 0:023 1:108 10 k

The convective coefficient is NuD k ð544:1Þð0:600Þ ¼ ¼ 8162 W=m2 C D 0:04 To get the exit temperature, we can use Eq. (6.131): h¼

ðTs Tm Þx2 ¼ ðTs Tm Þx1 e

hP mc ðx2 x1 Þ _p

Putting in the values, we have ð8162Þ½ð2Þð0:03þ0:06Þ

ð15Þ ð75 Tm Þx2 ¼ ð75 20Þx1 e ð5Þð4182Þ ¼ 19:2 C The exit temperature is Tm at x2 ; which is 75e19.2 ¼ 55.8 C. The fluid properties should be taken at the fluid mean bulk temperature. We took properties at 20 C, but with this result, we have a mean bulk temperature of (20 þ 55.8)/2 ¼ 37.9 C. We should do the problem again with water properties at a higher temperature. We did the problem again with properties at 40 C. We got a convective coefficient of 9970 W/m2 C and an exit temperature of 59.8 C. With this exit temperature, the mean bulk temperature is (20 þ 59.8)/2 ¼ 39.9 C. Properties were taken at 40 C so no further iteration is needed. In short, the Dittus-Boelter equation predicts that the exit temperature of the water is about 60 C. Let us look at this problem a little more. We used the Dittus-Boelter equation, which is the simplest of the three equations given for turbulent flow. However, the temperature difference between the tube surface and the fluid is considerable. The tube surface is at 75 C and the mean bulk water temperature is 40 C. And, some properties of water vary significantly with temperature, e.g., the Prandtl number and the viscosity. It might be interesting to see the result from the Sieder-Tate equation, which has a viscosity term to adjust for the difference between the surface and fluid temperatures. We did the problem using the Sieder-Tate equation, Eq. (6.136), and got h ¼ 11,460 W/m2 C, which is 15% higher than the 9970 W/m2 C value from Dittus-Boelter when we used properties at 40 C. With the Seider-Tate equation, the water exit temperature was 62.5 C. We also did the problem using the Gnielinski equation, Eq. (6.137). This equation has the friction factor f. We assumed that the tube had a fairly smooth surface equivalent to drawn tubing and solved for the friction factor using the Colebrook

6.4 Internal flow

255

Equation, Eq. (6.138). Using the Excel’s Goal Seek we found that the friction factor was 0.0164. Using this in the Gnielinski equation, we got a convective coefficient of 13,200 W/m2 C and a water exit temperature of 65.0 C. In short, the Dittus-Boelter equation gave a water exit temperature of about 60 C while the Sieder-Tate and Gnielinski equations gave the exit temperature as about 64 C. Because of the temperature-dependent nature of the properties of water, we will make an engineering judgment that the Sieder-Tate and Gnielinski equations are more accurate than the Dittus-Boelter equation for this problem. We conclude that the exit temperature of the water is about 64 C.

6.4.7 Annular flow We have discussed the equivalent diameter to use in correlation equations for flows through noncircular tubes. Flow through an annulus is a bit different as there may be two surfaces with heat transfer instead of one. Fig. 6.19 shows an annular flow area created by two concentric tubes. The area has an inside diameter D1, which is the outside diameter of the inner tube, and an outside diameter D2, which is the inside diameter of the outer tube. In a double-pipe heat exchanger, one fluid flows through the inner tube and the other flows through the annular area. The fluids have different temperatures, and heat is transferred from one fluid to the other through the tube wall of the inner tube. The hydraulic diameter for the annular flow area is the difference in the diameters, i.e., D2 D1 : hp i 2 D2 4 D 2 1 4A ¼ 4 ¼ D2 D1 (6.146) Dh ¼ P pD1 þ pD2 In most cases, one surface transfers heat and the other is perfectly insulated, i.e., adiabatic.

6.4.7.1 Fully developed laminar flow Table 6.5 [19] shows Nusselt numbers for fully developed laminar flow through an annulus when one surface is constant temperature and the other is adiabatic. Nui is the Nusselt number on the inner surface and Nuo is that on the outer surface.

hi ho

D1 D2

Annular Flow Area

FIGURE 6.19 Annular flow area.

256

Chapter 6 Forced convection

Table 6.5 Nusselt numbers for fully developed laminar flow in an annulus. D1/D2

Nui

Nuo

0.00 0.05 0.10 0.25 0.50 1.00

e 17.46 11.56 7.37 5.74 4.86

3.66 4.06 4.11 4.23 4.43 4.86

The convective coefficients hi and ho at the inner and outer surfaces of the annulus are hi ¼

Nui k Nuo k and ho ¼ Dh Dh

(6.147)

6.4.7.2 Fully developed turbulent flow For fully developed turbulent flow in an annulus, the convective coefficients at the two surfaces of the annulus are of the same magnitude. The above equations for turbulent flow in circular tubes may be applied with the hydraulic diameter of Eq. (6.146) used as the tube’s diameter. We conclude this chapter with an example that includes both internal and external convection.

Example 6.12 Heating of a warehouse Problem It is winter and workers in an unheated warehouse are cold and angry. They are freezing, and the people in the adjacent office are warm and comfy. There is a duct going through the warehouse which carries hot air to the office for heating. The warehouse workers have asked the management to provide diffusers in the duct to supply some hot air to the warehouse, but their request has been denied. The workers have decided to take matters into their own hands. They have decided to use fans to blow the warehouse air over the hot air duct and get some heating from the forced convection. The details are as follows: The duct is 10 inches diameter and the section in the warehouse is 50 feet long. The air enters the warehouse section at a rate of 1100 cfm and a temperature of 130 F. The air in the warehouse space is at 45 F. Fans will blow the air in crossflow over the duct at a velocity of 8 m/s. (a) What is the temperature of the air in the duct as it leaves the warehouse? (b) How much heating will be added to the warehouse air (Btu/hr)? (c) What are some practical problems with this proposed heating of the warehouse?

Solution We will assume that the duct wall is thin and a constant temperature surface. To get air property values, we need to make some assumptions on temperatures. We will check these assumptions after the calculations and iterate as necessary. We will assume that the air temperature drops by 20 F from the entrance to the exit. The air enters at 130 F (54.44 C) and leaves at 110 F (43.33 C). The average temperature of the air in the duct is then (54.44 þ 43.33)/2 ¼ 48.89 C. We will also assume that the temperature of the duct wall is Ts ¼ 85 F ¼ 29.44 C. We now proceed to get the convective coefficients hi and ho for the inside and outside surfaces of the duct.

6.4 Internal flow

257

Internal flow. The air enters at 1100 cfm and 130 F (54.44 C). For air at 54.44 C, r ¼ 1.075 kg/m3. Mass flow rate m_ ¼ rQ_ ¼ 1.075 kg=m3 1100 ft3 =min ð1 min = 60 sÞ ð1 m=3.2808 ftÞ3 ¼ 0.5581 kg=s D ¼ 10 inch ð1 m = 39.37 inchÞ ¼ 0.254 m The average duct air temperature is 48.89 C. For air at this temperature, r ¼ 1:096 kg=m3 ; cp ¼ 1007 J=kg C; k ¼ 0:0272 W=m C; y ¼ 17:87 106 m2 =s; Pr ¼ 0.722 i h m_ ¼ rAV ¼ 1:096 ðp = 4Þð0:254Þ2 V ¼ 0:5581 and V ¼ 10.05 m=s ReD ¼ Eq. (6.135):

VD ð10:05Þð0:254Þ ¼ ¼ 1:428 105 ðturbulentÞ y 17:87 106

0:3 5 0:8 ð0.722Þ0.3 ð0.0272 = 0.254Þ ¼ 29.70 W = m2 C hi ¼ 0:023Re0:8 D Pr ðk = DÞ ¼ ð0.023Þ 1.428 10 External flow. D ¼ 0.254 m and ¼ 8 m=s Eq. (6.95) with Table 6.2

Pr 1=4 ðk = DÞ Prs Prs is at the duct wall temperature of 29.44 C. Air at 29.44 C, Prs ¼ 0.728. All other air properties are at the warehouse air temperature of 7.22 C. The properties are 0:37 ho ¼ CRem D Pr

r ¼ 1:26 kg=m3 ; cp ¼ 1006 J=kg C; k ¼ 0:0242 W=m C; y ¼ 14 106 m2 =s; Pr ¼ 0.735 VD ð8Þð0:254Þ ¼ ¼ 1:451 105 y 14 106 For this Reynolds number, from Table 6.2, the values for Eq. (6.95) are C ¼ 0:26; m ¼ 0:6. Eq. (6.95): 1=4 0:37 Pr ho ¼ CRem ðk=DÞ D Pr Prs ReD ¼

0:6 ð0:735Þ0:37 ð0:735=0:728Þ1=4 ð0:0242=0:254Þ ¼ 27:70 W=m2 C ¼ ð0:26Þ 1:451 105 Back to the internal flow: We use Eq. (6.131) to get the exit temperature of the duct air. ðTs Tm Þx2 ¼ ðTs Tm Þx1 e

hP mc ðx2 x1 Þ _ p

ð29:70Þ½pð0:254Þ ð50=3:2808Þ ð0:5581Þð1007Þ

Putting in values, we have ð29:44 Tm Þx2 ¼ ð29:44 54:44Þe The exit temperature of the air is ðTm Þx2 ¼ 42:59 C ¼ 108.7 F. The heating provided to the warehouse is

_ p ðDTÞduct air ¼ ð0:5581Þð1007Þð54:44 42:59Þ ¼ 6660 W ¼ 22720 Btu=h q ¼ mc Let us now check the assumed duct wall temperature by doing an energy balance: hi Ai ðTduct air Ts Þ ¼ ho Ao ðTs Troom air Þ As the tube is thin, Ai ¼ Ao : Putting values into the equation; we have 54:44 þ 42:59 Ts ¼ 27:70ðTs 7:22Þ 29:70 2 Solving for Ts ; we get that the duct wall temperature is 28.6 C ¼ 83.5 F. Summarizing. We assumed the duct wall temperature was 85 F and we calculated it as 83.5 F. We assumed the exit temperature of the duct air was 110 F and we calculated it as 108.7 F. Our assumptions were close to the calculated values. We conclude that iteration is not necessary as it would have negligible impact on the results.

258

Chapter 6 Forced convection

Results (a) Duct air exits at 109 F. (b) Heating added to the warehouse is about 22,700 Btu/hr. (c) Practical problems include the following: The large number of fans needed. (This is not a problem if the facility is a fan factory.) Mounting of the fans is a major installation task. The noise due to the fans will be considerable. There will be considerable increased electrical load for the fans. The obtained warehouse heating is not very large considering the major installation requirements and costs. We calculated that the use of the fans added 22,700 Btu/hr to the warehouse air. Actually, the impact of the fans is even less than this. Without the fans, there would be natural convection from the hot duct. Example 7.4 estimates this natural convection as 5930 Btu/hr. Hence, the fans will actually only add 22,700 5930 ¼ 16,770 Btu/hr to the air. With such a low value, it is probably better to continue lobbying the management for diffusers in the hot air duct. Or, perhaps install gas-fired infrared heaters near the ceiling or unit heaters near the work stations in the warehouse.

6.5 Chapter summary and final remarks In this chapter, we first discussed the dimensionless parameters relevant to forced and natural convection: the Reynolds, Prandtl, and Nusselt numbers for forced convection and the Grashof, Prandtl, and Nusselt numbers for natural convection. We then proceeded to discuss forced convection for both external and internal flows. For external flows, we looked at the laminar and turbulent boundary layers on a flat plate and outlined the analytical solution of the flow and thermal equations for the laminar boundary layer. Besides the flat plate, we covered flows over cylinders and spheres. We also discussed flow through the tube banks of heat exchangers. For internal flows, we looked at flow through tubes and ducts. Correlation equations were presented for the different types of flows and different geometries. These equations, based on experimental studies, provide estimates for the convective coefficient h. Many of the equations have ranges of applicability for the Reynolds and Prandtl numbers. If a problem has parameters outside of the applicable ranges, then one should use the equation that most closely fits the problem. Of course, the best possible way to determine the h-value for an object is to perform an experiment on the actual object. This is usually not possible from an economic and/or practical basis, and correlation equations have to be used. One should keep in mind that the correlations are from carefully controlled lab experiments having conditions often quite different from those encountered in practice. Therefore, the h-value from a correlation equation may differ significantly from the actual h of an object, perhaps as much as 25% or more. (An aside to students who like to present answers with an extreme number of digits: Considering the uncertainty in the h-value, is this practice really appropriate?) We continue now to Chapter 7, which discusses natural convection.

6.6 Problems Notes: •

If needed information is not given in the problem statement, then use the Appendix for material properties and other information. If the Appendix does not have sufficient information, then use the Internet or other reference sources. Some good reference sources are mentioned at the end of Chapter 1. If you use the Internet, double-check the validity of the information by using more than one source.

6.6 Problems

• • •

259

Your solutions should include a sketch of the problem. Caution: Make sure that you use absolute temperatures (Kelvin or Rankine) if the problem involves radiation. In all problems, unless otherwise stated, the fluid pressure is atmospheric. Gas properties in the Appendix are at atmospheric pressure. If a problem has a gas at other than atmospheric pressure, the density and kinematic viscosity should be modified accordingly through use of the ideal gas law. 6-1 Fluid flows over a flat plate at 5 m/s. If the fluid is at 20 C, how far from the leading edge does transition from laminar flow to turbulent flow take place if (a) the fluid is air? (b) the fluid is water? (c) the fluid is engine oil? 6-2 Air at 40 C flows over a flat plate at 10 m/s. What is the boundary layer thickness 0.25 m from the leading edge? 6-3 Do Problem 6-2 if the air has a pressure of 2 atm. 6-4 Air at 20 C flows at 5 m/s over a square flat plate that is 5 cm by 5 cm. (a) What is the drag force on the plate? (b) Determine the velocity components u and v at a location 3 cm from the leading edge. Determine these components at two vertical distances from the plate: 1/3 and 2/3 of the distance from the plate surface to the edge of the boundary layer. 6-5 Water at 20 C flows over a rectangular heated flat plate that is at 80 C. The water velocity is 1.5 m/s. (a) What is the thickness of the hydrodynamic boundary layer 4 cm downstream of the leading edge? (b) What is the thickness of the thermal boundary layer 4 cm downstream of the leading edge? (c) If the plate is 5 cm wide and 20 cm long in the direction of flow, what is the rate of heat transfer from the plate to the water? 6-6 Engine oil at 40 C flows at a speed of 2 m/s over a flat plate that is 20 cm wide and 10 cm in the direction of flow. Five electric strip heaters, each 2 cm wide and 20 cm long, are placed sideby-side to create the flat plate. One side of the heaters is perfectly insulated and the other side heats the oil. What is the electric power (W) needed for each heater to produce a plate that has a uniform temperature of 200 C? 6-7 Air at 50 C and 2 atm pressure flows at a velocity of 4 m/s over a flat rectangular plate that is maintained at 100 C. The plate is 3 m long in the direction of flow. (a) What is the average convective coefficient for the plate? (b) What is the rate of convective heat transfer from the plate to the air per meter width of the plate? (c) If the plate’s surface has an emissivity of 0.6, what is the rate of radiative heat transfer to the 50 C surroundings per meter width of the plate? (d) Give a statement regarding the relative values of the Part (b) and (c) results. 6-8 Air at 20 C flows at a velocity of 0.5 m/s over a rectangular plate that is 15 cm wide and 25 cm in the direction of flow. The first 10 cm of the plate from the leading edge is perfectly insulated and the last 15 cm of the plate is maintained at a temperature of 150 C. What is the rate of heat transfer from the plate to the air?

260

Chapter 6 Forced convection

6-9 Air at 30 C flows over a horizontal flat plate at a speed of 2 m/s. The plate is square, 10 cm on a side. A thin electronic chip of size 1 cm by 1 cm is mounted on the plate about 3 cm downstream of the leading edge. The backside of the chip is perfectly insulated and the front surface convects to the air. If the chip produces 35 mW of power, what is the steady-state temperature of the chip? 6-10 Nitrogen at 20 C flows at a speed of 3 m/s over a flat plate that is 10 cm wide and 40 cm long in the direction of flow. The plate’s temperature is maintained at 80 C. (a) What is the drag force of the nitrogen on the plate? (b) What is the rate of heat transfer from the plate to the nitrogen? (c) Do Parts (a) and (b) assuming that there is a trip wire at the leading edge, which makes the flow turbulent over the entire plate. 6-11 Water at 20 C flows across a rectangular plate that is 10 cm wide and 20 cm in the direction of flow. The speed of the water is 1 m/s. The plate has an electric heater that produces a constant heat flux over the entire surface of the plate. If the plate temperature cannot exceed 70 C at any location, what is the maximum allowed power input to the heater? 6-12 Air at 25 C flows at a velocity of 40 m/s over a square plate that is 35 cm by 35 cm. The plate is at 80 C. (a) What is the drag force of the air on the plate? (b) What is the convective heat transfer from the plate to the air? (c) For the result of Part (b), how much heat transfer is from the laminar region and how much from the turbulent region? 6-13 Water at 15 C flows at a velocity of 2 m/s across an unheated flat plate that is 5 cm wide and 10 cm in the direction of flow. A thin electric strip heater that is 4 mm wide and 5 cm long is placed across the plate 5 cm from the leading edge. The strip is heated to a temperature of 40 C. What is the power to the heater? Assume that all the power goes into the water. 6-14 Air at 20 C flows at a velocity of 25 m/s over a thin horizontal circular plate of 0.5 m diameter. The plate is at 100 C. What is the rate of heat transfer from the plate to the air? 6-15 An electrical transmission line has an outside diameter of 2 cm. The current in the line is 250 A and the line has a resistance of 3 104 U per meter length. The line is outdoors and the wind is blowing over it at 20 miles per hour. If the air is at 15 C, what is the surface temperature of the line? 6-16 Steam at 1 atm and 100 C is flowing across a 2 cm diameter cylinder at a speed of 5 m/s. The surface of the cylinder is at 200 C. What is the rate of heat transfer per meter length of the cylinder to the steam? 6-17 Air at 20 C flows at a speed of 5 m/s across a 4 mm diameter wire. The surface of the wire is at 200 C. What is the heat transfer to the air per meter length of wire? 6-18 Air at 15 C flows across an electrical cylindrical heater at a velocity of 3 m/s. The heater has a diameter of 0.5 cm and a length of 4 cm. The emissivity of the heater’s surface is 0.9, and the surroundings are at 15 C. If the temperature of the heater’s surface cannot exceed 400 C, what is the maximum allowed power input to the heater? 6-19 An oil tank is 4 feet in diameter and 10 feet long. The tank is filled with hot oil. After the filling, the outer surface of the tank is at 140 C. To cool the oil, air at 20 C is blown over the tank’s cylindrical surface in the axial direction at a speed of 3 m/s. What is the rate of heat transfer to the air at the beginning of the oil cooling? Only include the heat transfer from the tank’s cylindrical surface. Do not include the tank ends.

6.6 Problems

261

6-20 Air at 20 C flows across a square cylinder. The sides of the cylinder are 3 cm wide and are at 180 C. The free-stream velocity of the air is 4 m/s. (a) What is the rate of heat transfer per meter length to the air if the air stream hits a flat surface of the cylinder? See Fig. P6.20A. (b) What is the rate of heat transfer per meter length to the air if the air stream hits a rotated square cylinder? See Fig. P6.20B.

(A)

3 cm

3 cm

u∞

(B) 3 cm

u∞ 3 cm

FIGURE P6.20 (A) Square cylinder. (B) Rotated square cylinder.

6-21 Air at 20 C flows over the rectangular cylinder shown in Fig. P 6.21. The cylinder has a length of 3 m and its surface is at 250 C. Iaf the speed of the air is 5 m/s, what is the rate of heat transfer to the air? 2 cm

u∞ = 5 m/s T∞ = 20 C

FIGURE P6.21

4 cm

262

Chapter 6 Forced convection

6-22 Air at 20 C flows over a thin flat plate that is positioned normal to the air stream. The plate is square, 5 cm on a side, and its temperature is 275 C. If the air velocity is 10 m/s, what is the rate of heat transfer to the air? 6-23 Air at 20 C flows at a velocity of 8 m/s over a 2 cm diameter aluminum sphere. The sphere’s surface is at 60 C. What is the rate of heat transfer to the air? 6-24 Water at 300 K flows at 4 m/s over a 4 cm diameter sphere whose surface temperature is 400 K. What is the rate of heat flow to the water? 6-25 An electronic device is cooled by blowing 30 C air over it. The device can be modeled as a sphere of 0.5 cm diameter. The power dissipated by the device is 5 W. What is the needed velocity of the air if the temperature of the outer surface of the device cannot exceed 200 C? Assume that conductive and radiative heat transfers are negligible. 6-26 It is desired to measure the temperature of hot air flowing through a duct. The air is flowing at 1.7 m/s. A thermistor is used as the thermometer. It is a 2 mm diameter sphere and has a surface emissivity of 0.85. The thermistor is inserted in the air stream, and it indicates an air temperature of 43 C when the duct walls are at 37 C. What is the actual temperature of the air? Assume that conduction through the leads of the thermistor is negligible. 6-27 Nitrogen at 100 C enters an inline tube bank with a free-stream velocity of 5 m/s. The tube bank has 48 tubes: eight rows in the direction of flow and six tubes per row. The tubes have an outside diameter of 2 cm and their centers are 3 cm apart. The outer surfaces of the tubes are at 200 C. What is the rate of heat transfer from the tube bank to the nitrogen per meter length of the tube bank? 6-28 Air flows through a tube bank. It enters with a temperature of 20 C and a free-stream velocity of 4 m/s. The tube bank is of staggered arrangement. The tubes are 1.25 cm OD and have a length of 1.5 m. The bank has SL ¼ 4 cm and ST ¼ 3 cm. There are 95 tubes with 10 rows in the flow direction. The outer surfaces of the tubes are at 120 C. (a) What is the rate of heat transfer to the air? (b) At what temperature does the air leave the tube bank? 6-29 Water at 20 C enters a tube bank with a free-stream velocity of 1.5 m/s. Each tube has an OD of 1 cm and a length of 1 m. The tube bank is of the inline arrangement and the tube centers are 2.5 cm apart in both the flow direction and the direction perpendicular to the flow. The surfaces of the tubes are at 150 C. It is desired to have the water leave the tube bank at 50 C. If there are five tubes in each row normal to the flow direction, how many rows of tubes are needed to achieve the desired water heating? 6-30 Fluid is flowing through a tube of pentagonal cross section. The sides of the flow area are of length “a.” What is the hydraulic diameter Dh ? 6-31 Fluid is flowing through a tube of hexagonal cross section. The sides of the flow area are of length “a.” What is the hydraulic diameter Dh ? 6-32 Fluid is flowing through a semicircular open channel of diameter “d.” The water has a maximum depth of d/4. What is the hydraulic diameter Dh ? 6-33 Water flows through a 2.5 cm diameter tube at a flow rate of 0.005 kg/s. The tube is 5 m long. The water enters at 20 C and the surface of the tube is maintained at a uniform temperature of 100 C. (a) What is the average convective coefficient? (b) At what temperature does the water leave the tube?

6.6 Problems

263

6-34 Water at a mean bulk temperature of 40 C flows at a velocity of 5 m/s through a heated 3/4 inch Type M copper tube. Determine the entrance length needed for the flow to become fully developed (a) hydrodynamically and (b) thermally. 6-35 Engine oil at a mean bulk temperature of 40 C flows at a mass flow rate of 1.4 kg/s through a heated 3/4 inch Type M copper tube. Determine the entrance length needed for the flow to become fully developed (a) hydrodynamically and (b) thermally. 6-36 Engine oil at 40 C enters a 1 cm diameter tube, which is maintained at a uniform temperature of 130 C. The tube is 50 m long and the oil leaves the tube at a temperature of 60 C. What is the mass flow rate of the oil? 6-37 Air at a pressure of 1000 kPa enters a rectangular duct that has a cross section of 12 cm by 7 cm. The duct is 5 m long. The mass flow rate of the air is 0.35 kg/s. The average temperature of the air in the duct is 250 C, and the duct wall is maintained at an average temperature of 200 C. What is the decrease in air temperature as it flows through the duct? 6-38 Air flows through a tube which has a square cross section of 1 cm by 1 cm. The tube is 20 cm long. The air enters the tube at 40 C and has a flow rate of 1.2 104 kg/s. The surface of the tube is maintained at 150 C. What is the exit temperature of the air? 6-39 Engine oil flows through a 2 cm diameter tube at a velocity of 0.5 m/s. The oil enters the tube at 60 C. The tube is 4 m long and its surface is maintained at 90 C. (a) What is the rate of heat transfer to the oil? (b) What is the exit temperature of the oil? 6-40 Hot water flows through a 1/2 inch nominal Sch 40 steel pipe at a rate of 3 gpm. The pipe is 10 feet long. The water enters the pipe at 120 F and the inside surface of the pipe is at a uniform temperature of 43 F. (a) What is the temperature of the water as it leaves the pipe? (b) If the flow rate of the water doubles to 6 gpm, what is the exit temperature of the water? 6-41 Water flows through a 3/4 inch nominal Sch 80 steel pipe at a rate of 4 gpm. The pipe is 5 feet long. The water enters at 20 C and leaves at 25 C. If the pipe surface has a uniform temperature, what is that temperature to achieve the stated heating of the water? Use Eq. (6.135) for the convective coefficient. 6-42 Do Problem 6-41, but use Eq. (6.137) for the convective coefficient. Also use Eq. (6.138) for the friction factor. 6-43 Air flows at a velocity of 6 m/s through a 20-m long annulus. The annulus has an inside diameter of 30 cm and an outside diameter of 45 cm. The air enters the annulus at 40 C and is heated by the inner surface of the annulus which is maintained at 150 C by condensing steam. The outer surface of the annulus is adiabatic. What is the convective coefficient for the inner surface of the annulus? 6-44 Ethylene glycol flows at a velocity of 8 m/s through an annulus formed by concentric circular tubes. The inner diameter of the annulus is 3.5 cm and the outer diameter is 5 cm. The glycol enters at 20 C and leaves at 45 C. The inner surface of the annulus, which heats the glycol, is maintained at 90 C. The outer surface is adiabatic. How long must the annulus be to achieve the desired heating of the glycol? 6-45 Air at 25 C flows across a flat plate that has a surface temperature of 125 C. The air velocity is 5 m/s. The plate is square, 0.5 m by 0.5 m. The drag force due to the air flow is measured and

264

6-46

6-47

6-48

6-49

6-50 6-51 6-52

6-53

6-54

Chapter 6 Forced convection

found to be 0.005 N. Using the Reynolds-Colburn analogy, estimate the convective coefficient h for the plate’s surface. Air at 20 C and a speed of 6 m/s flows across a flat plate whose surface temperature is 100 C. The plate is 1 m long in the direction of air flow and 0.5 m wide. Thermal measurements found that the average convective coefficient h for the plate’s surface is 8.2 W/m2 C. Using the Reynolds-Colburn analogy, estimate the drag force on the plate due to the air flow. In Section 6.3.3, data were given for factor F versus number of rows R in the tube bank. The data were given for 4, 5, 7, 10, 13, and 16 rows. A fourth degree polynomial was found to fit the data very well. Using the data in the section, determine a third degree polynomial fit for the data. Compare the values of F obtained from this third degree polynomial for rows 3 through 16 with those obtained from the fourth degree polynomial. Example 6.7 showed how the local convective coefficient varied for flow across a flat plate. We wish to investigate how the local drag coefficient Cfx varies with downstream distance x from the leading edge. Nitrogen at 50 C flows across a flat plate at a velocity of 12 m/s. The plate is at a uniform temperature of 100 C and is 5 m long in the direction of nitrogen flow. Plot Cfx versus x from the leading edge (x ¼ 0) to the downstream end of the plate (x ¼ L ¼ 5 m). A duct is 8 m long and has a rectangular cross section of 10 cm by 20 cm. Air flows through the duct, entering at 15 C and leaving at 80 C. The duct has a uniform surface temperature of 110 C. (a) What is the mass flow rate of the air (kg/s)? (b) What is the rate of heat transfer from the duct to the air? Engine oil at 25 C enters a 2.5 m long, 3 mm diameter tube at a velocity of 2.1 m/s. The temperature of the tube wall is uniform at 50 C. What is the exit temperature of the oil? Ethylene glycol flows at a velocity of 7.5 m/s through a 1.3 cm diameter tube that is 20 m long. The glycol enters at 70 C and leaves at 60 C. What is the needed uniform temperature of the tube wall to effect this cooling? Engine oil flows at a rate of 0.05 kg/s through a section of a 2.5 cm diameter tube. The oil enters the section at a temperature of 50 C. The tube wall is kept at 20 C, and it can be assumed that the flow through the section is fully developed. The section of the tube is 5 m long. (a) What is the exit temperature of the oil? (b) What is the heat transfer rate from the oil to the tube wall? A thin-walled pipe is 15 m long and has a diameter of 2.5 cm. Water is flowing through the pipe at a rate of 0.75 kg/s. The pipe wall imparts a uniform heat flux of 7 104 W/m2 to the water. The water enters the pipe at 20 C. Assume that the flow is fully developed. (a) What is the temperature difference between the local wall temperature and the local mean water temperature. (b) What is the exit temperature of the water from the pipe? Water flows through a 1/200 Sch 40 steel pipe at a rate of 2 gpm. The water enters at 70 F and leaves at 90 F. A cylindrical electrical heater jacket around the pipe keeps the inner surface of the pipe at 120 F. (a) How long does the pipe have to be to supply the specified water heating? (b) If the heater jacket has insulation at its outer surface so that 90% of its power output goes into the water and only 10% is lost to the room air, what is the needed power input to the heater? Assume the heater is 100% efficient.

References

265

6-55 Air flows through a duct that has a square cross section of 8 cm by 8 cm. The air enters a section of duct that is far enough downstream from the inlet so that the flow is fully developed. At the entrance to the section, the air is at 30 C and its velocity is 0.25 m/s. The section has a uniform wall heat flux to the air of 50 W/m2. (a) If the air exits the section at 100 C, how long is the section? (b) What is the temperature of the duct surface at the section’s exit? 6-56 Water is flowing at a rate of 0.04 kg/s through a thin-walled copper tube of 2 cm diameter and 5 m length. The water enters the tube at 90 C. Air at 20 C blows across the tube at a velocity of 7 m/s. (a) What is the temperature of the water as it exits the tube? (b) What is the heat transfer rate from the water to the air?

References [1] H. Blasius, Grenzschichten in Flussigkeiten mit kleiner Reibung, Z. Angew. Math. Phys. 56 (1908) 1e37. English translation in NACA TM 1256. [2] E. Pohlhausen, Der Warmeaustausch zwischen festen Korpern und Flussigkeiten mit kleiner Reibung und kleiner Warmeleitung, Z. Angew. Math. Mech. 1 (1921) 115e121. [3] S.W. Churchill, H. Ozoe, Correlations for laminar forced convection in flow over an isothermal flat plate and in developing and fully developed flow in an isothermal tube, ASME J. Heat Transf. 95 (1973) 416e419. [4] S.W. Churchill, A comprehensive correlation equation for forced convection from flat plates, AIChE J. 22 (2) (1976) 264e268. [5] W.M. Kays, M.E. Crawford, B. Weigand, Convective Heat and Mass Transfer, fourth ed., McGraw-Hill, 2005. [6] A. Zukauskas, Heat transfer from tubes in crossflow, Adv. Heat Tran. vol. 8 (1972) 93e160. [7] S.W. Churchill, M. Bernstein, A correlating equation for forced convection from gases and liquids to a circular cylinder in crossflow, J. Heat Transf. 99 (1977) 300e306. [8] E.M. Sparrow, J.P. Abraham, J.C.K. Tong, Archival correlations for average heat transfer coefficients for non-circular and circular cylinders and for spheres in cross- flow, Int. J. Heat Mass Transf. 47 (2004) 5285e5296. [9] G.C. Vliet, G. Leppert, Forced convection heat transfer from an isothermal sphere to water, ASME J. Heat Transf. 83 (1961) 163e175. [10] J.B. Will, N.P. Kruyt, C.H. Venner, An experimental study of forced convective heat transfer from smooth, solid spheres, Int. J. Heat Mass Transf. 109 (2017) 1059e1067. [11] S. Whitaker, Forced convection heat transfer correlations for flow in pipes, past flat plates, single cylinders, single spheres, and for flow in packed beds and tube bundles, AIChE J. 18 (2) (1972) 361e371. [12] A. Zukauskas, Heat transfer from tubes in crossflow, in: S. Kakac, Aung (Eds.), Handbook of Single Phase Convective Heat Transfer, Wiley Interscience, 1987. [13] H. Hausen, Darstellung des Warmeuberganges in Rohren Durch Vergallgemeinerte Potenzbeziehungen, Z. Ver. Deut. Ing. 4 (1943) 91e98. [14] E.N. Sieder, G.E. Tate, Heat transfer and pressure drop of liquids in tubes, Ind. Eng. Chem. 28 (1943) 1429e1435. [15] F.W. Dittus, L.M.K. Boelter, Heat transfer in automobile radiators of the tubular type, Univ. Calif. Berkeley Publ. Eng. 2 (1930) 443e461.

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[16] V. Gnielinski, New equations for heat and mass transfer in turbulent pipe and channel flow, Int. Chem. Eng. 16 (1976) 359e368. [17] L.F. Moody, Friction factors for pipe flow, Trans. ASME 66 (1944) 671e684. [18] C.F. Colebrook, Turbulent flow in pipes, with particular reference to the transition between the smooth and rough pipe laws, J. Inst. Civ. Eng. London 11 (1939) 133e156. [19] R.E. Lundberg, W.C. Reynolds, W.M. Kays, Heat Transfer with Laminar Flow in Concentric Annuli with Constant and Variable Wall Temperature and Heat Flux, 1963. NASA TN D-1972.

CHAPTER

Natural (free) convection

7

Chapter outline 7.1 Introduction ...............................................................................................................................267 7.2 Basic considerations ..................................................................................................................268 7.3 Natural convection for flat plates ................................................................................................ 271 7.3.1 Vertical plate..........................................................................................................271 7.3.1.1 Constant temperature surface............................................................................ 271 7.3.1.2 Constant heat flux surface ................................................................................. 273 7.3.2 Horizontal plate......................................................................................................274 7.3.2.1 Constant temperature surface............................................................................ 275 7.3.2.2 Constant heat flux surface ................................................................................. 275 7.3.3 Inclined plate.........................................................................................................277 7.4 Natural convection for cylinders..................................................................................................278 7.4.1 Horizontal cylinder .................................................................................................278 7.4.2 Vertical cylinder .....................................................................................................281 7.5 Natural convection for spheres....................................................................................................282 7.6 Natural convection for other objects............................................................................................ 285 7.7 Natural convection for enclosed spaces ...................................................................................... 286 7.7.1 Enclosed rectangular space .....................................................................................287 7.7.1.1 Horizontal rectangular enclosure ....................................................................... 288 7.7.1.2 Vertical rectangular enclosure............................................................................ 288 7.7.1.3 Inclined rectangular enclosure........................................................................... 289 7.7.2 Annular space between concentric cylinders .............................................................291 7.7.3 Space between concentric spheres...........................................................................292 7.8 Natural convection between vertical fins ..................................................................................... 294 7.9 Chapter summary and final remarks ............................................................................................. 297 7.10 Problems ...................................................................................................................................298 References ..........................................................................................................................................303

7.1 Introduction This chapter continues our discussion of convection, the mode of heat transfer between a surface and a fluid caused by their differences in temperature. Chapter 6 discussed forced convection, a category of convection in which the fluid has significant velocity, which is usually produced by fans, blowers, or Heat Transfer Principles and Applications. https://doi.org/10.1016/B978-0-12-802296-2.00007-X Copyright © 2021 Elsevier Inc. All rights reserved.

267

CHAPTER

Heat exchangers

8

Chapter outline 8.1 Introduction .................................................................................................................................305 8.2 Types of heat exchangers .............................................................................................................305 8.2.1 Temperature distribution in double-pipe heat exchangers.............................................308 8.3 The overall heat transfer coefficient .............................................................................................. 315 8.4 Analysis methods .........................................................................................................................318 8.4.1 Log mean temperature difference method...................................................................318 8.4.1.1 Double-pipe heat exchangers .............................................................................. 318 8.4.1.2 Nonedouble-pipe heat exchangers...................................................................... 320 8.4.2 Effectivenessenumber of transfer unit method ...........................................................326 8.5 Chapter summary and final remarks............................................................................................... 336 8.6 Problems .....................................................................................................................................336 References ..........................................................................................................................................340

8.1 Introduction Heat exchangers are devices used to transfer thermal energy between two fluids. As the fluids flow through an exchanger, one fluid gains heat and the other loses heat. If the process does not include evaporation or condensation, both fluids will experience a change in temperaturedone fluid increasing in temperature and the other decreasing in temperature. Heat exchangers have a myriad of applications. In power plants, they serve as steam condensers, feedwater heaters, steam generators, and preheaters. Heat exchangers see extensive use in chemical process applications. In addition, they are extensively used in the HVAC industry, where they are boilers, condensers, evaporators, heaters, air conditioners, radiators, and heating/cooling coils.

8.2 Types of heat exchangers There are three main types of heat exchangers: direct contact, regenerator, and recuperator. In a direct contact exchanger, the two inlet fluid streams combine and mix together. Examples of direct contact exchangers are cooling towers, open feedwater heaters in power plants, and chemical processes where fluids can be mixed together. There is no wall separating the fluids to corrode and foul

Heat Transfer Principles and Applications. https://doi.org/10.1016/B978-0-12-802296-2.00008-1 Copyright © 2021 Elsevier Inc. All rights reserved.

