Fundamentals of Heat Transfer: An Interdisciplinary Analytical Approach 9819909562, 9789819909568

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Moghtada Mobedi Gamze Gediz Ilis

Fundamentals of Heat Transfer An Interdisciplinary Analytical Approach

Fundamentals of Heat Transfer

Moghtada Mobedi · Gamze Gediz Ilis

Fundamentals of Heat Transfer An Interdisciplinary Analytical Approach

Moghtada Mobedi Department of Mechanical Engineering Shizuoka University Hamamatsu, Japan

Gamze Gediz Ilis Department of Mechanical Engineering Gebze Technical University Gebze, Türkiye

ISBN 978-981-99-0956-8 ISBN 978-981-99-0957-5 (eBook) https://doi.org/10.1007/978-981-99-0957-5 © The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2023 This work is subject to copyright. All rights are solely and exclusively licensed by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors, and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, expressed or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. This Springer imprint is published by the registered company Springer Nature Singapore Pte Ltd. The registered company address is: 152 Beach Road, #21-01/04 Gateway East, Singapore 189721, Singapore

To our precious: Azra, Ali, Ebru, Nima and Lara, Emin, Gediz Family

Preface I

Heat transfer is an interesting branch of science with wide application areas. Heat is a kind of energy faced everywhere; a universe without considering the transfer of heat cannot be imagined. This feature of heat transfer makes this branch of science multidisciplinary; almost in all branches of science and technology, heat transfer appears somehow. Heat transfer in buildings, drying, chemical reactions, use of renewable energy sources, biotechnology, medical science, high voltage cables and transformers, cooling of electronic equipment, global warming issues, or even the effect of climate on psychology are some branches of science and technology in which heat transfer is involved. Teaching the fundamentals of heat transfer to undergraduate and graduate students is important for their future careers. Many valuable books were written. However, the following remarks motivated authors to write the present book for graduate students, (a) In the past, the number of courses offered to graduate students was limited and they could take separate courses on conduction, convection, and radiation heat transfer. In recent years, many new courses in the universities or institutes have been suggested to graduate students. It is difficult for them to take 3 separate courses as conduction, convection and radiation heat transfer. The present book comprises 11 chapters and provides adequate fundamentals of heat transfer on conduction, convection, and radiation for researchers or graduate students. The authors believe that by learning the fundamental knowledge given in this book, which can be taught in a semester, readers can easily understand advanced heat transfer books by themselves if they need further information on heat transfer. (b) In the analysis of heat transfer, there are two main approaches: analytical and computational methods. Rapid developments in software and computer technologies have made computational heat transfer attractive; therefore, graduate students do not prefer to take courses relating to analytical methods in heat transfer. However, analyzing heat transfer problems computationally without an adequate heat transfer background is not advised. The analytical method provides a strong background of heat transfer, facilitating the correct interpretation of the results of computational heat transfer. The present book teaches

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analytical heat transfer methods for conduction, convection, and radiation by explanations of all necessary details. (c) At the undergraduate level, heat transfer is taught as a course in one semester in many universities. The given base may not be sufficient to understand many advanced heat transfer books. Furthermore, there are many students from different departments (such as food engineering, biotechnology, chemical engineering, etc.) who ask about an advanced heat transfer course that can be understood easily. The present book uses the math language and teaches the analytical methods of heat transfer to a graduate student or a researcher from any branch of engineering. It teaches what a graduate student (from different departments) should sufficiently know about heat transfer. (d) Due to the insufficient mathematics background of some graduate students, they cannot follow the derivation of equations and obtain the solution of problems. In this book, special attention is given to describing mathematical details that help students to learn how equations are solved and how the obtained solutions can be used. (e) Finally, nowadays there are universities in many countries whose medium of instruction is English. The English background of the students may make it difficult for them to follow advanced heat transfer books. The present book is written in a simple English without influencing the content. The aforementioned remarks motivated the authors to write the present book. The book consists of 11 chapters. It starts with a piece of brief information on common mathematical terminology and operations widely used in heat transfer in Chap. 1. Basic concepts such as ordinary and partial derivatives, coordinate systems, gradient and divergence operators are explained. Chapter 2 reviews the fundamentals of heat transfer as well as fluid dynamics. This chapter reminds the students about the important topics and terminology in heat transfer discussed in this book. Chapters 3 and 4 concern unsteady and steady-state conduction heat transfer in the Cartesian coordinate system while Chaps. 5 and 6 teach unsteady conduction heat transfer in cylindrical and spherical coordinate systems. The aim of Chaps. 3–6 is to provide adequate background on steady and unsteady conduction heat transfer in different coordinate systems. Convection heat transfer is discussed in Chaps. 7–10. In Chap. 7, classifications in convective heat transfer as well as governing equations are discussed. Chapter 7 provides sufficient basic knowledge that is needed for the next chapters. Chapter 8 teaches the concept of velocity and temperature boundary layers by analyzing forced convection heat transfer over a flat plate. Chapter 9 relates to heat and fluid flow in a channel. Important topics such as entrance and fully developed regions, energy balance in channels and determination of the Nusselt number are taught. Natural convection heat transfer over a vertical flat plate is discussed in Chap. 10. Students can understand the concept of natural convection by solving governing equations and applying an discussing the obtained results in this chapter. Thermal and velocity boundary layers in natural convection and the employed dimensionless numbers are discussed in Chap. 10. Finally, the radiation heat transfer is discussed in Chap. 11.

Preface I

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Basic concepts such as emission, absorption, reflection as well as radiosity, and view factor are explained. Radiation heat transfer between two surfaces is formulated. Authors believe that Chap. 11 provides sufficient background on radiation heat transfer to follow advanced radiation heat transfer books. As mentioned above, the present book provides an adequate basis on heat transfer in a semester enabling a graduate student to follow higher advanced heat transfer books, if she/he needs further detailed knowledge on conduction, convection, or radiation heat transfer. The authors would like to express their appreciation to all professors who taught heat transfer to authors during their university education especially Prof. Semra Ulku for her endless motivation, and to their families Ebru, Nima, Emin, Lara, and Gediz Family who always supported them and had been patient during the writing of this book. Hamamatsu, Japan Istanbul, Türkiye

Moghtada Mobedi Gamze Gediz Ilis

Preface II

The periods in which scientists involved in heat transfer live and the periods in which they present their inventions are subjects that we (as) engineers do not wonder about. We name them only by name, with very basic issues such as which laws they came up with or what formula they gave their name to. However, while examining these theories they gave us, thinking about the period in which they lived, the conditions under which they found these ideas and mathematical models, what makes them different, will ensure that you do not see the topics in this book as just formulas or equations. We hope this book in your hand will guide you so that you do not see the scientists who gave us theories and equations based on basic physics and engineering and specific to heat transfer, as only formulas. With this perspective, we are sure that heat transfer will become a more entertaining subject, which has a historical meaning and is easy to understand. The first physicist in history was Thales of Miletus (6th century BC). If a lot of things are happening in nature, the first person to ask is what could be the reason? He is the first person to try to explain nature with nature itself. He says that the building block of nature (Arche) is water. Then, Anaximenes (5th century BC) connects the physical diversity in nature to quantitative differences and says that the Arche of nature is air. Pythagoras (4th–5th century BC) says that all nature can be reduced to mathematics. It speaks of measures and proportions, and his Arche is numbers. Heraclitus (4th–5th century BC), on the other hand, discusses a process rather than existence. Although he did not directly mention the concept of time, he touched on beginning and end. His Arche is fire. For example, when the wood is burned, it turns into another substance, and this is a process. In fact, the foundations of science and heat transfer in our book began to be laid at that time. The thoughts of many philosophers, whom we have mentioned or not mentioned, are still based on researchers, and shed light on how we can solve the problems of nature. Mathematically solving heat problems, fluid mechanics, and boundary conditions have always been our legacy these days. Then, years later, Isaac Newton (17th century AD) appeared. Isaac Newton can be considered the pioneer of engineering. Let’s go back to the period in which he lived, the beginning of the 18th century. This unhappy child, whose father died before he was born, was born in the UK. He started xi

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his education at Trinity College in 1661. It was here that the thoughts of Aristotle, Descartes, Galileo, and Kepler began to shape his vision. Mathematics was a method of understanding everything for Newton as the philosophers we mentioned above. His laws of motion can be considered the basic concepts of engineering and fluid mechanics. If there is fluid and a temperature difference between two domains, we can discuss heat transfer then. Thus, especially his second law of motion is one of the basic equations in both fluid mechanics and heat transfer. After Newton, Swiss mathematician, and physicist Daniel Bernoulli (1700–1782) furthered his research on fluid mechanics and flow of blood in the body and began work on Conservation of Energy. It was known that if a moving object gains height, its kinetic energy turns into potential energy. He similarly noticed that the specific kinetic energy of a moving fluid changes with pressure. This is how the famous Bernoulli equation came about. Two centuries later, Irish-born Osborne Reynolds (1842–1912) broke new ground in aerodynamics with his famous unitless equation. The Reynolds unitless number, known by his name, gave us a shorthand answer as to whether the flow was in a laminar or turbulent region. Ludwig Prandtl (1875–1953), with his dimensionless number in his name, Prandtl was the one that brought together fluid mechanics and heat transfer. This number not only gives us the ratio of momentum dissipation to thermal dissipation, but also gives us another perspective. It also provides information about which of the thermal or momentum boundary conditions is fulfilled first. If Pr 1, heat is dissipated very quickly relative to momentum, and very slowly if Pr 1 (i.e. oils). That’s why we cool the parts exposed to friction with oil. The great philosophers, mathematicians, physicists, and scientists mentioned above are the people who led us to write this book on heat transfer. In this book, we have tried to explain what these great names have brought us, at a level that can be understood by all of us, that is to say, everyone working on engineering, whether mechanical engineer or electrical engineer. This book will hopefully show you that elusive heat transfer problems are not just a nightmare. Heat transfer will always exist thanks to the sun and curious people like you, and it will always remain open to development as a part of human curiosity about understanding our universe and nature. We are sure of that… Hamamatsu, Japan Istanbul, Türkiye

Moghtada Mobedi Gamze Gediz Ilis

Contents

1

Mathematics for Heat Transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 Importance of Mathematics in Heat Transfer . . . . . . . . . . . . . . . . 1.2 Vector and Scalar Quantities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3 Multiplication of Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4 Definition of Derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.5 Ordinary Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.6 Partial Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.7 Linear and Nonlinear Differential Equations . . . . . . . . . . . . . . . . 1.8 Coordinate Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.9 Gradient Operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.10 Divergence Operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.11 Vector Form Expression of Governing Equations . . . . . . . . . . . . 1.12 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.13 Review Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.14 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1 1 1 2 4 6 7 7 7 8 9 10 10 12 13

2

Basics of Heat Transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Heat Transfer Classification . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.1 Newtonian and Non-Newtonian Fluids . . . . . . . . . . . . . . 2.2.2 Compressible and Incompressible Flow . . . . . . . . . . . . . 2.2.3 Steady and Unsteady Heat Transfer . . . . . . . . . . . . . . . . . 2.2.4 Laminar and Turbulent Flow . . . . . . . . . . . . . . . . . . . . . . 2.2.5 Internal and External Flows . . . . . . . . . . . . . . . . . . . . . . . 2.2.6 One-, Two- and Three-Dimensional Heat Transfer . . . . 2.3 Thermodynamics Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.1 First Law of Thermodynamics . . . . . . . . . . . . . . . . . . . . . 2.3.2 Second Law of Thermodynamics . . . . . . . . . . . . . . . . . . 2.4 The Modes of Heat Transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4.1 Conduction Heat Transfer . . . . . . . . . . . . . . . . . . . . . . . . . 2.4.2 Convection Heat Transfer . . . . . . . . . . . . . . . . . . . . . . . . . 2.4.3 Radiation Heat Transfer . . . . . . . . . . . . . . . . . . . . . . . . . .

15 15 15 15 16 16 17 18 18 18 18 19 20 20 20 21 xiii

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2.5

Important Laws in Heat and Fluid Flow . . . . . . . . . . . . . . . . . . . . 2.5.1 Newton’s Second Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5.2 Fourier’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5.3 Newton’s Cooling Law . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5.4 Stefan–Boltzmann Law . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5.5 Newton’s Viscous Law . . . . . . . . . . . . . . . . . . . . . . . . . . . Common Boundary Conditions in Heat Transfer . . . . . . . . . . . . . 2.6.1 Boundary Conditions for Temperature . . . . . . . . . . . . . . 2.6.2 Boundary Conditions for Velocity . . . . . . . . . . . . . . . . . . Dimensionless Numbers in Heat Transfer . . . . . . . . . . . . . . . . . . . More Terminologies in Heat Transfer . . . . . . . . . . . . . . . . . . . . . . 2.8.1 Thermophysical Properties . . . . . . . . . . . . . . . . . . . . . . . . 2.8.2 Flow with Viscous Heat Dissipation . . . . . . . . . . . . . . . . 2.8.3 Heat Transfer Rate and Heat Flux . . . . . . . . . . . . . . . . . . 2.8.4 Volumetric Flow Rate and Mass Flow Rate . . . . . . . . . . 2.8.5 Pressure Drop . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.8.6 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Review Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

22 22 22 23 24 24 25 25 26 27 30 30 31 32 32 32 33 34 34

One-Dimensional Unsteady Heat Conduction in a Cartesian Coordinate System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Problem Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3 Assumptions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4 Implementation of Assumptions . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5 Analytical Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.6 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.7 Review Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.8 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

37 37 37 38 38 40 46 49 50

Two-Dimensional Steady Heat Conduction in Cartesian Coordinate System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 Problem Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3 Assumptions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4 Implementation of Assumptions . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.5 Analytical Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.6 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.7 Review Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.8 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

51 51 51 52 52 53 59 62 62

2.6

2.7 2.8

2.9 2.10 3

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One-Dimensional Unsteady State Heat Conduction in a Cylindrical Coordinate System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 Bessel Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3 Problem Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.4 Assumptions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.5 Implementation of Assumptions . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.6 Analytical Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.7 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.8 Review Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.9 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

65 65 65 67 68 68 69 74 77 77

One-Dimensional Unsteady State Heat Conduction in a Spherical Coordinate System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2 Problem Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3 Assumptions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.4 Implementation of Assumptions . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.5 Analytical Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.6 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.7 Review Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.8 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

79 79 79 79 80 82 85 88 89

Basics of Single Phase Convection Heat Transfer and Governing Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2 Single and Multi-phase Convection Heat Transfer . . . . . . . . . . . . 7.3 Single Phase Forced, Natural, and Mixed Convection Heat Transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.4 Governing Equations for Convection Heat Transfer . . . . . . . . . . 7.5 Difficulties in Solving Continuity, Momentum, and Energy Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.6 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.7 Review Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.8 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . External Flow: Heat and Fluid Flow Over a Flat Plate . . . . . . . . . . . . 8.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2 Fluid Flow over a Flat Plate (Isothermal Flow) . . . . . . . . . . . . . . 8.2.1 Definition of the Problem . . . . . . . . . . . . . . . . . . . . . . . . . 8.2.2 Assumptions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2.3 Similarity Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2.4 Solution of ODE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2.5 Boundary Layer Thickness . . . . . . . . . . . . . . . . . . . . . . . . 8.3 Heat and Fluid Flow over a Flat Plate . . . . . . . . . . . . . . . . . . . . . . 8.3.1 Definition of the Problem . . . . . . . . . . . . . . . . . . . . . . . . . 8.3.2 Energy Equation and Assumptions . . . . . . . . . . . . . . . . .