305

306

Chapter 8 Heat exchangers

and degrade the heat exchanger performance. In addition, the pressure drop of the exchanger is less than that of exchangers incorporating tubes. Finally, the direct contact exchanger can be considerably more economical than other types of exchangers. One major disadvantage, however, is that the fluids mix together. In many applications, contamination of a fluid by the other fluid is highly undesirable. For example, one fluid may be radioactive; the other nonradioactive. Or, one may be poisonous; the other nonpoisonous. In these examples, direct contact exchangers are definitely unsuitable. In a regenerator, heat is transferred to a storage medium, the “core” or “matrix” of the exchanger, by the hotter fluid, and then the cooler fluid gains heat from the storage medium. It is a periodic process, with the two fluids alternating in contact with the core. Regenerators are of the fixed-matrix or rotary types. Regenerators have a high surface area per volume, and the high surface area makes the exchanger particularly attractive for gas-to-gas applications. A disadvantage is that there can be minor mixing of the fluid streams. In recuperators, the two fluids are separated by a wall. The fluids transfer heat by convection to and from the wall and conduction through the wall. The major advantage of a recuperator is that the two fluids are indeed separated, with no cross-contamination if the wall maintains its integrity. Disadvantages include the higher capital cost of recuperators and the increased pressure drop, which results in higher pumping costs. Our discussion in this chapter is limited to recuperators. In particular, we will discuss double-pipe, shell-and-tube, and crossflow heat exchangers. The simplest type of recuperator is the double-pipe heat exchanger shown in Fig. 8.1A. It consists of two concentric tubes. One fluid flows in the inner tube and the other fluid flows in the annular area between the inner and outer tubes. If the fluids flow in the same direction, as they do in Fig. 8.1A, then the flow is said to be parallel flow. If the fluids flow in opposite directions, the flow is counterflow. Each fluid makes a single pass through the exchanger; they only go through the exchanger once. In a shell-and-tube heat exchanger, one fluid flows through the tubes, and the other fluid flows around the tubes, enclosed by the “shell” of the exchanger. Fig. 8.1B shows an exchanger with one shell pass and two tube passes. Exchangers can also have more than two tube passes and more than one shell pass. For example, Fig. 8.1C shows a two-shell pass, four-tube-pass exchanger. Baffles are placed in shell-and-tube heat exchangers to channel the flow of the shell fluid across the tubes. The baffles enhance contact of the shell fluid with the tubes. They minimize bypassing of tubes by the shell fluid. In crossflow heat exchangers, one fluid flows through the tubes and the other fluid flows crosswise to the tubes. The crossflow increases the heat transfer relative to flows that are longitudinal along the tubes, such as in the double-pipe exchanger. Fig. 8.1D shows two crossflow heat exchangers. In one exchanger, a fluid flows through the tubes and the other fluid flows unrestrained over the tubes. It is said that the tube fluid is “unmixed” as it is confined to the tubes. The other fluid is “mixed” as it is somewhat unconfined and can mix with itself. In the second exchanger, there are fins on the outside of the tubes. One fluid flows through the tubes and is “unmixed.” The other fluid flows through the channels formed by the fins. This fluid is also said to be “unmixed” as it is confined to the channels, which hinder its mixing. Figs. 8.2 and 8.3 are photographs of heat exchangers. Fig. 8.2A shows shell-and-tube heat exchangers used in a chemical plant to cool hot gases with water. Fig. 8.2B shows a tube bundle for a shell-and-tube heat exchanger. In the foreground is the tube sheet to which the tubes are attached. Metal plates, or “baffles,” are seen further back in the photo. These baffles direct the shell-side fluid around the outside of the tubes, assuring that the fluid contacts the tubes as much as possible.

8.2 Types of heat exchangers

(A)

307

(D) Double-Pipe

Crossflow Fluid A (tube) unmixed Fluid B (shell) mixed

Fluid B

(B) One-Shell-Pass, Two-Tube-Passes Shell-in Fluid A Tube-in Tube-out

Shell-out

Fluid A (tube) unmixed Fluid B (shell) unmixed

(C) Two-Shell-Passes, Four-Tube-Passes Shell-in

Fluid B

Tube-in

fins

Fluid A Tube-out

Shell-out

FIGURE 8.1 Types of heat exchangers.

Fig. 8.2C shows U-tubes for a one-shell-pass, two-tube-pass heat exchanger. In the left of the photo is a shell-and-tube heat exchanger. Finally, Fig. 8.2D is a crossflow heat exchanger. One fluid flows through the copper tubes and the other fluid flows over the tubes. The aluminum fins on the tubes inhibit mixing of the outer fluid. Hence, it is said that both fluids are “unmixed.” Fig. 8.3A and B show a small heat exchanger manufactured by Exergy LLC of Garden City, NY. The exchanger is 13.5 inches long, 3 inches diameter, and of nickel brazed 316L stainless steel. This model has 253 tubes, and its heat transfer area is 0.60 m2. Uses of the exchanger include chemical processing, water for injection for pharmaceutical plants, and semiconductor manufacturing. Many thanks to Exergy for providing the photos.

308

Chapter 8 Heat exchangers

FIGURE 8.2 (A) Shell-and-tube heat exchangers. (B) Heat exchanger tube sheet and tubes. (C) U-tube bundle. (D) Crossflow heat exchanger (both fluids unmixed).

Finally, Fig. 8.4 shows a badly corroded tube sheet and tubes. The effect of corrosion on heat transfer is discussed in Section 8.4 below. Corroded heat exchangers can often be repaired and returned to service through tube replacement. Several companies in the United States and overseas provide such service.

8.2.1 Temperature distribution in double-pipe heat exchangers Fig. 8.5 shows a double-pipe heat exchanger, which consists of two concentric pipes (or tubes). Fluid A flows through the inner tube, and Fluid B flows in the annular area between the tubes, which is called the “shell.” The outer surface of the outer tube is well-insulated so that heat transfer between the heat exchanger and the environment is insignificant compared with the heat transfer between the fluids. Fig. 8.5 shows both fluids flowing in the same direction, left to right. This is a double-pipe parallel flow exchanger. If the flow direction of one of the fluids was reversed, the exchanger would be a double-pipe counterflow exchanger.

8.2 Types of heat exchangers

309

FIGURE 8.3 (A) Small heat exchanger (outside). (B) Small heat exchanger (inside).

FIGURE 8.4 Corroded tube sheet and tubes.

Let us say that the temperatures of the fluids, TA and TB , are different, with TA being greater than TB: Then there will be heat transfer across the tube wall from Fluid A to Fluid B. As discussed Chapter 3, the rate of heat transfer q is DToverall TA TB q¼ P ¼ P R R

(8.1)

310

Chapter 8 Heat exchangers

Fluid B Shell Insulated

Fluid A

Tube

FIGURE 8.5 Double-pipe heat exchanger.

P where R is the sum of the resistances between fluids A and B. These resistances are the convective resistance at the inner surface of the tube, the conductive resistance of the tube’s wall, and the convective resistance at the outer surface of the tube. The rate of heat transfer between the fluids can also be expressed in terms of the overall heat transfer coefficient U. That is, q ¼ UAðDTÞavg

(8.2)

where ðDTÞavg is the “average” temperature difference between the fluids and A is the heat transfer area between the fluids. A double-pipe heat exchanger does not have a single, unique area A. It has the area of the inner surface of the tube and the area of the outer surface of the tube, which are different. The inner surface has area Ai ¼ 2pri L, where ri is the radius of the inner surface and L is the length of the tube. Similarly, the outer surface of the tube has an area Ao ¼ 2pro L, where ro is the radius of the outer surface and L is the tube’s length. As the areas are different, the U values must be referenced to a specific area: Ui is referenced to the inner surface area and Uo is referenced to the outer surface area. The rate of heat transfer in Eq. (8.2) is therefore q ¼ Ui Ai ðDTÞavg ¼ Uo Ao ðDTÞavg

(8.3)

From Eq. (8.3), it is seen that Ui Ai ¼ Uo Ao : As Ai ¼ 2pri L and Ao ¼ 2pro L, the coefficients are related by ri (8.4) Uo ¼ Ui ro Let us continue our discussion of the double-pipe heat exchanger. Fig. 8.6 shows two double-pipe exchangers. The one on the left is parallel flow; the one on the right is counterflow. Under each figure is a diagram showing how the fluid temperatures change as they travel through the exchanger. The hotter fluid enters the exchanger at Thi and leaves at Tho : The colder fluid enters at Tci and leaves at Tco : We will be calling the left end of the exchanger “Location 1” and the right end of the exchanger “Location 2.” Some interesting items: For parallel flow exchanger, the colder fluid cannot exit the

8.2 Types of heat exchangers

Parallel Flow

311

Counterflow

Fluid B

Fluid A

Fluid A

Fluid B

Thi

Thi Tho Tco

Tco

Tho Tci

Tci 1

2

1

2

FIGURE 8.6 Temperature distribution in double-pipe exchangers.

exchanger at a higher temperature than the hotter fluid exits. However, for the counterflow exchanger, the exit temperature of the colder fluid can be higher than the exit temperature of the hotter fluid. Also, for the counterflow exchanger, the temperature curves are not generally parallel. That is, the temperature difference between the hot and cold fluids usually varies with location in the exchanger. If, however, one fluid is evaporating at constant temperature and the other fluid is condensing at constant temperature, then the temperature difference will be constant throughout the exchanger. Such is also the case if the products of the mass flow rate and the specific heat for the two fluids are equal. Let us now determine the average temperature difference between the fluids, which is needed in Eqs. (8.2) and (8.3). Consider the parallel flow exchanger in Fig. 8.6. We will take a vertical slice of the exchanger between locations x and x þ dx. Fig. 8.7 shows control volumes for Fluids A and B on both sides of the tube wall. The area of the wall between x and x þ dx is dA. We will assume that the temperature of Fluid A is greater than the temperature of Fluid B. As the fluids flow through their respective control volumes, Fluid A gives heat to Fluid B and Fluid A’s enthalpy decreases. Fluid B experiences an increase in enthalpy due to the heat received from Fluid A. The various energy flows are shown in Fig. 8.7. For Fluid A, the energy balance is m_ A cA TA ¼ UdAðTA TB Þ þ m_ A cA ðTA þ dTA Þ

(8.5)

For Fluid B, the energy balance is m_ B cB TB þ UdAðTA TB Þ ¼ m_ B cB ðTB þ dTB Þ

(8.6)

312

Chapter 8 Heat exchangers

dq = U(dA)(TA–TB)

Fluid B ṁBcBTB

ṁBcB(TB+dTB) tube wall dA

ṁAcATA Fluid A

ṁAcA(TA+dTA) x

x+dx

FIGURE 8.7 Energy flows for differential control volumes.

In these two equations, m_ is the mass flow rate, c is the specific heat, T is the fluid temperature, U is the overall heat transfer coefficient of the exchanger, and dA is the differential wall area between the fluids. Subscripts indicate the particular fluid: the hot fluid (Fluid A) and the cold fluid (Fluid B). Rearranging Eq. (8.5), we get dTA ¼

UdAðTA TB Þ m_ A cA

(8.7)

Rearranging Eq. (8.6), we get dTB ¼

UdAðTA TB Þ m_ B cB

Subtracting Eq. (8.8) from Eq. (8.7), we have

1 1 þ dðTA TB Þ ¼ UdAðTA TB Þ m_ A cA m_ B cB Dividing both sides of Eq. (8.9) by ðTA TB Þ, we arrive at dðTA TB Þ 1 1 ¼ U þ dA TA TB m_ A cA m_ B cB

(8.8) (8.9)

(8.10)

Both sides of Eq. (8.10) can now be integrated over the entire length of the heat exchanger, from the left end (Location 1) to the right end (Location 2). Assuming that U, the mass flow rates, and the specific heats are constants, we have ðTA TB Þ2 1 1 þ ln ¼ UA (8.11) ðTA TB Þ1 m_ A cA m_ B cB As the differential heat flow dq for the infinitesimal control volumes shown in Fig. 8.7 is equal to UdAðTA TB Þ, Eq. (8.7) may be written as dq ¼ m_ A cA dTA

(8.12)

8.2 Types of heat exchangers

313

and Eq. (8.8) may be written as dq ¼ m_ B cB dTB

(8.13)

Integrating Eqs. (8.12) and (8.13) over the entire length of the heat exchanger; i.e., from the left end (Location 1) to the right end (Location 2), we get expressions for the total rate of heat flow q for the exchanger: q ¼ m_ A cA ðTA Þ1 ðTA Þ2 (8.14) (8.15) and q ¼ m_ B cB ðTB Þ2 ðTB Þ1 By using Eqs. (8.14) and (8.15) in Eq. (8.11), Eq. (8.11) then becomes a third expression for the heat flow q of the exchanger: ðTA TB Þ1 ðTA TB Þ2 q ¼ UA (8.16) ðTA TB Þ1 ln ðTA TB Þ2 Comparing this equation to Eq. (8.2), it is seen that the temperature factor is the average temperature difference between the two fluids in the heat exchanger. As the factor is the average (or mean) temperature difference and contains a log function, the temperature term is called the Log Mean Temperature Difference (LMTD). Summarizing, we have q ¼ UAðLMTDÞ where the LMTD is

ðTh Tc Þ1 ðTh Tc Þ2 LMTD ¼ ðTh Tc Þ1 ln ðTh Tc Þ2

(8.17) (8.18)

In Eq. (8.18), we changed the subscripts on the temperatures from A and B to the more commonly used subscripts h and c, respectively. We can do this as Fluid A is the hot fluid and Fluid B is the cold fluid. Calculation of the LMTD looks difficult, but it really is not. The term ðTA TB Þ1 is the temperature difference of the fluids at the left end of the exchanger and the term ðTA TB Þ2 is the temperature difference of the fluids at the right end. Eq. (8.18) says that LMTD is the temperature difference of the fluids at the left end of the exchanger minus the temperature difference of the fluids at the right end of the exchanger divided by the natural log of the ratio of the two temperature differences. Indeed, Eq. (8.18) can be written even more simply as LMTD ¼

ðDTÞ1 ðDTÞ2 ðDTÞ1 ln ðDTÞ2

(8.19)

We derived the above equations using a parallel flow double-pipe heat exchanger. If we had done the derivation for a counterflow double-pipe exchanger, and the cold fluid went right to left, the only change in the equations would be Eq. (8.15), which would be replaced with q ¼ m_ B cB ðTB Þ1 ðTB Þ2 (8.20)

314

Chapter 8 Heat exchangers

Just keep in mind that the temperature differences in the equations (and q) are positive. Caution: For a counterflow exchanger, DT1 can be equal to DT2 : If this is the case, Eq. (8.19) calculates the LMTD as 0/0. For this special case, the temperature difference between the fluids is the same throughout the exchanger, and DT1 ¼ DT2 should be used as the value for ðDTÞavg instead of the calculated LMTD. Example 8.1 shows how easy it is to calculate the LMTD for a double-pipe heat exchanger. It also shows that the LMTD for a parallel flow exchanger is different from the LMTD for a counterflow heat exchanger having the same entrance and exit fluid temperatures.

Example 8.1 LMTD for parallel and counterflow double-pipe exchangers Problem In a double-pipe heat exchanger, the hotter fluid enters at 150 C and leaves at 80 C and the colder fluid enters at 20 C and leaves at 60 C. Determine the LMTD if the heat exchanger is (a) parallel flow and (b) counterflow.

Solution Parallel flow. The first step in the solution is to draw a sketch of the temperature distribution of the exchanger and put the temperatures and flow directions on the sketch. Looking at Fig. 8.6, we have the parallel heat exchanger in the figure below. 150

80 60 20

From Eq. (8.19), we have, ðDTÞ1 ðDTÞ2 ð150 20Þ ð80 60Þ ¼ ¼ 58:8 C ðDTÞ1 150 20 ln ln 80 60 ðDTÞ2 Counterflow. Again using Fig. 8.6, the sketch for the counterflow exchanger is given below. LMTD ¼

150

60 80 20

ðDTÞ1 ðDTÞ2 ð150 60Þ ð80 20Þ ¼ ¼ 74:0 C ðDTÞ1 150 60 ln ln 80 20 ðDTÞ2 It is seen that the LMTD’s are different for the two different types of double-pipe heat exchangers even though the entering and leaving fluid temperatures are the same. Another interesting observation is that the LMTD for the counterflow exchanger is greater than the LMTD for the parallel flow exchanger. This is a general characteristic and the reason why counterflow exchangers are usually preferred over parallel exchangers. For the same U, A, and fluid entering and leaving temperatures, a counterflow exchanger will have a greater heat transfer rate q than a parallel flow exchanger. LMTD ¼

8.3 The overall heat transfer coefficient

315

8.3 The overall heat transfer coefficient Let us now look at the overall heat transfer coefficient in some detail. Fig. 8.8 shows a cross-sectional view of the inner tube of a double-pipe heat exchanger. The convective coefficient is hi at the inner surface of the tube and ho at the outer surface. The tube has inner and outer radii ri and ro and a length L perpendicular to the plane of the figure. Fluid A in inside the tube; Fluid B is in the shell; and Fluid A has a higher temperature than Fluid B. From Eqs. (8.1) through (8.3), we have DT q ¼ P ¼ Ui Ai DT ¼ Uo Ao DT R

(8.21)

Resistances were discussed in Chapter 3. At the inner surface of the tube, the convective resistance At the outer surface of the tube, the convective resistance is ho1Ao : lnðro =ri Þ , where k is the thermal conductivity of the tube And, the resistance of the tube wall is 2pkL material and L is the length of the tube. The circumferential areas in Eq. (8.21) are Ai ¼ 2pri L and Ao ¼ 2pro L: Inserting these resistances into Eq. (8.21), we have is

1 h i Ai :

Ui ¼

1 1 ri lnðro =ri Þ ri þ þ hi k ho r o

and Uo ¼

(8.22)

1 ro ro lnðro =ri Þ 1 þ þ k ho hi r i

(8.23)

In many cases, one term in the denominators of these equations predominates. The term associated with the resistance of the tube wall is usually much less than the other two terms. And, convective

Fluid B (shell)

insulation q

Fluid A (tube) ho

hi

ri ro

FIGURE 8.8 Convective coefficients for a double-pipe heat exchanger.

316

Chapter 8 Heat exchangers

Table 8.1 Typical ranges of the overall heat transfer coefficient U. Type of exchanger

U (W/m2 C)

Water-to-water Water-to-oil Gas-to-gas Gas-to-water Steam-to-fuel oil Oil-to-oil Feedwater heater Steam condenser Finned tube (water/air) Finned tube (steam/air)

850e2500 100e400 10e50 10e250 100e400 50e400 1000e8500 1000e6000 30e60 30e300

coefficients for liquids are usually quite higher than those for gases. Therefore, if one fluid is a liquid and the other is a gas, the term related to the liquid is often negligible. Typical ranges of the overall heat transfer coefficient are given in Table 8.1. The overall heat transfer coefficients in Eqs. (8.22) and (8.23) are for new, clean heat exchangers. With extended operation, the coefficients may appreciably decrease due to degradation of the heat transfer surfaces through corrosion, deposits, and possibly biological fouling such as algae. Both the inner and outer surfaces of the tube may be affected. Additional resistance terms may be added to the denominators of Eqs. (8.22) and (8.23) to account for the change in the overall heat transfer coefficient. The revised equations are Ui ¼

1 1 ri lnðro =ri Þ ri þ þ Rfi þ þ Rfo ðri =ro Þ hi k ho r o

and Uo ¼

1 ro ro lnðro =ri Þ 1 þ þ Rfo þ Rfi ðro =ri Þ þ k ho hi r i

(8.24)

(8.25)

where Rfi is the fouling factor (or resistance) on the inner surface of the tube and Rfo is the fouling factor (or resistance) on the outer surface of the tube. By comparing Eq. (8.22) to Eq. (8.24) and Eq. (8.23) to Eq. (8.25), we arrive at the following relations between the overall heat transfer coefficients for a new (clean) exchanger and one with fouling: 1 1 ri ¼ Rfi þ Rfo (8.26a) Ui ðfouledÞ Ui ðcleanÞ ro 1 1 ro ¼ Rfo þ Rfi (8.26b) Uo ðfouledÞ Uo ðcleanÞ ri Typical fouling factors are given in Table 8.2. More extensive information is given in Refs. [1,2].

8.3 The overall heat transfer coefficient

317

Table 8.2 Typical fouling factors Rf. Fluid

Rf (m2 C/W)

Distilled water and boiler feedwater Distilled water and boiler feedwater Water (river) Fuel Oil Oil (hydraulic, lubricating, and transformer) Ethylene glycol Steam Air Engine exhaust gases Refrigerants (liquid) Refrigerants (vapor)

0.0001 (below 50 C) 0.0002 (above 50 C) 0.0004 0.0009 0.0002

0.00035 0.0001 0.0004 0.0018 0.0002 0.0004

Example 8.2 Overall heat transfer coefficient and fouling factor Problem A refrigerant is flowing through the tube of a double-pipe counterflow heat exchanger. The tube is 1-inch Type L copper. Air flows in the annular region of the exchanger. The convective coefficient on the inner surface of the tube is 140 W/m2 C and the coefficient on the outer surface of the tube is 40 W/m2 C. (a) What is the overall heat transfer coefficient based on the area of the outer surface of the tube? (b) After a year of service, the inner surface of the tube has a fouling factor of 0.0002 m2 C/W and the outer surface has a fouling factor of 0.00015 m2 C/W. By what percentage has the overall heat transfer coefficient decreased from that calculated for the clean exchanger in Part (a)?

Solution From the Internet, the copper tubing has an outside diameter of 1.125 inch and a wall thickness of 0.05 inch. Also, copper has a thermal conductivity of 400 W/m C. Therefore, ro ¼ ð1:125 = 2Þ in

1m ¼ 0:01429 m 39:37 in

ri ¼ ro 0:05 ¼ ð1:125 = 2Þ 0:05 ¼ 0:5125 in

1m ¼ 0:01302 m 39:37 in

(a) We will use Eq. (8.23) for the overall heat transfer coefficient. 1 (8.23) ro ro lnðro =ri Þ 1 þ þ k ho hi ri From the problem statement, hi ¼ 140 W m2 C and ho ¼ 40 W m2 C. Putting values into Eq. (8.23), we have Uo ¼

Uo ðnewÞ ¼

1 1 ¼ 30:45 W=m2 C ¼ 0:01429 0:01429 lnð0:01429=0:01302Þ 1 0:03284 þ þ ð140Þð0:01302Þ 400 40

318

Chapter 8 Heat exchangers

The new exchanger has an overall heat transfer coefficient of 30.45 W/m2 C based on the outer surface area of the tube. (b) Eq. (8.25) includes terms for the fouling factors. 1 (8.25) ro ro lnðro =ri Þ 1 þ þ Rfo þ Rfi ðro =ri Þ þ k ho hi ri Comparing Eq. (8.25) to Eq. (8.23), it is seen that Eq. (8.25) has two additional resistance terms in the denominator. They are Uo ¼

Rfi ðro = ri Þ ¼ 0:0002ð0:01429 = 0:01302Þ ¼ 0:00022 and Rfo ¼ 0:00015: Adding these values to the denominator in the calculation for Uo ðnewÞ above, we have, for the year-old exchanger 1 ¼ 30:11 W=m2 C 0:03284 þ 0:00022 þ 0:00015 ð30:4530:11Þ 100 ¼ 1.12%. The % decrease in Uo is 30:45 It is seen that the effect of the fouling is very, very minimal. Uo ðyear oldÞ ¼

8.4 Analysis methods The two most common methods to solve heat exchanger problems are the LMTD method and the effectivenessenumber of transfer unit (ε-NTU) method. Although both methods can be used for a given problem, each method has its particular strong points. The LMTD method is especially useful in the design and sizing of heat exchangers. If inlet and outlet temperatures and the overall heat transfer coefficient are given, or are easily determined, then the LMTD can be readily calculated and the LMTD method can be used to determine the heat transfer rate and the required surface area for the exchanger. If, on the other hand, only the inlet temperatures and flow rates are known, then the LMTD cannot be easily determined. Use of the LMTD method will entail tedious iterations to ultimately determine the LMTD and achieve a successful problem solution. The ε-NTU method is greatly superior in these performance-type problems where the outlet temperatures are to be determined. We will now discuss the two methods and give examples to illustrate their use.

8.4.1 Log mean temperature difference method 8.4.1.1 Double-pipe heat exchangers The outer surfaces of heat exchangers are well-insulated and the heat transfer to the surrounding environment is very small. It is assumed that this heat transfer is negligible and that the total amount of heat leaving the hot fluid is gained by the cold fluid. There are three equations for q, which can be used to solve problems by the LMTD method: For the heat leaving the hot fluid, we have q ¼ m_ h cph ðDTÞh ¼ m_ h cph ðThi Tho Þ

(8.27)

For the heat gained by the cold fluid, we have q ¼ m_ c cpc ðDTÞc ¼ m_ c cpc ðTco Tci Þ

(8.28)

and the equation with the overall heat transfer coefficient U and the LMTD is q ¼ UAðLMTDÞ

(8.29)

8.4 Analysis methods

319

The temperature subscripts in Eqs. (8.27) and (8.28) are hi ¼ hot fluid entering (going in) ho ¼ hot fluid leaving (going out) ci ¼ cold fluid entering (going in) co ¼ cold fluid leaving (going out) Example 8.3 shows the application of Eqs. (8.27)e(8.29) in determining the required heat transfer area of an exchanger and the flow rate for one of the fluids.

Example 8.3 LMTD method for parallel and counterflow double-pipe exchangers Problem

A double-pipe heat exchanger heats water from 20 to 65 C. The heating is done by hot gases (cp ¼ 900 J/kg C) that enter the exchanger at 250 C and leave at 75 C. The flow rate of the water is 2 kg/s. The overall heat transfer coefficient of the heat exchanger is 280 W/m2 C. (a) What is the surface area required if the heat exchanger is parallel flow? (b) What is the surface area required if the heat exchanger is counterflow? (c) What is the flow rate of the hot gases?

Solution

(a) We first sketch the temperature diagram for a double-pipe parallel heat exchanger, putting the temperatures and flow directions on the sketch. 250 hot g

ases

75 65 20

water

Givens in this problem include U ¼ 280 W/m2 C and m_ c ¼ 2 kg=s. Using Eq. (8.19), LMTD ¼

ðDTÞ1 ðDTÞ2 ð250 20Þ ð75 65Þ ¼ ¼ 70:16 C ðDTÞ1 ð250 20Þ ln ln ð75 65Þ ðDTÞ2

q ¼ m_ c cpc ðDTÞc ¼ 2ð4180Þð65 20Þ ¼ 3:762 105 W q ¼ UAðLMTDÞ ¼ 280Að70:16Þ ¼ 3:762 105 Solving this for A, we have A [ 19.15 m2. (b) Similarly, we sketch the temperature diagram for a double-pipe counterflow heat exchanger. 250 ho

tg

65

wa

LMTD ¼

ter

as

es

75 20

ðDTÞ1 ðDTÞ2 ð250 65Þ ð75 20Þ ¼ ¼ 107:17 C ðDTÞ1 ð250 65Þ ln ln ð75 20Þ ðDTÞ2

320

Chapter 8 Heat exchangers

From Part (a), q ¼ 3.762 105 W q ¼ UA (LMTD) ¼ (280) A (107.17) ¼ 3.762 105 Solving for A, we have A [ 12.54 m2. It is seen that, for the same heat transfer, the double-pipe counterflow heat exchanger requires less heat transfer area than a parallel exchanger. (c) q ¼ m_ h cph ðDTÞh ¼ m_ h ð900Þð250 75Þ ¼ 3:762 105 The hot gases flow at a rate of 2.39 kg/s.

m_ h ¼ 2:39 kg=s

8.4.1.2 Nonedouble-pipe heat exchangers Not all heat exchangers are of the double-pipe type. For example, three nonedouble-pipe exchangers were shown in Fig. 8.1BeD. If a heat exchanger is not a double pipe, Eqs. (8.27) and (8.28) are still applicable. However, Eq. (8.29) has to be modified with the addition of a correction factor, F. The equation becomes q ¼ UAFðLMTDÞdpcf

(8.30)

Take special note of the subscript on the LMTD term. It stands for “double-pipe counterflow.” The correction factor F is a correction to the LMTD which we would have if the heat exchanger were of the double-pipe counterflow type rather than the nonedouble-pipe type. This is further illustrated in the examples presented below. The value of F is provided for some common types of heat exchangers in Figs. 8.9e8.11 below. In these graphs, F is a function of parameters P and R, which are P¼

to ti T i ti

(8.31)

1.0

Correction Factor F

0.9

0.8

R= 0.7

Ti − To t o − ti

4 3.5 3 2.5

2 1.8 1.6 1.4 1.2

1 0.9 0.8 0.7 0.6 0.5

0.4

0.3

0.2

0.1

0.6

0.5 0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

t −t P= o i Ti − ti

FIGURE 8.9 Log mean temperature difference (LMTD) correction factor F for one-shell-pass, multiple of two-tube-passes heat exchanger.

8.4 Analysis methods

321

1.0

Correction Factor F

0.9

0.8

R=

Ti − To t o − ti

4 3.5

3

2.5

2 1.8

1.6

1.4

1.2

1 0.9 0.8 0.7 0.6

0.5 0.4 0.3 0.2 0.1

0.7

0.6

0.5 0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

t −t P= o i Ti − ti

FIGURE 8.10 Log mean temperature difference (LMTD) correction factor F for two-shell-passes, multiple of four-tube-passes heat exchanger. 1.0

Correction Factor F

0.9

0.8

R= 0.7

Ti − To t o − ti

4 3.5 3 2.5

2 1.8 1.6 1.4 1.2 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3

0.2

0.1

0.6

0.5

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

t −t P= o i Ti − ti FIGURE 8.11 Log mean temperature difference (LMTD) correction factor F for crossflow heat exchanger: one fluid mixed and one fluid unmixed.

322

Chapter 8 Heat exchangers

Ti To (8.32) to ti In Eqs. (8.31) and (8.32), the t temperatures are for the tube-side fluid and the T temperatures are for the shell-side fluid. However, it makes no difference which fluid is assigned to the tube or to the shell. If the problem is not specific, you have the freedom to assign either of the fluids to the tube or the shell. The subscripts “i” and “o” refer to the fluid’s inlet and outlet. Functions for F for some common types of exchangers are given in Table 8.3. Graphs and functions for F were given in a pioneering 1940 paper by Bowman, Mueller, and Nagle [3]. More recent papers on the LMTD method and the correction factor are in Refs. [4e7]. Compilations of graphs and functions for F are in Refs. [8,9]. Please note that a graph and function have not been included above for crossflow exchangers with both fluids unmixed. Such exchangers are discussed in Refs. [3,5]. Problems involving a crossflow exchanger with both fluids unmixed may be treated using the ε-NTU method of Section 8.4.2. R¼

Table 8.3 Correction factor F for log mean temperature difference (LMTD) method. Parallel flow double-pipe Counterflow double-pipe One or both fluids condensing or evaporating

F¼1 F¼1 F¼1 In the below relations, Ti To i P ¼ Ttoit ti R ¼ to ti ðt ¼ tube; T ¼ shellÞ

One-shell-pass and multiple of two-tube-passes

F ¼

pﬃﬃﬃﬃﬃﬃﬃﬃﬃ 1þR2 1R

ln½ð1PRÞ=ð1PÞ pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 3 2 P 1 þ R 1 þ R2 ln4 pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 5 2 P 1 þ R þ 1 þ R2

2

pﬃﬃ 2 P " pﬃﬃﬃ# F ¼ 1P 2 P2þ 2 pﬃﬃﬃ ln 2 P2 2

Two-shell-passes and multiple of four-tube-passes

R ¼1

pﬃﬃﬃﬃﬃﬃﬃﬃﬃ 1þR2 2 F ¼ 2ð1RÞ

F ¼

Crossflow with one fluid mixed one fluid unmixed

for

pﬃﬃ 2 2

ln½ð1PRÞ=ð1PÞ pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ3 2 P 1 þ R 1 þ R2 þ 2 ð1 PÞð1 PRÞ 4 ln pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ5 2 P 1 þ R þ 1 þ R2 þ 2 ð1 PÞð1 PRÞ

P=ð1PÞ qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ3 pﬃﬃﬃ

2 62 P 2 2 þ 2 ð1 PÞ 7 qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ5 ln4 pﬃﬃﬃ

2 P 2 þ 2 þ 2 ð1 PÞ2

2

ln½ð1PRÞ=ð1PÞ ln½1þð1=RÞlnð1PRÞ P=ð1PÞ ln½1þlnð1PÞ for R

1 F ¼ 1R

F ¼

¼1

for

R ¼1

8.4 Analysis methods

323

There are a couple of special cases: First, if the heat exchanger is a double-pipe exchanger, then F ¼ 1 and the LMTD in Eq. (8.30) is the one appropriate for the type of double-pipe exchanger, i.e., parallel flow or counterflow. Also, if one or both of the fluids is condensing or evaporating, then F ¼ 1 and there is no need to use the F graphs or F equations for the specific type of heat exchanger.

Example 8.4 LMTD method with correction factor F Problem A shell-and-tube heat exchanger with one shell pass and two tube passes has water in both the shell and the tubes. The hot water enters at 90 C and is cooled to 60 C. The cold water enters at 10 C and is heated to 50 C. The rating of the heat exchanger is 80 kW and the overall heat transfer coefficient is 950 W/m2 C. (a) What is the needed surface area of the exchanger? (b) What are the flow rates of the two water streams?

Solution

This is a nonedouble-pipe exchanger so we need correction factor F. Graphs for F are based on a double-pipe counterflow exchanger with the same entering and leaving temperatures as those for the nonedouble-pipe exchanger of our problem. Therefore, we first sketch the temperature diagram for a double-pipe counterflow exchanger and put temperatures and flow directions on the diagram. 90 ho

tw

50 cold

wat

ate

r

60 er

10

(a) q ¼ UAFðLMTDÞdpcf ðDTÞ ðDTÞ2 ¼ 44:81 C ¼ ð9050Þð6010Þ From the diagram, LMTDdpcf ¼ 1 ðDTÞ1 ð90 50Þ ln ln ð60 10Þ ðDTÞ2 Factor F for this exchanger is in Fig. 8.9. Let us put the hot water in the tubes and the cold water in the shell. (Choice is arbitrary as it was not specified in the problem statement.) Ti To 6090 1050 i For Fig. 8.9, P ¼ Ttoit ti ¼ 1090 ¼ 0:38 and R ¼ to ti ¼ 6090 ¼ 1:33. For these P and R values, F ¼ 0.89. q ¼ UAFðLMTDÞdpcf ¼ 80; 000 ¼ ð950ÞðAÞð0:89Þð44:81Þ Solving for A, we get A [ 2.11 m2. (b) For the cold water, q ¼ m_ c cpc ðDTÞc ¼ 80; 000 ¼ m_ c ð4180Þð50 10Þ. Solving for m_ c , we get m_ c ¼ 0:478 kg=s. For the hot water, q ¼ m_ h cph ðDTÞh ¼ 80; 000 ¼ m_ h ð4180Þð90 60Þ. Solving for m_ h , we get m_ h ¼ 0:638 kg=s. The cold water flows at 0.478 kg/s and the hot water flows at 0.638 kg/s.

324

Chapter 8 Heat exchangers

Example 8.5 LMTD method with condensing fluid Problem It is desired to heat water using condensing steam in a crossflow, unfinned heat exchanger. The water is in the tubes and the steam flows across the tubes. The steam is at 130 C. The water flow rate is 3 kg/s and the water is heated from 20 to 90 C. The overall heat transfer coefficient is 3500 W/m2 C. What is the required surface area of the exchanger?

Solution

This is a nonedouble-pipe exchanger so the equation for the heat flow rate is q ¼ UAFðLMTDÞdpcf . Normally, we would have to use a graph or equation to find F for the crossflow heat exchanger. However, we have a fluid that is condensing. This is a special case, and F ¼ 1. The temperature diagram is given in the figure below. steam 130

130

90

r wate 20

LMTD ¼

ðDTÞ1 ðDTÞ2 ð130 20Þ ð130 90Þ ¼ ¼ 69:2 C ðDTÞ1 ð130 20Þ ln ln ð130 90Þ ðDTÞ2

q ¼ m_ c cpc ðDTÞc ¼ ð3Þð4180Þð90 20Þ ¼ 8:78 105 W q ¼ UAFðLMTDÞdpcf ¼ ð3500ÞðAÞð1Þð69:2Þ ¼ 8:78 105 Solving for A, we get that the required surface area is 3.63 m2.

Example 8.6 LMTD method with iteration Problem

A double-pipe counterflow heat exchanger uses exhaust gases (cp ¼ 1000 J/kg C) to heat 2 kg/s of water from 15 to 40 C. The gases enter the exchanger at 250 C. The overall heat transfer coefficient of the exchanger is 350 W/m2 C and the heat transfer area of the exchanger is 5 m2. What is the flow rate of the exhaust gases?

Solution The first step is to draw the temperature diagram for the exchanger. 250

ga

40

wat

se

s

er

Tho 15

We have three equations for the heat transfer rate q ¼ m_ c cpc ðDTÞc

(1)

q ¼ m_ h cph ðDTÞh

(2)

q ¼ UAðLMTDÞ

(3)

8.4 Analysis methods

325

We can get the heat transfer rate from Eq. (1): Using this in Eq. (2), we have

q ¼ m_ c cpc ðDTÞc ¼ 2ð4180Þð40 15Þ ¼ 2:09 105 W 2:09 105 ¼ m_ h ð1000Þð250 Tho Þ

The LMTD is LMTD ¼ Therefore, Eq. (3) becomes

(4)

ðDTÞ1 ðDTÞ2 ð250 40Þ ðTho 15Þ ¼ ðDTÞ1 250 40 ln ln Tho 15 ðDTÞ2 ð250 40Þ ðTho 15Þ 250 40 ln Tho 15

2:09 105 ¼ ð350Þð5Þ Simplifying Eq. (5), we get

(5)

225 Tho (6) 119:4 ¼ 210 ln Tho 15 Iteration is needed to solve Eq. (6) for the exit temperature of the exhaust gases. Once that is known, Eq. (4) can be used to obtain the desired flow rate of the gases. For the iteration, we could solve Eq. (6) by trial and error: Pick a value for Tho , evaluate the right side of Eq. (6) and see if it is equal to 119.4. If not, pick another value of Tho and repeat the process until the right side of Eq. (6) achieves 119.4. This can be very tedious. A better way is to use Excel’s Goal Seek to do the iterations and get an answer. On the Excel spreadsheet, put the initial guessed value for Tho in one cell, say cell A1. (Note: Tho has to be less than 250. We used a guess of 40. The value is not critical.) Put the right side of Eq. (6) in cell A2. That is, cell A2 will have [(225-A1)/LN(210/(A1-15)). Call up Goal Seek and tell it to make cell A2 equal to 119.4 by changing cell A1. From Goal Seek, we got that Tho ¼ 74.6 C. Using this in Eq. (4), we find that the flow rate of the exhaust gases is 1.192 kg/s.

Example 8.7 LMTD method when outlet temperatures are not given Problem Hot water is available at 75 C in a factory. It is desired to use the water in a double-pipe counterflow heat exchanger to heat cold water. The cold water enters the exchanger at 15 C and flows at a rate of 2 kg/s through the exchanger. The hot water flows at a rate of 1.5 kg/s. The overall heat transfer coefficient of the exchanger is 500 W/m2 C and the surface area of the exchanger is 8 m2. (a) What is the rating of the exchanger (kW)? (b) What are the exit temperatures of the two water streams?

Solution The first step is to draw the temperature diagram for the exchanger. 75 ho

tw

ate

r

Tco cold

wat

er

Tho 15

326

Chapter 8 Heat exchangers

We have three equations for the heat transfer rate: q ¼ m_ c cpc ðDTÞc

(1)

q ¼ m_ h cph ðDTÞh

(2)

q ¼ UAðLMTDÞ Putting the known values into these equations, we have

(3)

q ¼ 2ð4180ÞðTco 15Þ

(4)

q ¼ 1:5ð4180Þð75 Tho Þ

(5)

½ð75 Tco Þ ðTho 15Þ (6) 75 Tco ln Tho 15 We thus have three equations, Eqs. (4)e(6), to solve for the three unknowns q; Tco ; and Tho : Iteration is needed. We could solve the equations by trial and error. That is, guess Tco , use Eq. (4) to get q, put the q value into Eq. (5) to get Tho , use the Tco and Tho in Eq. (6) to calculate q, and see if that q value equals the q value of Eqs. (4) and (5). If not, pick another Tco and repeat the process until the q’s match. This is a potentially very tedious process. Fortunately, we can use software to do the iteration. One way is to use Excel’s Solver add-in program. To do this, we rearrange the equations to create functions. Let us call the functions f1 ; f2 ; and f3 : From Eq. (4), we have f1 ¼ q 2ð4180ÞðTco 15Þ ¼ 0. q ¼ ð500Þð8Þ

From Eq. (5), we have f2 ¼ q 1:5ð4180Þð75 Tho Þ ¼ 0. ¼ 0. co ÞðTho 15Þ From Eq. (6), we have f3 ¼ q ð500Þð8Þ ½ð75T 75 Tco ln Tho 15 If all three functions are zero, we have the solution to the three equations. And, making the quantity f12 þ f22 þ f32 equal to zero will assure that all three functions are indeed zero. In an Excel spreadsheet, we use three cells for the guesses for Tco ; Tho ; and q: Let us say we use cells A1, A2, and A3. (Reasonable guesses could be 40, 40, and 1e5.) In three other cells, say cells F1, F2, and F3, we put the three functions written with cell addresses. For example, cell F1 will have ¼ A3-24180(A1-15). Finally, there is a test cell, say cell F4, with ¼ F1^2 D F2^2 D F3^2 in it. We call up Solver and tell it to make cell F4 zero by changing cells A1 through A3. If there is no convergence to a solution, tell Solver to make cell F4 a minimum. The criterion that F4 is zero is often too stringent. From Solver we got the results q ¼ 1.54 105 W, Tco ¼ 33:4 C, and Tho ¼ 50:5 C. (a) The exchanger’s rating is 154 kW. (b) The cold water exits at 33.4 C and the hot water exits at 50.5 C.

8.4.2 Effectivenessenumber of transfer unit method In the last two examples, iteration was needed for problem solution. The LMTD method, although excellent for heat exchanger sizing problems, requires iteration for performance-type problems in which both the inlet and outlet temperatures are not given. One can only imagine the difficulty in solving iteration problems in the 1950s and earlier without the assistance of modern-day calculators and computers. Kays and London [10] developed the Effectiveness-Number of Transfer Units (ε-NTU) method that eliminates the need for iteration for most performance-type problems. The method is as follows: The “heat capacity rate” is Cc for the cold fluid and Ch for the hot fluid. It is defined as the product of the mass flow rate and the specific heat. For the cold fluid; Cc ¼ m_ c cpc

(8.33)

For the hot fluid; Ch ¼ m_ h cph

(8.34)

8.4 Analysis methods

327

For the two fluids in a heat exchanger, one has the higher heat capacity rate, Cmax , and the other has the lower rate, Cmin : The fluid with the higher heat capacity rate is called the “maximum fluid” and the fluid with the lower heat capacity rate is called the “minimum fluid”. The “effectiveness” of a heat exchanger is defined as ε¼

actual heat transfer q ¼ maximum heat transfer qmax

(8.35)

The actual heat transfer is given by Eqs. (8.27) and (8.28). q ¼ m_ c cpc ðTco Tci Þ ¼ m_ h cph ðThi Tho Þ

(8.36)

Using the definition of the heat capacity rate, Eq. (8.36) can be written as q ¼ Cc ðTco Tci Þ ¼ Ch ðThi Tho Þ

(8.37)

The maximum heat transfer is that where the minimum fluid undergoes the maximum possible temperature change in the exchanger. The maximum temperature in an exchanger is that of the incoming hot fluid and the minimum temperature in an exchanger is that of the incoming cold fluid. Therefore, the maximum possible temperature change in an exchanger is Thi Tci , and qmax ¼ Cmin ðThi Tci Þ

(8.38)

From Eqs. (8.35) and (8.38), we have ε¼

q q ¼ qmax Cmin ðThi Tci Þ

(8.39)

Looking at Eq. (8.39), we see that if the effectiveness ε, the heat capacity rate for the minimum fluid, and the incoming temperatures are known, then the actual heat transfer of the heat exchanger can be determined. The outlet temperatures of the fluids do not need to be known. Using Eqs. (8.37) and (8.39), the relation for the effectiveness can be written solely in terms of the fluid temperatures. If the cold fluid is the minimum fluid, then ε¼

Tco Tci Thi Tci

(8.40)

Thi Tho Thi Tci

(8.41)

If the hot fluid is the minimum fluid, then ε¼

Another important dimensionless parameter for the ε-NTU method is the “number of transfer units,” (NTU). This is defined as NTU ¼

UA Cmin

(8.42)

Looking at Eq. (8.42), we see that the NTU is directly proportional to the heat transfer area A. The larger the NTU, the larger the heat exchanger area A. The effectiveness ε for a given type of heat exchanger is a function of the ratio of heat capacity rates min Cratio ¼ CCmax and the number of transfer units NTU. Effectiveness graphs for various types of heat exchangers are in Figs. 8.12 through 8.17A and B. Effectiveness equations are in Table 8.4.

328

Chapter 8 Heat exchangers

Cratio =

1.0

0

0.9

0.1 0.2 0.3 0.4 0.5 0.6 0.8 1.0

0.8 0.7

Effectiveness ε

Cmin Cmax

0.6 0.5 0.4 0.3 0.2 0.1 0.0 0.0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

4.5

5.0

NTU FIGURE 8.12 Effectiveness ε for parallel flow.

Cratio =

0 0.2 0.4 0.6 0.8 1.0

1.0 0.9

Effectiveness ε

0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.0

0.0

0.5

1.0

1.5

2.0

2.5

NTU FIGURE 8.13 Effectiveness ε for counterflow heat exchanger.

3.0

3.5

4.0

4.5

5.0

Cmin Cmax

8.4 Analysis methods

Effectiveness ε

Cratio =

1.0

0

0.9

0.2

0.8

0.4

0.7

0.6 0.8 1.0

0.6

Cmin Cmax

0.5 0.4 0.3 0.2 0.1 0.0

0.0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

4.5

5.0

NTU FIGURE 8.14 Effectiveness ε for one-shell-pass, multiple of two-tube-passes heat exchanger.

Cratio =

0.8

0 0.2 0.4 0.6 0.8

0.7

1.0

1.0

Effectiveness ε

0.9

0.6 0.5 0.4 0.3 0.2 0.1 0.0 0.0

1.0

2.0

3.0

4.0

5.0

NTU FIGURE 8.15 Effectiveness ε for two-shell-passes, multiple of four-tube-passes heat exchanger.

Cmin Cmax

329

330

Chapter 8 Heat exchangers

Cratio =

0 0.2 0.4 0.6 0.8 1.0

1.0 0.9

Effectiveness ε

0.8 0.7

Cmin Cmax

0.6 0.5 0.4 0.3 0.2 0.1 0.0

0.0

1.0

2.0

NTU

3.0

4.0

5.0

FIGURE 8.16 Effectiveness ε for crossflow heat exchanger: both fluids unmixed.

Effectiveness ε

C

=

1.0

0

0.9

0.2

0.8

0.4 0.6 0.8 1.0

0.7 0.6

C C

0.5 0.4 0.3 0.2 0.1 0.0

0.0

1.0

2.0

3.0

4.0

5.0

NTU FIGURE 8.17A Effectiveness ε for crossflow heat exchanger: one fluid mixed and one fluid unmixed (Cmax fluid mixed, Cmin fluid unmixed).

8.4 Analysis methods

C

1.0

C C

0 0.2

0.9

Effectiveness ε

=

331

0.4

0.8

0.6

0.7

0.8

1.0

0.6 0.5 0.4 0.3 0.2 0.1 0.0 0.0

1.0

2.0

3.0

4.0

5.0

NTU FIGURE 8.17B Effectiveness ε for crossflow heat exchanger: one fluid mixed and one fluid unmixed (Cmin fluid mixed, Cmax fluid unmixed).

We now present a problem that would have required iteration if it had been solved by the LMTD method. The ε-NTU method does not require iteration.

Example 8.8 Effectivenessenumber of transfer unit method when outlet temperatures are not given Problem This is the same problem as Example 8.7. In that example, we showed that use of the LMTD method required iteration for a solution. In this example, we show how much easier it is to use the ε-NTU method when outlet temperatures are not given. Iteration is not needed for the solution. Please refer to the problem statement and temperature diagram for Example 8.7.

Solution

. . . Givens in this problem are m_ c ¼ 2 kg s m_ h ¼ 1:5 kg s cpc ¼ cph ¼ 4180 J kg C. Thi ¼ 75 C Tci ¼ 15 C U ¼ 500 W=m2 C A ¼ 8 m2 To find: q; Tco ; and Tho : We first determine the capacity rates: Ch ¼ m_ h cph ¼ 1:5ð4180Þ ¼ 6270 W=C Cc ¼ m_ c cpc ¼ 2ð4180Þ ¼ 8360 W=C It is seen that Cmin ¼ Ch ¼ 6270 and Cmas ¼ Cc ¼ 8360. Therefore, Cmin =Cmax ¼ 6270=8360 ¼ 0:75. UA ð500Þð8Þ ¼ 0:64 ¼ Cmin 6270 ¼ 0:75 and NTU ¼ 0:64; we use Fig. 8.13 and find that the effectiveness εx0:40. NTU ¼

With Cmin =Cmax

min Cratio ¼ CCmax

Double-pipe parallel flow

ratio Þ ε ¼ 1exp½NTUð1þC 1þCratio

Double-pipe counterflow

ε ¼

1exp½NTUð1Cratio Þ 1Cratio exp½NTUð1Cratio Þ

NTU ε ¼ NTUþ1

One-shell-pass and multiple of two-tube-passes

ε ¼

1þCratio þ

for

1þC ratio ð1þeG Þ=ð1eG Þ

where G ¼ NTU Two-shell-passes and multiple of four-tube-passes

Crossflow with both fluids unmixed

ε ¼

Cratio ¼ 1

2ﬃ pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 2

qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 1 þ C 2ratio

½ð1ε1 Cratio Þ=ð1ε1 Þ2 1 ½ð1ε1 Cratio Þ=ð1ε1 Þ2 Cratio

2ε1 ε ¼ 1þε for Cratio ¼ 1Where: ε1 is The effectiveness of each of the two shell passes. That is, ε1 is 1 the effectiveness given immediately above for a one-shell-pass exchanger. The NTU for the two-shellpass exchanger is twice the NTU for the one-shell-pass exchanger.

0:22 0:78 1 ε ¼ 1 exp NTU NTU exp C ratio Cratio

Crossflow with One fluid mixed One fluid unmixed

1 ð1 expf C For Cmax mixed and Cmin unmixedε ¼ Cratio ratio ½1 expð NTUÞgÞ

1 ½1 expð C For Cmin mixed and Cmax unmixedε ¼ 1 exp Cratio ratio NTUÞ

One or both fluids condensing or evaporating ðCratio [ 0Þ

ε ¼ 1 expð NTUÞ

Chapter 8 Heat exchangers

NTU ¼ CUA min

332

Table 8.4 Effectiveness ε for effectivenessenumber of transfer unit (ε-NTU) method.

8.4 Analysis methods

333

ð8360ÞðTco 15Þ c ðTco Tci Þ The effectiveness relation is ε ¼ CCmin ðThi Tci Þ ¼ ð6270Þð7515Þ ¼ 0:02222ðTco 15Þ and ε ¼ 0:4 ¼ 0:02222ðTco 15Þ: Solving for Tco, we get Tco ¼ 33.0 C. For the cold fluid, q ¼ Cc ðTco Tci Þ ¼ ð8360Þð33:0 15Þ ¼ 1:505 105 W. For the hot fluid, q ¼ Ch ðThi Tho Þ ¼ ð6270Þð75 Tho Þ ¼ 1:505 105 W. Solving for Tho ; we get Tho ¼ 51.0 C. In summary, the heat transfer rate of the exchanger is 1.505 3 105 W. The cold water leaves the exchanger at 33.0 C. The hot water leaves the exchanger at 51.0 C. These answers are almost identical to those obtained via the LMTD solution in Example 8.7. The differences are due to the minor inaccuracy of reading the graphs. It is seen that no iteration was needed for the ε-NTU solution.

We close this section with three more examples using the ε-NTU method.

Example 8.9 Effectivenessenumber of transfer unit method for determining surface area Problem

A crossflow heat exchanger uses exhaust gases (cp ¼ 1020 J/kg C) to heat water. The water flows through the tubes and the gases flow over the outside of the tubes, which are finned. The water flows at 1.5 kg/s. It enters the exchanger at 20 C and leaves at 70 C. The exhaust gases enter at 190 C and leave at 85 C. The overall heat transfer coefficient is 250 W/m2 C. What is the needed surface area?

Solution 190 ga

70

wa

se

s

85

ter

20

We will first find the flow rate of the exhaust gases, m_ h : Cc ¼ m_ c cpc ¼ ð1:5Þð4180Þ ¼ 6270 W=C q ¼ Cc ðDTÞc ¼ Ch ðDTÞh q ¼ ð6270Þð70 20Þ ¼ m_ h ð1020Þð190 85Þ Solving for m_ h , we get m_ h ¼ 2.93 kg/s and Ch ¼ m_ h cph ¼ ð2:93Þð1020Þ ¼ 2990. Cmin ¼ Ch ¼ 2990ðexhaust gasesÞ Cmax ¼ Cc ¼ 6270ðwaterÞ Cmin = Cmax ¼ 2990=6270 ¼ 0:48 As the minimum fluid is the exhaust gases, the equation for the effectiveness is Cc ðTco Tci Þ ð6270Þð70 20Þ ¼ 0:62 ¼ Cmin ðThi Tci Þ ð2990Þð190 20Þ With Cmin =Cmax ¼ 0:48 and ε ¼ 0:62; we go into Fig. 8.16 and find that NTU ¼ 1:3. 2 NTU ¼ CUA ¼ 1:3 ¼ ð250ÞA 2990 Solving for A, we get A ¼ 15.5 m . min 2 The needed surface area of the exchanger is 15.5 m . Note: As both the inlet and exit temperatures of the fluids were given, the LMTD method can also be readily used to solve this problem. The amount of work needed for both solutions is about the same. ε¼

334

Chapter 8 Heat exchangers

Example 8.10 Effectivenessenumber of transfer unit method for a steam condenser Problem A shell-and-tube heat exchanger uses condensing steam to heat water. Both the steam and the water make a single pass through the exchanger. The water flows at a rate of 4 kg/s and is heated from 25 to 70 C. The steam is condensing at 120 C. The overall heat transfer coefficient of the exchanger is 1100 W/m2 C. (a) What is the needed area of the exchanger? (b) If the flow rate of the water is reduced to 2 kg/s and everything else remains the same, what is the outlet temperature of the water?

Solution

(a) As always, we first draw the temperature diagram for the exchanger. steam 120

120 70 wate

r

25

For a condenser, Cmin =Cmax ¼ 0 and the noncondensing fluid is the minimum fluid. Cmin ¼ Cc ¼ m_ c cpc ¼ ð4Þð4180Þ ¼ 16720 To get the surface area, we need the number of transfer units NTU as NTU ¼ CUA : min

q c ðTco Tci Þ The effectiveness ε is ε ¼ qmax ¼ CCmin ðThi Tci Þ.

And, for a condenser, the relationship between ε and NTU is ε ¼ 1 eNTU . Using these three relationships, we have ε¼

ð16720Þð70 25Þ ¼ 0:4737 ¼ 1 eNTU ð16720Þð120 25Þ

eNTU ¼ 1 0:4737 ¼ 0:5263 Taking the natural log of both sides of the equation, we have lnðeNTU Þ ¼ NTU ¼ lnð0:5263Þ ¼ 0:642 and NTU ¼ 0.642 NTU ¼ Solving for A, we get A ¼ 9.76 m2. (b) With the reduction in flow rate, Cmin becomes

UA ð1100ÞA ¼ 0:642 ¼ Cmin 16720

Cmin ¼ Cc ¼ m_ c cpc ¼ ð2Þð4180Þ ¼ 8360 The number of transfer units is NTU ¼ CUA ¼ ð1100Þð9:76Þ ¼ 1:284. 8360 min The effectiveness is ε ¼ 1 eNTU ¼ 1 e1:284 ¼ 0:723.

q Cc ðTco Tci Þ ð8360ÞðTco 25Þ ¼ 0:723 ¼ ¼ qmax Cmin ðThi Tci Þ ð8360Þð120 25Þ Solving for Tco, we get. Tco ¼ 93:7 C. The outlet temperature of the water is 93.7 C. (Note: If we had solved this problem using the LMTD method, we would have needed iterations due to the log term in the LMTD.) ε¼

The final ε-NTU problem is a more complicated one. We mentioned that the ε-NTU method was especially useful if the LMTD solution for a problem required iteration. Well, here is an ε-NTU solution that also requires iteration.

8.4 Analysis methods

335

Example 8.11 Effectivenessenumber of transfer unit method involving iteration Problem Air is being heated by hot oil in a crossflow heat exchanger. The air enters at 20 C and 1 atm pressure and is heated to 40 C. The flow rate of the air is 1.3 m3/s. The oil (cp ¼ 2100 J/kg C) enters the exchanger at 120 C. The oil flows in the tubes of the exchanger and the air flows across the unfinned tubes. The overall heat transfer coefficient is 75 W/m2C and the heat transfer area is 12 m2. What is the heat transfer rate of the exchanger and the exit temperature of the oil?

Solution We first draw the temperature diagram. 120 oil

40

air

Tho 20

We then have to change the volumetric flow rate of the air to a mass flow rate. Using the ideal gas law, we have P 1:01325 105 ¼ ¼ 1:204 kg=m3 RT ð287Þð293:15Þ

m_ c ¼ rV_ ¼ 1:204 kg = m3 1:3 m3 = s ¼ 1.565 kg=s The hot fluid is the oil and the cold fluid is the air. r¼

ðairÞ Cc ¼ m_ c cpc ¼ ð1:565Þð1007Þ ¼ 1576 ðoilÞ Ch ¼ m_ h cph ¼ m_ h ð2100Þ The heat transfer rate is q ¼ Cc ðDTÞc ¼ ð1576Þð40 20Þ ¼ 3.152 104 W. As we do not have m_ h ; we do not know which fluid is the minimum fluid. We have the value of the capacity rate for the air, so let us assume the air is the minimum fluid. Then, Cmin ¼ Cc ¼ 1576. ¼ ð75Þð12Þ The number of transfer units is NTU ¼ CUA 1576 ¼ 0:571. min

q 3:15210 ¼ Cmin ðTqhi Tci Þ ¼ ð1576Þð12020Þ ¼ 0:2. The effectiveness is ε ¼ qmax 4

With these values, we go into Fig. 8.17B to get Cmin =Cmax : We find that there is no solution for our NTU and ε values. Hence, the oil, not the air, must be the minimum fluid. With the oil as the minimum fluid, we have NTU ¼

UA UA ð75Þð12Þ 0:4286 ¼ ¼ ¼ Cmin m_ h cph m_ h ð2100Þ m_ h

(1)

q q 3:152 104 0:1501 ¼ ¼ ¼ (2) qmax Cmin ðThi Tci Þ m_ h ð2100Þð120 20Þ m_ h To determine the flow rate of the oil, m_ h , we need an iterative process. We make an initial guess for m_ h and calculate NTU and ε from Eqs. (1) and (2). With these values of NTU and ε, we can go into Fig. 8.17A and determine the corresponding value of Cratio ¼ Cmin =Cmax : We can get m_ h from the value of this ratio as Cmax ¼ Cc ¼ 1576 and Cmin ¼ m_ h ð2100Þ: So, ε¼

Cmin m_ h ð2100Þ ¼ 1:332m_ h ¼ (3) 1576 Cmax We compare the calculated value for m_ h with the initial guessed value. If they agree, we have reached a solution. If not, we make another guess for m_ h and iterate again and again until the calculated value agrees with the guessed value. (Note: We are using Fig. 8.17A as the minimum fluid, the oil, is the unmixed fluid.) Cratio ¼

336

Chapter 8 Heat exchangers

Alternatively, we can use the equations in Table 8.4 to reach the solution. For a crossflow exchanger with one fluid unmixed and the other mixed, and with the unmixed fluid being the minimum fluid, the appropriate equation is

ε ¼ ð1 = Cratio Þ 1 exp Cratio 1 eNTU (4) The variables in this equation are all functions of m_ h as given in Eqs. (1)e(3). Therefore, Eq. (4) is a function of only one variable, m_ h : Eq. (4) is rearranged to

ε ð1 = Cratio Þ 1 exp Cratio 1 eNTU ¼0 (5) Eq. (5) can be solved by Matlab’s fzero function, Excel’s Goal Seek, or Excel’s Solver. The software iterates with changing values of m_ h until the left side of Eq. (5) is zero. It turns out that Eq. (5) has multiple solutions. We want the solution that hasε 1 and Cratio 1. If Excel is used, it turns out that Solver is better than Goal Seek as Solver allows constraints (for example, ε 1 and Cratio 1). Goal Seek does not have the capability of constraints. Depending on the initial guess of m_ h , the solution by Goal Seek may give a solution where ε and/or Cratio are greater than unity. Yes, it is a solution to Eq. (5), but it is not the one we are looking for. Using Solver, we got the result that m_ h ¼ 0.186 kg/s. Other parameters for the solution were Cratio ¼ Cmin =Cmax ¼ 0:248, ε ¼ 0.807, and NTU ¼ 2.3. For the oil, q ¼ m_ h cph ðDTÞh . 3:152 104 ¼ ð0:186Þð2100Þð120 Tho Þ Solving for Tho , we get Tho ¼ 39:3 C. The exit temperature of the oil is 39.3 C.

8.5 Chapter summary and final remarks In this chapter, we discussed the thermal aspects of some common types of heat exchangers. We discussed the determination of the overall heat transfer coefficient for an exchanger and the impact of the fouling of heat exchanger surfaces on the coefficient. We then outlined two methods of thermal analysis: the LMTD method and the ε-NTU method. Although either method could be used for thermal analysis, we found that the LMTD method was especially useful for design and sizing of the heat exchangers, while the ε-NTU method was especially useful for determining heat exchanger performance. There are several factors other than thermal, which must be considered in the selection of a heat exchanger for a particular purpose. Consideration must be given to the cost of the exchangereboth the capital cost and the operating costs. For example, there will be costs for cleaning of the exchanger to maintain the overall heat transfer coefficient, and electrical costs for pumping the fluids through the exchanger. An important factor is scheduling. Is the exchanger off-the-shelf or must it be customdesigned and custom-manufactured? Does its delivery schedule impact the overall schedule for the project? There are structural aspects such as the support for the exchanger and the design of the baffles to prevent or minimize vibration problems while still maintaining the desired fluid flow. The exchanger will also have physical constraints. It must fit in its allocated space in a plant. Indeed, there are many factors to be considered in the selection of a heat exchanger. Thermal design is a major factor, but there are other factors of significance.

8.6 Problems Notes: •

If needed information is not given in the problem statement, then use the Appendix for material properties and other information. If the Appendix does not have sufficient information, then use

8.6 Problems

• •

337

the Internet or other reference sources. Some good reference sources are mentioned at the end of Chapter 1. If you use the Internet, double-check the validity of the information by using more than one source. Your solutions should include a sketch of the problem. In all problems, unless otherwise stated, the fluid pressure is atmospheric. Gas properties in the Appendix are at atmospheric pressure. If a problem has a gas at other than atmospheric pressure, affected properties such as density and kinematic viscosity should be modified accordingly through use of the ideal gas law. 8-1 For a double-pipe heat exchanger: Hot water at an average temperature of 70 C flows through the tube and colder water at an average temperature of 30 C flows through the shell. The tube is a 1" Sch 10 carbon steel pipe. The convective coefficient is 5800 W/m2 C at the inner surface of the pipe and 2000 W/m2 C at the outer surface. What is the overall heat transfer coefficient of the exchanger? 8-2 After a few years of operation, the exchanger of Problem 8-1 has a fouling resistance of 0.0002 m2 C/W on its outer surface. What is the new overall heat transfer coefficient of the exchanger? 8-3 For a double-pipe heat exchanger: The tube is 1" Type M copper. The convective coefficient is 650 W/m2 C on the inside surface and 200 W/m2 C on the outside surface. What is the overall heat transfer coefficient of the exchanger? 8-4 After a period of use, the tube of Problem 8-3 has a fouling resistance of 0.002 m2 C/W on its inside surface. What is the new overall heat transfer coefficient of the exchanger? 8-5 A double-pipe heat exchanger heats water from 15 to 65 C. The heating is done by hot gases (cp ¼ 900 J/kg C) that enter the exchanger at 200 C. The flow rate of the water is 2 kg/s and the flow rate of the hot gases is 5 kg/s. The overall heat transfer coefficient is 310 W/m2 C. (a) What is the exit temperature of the hot gases? (b) What is the surface area required if the heat exchanger is parallel flow? (c) What is the surface area required if the heat exchanger is counterflow? 8-6 A double-pipe counterflow heat exchanger uses hot water to heat cold water. The cold water enters at 20 C and leaves at 50 C. The hot water enters at 80 C and leaves at 40 C. The mass flow rate of the cold water is 1.5 kg/s. If the overall heat transfer coefficient is 250 W/m2 C, (a) What is the required surface area of the exchanger? (b) What is the flow rate of the hot water (kg/s)? (c) What is the rate of heat transfer to the cold water? 8-7 A double-pipe parallel flow heat exchanger uses hot oil (cp ¼ 2400 J/kg C) to heat water. The water enters at 15 C and leaves at 80 C. The water has a flow rate of 1.5 kg/s. The oil enters at 150 C and has a flow rate of 3 kg/s. The overall heat transfer coefficient is 350 W/m2 C. (a) What is the exit temperature of the oil? (b) What is the needed surface area of the heat exchanger? 8-8 A double-pipe counterflow exchanger uses hot oil (cp ¼ 2000 J/kg C) to heat water from 30 to 80 C. The mass flow rate of the water is 0.8 kg/s. The oil flow rate is 0.9 kg/s and the oil enters the heat exchanger at 180 C. The overall heat transfer coefficient is 400 W/m2 C.

338

8-9

8-10 8-11

8-12

8-13

8-14

8-15 8-16 8-17

8-18

8-19

Chapter 8 Heat exchangers

(a) What is the exit temperature of the oil? (b) What is the required surface area of the exchanger? Water at a flow rate of 2.5 kg/s flows through the tube of a double-pipe counterflow heat exchanger. The water is heated from 20 to 70 C by condensing saturated steam at 100 C in the shell side of the exchanger. The overall heat transfer coefficient of the exchanger is 2500 W/m2 C. (a) What surface area is required to achieve the desired heating of the water? (b) What is the rate of condensation of the steam (kg/s)? Do Problem 8-9 if the exchanger were a single-shell-pass, two-tube-pass exchanger with the water in the tube and the steam in the shell. It is desired to heat water at a flow rate of 0.1 kg/s from 30 to 80 C. The heating will be done by hot oil (cp ¼ 2000 J/kg C) entering the exchanger at 190 C. The flow rate of the oil is 0.15 kg/s. The overall heat transfer coefficient is 400 W/m2 C. What heat exchanger area is needed? Hot water at a flow rate of 0.2 kg/s is needed in a factory. The water is heated from 15 to 45 C in a double-pipe counterflow heat exchanger. Heating is accomplished by hot air entering the exchanger at 250 C and flowing at a rate of 0.2 kg/s. Two heat exchangers are available in the factory: Exchanger 1 has U ¼ 450 W/m2 C and A ¼ 0.42 m2. Exchanger 2 has U ¼ 225 W/m2 C and A ¼ 0.55 m2. Which exchanger should be used? A double-pipe counterflow heat exchanger heats water from 15 to 45 C. The heating is accomplished by oil (cp ¼ 2100 J/kg C) entering the exchanger at 90 C and leaving at 60 C. The exchanger’s rating is 70 kW and its overall heat transfer coefficient is 360 W/m2 C. (a) What is the surface area of the exchanger? (b) What are the flow rates of the oil and the water? Exhaust gases (cp ¼ 1040 J/kg C) are used to preheat water in a single-pass, crossflow heat exchanger with one fluid mixed. The water flows through the tubes and the exhaust gases flow over the tubes. The water flows at 2 kg/s and is heated from 20 to 75 C. The gases are cooled from 250 to 90 C. The overall heat transfer coefficient is 220 W/m2 C. Using the LMTD method, determine the surface area of the exchanger. Do Problem 8-14 using the ε-NTU method. Condensing steam at 150 C heats oil (cp ¼ 2000 J/kg C) flowing at 2 kg/s from 15 to 80 C. The exchanger has two shell passes and four tube passes. The overall heat transfer coefficient is 2500 W/m2 C. What is the needed surface area of the exchanger? A double-pipe counterflow heat exchanger uses water to cool hot air. The water enters at a rate of 0.15 kg/s and a temperature of 25 C. The air enters at 125 C and is cooled to 75 C. The flow rate of the air is 1 kg/s. The overall heat transfer coefficient is 275 W/m2 C. (a) What is the surface area needed? (b) What is the temperature of the water leaving the exchanger? A crossflow heat exchanger with both fluids unmixed is used to heat water with engine oil. The water enters at 25 C and leaves at 70 C. The oil (cp ¼ 2300 J/kg C) enters the exchanger at 140 C. Both the water and the oil have a flow rate of 1.5 kg/s. The heat exchanger has a surface area of 50 m2. What is the overall heat transfer coefficient for the exchanger? A crossflow heat exchanger with both fluids unmixed is used to heat pressurized water with exhaust gases. The hot gases (cp ¼ 1040 J/kg C) enter the exchanger at 350 C with a flow rate

8.6 Problems

8-20

8-21

8-22

8-23 8-24

8-25

8-26

8-27

339

of 1 kg/s. The water enters at 40 C and leaves at 145 C. The water flow rate is 0.3 kg/s. The surface area of the exchanger is 4 m2. What is the overall heat transfer coefficient? A double-pipe counterflow heat exchanger has a cold fluid entering at 25 C and a hot fluid entering at 140 C. For the cold fluid, Cc ¼ 12,000 W/C, and for the hot fluid, Ch ¼ 9000 W/C. The surface area of the exchanger is 20 m2 and the overall heat transfer coefficient is 500 W/m2C. (a) What is the heat transfer rate for the exchanger? (b) What are the exit temperatures for the two fluids? For a double-pipe heat exchanger, air at an average temperature of 200 C flows through the tube and water at an average temperature of 60 C flows through the shell side. The tube is thin-walled and has a diameter of 5 cm. It is of stainless steel. The convective coefficient is 75 W/m2 C on the inside surface of the tube and 3500 W/m2 C on the outside surface. (a) What is the overall heat transfer coefficient of the exchanger? (b) After a long period of use, it is found that the overall heat transfer coefficient has decreased 20% from the value you calculated in Part (a). Assuming all the fouling is on the outside of the tube, what is the fouling factor? Water enters a heat exchanger at 40 C and has a flow rate of 4 kg/s. The water is heated by condensing steam at 1 atm pressure. The heat exchanger has an effectiveness of 80%, and its overall heat transfer coefficient is 950 W/m2 C. (a) What is the area of the exchanger? (b) What is the exit temperature of the water? A crossflow exchanger heats air from 20 to 50 C using water that enters the exchanger at 90 C and leaves at 65 C. Both fluids are unmixed. The flow rate of the water is 3 kg/s and the overall heat transfer coefficient is 75 W/m2 C. What is the surface area of the exchanger? A shell-and-tube heat exchanger has one shell pass and two tube passes. Oil (cp ¼ 2100 J/kg C) heats water. The oil enters the exchanger at 280 C and has a flow rate of 6 kg/s. The water enters the exchanger at 20 C and leaves at 85 C. The water has a flow rate of 8 kg/s. The overall heat transfer coefficient of the exchanger is 625 W/m2 C. What is the required surface area of the exchanger? A double-pipe counterflow heat exchanger heats 2.5 kg/s of water from 20 to 55 C. Heating is accomplished by oil (cp ¼ 2100 J/kg C) that enters the exchanger at 200 C and exits at 150 C. The overall heat transfer coefficient of the exchanger is 145 W/m2 C. (a) What is the flow rate of the oil? (b) If the water flow rate is reduced to 1 kg/s, what flow rate of oil is needed to keep the exit temperature of the water at 55 C? Assume that the overall heat transfer coefficient stays at 145 W/m2 C. A crossflow heat exchanger heats water from 25 to 50 C with oil (cp ¼ 2100 J/kg C) entering at 130 C. The oil flows in the tubes of the exchanger and the water flows over the outside of the tubes, which are unfinned. The flow rate of the oil is 1.5 kg/s and the water flow rate is 0.5 kg/s. The overall heat transfer coefficient of the exchanger is 225 W/m2 C. What is the required surface area to achieve the desired heating of the water? A crossflow heat exchanger heats water from 25 to 50 C with air entering at 200 C. The water flows in the tubes of the exchanger and the air flows over the outside of the tubes, which are unfinned. The flow rate of the water is 0.4 kg/s and the air flow rate is 0.4 kg/s. The overall heat

340

Chapter 8 Heat exchangers

transfer coefficient of the exchanger is 80 W/m2 C. What is the required surface area to achieve the desired heating of the water? 8-28 A crossflow heat exchanger heats air using hot water. The water is in the tubes and the air flows across the outside of the tubes. There are no fins on the tubes. The surface area of the exchanger is 4.5 m2 and the overall heat transfer coefficient is 75 W/m2 C. The water enters the exchanger at 95 C and the air enters at 20 C. Both the water and air flow rates are 0.25 kg/s. (a) At what temperatures do the fluids leave the exchanger? (b) What is the heat transfer rate for the exchanger? (Do this problem by the ε-NTU method.) 8-29 Do Problem 8-28 using the LMTD method. 8-30 A shell-and-tube counterflow heat exchanger with both fluids having a single pass cools water with ethylene glycol. The glycol has a flow rate of 1.2 kg/s and enters the exchanger at 15 C. The flow rate of the water is 0.8 kg/s and the water is cooled from 85 to 40 C. The overall heat transfer coefficient is 900 W/m2 C. (a) What is the required surface area of the exchanger to achieve the stated cooling of the water? (b) If the tubes are 3/4 inch Type L copper, and the exchanger must fit into a 30-foot space in the factory, how many tubes are needed? 8-31 A one-shell-pass, two-tube-pass heat exchanger cools water with oil (cp ¼ 2000 J/kg C). The oil has a flow rate of 1.2 kg/s and enters the exchanger at 20 C. The flow rate of the water is 0.8 kg/s, and the water enters the exchanger at 85 C. The overall heat transfer coefficient is 200 W/m2 C and the heat transfer area of the exchanger is 25 m2. (a) What are the exit temperatures of the water and the oil? (b) What is the heat transfer rate for the exchanger? 8-32 A shell-and-tube single-shell pass, two-tube pass heat exchanger is designed to heat 0.25 kg/s of water from 25 to 60 C. Heating is accomplished by hot oil (cp ¼ 2200 J/kg C) flowing through the exchanger at 0.2 kg/s. The oil enters the exchanger at 160 C. The design U of the exchanger is 320 W/m2 C. After considerable use, it is found that the water can only be heated to 50 C. What is the degraded value of U for the exchanger? 8-33 A crossflow heat exchanger uses air to cool water. The air is mixed; the water is unmixed. The air enters the exchanger at 30 C, leaves at 65 C, and has a flow rate of 3 kg/s. The water has a flow rate of 2 kg/s and enters the exchanger at 80 C. What is the UA value for the exchanger? 8-34 Water is used to heat oil. The water enters the heat exchanger at 90 C and leaves at 50 C. The oil is heated from 20 to 70 C. What is the effectiveness of the heat exchanger? 8-35 Hot water is used to heat cold water. The hot water enters at 80 C and leaves at 50 C. The cold water enters at 10 C and is heated to 50 C. What is the effectiveness of the heat exchanger?