91 91 91 92 93 95 96 99 99 103 103 103 103 105 107 111 112 112 112 113

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8.4 8.5 8.6 8.7 9

8.3.3 Simplification of the Energy Equation . . . . . . . . . . . . . . 8.3.4 Similarity Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.3.5 Solution of ODE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.3.6 Thermal Boundary Layer Thickness . . . . . . . . . . . . . . . . Comparison of Velocity and Thermal Boundary Layer Thickness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Review Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

114 114 117 117 118 119 123 123

Internal Flow: Heat and Fluid Flow in a Channel . . . . . . . . . . . . . . . . 9.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.2 Isothermal Flow in a Channel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.2.1 Definition of the Problem . . . . . . . . . . . . . . . . . . . . . . . . . 9.2.2 Entrance and Fully Developed Regions . . . . . . . . . . . . . 9.2.3 Governing Equations and Implementation of Assumptions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.2.4 Solution of ODE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.3 Non-isothermal Flow in a Channel . . . . . . . . . . . . . . . . . . . . . . . . . 9.3.1 Definition of the Problem . . . . . . . . . . . . . . . . . . . . . . . . . 9.3.2 Thermally Entrance and Fully Developed Regions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.3.3 Governing Equation and Implementation of Assumptions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.3.4 Five Important Points . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.3.5 Reduction of PDE to ODE . . . . . . . . . . . . . . . . . . . . . . . . 9.3.6 Solution of ODE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.3.7 Determination of the Nusselt Value . . . . . . . . . . . . . . . . . 9.4 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.5 Review Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.6 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

125 125 125 126 126

10 Natural Convection Over a Vertical Flat Plate . . . . . . . . . . . . . . . . . . . 10.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.2 Considered Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.3 Assumptions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.4 Occurrence of Natural Convection . . . . . . . . . . . . . . . . . . . . . . . . . 10.5 Governing Equations for Natural Convection Heat Transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.5.1 State Equation for Density . . . . . . . . . . . . . . . . . . . . . . . . 10.5.2 Boussinesq Approximation . . . . . . . . . . . . . . . . . . . . . . . . 10.6 Boundary Layer Concept and Boundary Layer Equations . . . . . 10.7 Boundary Layer Assumptions and Governing Equations . . . . . . 10.8 Similarity Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.9 Solution of ODE Momentum and Energy Equations . . . . . . . . . .

153 153 153 154 154

127 130 132 132 132 135 137 140 143 144 145 150 150

156 157 157 159 161 162 166

Contents

xvii

10.10 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 168 10.11 Review Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 170 10.12 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171 11 Radiation Heat Transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Emission . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.3 Blackbody Emission . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.4 Real Surface Emission . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.5 Irradiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.6 Surface Radiative Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.7 Opaque Surface, Gray Surface and Radiosity . . . . . . . . . . . . . . . . 11.8 View Factor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.9 Radiation Heat Transfer Between Two Black Surfaces . . . . . . . . 11.10 Radiation Heat Transfer Between Two Gray Surfaces . . . . . . . . . 11.11 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.12 Review Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.13 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

173 173 173 175 175 176 176 178 179 182 184 186 189 189

Appendix A: Finding Roots of Bessel Functions . . . . . . . . . . . . . . . . . . . . . . 193 Appendix B: Dimensionless Velocity Values in a Boundary Layer . . . . . . 195 Appendix C: Compatibility Relation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 197 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 199

Nomenclature

a A Bi c cn Cp D Dh E Eb F g G Gr h H I J J v, Y v k L Lc Le Lh Lt m m˙ Ma Nu P Pr

Acceleration, m2 /s Area, m2 Biot number Speed of light, m/s Coefficient Specific heat at constant pressure, J/kgK Diameter, m Hydraulic diameter, m Energy, J; emissive power, W/m2 Emitted radiation flux, W/m2 K Force, N; view factor Gravitational acceleration, m2 /s Irradiation, W/m2 Grashof number Enthalpy, J; convection heat transfer coefficient, W/m2 K Height of the duct, m Radiation intensity, W/m2 sr Radiosity, W/m2 Bessel functions Thermal conductivity, W/mK Length, m Characteristic length, m Entrance region length, m Hydraulic length, m Thermal entrance length, m Mass, kg Mass flow rate, kg/s Mach number Nusselt number Pressure, N/m2 ; perimeter, m Prandtl number xix

xx

Nomenclature

q q  q˙ Q r, θ, z r, θ, φ R Ra Re Rth Ri s t T u, v, w U∞ − → V W x, y, z

Heat transfer rate, W Heat flux, W/m2 Rate of energy generation per unit volume, W/m3 Heat energy, J Cylindrical coordinates Spherical coordinates Gas constant, J/kg K, radius, m Rayleigh number Reynolds number Thermal resistance Richardson number Entropy, J/K Time, sec Temperature, K Mass average fluid velocity components, m/s Free stream velocity, m/s Velocity vector Work, J Cartesian coordinates

Greek Letters α β βn ρ τ ν ψ ξ σ μ δ δt δx λ ε θ ∞

Thermal diffusivity, m2 /s; absorptivity Volumetric thermal expansion coefficient, 1/K Eigen functions Mass density, kg/m3 ; reflectivity Transmissivity Kinematic viscosity, m2 /s; frequency of radiation, 1/s Stream function, m2 /s Vorticity vector, rpm/s Stefan–Boltzmann constant, W/m2 K4 Viscosity, kg/ms Hydrodynamic boundary layer thickness, m Thermal boundary layer thickness, m Local boundary layer thickness, m Arbitrary constant; wavelength, m Emission, turbulent dissipation Dimensionless temperature Free stream conditions

Nomenclature

Subscripts abs blk cv f h i in irr m out rad ref s trn w

Absorbed Black body Control volume Film condition Hot condition Initial condition Inlet Irradiation Mean Outlet Radiation Reference; reflected Solid body Transmitted Wall

xxi

Chapter 1

Mathematics for Heat Transfer

1.1 Importance of Mathematics in Heat Transfer Heat transfer is one of the science branches requiring strong mathematics knowledge to establish or understand governing equations and solve them. That is why a separate chapter is provided in this book to give the necessary knowledge about mathematical concepts and operations widely used in heat transfer. This chapter starts by explaining simple definitions of scalar and vector quantities and finishes with the vector form of a governing equation in heat transfer.

1.2 Vector and Scalar Quantities The quantities in heat transfer can be classified into two groups: scalar and vector quantities. The difference between a scalar and a vector quantity is well known; however, it might be useful to mention it in this section, briefly. (a) A Scalar Quantity: A scalar quantity is expressed by its magnitude only. For instance, pressure and temperature are the quantities that can be evaluated by their magnitudes such as 20 °C or 20 kPa. (b) A Vector Quantity A vector quantity is expressed both by its magnitude and direction. Velocity is a vector quantity widely mentioned in convective heat transfer for which both the magnitude and direction should be known. In many cases, unit vectors are used to describe a vector. For instance, a velocity vector in a two-dimensional Cartesian coordinate system can be expressed as, − → − → − → V =u i +v j

© The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2023 M. Mobedi and G. Gediz Ilis, Fundamentals of Heat Transfer, https://doi.org/10.1007/978-981-99-0957-5_1

(1.1)

1

2

1 Mathematics for Heat Transfer

Fig. 1.1 Schematical definition of vector

− → − → where i and j are unit vectors in the x and y directions, respectively, and u and − → v are components of V in x and y directions, respectively as shown in Fig. 1.1. The direction of the vector can be easily found from Fig. 1.1, while its magnitude can be calculated as,  − → | V | = u 2 + v2 (1.2)

1.3 Multiplication of Vectors Two types of vector multiplication are widely used in heat transfer and fluid flow as “dot product” and “cross product”. In this book, the dot product is widely used, while the cross product is rarely used. However, brief information about both of them is given in this section. (a) Dot Product The result of the dot product of two vectors is a scalar quantity. Figure 1.2 shows the − → − → dot product of two vectors of V1 and V2 , and the result is, − →− → V1 . V2 = |V1 ||V2 |cosθ

(1.3)

− → − → − → − → where |V1 | and |V2 | are the magnitudes of V1 and V2 . The vectors of V1 and V2 can − → − → be defined by the unit vectors of i and j in the Cartesian coordinate system as, − → − → − → V1 = u 1 i + v1 j

(1.4)

− → − → − → V2 = u 2 i + v2 j

(1.5)

− → − → The magnitude of two vectors of V1 and V2 can be defined as, |V1 | = |V2 | =

 

u 21 + v12 u 22 + v22

(1.6)

1.3 Multiplication of Vectors

3

Fig. 1.2 Schematical definition of dot product

− → − → Furthermore, the dot product of V1 and V2 can also be expressed as, − → − → − → − → → − → − V1 . V2 = (u 1 i + v1 j ) . (u 2 i + v2 j )

(1.7)

or, − → − → V1 . V2 = u 1 u 2 + v1 v2

(1.8)

It might be useful to remember that based on Eq. 1.3, the following relationships are valid for unit vectors: − →− → − →− → i .i = j . j =1 (1.9) − →− → − →− → i . j = j .i =0

(1.10)

(b) Cross Product Cross product is completely different than dot product. The cross product of two vectors is a vector that is perpendicular to the surface in which two vectors exist, as shown in Fig. 1.3. The calculation of the cross product of two vectors has been discussed in many mathematics books [1]. If Fig. 1.3 is considered, the following cross products for the unit vectors in the Cartesian coordinate system can be written, − → − → − → i × j = k

(1.11)

− → − → − → j × i =−k

(1.12)

− → − → − → − → i × i = j × j =0

(1.13)

− → − → − → where i , j , and k are unit vectors in the x, y, and z directions, respectively. − → − → Therefore, the result of the cross products of two vectors such as V1 and V2 is defined by Eqs. 1.4 and 1.5 can be found as,

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Fig. 1.3 Schematic definition of cross product

− → − → − → − → − → − → V1 × V2 = (u 1 i + v1 j ) × (u 2 i + v2 j ) − → − → − → − → − → − → − → − → − → − → V1 × V2 = u 1 u 2 ( i × i ) + u 1 v2 ( i × j ) + v1 u 2 ( j × i ) + v1 v2 ( j × j ) (1.14) − → − → Considering Eqs. 1.11–1.13, the cross product of two velocity vectors of V1 and V2 can be obtained as, − → − → − → V1 × V2 = (u 1 v2 − u 2 v1 ) k

(1.15)

1.4 Definition of Derivative Before a brief explanation about the derivative, it may be useful to mention the concepts of “dependent variable” and “independent variable”. If a function such as T = f (x) exists (for instance Eq. 1.16), the variable T [which is equal to f (x)] is called a dependent variable, while the variable x is called an independent variable. T = f (x) = x 3 + x 2 + 1

(1.16)

Figure 1.4 shows the change in variable T or f (x) with respect to x. The value of T changes by changing the x values. The value of T depends on the value of x, while x is independent of T . For any value of the independent variable, the derivative can be simply defined as the rate of change of a dependent variable (for instance, T ) with respect to an independent variable (for instance, x). Figure 1.4 shows the change in f (x) defined by Eq. 1.16 with respect to x. As it can be seen, the derivative of T at the xi point is the slope of the tangential line at x = xi .

1.4 Definition of Derivative

5

Fig. 1.4 Schematic definition of derivative

dT T |x=xi = tan(α) = dx x

(1.17)

Two types of derivatives are widely used in heat transfer as ordinary and partial derivatives. (a) Ordinary Derivative: It shows a derivative when a dependent variable such as f (x) is the function of only one independent variable such as x. Assume the dependent variable of T which is function of x shown in Fig. 1.4, T = f (x) = x 5 + x 3 + 1

(1.18)

As is well known, the first and the second derivatives of the T function which only depends on x can be written as, d f (x) dT = = 5x 4 + 3x 2 dx dx

(1.19)

d2 f (x) d2 T = = 20x 3 + 6x dx 2 dx 2

(1.20)

The following style of writing for derivatives is also used. T  = f  (x) = 5x 4 + 3x 2

(1.21)

T  = f  (x) = 20x 3 + 6x

(1.22)

Equation 1.21 can be called the first-order ordinary derivative of T with respect to x, while Eq. 1.22 is the second-order ordinary derivative of T with respect to x. (b) Partial Derivative: A partial derivative of a function is used when a dependent variable is function of more than one independent variable. Consider the equation below,

6

1 Mathematics for Heat Transfer

T = f (x, y) = x 5 y + x y 3 + 1

(1.23)

The function of T depends on two independent variables: x and y. The partial derivative of T with respect to x and y can be written as, ∂ f (x, y) ∂T = = 5x 4 y + y 3 ∂x ∂x

(1.24)

∂T ∂ f (x, y) = = x 5 + 3x y 2 ∂y ∂x

(1.25)

Similarly, the second partial derivative can be written as, ∂2 T ∂ 2 f (x, y) = = 20x 3 y ∂x2 ∂x2

(1.26)

∂ 2 f (x, y) ∂2 T = = 6x y ∂ y2 ∂x2

(1.27)

Equation 1.24 is called the first-order partial derivative of T with respect to x and similarly, Eq. 1.25 is called the first order partial derivative of T with respect to y. By the same way, Eqs. 1.26 and 1.27 can be called the second partial derivative of T with respect to x and y, respectively. Rules about taking an ordinary derivative and partial derivatives are explained in many mathematics books (e.g., [1, 2]).

1.5 Ordinary Differential Equations If an equation involves one or more ordinary derivatives of a dependent variable, it can be called an ordinary differential equation (ODE). An example of an ordinary differential equation can be, dT d2 T +T =0 + 2 dx dx

(1.28)

The above ODE is called “the second-order ordinary differential equation”. The highest order of a derivative term in an ordinary differential equation is accepted as the order of ODE. It is necessary to define initial and/or boundary conditions for an ordinary differential equation to find the exact solution.

1.8 Coordinate Systems

7

1.6 Partial Differential Equations If an equation involves one or more partial derivatives of a dependent variable, it is called a partial differential equation (PDE). A two-dimensional heat conduction equation can be an example of a PDE. ∂2T ∂2T + =0 ∂x2 ∂ y2

(1.29)

The above equation is called “the second-order PDE”. It requires two boundary conditions for the x independent variable and two boundary conditions for the y independent variable to obtain the exact solution.

1.7 Linear and Nonlinear Differential Equations If a derivative term of an ordinary differential equation or a partial differential equation is multiplied with “the dependent variable” or “the derivative of the dependent variable”, the equation is called “a nonlinear ODE” or “a nonlinear PDE”. Examples of a nonlinear ODE and nonlinear PDE can be, dT d2 T +T =0 +T dx 2 dx

(1.30)

∂2T ∂ T ∂2T + =0 2 ∂x ∂ x ∂ y2

(1.31)

The second term of Eq. 1.30 is multiplied by T which is the dependent variable. That is why Eq. 1.30 is a nonlinear second-order ODE. Similarly, the second term of Eq. 1.31 is multiplied by the derivative of the dependent variable (∂ T /∂ x); therefore, Eq. 1.31 is also a nonlinear second-order PDE. The solution of a linear ODE or PDE is easier than a nonlinear ODE or PDE. Nonlinear terms make serious difficulties in solving heat transfer and fluid flow equations.