References [1] Standards of the Tubular Exchanger Manufacturers Association, seventh ed., Tubular Exchanger Manufacturers Association, New York, 1988. [2] E.F.C. Somerscales, J.G. Knudsen, Fouling of Heat Transfer Equipment, Hemisphere Publishing, 1981.

References

341

[3] R.A. Bowman, A.C. Mueller, W.M. Nagle, Mean temperature difference in design, Trans. ASME 62 (1940) 283e294. [4] K. Gardner, J. Taborek, Mean temperature difference: a reappraisal, AIChE J. 23 (1977) 777e786. [5] A.S. Tucker, The LMTD correction factor for single-pass crossflow heat exchangers with both fluids unmixed, J. Heat Transf. 118 (1996) 488e490. [6] A. Fakheri, Log mean temperature correction factor: an alternative representation, Proc. ASME IMECE (2002) 111e115. New Orleans, 2002. [7] A. Fakheri, A general expression for the determination of the log mean temperature correction factor for shell and tube heat exchangers, J. Heat Transf. 125 (2003) 527e530. [8] L.C. Thomas, Heat Transfer e Professional Version, second ed., Capstone Publishing, 1999. [9] R.K. Shah, D.P. Sekulic, Fundamentals of Heat Exchanger Design, Wiley, 2003. [10] W. Kays, A.L. London, Compact Heat Exchangers, McGraw-Hill, 1955.

268

Chapter 7 Natural (free) convection

pumps. This chapter discusses natural or free convection in which the fluid has much gentler motion caused by density gradients in the fluid. Natural convection is an important topic. It has many applications, such as radiators for heating a room, refrigeration coils, multiglazed windows, transmission lines, electric transformers, immersion heaters, and cooling of electronic devices. As flow velocities are small, in many cases natural convection may be insignificant. However, in other situations, it may be the only significant mechanism of heat transfer. Our main objective, both in Chapter 6 and this chapter, is to determine the convective coefficient, h, for a variety of heat transfer situations and geometries. Much experimentation has been performed in the field of convective heat transfer. The results of these investigations have been correlated by researchers using dimensionless numbers. We will first discuss the dimensionless numbers associated with natural convection. Following that, we will present correlations for a variety of geometries: vertical plates, horizontal plates, inclined plates, cylinders, and spheres. We also will discuss natural convection in enclosed spaces such as rectangular spaces relevant to double-glazed windows and solar collectors, annular spaces between concentric cylinders, and spaces between concentric spheres.

7.2 Basic considerations As discussed in Chapter 6, there are three major dimensionless numbers associated with forced convectiondthe Nusselt, Reynolds, and Prandtl numbers. The convective coefficient h is found in the Nusselt number, Nu. hL k and the flow velocity is in the Reynolds number, Re. Nu ¼

Re ¼

rVL VL ¼ m y

(7.1)

(7.2)

Another relevant dimensionless number is the Prandtl number, Pr, which contains fluid properties. cp m Pr ¼ (7.3) k In these numbers, h ¼ convective coefficient. L ¼ a characteristic length for the problem. k ¼ thermal conductivity of the fluid. V ¼ fluid velocity. r ¼ density of the fluid. m ¼ absolute viscosity of the fluid. y ¼ kinematic viscosity of the fluid. cp ¼ specific heat at constant pressure of the fluid. For forced convection, correlation of experimental data gives the Nusselt number as a function of the Reynolds and Prandtl numbers. That is, Nu ¼ fðRe; PrÞ

(7.4)

After the Nusselt number has been determined, the convective coefficient can be obtained from it.

7.2 Basic considerations

269

k Nu (7.5) L For natural convection, fluid flow is caused by density gradients in the fluid. The flow velocity is significantly less than for forced convection. Therefore, the Reynolds number is no longer relevant. It is replaced with the Grashof number, GrL , which contains the acceleration of gravity, g, and the volumetric coefficient of thermal expansion, b. h ¼

g b ðTs TN Þ L3 (7.6) y2 Ts and TN are the temperatures of the surface and fluid respectively. The volumetric coefficient of Gr ¼

expansion is defined as b ¼ V1

vV vT

p

. For ideal gases, b ¼ T1 , where T is the absolute temperature of

the gas. For nonideal gases and liquids, b is obtained from property tables. An Important Note: The Grashof number is positive. It contains the term ðTs TN Þ This probably should have been written jTs TN j. However, carrying around the absolute value function is unwieldy. So, just remember that Gr is positive and when the fluid temperature is higher than the surface temperature, make the term ðTN Ts Þ rather than ðTs TN Þ For natural convection, experimental data are correlated using the Grashof, Prandtl, and Nusselt numbers: hL ¼ fðGr; PrÞ (7.7) k Once the Nusselt number for a problem is found, the convective coefficient may be obtained from Eq. (7.5). Another dimensionless number found in natural convection correlations is the Rayleigh Number, Ra. This is the product of the Grashof and Prandtl numbers. Nu ¼

Ra ¼ Gr Pr ¼

gbðTs TN ÞL3 gbðTs TN ÞL3 Pr ¼ 2 y ya

(7.8)

where a ¼ thermal diffusivity of the fluid ¼ rck p : Like forced convection, hydrodynamic and thermal boundary layers also form on surfaces associated with natural convection. Figs. 7.1 and 7.2 show the hydrodynamic boundary layer for a vertical surface. Fig. 7.1 is for a surface that is hotter than the fluid, and Fig. 7.2 is for a surface that is colder than the fluid. In both cases, the boundary layer starts at the leading edge x ¼ 0. The boundary layer is first laminar, and then, if the plate is long enough, the boundary layer turns turbulent. For natural convection on a vertical surface, the transition from laminar to turbulent occurs at critical distance xc where the local Rayleigh number is about 109. g b ðTs TN Þ x3c z 109 (7.9) ya Both figures show the velocity distribution in the laminar boundary layer. The velocity component in the x direction is u. It is seen that u ¼ 0 at both the surface and the edge of the boundary layer. Raxc ðvertical surfaceÞ ¼

270

Chapter 7 Natural (free) convection

Turbulent

u Surface at Ts (Ts > T∞)

Fluid at T∞

x

ḡ

Laminar

y Leading Edge

FIGURE 7.1 Hydrodynamic boundary layer on a vertical heated surface.

Leading Edge y ḡ

Laminar x Fluid at T∞ u

Surface at Ts (Ts < T∞)

FIGURE 7.2 Hydrodynamic boundary layer on a vertical cooled surface.

Turbulent

7.3 Natural convection for flat plates

271

Many experimental studies have been performed to determine the convective heat transfer for different geometries. Some correlation equations from these studies are given below. We include correlations for flat plates, cylinders, spheres, and enclosed spaces.

7.3 Natural convection for flat plates The following sections include correlations for vertical, horizontal, and inclined flat plates.

7.3.1 Vertical plate For a vertical plate, the characteristic length, L, is the height of the plate.

7.3.1.1 Constant temperature surface For a constant temperature surface, Churchill and Chu [1] recommend the following correlation equation if there is only a laminar region on the plate, i.e., if RaL < 109 : 1=4

Nu ¼

0:670 RaL hL ¼ 0.68 þ h i4=9 k 1 þ ð0:492=PrÞ9=16

(7.10)

For a constant temperature surface, Churchill and Chu [1] recommend the following correlation equation if there are both laminar and turbulent regions on the plate, i.e., if RaL > 109 : ( )2 1=6 0:387 RaL hL ¼ 0.825 þ h Nu ¼ (7.11) i8=27 k 1 þ ð0:492=PrÞ9=16 Simpler equations in common use are [2e4]. hL 1=4 ¼ 0:59 RaL for 104 < RaL < 109 (7.12) k hL 1=3 ¼ 0:10 RaL for 109 < RaL < 1013 (7.13) and Nu ¼ k For Eqs. (7.10) through (7.13), fluid properties should be evaluated at the film temperature Tf ¼ ðTs þTN Þ=2: Nu ¼

Example 7.1 Vertical plate with constant surface temperature Problem A vertical plate is in 25 C water. It is 10 cm wide and 20 cm high. A surface of the plate has a constant temperature of 75 C. What is the rate of heat transfer from the surface?

Solution The characteristic length L is the height of the plate, 20 cm. From Eq. (7.8), the Rayleigh number based on L is gbðTs TN ÞL3 Pr y2 Fluid properties are taken at the film temperature, Tf ¼ ðTs þTN Þ=2 ¼ ð75 þ25Þ=2 ¼ 50 C For water at 50 C, k ¼ 0:644 W m C; y ¼ 5:536 107 m2 s; Pr ¼ 3.55; b ¼ 0:451 103 =K Also, the acceleration of gravity g ¼ 9.807 m/s2. RaL ¼ GrL Pr ¼

272

Chapter 7 Natural (free) convection

The Rayleigh number is therefore ð9:807Þ 0:451 103 ð75 25Þð0:2Þ3

ð3:55Þ ¼ 2:049 1010 ð5:536 107 Þ2 Eq. (7.11) is an appropriate equation for convective coefficient h. )2 ( 1=6 0:387 RaL k 0.825 þ h h¼ i8=27 L 1 þ ð0:492=PrÞ9=16 Putting values in this equation, we have 1=6 )2 ( 0:387 2:049 1010 0:644 0.825 þ h ¼ 1216 W=m2 C h¼ i8=27 0:2 9=16 1 þ ð0:492=3:55Þ The rate of heat transfer from the surface is RaL ¼

q ¼ hAðTs TN Þ ¼ ð1216Þ½ð0:1Þð0:2Þð75 25Þ ¼ 1216 W It is interesting (and somewhat upsetting) to note that the correlation of Eq. (7.13) gives a result of 881 W, which is 28% different from the 1216 W value we calculated using Eq. (7.11). This gives emphasis to the fact that results from existing correlations are only approximate. Accurate results can often only be obtained through experimentation on the actual surfaces in question.

Example 7.2 Vertical plate with constant heat flux Problem A thin vertical plate is 1.2 m high and 0.5 m wide. The left side of the plate is insulated and the right side absorbs a constant heat flux of 200 W/m2. There is convection from the right side of the plate to the adjacent 20 C air. Assume that radiation from the plate is negligible. (a) What is the temperature at the midheight of the plate, i.e., at x ¼ L=2 ¼ 0:6 m ? (b) What is the temperature at the top of the plate, i.e., at x ¼ L ¼ 1:2 m ?

Solution

(a) To start the solution, we have to make an educated guess of the midheight surface temperature. Convective coefficient values for natural convection in air are small, say on the order of magnitude of 5 W/m2 C. Using Eq. (7.16), 0 we have Ts ¼ TN þ qhsx ¼ 20 þ 200 5 ¼ 60 C. The film temperature is then Tf ¼ ðTs þTN Þ=2 ¼ ð60 þ20Þ=2 ¼ 40 C. Air properties at 40 C are k ¼ 0:027 W m C; y ¼ 1:703 105 m2 s; Pr ¼ 0.710 We will use Eq. (7.17) for the convective coefficient at x ¼ L=2 ¼ 0:6 m. 8 9 1=4 < = 0:670 RaL=2 k 0.68 þ h (7.17) hx ¼ hL=2 ¼ i 4=9 ; : L=2 1 þ ð0:437=PrÞ9=16 gbðTs TN ÞðL=2Þ3 Pr y2 where g ¼ 9.807 m/s2; b ¼ 1=Tf ¼ 1=ð40 þ273:15Þ ¼ 0:003193 = K; and L ¼ 1.2 m. Putting values into Eq. (7.20), we have Rax ¼ RaL=2 ¼ GrL=2 Pr ¼

RaL=2 ¼

ð9:807Þð0:003193Þð60 20Þð0:6Þ3 2

ð1:703 105 Þ

ð0:710Þ ¼ 6:623 108

Putting values into Eq. (7.17), we have 1=4 ) ( 0:670 6:623 108 0:027 2 0.68 þ h ¼ i4=9 ¼ 3:80 W=m C 0:6 1 þ ð0:437=0:710Þ9=16

hL=2

(7.20)

7.3 Natural convection for flat plates

273

Using Eq. (7.16), we calculate the surface temperature at x ¼ L=2 ¼ 0:6 m as q0s 200 ¼ 72:6 C ¼ 20 þ 3:80 hL=2 This answer is quite different from our initial guess of 60 C, so another iteration is appropriate. Let us do the calculations again with Ts ¼ 70 C.The film temperature is then Tf ¼ ðTs þTN Þ=2 ¼ ð70 þ20Þ=2 ¼ 45 C. The properties of air at 45 C are k ¼ 0:0274 W m C; y ¼ 1:750 105 m2 s; Pr ¼ 0.710; b ¼ 0.003143 K We do the calculations as before, and get RaL=2 ¼ 7:718 108 and hL=2 ¼ 4:00 W m2 C. The surface temperature at the midheight is Ts ðxÞ ¼ Ts ðL = 2Þ ¼ TN þ

qs 0 200 ¼ 70:0 C ¼ 20 þ 4:00 hL=2 Our starting surface temperature was 70 C and we calculated 70.0 C. No further iteration is necessary. The surface temperature at the midheight of the plate is 70.0 C. (b) To get the temperature at the top of the plate, we need hx at x ¼ L: From Eq. (7.17), ) ( 1=4 0:670 RaL k 0.68 þ h hL ¼ (7.21) i4=9 L 1 þ ð0:437=PrÞ9=16 Ts ðL = 2Þ ¼ TN þ

gbðTs TN ÞðLÞ3 Pr (7.22) y2 ¼ ð70 þ20Þ=2 ¼ 45 C. The air properties at 45 C are given below: k ¼ The film temperature is Tf ¼ ðTs þTN Þ=2 0:0274 W m C; y ¼ 1:750 105 m2 s; Pr ¼ 0.710. Putting values into Eq. (7.22), we have RaL ¼ GrL Pr ¼

RaL ¼

ð9:807Þð1=ð45 þ 273:15ÞÞð70 20Þð1:2Þ3 2

ð1:750 105 Þ

ð0:710Þ ¼ 6:174 109

Putting values into Eq. (7.21), we have 1=4 ) ( 0:670 6:174 109 0:0274 2 0.68 þ h hL ¼ i4=9 ¼ 3:35 W=m C 1:2 9=16 1 þ ð0:437=0:710Þ Finally; Ts ðLÞ ¼ TN þ

qs 0 200 ¼ 79:7 C ¼ 20 þ 3:35 hL

The surface temperature at the top of the plate is 79.7 C. Note: We used the Churchill and Chu correlation equation, Eq. (7.17), for this solution. Alternatively, we could have used the Fujii and Fujii equation, Eq. (7.18). If the latter equation had been used, the results for the midheight and top surface temperatures would have been 77.5 and 86.0 C, respectively. The Churchill and Chu correlation gives temperatures about 7 C lower than the Fujii and Fujii correlation. Without experimentation on the actual plate, it is not possible to make a judgment of which correlation is more accurate.

7.3.1.2 Constant heat flux surface For a constant heat flux surface, we are interested in determining the temperature distribution of the plate’s surface, that is, we want to find Ts ðxÞ: For convective heat transfer, we have the basic equation q ¼ hAðTs TN Þ: With a constant heat flux q0s , this equation becomes q qs 0 ¼ ¼ hðTs TN Þ (7.14) A Written for location x, Eq. (7.14) is qs 0 ¼ hx ½Ts ðxÞ TN Rearranging Eq. (7.15), we get an expression for the surface temperature distribution: qs 0 Ts ðxÞ ¼ TN þ hx

(7.15)

(7.16)

274

Chapter 7 Natural (free) convection

To determine the surface temperature distribution from Eq. (7.16), we need an equation for local convective coefficient hx . For the laminar region, Rax T∞ Fluid at T∞

Ts Ts Hot Surface Facing Downwards

Fluid at T∞

FIGURE 7.3 Natural convection from horizontal hot surfaces.

Hot Surface Facing Upwards

7.3 Natural convection for flat plates

275

The upper surface of the plate shown in Fig. 7.3 is called a “hot surface facing upward,” and the lower surface is called a “hot surface facing downward.” Fluid motion at the two surfaces is quite different. At the upper surface, the heating of the fluid causes a decrease in the density of the fluid adjacent to the plate. This results in the rising of a heated plume through the colder fluid above it. If the temperature difference between the plate surface and the fluid above the plate is great enough, pockets of heated fluid will be created. These pockets move through the colder fluid above them, causing considerable movement and fluid mixing. At the lower surface, the heating of the fluid also causes a decrease in density of the adjacent fluid. The fluid wants to rise, but the plate is in the way. Hence, there is less motion at the lower surface than at the upper surface. This difference in fluid motion results in a difference in the convective coefficients for the two surfaces. The rate of heat transfer at the upper surface is greater than that at the lower surface. For cold surfaces (i.e., for Ts < TN ), the heat transfer of a cold surface facing downward is the same as that of a hot surface facing upward. And, the heat transfer of a cold surface facing upward is the same as that of a hot surface facing downward. For a horizontal surface, the characteristic length is L¼

A P

(7.23)

where A is the area of the surface, and P is the perimeter of the surface. a2 ¼ 1 a: For a rectangular area of sides “a” and For example, for a square area of side “a,” L ¼ 4a 4 “b,” L ¼ 2

a b : ðaþbÞ

For a circular area of diameter “D,” L ¼ pDpD=4 ¼ 14 D: 2

7.3.2.1 Constant temperature surface Much experimentation has been performed for natural convection from horizontal plates [2,7,8]. Correlation equations are: For a hot surface facing upward or a cold surface facing downward, hL 1=4 ¼ 0:54 RaL for 2 104 < RaL < 8 106 k hL 1=3 ¼ 0:15 RaL for 8 106 < RaL < 2 109 Nu ¼ k For a hot surface facing downward or a cold surface facing upward, Nu ¼

hL 1=4 ¼ 0:27 RaL for 105 < RaL < 1010 k Fluid properties are taken at the film temperature Tf ¼ ðTs þTN Þ=2: Nu ¼

(7.24) (7.25)

(7.26)

7.3.2.2 Constant heat flux surface Fujii and Imura [7] recommend the following correlations: For a hot surface facing upward, Nu ¼

hL 1=3 ¼ 0:16 RaL k

for RaL < 2 108

(7.27)

276

Chapter 7 Natural (free) convection

hL 1=3 ¼ 0:13RaL k For a hot surface facing downward, Nu ¼

for

5 108 < RaL < 1011

(7.28)

hL 1=5 ¼ 0:58 RaL for 106 < RaL < 1011 (7.29) k In Eqs. (7.27)e(7.29), fluid properties, with the exception of b, are taken at the temperature Ts 0:25 ðTs TN Þ: b is evaluated at temperature TN þ 0:25 ðTs TN Þ: Nu ¼

Example 7.3 Heat flow from a heating duct Problem A horizontal heating duct is 16 inches wide and 8 inches high. The outer surface of the duct is at 110 F and the room air is at 70 F. What is the rate of convective heat transfer to the room air per foot length of duct?

Solution We can either work the problem using English units or SI units. Let us do the latter. Duct width ¼ 16 in x (2.54 cm/in) ¼ 40.64 cm. Duct height ¼ 8 inch x (2.54 cm/in) ¼ 20.32 cm Ts ¼ 110 F ¼ 5=9 ð110 32Þ ¼ 43.3 C TN ¼ 70 F ¼ 5=9 ð70 32Þ ¼ 21:1 C þ21:1Þ=2 ¼ 32:2 C. Fluid properties are taken at the film temperature: Tf ¼ ðTs þTN Þ=2 ¼ ð43:3 At this temperature, for air, k ¼ 0:0264 W m C; y ¼ 1:62 105 m2 s; Pr ¼ 0.711 Also, g ¼ 9.807 m/s2 and b ¼ 1=Tf ¼ 1=ð32:2 þ273:15Þ ¼ 0:003275 = K For the upper and lower surfaces,. A ð0:4064Þ ðlengthÞ 0:4064 z 0:2032 m ¼ ¼ 0:4064 P ½2 ð0:4064Þ þ 2ðlengthÞ þ 1 2 length We have reasonably assumed that the duct’s length is much greater than its width. L¼

RaL ¼ GrL Pr ¼

gbðTs TN ÞðLÞ3 ð9:807Þð0:003275Þð43:3 21:1Þð0:2032Þ3 Pr ¼ ð0:711Þ 2 2 y ð1:62 105 Þ

RaL ¼ 1:621 107 The upper surface of the duct is a hot surface facing upward. Using Eq. (7.25), we have 1=3 hL 1=3 ¼ 0:15 RaL ¼ 0:15 1:621 107 ¼ 37:96 k k 0:0264 ð37:96Þ ¼ 4:93 W = m2 C Nu ¼ h¼ L 0:2032

Nu ¼

The heat flow per meter length is q = length ¼ hðwidthÞðTs TN Þ ¼ 4:93ð0:4064Þð43:3 21:1Þ ¼ 44:48 W=m length The lower surface of the duct is a hot surface facing downward. Using Eq. (7.26), we have 1=4 hL 1=4 ¼ 0:27 RaL ¼ 0:27 1:621 107 ¼ 17:13 k k 0:0264 Nu ¼ ð17:13Þ ¼ 2:23 W=m2 C h¼ L 0:2032

Nu ¼

7.3 Natural convection for flat plates

277

The heat flow per meter length is q = length ¼ hðwidthÞðTs TN Þ ¼ 2:23ð0:4064Þð43:3 21:1Þ ¼ 20:12 W=m length For the sides of the duct, L ¼ height of duct ¼ 0.2032 m This characteristic length happens to be the same as that for the upper and lower surfaces, so we can use the result of the RaL calculation above. RaL ¼ 1:621 107 The sides are vertical and RaL < 109 . Therefore we use Eq. (7.10) for Nu and h. 1=4 1=4 0:670 1:621 107 0:670 RaL hL Nu ¼ ¼ 0.68 þ h ¼ 0.68 þ i4=9 h i4=9 ¼ 33:32 k 1 þ ð0:492=PrÞ9=16 1 þ ð0:492=0:711Þ9=16 k 0:0264 Nu ¼ ð33:32Þ ¼ 4:33 W=m2 C h¼ L 0:2032 The heat flow for each side per meter length is q = length ¼ hðheightÞðTs TN Þ ¼ 4:33ð0:2032Þð43:3 21:1Þ ¼ 19:53 W=m length Total heat flow from the duct to the air is 44.48 þ 20.12 þ 2 (19.53) ¼ 103.7 W/m length. The problem was posed in English units so we give the result as 103:7 W=m

1m 3:412 Btu = h ¼ 107.8 Btu=h per foot length 3:2808 ft W

7.3.3 Inclined plate Fig. 7.4 shows two orientations of an inclined flat plate. In (A), the hot surface is facing downwards and the cold surface upwards. For this case, the angle q from the vertical is positive. In (B), the hot surface is upwards and the cold surface is downwards. The angle q is negative. The length of the plate is L. Surface at Ts Fluid at T∞

cold

hot θ=–

θ=+

L

(A) FIGURE 7.4 Inclined plate.

hot

cold

L

(B)

278

Chapter 7 Natural (free) convection

Fujii and Imura [7] recommend the following correlations: For the hot surface facing downward or cold surface facing upward, Nu ¼ for

hL ¼ 0:56 ðRaL cos qÞ1=4 k

105 < RaL cos q < 1011

and

0 < q < 88

For the hot surface facing upward or cold surface facing downward, h i hL ¼ 0:145 ðGrL PrÞ1=3 ðGrc PrÞ1=3 þ 0:56ðGrL Pr cos qÞ1=4 Nu ¼ k for

105 < GrL Pr cos q < 1011

and

(7.30)

(7.31)

75 < q < 15

If GrL < Grc , then the first term in Eq. (7.31) is deleted. In Eq. (7.31), the subscript “c” relates to where Nu starts to separate from the characteristic of the laminar region. The values of Grc Pr depend on the angle of inclination q: For angles of 15, 30, 45, 60, and 75 , the Grc Pr values are, respectively, about 5 109, 2 109, 109, 108, and 106. In Eqs. (7.30) and (7.31), fluid properties, with the exception of b, are taken at the temperature Ts 0:25 ðTs TN Þ: b is evaluated at temperature TN þ 0:25 ðTs TN Þ:

7.4 Natural convection for cylinders The following sections include correlations for horizontal and vertical cylinders. These geometries have wide applications such as steam heating pipes for buildings, refrigeration lines, pin fins, and immersion heaters.

7.4.1 Horizontal cylinder Fig. 7.5 shows natural convection from a horizontal cylinder of diameter D. The surface temperature of the cylinder is Ts , and the adjacent fluid is at TN : The gravity vector is downwards. Much experimentation has been performed for natural convection from cylinders. We provide two correlations from Churchill and Chu [9] for the average Nusselt Numbers at the surface. The correlations give a good fit to the experimental data over a wide range of Rayleigh Numbers. We also

ḡ Fluid at T∞ D Ts

FIGURE 7.5 Natural convection from a horizontal cylinder.

7.4 Natural convection for cylinders

279

include a simpler correlation equation from Morgan [10]. The characteristic length for the correlations is D, the diameter of the cylinder. Eqs. (7.32) and (7.33) are from Churchill and Chu. Nu ¼

h D ¼ k

1=4

0.36 þ h

0:518 RaD

1 þ ð0:559=PrÞ

( h D ¼ Nu ¼ k

9=16

i4=9

0.60 þ h

1 þ ð0:559=PrÞ

9=16

106

1708: For lower values of RaL , the buoyant forces do not overwhelm the viscous forces in the fluid. McAdams [2] recommends the following equations for air with the hot plate on the bottom: hL 1=4 ¼ 0:21RaL for 104 < RaL < 3.2 105 (7.48) k hL 1=3 ¼ 0:075RaL for 3.2 105 < RaL < 107 (7.49) Nu ¼ k Hollands, Unny, Raithby, and Konicek [21] proposed the following equation for air in a horizontal enclosure with a large aspect ratio (H=L > 12) with the hot plate on the bottom: " # hL 1708 RaL 1=3 ¼ 1 þ 1:44 1 Nu ¼ þ 1 for RaL < 108 (7.50) k RaL 5830 Nu ¼

Important Note: For the terms marked ½ in Eq. (7.50): If the value inside the bracket is negative, the term should be set to zero. Globe and Dropkin [22] proposed the following equation for liquids (0.02 < Pr < 8750) in a horizontal enclosure with the hot plate on the bottom: Nu ¼

hL 1=3 ¼ 0:069RaL Pr0:074 k

for

1.5 105 < RaL < 6.8 108

(7.51)

7.7.1.2 Vertical rectangular enclosure McAdams [2] recommends the following equations for a vertical enclosed air space: 1=9 hL H 1=4 ¼ 0:20 Nu ¼ RaL for 2 104 < RaL < 2.1 105 k L 1=9 hL H 1=3 ¼ 0:071 Nu ¼ RaL for 2.1 105 < RaL < 1.1 107 k L

(7.52)

(7.53)

ElSherbiny, Raithby, and Hollands [23] recommend the following equations for a vertical enclosed air space for different aspect ratios (H/L). Aspect Ratio ðH = LÞ ¼ 5 and RaL < 108 82 9 !3 31=3 1=4 < = 0:193 RaL h L 5 ; 0:0605 Ra1=3 ¼ 41 þ Nu ¼ L ; : k 1 þ ð1800=RaL Þ1:289

(7.54) max

7.7 Natural convection for enclosed spaces

289

There are two expressions inside the parentheses. Use the larger value. Aspect Ratio ðH = LÞ ¼ 10 and RaL < 9:7 106 nh o 9 i1=9 h L 1=3 ¼ 1 þ 0:125 Ra0:28 Nu ¼ ; 0:061 Ra L L max k There are two expressions inside the parentheses. Use the larger value. Aspect Ratio ðH = LÞ ¼ 20 and RaL < 2 106

1=6:5 h L 1=3 6:5 ¼ 1 þ 0:064 RaL Nu ¼ k Aspect Ratio ðH = LÞ ¼ 40 and RaL < 2 105 h 11 i1=11 h L ¼ 1 þ 0:0303 Ra0:402 Nu ¼ L k Aspect Ratio ðH = LÞ ¼ 80 and RaL < 3 104 h 18 i1=18 h L ¼ 1 þ 0:0227 Ra0:438 Nu ¼ L k Aspect Ratio ðH = LÞ ¼ 110 and RaL < 1:2 104

1=18 h L 1=3 18 ¼ 1 þ 0:0607 RaL Nu ¼ k

(7.55)

(7.56)

(7.57)

(7.58)

(7.59)

The above correlations are for vertical spaces with air. The following two correlations for constant heat flux on the hot surface are recommended by MacGregor and Emery [24] for fluids with higher Prandtl numbers than air, e.g., liquids. 0:3 hL H 1=4 ¼ 0:42 RaL Pr0:012 Nu ¼ (7.60) k L 104 < RaL < 107 Eq. (7.60) is for 10 < H L < 40 1 < Pr < 2 104

Eq. (7.61) is for 1 < H L < 40

hL 1=3 ¼ 0:46 RaL k 1 < Pr < 20 106 < RaL < 109

Nu ¼

(7.61)

7.7.1.3 Inclined rectangular enclosure Based on their experiments, Hollands, Unny, Raithby, and Konicek [21] proposed the following equation for air in an inclined rectangular enclosure of large aspect ratio (H=L > 12) with the hot plate

290

Chapter 7 Natural (free) convection

on the bottom. It is valid for angles q 60 and should give good results for angles as large as 75 degrees. ! hL 1708 1708ðsin 1:8 qÞ1:6 ¼ 1 þ 1:44 1 Nu ¼ 1 k RaL cos q RaL cos q # " 1=3 RaL cos q 1 (7.62) þ 5830 for RaL < 105 Important Note: For the terms marked ½ in Eq. (7.62): If the value inside the bracket is negative, the term should be set to zero.

Example 7.6 Solar collector Problem The flat plate solar collector shown in Fig. 7.9 consists of a glass cover and an absorber plate. The collector is 2 m high and 1.5 m wide and is tilted at an angle of 60 degrees from the horizontal. The air space between the cover and the absorber plate is 3 cm thick. The absorber plate temperature is 75 C and the cover temperature is 35 C. What is the heat transfer rate by natural convection from the absorber plate?

Solution We will use Eq. (7.62) for the solution. ! " # hL 1708 1708ðsin 1:8 qÞ1:6 RaL cos q 1=3 ¼ 1 þ 1:44 1 1 þ 1 k RaL cos q 5830 RaL cos q The average temperature of the cover and absorber plate is (35 þ 75)/2 ¼ 55 C. Air at 55 C has the following properties: k ¼ 0:028 W mC; y ¼ 1:84 105 m2 s; Pr ¼ 0.708. The Rayleigh number based on the air gap L is Nu ¼

35 C

Solar

75 C air space

insulation

cover

60°

absorber plate

FIGURE 7.9 Solar collector for Example 7.6.

(7.62)

7.7 Natural convection for enclosed spaces

291

gbðTs TN ÞðLÞ3 ð9:807Þ½1=ð55 þ 273:15Þð75 35Þð0:03Þ3 Pr ¼ ð0:708Þ ¼ 6:750 104 2 y2 ð1:84 105 Þ Looking at Eq. (7.62), we need values of RaL cos 60 ¼ 6:750 104 ð0:5Þ ¼ 3:375 104 sinð1:8qÞ ¼ sin½ð1:8Þð60Þ ¼ sin 108 ¼ 0:951 Putting the values into Eq. (7.62), we have # ! " 1=3 hL 1708 1708ð0:951Þ1:6 3.375 104 1 ¼ 3.0989 Nu ¼ ¼ 1 þ 1:44 1 1 þ k 3.375 104 3.375 104 5830 RaL ¼

h¼ The natural convection heat transfer is

Nu k ð3:0989Þð0:028Þ ¼ ¼ 2:892 W=m2 C L 0:03

qconv ¼ h A ðT1 T2 Þ ¼ ð2:892Þ ½ð2Þð1:5Þ ð75 35Þ ¼ 347 W

7.7.2 Annular space between concentric cylinders Concentric cylinders are shown in Fig. 7.10. The inner cylinder has an outer radius r1 , and the outer cylinder has an inner radius r2 : The surfaces of the cylinders are at temperatures T1 and T2 ; respectively. The annular space between the cylinders is filled with a gas or a liquid. The cylinders are horizontal and have a length L. If the space between the cylinders contained a solid, we showed in an earlier chapter that the conduction through a cylindrical shell is q¼

2pkL ðT1 T2 Þ ln ðr2 =r1 Þ

(7.63)

With gas or liquid in the annular space, we have motion and convection rather than conduction. In Eq. (7.63), the thermal conductivity k of the motionless fluid is replaced by an effective conductivity keff for the fluid in motion.

qconv fluid r1

T1

T2

r2 L

FIGURE 7.10 Concentric cylinders.

292

Chapter 7 Natural (free) convection

Raithby and Hollands [26] recommend the following equation for the effective conductivity for use with Eq. (7.63): 1=4 keff Pr 1=4 ¼ 0:386 Racyl (7.64) 0:861 þ Pr k 2 ½ln ðr2 =r1 Þ4=3 where : The length to use in Racyl is Lcyl ¼

3=5 3=5 5=3 þ r2 r1

(7.65)

Eq. (7.64) should give good results for 0:7 Pr 6000 and Racyl 107 : As keff has to be greater than k: If keff =k calculated by Eq. (7.64) is less than unity, keff should be made equal to k. Fluid properties are taken at the average temperature of the surfaces, i.e., ðT1 þT2 Þ=2:

7.7.3 Space between concentric spheres Concentric spheres are shown in Fig. 7.11. The inner sphere has an outer radius r1 and the outer sphere has an inner radius r2 : The surfaces of the spheres are at temperatures T1 and T2 ; respectively. The annular space between the spheres is filled with a gas or a liquid. If the space between the spheres had been solid, we showed in Chapter 3 that the conduction through a spherical shell is q¼

4pk ðT1 T2 Þ ð1=r1 Þ ð1=r2 Þ

(7.66)

With gas or liquid in the annular space, we have motion and convection rather than conduction. In Eq. (7.66), the thermal conductivity k of the motionless fluid is replaced by an effective conductivity keff for the fluid in motion. Raithby and Hollands [26] recommend the following equation for the effective conductivity for use with Eq. (7.66): 1=4 keff Pr 1=4 ¼ 0:74 Rasph (7.67) 0:861 þ Pr k

qconv T1

fluid r1

T2

r2

FIGURE 7.11 Concentric spheres.

7.7 Natural convection for enclosed spaces

where : The length to use in Rasph is Lsph ¼

½ð1=r1 Þ ð1=r2 Þ4=3

7=5 7=5 5=3 21=3 r1 þ r2

293

(7.68)

Eq. (7.67) should give good results for 0:7 Pr 4000 and Rasph 104 : As keff has to be greater than k: If keff =k calculated by Eq. (7.67) is less than unity, keff should be made equal to k. Fluid properties are taken at the average temperature of the surfaces, i.e., ðT1 þT2 Þ=2:

Example 7.7 Double-walled spherical container Problem A liquid at 50 C is stored in a double-walled spherical container. The inner sphere is thin-walled and has a diameter of 1.5 m. The outer sphere is at 20 C. The space between the spheres is 5 cm thick and is filled with air at 0.5 atm. What is the rate of natural convection heat flow out of the liquid?

Solution r1 ¼ D1 =2 ¼ 1:5=2 ¼ 0:75 m and r2 ¼ r1 þ thickness of air gap ¼ 0.75 þ 0.05 ¼ 0.80 m The properties of the air in the gap are taken at the average temperature Tavg ¼ ð50 þ20Þ=2 ¼ 35 C and 0.5 atm pressure. Air properties are given in the tables for 1 atm pressure. The thermal conductivity and absolute viscosity are not significantly affected by pressure. However, the kinematic viscosity is affected as its definition includes density. The air may be considered to be an ideal gas. The ideal gas law is P ¼ rRT. This shows that, with a constant temperature, the pressure is directly proportional to the density. In our case, the pressure has been halved. Therefore, the density at 0.5 atm is half of the density at 1 atm. m Regarding the kinematic viscosity y : y at 0.5 atm ¼ r at 0.5 ¼ ð0:5Þ ðr mat 1 atmÞ atm Putting in values for m and r for 1 atm and 35 C; we have m m 1:89 105 ¼ ¼ ¼ 3:287 105 m2 =s r at 0.5 atm ð0:5Þ ðr at 1 atmÞ ð0:5Þð1:150Þ The other properties of air at 35 C are k ¼ 0.0267 W/m C and Pr ¼ 0.711. We will use Eqs. (7.66), (7.67), and (7.68). y at 0.5 atm ¼

From Eq. (7.68), Lsph ¼

½ð1=0:75Þð1=0:80Þ4=3 5=3

21=3 ½ð0:75Þ7=5 þð0:80Þ7=5

¼ 0:005006

Using this value to get Rasph , Rasph ¼

gbðTs TN ÞðLsph Þ3 ð9:807Þ½1=ð35 þ 273:15Þð50 20Þð0:005006Þ3 Pr ¼ ð0:711Þ ¼ 78:82 2 y2 ð3:287 105 Þ

From Eq. (7.67), keff ¼ 0:74 k Finally, from Eq. (7.66), qconv ¼

1=4 1=4 Pr 0:711 1=4 Rasph ¼ 0:74 ð78:82Þ1=4 ¼ 1:808 0:861 þ Pr 0:861 þ 0:711

4pkeff 4p½ð1:808Þð0:0267Þ ð50 20Þ ¼ 218 W ðT1 T2 Þ ¼ ð1=0:75Þ ð1=0:80Þ ð1=r1 Þ ð1=r2 Þ

294

Chapter 7 Natural (free) convection

7.8 Natural convection between vertical fins The addition of fins to a surface enhances the heat transfer for the surface. Fins were discussed in Chapter 3. However, that chapter only discussed single fins convecting to an open adjacent fluid. This section discusses fins that are closely spaced. In this situation, the flow field in the fluid adjacent to a fin impacts the flow field of the fin next to it. Such interaction affects the convective coefficient for the fins and hence affects the heat transfer. In particular, we will be looking at vertical straight rectangular fins on a vertical surface. A picture of such fins were shown in Fig. 3.27; specifically, the picture of the fins on the power transformer. Fig. 7.12 shows three vertical fins attached to a vertical surface. Only three fins are shown, but there can indeed be many fins on the surface. The fins have thickness t, and the space between the fins is S. The fins are L high by b wide. The surface to which the fins are attached is L high and W wide. Elenbaas [27] considered the heat transfer in vertical channels both analytically and experimentally. Bar-Cohen and Rohsenow [28] later provided the following Nusselt Number correlation equation for isothermal vertical plates: !1=2 576 2:873 Nus ¼ þ (7.69) ½Ras ðS=LÞ2 ½Ras ðS=LÞ1=2 where Nus ¼

h S k

(7.70)

b

W

L

t Fin Surfaces at Ts

FIGURE 7.12 Vertical rectangular fins on a surface.

S Fluid at T∞

7.8 Natural convection between vertical fins

295

g b ðTs TN Þ S3 Pr (7.71) y2 The material properties in Eqs. (7.69)e(7.71) are those of the fluid. They should be taken at the film N temperature Tf ¼ Ts þT 2 : After obtaining h from Eq. (7.70), the convective heat transfer for an array of N fins on a surface is Ras ¼

q

¼ h ½ð2NÞðbÞðLÞ ðTs TN Þ

(7.72)

(Note: Eq. (7.72) ascribes the same convective coefficient to all surfaces of the fins in the array. Actually, the two outer surfaces of the fin array would have a different h value than the surfaces between fins. Assuming that the two outer surfaces have the same h as the other surfaces should, in most cases, have insignificant impact on the result.) Bar-Cohen and Rohsenow also determined the optimum spacing, Sopt , between fins. This is the spacing that maximizes the heat transfer from an array of fins. The optimum spacing is 1=4 Ras Sopt ¼ 2:714 (7.73) S3 L We close this section with two examples for fins.

Example 7.8 Fins on a tank wall Problem Fins have been added to the vertical wall of a small tank to assist in cooling the liquid inside the tank. Eleven vertical fins have been added to a square section of the tank wall that is 10 cm by 10 cm. The fins are 0.8 mm thick, 10 cm high, and 2 cm wide. They are equally spaced on the total width of the wall section. The fins are at a temperature of 130 C, and the adjacent air is at 30 C. What is the rate of heat removal by the fins?

Solution Ts þ TN 130 þ 30 ¼ 80 C ¼ Tf ¼ 2 2 1 For air at 80 C, k ¼ 0:0295 W m C; y ¼ 21:01 106 m2 s; Pr ¼ 0.715; b ¼ T1f ¼ 80þ273:15 ¼ 0:00283 K There are 11 fins equally spaced over the width of the wall section that is 10 cm wide. The spacing between fins is S¼ From Eq. (7.71), Ras ¼

W N t 0:10 11 ð0:0008Þ ¼ ¼ 0:00912 m N 1 10

gbðTs TN ÞS3 ð9:807Þð0:00283Þð130 30Þð0:00912Þ3 Pr ¼ ð0:715Þ ¼ 3412 2 y2 ð21:01 106 Þ

From Eq. (7.69), Nus ¼

576

þ

2:873

!1=2

½Ras ðS=LÞ2 ½Ras ðS=LÞ1=2 For this problem, Ras ¼ 3412; S ¼ 0:00912 m; and L ¼ 0:10 m. Putting these values into Eq. (7.69), we get Nus ¼ 2:434. From Eq. (7.70), h ¼ k SNus ¼ ð0:0295Þð2:434Þ ¼ 7:88 W m2 C. 0:00912

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Chapter 7 Natural (free) convection

Finally, we get the heat transfer rate from Eq. (7.72): q ¼ h½ð2NÞðbÞðLÞðTs TN Þ ¼ ð7:88Þð2Þð11Þð0:02Þð0:10Þð130 30Þ ¼ 34:7 W The rate of heat removal by the 11 fins is 34.7 W.

The second fin example deals with designing a fin array for removing heat at a specified rate from a plate.

Example 7.9 Fins for cooling of electronic components Problem Heat-producing electronic components are mounted on one side of a vertical plate. Vertical fins are to be added on the other side of the plate to increase the convective heat transfer from the plate to the adjacent air. The fins need to remove heat at a rate of 25 W from the plate. The plate is 8 cm wide and 12 cm high. The fins have a thickness of 1 mm and are 12 cm high and 4 cm wide. They are equally spaced. The plate and the fins are at 155 C and the adjacent air is at 25 C. (a) How many fins are needed to remove the specified heat, and what is the spacing S between them? (b) If the fins had not been added to the plate, what would have been the natural convection heat transfer rate from the surface?

Solution Tf ¼ For air at 90 C,

Ts þ TN 155 þ 25 ¼ 90 C ¼ 2 2

k ¼ 0:0302 W=m C; y ¼ 22:02 106 m2 =s; Pr ¼ 0.713; b ¼

1 1 ¼ 0:00275=K ¼ Tf 90 þ 273:15

The heat transfer rate due to the fins is given by Eq. (7.72): q ¼ h ½ð2NÞðbÞðLÞ ðTs TN Þ For this problem, q ¼ 25 W; b ¼ 0:04 m; L ¼ 0:12 m; and ðTs TN Þ ¼ 155 25 ¼ 130 C. Putting the values into Eq. (7.72), we get the requirement

(7.72)

h N ¼ 20:03 (7.74) The convective coefficient h is a function of the spacing S between fins. A spacing that results in a higher convective coefficient will require a lesser number of fins to achieve the desired heat transfer q. Let us continue with our analysis: From Eq. (7.71), Ras ¼

gbðTs TN ÞS3 ð9:807Þð0:00275Þð155 25ÞS3 Pr ¼ ð0:713Þ 2 2 y ð22:02 106 Þ and Ras ¼ 5:162 109 S3

(7.75)

Combining Eqs. (7.69) and (7.70), we have !1=2 k (7.76) þ h¼ S ½Ras ðS=LÞ2 ½Ras ðS=LÞ1=2 For this problem, k ¼ 0.0302 W/m C and L ¼ 0.12 m. Using these values and Eqs. (7.74) and (7.75), Eq. (7.76) can be changed to !1=2 0:0302N 576 2:873 20:03 ¼ þ (7.77) S ð4.302 1010 S4 Þ2 ð4.302 1010 S4 Þ1=2 576

2:873

For a desired number of fins N, Eq. (7.77) may be solved for the required spacing S between the fins. Of course, we have to make sure that the overall width of the fin array is less than or equal to the width W of the plate. That is, we need to verify that ðN 1Þ S

þ

Nt

W

(7.78)

7.9 Chapter summary and final remarks

297

Summarizing: There are multiple answers to Part (a) of this problem. We first decide how many fins we want. Then we solve Eq. (7.77) for the required fin spacing that will give us the specified heat removal rate of 25 W. For example, let us say that we want 15 fins on the wall. Then N ¼ 15 and Eq. (7.77) becomes !1=2 0:0302ð15Þ 576 2:873 þ (7.79) 20:03 ¼ S ð4.302 1010 S4 Þ2 ð4.302 1010 S4 Þ1=2 Eq. (7.79) can be solved for spacing S. One way to do this is by using Excel Goal Seek. If we wish to use Goal Seek, we rearrange Eq. (7.79) to the form !1=2 0:0302ð15Þ 576 2:873 20:03 þ ¼0 S ð4.302 1010 S4 Þ2 ð4.302 1010 S4 Þ1=2

(7.80)

We tell Goal Seek to make the left side of Eq. (7.80) zero by changing the value of S. When we did this, we got the result S ¼ 0.002924 m ¼ 2.924 mm. Using Eq. (7.78), we find that the total width of the fin array is 5.3 cm and the fin array will indeed fit on the 8-cm-wide plate. We have seen that an array of 15 fins will remove the specified 25 W from the plate. But there are several possible solutions to Part (a) of this problem. Indeed, we could have any number of fins from 3 to 25 to remove the specified heat. We would just input the number N of the fins into Eq. (7.77) and solve the equation for the needed fin spacing S. We did this for several numbers of fins and got the results: For 3 fins, the needed spacing is 6.009 mm; for 5 fins, the spacing is 4.398 mm; for 10 fins, the spacing is 3.367 mm; and for 20 fins, the spacing is 2.652 mm. It turns out that 25 fins is the maximum number of fins that can be used. A greater number of fins give a fin array width that, per Eq. (7.78), is greater than the width of the plate. (b) We can use Eq. (7.10) for solution of Part (b): 1=4

0:670 RaL 0.68 þ h i4=9 1 þ ð0:492=PrÞ9=16 For this problem, L ¼ 0.12 m. Calculating RaL , we have Nu ¼

h L ¼ k

(7.10)

gbðTs TN ÞL3 Pr ð9:807Þð0:00275Þð155 25Þð0:12Þ3 ð0:713Þ ¼ ¼ 8:909 106 2 y2 ð22:02 106 Þ From Eq. (7.10), we have " " # 1=4 # 1=4 ð0:670Þ 8:909 x 106 0:670 RaL k 0:0302 ¼ 0.68 þ h ¼ 0.68 þ h i4=9 L h i4=9 0:12 1 þ ð0:492=PrÞ9=16 1 þ ð0:492=0:713Þ9=16 RaL ¼

h ¼ 7:245 W=m2 C The heat flow from the unfinned plate is q ¼ h A ðTs TN Þ ¼ ð7:245Þ½ð0:08Þ ð0:12Þ ð155 25Þ ¼ 9:04 W Without fins, the heat transfer by natural convection from the plate would have been 9.04 W.

7.9 Chapter summary and final remarks In this chapter, we discussed natural (free) convection for a variety of surfaces and enclosures. Much experimentation and analysis has been done on this topic, and we have provided several equations for predicting the convective coefficients. One must keep in mind, however, that the correlation equations come from carefully-controlled laboratory experiments. Laboratory conditions, in most cases, are not duplicated in practice. Hence, the results from the correlation equations may differ by as much as 25% or more from results actually obtained in practice. The best situation, of course, would be to obtain

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Chapter 7 Natural (free) convection

convective coefficients from experimentation on the actual objects of interest. This, however, is usually not possible due to cost, schedule, and other constraints. Although the results from the correlation equations are “rough,” they do indeed provide valuable information for conceptual and design purposes. We did not include a discussion of combined free and forced convection. In some situations, fluid flow is such that convection is not entirely natural (free) or not entirely forced. Flow velocities may be in excess of those caused by the density gradients of natural convection but may not be large enough to completely overwhelm the effects of the density gradients. In these cases of mixed convection, the Nusselt Numbers for sole natural convection and sole forced convection have to be combined in some fashion to accurately model the situation. GrL In general, combined free and forced convection should be considered if Re 2 z 1: L GrL the Nusselt number is a function of the Reynolds and If Re2 1; free convection is negligible and L Prandtl numbers. That is NuL ¼ f ðReL ; Pr GrL If Re 2 [1; forced convection is negligible and the Nusselt number is a function of the Grashof and L Prandtl numbers. That is, NuL ¼ f ðGrL ; Pr Combined free and forced convection is discussed in Refs. [29e35]. We now move on to the next chapter that deals with heat exchangers. The chapter is very application-oriented and should be of much interest to many of you.

7.10 Problems Notes: •

• • •

If needed information is not given in the problem statement, then use the Appendix for material properties and other information. If the Appendix does not have sufficient information, then use the Internet or other reference sources. Some good reference sources are mentioned at the end of Chapter 1. If you use the Internet, double-check the validity of the information by using more than one source. Your solutions should include a sketch of the problem. Caution: Make sure that you use absolute temperatures (Kelvin or Rankine) if the problem involves radiation. In all problems, unless otherwise stated, the fluid pressure is atmospheric. Gas properties in the Appendix are at atmospheric pressure. If a problem has a gas at other than atmospheric pressure, affected properties such as density and kinematic viscosity should be modified accordingly through use of the ideal gas law. 7-1 A thin vertical flat plate is 50 cm high and 20 cm wide. Its two surfaces are at 200 C and the plate convects to the surrounding 30 C air. What is the heat transfer from the two sides of the plate to the air? 7-2 A thin, square electrical heater is vertical and is 20 cm by 20 cm. One side of the heater is perfectly insulated and the other side has convection to the room air. The air is at 25 C and the heater dissipates 40 W. The active surface of the heater can be considered as a constant flux surface. (a) What is the convective coefficient at 10, 15, and 20 cm from the bottom of the plate? (b) What is the average convective coefficient for the plate?

7.10 Problems

299

7-3 A vertical square plate is 20 cm by 20 cm. Its two surfaces are at 80 C. The plate is in a tank of 30 C water. What is the rate of heat transfer from the plate to the water? 7-4 A square flat plate is 1.5 m by 1.5 m. Its two surfaces are maintained at 120 C. The plate convects from its two surfaces to the room air which is at 20 C. (a) What is the rate of heat transfer to the air if the plate is vertical? (b) What is the rate of heat transfer to the air if the plate is inclined 45 degrees with the horizontal? 7-5 A thin flat plate is 10 cm by 20 cm. It is in a tank of 20 C water. The 10 cm edge is horizontal and the plate is inclined at an angle of 30 degrees with the horizontal. The plate’s two surfaces are at 90 C. What is the rate of convective heat transfer from the plate to the water? 7-6 A vertical cylinder of 5 cm diameter and 20 cm length has a surface temperature of 150 C. It is in room air at 20 C. (a) In determining the heat transfer from the cylinder, can the surface of the cylinder be treated as a vertical flat plate? (b) What is the rate of heat transfer from the cylinder to the air? 7-7 A circular plate, 20 cm in diameter, is suspended horizontally in 20 C air. If the top and bottom of the plate are both at 100 C, what is the rate of heat transfer to the air? 7-8 A square flat plate is 15 cm by 15 cm. It is oriented horizontally in 20 C air. The top and bottom of the plate are at 80 C. What is the rate of heat transfer from the plate to the air? 7-9 A thin flat plate has the shape of an equilateral triangle with sides of length 15 cm. The plate is suspended horizontally in a tank of 40 C water. The top of the plate is at 80 C and the bottom is at 60 C. What is the rate of heat transfer from the plate to the water? 7-10 A circular flat plate heater has a diameter of 25 cm. It is suspended vertically in 30 C water. If the surface of the heater cannot exceed 95 C, what is the maximum allowed power to the heater? 7-11 A square plate is 1 m by 1 m. It is inclined 30 degrees with the horizontal. The downward facing surface of the plate is perfectly insulated and the upward facing surface receives solar radiation. At equilibrium, the upper surface convects to the surrounding 20 C air with a constant flux of 700 W/m2. What is the equilibrium mean temperature of the upper surface? 7-12 A horizontal plate is 30 cm by 50 cm. Its bottom surface is perfectly insulated and its top surface is a constant flux surface convecting to the adjacent 25 C air. The mean temperature of the top surface of the plate is 55 C. What is the rate of convective heat transfer from the plate to the air? 7-13 A metal street sign is about 0.5 m by 0.5 m. Solar radiation at a rate of 250 W/m2 hits the sign and heats it. If the sign has an absorptivity of 0.6 for the solar radiation and an emissivity of 0.7, what is the surface temperature of the sign? Assume that the air is quiet and the air and surroundings are at 30 C? 7-14 A horizontal cylinder has a diameter of 1.5 cm and is in water at 25 C. The surface of the cylinder is at 90 C. What is the heat transfer from the cylinder to the water per meter length of cylinder? 7-15 Steam is flowing through a horizontal 200 Sch 40 steel pipe. The outer surface of the pipe has a temperature of 225 F. The pipe is used to help heat a warehouse. (a) If the pipe is 70 feet long, estimate the rate of convective heat transfer to the 65 F room air.

300

7-16 7-17

7-18

7-19 7-20

7-21 7-22 7-23 7-24

7-25

7-26

Chapter 7 Natural (free) convection

(b) If the pipe’s outer surface has an emissivity of 0.5, what is the rate of radiative heat transfer to the room? Assume the room walls are at the same temperature as the room air, that is, 65 F. Oil at 60 C flows through a horizontal thin-walled copper tube of 1.9 cm outside diameter. The tube is in 25 C water. What is the rate of heat transfer to the water per meter length of tube? 00 00 A natural gas hot water heater is in the basement of a house. The heater is 18 diameter and 55 high. The air in the basement is at 65 F. If the top and side surfaces of the heater are at 95 F, estimate the heat loss from the heater. Only include heat transfer from the top and side of the heater. The heat transfer from the bottom of the heater is negligible. A cylindrical immersion heater has a rating of 700 W. It has a diameter of 2.5 cm and a length of 40 cm and is oriented horizontally in a large tank of water. The water is at 20 C. (a) At steady state, what is the surface temperature of the heater? (b) If the power to the heater is accidentally turned on when the heater is in 20 C air, what would be the steady-state temperature of the heater’s surface? An exhaust duct from a furnace is uninsulated and provides heating to a factory. The duct is horizontal and of 20 cm diameter. The gases inside the duct are at 250 C and the room air is at 15 C. What is the rate of convective heat transfer to the air per meter length of duct? An incandescent light bulb is very inefficient. Only about 5% of the input power produces light. The remainder is absorbed by the glass bulb, which then dissipates the heat by convection to the surrounding air and by radiation to the surroundings. A 75 W light bulb is modeled as a 10 cm diameter sphere. The emissivity of the glass is about 0.9, and both the air and the walls of the room are at 25 C. Estimate the temperature of the glass bulb. A 5 cm diameter sphere has a surface temperature of 50 C. It is in 20 C water. What is the rate of heat transfer from the sphere to the water? An electric heater is inside a hollow aluminum sphere. The sphere’s outside diameter is 3 cm, and the sphere is in air at 20 C. What electric power input to the heater is needed for the sphere’s outer surface to be at 80 C? Do Problem 7-22 if the sphere is in 20 C water rather than air. An incandescent light bulb can be modeled as a sphere of 10 cm diameter. Operating in 20 C air, the outer surface of the light bulb is at 150 C. Assuming that the emissivity of the bulb’s surface is 0.8 and that the walls of the room are at 20 C, what is the rate of heat transfer to the room air and surroundings? Include both convective and radiative modes. An electric heater is inside an aluminum cube that is 8 cm by 8 cm by 8 cm. The cube is on a table and it may be assumed that heat transfer from the cube’s bottom surface is negligible. Power to the heater causes all five exposed surfaces of the cube to be at 150 C. The room air is at 15 C. What is the rate of heat transfer from the cube to the air? Thin electronic components that generate heat are sandwiched between two aluminum plates. The plates are rectangular, 20 cm by 30 cm. The assembly is positioned horizontally. As the plates are of high conductivity, we will assume that both plates and the components are at the same temperature. The components will fail if their temperature reaches 80 C. (a) If the assembly is in 20 C air, what is the maximum allowable power to the components? (b) If the assembly is in 20 C water, what is the maximum allowable power to the components?

7.10 Problems

301

7-27 A horizontal heating duct is 30 cm wide and 15 cm high. The outer surface of the duct is at 65 C and the temperature of the room air is 22 C. What is the rate of convective heat transfer to the room air per meter length of duct? 7-28 A cube of copper, 5 cm by 5 cm by 5 cm, cools in a tank of 20 C water. What is the rate of heat transfer from the cube to the water when the cube’s surface temperature is 80 C? 7-29 In the winter, a homeowner adds storm windows to his house. Consider a vertical window that is 0.9 m wide and 1.5 m high, which has a storm window on it. The window/storm window combination forms an enclosed cavity having a 5 cm air gap between the windows. The inside of the storm window is at 5 C, and the outside of the house window is at 15 C. What is the rate of convective heat transfer through the air gap? 7-30 Instead of using a storm window, the homeowner of Problem 7-29 decides to replace the house window with a double-glazed window. The space between the glass panes is 1.25 cm and it is filled with argon. For the same temperature conditions as Problem 7-29, determine the rate of convective heat transfer through the argon space if (a) the argon is at 1 atm pressure and (b) the argon is at 0.5 atm pressure. 7-31 A solar collector consists of a glass cover plate and an absorber plate separated by a 4 cm thick air space. The assembly is a sealed square box of sides 2 m by 2 m. The collector is inclined at an angle of 50 degrees from the horizontal. If the temperature of the cover plate is 35 C and the absorber plate is at 65 C, what is the rate of convective heat transfer across the air space? 7-32 A rectangular sealed assembly has two parallel square plates (60 cm by 60 cm) separated by an air space that is 4 cm thick. The hotter plate is at 70 C and the cooler plate is at 0 C. Determine the convective heat transfer between the plates if the assembly is (a) vertical. (b) horizontal with the cooler plate on the bottom. (c) horizontal with the hotter plate on the bottom. 7-33 A vertical wall section of a house consists of two plywood layers separated by a 9 cm thick air gap. The wall section is 2.5 m high and 5 m wide. The temperature of the two surfaces in contact with the air gap are 20 C and 5 C, respectively. (a) What is the rate of convective heat transfer across the air gap? (b) Let us say that a horizontal baffle is placed in the wall section halfway up the wall. The baffle changes the air gap from a single air gap 2.5 m high and 5 m wide to two gaps that are each 1.25 m high and 5 m wide. What is the new heat transfer rate if temperatures remain the same? 7-34 A solar collector has a glass cover plate and an absorber plate separated by an air space. The collector is square, 1 m by 1 m, and is tilted at an angle of 20 degrees from the horizontal. The cover plate is at 30 C and the absorber plate is at 60 C. What is the rate of convective heat transfer across the air space if (a) the air space is 2 cm thick? (b) the air space is 4 cm thick? 7-35 An annulus is formed by two concentric horizontal cylinders. The outer diameter of the inner cylinder is 70 mm and the inner diameter of the outer cylinder is 100 mm. The space between the cylinders is filled with water. The inner and outer surfaces of the annulus are at 20 and 80 C, respectively. What is the rate of convective heat transfer per meter length of the cylinders?

302

Chapter 7 Natural (free) convection

7-36 A solar collector consists of a glass outer tube and a concentric inner absorber tube. The collector is horizontal, with the inner tube having an outer diameter of 10 cm and the outer tube having an inner diameter of 14 cm. The solar collector is 2 m long. The glass tube has a temperature of 25 C and the absorber tube has a temperature of 75 C. If air fills the annular space, what is the rate of convective heat transfer through the air? 7-37 A container for liquid oxygen consists of two concentric spheres. The inner sphere, which holds the LO2, is thin-walled and has a diameter of 1 m. The outer sphere has an inner diameter of 1.05 m. The space between the spheres is filled with air. With the container filled with LO2, the inner sphere has a temperature of 90.2 K and the inner surface of the outer sphere has a temperature of 287 K. The latent heat of vaporization of oxygen is 2.14 105 J/kg. What is the rate (kg/s) at which gaseous oxygen leaves the container through a small vent on the top of the container? 7-38 The space between two thin concentric spheres contains air at 5 atm pressure. The spheres have diameters of 20 and 25 cm and respective temperatures of 40 and 100 C. What is the rate of convective heat transfer through the air space? 7-39 A house has a gable roof. Each of the two roof sections is 20 feet by 30 feet. The 20 foot edge is parallel to the ground and the 30 foot edge is inclined 25 from the horizontal. The outside air is quiet and at 10 F. The outer surface of the roof is at 40 F. What is the rate of convective heat loss from the roof? 7-40 A wire of 0.3 mm diameter is horizontal in a tank of water which is at 25 C. A current passing through the wire heats the outer surface of the wire to 90 C. What is the rate of heat convection to the water per meter length of wire? 7-41 Water at 40 C flows over a 10 cm diameter sphere at a velocity of 0.1 m/s. The sphere’s surface is at 80 C. Should both free and forced convection be considered or is one insignificant compared with the other? 7-42 Air at 20 C flows across the top surface of a horizontal heated plate at a free stream velocity of 0.5 m/s. The plate is 10 cm wide and 20 cm in the direction of flow. The temperature of the top of the plate is 120 C. Should both free and forced convection be considered or is one insignificant compared with the other? 7-43 A vertically-mounted aluminum heat sink is 7.5 cm high and 15 cm wide. The vertical fins are equally spaced across the 15 cm width of the heat sink. There are 20 vertical fins forming 19 vertical channels for air flow. Each fin is 0.5 mm thick, 7.5 cm high, and 2.5 cm wide. The fins are at 80 C and the adjacent air is at 20 C. What is the heat transfer from the fins to the air (W)? 7-44 A vertically-mounted aluminum heat sink is 9 cm high and 12 cm wide. There are equally spaced vertical fins across the entire 12 cm width of the heat sink, and the fins are spaced at their optimal spacing for maximum heat transfer. Each fin is 0.5 mm thick, 9 cm high, and 1.5 cm wide. The fins are at 80 C and the adjacent air is at 20 C. (a) How many fins are there? (b) What is the heat transfer from the fins to the air? 7-45 It is desired to remove 55 W of heat from a tank wall through the use of vertical fins. There are 17 fins and they are equally spaced at a spacing of 6 mm. The fins are 0.7 mm thick and 3 cm wide. The fins are at 80 C and the adjacent air is at 10 C. What is the needed vertical height of the fins for the required 55 W heat removal?

References

303

7-46 A student is performing an experiment in a thermal lab. He is measuring the natural convective heat transfer from a horizontal section of duct that has an equilateral triangular cross section. The measured heat transfer will be compared with the predictions from existing correlation equations. The duct section is 2 m long, and each of the three sides of the duct is 25 cm wide. There is an electric heater inside the duct, which keeps the duct wall at 90 C. The room air is at 25 C. (a) If the duct is positioned with one of its flat surfaces facing up, what is the predicted rate of natural convection to the room air? (b) If the duct is positioned with one of its flat surfaces facing down, what is the predicted rate of natural convection to the room air? (c) Compare the results of Part (a) and Part (b) with the predicted rate of natural convection from a horizontal cylindrical duct section having the same surface area as the triangular duct section. 7-47 Water is flowing at a rate of 0.04 kg/s through a horizontal, thin-walled copper tube of 2 cm diameter and 5 m length. The water enters the tube at 90 C. The room air is at 20 C, and the tube transfers heat to the room air by natural convection. (a) What is the temperature of the water as it exits the tube? (b) What is the heat transfer rate from the water to the room air?

References [1] S.W. Churchill, H.H.S. Chu, Correlation equations for laminar and turbulent free convection from a vertical plate, Int. J. Heat Mass Transf. 18 (1975) 1323e1328. [2] W.H. McAdams, Heat Transmission, 3rd ed., McGraw-Hill, 1954. [3] C.Y. Warner, V.S. Arpaci, An experimental investigation of turbulent natural convection in air at low pressures for a vertical heated flat plate, Int. J. Heat Mass Transf. 11 (1968) 397. [4] F.J. Bayley, An analysis of turbulent free convection heat transfer, Proc. Inst. Mech. Eng. 169 (1955) 361. [5] T. Fujii, M. Fujii, The dependence of local Nusselt number on Prandtl number in the case of free convection along a vertical surface with uniform heat flux, Int. J. Heat Mass Transf. 19 (1976) 121e122. [6] E.M. Sparrow, J.L. Gregg, Laminar free convection from a vertical plate with uniform surface heat flux, Trans. ASME 78 (1956) 435e440. [7] T. Fujii, H. Imura, Natural convection heat transfer from a plate with arbitrary inclination, Intl. J. Mass Heat Transf. 15 (1972) 755e767. [8] J.R. Lloyd, W.R. Moran, Natural convection adjacent to horizontal surface of various planforms, J. Heat Transf. 96 (1974) 443e447. [9] S.W. Churchill, H.H.S. Chu, Correlating equations for laminar and turbulent free convection from a horizontal cylinder, Int. J. Heat Mass Transf. 18 (1975) 1049e1053. [10] V.T. Morgan, The overall convective heat transfer from smooth circular cylinders, Adv. Heat Transf. 11 (1975) 199e264. [11] T. Yuge, Experiments in heat transfer from spheres including combined natural and forced convection, J. Heat Transf. 82 (1960) 214e220. [12] W.S. Amato, C. Tien, Free convection heat transfer from isothermal spheres in water, Int. J. Heat Mass Transf. 15 (1972) 327e339. [13] S.W. Churchill, Free convection around immersed bodies, in: G.F. Hewitt (Ed.), Heat Exchanger Design Handbook, Hemisphere Publishing, Washington, D. C., 1983, pp. 2.5.7e24.

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Chapter 7 Natural (free) convection

[14] E.M. Sparrow, A.J. Stretton, Natural convection from variously oriented cubes and from other bodies of unity aspect ratio, Int. J. Heat Mass Transf. 28 (4) (1985) 741e752. [15] J.H. Lienhard, On the commonality of equations for natural convection from immersed bodies, Int. J. Heat Mass Transf. 16 (1973) 2121e2123. [16] E.M. Sparrow, M.A. Ansari, A refutation of King’s rule for multi-dimensional external natural convection, Int. J. Heat Mass Transf. 26 (1983) 1357e1364. [17] E.M. Sparrow, J.L. Gregg, Laminar free convection from the outer surface of a vertical cylinder, Trans. ASME 78 (1956) 1823e1829. [18] W.J. Minkowycz, E.M. Sparrow, Local nonsimilar solutions for NAtural convection on a vertical cylinder, J. Heat Transf. 96 (1974) 178e183. [19] T. Cebeci, Laminar free convective heat transfer from the outer surface of a vertical slender circular cylinder, in: Proc. Fifth Intl. Heat Transfer Conference, Tokyo, Paper NC 1.4, September 1974, pp. 15e19. [20] E.J. LeFevre, A.J. Ede, Laminar free convection from the outer surface of a vertical circular cylinder, in: Proc. Ninth Intl. Congress Applied Mechanics, Brussels, Vol. 4, 1956, pp. 175e183. [21] K.G.T. Hollands, T.E. Unny, G.D. Raithby, L. Konicek, Free convective heat transfer across inclined air layers, J. Heat Transf. 98 (1976) 189e193. [22] S. Globe, D. Dropkin, Natural-convection heat transfer in liquids confined by two horizontal plates and heated from below, J. Heat Transf. 81 (1959) 24e28. [23] S.M. ElSherbiny, G.D. Raithby, K.G.T. Hollands, Heat transfer by natural convection across vertical and inclined air layers, J. Heat Transf. 104 (1982) 96e102. [24] R.K. MacGregor, A.P. Emery, Free convection through vertical plane layers: moderate and high Prandtl number fluids, J. Heat Transf. 91 (1969) 391e403. [25] I. Catton, Natural convection in enclosures, in: Proc. Sixth Intl. Heat Transfer Conference, Toronto, Canada, vol. 6, 1978, pp. 13e31. [26] G.D. Raithby, K.G.T. Hollands, A general method of obtaining approximate solutions to laminar and turbulent free convection problems, Adv. Heat Transf. 11 (1975) 265e315. [27] W. Elenbaas, Heat dissipation of parallel plates by free convection, Physica 9 (1942) 1e28. [28] A. Bar-Cohen, W.M. Rohsenow, Thermally optimum spacing of vertical natural convection cooled parallel plates, J. Heat Transf. 106 (1984) 116e123. [29] B. Metais, E.R.G. Eckert, Forced, mixed, and free convection regimes, J. Heat Transf. 86 (1964) 295. [30] C.K. Brown, W.H. Gauvin, Combined free and forced convection I, II, Can. J. Chem. Eng. 43 (6) (1965) 306e313. [31] C.A. Depew, J.L. Franklin, C.H. Ito, Combined free and forced convection in horizontal, uniformly heated tubes, ASME Paper 75-HT-19 (August 1975). [32] S.W. Churchill, Combined free and forced convection around immersed bodies, in: Heat Exchanger Design Handbook, Section 2.5.9, Hemisphere Publishing, New York, 1983. [33] S.W. Churchill, Combined free and forced convection in channels, in: Heat Exchanger Design Handbook, Section 2.5.10, Hemisphere Publishing, New York, 1983. [34] D.G. Osborne, F.P. Incropera, Experimental study of mixed convection heat transfer for transitional and turbulent flow between horizontal, parallel plates, Int. J. Heat Mass Transf. 28 (1985) 1337. [35] J.R. Lloyd, E.M. Sparrow, Combined forced and free convection flow on vertical surfaces, Int. J. Heat Mass Transf. 13 (1970) 434e438.

CHAPTER

Radiation heat transfer

9

Chapter outline 9.1 9.2 9.3 9.4 9.5

Introduction .................................................................................................................................343 Blackbody emission......................................................................................................................344 Radiation properties .....................................................................................................................349 Radiation shape factors ................................................................................................................353 Radiative heat transfer between surfaces....................................................................................... 359 9.5.1 Radiation heat transfer for a two-surface enclosure......................................................362 9.5.1.1 For surface 1 ...................................................................................................... 362 9.5.1.2 For surface 2 ...................................................................................................... 362 9.5.2 Radiation heat transfer for a three-surface enclosure ...................................................367 9.5.2.1 For surface 1 ...................................................................................................... 367 9.5.2.2 For surface 2 ...................................................................................................... 367 9.5.2.3 For surface 3 ...................................................................................................... 368 9.5.2.4 Three-surface enclosure with an insulated surface ............................................... 370 9.6 Radiation shields .........................................................................................................................373 9.7 Sky radiation and solar collectors .................................................................................................380 9.8 Chapter summary and final remarks............................................................................................... 381 9.9 Problems .....................................................................................................................................381 References ..........................................................................................................................................389 Further reading ....................................................................................................................................389

9.1 Introduction The three modes of heat transfer are conduction, convection, and radiation. We discussed details of conduction and convection in previous chapters and briefly discussed radiative heat transfer in Chapter 1. We presented the following equation for radiation from a surface at temperature Ts to a large surrounding enclosure at Tsurr : 4 (9.1) qrad ¼ εsA Ts4 Tsurr

Heat Transfer Principles and Applications. https://doi.org/10.1016/B978-0-12-802296-2.00009-3 Copyright © 2021 Elsevier Inc. All rights reserved.

343

344

Chapter 9 Radiation heat transfer

where ε is the emissivity of the surface A is the area of the surface s is the Stefan Boltzmann constant ¼ 5.67 108 W=m2 K4 ¼ 0:1714 108 Btu=h ft2 R4 : Temperatures Ts and Tsurr must be in absolute temperature units (Kelvin or Rankine). In this chapter, we discuss radiative heat transfer in considerably more detail. Topics covered include blackbody radiation, radiation properties, radiation shape factors, exchange of radiant energy between surfaces, and radiation shields.

9.2 Blackbody emission The surfaces of all objects emit thermal radiation by virtue of their temperature being above absolute zero. This radiation is emitted as electromagnetic radiation in the wavelength range of about 0.1e100 mm (1 mm ¼ 106 m ¼ 1 micron). Some other areas of the electromagnetic spectrum are visible light (about 0.4e0.8 mm), X-rays (about 1011 m to 2 108 m), microwaves (about 1 mme10 m), and radio waves (about 10 me30 km). Unlike conduction and convection, which require a medium for their transport, radiation is transported unhindered through a vacuum. Air, oxygen, and nitrogen are also essentially transparent to radiation. Some other gases, however, (e.g., water vapor, carbon dioxide, and some hydrocarbon gases), have appreciable absorption of radiation in specific wavelength regions. We will not be considering radiation transport through these latter types of gases. The wavelength l of the radiation is related to the frequency y of the radiation by the speed of light co :

co ¼ ly

(9.2)

where co ¼ 2:9979 108 m s in a vacuum. A “blackbody” has a surface that emits radiation at the maximum possible rate. The emission is over all wavelengths and is given by Planck’s law, which is Ebl ðl; TÞ ¼

2phc2o l5 hco exp 1 lkT

(9.3)

where h ¼ Planck’s constant ¼ 6.62607 1034 J$s k ¼ Boltzmann’s constant ¼ 1.38065 1023 J=K Ebl ¼ spectral emissive power of a blackbody The spectral emissive power is the rate of thermal emission per unit surface area per unit wavelength. If the wavelength of the radiation is in mm, then the units of the spectral emissive power are W/ (m2 mm).

9.2 Blackbody emission

10

345

Spectral Emissive Power Ebλ [W / (m2 μ m)]

9

5800 K

108

10 7 10 6 105

1000 K

10 4 103

10 2

300 K

101 10 0 10 −1 10 −2 10 −3

0.1

1

10

Wavelength λ ( μm)

100

FIGURE 9.1 Blackbody spectral emissive power.