1.8 Coordinate Systems In order to define a point in a two-dimensional (i.e., 2D) or three-dimensional (i.e., 3D) domain, three coordinate systems as Cartesian, cylindrical, and spherical coordinate systems are widely used. For a three-dimensional domain, the coordinates of a point in the Cartesian coordinate system are generally shown by x, y and z; for the cylindrical coordinate system, the coordinates are generally illustrated by r, θ , and z. Finally, the symbols r, θ , and φ are used to define coordinates of a point in

8

1 Mathematics for Heat Transfer

Fig. 1.5 Coordinates of a point in Cartesian, cylindrical, and spherical coordinate systems

a spherical coordinate system. Therefore, a function in a Cartesian, cylindrical, or spherical coordinate system can generally be expressed as A function in the Cartesian coordinate system: T = f (x, y, z) A function in the cylindrical coordinate system: T = f (r, θ, z) A function in the spherical coordinate system: T = f (r, θ, φ). The above coordinate systems are widely used in heat transfer problems; the coordinates of a point such as P in each coordinate system are shown in Fig. 1.5.

1.9 Gradient Operator The gradient is a mathematical operator implemented on a scalar quantity. Simply, the gradient of a scalar quantity is a vector quantity that shows the rate of change of the scalar quantity in the independent variable direction. Generally, the symbol − → − → of ∇ (i.e., nabla) is used to show the gradient operator. The operator of ∇ in a three-dimensional Cartesian coordinate system can be defined as, ∂ − ∂ − ∂ − → → → − → i + j + k ∇ = ∂x ∂y ∂z

(1.32)

− → − → − → where i , j , and k are unit vectors in the x, y, and z directions, respectively. The first term on the right side of Eq. 1.32 shows the change in the scalar quantity in x and, similarly, the second and third terms on the right side of Eq. 1.32 refer to the change in the scalar quantity in the y and z directions, respectively. For instance, considering the function of T (x, y, z) in the Cartesian coordinate system, the change in T in the x, y, and z directions can be expressed as ∂T − → ∂T − → ∂T − → − → i + j + k ∇T = ∂x ∂y ∂z

(1.33)

It should be kept in mind that the gradient of a scalar quantity is a vector quantity. Similarly, the gradient of temperature in the cylindrical coordinate system is,

1.10 Divergence Operator

9

∂T − → 1 ∂T − → ∂T − → − → ir + iθ + iz ∇T = ∂r r ∂θ ∂z

(1.34)

− → − → − → where ir , i θ , and i z are unit vectors in the r , θ , and z directions, respectively. For the spherical coordinate system, the gradient of temperature can be written as, ∂T − 1 ∂T − → 1 ∂T − → → − → ir + iθ + iφ ∇T = ∂r r ∂θ r sin(φ) ∂φ

(1.35)

− → − → − → where ir , i θ , and i φ are unit vectors of the spherical coordinate system.

1.10 Divergence Operator The divergence operator shows the total rate of change of a vector quantity in a domain. It is the rate of change of a vectorial dependent variable with respect to all its independent variables. The divergence of a vector quantity can be found by − → the dot product of the nabla operator (i.e., ∇ ) and the vector quantity. For instance, − → the divergence of the V velocity vector represented by the u, v, and w velocity components in the Cartesian coordinate system can be obtained as, → − →− ∇.V =



 ∂ − ∂ − ∂ − → → → − → − → − → i + j + k . (u i + v j + w k ) ∂x ∂y ∂z ∂v ∂w − → − → ∂u →− + + div V = ∇ . V = ∂x ∂y ∂z

(1.36)

(1.37)

Similarly, if a velocity vector is defined in cylindrical and spherical coordinates, − → − → − → − → V = u r ir + u θ i θ + u φ i z

(1.38)

− → − → − → − → V = u r ir + u θ i θ + u φ i φ

(1.39)

− → by using the ∇ operator in cylindrical and spherical coordinates, the divergence − → of V in cylindrical and spherical coordinate systems can be obtained as, 1 ∂u θ ∂u φ → 1 ∂r u r − →− ∇.V = + + r ∂r r ∂θ ∂z

(1.40)

10

1 Mathematics for Heat Transfer

∂ 1 1 ∂u φ 1 ∂ → − →− ∇ . V = 2 (u r r 2 ) + (u θ Sinθ ) + r ∂r r Sinθ ∂θ r Sinθ ∂φ

(1.41)

1.11 Vector Form Expression of Governing Equations − → The ∇ operator is widely used in heat transfer and fluid flow. Rules have been developed on gradient and divergence of quantities, and those rules facilitate the exploration of new and more general (independent from the coordinate system) concepts in heat transfer. The nabla operator also makes the size of the equations smaller. It provides the writing of governing equations independent of a coordinate system. Therefore, vector forms of continuity, momentum, and energy equations are widely used in many scientific papers. The open form and vector form of the ODE and PDE equations are identical; however, the type of writing is different. For instance, the continuity equation for an incompressible flow without mass generation/dissipation in vector form can be written as, → − →− ∇.V = 0

(1.42)

By using Eq. 1.36, the same equation in open form for the Cartesian coordinate system can be written as, ∂v ∂w ∂u → − →− + + =0 → ∇ . V = 0 − ∂x ∂y ∂z vector form  

(1.43)

open form

As can be seen, the vector form of the equation is shorter and can be used for Cartesian, cylindrical, and spherical coordinate systems.

1.12 Example Assume a two-dimensional flow in the Cartesian coordinate system. A velocity vector in this field can be defined as, − → − → − → V =u i +v j where u and v are velocity components in the x and y directions, respectively. They can be written in terms ψ (a function called the stream function) as,

1.12 Example

11

u=

∂ψ ∂ψ ; v=− ∂y ∂x

Additionally, a vector called a vorticity vector can be defined as, → − → − ξ = −∇ × V prove that the following relation is valid between ψ and ξ , − → ∇ 2 ψ = −ξ Solution: The definition of cross product is given in Sect. 1.3. For the cross product of two vectors in the Cartesian coordinate system, the following equalities are valid, − → − → − → i × j = k − → − → − → j × i =−k − → − → − → − → i × i = j × j =0 − → − → The definition ∇ and V are, − → − → − → V =u i +v j ∂ − ∂ − → → − → ∇ = i + j ∂x ∂y Substituting the above equations in the definition of vorticity yields, → − → − ξ = −∇ × V

  ∂ − ∂ − → → → − → − → − → − ξ = −∇ × V = − i + j × (u i + v j ) ∂x ∂y   ∂v − → ∂u − ξ= k ∂y ∂x If the definitions of u and v velocity components in terms of stream function (ψ) are substituted into the above equation,  ∂u ∂v − ∂x ∂y ⎞ ⎛ ∂ψ ∂ψ ∂ − ∂ ∂x ∂y − →⎝ ⎠ ξ =−k − ∂x ∂y

− → ξ= k



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1 Mathematics for Heat Transfer

The definition of vorticity can be written as,   ∂ 2ψ − → ∂ 2ψ + ξ= k ∂x2 ∂ y2 − → The definition of ( ∇ 2 or ∇ 2 ) in Cartesian coordinate system can be found as,     ∂ − ∂ − ∂ − ∂ − → → → → 2 i + j . i + j ∇ = ∂x ∂y ∂x ∂y or it can be simplified as , ∇2 =

∂2 ∂2 + ∂x2 ∂ y2

Therefore, it is proved that, ξ = −∇ 2 ψ

1.13 Review Questions

Questions (Q1) In order to define a vector quantity, the use of the vector’s magnitude is sufficient. (Q2) Coordinates of a point in a cylindrical coordinate system can be represented by r , θ and z . (Q3) Coordinates of a point in a spherical coordinate system can be represented by r , θ and φ . (Q4) In order to define a scalar quantity, the use of the magnitude of the scalar quantity is sufficient. (Q5) Pressure and temperature are scalar quantities, while velocity is a vector quantity. (Q6) The result of the dot product of two velocity vectors is a scalar quantity. (Q7) The result of the cross product of two velocity vectors is a scalar quantity. (Q8) Derivative of f (x) at x1 is the slope of the line which is tangent to the curve of function at the point of x1 , f (x1 ). (Q9) Derivative of f (x) function at a point such as x0 shows the change of f (x) function with respect to x at point of x0 . (Q10) In the ordinary derivative of a function, the function depends on many independent variables. (Q11) In the partial derivative of a function, the function depends on more than one independent variable. (Q12) In an ordinary differential equation, the unknown function depends on only one independent variable. (Q13) In a partial differential equation, the unknown function depends on only one independent variable. (Q14) Nonlinearity of a differential equation can be due to the multiplication of a derivative of differential equation with the dependent variable. (Q15) Solving the nonlinear differential equation is easier than the linear differential equation. (Q16) Gradient operator can be applied to a scalar quantity, and the result is a vector quantity. (Q20) Divergence operator can be applied into a scalar quantity, and result is a vector quantity. (Q21) Writing of vector form of continuity, momentum, and energy equations make the size of equations shorter.

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1.14 Problems

13

1.14 Problems − → (1) u and v are the components of a velocity vector (i.e., V ) defined in a two− → − → − → dimensional Cartesian coordinate system with unit vectors of i and j . Find S vector considering the following equation, − → − →− → →− S = (V .∇ )V (2) For one-dimensional heat transfer in the radial direction (i.e., r direction) of a cylindrical coordinate system, the variation of heat diffusion can be expressed as, → − →− D = ∇ . q  = k∇ 2 T prove that the above term can also be written as,  D=k

1 ∂T ∂2T + ∂r 2 r ∂r



(3) Consider the vector form of the following momentum equation, 1− − → − →− → →− → ( V . ∇ ) V = − ∇ P + ν∇ 2 V ρ where ρ and ν are density and kinematics viscosity which are scalar quantities and constant. For two-dimensional Cartesian coordinate system with x and y axes and − → − → − → velocity vector of V = u i + v j , prove that the above equation can be written as two separate PDEs shown below.   2 ∂u 1 ∂P ∂ u ∂ 2u ∂u +v =− +ν + 2 u ∂x ∂y ρ ∂x ∂x2 ∂y   2 ∂u 1 ∂P ∂ v ∂v ∂ 2v +v =− +ν u + ∂x ∂y ρ ∂y ∂x2 ∂ y2 For the above problem, the flow is incompressible. Therefore, → − →− ∇.V = 0

Chapter 2

Basics of Heat Transfer

2.1 Introduction Before starting the analysis of conduction, convection, and radiation heat transfer in this book, it might be useful to recall the basic concepts and terminologies in heat transfer and fluid flow. This chapter is important for readers since many heat transfer laws and terminologies used in this book are briefly explained in this chapter.

2.2 Heat Transfer Classification It is possible to classify heat transfer from different aspects. These classifications are important since the governing equations change for each classified group. In order to write appropriate governing equations for a problem, the type of fluid and flow as well as heat transfer must be known. In this section, significant classifications that must be definitely known before writing the governing equations for a heat transfer problem are described.

2.2.1 Newtonian and Non-Newtonian Fluids A fluid is called N ewtonian f luid when its shear stress is linearly proportional to the deformation rate (velocity gradient). Fluids such as water, air, oil, and gasoline are accepted as Newtonian fluids. Newtonian fluids follow the “Newtonian law of viscosity” which will be described later in this chapter. For a Newtonian fluid, a linear change exists between the fluid shear stress and deformation rate, and the line representing this variation passes from the origin, as seen in Fig. 2.1.

© The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2023 M. Mobedi and G. Gediz Ilis, Fundamentals of Heat Transfer, https://doi.org/10.1007/978-981-99-0957-5_2

15

16

2 Basics of Heat Transfer

Fig. 2.1 Relation between shear stress and deformation rate of various fluids [3]

All fluids that are not Newtonian fluids are called non-Newtonian fluids. They do not obey the Newtonian law of viscosity. All fluids considered in this book are Newtonian fluids.

2.2.2 Compressible and Incompressible Flow If the density of a fluid does not change in a process of heat transfer, the flow of fluid is accepted as an incompressible flow. If the density changes in a process and when this change must be taken into account, the flow is called compressible flow. Density can change by changing the temperature for liquids and gases, and it also changes with pressure for gases. Therefore, for a compressible flow, a state equation is needed to establish a relation between temperature, pressure, and density. The ideal gas equation is an equality that can be used as a state equation for the determination of the density of gases having thermodynamical behavior close to an ideal gas, P = ρ RT

(2.1)

It should be mentioned that when the velocity of a gas flow is low (i.e., Ma < 0.3), the assumption of incompressibility for a gas flow is valid, and generally, this assumption can yield accurate results.

2.2.3 Steady and Unsteady Heat Transfer If a heat transfer process does not depend on time, the process of heat transfer is called the “steady state”. In steady state heat transfer, temperature, pressure, or velocity does not change with time.

2.2 Heat Transfer Classification

17

If a dependent variable of a heat transfer problem (such as temperature, pressure, or velocity) changes with time, the process of heat transfer is called “unsteady state” or “transient heat transfer”. Before starting to solve a heat transfer problem, it is necessary to know whether the process is a steady or unsteady state.

2.2.4 Laminar and Turbulent Flow The definition of the Reynolds number is given later in this chapter. Briefly, the Reynolds number can be defined as the ratio of inertia forces to viscous forces. For low values of the Reynolds number (such as Re < 5 × 105 for forced convection flow over a flat plate), viscous forces are dominant compared to inertia forces; therefore, smooth flow patterns with no (or negligible) mixing with adjacent layers are observed (Fig. 2.2a). For high values of the Reynolds number (such as Re > 5 × 105 for forced convection flow over a flat plate), the flow is associated with random fluctuation (mixing of fluid at the micro-scale) causing the emergence of a new stress called the Reynolds stress (Fig. 2.2b). An additional diffusivity called as Eddy diffusivity is used to formulate Reynolds stresses. The value of eddy diffusivity (changing by location and depending on many parameters) can be obtained by using turbulent models. For instance, the k − ε turbulent model is one of the models that is widely used for solving turbulent heat and fluid flow problems. The equations for k (turbulent kinetic energy) and ε (turbulent dissipation) must be solved to determine the value of eddy diffusivity used in the momentum and energy equations.

Fig. 2.2 Character of flow at a point, a laminar flow with no fluctuation, b turbulent flow always involving 3D fluctuation

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2 Basics of Heat Transfer

Fig. 2.3 Type of flows, a internal flow, b external flow

2.2.5 Internal and External Flows Briefly, when a fluid flows over a body, the flow can be called as external flow; Fig. 2.3 shows flow over a cylinder, which is an example of external flow. When a fluid flows inside a body, the flow is called an internal flow such as the flow in a tube.

2.2.6 One-, Two- and Three-Dimensional Heat Transfer Heat transfer in a domain can be “one dimensional” (i.e., 1D), “two dimensional” (i.e., 2D), or “three dimensional” (i.e., 3D). For instance, if a domain in the Cartesian coordinate system is considered, for one-dimensional heat transfer in the x direction, the unknown dependent variable changes only in the x direction [i.e., T = f (x), − → − → and/or V = f (x) and/or p = f (x)]. Similarly, if T = f (x, y) and/or V = f (x, y) and/or p = f (x, y), the heat and fluid flow is two dimensional. Finally, if dependent variable(s) of temperature and/or velocity and/or pressure are functions of x, y, and z, the heat transfer problem is called three-dimensional heat transfer.