Putting in the values for the various parameters, Eq. (9.3) can be expressed as Ebl ðl; TÞ ¼

C1 l5 expðC2 =lTÞ 1

(9.4)

where C1 ¼ 3:7417 108 ðW=m2 $ðmmÞ4 and C2 ¼ 1:4388 104 mm$K. The blackbody spectral emissive power is shown in Fig. 9.1 for blackbodies having temperatures of 300 K, 1000 K, and 5800 K. The 300 K curve is for bodies at typical room temperature and the 5800 K curve approximates the emission from the sun. The spectral emissive power of a blackbody given in Eq. (9.4) is a function of both the wavelength of the radiation and the absolute temperature of the blackbody. The blackbody emits over all wavelengths. If we integrate spectral emissive power Ebl over all wavelengths, we get the total emissive power Eb which is a function of only the temperature T of the blackbody. Z N Ebl ðl; TÞdl ¼ sT 4 (9.5) Eb ðTÞ ¼ 0

108

2

where s ¼ 5:670 W=m ¼ Stefan Boltzmann constant Eq. (9.5) is called the StefaneBoltzmann law. We have used the adjectives “spectral” and “total”. “Spectral” is for a parameter that depends on the wavelength. A parameter with “total” is independent of wavelength. K4

346

Chapter 9 Radiation heat transfer

Looking at Fig. 9.1, it is seen that the maximum (or peak) emission from a blackbody occurs at lower wavelengths for higher temperatures. For 5800 K, the peak emission is at about 0.5 mm. For 1000 K, it is at about 3 mm. And, for 300K, it is about 9 mm. This is Wien’s displacement law, which is lmax T ¼ 2898 mm$K

(9.6)

where lmax is the wavelength of peak emission. Using Eq. (9.6), we get the following wavelengths of peak emission that we previously estimated by looking at Fig. 9.1: For 5800 K; lmax ¼ 2898=5800 ¼ 0:500 mm For 1000 K; lmax ¼ 2898=1000 ¼ 2:90 mm For 300 K; lmax ¼ 2898=300 ¼ 9:66 mm This shifting of wavelengths with temperature causes the “greenhouse effect” We have all been in hot, uncomfortable cars and rooms on hot, sunny days. This is primarily due to the transmissivity property of window glass. Glass has high transmission at the low wavelengths of solar radiation (5800 K blackbody) and very low transmission at the higher wavelengths of the 300 K radiation inside the car or room. The solar radiation easily goes through the glass into the inside air space, but it cannot leave the space through the essentially opaque glass. The temperature of the space increases to a very uncomfortable level if air conditioning is not provided. We have seen that blackbody emission depends on wavelength and temperature. Sometimes we are interested in finding the emissive power from a blackbody for a given wavelength band. This can be determined by integrating the spectral emissive power over the wavelength range. For example, if we want the total emissive power for a blackbody at temperature T for the wavelength band l1 to l2 , we have Z l2 Eb ðl1 / l2 ; TÞ ¼ Ebl ðl; TÞdl (9.7) l1

If we want the fraction of the radiation emitted by a blackbody that is in wavelength band l1 to l2 , we have the total emissive power for the wavelength band divided by the total emissive power for all wavelengths. That is, R l2 l Ebl ðl; TÞdl Fl1 /l2 ¼ R N1 (9.8) 0 Ebl ðl; TÞdl where Fl1 /l2 is the fraction of blackbody emission that is in wavelength range l1 to l2 . From the StefaneBoltzmann Law, the denominator is equal to sT 4 , so Eq. (9.8) may be written as R l2 l Ebl ðl; TÞdl Fl1 /l2 ¼ 1 (9.9) sT 4 Using Eq. (9.4), Eq. (9.9) becomes R l2 Fl1 /l2 ¼

l1

C1 l5 dl expðC2 =lTÞ 1 sT 4

(9.10)

9.2 Blackbody emission

347

The integral in Eq. (9.10) may be broken into two integrals, each starting at l ¼ 0: That is, R l1 C1 l5 C1 l5 dl dl 0 expðC2 =lTÞ 1 expðC2 =lTÞ 1 Fl1 /l2 ¼ F0/l2 F0/l1 ¼ (9.11) sT 4 sT 4 The integrals in Eqs. (9.10) and (9.11) are over wavelengths l for a given blackbody temperature T. The variable of integration can be changed from l to lT, which modifies Eq. (9.11) to Z l2 T C1 =s Fl1 T/l2 T ¼ F0/l2 T F0/l1 T ¼ dðlTÞ 5 ðlTÞ ½expðC2 =lTÞ 1 0 Z l1 T C1 =s dðlTÞ (9.12) 5 ðlTÞ ½expðC2 =lTÞ 1 0 R lT C1 =s Table 9.1 gives the blackbody radiation function F0/lT ¼ 0 dðlTÞ for 5 R l2 0

ðlTÞ ½expðC2 =lTÞ1

different values of lT. The use of Table 9.1 in determining the fraction of blackbody radiation in the wavelength band of l1 to l2 is illustrated in Example 9.1.

Table 9.1 Blackbody radiation function. lT (mm$K)

F0-lT

lT (mm$K)

F0-lT

lT (mm$K)

F0-lT

1000 1200 1400 1600 1800 2000 2200 2400 2600 2800 3000 3200 3400 3600 3800 4000 4200 4400 4600 4800

0.0003 0.0021 0.0078 0.0197 0.0393 0.0667 0.1009 0.1403 0.1831 0.2279 0.2733 0.3181 0.3618 0.4036 0.4434 0.4809 0.5161 0.5488 0.5793 0.6076

5000 5200 5400 5600 5800 6000 6200 6400 6600 6800 7000 7200 7400 7600 7800 8000 8500 9000 9500 10,000

0.6338 0.6580 0.6804 0.7011 0.7202 0.7379 0.7542 0.7692 0.7833 0.7972 0.8082 0.8193 0.8296 0.8392 0.8481 0.8563 0.8747 0.8901 0.9031 0.9143

10,500 11,000 11,500 12,000 13,000 14,000 15,000 16,000 18,000 20,000 25,000 30,000 40,000 50,000 75,000 100,000

0.9238 0.9320 0.9390 0.9452 0.9552 0.9630 0.9691 0.9739 0.9809 0.9857 0.9923 0.9954 0.9981 0.9998 0.9998 1.0000

348

Chapter 9 Radiation heat transfer

Example 9.1 Blackbody emission in a wavelength band Problem What fraction of the emission from a blackbody at 5800 K lies in the visible range of the electromagnetic spectrum, i.e., in the range from 0.4 to 0.8 mm?

Solution There are different ways to solve this problem. We will first solve it using Table 9.1. The fraction in the band from l1 ¼ 0:4 mm to l2 ¼ 0:8 mm is F0/l2 T F0/l1 T l1 T ¼ ð0:4Þð5800Þ ¼ 2320 l2 T ¼ ð0:8Þð5800Þ ¼ 4640 From Table 9.1; F0/l2 T ¼ 0:5850 and F0/l1 T ¼ 0:1245 Fraction ¼ F0/l2 T F0/l1 T ¼ 0:5850 0:1245 ¼ 0:4605 46.1% of the radiation from a blackbody at 5800 K lies in the wavelength band from 0.4 to 0.8 mm. This problem can also be solved without using Table 9.1. For example, the quad function of Matlab can be used to determine the integral in Eq. (9.10) directly. For this problem, we have C1 l5 dl expðC2 =lTÞ 1 Fl1 /l2 ¼ 4 sT Putting in the values for the limits and constants for this problem, we have R l2 l1

R 0:8 F0:4/0:8 mm ¼

0:4

3:7417 108 l5 dl expð1:4388 104 =lð5800ÞÞ 1 4 ð5:67 108 Þð5800Þ

(9.10)

(9.13)

Eq. (9.13) simplifies to Z

l5 dl expð2:4807=lÞ 1 0:4 The integral in Eq. (9.14) can be determined using the following Matlab statement interactively: [S ¼ quad 0 x:^ð 5Þ: = ðexpð2:4807:=xÞ 1Þ0 ; 0:4; 0:8 (Note that the periods for element-by-element operations are needed in the Matlab statement.) The result is S ¼ 0.07907, which, using Eq. (9.14), gives the same result as we got using Table 9.1. F0:4/0:8 mm ¼ 5:8314

0:8

(9.14)

F0:4/0:8 mm ¼ ð5:8314Þð0:07907Þ ¼ 0:461 Excel could also have been used to determine the integral in Eq. (9.14). Indeed, the spreadsheet could have been programmed to determine the integral using, for example, Simpson’s rule. Such programming, however, is a bit involved, and this is a study of heat transfer, not computer programming. Therefore, we will determine the integral in an alternative way, as follows: Open an Excel spreadsheet. Our wavelength limits are 0.4e0.8. Using an increment of 0.02, we put the wavelength values in Column A, cells A1 through A21. In Column B, we put the values of the integrand in Eq. (9.14) corresponding to the wavelength values of Column A. The integrand values are in cells B1 through B21. We highlight Columns A and B and do Insert of a scatter chart. Clicking on the plot area, using the Layout tab, we Insert a trendline of a polynomial of order 3. We click the appropriate boxes to display the trendline equation and the R-squared value on the chart. Doing this, we got the trendline equation y ¼ 3:5663x3 7:3632x2 þ 4:7057x 0:7328 and an R-squared value of 0.9991, which showed that the third order polynomial was an excellent fit for the integrand. We chose the polynomial form as a polynomial is easy to integrate by hand. Integrating the integrand function, we get Z0:8 ydx ¼ 0:4

7:3632 3 4:7057 2 3:5663 4 0:8 0:44 0:8 0:43 þ 0:8 0:42 0:7328ð0:8 0:4Þ ¼ 0.0791 4 3 2

From Eq. (9.14), F0:4/0:8 mm ¼ ð5:8314Þð0:0791Þ ¼ 0:461. In conclusion, the same result was obtained using Table 9.1, Matlab, and Excel.

9.3 Radiation properties

349

9.3 Radiation properties In this section, we define the following radiation properties: emissivity (or “emittance”), reflectivity (or “reflectance”), absorptivity (or “absorptance”), and transmissivity (or “transmittance”). The surface of an ideal blackbody absorbs all radiation that hits it and emits the maximum possible radiant energy. A real surface emits less than a black surface at the same temperature. The total emissivity ε of a real surface is the ratio of the emissive power E of the surface to the emissive power Eb of a black surface at the same temperature. The emissivity of a black surface is 1. The emissivity of real surfaces is less than 1. RN EðTÞ 0 El ðl; TÞdl ¼ RN Total Emissivity ¼ εðTÞ ¼ (9.15) Eb ðTÞ 0 Ebl ðl; TÞdl This is the “total” emissivity of the surface as the emissive powers have been integrated over all wavelengths. The spectral emissivity of the surface is the ratio of the spectral emissive power of the real surface to the spectral emissive power of a black surface for the same wavelength l and same temperature T. That is, El ðl; TÞ Spectral Emissivity ¼ εl ðl; TÞ ¼ (9.16) Ebl ðl; TÞ Fig. 9.2 shows the spectral emissive powers for black, gray, and real surfaces at 2000 K. The gray surface in the figure has an emissivity of 0.6. For a real surface at a given temperature, the emissivity varies with wavelength. For a gray surface, which is an idealization like the black surface, the emissivity does not vary with wavelength. The spectral emissive power of a gray surface at all wavelengths is a constant fraction of the emissive power of the black surface at the corresponding wavelength. That is, For a gray surface: El ðl; TÞ ¼ ε Ebl ðl; TÞ (9.17) where emissivity ε is a constant. Spectral Emissive Power Eλ [W / (m 2 μ m)] 14000 12000

Black (ε =1)

10000

Gray (ε =0.6)

8000 6000

Real

4000 2000 0 0

2

4

6

Wavelength λ ( μ m)

FIGURE 9.2 Spectral emissive power for black, gray, and real surfaces.

8

10

350

Chapter 9 Radiation heat transfer

G (incident)

ρ G (reflected)

α G (absorbed)

τ G (transmitted)

FIGURE 9.3 Radiation incident on an object.

Fig. 9.3 shows radiation hitting an object. The incident radiation flux is G (W/m2). Some of the incident radiation is reflected from the surface, some is absorbed by the body, and some might be transmitted through the object. The reflectivity r is the fraction of incident radiation that is reflected. The absorptivity a is the fraction of incident radiation that is absorbed. And, the transmissivity s is the fraction of incident radiation that is transmitted through the object. Therefore, if the incident flux is G, the amount rG is reflected, aG is absorbed, and sG is transmitted. The energy balance is G ¼ rG þ aG þ sG (9.18) Factoring out G, we have 1¼r þ a þ s

(9.19)

The radiation properties in Eq. (9.19) are “total” properties, i.e., properties for all wavelengths. Eq. (9.19) could also be written for specific wavelengths. If Gl is the incident radiation flux at wavelength l, the corresponding equations for Eqs. (9.18) and (9.19) are (9.20) Gl ¼ rl Gl þ al Gl þ sl Gl and 1 ¼ rl þ al þ sl

(9.21)

If no radiation is transmitted through the object, the object is said to be “opaque” and transmissivities s and sl are zero. We now discuss Kirchoff’s law, which relates the emissivity and absorptivity of a surface. Let us put a small object inside a black enclosure that is at temperature T. When thermal equilibrium is reached, both the object and the enclosure are at temperature T. For this equilibrium, the object’s emission at a given wavelength must equal the radiation the object absorbs at that wavelength. That is, El ¼ al Gbl

(9.22)

where Gbl is the incident radiation on the object from the walls of the black enclosure. From Eq. (9.16), the spectral emissive power of the object is El ¼ εl Ebl so Eq. (9.22) becomes εl Ebl ¼ al Gbl

(9.23)

If the object inside the enclosure is a blackbody, then El ¼ Ebl and al ¼ 1 as a black surface absorbs all incident radiation. Eq. (9.22) becomes Ebl ¼ Gbl

(9.24)

9.3 Radiation properties

351

Putting Eq. (9.24) into Eq. (9.23), we arrive at εl ðl; TÞ ¼ al ðl; TÞ

(9.25)

The spectral emissivity of a surface at a given wavelength and temperature equals the spectral absorptivity of the surface at the same wavelength and temperature. For a gray surface, defined above, radiation properties are independent of wavelength. For such a surface, Eq. (9.25) becomes εðTÞ ¼ aðTÞ

(9.26)

Eq. (9.26) is also valid when the incident and emitted radiation have the same spectral distribution.

Example 9.2 Equilibrium temperature of a plate Problem

Sunlight is incident on one side of a thin metal plate at the rate of 800 W/m2. The other side of the plate is perfectly insulated. The plate has an absorptivity of 0.85 for solar wavelengths and 0.1 for long wavelengths. The air adjacent to the plate is at 15 C, and the surroundings are also at 15 C. The convective coefficient at the plate’s surface is 20 W/m2 C. What is the equilibrium temperature of the plate?

Solution

Given: asolar ¼ 0:85, along ¼ 0:1, and G ¼ 800 W = m2 . As radiation is involved, we need to use absolute temperature units. Therefore, TN ¼ Tsurr ¼ 15 C ¼ 15 þ 273.15 ¼ 288.15 K. For equilibrium, an energy balance on the plate gives: Energy Flux In ¼ Energy Flux Out Energy Flux In ¼ radiant energy absorbed by the plate ¼ asolar G Energy Flux Out ¼ convection to surrounding air þ radiation to surroundings 4 Energy Flux Out ¼ ðT TN Þ þ εs T 4 Tsurr From the energy balance, we have 4 asolar G ¼ hðT TN Þ þ εs T 4 Tsurr (9.27) The emission from the plate surface is at long wavelengths, so we have ε ¼ εlong , and by Kirchoff’s Law, Eq. (9.25), εlong ¼ along , so Eq. (9.27) becomes 4 asolar G ¼ hðT TN Þ þ along s T 4 Tsurr (9.28) Putting values into Eq. (9.28) and rearranging the equation, we get (9.29) 680 20ðT 288:15Þ ð0:1Þ 5:67 108 T 4 288:154 ¼ 0 Eq. (9.29) can be solved for the plate temperature T by trial and error, Matlab’s fzero function, or Excel’s Goal Seek. The plate temperature is found to be 321.1 K ¼ 47.9 C.

So far, we have discussed “spectral” radiation properties and “total” radiation properties. Spectral properties are wavelength-dependent. Total properties are wavelength-independent and are obtained by integrating the spectral property over a wavelength range. Total emissivity was defined in Eq. (9.15). Total absorptivity is defined as RN al Gl dl Total Absorptivity ¼ a ¼ 0R N (9.30) 0 Gl dl

352

Chapter 9 Radiation heat transfer

where al is the spectral absorptivity. Gl is the spectral distribution of the incident radiation (W/m2 mm). Total reflectivity and transmissivity are defined similarly.

Example 9.3 Total absorptivity of a plate Problem Radiation is incident on a plate’s surface with the following intensity and spectral distribution: Gl ¼ 300 W=m2 mm

for 0 < l < 2.5 mm

1200 W=m2 mm

for 2.5 mm < l < 5 mm

500 W=m2 mm

for 5 mm < l < 10 mm for l > 10 mm

0 The spectral absorptivity of the surface is al ¼ 0

for 0 < l < 1 mm

0.73

for 1 mm < l < 4 mm

0.24

for l > 4 mm

(a) What is the total absorptivity of the surface? (b) What is the rate of radiant energy absorption by the surface for the given incident radiation (W/m2)?

Solution From Eq. (9.30),

RN al Gl dl A ¼ a ¼ 0R N B 0 Gl dl

where Z

1

A¼

Z

2:5

ð0Þð300Þdl þ

0

Z

Z ð0:73Þð300Þdl þ

1 5

þ

Z

ð0:73Þð1200Þdl þ /

2:5

ð0:24Þð1200Þdl þ

4

4

10

Z

ð0:24Þð500Þdl þ

5

N

ð0Þð0:24Þdl

10

¼ 0 þ ð0:73Þð300Þð2:5 1Þ þ ð0:73Þð1200Þð4 2:5Þ þ ð0:24Þð1200Þð5 4Þ þ / Z B¼

þ ð0:24Þð500Þð10 5Þ þ 0 ¼ 2530:5 Z 5 Z 10 ð300Þdl þ ð1200Þdl þ ð500Þdl

2:5

0

2:5

5

¼ ð300Þð2:5 0Þ þ ð1200Þð5 2:5Þ þ ð500Þð10 5Þ ¼ 6250 (a) a ¼ AB ¼ 2530:5 6250 ¼ 0.405 (b)

q A

. ¼ aG ¼ ð0:405Þð6250Þ ¼ 2530 W m2

Example 9.4 Radiant transmission through a window Problem A kiln has a 6 cm diameter window in one of its walls. The kiln environment is at 1200 C and can be assumed to be black. The window has the following spectral transmissivity to radiation from a black environment at 1200 C:

9.4 Radiation shape factors

sl ¼ 0:6

353

for 0 < l < 2 mm

0.75

for 2 mm < l < 4 mm

0

for l > 4 mm

(a) What is the total transmissivity of the glass window for black radiation at 1200 C? (b) What is the rate of radiant energy transmitted through the window to the outside of the kiln?

Solution The total transmissivity for incident black radiation is R 2 mm R 4 mm RN ð0:6Þ 0 Ebl dl þ ð0:75Þ 2 mm Ebl dl sl Ebl dl s ¼ R0 N ¼s¼ sT 4 0 Ebl dl T ¼ 1200 C þ 273.15 ¼ 1473.15 K l1 T ¼ ð2Þð1473:15Þ ¼ 2946 mm K l2 T ¼ ð4Þð1473:15Þ ¼ 5893 mm K From Table 9.1; F0/l1 T ¼ 0:2610 and F0/l2 T ¼ 0:7284 (a) s ¼ ð0:6Þð0:2610Þ þ ð0:75Þð0:7284 0:2610Þ ¼ 0.507 i h (b) q ¼ sAsT 4 ¼ ð0:507Þ ðp =4Þð0:06Þ2 5:67 108 ð1473:15Þ4 ¼ 383 W

Radiation properties can also be directional, varying with the direction of the radiation to and from a surface. For example, we can have “normal” emission or reflection in a direction normal, or perpendicular, to a surface. There are also “hemispherical” radiation properties, with the properties integrated over the hemisphere from the surface, that is, in all directions from or to the surface. In this text, we will not consider directional characteristics of radiation. We will assume that radiation is hemispherical. For example, when we use the symbol εl , we mean the hemispherical spectral emissivity and when we use ε, we mean the hemispherical total emissivity. Typical hemispherical emissivities for various surfaces are given in Appendix F.

9.4 Radiation shape factors Radiative heat transfer between surfaces will be discussed in the next section. However, before doing so, we must discuss the Radiation Shape Factor, also called the Angle Factor, View Factor, or Configuration Factor. Radiation leaves surfaces by both emission and the reflection of incident radiation. Reflection from a surface can be specular, like reflection from a mirror, or diffuse with reflection over all directions. Or, it can be partially specular and partially diffuse. We will assume that both emission and reflection from a surface is diffuse, with radiation leaving a surface over all the angles of a hemisphere. In determining the transfer of radiation between surfaces, one major factor is the portion of the radiation leaving a surface that is intercepted by the other surface. This is a geometric factor that depends on the relative orientations of the two surfaces and their sizes. The radiation shape factor Fij is defined as the fraction of radiant energy leaving Surface i that hits Surface j directly. Sometimes determination of shape factors is quite easy. For example, if two infinite parallel plates (Plates 1 and 2) are exchanging radiation, then F12 ¼ 1 and F21 ¼ 1 as all the radiation from one plate will hit the other plate. As another example: Say we have two infinitely long concentric cylinders and we have radiation between the outer surface of the inner cylinder (Surface 1) and the inner surface of the outer cylinder (Surface 2). All of the radiation from Surface 1 will hit Surface 2 and F12 ¼ 1:

354

Chapter 9 Radiation heat transfer

However, not all of the radiation from Surface 2 will hit Surface 1 as some of the radiation from Surface 2 will hit Surface 2. Therefore, F21 s1: The actual value of F21 depends on the diameters of the cylinders. More complex shape factors are obtained through double integration over the two involved surfaces. Figs. 9.4e9.7 give radiation shape factors for parallel rectangles, perpendicular rectangles with a common edge, parallel coaxial disks, and concentric cylinders. Table 9.2 gives the relations that were used to generate Figs. 9.4e9.7. There is another item that needs to be discussed before we consider radiative transfer between surfaces, as follows: Let us say we have two surfaces, i and j, which are exchanging radiation. There is a relationship between the areas of the surfaces and the respective shape factors of the surfaces. This relationship, called the reciprocity theorem, is Ai Fij ¼ Aj Fji

(9.31)

b/c

1

10 4 2 1.5 1 0.8 0.6

b

0.4

a c

F1−2

0.3 0.1

0.2

2

0.15

1

0.1

0.01 0.1

1

a/c FIGURE 9.4 Radiation shape factor for parallel rectangles.

10

9.4 Radiation shape factors

c/b

0.5

b

F1−2

2 a

355

0.45

0.02 0.06 0.1

0.4

0.2

0.35

0.4

0.3

0.6

0.25

0.8 1

0.2

1.5

1 c

2 0.15

3 4 6 10

0.1 0.05 0 0.1

1

10

a/b

FIGURE 9.5 Radiation shape factor for perpendicular rectangles with a common edge. 1

8

0.9 0.8

r2

L

F1−2

5

4 3.5

3

2.5

2

0.7

2

6

0.6 0.5

1.8 1.6 1.4 1.2

0.4

1

r1

r2 /L

1 0.9 0.8 0.7 0.6 0.5 0.4 0.3

0.3 0.2 0.1 0 0.1

1

L/r1 FIGURE 9.6 Radiation shape factor for two parallel coaxial disks.

10

356

Chapter 9 Radiation heat transfer

1 0.9 0.8 0.7

F2−1

0.6

L/r2

0.5 0.4 0.3

r2 2 1

0.2

∞

3 2 1.5 1 0.8 0.6 0.4 0.3 0.2

0.1

0.1

r1

0 0

L

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0.5

0.6

0.7

0.8

0.9

1

r1/r2

1 0.9

L/r2

0.8

∞ 10

0.7

4 3

0.6

F2−2 0.5

2 1.5

0.4

1 0.8 0.6 0.4 0.3 0.2

0.3 0.2 0.1 0 0

0.1

0.2

0.3

0.4

r1/r2 FIGURE 9.7 Radiation shape factors for concentric cylinders.

Figs. 9.4e9.7 are very useful in common problems involving radiative transfer. The following example shows the usefulness of the shape factor figures for determining shape factors for other geometries. In particular, we will be using Fig. 9.5 to determine the shape factor for perpendicular, unconnected rectangles.

9.4 Radiation shape factors

Table 9.2 Radiation shape factor relations for Figs. 9.4e9.7. Fig. 9.4dParallel rectangles x ¼ a=c y ¼ b=c 8 !9 1=2 > > 1=2 1 1 þ x2 1 þ y2 x > > 2 > > > > þx 1þy tan ln > > 2 2 1=2 > > 2 1þx þy > > 1 þ y < = 2 F12 ¼ ! > pxy > > > > > > > y 2 1=2 1 1 1 > > > > þy 1 þ x tan ðxÞ ytan ðyÞ xtan > > 1=2 : ; 2 1þx Fig. 9.5dPerpendicular rectangles with common edge x ¼ a=b y ¼ c=b 0 1

F12

1

2 1=2 1 tan 1 x x þ y2

! 1

1

C B ytan ð1=yÞ þ xtan 1=2 C B x2 þ y2 C 1 B C B ¼ C py B 2 2 C B

y x 2 2 2 2 2 2 2 2 A @ 1 y 1þy þx x 1þx þy 1þy 1þx þ ln 4 ð1 þ y2 Þðy2 þ x2 Þ ð1 þ x2 Þðx2 þ y2 Þ 1 þ y2 þ x2

Fig. 9.6dParallel coaxial disks x ¼ L r1 y ¼ r2 L S ¼ 1 þ x2 1 þ y2 h i1=2 o 1n S S2 4ðxyÞ2 F12 ¼ 2 Fig. 9.7dConcentric cylinders x ¼ r1 =r2

F21

F22

y ¼ L=r2 8 8 " #1=2 99 2 > > > > 1 y2 1 y2 4 > > > > > > > > > > > > þ 21 þ4 2þ 21 2þ4 > > > > 2 > > > > x x x x x > > > > > > > > > > > > > > > > > > > > > > > > 2 < < = = 2 2

2 2 x x þ y 1 x 1 x x þ y 1 y 1 ¼x cos 1 1 þ 2 2 þ 1 sin ðxÞ > cos > p> 2y > 1 þ y2 x2 > > > > > > > > x 1 þ y2 x2 x > > > > > > > > > > > > > > > > > > > > > > > > 2 > > > > > > > > 1 y > > > > : : p ; ; þ 1 2 2 2 x x 1 2x2 3 9 2 1 x2 8 1=2 þ y2 4 2 > > 2 > 4þy x 1 4 x x2 > > 5> > > sin > > " 2 2 > > 1=2 # 2 y x = < y 1 x 2 y þ 4 2x 1x x 1 2 ¼ 1 x þ p tan 2 x2 x y 2p > x > > > > > # " > > 1=2 > > 2 > > p x 4 þ y ; : 1 2 sin 1 2x þ 1 2 2 y x

357

358

Chapter 9 Radiation heat transfer

Example 9.5 Radiation shape factor for perpendicular, unconnected rectangles Problem Determine the shape factor F14 for the Surfaces 1 and 4 in the following sketch:

6 4 3 4

2

1

2 5

8

Solution Fig. 9.5 is for perpendicular rectangles with a common edge, so it cannot be applied directly to our problem as Surfaces 1 and 4 do not have a common edge. However, the figure can be used along with shape factor algebra to determine the desired shape factor. From the definition of shape factor, F14 is the fraction of radiation from Surface 1 that hits Surface 4 directly. The fraction of radiation from Surface 1 that hits the vertical surface of Surfaces 3 and 4 is F1ð3þ4Þ , which is equal to the fraction of radiation from Surface 1 that hits Surface 3 plus the fraction of radiation from Surface 1 that hits Surface 4. That is, F1ð3þ4Þ ¼ F13 þ F14

(a)

Rearranging; we get F14 ¼ F1ð3þ4Þ F13 (b) To determine F14 , we need to get the shape factors on the right side of Eq. (b). These shape factors are not for surfaces sharing a common edge. Therefore, Fig. 9.5 cannot be used directly to get them. However, through shape factor algebra, we can convert Eq. (b) into an equation that has only the unknown factor F14 and shape factors that can be obtained from Fig. 9.5. Note that Eqs. (a) and (b) do not contain the areas of the surfaces. Such is not the case if a surface comprised of more than one surface radiates to a single surface. For example, let us consider the radiation from the combined surface of Surfaces 3 and 4 to Surface 1. That is, we want Fð3þ4Þ1 : Multiplying Eq. (a) by the area A1 of Surface 1, we get A1 F1ð3þ4Þ ¼ A1 F13 þ A1 F14

(c)

A1 F1ð3þ4Þ ¼ Að3þ4Þ Fð3þ4Þ1

(d)

From reciprocity, A1 F13 ¼ A3 F31 A1 F14 ¼ A4 F41 Substituting Eq. (d) into Eq. (c), we have Að3þ4Þ Fð3þ4Þ1 ¼ A3 F31 þ A4 F41 (e) That is, the radiation from Surfaces 3 and 4 to Surface 1 equals the radiation from Surface 3 to Surface 1 plus the radiation from Surface 4 to Surface 1. Following Eq. (e), we have, for the radiation from Surfaces 3 and 4 to Surfaces 1 and 2, Að3þ4Þ Fð3þ4Þð1þ2Þ ¼ A3 F3ð1þ2Þ þ A4 F4ð1þ2Þ

(f)

9.5 Radiative heat transfer between surfaces

359

Similarly, for the combined horizontal Surfaces 1 and 2 radiating to surface 4, we have Að1þ2Þ Fð1þ2Þ4 ¼ A1 F14 þ A2 F24

(g)

Also, F2ð3þ4Þ ¼ F23 þ F24 Substituting F24 from Eq. (h) into Eq. (g), we have Að1þ2Þ Fð1þ2Þ4 ¼ A1 F14 þ A2 F2ð3þ4Þ F23 Rearranging Eq. (f), we have

(h) (i)

(j) A4 F4ð1þ2Þ ¼ Að3þ4Þ Fð3þ4Þð1þ2Þ A3 F3ð1þ2Þ Because of reciprocity, the left sides of Eqs. (i) and (j) are equal. Hence, we can equate the right sides of the equations, which gives A1 F14 þ A2 F2ð3þ4Þ F23 ¼ Að3þ4Þ Fð3þ4Þð1þ2Þ A3 F3ð1þ2Þ (k) Rearranging Eq. (k) for the desired F14 , we get i 1 h F14 ¼ Að3þ4Þ Fð3þ4Þð1þ2Þ A3 F3ð1þ2Þ A2 F2ð3þ4Þ F23 (l) A1 All of the shape factors in Eq. (l) are for two surfaces having a common edge. Hence, Fig. 9.5 can be used for their evaluation. In summary: Through the use of shape factor algebra, we have been able to obtain an expression for the desired F14 in terms of shape factors that can be evaluated using Fig. 9.5. Looking at the figure for this problem, A1 ¼ 18; A2 ¼ 30; A3 ¼ 12; and Að3þ4Þ ¼ 24: ForshapefactorFð3þ4Þð1þ2Þ using Fig. 9.5, a ¼ 8; b ¼ 6; c ¼ 4: c=b ¼ 0:6667; a=b ¼ 1:3333; and Fð3þ4Þð1þ2Þ ¼ 0.27 from the figure. The other shape factors, from Fig. 9.5, are F3ð1þ2Þ ¼ 0:35; F2ð3þ4Þ ¼ 0:19; and F23 ¼ 0:13. Putting these values into Eq. (l), we get F14 ¼

1 ½24ð0:27Þ 12ð0:35Þ 30ð0:19 0:13Þ ¼ 0.027 18

9.5 Radiative heat transfer between surfaces Let us now discuss the radiation coming into a surface and the radiation leaving a surface. The incoming radiation is irradiation G and the outgoing radiation is radiosity J. The surface is named “Surface i.” Gi ¼ irradiation ¼ rate at which radiation is incident on surface i per unit surface area The radiation leaving a surface consists of two parts: the portion of the incident radiation that is reflected from the surface and the radiation emitted by the surface. That is, Ji ¼ ri Gi þ εi Ebi where Ji ¼ radiosity of surface i. Gi ¼ irradiation of surface i. ri ¼ reflectivity of surface i. εi ¼ emissivity of surface i. Ebi ¼ blackbody emissive power of surface i ¼ s Ti4 . The net rate of heat loss from surface i is

(9.32)

360

Chapter 9 Radiation heat transfer

qi ¼ Ai ðJi Gi Þ

(9.33)

Ji εi Ebi ri

(9.34)

where Ai is the area of surface i. Rearranging Eq. (9.32), we get Gi ¼

Substituting Gi from Eq. (9.34) into Eq. (9.33), we get Ji εi Ebi qi ¼ Ai J i ri As we have a gray surface, ri ¼ 1 εi , and Eq. (9.35) becomes Ji εi Ebi qi ¼ Ai J i 1 εi

(9.35)

(9.36)

Rearranging Eq. (9.36), we finally arrive at qi ¼

Ai ε i ðEbi Ji Þ 1 εi

(9.37)

Let us look again at Eq. (9.33): qi ¼ Ai ðJi Gi Þ

(9.33)

We would like to get Gi in terms of Ji : Let us consider the enclosure of N surfaces shown in Fig. 9.8. The N surfaces are exchanging radiant energy with each other. (It should be mentioned that these surfaces do not need to be solid surfaces. Some could be open or closed windows through which radiation passes. If such is the case, these windows would be assigned equivalent blackbody temperatures corresponding to the rate of radiation going through them.) Let us look at the radiant interactions for one of the surfaces of the enclosure, i.e., the interactions for surface i. In Section 9.4, we discussed radiation shape factors. Shape factor Fij is the fraction of radiant energy leaving surface i that hits surface j. We assume that radiation from a surface is diffuse and the radiation leaves a surface over the entire hemisphere. Fig. 9.9 shows three different types of surfaces: flat, convex, and concave. If a surface is flat or convex, none of the radiation leaving it will return to hit the surface. However, if a surface is concave, some of the radiation leaving the surface will hit the same surface. That is, for flat or convex surfaces, Fii ¼ 0. For concave surfaces, Fii s0. N

5 4

1

3 2

FIGURE 9.8 An enclosure of N surfaces.

9.5 Radiative heat transfer between surfaces

Flat

Convex

361

Concave

FIGURE 9.9 Radiation from flat, convex, and concave surfaces.

There is a very useful relation between the shape factors of the surfaces of an enclosure, as follows: Radiation leaving a surface of an enclosure will hit all of the surfaces of the enclosure that have a nonzero shape factor with the surface sending out the radiation. That is, if the enclosure has N surfaces, and Surface i is sending out the radiation, the relation is Fi1 þ Fi2 þ Fi3 þ / þ FiN ¼ 1 Using the summation symbol, this relation is

N P

Fij ¼ 1.

j¼1

Let us now look at the radiation coming into and leaving one of the surfaces of the enclosure. From previous definitions, Gi is the rate at which radiant energy is incident on Surface i per unit area of surface, and Ji is the rate at which radiant energy leaves Surface i per unit area of surface. Surface i has a surface area of Ai . Therefore, Gi Ai is the rate at which radiation is incident on Surface i. Consider the enclosure of N surfaces. Surface i gets radiation from the other surfaces and, if concave, can get radiation from itself. From the definition of radiosity, the rate of radiation leaving Surface 1 is J1 A1 , and, from the definition of shape factor, the rate at which the radiation from Surface 1 hits surface i is J1 A1 F1i . Therefore, looking at radiation coming in from all N surfaces of the enclosure to surface i, we have Gi Ai ¼ J1 A1 F1i þ J2 A2 F2i þ J3 A3 F3i þ / þ JN AN FNi

(9.38)

Using the summation symbol, Eq. (9.38) is G i Ai ¼

N X

Jj Aj Fji

(9.39)

j¼1

Substituting this into Eq. (9.33), we get qi ¼ Ai J i

N X

Jj Aj Fji

(9.40)

j¼1

We discussed earlier the Reciprocity Theorem for two Surfaces i and j, namely, Ai Fij ¼ Aj Fji Using this in the summation term of Eq. (9.40), the equation becomes

(9.41)

362

Chapter 9 Radiation heat transfer

qi ¼ Ai J i

N X

Jj Ai Fij

(9.42)

j¼1

Factoring out Ai , Eq. (9.42) becomes

0

qi ¼ Ai @Ji

N X

1 Jj Fij A

(9.43)

j¼1

From above, we also had Eq. (9.37) qi ¼

Ai ε i ðEbi Ji Þ 1 εi

(9.37)

where Ebi ¼ total emissive power of a black surface ¼ sT4i . Eqs. (9.43) and (9.37) may be written for all N surfaces of the enclosure. We have N equations from Eq. (9.43) and N equations from Eq. (9.37). These 2N simultaneous equations may be solved for the N qi ; s and the N Ji ; s. We could decrease the number of simultaneous equations from 2N to N by doing the following: each surface has two equations for q: Eqs. (9.37) and (9.43). If we equate the right sides of these equations, we get N equations for the Ji : After we solve the N equations and get the Ji , we can then use either Eq. (9.43) or Eq. (9.37) to get the heat flows qi :

9.5.1 Radiation heat transfer for a two-surface enclosure If an enclosure has only two surfaces exchanging radiation, Eqs. (9.43) and (9.37) are

9.5.1.1 For surface 1 0 q1 ¼ A1 @ J 1

2 X

1 Jj F1j A ¼ A1 ðJ1 J1 F11 J2 F12 Þ

(9.44)

j¼1

q1 ¼

A1 ε 1 ðEb1 J1 Þ 1 ε1

(9.45)

9.5.1.2 For surface 2 0 q2 ¼ A2 @ J 2

2 X

1 Jj F2j A ¼ A2 ðJ2 J1 F21 J2 F22 Þ

(9.46)

j¼1

q2 ¼

A2 ε 2 ðEb2 J2 Þ 1 ε2

(9.47)

9.5 Radiative heat transfer between surfaces

363

Eqs. (9.44) through (9.47) may be solved simultaneously for q1 ; q2 ; J1 ; and J2 : The results for the heat flows are s T14 T24 and q2 ¼ q1 (9.48) q1 ¼ 1 ε1 1 1 ε2 þ þ A1 F12 ε 1 A1 ε2 A 2 Note that Eq. (9.48) only contains one shape factor. The other shape factors have been eliminated by applying the shape factor relations F11 þ F12 ¼ 1 and F21 þ F22 ¼ 1. The reciprocity theorem A1 F12 ¼ A2 F21 and the StefaneBoltzmann Law Eb ¼ sT 4 were also used in reaching the final Eq. (9.48). As a special case, if shape factor F12 ¼ 1, then Eq. (9.48) becomes sA1 T14 T24 q1 ¼ (9.49) 1 A1 1 ε2 þ ε1 A2 ε2 Common applications of Eq. (9.49) are the following: (A) Small object in a large enclosure

q1

A1 ε1 T1

Conditions : A1 A2 and F12 ¼ 1 or F12 ¼ 1 and ε2 ¼ 1ðblack surfaceÞ q1 ¼ ε1 sA1 T14 T24 (B) Infinite parallel plates q1

A2 ε 2 T2 A1 ε 1 T1

Conditions : A1 ¼ A2 ¼ A and F12 ¼ 1

(9.50)

364

Chapter 9 Radiation heat transfer sA T14 T24 q1 ¼ 1 1 þ 1 ε1 ε2

(9.51)

(C) Infinitely long concentric cylinders r2

A2 ε2 T2

r1

q1

A1 ε1 T1

Conditions : A1 = A2 ¼ ð2pr1 LÞ=ð2pr2 LÞ ¼ r1 =r2 and F12 ¼ 1 A1 ¼ 2pr1 L where L ¼ length of cylinders sA1 T14 T24 q1 ¼ 1 r1 1 ε2 þ ε1 r2 ε2

(9.52)

(D) Concentric spheres q1

r1 r2

A2 ε2 T2

A1 ε1 T1

Conditions : A1 = A2 ¼ 4pr12 = 4pr22 ¼ ðr1 =r2 Þ2 F12 ¼ 1 A1 ¼ 4pr12 sA1 T14 T24 q1 ¼ 2 1 r1 1 ε2 þ ε1 r2 ε2

(9.53)

9.5 Radiative heat transfer between surfaces

365

Example 9.6 Radiant heat transfer for a cubical storage container Problem A thin-walled cubical container that is 0.5 m by 0.5 m by 0.5 m is filled with ice water. The container is inside another cubical container and the 3 cm gap between the containers is a vacuum. The outer surface of the inner container has an emissivity of 0.1 and a temperature of 1 C. The inner surface of the outer container has an emissivity of 0.3 and a temperature of 19 C. (a) What is the rate of heat transfer to the ice water? (b) If the container initially contains 1/2 ice and 1/2 water by volume, how long will it take for all the ice to be melted?

Solution This is a two-surface radiation problem. Let Surface 1 be the outer surface of the inner cube and Surface 2 be the inner surface of the outer cube. A1 ¼ ð6Þð0:5Þ2 ¼ 1:5 m2 A2 ¼ ð6Þ½0:5 þ 2ð0:03Þ2 ¼ 1:8816 m2 ε1 ¼ 0:1 and T1 ¼ 1 C ¼ 274.15 K ε2 ¼ 0:3 and T2 ¼ 19 C ¼ 292.15 K (a) All of the radiation leaving the inner cube will hit the outer cube. Therefore F12 ¼ 1. Hence, Eq. (9.49) is the appropriate equation for this two-surface problem. sA1 T14 T24 (9.49) q1 ¼ 1 A1 1 ε2 þ ε1 A2 ε2 Putting values into Eq. (9.49), we have sA1 T14 T24 5:67 108 ð1:5Þ 274:154 292:154 ¼ ¼ 11:73 W q1 ¼ 1 A1 1 ε2 1 1:5 1 0:3 þ þ ε1 0:1 1:8816 0:3 A2 ε2 The minus sign for q1 means that the heat flow is into Surface 1, that is, into the ice water. The rate of heat transfer to the ice water is 11.73 W. (b) The volume of the container is (0.5 m)3 ¼ 0.125 m3. The volume of the ice is half of this or 0.0625 m3. The density of ice is 916.7 kg/m3, so the initial mass of ice in the container is m1ce ¼ 916:7 kg=m3 0:0625 m3 ¼ 57:3 kg The heat of fusion of ice is 333.6 kJ/kg, so the amount of heat needed to melt all of the ice in the container is (333.6 kJ/kg) x (57.3 kg) ¼ 19115 kJ. From Part (a), q ¼ 11:73 W ¼ 11:73 J=s. The time need to melt all the ice is 19115 103 J 1h 1 day ¼ 1:630 106 s ¼ 18:9 days 11:73 J=s 3600 s 24 h It will take about 19 days to melt all the ice. An alternative way to get equations to solve for the radiosities and heat flows is to write nodal equations using the electrical circuit diagram. We assume that heat flows are directed into the nodes, and the sum of the heat flows into a node has to equal zero for steady state. Looking at Fig. 9.10, we have the following equations for nodes J1 and J2: t¼

Eb1 J1 J J1 þ 2 ¼0 Node J1 : 1 ε1 1 A1 F12 ε1 A1 Eb2 J2 J J2 þ 1 ¼0 Node J2 : 1 ε2 1 A1 F12 ε2 A2

(9.54)

(9.55)

366

Chapter 9 Radiation heat transfer

q1

Eb1 = σ T14

J1

Eb2 = σ T24 q2

J2

1– ε 1

1

1– ε 2

ε 1A1

A1F1–2

ε 2A2

FIGURE 9.10 Resistive circuit for a two-surface enclosure. Eq. (9.54) and (9.55) can be solved simultaneously for J1 and J2 : Eq. (9.44) or Eq. (9.45) may then be used to get q1 : Or Eq. (9.46) or Eq. (9.47) may be used to get q2 : When the heat flow from one surface is known, it is easy to get the heat flow for the other surface, as, for a two-surface enclosure, q1 ¼ q2 :

The Electric-Heat Analogy and the Resistance Concept were discussed in Chapter 3. Radiant exchange in a two-surface enclosure can be modeled by the resistive circuit shown in Fig. 9.10. The end potentials are Eb1 ¼ sT14 and Eb2 ¼ sT24 : Intermediate potentials are J1 and J2 :

Example 9.7 Radiant heat transfer for a semicylindrical furnace Problem A very long semicylindrical furnace has a radius of 2 m. The curved top surface, whose emissivity is 0.75, has built-in radiant heaters that provide a uniform heat flux of 8000 W/m2 to the furnace space. The bottom surface is at 150 C and has an emissivity of 0.9. (a) What is the temperature of the curved top surface? (b) What is the rate of heat flow to the bottom surface per meter length of the furnace? q1 = 8000 W/m2 A1 ε 1 = 0.75 1 ε 2 = 0.9

T2 = 150 C 2 4m Conditions: A1 A2 and F12 ¼ 1 or F12 ¼ 1 and ε2 ¼ 1ðblack surfaceÞ q1 ¼ ε1 sA1 T14 T24

Solution This is a two-surface problem as the furnace is very long. If the furnace had been short, some radiation would have escaped through the ends of the furnace, and the problem would have been a three-surface or four-surface problem. (a) The applicable equation for this two-surface problem is Eq. (9.48). s T14 T24 (9.48) q1 ¼ 1 ε1 1 1 ε2 þ þ A1 F12 ε1 A1 ε2 A2 For this problem, ε1 ¼ 0:75, ε2 ¼ 0:9, and T2 ¼ 150 C ¼ 423.15 K. q1 = A1 ¼ 8000 W=m2 The surface areas are A1 ¼ prL and A2 ¼ 2rL.where r ¼ radius and L ¼ length of the furnace.

9.5 Radiative heat transfer between surfaces

367

p Hence, A1 A2 ¼ prL 2rL ¼ 2 ¼ 1:5708. Looking at Eq. (9.48), we need shape factor F12 : For radiation from the bottom Surface 2, we have F21 þ F22 ¼ 1: However F22 ¼ 0 as Surface 2 is flat. Therefore, F21 ¼ 1: From reciprocity, A1 F12 ¼ A2 F21 and A2 1 ð1Þ ¼ 0:6366 F21 ¼ 1:5708 A1 Dividing both sides of Eq. (9.48) by A1, we have F12 ¼

s T14 T24 q1 ¼ 1 1 ε2 A1 A1 1 ε1 þ þ F12 ε1 ε2 A2

(9.56)

Putting values into Eq. (9.56), we have 8000 ¼

h i 5:67 108 T14 ð423:15Þ4 1 0:75 1 1 0:9 þ þ ð1:5708Þ 0:75 0:6366 0:9

(9.57)

Solving Eq. (9.57) for T1 , we get T 1 ¼ 755 K ¼ 482 C. The top surface of the furnace is at 482 C. (b) The heat flux from Surface 1 is Aq11 ¼ 8000 W m2 and the area of Surface 1 is A1 ¼ prL ¼ pð2ÞL ¼ 6:2832L. q1 ¼ ð8000Þð6:2832Þ ¼ 50265 W m length. Therefore, L By energy conservation, the rate of heat flow from Surface 1 is equal to the rate of heat flow to Surface 2. Hence, the bottom surface of the furnace (Surface 2) receives a heat flow of 50,265 W/m length.

9.5.2 Radiation heat transfer for a three-surface enclosure If an enclosure has three surfaces exchanging radiation, Eqs. (9.43) and (9.37) are

9.5.2.1 For surface 1 0 q1 ¼ A1 @ J 1

3 X

1 Jj F1j A ¼ A1 ðJ1 J1 F11 J2 F12 J3 F13 Þ

(9.58)

j¼1

q1 ¼

A1 ε 1 ðEb1 J1 Þ 1 ε1

(9.59)

9.5.2.2 For surface 2 0 q2 ¼ A2 @ J 2

3 X

1 Jj F2j A ¼ A2 ðJ2 J1 F21 J2 F22 J3 F23 Þ

(9.60)

j¼1

q2 ¼

A2 ε 2 ðEb2 J2 Þ 1 ε2

(9.61)

368

Chapter 9 Radiation heat transfer

9.5.2.3 For surface 3 0 q3 ¼ A3 @J3

3 X

1 Jj F3j A ¼ A3 ðJ3 J1 F31 J2 F32 J3 F33 Þ

(9.62)

j¼1

q3 ¼

A3 ε 3 ðEb3 J3 Þ 1 ε3

(9.63)

Eqs. (9.58) through (9.63) may be solved simultaneously for q1 ; q2 ; q3 ; J1 ; J2 ; and J3 : Radiant exchange in a three-surface enclosure can be modeled by the resistive circuit shown in Fig. 9.11. The end potentials are Eb1 ¼ sT14 ; Eb2 ¼ sT24 ; and Eb3 ¼ sT34 : Intermediate potentials are J1 ; J2 ; and J3 : An alternative way to get equations to solve for the radiosities and heat flows is to write nodal equations using the electrical circuit diagram. We assume that heat flows are directed into the nodes, and the sum of the heat flows into a node has to equal zero for steady state. Looking at Fig. 9.11, we have the following equations for nodes J1 ; J2 ; and J3 : Eb1 J1 J J1 J J1 þ 2 þ 3 ¼0 Node J1 : 1 ε1 1 1 A1 F12 A1 F13 ε 1 A1 Eb2 J2 J J2 J J2 þ 1 þ 3 ¼0 Node J2 : 1 ε2 1 1 A1 F12 A2 F23 ε 2 A2 Eb3 J3 J J3 J J3 þ 1 þ 2 ¼0 Node J3 : 1 ε3 1 1 A1 F13 A2 F23 ε 3 A3

(9.64)

(9.65)

(9.66)

Eqs. (9.64) through (9.66) can be solved simultaneously for J1 ; J2 ; and J3 : Eqs. (9.58) through (9.63) may then be used to get the heat flows q1 ; q2 ; and q3 :

q1

Eb1 = σ T14

J1

1 A1 F1–2

1– ε 1

ε 1A1

Eb2 = σ T24 q2

J2 1– ε 2

1 A1 F1–3

J3

ε 2A2 1 A2 F2–3 1– ε 3

ε 3A3 Eb3 = σ T34 q3

FIGURE 9.11 Resistive circuit for a three-surface enclosure.

9.5 Radiative heat transfer between surfaces

369

Example 9.8 Heat transfer from a radiant heater Problem A radiant heater on the ceiling of a factory is used to heat a work table directly under it. Both the heater and table top are 2 m by 2 m. The heater is at 500 C and has an emissivity of 0.9. The table top is at 30 C and has an emissivity of 0.5. The temperature of the factory is 15 C. The distance between the heater and the table top is 3.6 m. (a) What is the required power output of the radiant heater? (b) What is the radiant heat flux received by the table top?

Solution This is a three-surface problem. Surface 1 is the heater; Surface 2 is the table top; and Surface 3 is the factory environment. The electrical circuit diagram is

Eb1

J1

1 A1 F1–2

Eb2 1– ε 2

1– ε 1

ε 1A1

J2

1 A1 F1–3

ε 2A2 1 A2 F2–3 J3 = Eb3

(Note: Fig. 9.11 shows that the resistance between J3 and Eb3 is normally ð1 ε3 Þ=ðε3 A3 Þ: However, in this problem, the area of Surface 3 is very large. Hence, the resistance between J3 and Eb3 is negligible. Hence, J3 ¼ Eb3 . Alternatively, we could have considered the factory environment, i.e., Surface 3, to be black.) Eb1 ¼ sT14 ¼ 5:67 108 ð500 þ 273:15Þ4 ¼ 20260 Eb2 ¼ sT24 ¼ 5:67 108 ð30 þ 273:15Þ4 ¼ 478:9 Eb3 ¼ sT34 ¼ 5:67 108 ð15 þ 273:15Þ4 ¼ 390:9 1 ε1 1 0:9 ¼ 0:02778 ¼ ε1 A1 ð0:9Þð4Þ

1 ε2 1 0:5 ¼ 0:25 ¼ ð0:5Þð4Þ ε2 A2

We can get shape factor F12 from Fig. 9.4 with a ¼ b ¼ 2 and c ¼ 3.6. From the figure, we get F12 ¼ 0:080. From reciprocity, F21 ¼ 0:080: Looking at the electrical circuit, we also need F13 and F23 : Shape factor equations for the three-surface enclosure are F11 þ F12 þ F13 ¼ 1 F21 þ F22 þ F23 ¼ 1 F31 þ F32 þ F33 ¼ 1 From the first equation, F11 ¼ 0 as the surface is flat; F13 ¼ 1 F12 ¼ 1 0:08 ¼ 0:92. From the second equation, F22 ¼ 0 as the surface is flat; F23 ¼ 1 F21 ¼ 1 0:08 ¼ 0:92. Looking at the electrical circuit, we see that we need three additional resistances. They are 1 1 1 1 1 ¼ 0:272 and ¼ 3:125 ¼ ¼ ¼ A1 F13 A2 F23 ð4Þð0:92Þ A1 F12 ð4Þð0:08Þ Redrawing the electrical circuit to show the resistance values, we have

370

Chapter 9 Radiation heat transfer

0.02778 J1

3.125

J2

0.25

Eb1 = 20260

Eb2 = 478.9 0.272

0.272 Eb3 = J3 = 390.9

The nodal equations are 20260 J1 J2 J1 390:9 J1 þ þ ¼0 0:02778 3:125 0:272 J1 J2 478:9 J2 390:9 J2 þ þ ¼0 For Node J2 : 3:125 0:25 0:272 Solving the nodal equations simultaneously for J1 and J2 ; we get For Node J1 :

J1 ¼ 18280 and J2 ¼ 1150 Eb1 J1 20260 18280 ¼ 71270 W ¼ 0:02778 0:02778 The required power output of the radiant heater is 71.3 kW. (a) q1 ¼

Eb2 J2 478:9 1150 ¼ 2684 W ¼ 0:25 0:25 (Note: Negative value means that heat flow is inward to the surface.) (b) q2 ¼

q2 2684 ¼ 671 W=m2 ¼ 4 A2 The radiative heat flux to the table top is 671 W/m2.

9.5.2.4 Three-surface enclosure with an insulated surface Sometimes surfaces of an enclosure are perfectly insulated on their back side. For these surfaces, the energy incident on the surface is equal to the energy leaving it. That is, the irradiation G equals the radiosity J. As shown by Eq. (9.33), the heat loss q from the surface is zero. This adiabatic surface is also called a reradiating or refractory surface. If surface i is the adiabatic surface, then Gi ¼ Ji : The definition of radiosity Ji is Ji ¼ ð1 εi ÞGi þ εi Ebi . As Gi ¼ Ji , it follows that Ji ¼ Ebi ¼ sTi4 for the adiabatic surface. Fig. 9.12 shows the electrical circuit diagram for a three-surface enclosure with Surface 3 being the perfectly insulated surface.

Eb1 = σ T14

J1

1 A1F1–2

1– ε 1

ε 1A1

Eb2 = σ T24

J2 1– ε 2

1

1

A1F1–3

A2F2–3

ε 2A2

J3 = Eb3 = σ T34

FIGURE 9.12 Resistive circuit for a three-surface enclosure with Surface 3 adiabatic.

9.5 Radiative heat transfer between surfaces

371

The nodal equations (Eqs. 9.64e9.66) still apply, with the first term of Eq. (9.66) deleted as J3 ¼ Eb3 : Once the equations are solved and J3 has been determined, the temperature of the insulated surface may be found as J3 ¼ Eb3 ¼ sT34 : It should be noted that the emissivity of the insulated surface is irrelevant and does not factor into the calculations. Looking at the electrical circuit of Fig. 9.12, one can obtain the following expression for the heat flow from Surface 1 to Surface 2: s T14 T24 q1 ¼ q2 ¼ (9.67) 1 ε1 1 1 ε2 þ þ ε 1 A1 ε 2 A2 A1 F12 þ 1 þ1 1 A1 F13

A2 F23

Example 9.9 illustrates the solution of a problem involving three surfaces, with one of the surfaces being perfectly insulated.

Example 9.9 Radiant heat transfer for a truncated cone with an insulated surface Problem A truncated cone has a height of 4 cm. The closed bottom of the cone has a diameter of 4 cm, a temperature of 800 C, and an emissivity of 0.8. The top surface has a diameter of 3.5 cm and is open to a large room that is at 25 C. The side wall of the cone is perfectly insulated and has an emissivity of 0.2. (a) What is the temperature of the side wall? (b) What is the rate at which heat is transferred into the room?

Solution This is a problem involving a three-surface enclosure. The bottom and side surfaces are closed surfaces, and the top surface is an open window. The side surface is a perfectly insulated or “refractory” surface.

d2 2

3 H

d1 d1 ¼ 0:04 m Shape factor F12

ε1 ¼ 0:8

1 T1 ¼ 800 C ¼ 1073.15 K

d2 ¼ 0:035 m H ¼ 0:04 m is determined from Fig. 9.6. For this figure,

ε3 ¼ 0:2

r1 ¼ 0:02 r2 ¼ 0:0175

L ¼ 0:04

L=r1 ¼ 0:04=0:02 ¼ 2

r2 =L ¼ 0:0175=0:04 ¼ 0:44 and F12 ¼ 0:14

F11 þ F12 þ F13 ¼ 1 F11 ¼ 0 since flat surface; and F13 ¼ 1 F12 ¼ 1 0:14 ¼ 0:86

372

Chapter 9 Radiation heat transfer

p p A1 ¼ d12 ¼ ð0:04Þ2 ¼ 1:257 103 m2 4 4 p 2 p A2 ¼ d2 ¼ ð0:035Þ2 ¼ 9:621 104 m2 4 4 qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ A3 ¼ pðr1 þ r2 Þ H 2 þ ðr1 r2 Þ2 ¼ 4:722 103 m2 From reciprocity, A2 F21 ¼ A1 F12 A1 1:257 103 F12 ¼ ð0:14Þ ¼ 0:183 A2 9:621 104 F21 þ F22 þ F23 ¼ 1 F21 ¼

F22 ¼ 0 as flat surface; and F23 ¼ 1 F21 ¼ 1 0:183 ¼ 0:817 The electrical circuit diagram for this problem is

Eb1

J1

1 A1F1–2

J2 = Eb2 = σT24

1– ε 1

ε 1A1

1 A1F1–3

1 A2F2–3 J3 = Eb3 = σ T34

Eb1 ¼ sT14 ¼ 5:67 108 ð800 þ 273:15Þ4 ¼ 75201 Eb2 ¼ sT24 ¼ 5:67 108 ð25 þ 273:15Þ4 ¼ 448 1 ε1 1 0:8 ¼ 198:9 ¼ ε1 A1 ð0:8Þð1:257 103 Þ 1 1 ¼ 925 ¼ A1 F13 ð1:257 103 Þð0:86Þ 1 1 ¼ 5682 ¼ A1 F12 ð1:257 103 Þð0:14Þ 1 1 ¼ 1272 ¼ A2 F23 ð9:621 104 Þð0:817Þ Nodal Equations. Eb1 J1 J J1 J J1 þ 2 þ 3 ¼0 Node J1 : 1 ε1 1 1 A1 F12 A1 F13 ε1 A1 J1 J3 J J3 þ 2 ¼0 1 1 A1 F13 A2 F23

Node J3 :

(a)

(b)

Putting in the known values, For Eq. (a), 75201 J1 448 J1 J3 J1 þ þ ¼0 198:9 5682 925

(c)

9.6 Radiation shields

373

For Eq. (b), J1 J3 448 J3 þ ¼0 (d) 925 1272 Solving Eq. (c) and (d) simultaneously, we get J1 ¼ 66863 and J3 ¼ 38901. 0:25 0:25 (a) 38901 J3 ¼ Eb3 ¼ sT34 T3 ¼ Js3 ¼ 5:6710 ¼ 910:1 K ¼ 637 C 8 The side wall is at 637 C. (b) The side wall is adiabatic. Therefore, the heat flow into the room equals the heat flow from the bottom surface of the truncated cone. This heat flow is Eb1 J1 75201 66863 ¼ ¼ 41:9 W q1 ¼ 1 ε1 198:9 ε1 A1 The rate of radiant heat flow into the room is 41.9 W. Below is another circuit diagram for this problem. We have put the resistance and potential values on the diagram. We have also included the heat flows for the various surfaces. It is seen that the side wall is indeed adiabatic, with heat in ¼ heat out. Alternatively, Eq. (9.67), with the last term in the denominator deleted, could have been used to get this result.

66863 5682 J1

(cone bottom) 198.9

11.69 .2 30

30.2 3

925

Eb2 = 448 (room) 3

Eb1 = 75201 41.92

1272

J3 = Eb3 = 38901 (side wall)

9.6 Radiation shields Sometimes we want to decrease the radiant heat transfer between two surfaces. This can be done by inserting a thin sheet of material between the two surfaces. The material can be in the form of a plate, cylinder, sphere, or any other shape appropriate to the surfaces being affected. Usually, the added sheet has low emissivities on its two surfaces. The inserted sheet is called a radiation shield. Multiple layers of shields are used in insulations for cryogenic fluid containers. Shields are also used to improve the accuracy of temperature measurement. For example, say we are measuring the temperature of a gas flowing through a pipe. If the temperature of the pipe wall is different from that of the gas, the measured gas temperature may be significantly in error due to radiative transfer between the pipe wall and the temperature sensor. Placing a radiation shield between the sensor and the pipe wall will reduce the measurement error. Let us look at a couple of examples that illustrate the effectiveness of radiation shields in decreasing radiant transfer between surfaces. Fig. 9.13 shows a radiation shield between two large plane surfaces. The left surface (Surface 1) has a temperature T1 and an emissivity ε1 : The right surface (Surface 2) has temperature T2 and emissivity ε2 : The radiation shield has emissivity ε31 on its side facing Surface 1 and emissivity ε32 on its side facing Surface 2.

374

Chapter 9 Radiation heat transfer

Radiation Shield

Plate1

Plate2

3

1

ε1 T1

2

ε2 T2

T3

ε 31

ε 32

FIGURE 9.13 Radiation shield between two large plates.

Let us look at the impact of the radiation shield on the radiative transfer between Surfaces 1 and 2. The heat transfer from Surface 1 to Surface 2 without the shield in place is given by Eq. (9.51). sA T14 T24 q1 ¼ (9.51) 1 1 þ 1 ε1 ε2 For steady heat transfer, with the shield in place, the heat transfer from Surface 1 to the shield equals the heat transfer from the shield to Surface 2. That is, q1/3 ¼ q3/2 ¼ q1/2

(9.68)

sA T14 T34 sA T34 T24 q1/3 ¼ ¼ q3/2 ¼ 1 1 1 1 þ 1 þ 1 ε1 ε31 ε32 ε2

(9.69)

Using Eq. (9.51), we have

With temperatures T1 and T2 and all the emissivities known, Eq. (9.69) may be solved for unknown temperature T3 . Through algebra, it is found that T3 is 0:25 1 1 1 3 2 1 4 þ 1 T1 þ þ 1 T24 ε ε2 ε1 ε31 5 (9.70) T3 ¼ 4 32 1 1 1 1 þ þ þ 2 ε1 ε2 ε31 ε32 After T3 is determined, the heat flow for the system can be obtained from Eq. (9.69). If there are multiple radiation shields, equations like Eq. (9.69) can be written between adjacent shields and the resulting equations can be solved simultaneously for the temperatures of the shields. Then one of the heat flow equations can be used to get the heat flow through the system.

9.6 Radiation shields

q1

Eb1

J1

1– ε 1

ε 1A1

J31 1

Eb3 1– ε 31 1– ε 32

A1F1–3 ε 31A3

J32

J2 1

Eb2

375

q2

1– ε 2

ε 32A3 A3F3–2 ε 2A2

FIGURE 9.14 Circuit diagram for a one radiation shield system.

Another way to get the heat flow is through use of the electrical circuit for the system. The circuit has resistances in series as the heat flow is the same through all segments of the system. Fig. 9.10 showed the circuit diagram for radiative transfer between two surfaces. There are two surface resistances and one space resistance. Addition of a radiation shield between the surfaces adds two surface resistances and one space resistance to the diagram. The resulting circuit for one radiation shield is shown in Fig. 9.14. If there are more than one shield between the end surfaces, additional surface and space resistances can be added to the circuit for the additional shield(s). We can use this circuit diagram like we previously used an electrical circuit diagram when we discussed conduction through composite materials. Using the circuit diagram of Fig. 9.14, the heat flow is the overall difference in potential (i.e., difference in emissive power Eb1 Eb2 ¼ s T14 T24 ) divided by the sum of the resistances between Eb1 and Eb2 . That is, s T14 T24 (9.71) q1/2 ¼ 1 ε1 1 1 ε31 1 ε32 1 1 ε2 þ þ þ þ þ A1 F13 A3 F32 ε 1 A1 ε31 A3 ε32 A3 ε 2 A2 Eq. (9.71) may also be used for concentric cylinders and concentric spheres. For parallel plates, all of the areas are equal and all the shape factors in Eq. (9.71) are unity. Eq. (9.71) then reduces to sA T14 T24 (9.72) q1/2 ¼ 1 1 1 1 þ 1 þ þ 1 ε1 ε2 ε31 ε32 Let us say we want to find the temperature T3 of the shield. The heat flow is the same through all segments of the system and we can write the resistance equation for q between any two points in the circuit of Fig. 9.14. Therefore, if we want the temperature of the shield, we can write an equation between the unknown potential Eb3 and one of the known end potentials (Eb1 or Eb2 ). For example, between Eb1 and Eb3 we have s T14 T34 (9.73) q1/2 ¼ 1 ε1 1 1 ε31 þ þ A1 F13 ε1 A 1 ε31 A3 and between Eb3 and Eb2 , we have q1/2 ¼

s T34 T24 1 ε32 1 1 ε2 þ þ A3 F32 ε32 A3 ε 2 A2

(9.74)

376

Chapter 9 Radiation heat transfer

Equating the right sides of Eqs. (9.73) and (9.74), we have s T14 T34 s T34 T24 ¼ 1 ε1 1 1 ε31 1 ε32 1 1 ε2 þ þ þ þ A1 F13 A3 F32 ε 1 A1 ε31 A3 ε32 A3 ε 2 A2

(9.75)

Eq. (9.75) may be solved for its only unknown, the shield temperature T3 : Example 9.10 illustrates the reduction of radiant heat transfer between surfaces through the use of radiation shields.

Example 9.10 Radiant heat transfer for a parallel plate system with a radiation shield Problem Two large parallel plates are exchanging radiation. One plate has a temperature of 1000 K and an emissivity of 0.8. The other plate is at 350 K and has an emissivity of 0.4. (a) What is the radiant heat flux between the plates? (b) It is desired to reduce the heat transfer between the plates by inserting a thin radiation shield between them. The shield has an emissivity of 0.1 on one side and 0.05 on the other. What is the new rate of heat transfer after the shield is in place, and what is the temperature of the shield? (c) An additional shield like the one of Part (b) is placed in the system to further reduce the heat transfer. What is the resulting heat flux, and what are the temperatures of the two shields?

Solution

(a) ε1 ¼ 0:8; ε2 ¼ 0:4; T1 ¼ 1000 K; T2 ¼ 350 K. Using Eq. (9.51), 5:67 108 10004 3504 q1 s T14 T24 ¼ 20310 W=m2 ¼ ¼ 1 1 1 1 A þ 1 þ 1 ε1 ε2 0:8 0:4 (b) Consider Fig. 9.13. For this problem, ε1 ¼ 0:8; ε2 ¼ 0:4; ε31 ¼ 0:1; and ε32 ¼ 0:05. T1 ¼ 1000 K and T2 ¼ 350 K. Using Eq. (9.70), 1 1 1 1 1 1 2 1 30:25 " 1 þ 1 10004 þ þ 1 3504 #0:25 þ 1 T14 þ þ 1 T24 ε ε2 ε1 ε31 0:05 0:4 0:8 0:1 5 ¼ T3 ¼ 4 32 1 1 1 1 1 1 1 1 þ þ þ 2 þ þ þ 2 ε1 ε2 ε31 ε32 0:8 0:4 0:1 0:05 T3 [ 908.8 K Now that the shield temperature is known, one of the relations in Eq. (9.69) may be used to get the heat flux. s T34 T24 q1/2 s T14 T34 ¼ ¼ 1 1 1 1 A þ 1 þ 1 ε1 ε31 ε32 ε2 Using the first relation, we have q1/2 5:67 108 10004 908:84 ¼ 1760 W=m2 ¼ 1 1 A þ 1 0:8 0:1

9.6 Radiation shields

377

Alternatively, Eq. (9.72), which was obtained from the circuit diagram, could have been used to get the heat flux without having to first find the shield temperature. s T14 T24 5:67 108 10004 3504 q1/2 ¼ 1760 W=m2 ¼ ¼ 1 1 1 1 1 1 1 1 A þ 1 þ þ 1 þ 1 þ þ 1 ε1 ε2 ε31 ε32 0:8 0:4 0:1 0:05 The temperature of the shield is 908.8 K and the heat flux is 1760 W/m2. (c) A two-shield system is shown in Fig. 9.15 and the circuit diagram for the system is shown in Fig. 9.16. Let us look at Fig. 9.15. For this problem, ε1 ¼ 0:8; ε2 ¼ 0:4; ε31 ¼ 0:1; ε32 ¼ 0:05; ε41 ¼ 0:1; and ε42 ¼ 0:05. T1 ¼ 1000 K and T2 ¼ 350 K. We can use the circuit diagram of Fig. 9.16 to get equations to solve for q1/2 ; T3 ; and T4 : The heat flow is the difference in overall Eb divided by the resistances between the end Eb s. That is, s T14 T24 (9.76) q1/2 ¼ 1 ε1 1 1 ε31 1 ε32 1 1 ε41 1 ε42 1 1 ε2 þ þ þ þ þ þ þ þ A1 F13 A3 F34 A4 F42 ε1 A1 ε31 A3 ε32 A3 ε41 A4 ε42 A4 ε2 A2 For this problem, A1 ¼ A2 ¼ A3 ¼ A4 ¼ A and all the shape factors are unity. Eq. (9.76) then reduces to sA T14 T24 (9.77) q1/2 ¼ 1 1 1 1 1 1 þ þ þ þ þ 3 ε1 ε31 ε32 ε41 ε42 ε2

Plate 1

Radiation Shields

3

Plate 2

4

2

1

ε1 T1

T4

T3

ε 31

ε 32

ε 41

ε2 T2 ε 42

FIGURE 9.15 Two-shield system. Eb1 q1

J1

J31

Eb3

J32

J41

1– ε 1

1

1– ε 31

1– ε 32

ε 1A1

A1F1–3

ε 31A3

ε 32A3 A3F3–4

FIGURE 9.16 Circuit diagram for two-shield system of Fig. 9.15.

1

Eb4

J42

J2

Eb2

1– ε 41

1– ε 42

1

1– ε 2

ε 41A4

ε 42A4

A4F4–2

ε 2A2

q2

378

Chapter 9 Radiation heat transfer

2 Putting in values for the parameters of Eq. (9.77), we get q1/2 A ¼ 919 W m . We can also use Fig. 9.16 to get the shield temperatures. For T3 , we can write an equation either between nodes Eb1 and Eb3 or between nodes Eb3 and Eb2 : The equation between nodes Eb1 and Eb3 is s T14 T34 (9.78) q1/2 ¼ 1 ε1 1 1 ε31 þ þ A1 F13 ε1 A1 ε31 A3 With equal areas and unity shape factors, Eq. (9.78) becomes q1/2 s T14 T34 ¼ (9.79) 1 1 A þ 1 ε1 ε31 Using the values we have for q1/2 A , T1 , ε1 ; and ε31 , we can solve Eq. (9.79) for shield temperature T3 : It is found that T3 ¼ 955:6 K. For shield temperature T4 , we can write an equation between nodes Eb4 and Eb2 of Fig. 9.16. This equation is s T44 T24 (9.80) q1/2 ¼ 1 ε42 1 1 ε2 þ þ A4 F42 ε42 A4 ε2 A2 With equal areas and unity shape factors, Eq. (9.80) becomes q1/2 s T44 T24 ¼ (9.81) 1 1 A þ 1 ε42 ε2 Using the values we have for q1/2 A , T2 , ε42 ; and ε2 , we can solve Eq. (9.81) for shield temperature T4 : It is found that T4 ¼ 776:5 K. For the two-shield system, the heat flux is 919 W/m2 and the shield temperatures are 955.6 and 776.5 K. To summarize: Without radiation shields, the heat flux between the two plates is 20310 W/m2. With one shield, the heat flux is 1760 W/ m2. With two shields, the heat flux is 919 W/m2. It is seen that use of one shield greatly decreases the heat flux. Use of an additional shield further decreases the heat flux but not to as great an extent as achieved by the first shield.

We mentioned at the beginning of this section that radiation shields are effective in increasing the accuracy of temperature measurements in certain situations. Example 9.11 illustrates this.

Example 9.11 Radiation shield for temperature sensor Problem (a) A thermocouple is used to measure the temperature of hot air flowing through a large duct. The emissivity of the thermocouple is 0.6 and the duct walls are at a temperature of 500 K. The convective coefficient at the thermocouple’s surface is 100 W/m2 C. The thermocouple indicates an air temperature of 1025 K. What is the actual temperature of the air?

thermocouple Ttc = 1025 K Air Tair = ?

500 K

9.6 Radiation shields

379

Solution (a) From an energy balance on the thermocouple at steady state: The heat gained by convection from the air equals the heat lost by radiation to the duct walls. That is, 4 hAðTair Ttc Þ ¼ εsA Ttc4 Tduct Putting in values, we have 100ðTair 1025Þ ¼ 0:6 5:67 108 10254 5004 Solving this we get Tair ¼ 1379 K. The actual air temperature is 1379 K, but the thermocouple indicates a temperature of 1025 K. This difference of 354 K is due to the duct walls being at a much lower temperature than the air There is radiative transfer from the thermocouple to the duct walls, thereby lowering the thermocouple’s temperature. The relatively high emissivity of the thermocouple also contributes to the large difference between the measured and actual air temperature.

Problem (b) A thin cylindrical shield is placed over the thermocouple to shield the thermocouple from the duct walls. The shield has an emissivity of 0.1 on both of its surfaces, and the convective coefficient is 50 W/m2 C on both surfaces of the shield. Determine the effect of the shield on the accuracy of the thermocouple reading.

Shield

Air Tair = 1379 K

500 K

thermocouple Ttc = ?

Solution (b) We will realistically assume that the surface area of the thermocouple is very small compared with the inner surface area of the shield and that the outer surface area of the shield is small compared with the surface area of the duct walls. The heat balance on the shield is 4 εtc sAtc Ttc4 Ts4 þ 2hs As ðTair Ts Þ ¼ εs sAs Ts4 Tduct where As is the surface area of one side of the shield. Dividing this equation by area As , we have 4 εtc sðAtc = As Þ Ttc4 Ts4 þ 2hs ðTair Ts Þ ¼ εs s Ts4 Tduct As Atc As , the first term of this equation can be deleted, which leaves us with 4 2hs ðTair Ts Þ ¼ εs s Ts4 Tduct Putting in values, we have 2ð50Þð1379 Ts Þ ¼ 0:1 5:67 108 Ts4 5004 . Solving for shield temperature Ts , we get Ts ¼ 1246 K. The heat balance on the thermocouple is htc Atc ðTair Ttc Þ ¼ εtc sAtc Ttc4 Ts4 Putting in values, we have ð100Þð1379 Ttc Þ ¼ ð0:6Þ 5:67 108 Ttc4 12464 . Solving for the thermocouple temperature, we get Ttc ¼ 1281 K. The actual air temperature is 1379 K and the indicated temperature by the thermocouple is 1281 K, giving an error of 98 K. Use of the radiation shield reduced the thermocouple error from 354 to 98 K. This is a very significant improvement.