2.3 Thermodynamics Laws Thermodynamics concerns heat, work, and energy. It has four laws called the zeroth, first, second, and third laws. The first and second laws of thermodynamics play an important role in establishing governing equations in heat transfer. Therefore, the first and second laws of thermodynamics are briefly explained in this subsection.

2.3.1 First Law of Thermodynamics The first law of thermodynamics can also be called the “principle of energy conservation”. The first law of thermodynamics states that energy cannot be created

2.3 Thermodynamics Laws

19

or destroyed; it can only change its form. There exists a transition from one type of energy to another type, such as work done on a system that can be converted to thermal energy. For a closed system, the following relation between work and heat based on the first law can be written, E = Q − W

(2.2)

where E is the change in the total energy stored in the system, Q and W are the net heat transferred to the system, and the net work done by the system, respectively. The first law of thermodynamics for an open system can be mathematically written as,     vi2 dE cv vo2 hi + ho + = Q−W + + zo − + zi dt 2 2

(2.3)

where h is the enthalpy, v is the velocity, and z shows the elevation from a reference plane. The term on the left side shows the change in energy in the control volume over time. The first two terms on the right side refer to heat transfer to the system and work done by the system. The last two summation terms on the right side are the total energy entering and leaving the system by inlet and outlet mass. Particularly, the first law of thermodynamics is widely used to derive governing equations for conduction, convection, and radiation heat transfer.

2.3.2 Second Law of Thermodynamics The first law of thermodynamics relates to the balance between energy and work. It does not consider irreversibility. The second law of thermodynamics focuses on the direction of heat transfer. It says that heat is always transferred from a high temperature to a low-temperature heat source. Therefore, heat transfer is an irreversible process. In the second law of thermodynamics, entropy is defined as a measure of reversibility. For a reversible process, entropy is defined as, ds =

δ Q rev T

(2.4)

where s is entropy. Since heat can only be transferred from high to low-temperature heat sources, the total entropy change of the system and environment is always positive.

20

2 Basics of Heat Transfer

1 dQ ds > dt T dt

(2.5)

The left term shows the change in entropy during the dt interval.

2.4 The Modes of Heat Transfer As it is known well, there are three modes of heat transfer: conduction, convection, and radiation. In this section, these modes of heat transfer are summarized.

2.4.1 Conduction Heat Transfer Conduction heat transfer occurs in three phases of materials: solid, liquid, and gas. Most practical problems with conduction heat transfer relate to conduction heat transfer in solids; however, conduction occurs in liquids and gases (even if fluid flows). Briefly, conduction heat transfer is the transfer of heat from high to low temperatures in a medium without considering any macroscopic motion. In other words, conduction heat transfer does not occur due to the macroscopic motions in the medium; it occurs due to diffusion through a medium. As mentioned before, the medium can be solid, liquid, or gas. Heat transfer in a cylinder in the radial direction shown in Fig. 2.4a is an example of conduction heat transfer. The most important law of conduction heat transfer is the Fourier law which is explained in Sect. 2.5.2. The term Diffusion is also used to express conduction. Diffusion heat transfer or conduction heat transfer are two statements that are equivalent in many heat transfer reports and papers.

2.4.2 Convection Heat Transfer Convection heat transfer occurs in liquid and gas phases. This occurs due to the motion of fluid transferring heat from one point to another point in a fluid domain. When there is no motion in a fluid domain, no convective heat transfer occurs. Convection heat transfer over a cylinder whose surface is at higher temperature than the fluid can be an example of convection heat transfer. Heat is transferred from the hot cylinder to the fluid, and the fluid carries the energy from one point to another point of the domain (Fig. 2.4b). The most important law of convective heat transfer is Newton’s cooling law which is explained in Sect. 2.5.3. Practically, there are four types of convective heat transfer, which are briefly explained below:

2.4 The Modes of Heat Transfer

21

Fig. 2.4 Examples of modes of heat transfer, a conduction heat transfer (conduction in a solid conduit), b convection heat transfer (convection around a cylinder), and c radiation heat transfer (radiation heat transfer in a space between two cylinders)

• Forced convection: The motion of the fluid is provided by using an external power such as a power provided by fan or pump. • Natural convection: The motion of the fluid is caused by the density difference of the fluid as well as the effect of gravity (or any body force). Due to Archimedes’s law, the light region in the fluid domain moves up when the buoyancy force exists. Natural convection is also called free convection in the literature. • Mixed convection: This is a forced convection flow for which the effect of natural convection cannot be neglected. Therefore, the effect of the buoyancy force is included in the governing equations. • Boiling: It occurs due to the phase change of a liquid to gas (or vapor). When the temperature of a liquid exceeds the boiling temperature, a phase change occurs. In boiling heat transfer, in addition to heat consumed to increase the temperature of the liquid (sensible heat), an extra heat due to the phase change called “Latent Heat” is transferred to the liquid domain. • Condensation: This occurs due to the phase change of gas into a liquid. Condensation occurs when the temperature of the gas falls below the condensation temperature (gas-liquid phase change temperature). Similar to boiling, in addition to the heat transferring from the gas to reduce its temperature (sensible heat), an additional heat (latent heat)emerging during condensation must be taken away. This book focuses on two types of convection: forced and natural convection.

2.4.3 Radiation Heat Transfer The energy emitted by matter at a nonzero temperature (in Kelvin) is called thermal radiation. Emissions occur from solid surfaces, liquids, and gases. Electromagnetic waves (or photons) carry the radiation energy. Contrary to conduction and

22

2 Basics of Heat Transfer

convection heat transfer, radiation heat transfer can occur without the need for a material medium. That is why radiation heat transfer can take place in a “vacuum environment”. Figure 2.4c shows radiation heat transfer between two surfaces that are at different temperatures. The surfaces A and B see each other, and there is a net heat transfer rate between two surfaces transferring energy from the hot to the cold surface.

2.5 Important Laws in Heat and Fluid Flow Similar to other branches of science, there are some laws on which heat transfer engineering is established. Without mentioning these laws, starting other chapters of this book is impossible. Therefore, the scientific laws widely used in heat transfer are explained below.

2.5.1 Newton’s Second Law As it is well known, the second law of Newton states that if a force acts on an object, it will be accelerated or decelerated. Mathematically, Newton’s second law can be written as: − → → F = m.− a (2.6) where F with unit of N (Newton) is the force acting on a mass m, and a is acceleration whose unit is m/s2 . Since acceleration can be defined as the change in velocity in a time period, it is possible to write that a = V /t. Substituting this definition into Eq. 2.6 yields, − → − → (m V ) V − → − → = F = m. a = m t t

(2.7)

− → m V is known as momentum. Equation 2.7 states that if no force is applied to an object, its moment never changes. The definition of Newton’s second law in terms of momentum is also widely used for the derivation of momentum equations used in heat and fluid flow problems.

2.5.2 Fourier’s Law Fourier’s law is the basis of heat conduction. Briefly, for a one-dimensional plane wall with a thickness of L whose surface temperatures are at Th and Tc , the following equation can be written:

2.5 Important Laws in Heat and Fluid Flow

qx = −k

23

T L

(2.8)

where q  is the heat flux with unit of (W/m2 ) through the wall, k is the thermal conductivity of the wall, and T = Th − Tc . As it can be seen, heat flux through the wall proportionally changes with the thermal conductivity and the difference in surface temperatures, while it has a reverse relationship with L, which is the thickness of the wall. The minus sign is due to the heat being transferred from high to low temperature causing a negative slope for temperature gradient. For an infinitesimal control volume with a thickness of d x in a wall, Eq. 2.8. takes the following form, q  = −k

dT dx

(2.9)

where qx is the heat flux in the x direction. If a three-dimensional heat transfer exists in a body located in the Cartesian coordinate system, the following equation can be written when the thermal conductivity of the body is constant, − → q = −k



 ∂T − → ∂T − → ∂T − → i + j + k ∂x ∂y ∂z

(2.10)

The above form of Fourier’s law can be written in vector form valid for all coordinate systems, − → − → q = −k ∇ T

(2.11)

The right term is the multiplication of thermal conductivity with the gradient of − → temperature yielding heat flux. Substituting the definition of gradient ( ∇ ) in different  coordinate systems yields the definition of q for the desired coordinate systems.

2.5.3 Newton’s Cooling Law Newton’s cooling law is the fundamental law of convective heat transfer. By using this law, it is possible to calculate the heat transfer rate from a surface to a fluid. Newton’s cooling law can be expressed as, q  = h(Tw − T∞ )

(2.12)

where Tw is the temperature of a solid touching a fluid at T∞ . h with unit of (W/m2 K) is termed as a “convection heat transfer coefficient”. The main difficulty of the use of Newton’s cooling law is the convection heat transfer coefficient

24

2 Basics of Heat Transfer

(i.e., h). Each fluid flows has its own convection heat transfer coefficient. It can be calculated by using correlations suggested in the literature or found computationally directly. By changing the shape of the solid surface (or position of solid surfaces with respect to each other), heat transfer coefficients change. That is why there are many correlations in the literature for many solid shapes or configurations. The convective heat transfer coefficient also depends on other parameters, such as fluid material and type of flow as well as fluid velocity.

2.5.4 Stefan–Boltzmann Law Stefan–Boltzmann is an important law in radiation heat transfer. Radiation heat transfer is a surface phenomenon and the heat transfer rate depends on surface radiative properties. The highest absorption and emission are observed for the black surface. For a black surface, the following equation can be written, E b = σ Ts4

(2.13)

where E b is the emitted radiation flux with unit of W/m2 K, σ is the Stefan–Boltzmann constant whose value is 5.67×10−8 with unit of W/m2 K4 and Ts is the temperature of the black surface with unit of K .

2.5.5 Newton’s Viscous Law Newton’s viscous law states that shear stress applied to a fluid element is proportional to the rate of shear strain (deformation). Assume a two-dimensional flow in which fluid flows in the x direction. If the velocity of the top layer is greater than that of the bottom layer, shear stress emerges in the control volume in the y direction, τ∝

du dy

(2.14)

In order to make Eq. (2.14) equality, a coefficient called “viscosity” can be defined and added to the above equity. Therefore, τ =μ

du dy

(2.15)

2.6 Common Boundary Conditions in Heat Transfer

25

μ with unit of kg/m s is called the dynamics viscosity which is a measure of the resistance of fluid against flowing. In addition to dynamics viscosity, kinematics viscosity shown by ν with unit of m2 /s and definition of ν = μ/ρ is also widely used in heat and fluid flow problems.

2.6 Common Boundary Conditions in Heat Transfer Heat and fluid flow problems are generally expressed in differential equation forms and the solution of those differential equations yields temperature, pressure, and/or velocity. There are many kinds of boundary conditions in heat transfer problems. A boundary condition may be dependent or independent of time, and it might change with location sinusoidally or change as a step function. However, basically, there are three kinds of boundary conditions for heat and fluid flow problems, particularly if the analytical solution is asked. In this section, classical boundary conditions for temperature and velocity are described.

2.6.1 Boundary Conditions for Temperature There are three common boundary conditions for temperature as shown in Fig. 2.5. • Constant surface temperature The temperature at the wall is specified for the problem; for instance, the temperature can be 10 or 90 °C. For Fig. 2.5a and a steady state heat transfer, T (0, y) = T (x, y)|x=0 = Tc

(2.16)

• Constant heat flux Heat flux at a surface is also defined in some heat transfer problems. For instance, a heat flux with 100 W/m2 (i.e., q  = 100 W/m2 ) can be imposed on a surface. The boundary condition for Fig 2.5b can be written as, q  (0, y) = −ks

∂T |x=0 = qs ∂x

(2.17)

It should be mentioned that when the heat flux imposed on the surface is 0 W/m2 , (i.e., q  = 0 W/m2 ) the surface is called adiabatic or it can be said that “the surface is insulated”, meaning that there is no heat transfer at the surface. • Convection heat transfer Heat is transferred from a solid surface to a fluid with constant temperature contacting the surface. Considering Fig. 2.5c, the boundary condition can mathematically be written by using Newton’s cooling law,

26

2 Basics of Heat Transfer

Fig. 2.5 Three types of boundary conditions for the temperature at a solid surface

q  (0, y) = −ks

∂T |x=0 = h(T (0, y) − T∞ ) ∂x

(2.18)

where h is the convective heat transfer coefficient with units of W/m2 K and T∞ is the temperature of the fluid contacting the wall surface. ks is the thermal conductivity of the wall.

2.6.2 Boundary Conditions for Velocity There are many types of boundary conditions for velocity (which is an unknown parameter in many heat transfer problems), such as time-dependent inlet velocity, parabolic inlet velocity, symmetry boundary, periodic conditions, etc. In this section, two common boundary conditions for velocity are mentioned, (1) No-slip surface No-slip boundary condition is observed for the velocity at a solid surface that is in a − → stationary state. All components of the velocity vector (i.e., V ) are zero at the solid surface. (2) Uniform inlet velocity There is a uniform velocity profile at the inlet of a domain. If Fig. 2.6 which is a domain in the Cartesian coordinate system is considered, all components of the velocity vector except the component perpendicular to the inlet surface are zero. The perpendicular velocity component to the inlet surface takes the value of the inlet velocity.

2.7 Dimensionless Numbers in Heat Transfer

27

Fig. 2.6 Two common types of boundary conditions for velocity

2.7 Dimensionless Numbers in Heat Transfer In heat and fluid flow problems, the size, shape, configuration, and working conditions are widely changed. Each problem has its own thermal behavior. In order to generalize a solution that can be used for wide ranges of sizes and operation conditions, the governing equations are made dimensionless. Solving dimensionless governing equations yields dimensionless results. The dimensionless results can be used for wide ranges of sizes and operating temperatures. The dimensional results for any considered case can be easily obtained from the dimensionless solution. The dimensionless parameters appear during the non-dimensionalization of the governing equations. These numbers do not have any units, and each has its own definition. The dimensionless number used in this book and their definitions are given below: • Reynolds number: It is defined as the ratio of inertia to viscous forces, Re =

u Lc ν

(2.19)

− → where u is the scalar quantity and can be any component of the velocity vector ( V ) based on the problem. L c is the characteristic length of the problem and its definition varies with the problem. Finally, ν is the kinematic viscosity. The Reynolds number is widely used in forced convection heat transfer. For instance, it is a parameter used to decide on the type of flow as laminar or turbulent. • Prandtl number: It compares the propagation of momentum and heat in a domain, and it is factional of kinematics viscosity and thermal diffusivity. Pr =

ν α

(2.20)

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2 Basics of Heat Transfer

Fig. 2.7 The effect of Bi number

For mercury, the value of the Prandtl number is around 0.015 showing that the diffusion of heat is faster than momentum propagation in a domain. It is around 10,000 for glycerol, indicating low propagation of heat in a domain comparing momentum transfer. • Biot number: Biot number is the ratio of conduction thermal resistance inside a solid body and convection thermal resistance at the surface. The Biot number is defined as, Bi =

h Lc ks

(2.21)

where h is the convective heat transfer coefficient, L c is the characteristic length, and ks is the thermal conductivity of the solid body. For instance, the right surface of the wall shown in Fig. 2.7 contacts a fluid that is at T∞ . As seen from this figure, T1 > T2 and heat is transferred to the right surface of the fluid with a convection heat transfer coefficient of h. The characteristic length of this problem is L (i.e., L c = L). Figure 2.7 shows the change in the temperature profile in the wall by changing the Biot number. For very small values of Bi (Bi < 0.01), the convective thermal resistance at the surface may be very large, and consequently, a uniform temperature inside the body appears. However, for large values of the Biot number, the convective thermal resistance is smaller than the conduction thermal resistance inside the wall, and therefore, considerable temperature variation inside the wall can be observed.