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Chapter 9 Radiation heat transfer

9.7 Sky radiation and solar collectors Radiant emission from the sun has a spectral distribution similar to that of a blackbody at about 5800 K. The sun radiates about 3.8 1023 kW of energy, of which about 1.7 1014 kW is intercepted by the earth. At the outer edge of the earth’s atmosphere, the solar irradiance, at normal incidence, is about 1360 W/m2. This is called the “Solar Constant.” As the radiation goes through the atmosphere, its intensity is decreased due to absorption and scattering by gases, particulates, and water vapor. On clear, sunny days, the solar insolation at the surface of the earth can be as high as about 1000 W/m2. It is much lower on cloudy or smoggy days. Let us now briefly discuss solar collectors. A solar collector will typically have a cover plate in contact with the environment and an absorber plate deeper within the device. We want as much radiation as possible to enter the collector and the least possible amount to leave it. For the cover plate, therefore, we want high transmittance for solar radiation and a low transmittance for radiation at the infrared wavelengths of the collector. Fortunately, several plastics and glasses have these characteristics, including polycarbonate, plexiglass, clear low-iron oxide glass. For float glass and tempered glass, the solar transmittance is about 0.8 and the infrared transmittance is estimated at 0.2 [1]. For the absorber plate, we want high absorptance and low emittance. This is often achieved by putting coatings on the absorber plates. One very good coating is black chrome on copper, aluminum, or stainless steel. This coating has an absorptance of about 0.87 and an emittance of about 0.08. The coating must be commercially applied. However, there are spray coatings available, which can be applied by the user, which also have high absorptance and low emittance. One might think that flat black paint would be a very good coating for an absorber plate. Such is not the case. Indeed, the paint has a high absorptance (about 0.95), but it unfortunately also has a high emittance (about 0.9) [1]. In predicting the performance of solar collectors and systems, it is necessary to have data on the amount of solar radiation reaching a particular location. Such data are available from the National Renewable Energy Laboratory in Golden, Colorado. One of its publications [2] gives the solar radiation incident on many US cities. The publication gives incident solar radiation (kWh/m2/day) for several types of solar collectors: flat-plate collectors (fixed, 1-axis tracking, 2-axis tracking) and concentrating collectors. As an example for New York City, for a fixed flat-plate collector facing south, the average solar radiation is 1.6 kWh/m2 for a day in December and 6.0 kWh/m2 for a day in June. We have mentioned scattering and absorption of solar radiation by gases and particulates. Such materials also emit radiation. The spectral emission is unlike blackbody radiation, but it is often useful to treat the atmosphere as a blackbody having a temperature called the “effective sky temperature.” This temperature depends on parameters such as altitude, humidity, cloud coverage, and particles in the air such as dust and pollution. The sky temperature is lower than the air temperature at the earth’s surface. The sky temperature is typically between about 230 and 290 K. The lower temperatures are for cold, cloudless conditions while the higher temperatures are for warm, cloudy conditions. Discussion of an interesting (and fun) phenomenon closes this chapter. The author has observed ice in the street in the morning when the air temperature overnight has been above freezing. This can be explained by radiative heat transfer. The phenomenon is illustrated in Example 9.12.

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Example 9.12 Water freezing in street Problem There are shallow puddles of water in the street during daytime. At night, the average air temperature is 5 C and the effective sky temperature is 250 K. The convective coefficient between the water and the air is 10 W/m2 C and the emissivity of the water surface is about 0.95. Will the water freeze overnight?

Solution For equilibrium: Heat into the water by convection from the air ¼ heat out of the water by radiation to the sky 4 4 Tsky hAðTair Twater Þ ¼ εsA Twater Putting in values, we have 4 2504 ð10Þð278:15 Twater Þ ¼ ð0:95Þ 5:67 108 Twater Solving for the water temperature, we get Twater ¼ 270:4 K. As Twater < 273:15 K; the water will freeze.

9.8 Chapter summary and final remarks In this chapter, we discussed the fundamentals of radiation heat transfer. We first considered emission from blackbodies. We looked at the wavelength distribution of the emission as given by Planck’s law. We discussed Wien’s law and its importance with respect to the “greenhouse effect.” We defined radiation properties (emissivity, reflectivity, absorptivity, and transmissivity), irradiation, radiosity, and the gray body. We discussed radiation shape factors that are used in determining the radiant heat transfer between surfaces. We covered radiant exchange between gray, diffuse surfaces of an enclosure, with special emphasis on two- and three-surface enclosures. We also observed the usefulness of electrical circuit diagrams in determining the equations for such radiant exchange. We discussed the use of radiation shields to decrease radiative transfer between surfaces. Finally, we briefly discussed sky radiation and solar collectors. In short, we covered a lot of material. However, there are more-advanced topics of radiation heat transfer that were not covered. These include the angular dependence of radiation properties, radiative exchange for specular surfaces, gaseous radiation, and radiation transmission through media that absorb and scatter radiation. Information on these topics may be obtained from the references in the "Further reading" section at the end of the chapter.

9.9 Problems Notes: •

• •

If needed information is not given in the problem statement, then use the Appendix for material properties and other information. If the Appendix does not have sufficient information, then use the Internet or other reference sources. Some good reference sources are mentioned at the end of Chapter 1. If you use the Internet, double-check the validity of the information by using more than one source. Your solutions should include a sketch of the problem. Unless specifically mentioned, the problems only involve radiant heat transfer, and there is no convection heat transfer.

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9-1 What is the total emissive power of a blackbody at 1500 C? 9-2 What percentage of the emission from a blackbody at 2000 K lies in the wavelength band of 2 to 6 mm? 9-3 What is the rate of emission from a 85 F Gy body with ε ¼ 0.5 between wavelengths 3 to 15 mm? 9-4 Radiation from a 600 C blackbody is incident on a window. The window transmits 85% of incident radiation between the wavelengths 0.3 to 3.5 mm. What heat flux (W/m2) will be transmitted through the window? 9-5 A surface at 1500 K has the following spectral emissivities: For 0 < l < 1.5 mm, εl ¼ 0.2, For 1.5 < l < 4 mm, εl ¼ 0.45, For 4 < l < N mm, εl ¼ 0.65. (a) What is the total emissivity ε of the surface? (b) What is the emissive power of the surface? 9-6 Obtain Wien’s law (Eq. 9.6) by differentiating Eq. (9.4) with respect to wavelength l and setting the result to zero. 9-7 A square black surface is at 800 C. It is 40 cm by 40 cm. (a) What is the rate (W) at which the surface emits thermal radiation? (b) What is the spectral emissive power of the surface at a wavelength of 8 mm? 9-8 Three possible glass cover plates are available for a solar collector. Plate A has a transmissivity of 0.85 for wavelengths of 0.25 to 1.5 mm and zero transmission for wavelengths outside this range. Plate B has a transmissivity of 0.95 for wavelengths of 0.4 to 1 mm and zero transmission for wavelengths outside this range. Plate C has a transmissivity of 0.72 for all wavelengths. Which plate will have the greatest transmission of solar radiation into the collector? Assume the sun is a blackbody at 5800 K. 9-9 A glass window has a transmissivity of 0.87 for wavelengths between 0.2 and 3.5 mm. It is opaque outside this wavelength range. For a blackbody source at 1500 K, what is the radiant heat flux transmitted through the window? 9-10 A thermos bottle consists of two thin concentric cylindrical walls with a vacuum between them. The inner cylinder has a diameter of 10 cm and the outer cylinder has a diameter of 11.5 cm. The thermos is 30 cm high. The inner cylindrical container holds coffee at 70 C, and the outer cylinder is essentially at the room temperature of 20 C. Both wall surfaces facing the vacuum space have an emissivity of 0.1. What is the rate of heat transfer from the coffee? 9-11 A thermistor temperature sensor is spherical in shape, 4 mm diameter, and is used to measure the temperature of dry air flowing in a 25 cm diameter pipe. The thermistor’s surface has an emissivity of 0.9 and the emissivity of the pipe wall is 0.8. The convective coefficient between the thermistor and the air is 50 W/m2 C. The pipe wall is at 250 C, and the sensor’s temperature is 540 C. What is the temperature of the air? 9-12 A flat-plate solar collector has an absorptivity of 0.91 and an emissivity of 0.12. The plate is square, 5 m by 5 m, and the convective coefficient at the plate’s surface is 4 W/m2 C. The solar insolation on the plate is 750 W/m2, the plate’s temperature is 60 C, and the air temperature is 16 C. (a) What is the rate of energy absorbed by the plate? (b) What is the plate’s collection efficiency?

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383

9-13 A gray body having area A1 ¼ 1:5 m2 and emissivity ε1 ¼ 0:8 is being heated in a rectangular furnace having dimensions 3 m by 3.5 m by 2.5 m. The inside surface of the furnace has an emissivity ε2 ¼ 0:65: The surface of the body has a temperature of 750 C, and the inside surface of the furnace has a temperature of 250 C. The shape factor between the body’s surface and the enclosure is 1. (a) What is the rate of radiant heat transfer from the gray body to the enclosure? (b) What would be the % error if we had assumed that A1 A2 in calculating the solution to Part (a)? 9-14 A furnace is cylindrical in shape: 2 m long and 1.5 m diameter. There is a window at one end of the furnace. The window is on the axis of the furnace and is 25 mm diameter. A pyranometer placed at the window receives 2.3 W of radiation from the opposite end of the furnace. The temperature of the pyranometer is 60 C. What is the temperature of that opposite end of the furnace? 9-15 A cubic oven has inside dimensions 0.45 m by 0.45 m by 0.45 m. The emissivity of all six walls of the oven is 0.7. One wall of the oven is at 300 C. The other walls of the oven are at 150 C. What is the rate of radiant heat transfer from the 300 C wall? 9-16 A square room has floor and ceiling dimensions of 3.5 m by 3.5 m and a height of 2.5 m. The floor is at 300 K, the ceiling is at 285 K, and the walls are perfectly insulated. The emissivity of all surfaces is 0.7. (a) What is the rate of radiant heat transfer from the floor? (b) What is the rate of radiant heat transfer to the ceiling? (c) What is the temperature of the walls? 9-17 An environmental testing chamber is rectangular, with a 4 m by 5 m floor and walls 3 m high. One 5 m by 3 m side wall is at 250 C and the floor is at 80 C. The other four surfaces of the chamber are perfectly insulated. If all the surfaces of the chamber are black, what is the net radiant heat transfer between the heated wall and the floor? 9-18 A long horizontal pipe 10 cm in diameter runs through a space with air at 20 C. The pipe surface is at 90 C and has an emissivity of 0.65. What is the radiant heat loss from the pipe per meter length? 9-19 A horizontal pipe of 6.35 cm outer diameter and 8 m length runs through a large room where the air temperature is 20 C. The outer surface of the pipe is at 110 C. The walls of the room are at 25 C and have an emissivity of 0.3. The convective coefficient at the outer surface of the pipe is 3 W/m2 C and the surface has an emissivity of 0.5. What is the total rate of heat loss from the pipe by convection and radiation? 9-20 A long solid cylinder has a diameter of 2.5 cm. Its temperature is 500 C and its surface has an emissivity of 0.3. A long thin-walled cylinder is placed concentrically over the first cylinder. This second cylinder has a diameter of 6 cm and an emissivity of 0.15 on both of its surfaces. The two-cylinder arrangement is in a vacuum chamber whose surfaces are at 28 C. (a) What is the rate of energy lost by the inner cylinder? (b) What is the equilibrium temperature of the outer cylinder? 9-21 Energy is being transferred from one spaceship to another. This is done by two square plates 2 m by 2 m; one plate on each spaceship. The ships are maneuvered so that the plates are parallel and 40 cm apart. One plate has emissivity 0.6 and temperature 800 C. The other plate

384

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9-23

9-24

9-25

9-26

9-27

9-28

Chapter 9 Radiation heat transfer

has emissivity 0.85 and temperature 250 C. Assume that the space environment is equivalent to a blackbody at 0 K. (a) What is the net rate of energy transferred from one spaceship to the other? (b) What is the rate of energy loss from the hot plate? A vertical cylindrical furnace has dimensions of 80 cm diameter by 50 cm high. A heater in the bottom of the furnace produces a heat input of 3000 W to the furnace. The bottom surface of the furnace has an emissivity of 0.8. The top surface of the furnace has an emissivity of 0.55 and is maintained at 425 K. The side wall of the furnace is nonconducting and reradiating. What are the temperatures of the bottom and the side of the furnace? Two steel plates are welded together, perpendicular to each other, with a common edge. Plate No. 1 is 25 cm by 50 cm and has a temperature of 600 C and an emissivity of 0.55. Plate No. 2 is 40 cm by 50 cm, has an emissivity of 0.4, and is perfectly insulated on its back side. The common edge between the plates is 50 cm long. The inner surfaces of the plates face into a large room at 25 C. (a) What is the rate of heat lost by Plate No. 1? (b) What is the temperature of Plate No. 2? Two large plates are parallel and facing each other. One plate has temperature 1000 K and emissivity 0.3. The other plate has temperature 400 K and emissivity 0.6. A thin radiation shield having an emissivity of 0.04 on both sides is placed between the two plates. (a) What is the heat flux (W/m2) between the plates without the shield? (b) What is the heat flux with the shield in place? (c) What is the temperature of the shield when it is between the plates? A long duct has a rectangular cross section. The height is 30 cm and the width is 50 cm. The top surface has a heat flux of 1000 W/m2 and an emissivity of 0.3, and the bottom is perfectly insulated. One side wall has temperature 250 C and emissivity 0.45. The other side wall has temperature 150 C and emissivity 0.7. (a) What are the temperatures of the top and bottom surfaces? (b) What are the heat fluxes for the side walls? An environmental test chamber is cubic in shape, 7 m by 7 m by 7 m. The floor is electrically heated, with a uniform heat flux of 500 W/m2 and emissivity 0.8. The ceiling is perfectly insulated. The side walls are all at 300 K and have an emissivity of 0.3. (a) What are the temperatures of the floor and ceiling? (b) What are the heat fluxes for the four side walls? A long duct has a cross section of an equilateral triangle. The sides of the triangle are 1.5 feet long. One side of the duct is perfectly insulated. Another side has emissivity 0.2 and temperature 400 F. The third side has emissivity 0.5 and temperature 800 F. (a) What is the temperature of the insulated side? (b) What are the heat flows per foot length for the two other sides? A room is rectangular in shape, with the floor and ceiling being 3 m by 5 m and the height being 2.5 m. The floor is electrically heated and has a temperature of 35 C and an emissivity of 0.75. The side walls are perfectly insulated and have an emissivity of 0.8. The ceiling has an emissivity of 0.9. It is poorly insulated and has a temperature of 15 C. (a) What is the temperature of the side walls? (b) What is the required power input to the floor heater to keep the floor at 35 C?

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385

9-29 Two very long parallel plates are facing each other. Plate 1 is at 600 K and the surface facing Plate 2 has an emissivity of 0.6. Plate 2 is at 400 K and its surface facing Plate 1 has an emissivity of 0.3. A thin radiation shield (Plate 3) is placed between the other two plates. This plate has sides A and B, which are of different emissivities. When side A faces Plate 1, Plate 3 has a temperature of 520 K. When side B faces Plate 1, Plate 3 has a temperature of 567 K. (a) What is the heat flux between Plates 1 and 2 without the radiation shield in place? (b) What are the emissivities of sides A and B of the radiation shield? (c) When side A of the shield is facing Plate 1, what is the heat flux for the three-plate system? (d) When side B of the shield is facing Plate 1, what is the heat flux for the three-plate system? 9-30 A plane surface having temperature T1 and emissivity ε1 is radiating to a large enclosure at temperature T2 : A thin, plane radiation shield is placed between the plane surface and the enclosure’s surface. The side of the shield facing the plane surface has emissivity ε2 and the side of the shield facing the enclosure’s surface has emissivity ε3 : (a) Show that, with the shield in place, the heat flux from the plane surface to the enclosure is s T14 T24 q12 ¼ ð1=ε1 Þ þ ð1=ε2 Þ þ ð1=ε3 Þ 1 A

9-31

9-32

9-33

9-34

9-35

(b) If ε1 ¼ 0:4; ε2 ¼ 0:15; ε3 ¼ 0:05; T1 ¼ 500 C; and T2 ¼ 25 C; what is the heat flux from the plane surface to the enclosure and what is the temperature of the shield? A rectangular oven has a floor and ceiling 3.75 m by 5.25 m and walls of 4 m height. The upper half of the four walls are radiant heaters having temperature 600 C and emissivity 0.9. The floor has a temperature of 100 C and an emissivity of 0.8. The ceiling and the lower unheated halves of the walls are perfectly insulated. What is the rate of radiant heat transfer to the floor? Two concentric, parallel disks having diameters of 0.5 m are 10 cm apart and are in a large room that is at 20 C. One disk has a temperature of 250 C and an emissivity of 0.7. The other disk is perfectly insulated on its back side. (a) What is the rate of radiant heat transfer from the heated disk? (b) What is the temperature of the insulated disk? A long electric cylindrical heater has a diameter of 1.5 cm and produces 40 W per meter length. It is on the centerline of a concentric tube of 2.5 cm inside diameter that is at 30 C. The space between the heater and the tube is evacuated so there is only radiant heat transfer. The surface of the heater has an emissivity of 0.8 and the inner surface of the tube has an emissivity of 0.7. What is the surface temperature of the heater? A thin-walled spherical tank has an outer diameter of 1 m and is filled with liquid nitrogen at 77 K. The tank is enclosed within another spherical tank of inner diameter 1.04 m, and the space between the tanks is a vacuum so that radiative heat transfer is the only mode of heat transfer between the tanks. The emissivities of the outer surface of the inner tank and the inner surface of the outer tank are both 0.08. The temperature of the outer tank is 20 C. (a) What is the rate of heat transfer to the LN2? (b) What is the rate of evaporation of the LN2 if the heat of vaporization of the liquid nitrogen is 200 kJ/kg? A long steam pipe passes through the center of a long rectangular duct. The outer surface of the pipe is at 200 C and has an emissivity of 0.8. The duct walls are at 25 C and have an emissivity of 0.6. The OD of the pipe is 10 cm and the duct cross section is 45 cm by 60 cm. What is the radiation loss from the pipe per meter length?

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9-36 An incandescent light bulb has a tungsten filament that can be modeled as a gray body at 2800 K. What fraction of the radiant energy emitted by the filament is in the visible wavelengths of 0.4e0.8 mm? (Note: This problem shows the inefficiency of incandescent light bulbs. Most of the energy emitted by the tungsten filament of the bulb is not in the visible region of the electromagnetic spectrum. Rather, it is in the infrared region.) 9-37 Two concentric, parallel disks are radiating between themselves and the space between them. The disks are 0.3 m apart. Disk 1 has a diameter of 0.7 m, temperature 700 K, and emissivity 0.9. Disk 2 has a diameter of 0.5 m, temperature 450 K, and emissivity 0.6. The space between them may be considered as a black medium at 300 K. (a) What is the rate of radiative heat transfer from Disk 1 to Disk 2? (b) What is the rate of heat transfer to the environment between the disks? 9-38 A thin flat plate is 1 m by 1 m. One side of the plate is exposed to solar radiation and also convection to the adjacent 300 K environment with a convective coefficient of 15 W/m2 C. The other side of the plate is perfectly insulated. The side of the plate facing the sun has an absorptivity of 0.7 for long-wavelength radiation and an absorptivity of 0.93 for solar radiation. If the solar irradiation of the plate is 1000 W/m2, what is the temperature of the plate? 9-39 A shallow pan of water is placed outdoors on a night when the effective sky temperature is 230 K. The air temperature is 5 C, which is above the freezing temperature of 0 C for water. The water’s surface has a convective coefficient of 30 W/m2 C and an emissivity of 0.8. Will the water in the pan freeze? 9-40 An object has an emissivity of 0.5 between wavelengths 0.1 and 3 mm and an emissivity of 0.2 between wavelengths 3 and 10 mm. If the object is at 4000 K, what is its total emissive power? 9-41 Regular window glass has a transmissivity of 0.9 for wavelengths from 0.2 to 3 mm. Tinted glass has a transmissivity of 0.9 for wavelengths from 0.5 to 1 mm. Assume that the transmissivity outside these wavelength ranges is zero. Consider the sun to be a blackbody at 5800 K. (a) What is the fraction of incident solar energy that is transmitted through each glass? (b) What is the fraction of incident solar energy in the visible range of 0.4 to 0.8 mm that is transmitted through each glass? 9-42 Hot air is flowing through a 15 cm diameter duct. A thermistor is used to measure the air temperature. The thermistor is spherical in shape. It has a diameter of 2 mm and an emissivity of 0.8. The surface of the thermistor has a convective coefficient of 100 W/m2 C. The surface of the duct has a temperature of 150 C and an emissivity of 0.7. (a) If the thermistor indicates a temperature of 325 C, what is actual temperature of the air? (b) It is decided to place a radiation shield between the thermistor and the duct wall to decrease the error between the thermistor reading and the actual air temperature. The shield is a thin-walled cylinder of 2 cm diameter and 4 cm length. Both sides of the shield have an emissivity of 0.15 and a convective coefficient of 80 W/m2 C. Assuming that the shield does not affect the convective coefficient for the thermistor, what will be the thermistor reading for the actual gas temperature found in Part (a)?

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9-43 A hemispherical furnace has a flat bottom surface of 3 m diameter. The curved top surface has a temperature of 800 C and an emissivity of 0.9. The bottom surface has a temperature of 100 C and an emissivity of 0.6. What is the rate of radiative heat transfer from the top surface to the bottom surface? 9-44 A horizontal disk has a diameter of 8 cm. There is a horizontal ring parallel to and above the disk. The ring is concentric with the disk and 10 cm above it. The ring has an inner diameter of 12 cm and an outer diameter of 14 cm. If the disk is Surface 1 and the ring is Surface 2, what is the radiation shape factor F12 ? 9-45 An evacuated multilayered insulation is used for a liquid nitrogen tank. The outer layers of the insulation are aluminum with an emissivity of 0.15. They are at temperatures 80 and 300 K. There are four internal polished aluminum shields, each of which has an emissivity of 0.04 on both sides. The shields are 1.5 mm apart and the total thickness of the insulation is 7.5 mm. (a) What is the radiant heat transfer across the insulation? (b) What is the effective conductivity and R-value for the insulation? 9-46 A long duct has a square cross section, 45 cm on a side. The four inside surfaces of the duct have different boundary conditions as follows: Surface 1 has a temperature of 750 K and an emissivity of 0.75. Surface 2 is adjacent to Surface 1 and has a temperature of 500 K and an emissivity of 0.6. Surface 3, which is across from Surface 1, has a heat flux of 450 W/m2 and an emissivity of 0.4. Surface 4, which is across from Surface 2, is perfectly insulated. (a) What are the heat flows per meter length of duct for Surfaces 1 and 2? (b) What are the temperatures of Surfaces 3 and 4? 9-47 A solar collector has a metal absorber plate in contact with the atmospheric air with water tubes directly behind the absorber plate. Solar radiation is incident on the plate at a rate of 700 W/m2. The solar absorptance of the plate is 0.89 and the emittance is 0.12. The adjacent air is at 22 C and the effective sky temperature is 15 C. The convective coefficient for the plate is 8 W/m2 C. The absorber plate’s temperature is 65 C. What is the rate of heat transfer to the tubes behind the absorber plate per square meter of plate surface area? 9-48 The solar collector of Problem 9-47 is not in service and no heat is being removed by the water in the tubes behind the plate. What is the temperature of the absorber plate? 9-49 A glass window has a transmissivity of 0.91 for wavelengths of 0.25 to 4 mm. It has zero transmissivity for other wavelengths. (a) What is the total transmissivity of the window for solar radiation? Assume blackbody radiation at 5800 K. (b) What is the total transmissivity of the glass for room temperature radiation? Assume blackbody radiation at 300 K. (c) If solar radiation is incident on the window at a rate of 700 W/m2, what is the rate of radiation transmission through the window? 9-50 Radiation incident on a surface has the following intensity: Gl ¼ 400 W=m2 mm

for 0 < l < 2 mm

1500 W=m2 mm

for 2 mm < l < 4 mm

600 W=m mm 0

for 4 mm < l < 7 mm for l > 7 mm

2

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The spectral absorptivity of the surface is al ¼ 0 0.85 0.12

for 0 < l < 3 mm for 3 mm < l < 5 mm for l > 5 mm

(a) What is the total absorptivity of the surface? (b) What is the rate of radiant energy absorption by the surface for the given incident radiation (W/m2)? 9-51 A kiln has a 7 cm diameter window in one of its walls. The kiln environment is at 1800 C and may be assumed to be black. The room environment is at 23 C and may be assumed to be black. The window has the following radiation properties: For l < 3 mm; εl ¼ 0:1; sl ¼ 0:88; rl ¼ 0:02 For l > 3 mm; εl ¼ 0:6; sl ¼ 0:15; rl ¼ 0:25 (a) Assuming conduction and convection are negligible, what is the temperature of the window? (b) What is the net radiant heat transfer through the window from the furnace to the room? 9-52 It is desired to pick the glass for the cover of a solar collector that has the greatest transmission of solar radiation. The sun is modeled as a blackbody at 5800 K. The two options are as follows: Glass A sl ¼ 0

for 0 < l < 0:3 mm

0.9

for 0.3 mm < l < 3 mm

0

for l > 3 mm

Glass B sl ¼ 0

for 0 < l < 0:5 mm

0.95

for 0.5 mm < l < 5 mm

0

for l > 5 mm

Which glass should be chosen? 9-53 It is desired to pick the coating for the absorber plate of a solar collector that has the greatest absorption of solar radiation. The sun is modeled as a blackbody at 5800 K. The three options are as follows: Coating A al ¼ 0:9 0.15 0

for 0 < l < 3 mm for 3 mm < l < 8 mm for l > 8 mm

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Coating B al ¼ 0:15 0.9 0

for 0 < l < 3 mm for 3 mm < l < 8 mm for l > 8 mm

Coating C al ¼ 0:72 for all wavelengths Which coating should be chosen?

References [1] F. Kreith, A. Rabl, Solar energy, in: W.M. Rohsenow, J.P. Hartnett, E.N. Ganic (Eds.), Handbook of Heat Transfer Applications, second ed., McGraw-Hill, 1985. Chapter 7. [2] Solar Radiation Data Manual for Flat-Plate and Concentrating Collectors, National Renewable Energy Laboratory (Golden, CO), NREL/TP-463-5607, April 1994. Available online, rredc.nrel.gov/solar/pubs/ redbook/.

Further reading [1] [2] [3] [4] [5] [6]

R. Siegel, J.R. Howell, Thermal Radiation Heat Transfer, third ed., Hemisphere, 1993. E.M. Sparrow, R.D. Cess, Radiation Heat Transfer, Hemisphere, 1978. H.C. Hottel, in: W.C. McAdams (Ed.), Heat Transmission, third ed., McGraw-Hill, 1954. Chpt. 4. H.C. Hottel, R.B. Egbert, Radiant heat transmission from water vapor, AIChE Trans 38 (1942) 531e565. C.L. Tien, Thermal radiation properties of gases, Adv. Heat Tran. 5 (1968) 254e321. D.K. Edwards, R. Matovosian, Scaling rules for total absorptivity and emissivity of gases, J. Heat Transf. 106 (1984) 685e689. [7] E.R.G. Eckert, Radiation relations and properties, in: W.M. Rohsenow, J.P. Hartnett (Eds.), Handbook of Heat Transfer, Section 15, McGraw-Hill, 1973.

CHAPTER

Multimode heat transfer

10

Chapter outline 10.1 10.2 10.3 10.4 10.5

Introduction ...............................................................................................................................391 Procedure for solution of multimode problems ............................................................................. 392 Examples ...................................................................................................................................392 Chapter summary and final remarks ............................................................................................. 422 Problems ...................................................................................................................................423

10.1 Introduction Like essentially all heat transfer texts, the organization of this text is segmental. The first chapter is general, and presents an introduction to heat transfer and its three modes: conduction, convection, and radiation. Next are three chapters concentrating on conduction. There are two later chapters on convection, and then a chapter on radiation. This organization is very good from a pedagogical viewpoint. However, concentrating on a single mode at a given time is not very realistic. Most problems involve more than one mode of heat transfer, and these modes act simultaneously. For example, heat transfer through a building wall involves conduction through the wall, convection at the surfaces of the wall, and sometimes radiation from the wall surfaces, all acting together. At the time of the early chapters on conduction, the details of convection had not yet been discussed. Therefore, values of convection coefficients had to be given in the problem statements. Examples and problems in this chapter are designed to be much more realistic, and convective coefficients are not provided. Indeed, for real engineering problems, an engineer has to determine the convective coefficients by his/her self. If quick, approximate answers are sufficient, the engineer can use estimates based on prior experience. If more accurate results are needed, then the appropriate equations can be used to determine the coefficients. At this point in the text, the reader has the capability to determine convective coefficients by his/her self. Problems can be more like “real world” problems. The problems in this chapter involve multiple modes of heat transfer. The reader has to determine which modes of heat transfer are relevant for a given problem. He/she also has to somehow obtain any needed information that is lacking in the problem statement. In addition, a decision must be made on the method to be used for the solution of the problem.

Heat Transfer Principles and Applications. https://doi.org/10.1016/B978-0-12-802296-2.00010-X Copyright © 2021 Elsevier Inc. All rights reserved.

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Chapter 10 Multimode heat transfer

Good engineers are resourceful and persistent. This chapter should enhance the reader’s problemsolving abilities. It should also increase the reader’s confidence in being able to solve heat transfer problems. Finally, the examples and problems in this chapter should provide the reader with a worthwhile review of material covered earlier in the book.

10.2 Procedure for solution of multimode problems In solving the problems, the following questions must be answered: What modes of heat transfer (conduction, convection, and/or radiation) are significant? If convection is significant, is it natural convection or is it forced convection? Is the problem steady-state or transient? Does the problem statement contain all necessary information? If not, where will the missing information be obtained? How should the problem be solved? Should it be solved analytically or numerically? Does the problem involve differential equations? Does the problem involve simultaneous equations? What software, if any, should be used in the solution? After these questions are answered, the solution can proceed in earnest.

10.3 Examples The following eight examples each incorporate more than one mode of heat transfer. The examples are based on earlier examples and problems in this book. At the time they were earlier discussed, concentration was focused on one specific mode of heat transfer. For example, Chapters 2, 3, and 4 dealt with conduction. For problems in these chapters involving convection boundary conditions, the values of the convective coefficients were givens in the problems. Now, with the experience of Chapters 6 and 7, the reader has the ability to determine convective coefficients by his/her self. There is no need for the convective coefficients to be given. The below examples involve the solution of nonlinear simultaneous equations. We have found that Microsoft Excel’s Solver add-in is very useful. (This add-in program comes with Excel. It just has to be enabled.) Sometimes convergence depends on the initial guesses of the variables. However, we have found Solver to be successful in almost all situations. We have provided details regarding the Excel spreadsheets used in the solution of the examples. We have also provided listings of the Matlab programs used for solution of some of the examples. Property information was needed for the determination of forced and natural convective coefficients for air. In particular, we needed the temperature variation of air’s thermal conductivity, kinematic viscosity, and Prandtl number. We took tabular data for these parameters and fit the data to polynomial functions. This was done through use of Excel spreadsheets. Tabular data were input to the spreadsheet, scatter charts were inserted, and polynomial trendlines and equations were obtained. We found that third-degree polynomial fits were excellent for thermal conductivity and kinematic

10.3 Examples

393

viscosity. A fourth-degree polynomial was needed to accurately fit the data for the Prandtl number. The obtained equations were: Thermal conductivity of air (0e1000 C): kðW = m CÞ and TðCÞ: k ¼ 4:629 1012 T 3 2:520 108 T 2 þ 7:561 105 T þ 0:02364 (10.1) kðW = mKÞ and TðKÞ: k ¼ 4:629 1012 T 3 2:899 108 T 2 þ 9:041 105 T þ 0:001009 (10.2) Kinematic viscosity of air (0e1000 C): yðm2 = sÞ and TðCÞ: y ¼ 1:725 1014 T 3 þ 8:853 1011 T 2 þ 8:947 108 T þ 1:329 105 (10.3) yðm2 = sÞ and TðKÞ: y ¼ 1:725 1014 T 3 þ 1:027 1010 T 2 þ 3:725 108 T 4:196 106 (10.4) Prandtl number of air (0e1000 C): TðCÞ: Pr ¼ 2:916 1013 T 4 8:291 1010 T 3 þ 8:653 107 T 2 3:387 104 T þ 0:7371 (10.5) TðKÞ: Pr ¼ 2:916 1013 T 4 1:148 109 T 3 þ 1:675 106 T 2 1:021 103 T þ 0:9126 (10.6)

Example 10.1 Cylindrical heater Problem A horizontal electric cylindrical heater has an output of 50 W. The heater has a diameter of 4 cm and a length of 30 cm. The ends of the cylinder are insulated so all of the heat output leaves through the cylindrical side of the heater. The heater is in a large room whose walls are at 30 C, and the room air is at 20 C. The surface of the heater has an emissivity of 0.5. (a) What is the temperature of the surface of the heater? (b) What are the respective heat flows by convection and radiation from the surface of the heater?

h,T∞ = 20 C

ε = 0.5 Tsurr = 30 C

r = 2 cm qhtr = 50 W 30 cm

Solution Preliminary: There is convection and radiation at the surface of the cylinder. The convection is natural convection. We have steady state. All necessary information is given. The heater output equals the convection and radiation from the surface.

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Chapter 10 Multimode heat transfer

Because radiation is involved, temperatures must be in absolute temperature units. Eqs. (10.2), (10.4), (10.6) will be used for the properties of air. The energy conservation equation at the cylinder’s surface is Heater output ¼ Natural convection and radiation from the surface 4 (10.7) q ¼ 50 W ¼ hAðTs TN Þ þ εsA Ts4 Tsurr where Ts ¼ surface temperature of the cylinder

Also,

A ¼ pDL ¼ pð0:04Þð0:3Þ ¼ 0:0377 m2 ε ¼ 0:5 s ¼ 5:67 108 W=m2 K4 TN ¼ 20 þ 273:15 ¼ 293:15 K Tsurr ¼ 30 þ 273:15 ¼ 303:15 K An appropriate equation for the natural convection is Eq. (7.32): Nu ¼

hD ¼ 0.36 þ h k

1=4

0:518RaD

9=16

1 þ ð0:559=PrÞ

i4=9

for 106 < RaD < 109

(7.32)

where gbðTs TN ÞD3 Pr RaD ¼ GrD Pr ¼ y2 (Note: At the end of the solution, we will check that the RaD limits for Eq. (7.32) have been satisfied.) N The air properties are taken at the film temperature Tfilm ¼ Ts þT and 2 1 Tfilm We can use a Microsoft Excel spreadsheet for the solution. In particular, we can use Excel’s Goal Seek program. Rearranging Eq. (10.7) for such a solution, we get function f1 : 4 ¼0 (10.8) f1 ¼ 50 hAðTs TN Þ εsA Ts4 Tsurr Putting the various constants and parameters into this equation and getting convective coefficient h from Eq. (7.32), we have a function that only depends on the unknown surface temperature Ts : If function f1 equals zero, we have a solution. In the spreadsheet, we give a guess for the value of Ts : Goal Seek iterates on this value until a value that satisfies f1 ¼ 0 is hopefully obtained. We now give a detailed description of the Excel spreadsheet. Readers uninterested in the details can go directly to the end of this description for the results. g ¼ 9:807 m=s2

b¼

Excel spreadsheet for Example 10.1 The general approach is to put labels in one column and their values/expressions in the adjacent column. Entries are in SI units and temperatures in Kelvin. 1

A Ts

B 400 (initial guess)

D ¼0.518B16^0.25

2 3 4 5 6 7 8 9 10 11

Tinf Tsurr D L q htr emiss sigma g Tfilm k

¼1þ(0.559/B13)^(9/16)

12

nu

293.15 303.15 0.04 0.3 50 0.5 5.67E-8 9.807 ¼(B1þB2)/2 ¼4.629E-12B10^32.899E-8B10^2 þ9.041E-5B10þ0.001009 ¼1.725E-14B10^3þ1.027E-10B10^2 þ3.725E-8B10-4.196E-6

F ¼B6-B20-B21

10.3 Examples

13

Pr

14 15 16 17 18 19 20 21

beta Gr Ra Nu h A qconv qrad

395

¼2.916E-13B10^4-1.148E9B10^3þ1.675E-6B10^21.021E3B10þ0.9126 ¼1/B10 ¼B9B14(B1B2)B4^3/(B12^2) ¼B15B13 ¼0.36þD1/D2^(4/9) ¼B11/B4B17 ¼PI()B4B5 ¼B18B19(B1B2) ¼B7B8B19(B1^4B3^4)

Calling up Goal Seek, we tell it to change cell B1 until cell F1 is zero. Convergence is reached, and the value in B1 becomes 404.977 K. Cell B16 is checked, and its value is within the appropriate limits for Eq. (7.32). The values in Cells B20 and B21 are, respectively, the heat flows by convection and radiation. The value in B20 is 30.28 and the value in B21 is 19.72. (a) The temperature of the surface of the cylinder is 405.0 K ¼ 131.9 C. (b) The convective heat flow at the surface is 30.28 W, and the radiative heat flow at the surface is 19.72 W. When the program is executed, the values are found in cells B20 and B21, respectively. Note: In Chapter 3, we discussed combined convection and radiation from a surface. We defined a radiative coefficient hrad that could be added to the convective coefficient h to obtain a combined coefficient hcombined ¼ h þ hrad . This combined coefficient could then be used to get the total heat transfer from the surface. That is q ¼ hcombined AðTs TN Þ. Eq. (3.35) 2 defined the radiative coefficient as hrad ¼ εs Ts2 þTsurr ðTs þTsurr Þ. This technique of combining the convective and the radiative coefficients gives exact results if TN ¼ Tsurr . It can be applied when TN sTsurr , but there will be some error. In this problem, TN and Tsurr were different. Specifically, TN ¼ 20 C and Tsurr ¼ 30 C. We modified the above Excel program slightly to solve the problem using the combined-coefficient approach and got results very close to the above exact results. The surface temperature was 401.8 K, the convective heat transfer was 29.26 W, and the radiative heat transfer was 20.74 W.

Matlab program for Example 10.1 % Example 10.1 % Cylindrical Heater clear, clc % Kelvin temperatures since we have radiation tsurr=303.15; tinf=293.15; D=0.04; L=0.3; A=piDL; g=9.807; sigma=5.67e-8; eps=0.5; q_heater=50; % Begin looping on ts; Each pass increases the surface temperature by 0.01 K for ts=350:0.01:450 tfilm=(tsþtinf)/2; kair=4.629e-12tfilm^3-2.899e-8tfilm^2þ9.041e-5tfilmþ.001009; nuair=-1.725e-14tfilm^3þ1.027e-10tfilm^2þ3.725e-8tfilm-4.196e-6; Pr=2.916e-13tfilm^4-1.148e-9tfilm^3þ1.675e-6tfilm^2-1.021e-3tfilmþ0.9126; beta=1/tfilm; Gr=gbetaabs(ts-tinf)D^3/nuair^2;

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Chapter 10 Multimode heat transfer

Ra=GrPr; d1=(1þ(.559/Pr)^(9/16))^(4/9); Nu=0.36þ0.518(Ra)^0.25/d1; h=Nukair/D; f1=q_heater-hA(ts-tinf)-epssigmaA(ts^4-tsurr^4) % Stop program when f1