2.7 Dimensionless Numbers in Heat Transfer

29

• Grashof number: It relates to the ratio of buoyancy to viscous forces. If a fluid domain on which gravity acts downwardly is considered, the buoyancy force can be defined as a force causing the motion of fluid upwardly. The Grashof number is defined as, Gr =

gβT L 3c ν2

(2.22)

where β is the coefficient of thermal expansion of the fluid, g is gravity, T is the temperature difference and L c is the characteristic length. The definition of the Grashof number is discussed in Chap. 10. It might be useful to mention that the definitions of T and L c change with the geometry and operating temperature of the problems. For instance, for natural convection over a vertical plate, L c is the length of the plate, while for natural convection around a cylinder L c is defined as the diameter of the cylinder. • Rayleigh number: Rayleigh number is the product of Grashof and Prantl numbers. Ra = Gr.Pr

(2.23)

or Ra =

gβT L 3c να

(2.24)

The Grashof number compares buoyancy and viscous forces; however, the Rayleigh number additionally involves thermal diffusivity (since velocity and temperature are coupled in natural convection). Therefore, it becomes a more generalized dimensionless quantity than the Grashof number, and it is used widely in natural convection problems. It is a dimensionless number to specify the type of natural convection flow (laminar or turbulent) for any fluid. For instance, for a natural convection flow over a vertical isothermal flat plate, the flow is accepted as laminar when Ra L < 109 . • Nusselt number: Nusselt number compares convection heat transfer with conduction heat transfer at a solid surface. Nu =

h Lc kf

(2.25)

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2 Basics of Heat Transfer

where h is the convective heat transfer coefficient and kf is the thermal conductivity of the fluid. L c is the characteristic length depending on the problem. For high Nusselt numbers (such as Nu = 1000 for heat transfer over a flat plate), a strong convective heat transfer exists at the surface. For heat transfer from a solid surface to a stagnant fluid, the Nusselt number is 1 (Nu = 1) showing that all heat is transferred to the fluid by conduction. • Richardson number: Richardson number is defined by a combination of Grashof and Reynolds numbers.

Ri =

Gr gβT L = 2 u2 Re

(2.26)

where g is the gravitational acceleration, β is the thermal expansion coefficient, T is the reference temperature difference, L is the characteristic length, and u is a component of the velocity vector (a scalar quantity) and represents the characteristic velocity. In convective heat transfer, the Richardson number compares the importance of natural convection. Therefore, the Richardson number takes small values when forced convection is dominant and its value is high for flows with strong natural convection.

2.8 More Terminologies in Heat Transfer In the above subsections, many terminologies used in heat and fluid flow problems are explained and important concepts of heat transfer are discussed. However, there are some additional terminologies that might be useful to mention before finishing this chapter.

2.8.1 Thermophysical Properties Thermophysical properties are physical properties concerning heat and fluid flow. The thermophysical properties of materials have significant effects on the results of a problem. In order to find an accurate result for a heat transfer problem, accurate thermophysical properties must be assigned. The thermophysical properties that are used in this book with their units are given in Table 2.1. In order to find the accurate value of a thermophysical property of a material for a heat transfer problem, the range of temperature and/or pressure (maximum and minimum values of temperature and/or pressure) for the problem must be known.

2.8 More Terminologies in Heat Transfer

31

Table 2.1 Important thermophysical properties in heat transfer Property Abbreviation Thermal conductivity Density Specific heat Dynamics viscosity Kinematic viscosity Thermal diffusivity

k ρ cp μ ν α=

Thermal expansion coeff.

β

Unit (SI) W/mK kg/m3 J/kgK kg/m s m2 /s m2 /s

k ρC p

1/K

In many heat transfer correlations, the temperature for which the thermophysical properties must be found is specified. For instance, for forced convection heat transfer over an isothermal flat plate, film temperature is generally used, Tf =

Tw + T∞ 2

(2.27)

where Tf , Tw , and T∞ are the fluid, wall, and free stream fluid temperatures, respectively. For fluid flow in a pipe whose inlet and outlet temperatures are known, the mean temperature might be meaningful for the determination of fluid thermophysical properties, Tm =

Tin + Tout 2

(2.28)

where Tm , Tin , and Tout are the mean, inlet, and outlet temperatures of flow in the pipe, respectively. There is no specific rule for the temperature at which the thermophysical properties must be found. Generally, each heat transfer correlation has a suggested equation or specified temperature for the determination of the temperature at which thermophysical properties should be calculated. If there is no specified temperature for the determination of the thermophysical properties of a heat transfer problem, the researcher should decide on an acceptable reference temperature.

2.8.2 Flow with Viscous Heat Dissipation The friction between the adjacent fluid layers in a flow, as well as friction between the fluid and solid surface, generates heat which should be taken into account in some heat and fluid flow problems. The viscosity and gradient of velocity in a flow domain are two parameters that affect the generation of heat in a flow. The effect of viscous dissipation is not considered in the problems studied in this book.

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2 Basics of Heat Transfer

2.8.3 Heat Transfer Rate and Heat Flux In many heat transfer problems, the heat transfer rate is defined as the amount of total heat transferred from a surface per unit of time. The unit of heat transfer rate is W . For instance, if 100 J is transferred from a surface in 10 s, the heat transfer rate is, q=

100(J) = 10 W 10(s)

(2.29)

where q is the heat transfer rate. Heat flux is the amount of heat that is transferred from the unit area of a surface per unit of time. Therefore, the following equation can be written between heat transfer rate and heat flux, q  = q/A

(2.30)

The unit of heat flux is W/m2 . For instance, if a total of 100 J of heat is transferred from a surface with an area of 10 m2 in 10 s, the heat flux for the surface is 1 W/m2 .

2.8.4 Volumetric Flow Rate and Mass Flow Rate The volumetric flow rate is the total volume of a fluid passing through a surface in units of time; consequently, the unit of volumetric flow rate is m3 /s. Similarly, the mass flow rate is the total mass passing through a surface in unit of time. The unit of mass flow rate is kg/s. The following relationship can be written between mass and volumetric flow rates, Q = mρ ˙ A

(2.31)

where Q and m˙ are volumetric and mass flow rates, respectively. A and ρ are the areas of the considered surface which is perpendicular to flow and density of flowing fluid, respectively.

2.8.5 Pressure Drop By flowing a fluid, its pressure decreases due to friction that takes place between the infinitesimal adjacent layers of fluid as well as between the fluid and solid surface. The decrease in total pressure from a point in a flow domain to a forward point is called a pressure drop. It is clear that the unit of pressure drop is Pascal in the SI unit system.

2.8 More Terminologies in Heat Transfer

33

2.8.6 Example Assume a cylinder with a length of 1 m and diameter of 10 mm whose surface temperature is 100 ◦ C. Air at 20 ◦ C flows around the cylinder at a velocity of 1 m/s. (a) Use the film temperature and calculate all thermophysical properties of the fluid. (b) Calculate the Reynolds number for the flow. (c) The convective heat transfer coefficient has been calculated as 250 W/m2 K, calculate the heat flux and heat transfer rate from the cylinder by convection mode of heat transfer. (d) If it is assumed that the outer surface of the cylinder is an ideal black surface and the environment is at 20 ◦ C, calculate heat flux and total heat transfer rate from the cylinder by radiation mode of heat transfer Solution: (a) As mentioned, the film temperature is the average temperature of the wall and the flow over the cylinder. Therefore, by using Eq. 2.27, the film temperature can be found as: 100 + 20 Tw + T∞ = = 60 ◦ C Tf = 2 2 Based on this temperature, the thermophysical properties can be found from the thermophysical properties of various heat transfer books such as [6] for air as: ρ = 1.06 kg/m3 , μ = 18.86 × 10−6 m2 /s, k = 0.02808 W/mK. (b) The Re number can be found by using Eq. 2.19: Re =

1.0610.01 ρu D = = 562 μ 18.86 × 10−6

(c) As mentioned in Sect. 2.5.3, Newton’s cooling law equation can be used to determine convective heat transfer from a solid surface, therefore the heat flux from the surface of the cylinder can be calculated easily: q  = h(Tw − T∞ ) = 250(100 − 20) = 20 kW/m2 If this value is multiplied by the total surface area for the cylinder with a 1 m length (which can be found as 2πr L = 2 × 3.14 × 0.005 × 1 = 0.0314 m2 ), the total heat transfer from this cylinder is 0.628 kW. (d) For the determination of radiation heat transfer from a black surface to the surroundings, Eq. 2.13 can be used. So, E b = σ Ts4 = 5.6710−8 (20 + 273)4 = 417.88 W/m2

34

2 Basics of Heat Transfer

The total heat transfer by radiation from the cylinder can be found by multiplication of E b and the total area of the cylinder, which is 2πr L, therefore  = 2 × 3.14 × 0.005 × 1 × 417.88 = 13.12 kW qrad

2.9 Review Questions

Questions (Q1) There is a linear variation between strain and stress for Newtonian fluids (Q2) When heat transfer in a process does not change with time, the process is called a steady state process (Q3) In laminar flows, there is an additional stress in the flow due to 3D timedependent small fluctuations of fluid Q4) The first law of thermodynamics relates to the conservation of energy, while the second law is about the direction of heat transfer (Q5) Heat transfer can be classified into three modes: conduction, convection, and radiation heat transfer (Q6) Conduction heat transfer can occur without the presence of a material (Q7) Fourier’s law is the basis of conduction heat transfer (Q8) Stefan–Boltzmann’s law is an important law relating to convective heat transfer (Q9) Viscosity is a parameter referring to resistance against fluid flow (Q10) For low values of the Biot number (Bi  1, there is a large change in temperature in the solid body (Q11) Rayleigh number relates to forced convection, while Reynolds number is used in natural convection (Q12) Generally, for convection boundary conditions, the temperature of the surface is known (Q13) Reynolds number is the ratio of inertia forces and viscous forces (Q14) Richardson number is the combination of Grashof and Rayleigh numbers (Q15) Biot number is the ratio of conduction thermal resistance inside a solid body and convection thermal resistance at the surface (Q16) The unit of heat flux is W , while the unit of heat transfer rate is W/m 2 (Q17) Nusselt number is the ratio of conduction and convection thermal resistances of fluid at the surface (Q18) The unit of volumetric flow rate is kg/s, while the unit of mass flow rate is m3 /s

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2.10 Problems Problem 1: Consider the following figure showing an interface at a temperature of Ts between solid and fluid regions. The surface touching the fluid is accepted as

2.10 Problems

35

Fig. 2.8 Conduction, convection, and radiation heat transfer at an interface between a solid and a fluid for Problem 1

a black surface. Heat is transferred by conduction to the interface, while it leaves the interface by convection and radiation modes. Based on the heat transfer laws explained in this chapter,    (a) Define qx1 , qx2 and qx3 , (b) Write an equation concerning the energy balance between heat fluxes at the interface (Fig. 2.8).

Problem 2: Air at 20 ◦ C flows over a flat plate that is 100 ◦ C and length of 1 m. Assume that the depth of the plate is 1 m. (a) Find the values of dynamic viscosity, density, thermal conductivity, and specific heat. Use the film temperature for finding values of thermophysical properties. Do not forget to write the units. (b) Calculate values of kinematics viscosity, thermal diffusivity, and Prandtl number from the above values. (c) What is the Reynolds number for the plate? (d) If the average convective heat transfer coefficient for forced convection heat transfer from the plate is 100 W/m2 K, find the total heat transfer rate from the surface. Problem 3: Consider a vertical plate with a height of 0.1 m and depth of 1 mm which is at 80 ◦ C. The vertical plate is in contact with air at 50 ◦ C. The average heat transfer coefficient for natural convection heat transfer from the plate is 20 W/m2 k. Assume that the depth of the plate is 1 m. (a) Find the values of dynamics and kinematics viscosity, density, specific heat, thermal conductivity, and diffusivity of air at film temperature. (b) Calculate the values of Grashof, Rayleigh, and Nusselt numbers for the plate. (c) What is the total heat transfer and heat flux between the plate and air?

Chapter 3

One-Dimensional Unsteady Heat Conduction in a Cartesian Coordinate System

3.1 Introduction As mentioned in Chap. 2, conduction heat transfer is one mode of heat transfer. In this section, a problem of conduction heat transfer in the Cartesian coordinate system for a one-dimensional unsteady state case is introduced and solved. The solution of one-dimensional steady state heat conduction, as well as lumped analysis (one dimensional in time), can be found in many undergraduate heat transfer books (e.g. [4–6]); therefore, those problems are not discussed in this book. In the problem of this chapter, a constant temperature boundary condition is applied on both surfaces of the wall; however, a solution for further one-dimensional unsteady heat conduction problems (such as unsteady state conduction heat transfer in a wall with constant heat flux or convection heat transfer boundary conditions or for conduction heat transfer in a wall with heat generation) can be found in Refs. [7, 10].

3.2 Problem Definition One of the classical conduction heat transfers is the heat transfer in a wall whose temperature depends on location and time, as shown in Fig. 3.1. The height and depth of the wall are too long; therefore, heat transfer occurs only in the x direction. Initially, the entire wall is at a temperature of Ti , and suddenly, the temperatures of the right and left sides of the wall change to 0 (the unit of this temperature depends on the reader). The temperature inside the wall is a function of two independent variables x and t (location and time). The temperature in the wall changes from point to point, and it varies with time. After a long period of time (t → ∞), a uniform temperature in the entire wall is expected which is equal to the temperature of the boundaries (i.e., 0). Our aim in this chapter is to find an analytical function in terms of location and time (i.e. x and t) yielding temperature at any point and at any time in the wall.

© The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2023 M. Mobedi and G. Gediz Ilis, Fundamentals of Heat Transfer, https://doi.org/10.1007/978-981-99-0957-5_3

37

38

3 One-Dimensional Unsteady Heat Conduction in a Cartesian …

Fig. 3.1 Considered wall and cross-section of the wall with the defined initial and boundary conditions

3.3 Assumptions In this chapter, the analytical solution for the problem explained above is obtained under the following assumptions: (a) The problem is unsteady one-dimensional conduction heat transfer in the Cartesian coordinate system. (b) All thermophysical properties of the wall are constant. (c) There is no heat generation in the wall.

3.4 Implementation of Assumptions The vector form of the heat conduction equation is derived in many heat transfer books (e.g., [6, 7, 10]) and can be written as, ∂(ρcp )T − → − → = ∇ . (k ∇ T ) + q˙ ∂t

(3.1)

3.4 Implementation of Assumptions

39

− → The definition of ∇ in the Cartesian coordinate system is ∂ − ∂ − ∂ − → → → − → i + j + k ∇ = ∂x ∂y ∂z

(3.2)

Substituting Eq. 3.2 into Eq. 3.1 yields the open form of the heat conduction equation in Cartesian coordinates: ∂(ρcp )T ∂ = ∂t ∂x

 k

∂T ∂x

 +

∂ ∂y

    ∂T ∂ ∂T k + k + q˙ ∂y ∂z ∂z

(3.3)

Based on the assumptions explained in the previous subsection, all thermophysical properties of the wall are constant, therefore, ρ, cp , and k can be removed from the derivatives. The last term in Eq. 3.3 can be removed since there is no heat generation in the wall. Based on these assumptions, the conduction heat transfer equation for the considered wall takes the following form, ∂2T 1 ∂T ∂2T ∂2T = + + α ∂t ∂x2 ∂ y2 ∂z 2

(3.4)

where α = ρck p is called the thermal diffusivity with units of m2 /s. The temperature in the wall changes only in the x direction since the wall is too long in the y and z directions (or the wall may also be insulated from side surfaces in the y and z directions). Since the problem is in an unsteady state, the temperature in the wall also changes with time. Mathematically, it can be said that T = f (x, t), therefore, ∂2T ∂ 2 T ∂ 2 T 1 ∂T = + 2 + 2 2 α ∂t ∂x ∂z  ∂ y

(3.5)

The simplified form of the heat conduction equation for this problem is, ∂2T 1 ∂T = α ∂t ∂x2

(3.6)

Equation 3.6 is the first-order partial differential equation in time and second-order in space (x direction). Therefore, an initial condition and two boundary conditions are required to obtain the solution. As seen from Fig. 3.1, the right and left surfaces of the wall are at 0 and the wall initially is at Ti . Mathematically, the boundary conditions can be written as follows: x =0→T =0 x =L→T =0

(3.7)

40

3 One-Dimensional Unsteady Heat Conduction in a Cartesian …

The initial condition is: t = 0 → T = Ti

(3.8)

3.5 Analytical Solution There are analytical methods for solving Eq. 3.6; however, in this book, we use the separation of variables method to solve the heat conduction equation for this problem. Figure 3.2 is used to describe the separation of variables method for this problem. As seen, since the solution of Eq. 3.6 is difficult, the temperature function can be divided into two functions as X (x) and (t) such that the multiplication of two functions yields T (x, t), mathematically it can be written as, T (x, t) =

X (x)   

.

(t) 

function of x only function of t only

Fig. 3.2 Description of separation of variables method for the problem of this chapter

(3.9)

3.5 Analytical Solution

41

If we find analytical expressions for the functions of X (x) and (t), they can be multiplied by each other, which will yield T (x, t). In order to find X (x) and (t), the derivatives of T (x, y) defined by Eq. 3.9 with respect to x and t can be taken as shown below: ∂T d(t) = X (x). = X (x).  (t) ∂t dt dX (x) ∂T = .(t) = X  (x).(t) ∂x dx ∂2T ∂ 2 X (x) = .(t) = X  (x)(t) ∂x2 ∂x2

(3.10) (3.11) (3.12)

By substituting Eqs. 3.10 and 3.12 into Eq. 3.6, we obtain: 1 X (x)  (t) = X  (x).(t) α

(3.13)

which can also be written as: 1   (t) X  (x) = α (t) X (x)       function of time

(3.14)

function of x

As seen from the above equation, the left side of the equation is in terms of t, while the right side is the function of x. They can be equal to each other when they are equal to a constant such as −β 2 . 1   (t) X  (x) = = −β 2 α (t) X (x)

(3.15)

Therefore, it is possible to separate the above equation into two ordinary differential equations as, d(t) + αβ 2 (t) = 0 dt

(3.16)

d2 X (x) + β 2 X (x) = 0 dx 2

(3.17)

Now, we have two ordinary differential equations. Eq. 3.16 is the first-order ODE with an independent variable of t, while Eq. 3.17 is a second-order ODE with an independent variable of x. If the solution of these ODEs is found and multiplied with each other, the solution for T (x, t) will be found. In order to find the solution of

42

3 One-Dimensional Unsteady Heat Conduction in a Cartesian …

Eqs. 3.16 and 3.17, it is necessary to write the initial and boundary conditions for functions of X (x) and (t). Two boundary conditions for Eq. 3.17: Function

 x = 0 → T (0, t) = X (0). (t) = 0 → X (0) = 0

(3.18)

Function

 x = L → T (L , t) = X (L). (t) = 0 → X (L) = 0

(3.19)

The following form of the initial condition will be useful to find the solution for Eq. 3.16, t = 0 → T = Ti → T (x, 0) = X (x).(0) = Ti

(3.20)

The second-order partial differential equation (Eq. 3.6) and its initial and boundary conditions (Eqs. 3.7 and 3.8) are divided into two independent ODEs. Now, it is time to solve the derived ODEs. Solution of the first-order ODE: The function of (t) must be found as an analytical expression for T (x, t). If Eq. 3.16 is considered and both sides of the equation are divided by (t) and multiplied by dt, to the following equation can be obtained. d(t) = −αβ 2 dt (t)

(3.21)

Taking the integral from the above equation yields the function of (t). The process of integration is shown below. We should not forget that α and β are constants. 

 d(t) = −αβ 2 dt dt ln (t) = −αβ 2 t + ln c

ln

(3.22)

(t) 2 = −αβ 2 t → (t) = ce−αβ t c

Therefore, the general solution of 3.16 can be written as, (t) = ce−αβ

2

t

(3.23)

3.5 Analytical Solution

43

Solution for the second-order ODE: Equation 3.17 is a second-order ordinary differential equation. The solution of this equation can be found in many differential equation books [1, 2]. X (x) = C1 cos βx + C2 sin βx

(3.24)

C1 and C2 are constants and their values should be obtained from the boundary conditions. The first boundary condition (Eq. 3.18) at X = 0 can be used as: X = 0 → X (0) = 0 C1 cos(0) + C2 sin(0) = 0 C1 = 0

(3.25)

Therefore, Eq. 3.24 takes the following form, X (x) = C2 sin βx

(3.26)

Now, it is time to find the second constant which is C2 by using the second boundary condition (Eq. 3.19) at X = L, X = L → X (L) = 0 → X (L) = C2 sin β L = 0

(3.27)

There are two choices that make the value of X (x) zero. The first possibility is that C2 can be zero, and another choice is that sin β L can be zero. Keep in mind that based on the separation of the variable method, T (x, t) is the multiplication of X (t) and (t). C2 cannot be zero since in that case X (x) = 0 , which makes the function of T (x, t) = 0. Therefore, the term sin β L must be 0. sin β L = 0

(3.28)

sin β L can be zero, when the value of β L is 0, π , 2π , 3π ... . In other words, sin β L = nπ → n = 0, 1, 2, ... β L = nπ

(3.29)

Therefore, β can take infinite values. The value of βn can be calculated from the following equality, βn =

nπ L

(3.30)

44

3 One-Dimensional Unsteady Heat Conduction in a Cartesian …

where n is an integer number and its value changes from 0 to ∞. Therefore, there is no one solution for the second-order ODE (i.e., Eq. 3.17), since βn has many values. The parameter of βn is called the eingen value and the solution of X (x) is named the eigen function. The complete solution for the ODE of Eq. 3.17 can be the summation of all terms involving all values of βn , X (x) =

∞

cn sin(βn x)

(3.31)

n=0

Equations 3.23 and 3.31 are the solutions of X (x) and (t), which can provide the solution of T (x, t), which is the product of these two functions. T (x, t) = X (x).(t) ∞ 2 cn e−αβn t sin(βn x) T (x, t) =

(3.32) (3.33)

n=0

Now, the only difficulty is the coefficient cn , which is unknown. Two boundary conditions were used to find the solution of X (x) but the initial condition has not yet been used. Hence, it is possible to find the value of cn by using the initial condition. t = 0 → T = Ti → Ti =

∞

cn .e−αβ

2

.0

sin(βn x)

(3.34)

n=0

Therefore, t = 0 → Ti =

∞

cn sin(βn x)

(3.35)

n=0

Finding cn from the above equation seems slightly difficult. Fortunately, there is a theorem in mathematics called the orthogonality theorem, and mathematically, it is defined below, L 0

⎧ ⎪ ⎨0 for n = m sin(βn x). sin(βm x)dx = L 2 ⎪ ⎩ (sin(βn x)) for n = m

(3.36)

0

On the other hand, Eq. 3.36 can be expanded and written as follows: Ti = c1 sin β0 x + c2 sin β1 x + c3 sin β2 x + · · · + cn sin βn x + · · ·

(3.37)

Let us use the orthogonality theorem by multiplying of both sides of Eq. 3.37 by sin(βn x) and take the integral of both sides of equality from 0 to L.

3.5 Analytical Solution

45

L

L Ti sin(βn x)dx =

0

[c1 sin β0 x sin βn x + c2 sin β1 x sin βn x + · · · 0

+ · · · + cn sin βn x sin βn x + · · · ]dx Based on the orthogonality theorem, all terms of the right side of the above equation are zero since the subindices are different except when both subindices are n. Therefore, the above equality takes the following form, L

L Ti (sin βn x)dx =

0

cn sin2 (βn x)dx

(3.38)

0

The integral of the right and left sides of Eq. 3.38 can be found as, L Ti sin(βn x)dx = 0

Ti [1 − (−1)n ] βn

(3.39)

L cn sin2 (βn x)dx = cn L/2

(3.40)

0

If we substitute the results of the above integrals into Eq. 3.38, the coefficient of cn will be obtained, Ti [1 − (−1)n ] cn L = βn 2 2Ti [1 − (−1)n ] cn = βn L

(3.41) (3.42)

By substituting the value of constant cn into Eq. 3.33, a function for T (x, t) can be obtained. T (x, t) =

∞ 2Ti [1 − (−1)n ] n=0

βn L

e−αβn t sin(βn x) 2

(3.44)

It is possible to simplify the above expression since the value of cn is 0 for even numbers and 1 for odd numbers,

46

3 One-Dimensional Unsteady Heat Conduction in a Cartesian …

n = 0 → [1 − (−1)0 ] = 0 n = 1 → [1 − (−1)1 ] = 2 n = 2 → [1 − (−1)2 ] = 0 n = 3 → [1 − (−1)3 ] = 2

(3.45)

n = 4 → [1 − (−1)4 ] = 0 n = 5 → [1 − (−1)5 ] = 2 Therefore, the last form of the solution of Eq. 3.6 with initial and boundary conditions given by Eqs. 3.7 and 3.8 becomes, T (x, t) =

∞

4Ti −αβn2 t e sin(βn x) β L n=1,3,5... n

(3.46)

where βn =

nπ L

(3.47)

By substituting the values of t and x into the above equation and by considering sufficient terms, the value of temperature at any point in the wall and at any time can be easily calculated.

3.6 Example Consider a long and wide plate with a thickness of 100 mm. The material of the wall is plain steel (AISI 1010) as shown in Fig. 3.3. The initial temperature of the wall is 100 °C. Suddenly the temperature of the side surfaces of the wall drops to 0 °C. Find the temperature of the plate at t = 1, 3, 7 and 10 s.

Fig. 3.3 Considered wall in the Example

3.6 Example

47

Solution: In order to find the solution to this problem, the following steps are suggested, Step 1: Necessary assumptions Based on the explained problem, the following assumptions can be written. (a) Nothing is mentioned about heat generation in the problem; therefore, there is no heat generation. (b) Nothing is mentioned about the change in thermophysical properties with temperature; therefore, all thermophysical properties are constant. (c) The plate is long and wide; therefore, most of the heat is transferred in one direction (see Fig. 3.3). Step 2: Governing equation, initial and boundary conditions The governing equations for the problem is heat conduction equation. The problem is one-dimensional steady conduction heat transfer. The temperature in the wall depends on the location and time. Therefore, the simplified form of the governing equation for this problem can be written as ∂2T 1 ∂T = α ∂t ∂x2 Based on the explanation of the problem, the initial and boundary conditions for the problem can be written as x =0→T =0 x = 100 mm → T = 0 Initial condition: t = 0 → T = 100 ◦ C Step 3: Thermophysical properties It seems that the values of all necessary parameters are given in the problem except thermal diffusivity. Sometimes, the value of thermal diffusivity is directly given in the tables, but sometimes it should be calculated. The thermal properties of plain Carbon AISI 1010 can be found as k = 60.5 W/mK, ρ = 7854 kg/m3 , C p = 434 J/kgK. Therefore, the value of thermal diffusivity (α = k/ρcp ) can be calculated as 1.775 × 10−5 m2 /s. Step 4: Solution of the heat conduction PDE The solution of the one-dimensional unsteady heat conduction equation is given by Eq. 3.46. Substituting the values of the thermophysical properties, geometrical parameters, and initial condition in Eq. 3.46 yields a function in terms of x and t. The temperature in the wall at any point and any time can be calculated by using the equation below,

48

3 One-Dimensional Unsteady Heat Conduction in a Cartesian …

T (x, t) =

∞

4 × 100 −1.775x10−5 βn2 t e sin(βn x) β × 0.1 n=1,3,5... n

Considering Eq. 3.47, the values of βn are π /0.1, 3π /0.1, 5π /0.1, 7π /0.1, 9π /0.1 .... or βn = 31.4159, 94.2478, 157.0796, 219.9115, 282.7433.... Step 5: Demonstration of the results Demonstration of the results can be done by plotting temperature profiles at different instants. The temperature profile at different instants can be plotted by using Eq. 3.46. The only difficulty is that n starts from 1 and continues to ∞. Since it is impossible to calculate infinite terms, the temperature distribution should be plotted for a sufficient number of terms. Figure 3.4 shows results for the cases by considering 3 and 21 terms for t = 1 s. As seen, by the increasing number of terms in the series of Eq. 3.46, smoother curves for t = 1 s can be obtained. It was observed that n = 11 is sufficient to remove any fluctuation at the beginning of the process (t = 2 s). Initially, the wall is at 100 °C, and with increasing time, the temperature in the wall decreases, and finally, the temperature in the wall drops to the surrounding temperature. Therefore, by using Eq. 3.46, it is possible to find the temperature at any point in the wall and for any required instant. Figure 3.5 shows the temperature profiles for different instants in the wall when n = 11.

Fig. 3.4 Temperature profiles for t = 1 s. when the number of considered terms in the series of Eq. 3.46 is 3 and 21

3.7 Review Questions

49

Fig. 3.5 Temperature profiles for different instants in the wall when n = 11

3.7 Review Questions

Question (Q1) The problem studied in this chapter is an unsteady one-dimensional heat conduction problem (Q2) In the problem of this chapter, the temperature is the function of x and t (Q3) The heat conduction equation solved in this chapter is a first-order ordinary differential equation (Q4) The heat conduction equation of this chapter is a second-order partial differential equation (Q5) The heat conduction equation of this chapter is solved numerically (Q6) One boundary condition and one initial condition are sufficient to solve the heat conduction equation of this chapter and find results (Q7) Two boundary conditions and one initial condition are sufficient to find the solution for the heat conduction equation of this chapter (Q8) The separation of variables method is used to solve the heat conduction equation of this chapter (Q9) In the separation of the variable method used in this chapter, the function of T (x, t) is separated into two equations as X (x), T (t) (Q10) In the separation of the variable method used in this chapter, the function of T (x, t) is divided into two equations as X (x), and (t) (Q11) The differential equations of the functions of X (x) and (t) are the second-order ordinary differential equations (Q12) Differential equation of function of X (x) is a second-order ODE while the differential equation of (t) is the first-order ODE (Q13) In order to find a solution for the second-order ODE, boundary conditions should also be defined in terms of X (x) (Q14) The solution of the heat conduction equation consists of only one term (Q15) The solution of the heat conduction equation consists of the summation of infinite terms (Q16) An appropriate result can be found when sufficient terms from infinite terms of the solution will be taken into account

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3 One-Dimensional Unsteady Heat Conduction in a Cartesian …

Fig. 3.6 Figure of Problem 4

3.8 Problems For the below problems, follow the solution style of the example in this chapter. All steps mentioned in the Solution part of the example should be followed one by one. Problem 1) A long wall is initially at 0 °C, suddenly the temperature of the outer surfaces of the wall increases to 50 °C. The material of the wall is plastic with a thermal diffusivity of 10−5 m2 /s. The thickness of the wall is 100 mm. Find the temperature at t = 10, 30, and 60 min after starting the heating process. Problem 2) There is a long wall with an initial temperature of 80 °C and a thickness of 10 mm. The left side of the wall is thermally insulated while the temperature of the right side can be controlled. Suddenly, the temperature of the right side is reduced to 10 °C. The material of the wall is aluminum with α = 10−4 m2 /s, find temperature profiles in the wall at t = 1, 20, and 60 s. Problem 3) There are two walls with a thickness of 5 mm, one from concrete with thermal diffusivity of 6 × 10−6 m2 /s and the other from copper with a thermal diffusivity of 1.17 × 10−4 . The wall is at an initial temperature of 30 °C and suddenly the outer surfaces of both walls are reduced to 0 °C. Draw the change of center temperature of both walls with respect to time. Problem 4) Assume a rod with a length of 1 m and a cross-section of 50×50 mm2 . It is made of wood and is accepted as an isotropic material. The thermal diffusivity of wood can be accepted as 3 × 10−7 m2 /s. Initially, the entire wood is at 20 °C, but suddenly the temperature of right and left surfaces decrease to 0 °C. Find the temperature change at point A with respect to time. The coordinate of point A is (40 mm, 40 mm). See Fig. 3.6.

Chapter 4

Two-Dimensional Steady Heat Conduction in Cartesian Coordinate System

4.1 Introduction Unsteady state heat transfer in a wall was studied in Chap. 3. However, there are many industrial heat transfer processes in which heat transfer is steady state. That is why in this chapter, a steady state conduction heat transfer is solved. In steady state heat transfer problems, the temperature does not change with time and depends only on the location in the considered domain. One-dimensional steady state conduction heat transfer in a wall can be found in many undergraduate heat transfer textbooks [4–6]. In this chapter, two-dimensional steady state conduction heat transfer is studied for a long bar or a bar in which the front and backsides are thermally insulated. The boundaries of the studied 2D domain are at 0 (the unit depends on the problem) except for one boundary which is at a nonzero constant temperature. Solutions for problems with other kinds of thermal boundary conditions (such as constant heat flux or heat transfer by convection) can be found in other heat transfer textbooks [7, 10].

4.2 Problem Definition The problem is two-dimensional heat transfer in a long rod with a square crosssection. Since the problem is a steady state problem, the temperature changes only with the location. The considered coordinate system is Cartesian, and it is assumed that temperature is a function of x and y as shown in Fig. 4.1. The direction for depth is shown by z and the temperature does not change in this direction. This chapter aims to easily find an analytical expression for temperature in terms of x and y yielding temperature at any point of the bar.

© The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2023 M. Mobedi and G. Gediz Ilis, Fundamentals of Heat Transfer, https://doi.org/10.1007/978-981-99-0957-5_4

51

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4 Two-Dimensional Steady Heat Conduction in Cartesian Coordinate …

Fig. 4.1 Definition of the problem: a the considered bar in the Cartesian coordinate system, b the studied two-dimensional steady state conduction heat transfer problem

4.3 Assumptions The analytical solution for this problem is obtained under the following assumptions: (a) The problem is at a steady state, and the temperature does not change over time. (b) The problem is two dimensional, and the temperature does not change in the depth of the bar. Temperature changes in the x and y directions; however, it does not vary in the z direction. (c) All thermophysical properties of the long bar are constant and do not change with temperature. (d) There is no heat generation in the considered long bar.

4.4 Implementation of Assumptions As mentioned before, the vector form of the heat conduction equation can be written as ∂(ρcp )T − → − → = ∇ . (k ∇ T ) + q˙ ∂t

(4.1)

When the thermophysical properties are constant, the open form of the equation in the Cartesian coordinate system can be written by Eq. 4.2. ∂2T 1 ∂T ∂2T ∂2T q˙ = + + 2 + 2 2 α ∂t ∂x ∂y ∂z k

(4.2)

4.5 Analytical Solution

53

Based on the explained assumptions, the first term on the left side and the last two terms on the right side are zero since temperature does not change by time and also in the z direction and additionally there is no heat generation. ∂2T ∂2T ∂2T q˙ 1 ∂T = + + + 2 2 2 ∂x ∂y ∂z k α ∂t   =0

=0

(4.3)

=0

Therefore, the simplified form of the governing equation that should be solved to find the temperature for this problem is, ∂2T ∂2T + =0 ∂x2 ∂ y2

(4.4)

The dependency of temperature on x and y can be seen from the above partial differential equation. Since the problem is in a steady state, there is no initial condition. The square bar is insulated at the left and right walls as well as at the top surface. The bottom surface of the square bar is maintained at a constant temperature. By considering Fig. 4.1, the boundary conditions for Eq. 4.4 are, x = 0 → T (0, y) = 0 x = L → T (L , y) = 0 Y = L → T (x, L) = 0

(4.5)

Y = 0 → T (x, 0) = Tw where Tw is the temperature of the bottom surface.

4.5 Analytical Solution The “separation of variables” method is used to solve the above partial differential equation (Eq. 4.4) under the given boundary conditions (Eq. 4.5). The function of temperature can be split into two independent functions as X (x) and Y (y) such that their multiplication yields T (x, y). Based on this assumption, Eq. 4.4, which is a PDE, can also be separated into two ODEs in terms of X (x) and Y (y). If analytical solutions for X (x) and Y (y) are found, an expression for T (x, y) will be obtained easily. Figure 4.2 shows the application procedure of “separation of variables” for the problem of this chapter and this procedure is followed in this section. As shown in Fig. 4.2, it is possible to split T (x, y) into two functions: X (x) and Y (y). The multiplication of X (x) and Y (y) yields the function of T (x, y),

54

4 Two-Dimensional Steady Heat Conduction in Cartesian Coordinate …

Fig. 4.2 Application procedure of “separation of variables” for the problem of this chapter

T (x, y) = X (x).Y (y)

(4.6)

The second derivative of T (x, y) in Eq. 4.6 with respect to x and y can be obtained easily, ∂2T = X  (x).Y (y) ∂x2 ∂2T = X (x).Y  (y) ∂ y2

(4.7) (4.8)

By substituting these derivatives into Eq. 4.4, the following equation can be obtained. X  (x).Y (y) + X (x).Y  (y) = 0

(4.9)

If both sides of the above equation are divided by X (x).Y (y), the following equation will appear. The first term depends on x only, while the second term of Eq. 4.10 depends on y. They can be equal to each other when they are equal to a constant such as −β 2 .

4.5 Analytical Solution

55

X  (x) Y  (y) =− = −β 2 X (x) Y (y)       function of x

(4.10)

function of y

The above mathematical operation yields two separate second-order ordinary differential equations, X  (x) + β 2 X (x) = 0 

(4.11)

Y (y) − β Y (y) = 0 2

(4.12)

If the solution of these ODEs is obtained, they can be multiplied by each other and an analytical expression for T (x, y) will be found. In order to solve Eqs. 4.11 and 4.12, their boundary conditions are also needed. The boundary conditions for the problem given by Eq. 4.5 must be written in terms of X (x) and Y (y). The boundary conditions for the ODE in terms of x can be found as, a value a function

      x = 0 → T (0, y) = 0 → T (0, y) = X (0) . Y (y) = 0 ↓ X (0) = 0 a value

(4.13)

a function

      X = L → T (L , y) = 0 → T (L , y) = X (L) . Y (y) = 0 ↓

(4.14)

X (L) = 0 Therefore, in order to find an expression for X (x), the following second-order ODE under the obtained boundary conditions must be solved. X  (x) + β 2 X (x) = 0 x = 0 → X (0) = 0 x = L → X (L) = 0

(4.15)

Similarly, the boundary conditions for the second ODE (Eq. 4.12) in terms of Y (y) are also obtained as follows: y = L → T (x, L) = 0 → X (x).Y (L) = 0 → Y (L) = 0 y = 0 → T (x, 0) = X (x).Y (0) = Tw

(4.16)

Therefore, the second ODE in terms of Y (y) and its boundary conditions that should be solved are,

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4 Two-Dimensional Steady Heat Conduction in Cartesian Coordinate …

Y  (y) − β 2 Y (y) = 0 y = L → Y (L) = 0 y = 0 → X (x).Y (0) = Tw

(4.17)

Solution for the first ODE The general solution of the first ODE (Eq. 4.11) is given in many differential equation textbooks [1, 2]. (4.18) X (x) = c1 cos(βx) + c2 sin(βx) where C1 and C2 are constants of ODE and should be determined by using boundary conditions. From the first boundary condition of Eq. 4.15, x = 0 → c1 cos(0) + c2 sin(0) = 0 → c1 = 0

(4.19)

Therefore, the solution of the first ODE takes the following form: X (x) = c2 sin(βx)

(4.20)

The second boundary condition (see Eq. 4.15) can be applied, x = L → X (L) = 0 → c2 sin(β L) = 0

(4.21)

There are two possibilities for Eq. 4.21 to be zero, (a) the value of c2 can be zero (i.e., c2 = 0), in that case c2 = 0 → X (x) = 0 → T (x, y) = X (x).Y (y) = 0

(4.22)

As seen, when the value of c2 = 0, the function of X (x) becomes zero; therefore, the values of temperature in the entire domain must be zero which is impossible; therefore, the value of c2 cannot be zero. (b) The second possibility is that sin(β L) = 0

(4.23)

The value of L is constant; therefore, in order to make sin(β L) = 0, the value of β must be found. sin(β L) = 0 → sin(β L) = sin(nπ ) (4.24) where n = 0, 1, 2, 3.... Therefore, the value of β can be expressed as,

4.5 Analytical Solution

57

βn =

nπ L

(4.25)

where n is an integer whose values change from 0 to ∞. Therefore, the general solution for the first ODE (Eq. 4.15) with its boundary conditions can be written as, X (x) = c sin(βn x)

βn =

nπ L

(4.26)

The solution of this ODE has many terms since n has many values (n = 0, 1, 2, 3, ....); therefore, the solution can be written as the summation of the solution for each n. X (x) = c1 sin(β1 x) + c2 sin(β2 x) + · · ·

(4.27)

or mathematically the solution of Eq. 4.15 can be written as X (x) =

∞ 

cn sin(βn x)

βn =

n=o

nπ L

(4.28)

Solution for the second ODE The general solution of the second ODE given by Eq. 4.17 can also be found in many differential equation textbooks [1, 2]. In order to find the solution, the roots of the following equation should be found as (D 2 − β 2 )Y = 0 → D = ±β

(4.29)

Therefore, the general solution for the second-order ODE in terms of Y (y) can be written as, Y (y) = c1 e+βy + c2 e−βy

(4.30)

The coefficients of c1 and c2 are constants, and their values should be found by using boundary conditions. If the boundary condition at y = L is applied, Y (L) = c1 e+β L + c2 e−β L = 0 → c1 = −c2

e−β L e+β L

(4.31)

The coefficient of c1 can be substituted in Eq. 4.30 and then the following equality is obtained,   2c2 e−β(y−L) − eβ(y−L) (4.32) y(x) = β L e 2

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4 Two-Dimensional Steady Heat Conduction in Cartesian Coordinate …

The coefficient of 2c2 /eβ L has an arbitrary value and can be renamed as C, hence C = 2c2 /eβ L . The definition of the hyperbolic sine function can be used to make the solution equation shorter, eβ(L−y) − e−β(L−y) (4.33) sinh(β(L − y)) = 2 Finally, the general solution of the second ODE takes the following form, Y (y) = C sinh([β(L − y)])

(4.34)

As we mentioned before, the solution of T (x, y) is the multiplication of two functions of X (x) and Y (y); therefore, T (x, y) =

∞ 

Cn sinh[βn (L − y)] sin(βn x)

(4.35)

n=0

where Cn is the product of C (from Eq. 4.34) and cn (from Eq. 4.28). The only problem that remains is the value of Cn . Similar to the unsteady one-dimensional conduction heat transfer described in the previous chapter, the theory of orthogonality can be used to find an equation for Cn . Let’s rewrite the orthogonality theorem as, L 0

⎧ ⎪ ⎨0 n  = m sin(βn x). sin(βm x)dx = L 2 ⎪ ⎩ (sin(βn x)) n = m

(4.36)

0

The boundary condition at y = 0 can be applied to Eq. 4.35 which is a general solution, T (x, 0) = Tw =

∞ 

Cn sin(βx). sinh(β L)

(4.37)

n=0

Both sides of Eq. 4.37 can be multiplied by sin(βm x), and then the integral of both sides in the x direction from 0 to L can be taken, L

L

Tw sin(βm x)dx = 0

∞ 

Cn sin(βm x) sin(βn x) sinh(β L)dx

(4.38)

n=0 0

Based on Eq. 4.36, the integral value of all terms on the right side of Eq. 4.38 is zero except for a term in which n = m. Therefore, a solution exists when n = m and Eq. 4.38 can be simplified to the following form,

4.6 Example

59

L

L Tw sin(βn x)dx = Cn . sinh(βn L).

0

(sin(βn x))2 dx

(4.39)

0

The integral of the left side of Eq. 4.39 can be found as, L Tw sin(βn x)dx = 0

Tw [1 − (−1)n ] βn

(4.40)

The integral of the right side of Eq. 4.39 can also be found as, L (sin(βn x))2 dx = Cn sinh(βn L)

L 2

(4.41)

0

By substituting Eqs. 4.40 and 4.41 into Eq. 4.39, the value of the Cn can be obtained, Cn =

2Tw [1 − (−1)n ] βn L sinh(βn L)

(4.42)

For the even values of n, the value of Cn is zero, and for odd values, it takes the following form: Cn =

4Tw βn L sinh(βn L)

(4.43)

Substituting Cn into Eq. 4.35 yields the solution for the heat conduction equation of this chapter (Eq. 4.4). T (x, y) =

∞ 

4Tw sin(βn x) sinh(βn (L − y)) β L sinh(β n n L) n=1,3,5...

(4.44)

By putting the values of x and y in the above equation, it is possible to easily find the temperature at any point in the square bar of Fig. 4.1.

4.6 Example The cross-section of a long bar is illustrated in Fig. 4.3. All walls are at 0 °C except the bottom wall which is at 50 °C. The width and height of the bar are 100 mm. Draw temperature profiles at y = 20, 40, 60 and 80 mm.

60

4 Two-Dimensional Steady Heat Conduction in Cartesian Coordinate …

Fig. 4.3 Cross-section of the square bar and boundary conditions

Solution: In order to solve this problem, the following steps can be followed, Step 1: Necessary assumptions Based on the explanation of this problem, the following assumptions can be written, (a) Nothing is mentioned about the change in temperature with time; therefore, the problem is a steady state. (b) Nothing is mentioned about heat generation in the bar; therefore, there is no heat generation. (c) Nothing is mentioned about the change in thermophysical properties with temperature, therefore all thermophysical properties are constant. (d) The bar is long (in the z direction); therefore, is no heat transfer in the z direction and the heat flux changes in the x and y directions. Step 2: Governing equation, initial and boundary conditions The governing equation for the problem is the heat conduction equation. The problem is a two-dimensional steady conduction heat transfer. The temperature in the bar depends only on the location. Therefore, the simplified form of the conduction heat transfer equation for this problem can be written as, ∂2T ∂2T + =0 ∂x2 ∂ y2 Considering Fig. 4.3, the boundary conditions for the problem mathematically can be written as, x = 0 → T (0, y) = 0 x = L → T (L , y) = 0 Y = L → T (x, L) = 0 Y = 0 → T (x, 0) = 50 ◦ C Step 3: Thermophysical properties As can be seen from the heat conduction equation and initial and boundary conditions of this problem, there is no need to use any thermophysical properties for this problem and the temperature solution does not depend on the material.

4.6 Example

61

Fig. 4.4 Temperature profiles for different values of y

Step 4: Solution of the heat conduction PDE The solution of the simplified heat conduction equation is given by Eq. 4.44, the only thing that should be done is to substitute the values of variables (x and y) into that equation. If it is done, the following equation will be found, T (x, y) =

∞ 

4.50 sin(βn x) sinh(βn (0.1 − y)) 0.1β sinh(0.1β n n) n=1,3,5...

where βn = nπ/0.1. There are infinite terms in the above series. Accepting sufficient term 1 < n < 21 can provide an accurate solution for this problem. The values of βn are 31.42, 94.25, 157.08, 219.91, etc. Step 5: Demonstration of the results Demonstration of the results can be done by plotting temperature profiles in the cross-section of the bar as shown in Fig. 4.4. Two important points take attention, (a) The hottest temperature is seen on the bottom of the bar as expected. (b) The temperature profiles are symmetric with respect to the line of x = 50 mm since the boundary conditions on the left and right walls of the domain are identical.

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4 Two-Dimensional Steady Heat Conduction in Cartesian Coordinate …

4.7 Review Questions

Question (Q1) The problem studied in this chapter is two-dimensional steady state conduction heat transfer (Q2) In the problem of this chapter, the temperature is a function of x and y (Q3) The heat conduction equation of this chapter is a first-order partial differential equation (Q4)The heat conduction equation solved in this chapter is a second-order partial differential equation (Q5) The heat conduction equation of this chapter is solved analytically (Q6) Two boundary conditions and two initial conditions are sufficient to find the solution to the heat conduction equation of this chapter (Q7) Two boundary conditions for the x direction and two boundary conditions for the y direction are sufficient to solve the heat conduction equation of this chapter and find results (Q8) The Laplace transform method is used to solve the heat conduction equation of this chapter (Q9) In the separation of variable method used in this chapter, the function of T (x, y) is divided into two equations as X (x), Y (y) (Q10) In the method used in this chapter, the function of T (x, t) is divided into two functions as X (x), (t) (Q11) The differential equations of X (x) and Y (y) are the secondorder ordinary differential equations (Q12) Differential equation of X (x) is a second-order ODE while the differential equation of Y (y) is the first-order ODE (Q13) In order to find a solution for the second-order ODEs of X (x) and Y (y), boundary conditions should also be defined in terms of X (x) and Y (y) (Q14) The solution of the heat conduction equation consists of only two terms (Q15) The solution of the heat conduction equation is the summation of infinite terms (Q16) An appropriate result can be found when a sufficient number of terms from infinite terms of the solution will be taken into account

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4.8 Problems For the problems below, follow the solution style of the example explained in this chapter. All steps mentioned in the solution part of the example should be followed one by one. Problem 1) Consider a long bar with a square cross-section. There is no heat transfer in depth. The height and width of the bar are 50 mm. Three sides of the bar are at

4.8 Problems

63

Fig. 4.5 Cross-section of the Problem 2

50 °C, while the bottom wall is at 80 °C. Write an expression for T (x, y) and draw temperature profiles for the lines of y = 0, 15, 25, 35, and 50 mm. Problem 2) Consider a cross-section of a long bar. The size and boundary conditions are shown in Fig. 4.5. Draw temperature profile for the line of x = 15 mm. Problem 3) There is a bar with a square cross-section. There is no heat transfer in depth which is in the z direction. Heat is transferred in the x and y directions. The width and height of the square channel are 30 mm. x shows the horizontal axis, while y shows vertical axis. Three sides of the cross-section are at 0 °C, while the temperature at x = 0 is at 100 °C. Draw temperature profiles for lines of (a) x = 15 mm (b) y= 15 mm (c) x = y (d) x = 1 − y.

Chapter 5

One-Dimensional Unsteady State Heat Conduction in a Cylindrical Coordinate System

5.1 Introduction In this chapter, a one-dimensional unsteady state heat conduction equation in a cylindrical coordinate system is considered. Temperature changes by time in the radial direction. Bessel functions are widely used for the analytical analysis of heat conduction equations in the cylindrical coordinate system. That is why at the beginning of this chapter, brief information about the Bessel function is given, and after that, the considered problem is explained and solved analytically. The boundary condition at the outer surface is the first kind, which is a constant temperature. Analytical solutions for other types of boundary conditions and conduction heat transfer with heat generation in a cylindrical coordinate system can be found in Refs. [7, 10].

5.2 Bessel Function In the analytical solution of heat conduction in a cylindrical coordinate system, a function called the Bessel function is used. In this subsection, brief information about the Bessel function is given. We know the definitions of cos(x) or sin(x) from high school knowledge. The values of those functions depend on x which is the independent variable, and it changes between − 1 and 1. Simply, Bessel functions are similar to cos(x) and sin(x), but they have their own definitions. Their values depend on two independent variables can be shown by ν (order of Bessel function) and x (independent variable). By changing of ν and x, the value of a Bessel function changes. We have different kinds of Bessel functions, such as the first kind, the second kind, and modified Bessel functions. In this book, we will use the first and the second kinds of Bessel functions, which are shown as Jν (x) and Yν (x) in which x is the independent variable and ν shows the order of the Bessel function. The expressions and also tables for the calculation of Bessel functions can be found in many mathematics and heat transfer books (e.g. [7]). The Bessel function can be © The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2023 M. Mobedi and G. Gediz Ilis, Fundamentals of Heat Transfer, https://doi.org/10.1007/978-981-99-0957-5_5

65

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5 One-Dimensional Unsteady State Heat Conduction in a Cylindrical …

Fig. 5.1 Calculation values of Bessel functions by using Excel

easily calculated by Excel as shown in Fig. 5.1, and its variation with x for any order can be easily plotted. The importance of Bessel functions: Bessel functions are used in many engineering applications such as the analysis of hydrodynamics and electromagnetic waves as well as heat transfer problems. The Bessel function can be used to find the solution of the second-order differential equation given below,   d2 y 1 dy v2 2 y=0 + − + m dx 2 x dx x2

(5.1)

In the above second-order ODE, the dependent variable is y while the independent variable is x. The parameters of m and v are constants. The analytical solution of this second-order differential equation can be expressed as [7, 10], y(x) = C1 Jv (mx) + C2 Yv (mx)

(5.2)

where Jv (mx) and Yv (mx) are Bessel functions. Jv (mx) is the first kind of Bessel function and Yv (mx) is the second kind of Bessel function with an independent variable of x. The parameter ν shows the order of the Bessel function of Jv (mx) and Yv (mx) in Eq. 5.2. The variation, n Jv (mx) and Yv (mx) with x for different orders (ν = 1, 2 and 3) when m = 1 is shown in Fig. 5.2.

Fig. 5.2 Change in Jv (x) and Yv (x) with x for different orders

5.3 Problem Definition

67

Example: Consider the following second-order ODE. The general solution of this equation is asked. d2 y dy (5.3) x2 2 + x + (x 2 − 1)y = 0 dx dx Solution: The above equation can be written in the following form, 1 1 dy d2 y + (1 − 2 )y = 0 + dx 2 x dx x

(5.4)

The comparison of Eqs. 5.1 and 5.4 shows that both values of m and ν are 1 [i.e., m = 1 and ν = 1]; therefore, based on the solution represented by Eq. 5.2, the general solution of Eq. 5.3, which is a second-order ODE can be written as, Y (x) = C1 J1 (x) + C2 Y1 (x)

(5.5)

where J1 (x) and Y1 (x) are the first and second kinds of Bessel functions with an order of 1. The values of C1 and C2 can be found by using boundary conditions.

5.3 Problem Definition After a brief discussion of the Bessel functions and their use in the analytical solution of the second-order ordinary differential equations, the problem of this chapter which is an unsteady state one-dimensional heat conduction in a cylinder can be explained. The problem of this chapter is shown in Fig. 5.3. Initially, the cylinder is at Ti temperature and suddenly the temperature of the outer surfaces of the cylinder drops to zero (the unit depends on the reader). The cylinder is too long and the temperature depends both on location in radial direction and time (i.e., T = f (t, r ))). The temperature in the entire cylinder changes with time, and the final temperature value will be zero since the boundaries are kept at zero. Again, our aim in this chapter is to find a mathematical expression yielding the temperature at any point of the cylinder and at any time.

68

5 One-Dimensional Unsteady State Heat Conduction in a Cylindrical …

Fig. 5.3 Schematic view of the cylinder analyzed in this study

5.4 Assumptions The following assumptions are used to find the temperature inside the cylinder in this chapter, (a) The problem is one-dimensional and unsteady and heat changes only in the radial direction with time. (b) The thermophysical properties of the cylinder are constant and do not change with time or location. (c) There is no heat generation inside the wall.

5.5 Implementation of Assumptions The vector form of the heat conduction equation is, ∂(ρc p )T − → − → = ∇ . (k ∇ T ) + q˙ ∂t

(5.6)

For a cylindrical system when all thermophysical properties are constant, the general form of the heat conduction equation can be expressed as, 1 ∂T 1 ∂T ∂2T q˙ ∂2T 1 ∂2T + + + = + 2 2 2 2 α ∂t ∂r r ∂r r ∂φ ∂z k

(5.7)

5.6 Analytical Solution

69

where α is the thermal diffusivity in units of m2 /s (i.e., α = k/ρcp ). As seen from the above equation, the temperature of T (i.e., dependent variable) depends on four independent variables which are r, θ, z, and t. However, the problem of this chapter depends on time (i.e., t) and radius (i.e., r ) only as well as there is no heat generation and all thermophysical properties are constant. Therefore, Eq. 5.7 can be simplified into the following form ∂2T 1 ∂2T 1 ∂T ∂2T q˙ 1 ∂T = + + + + α ∂t ∂r 2 r ∂r r 2 ∂φ 2 ∂z 2  k     =0

=0

(5.8)

=0

The simplified form of Eq. 5.7 for this problem becomes as, ∂2T 1 ∂T 1 ∂T = + α ∂t ∂r 2 r ∂r

(5.9)

In order to solve Eq. 5.9, we need to specify the initial and boundary conditions. Initially, the cylinder is at Ti , and suddenly, the outer surface temperature of the cylinder drops to zero. Mathematically, these thermal conditions can be written as, ∂T =0 ∂r r =a→T =0 t = 0 → T (r, 0) = Ti

r =0→

(5.10)

5.6 Analytical Solution The separation of variables method is used to solve Eq. 5.9 under boundary conditions written by Eq. 5.10. This means that the function of T (r, t) can be separated into two functions such as R(r ) and (t), and two ordinary differential equations can be obtained, as shown in Fig. 5.4. The multiplication of the solution of two differential equations yields the main function which is T (r, t). The procedure for application of the separation of variable method for this chapter is shown in Fig. 5.4. Based on Fig. 5.4, the following equation can be written, T (r, t) = (t)R(r )

(5.11)

70

5 One-Dimensional Unsteady State Heat Conduction in a Cylindrical …

Fig. 5.4 Procedure for the use of the separation of the variables method to solve the heat conduction equation of this chapter

Let’s take the first derivative of Eq. 5.11 with respect to t and the second derivative with respect to r . ∂T = (t)R  (r, t), ∂r ∂2T = (t)R  (r, t) ∂r 2 ∂T = R(r, t)  (t) ∂t

(5.12)

By substituting Eq. 5.12 into Eq. 5.11, the following equation is obtained, R  (r, t)(t) +

1  1 R (r, t)(t) =   (t)R(r, t) r α

(5.13)

Rearranging the above equation yields the equation below, R  1   (t) 1 R = = −β 2 + R r R α (t)       f (r )

f (t)

(5.14)

5.6 Analytical Solution

71

The right side of the above equation depends on r , while the left side depends on t. They can be equal to each other when they are equal to a constant such as −β 2 . It is possible to separate the above equality (Eq. 5.14) into two ordinary differential equations as follows: 1   (t) = −β 2 α (t) 1 R R  + = −β 2 R r R

  (t) + β 2 α(t) = 0

→ →

R  (r, t) +

(5.15)

1  R (r, t) + β 2 R(r, t) = 0 r (5.16)

Now, instead of a partial differential equation (Eq. 5.9), two ordinary differential equations can be solved. Equation 5.16 is a second-order ODE while Eq. 5.15 is the first-order ODE. If we solve these ODEs and multiply the solutions by each other, the main unknown which is T (r, t) can be obtained. In order to solve these equations, the initial and boundary conditions in terms of R(r ) and (t) should also be written. r =0→

∂T =0→ ∂r

r = a → T (a, t) = 0 →

(t)  It is a function

R(a) .   

.

∂ R(r ) ∂ R(r ) =0 =0→ ∂r  ∂r   It has a value

(t) 

= 0 → R(a) = 0

(5.17)

It has a value It is a function

t = 0 → T (r, 0) = Ti → R(r ).(0) = Ti It might be useful to rewrite the above boundary conditions again. ∂ R(r ) =0 ∂r r = a → R(a) = 0 t = 0 → R(r ).(0) = Ti

r =0→

(5.18)

Solution of the first-order ODE The solution of the first-order differential equation (Eq. 5.15) can be found easily similar to what is explained in Chap. 3.   (t) + β 2 α(t) = 0

→

(t) = ce−αβ

2

where c is the integral constant and its value will be found later.

t

(5.19)

72

5 One-Dimensional Unsteady State Heat Conduction in a Cylindrical …

Solution of the second-order ODE The second-order ODE is Eq. 5.16. This equation can also be written in the following form,   1 0 (5.20) R  (r, t) + R  (r, t) + β 2 − 2 R(r, t) = 0 r r Equation 5.20 is similar to the second-order ODE of Eq. 5.1; therefore, it is possible to write the same solution of Eq. 5.1 for the above equation (i.e., Eq. 5.20). R(r ) = c1 J0 (βr ) + c2 Y0 (βr )

(5.21)

where c1 and c2 are the constants whose values should be found by using the boundary conditions. Let’s apply the first boundary condition as, r =0

R(0) = c1 J0 (β0) + c2 Y0 (β0)

→

(5.22)

If we look at Fig. 5.2b, it is seen that the value of Y0 (0) is infinite. It is impossible to have an infinite temperature at the center; thus, there is no other choice except to make the value of c2 zero, in other words, r =0

→

R(r ) = c1 J0 (βr ) +

c2 

Y0 (βr )

(5.23)

must be zero

Therefore, the solution of the second-order ODE (Eq. 5.16) can be simplified as, R(r ) = C J0 (βr )

(5.24)

The next step is to apply the second boundary condition which is the temperature at the cylinder surface (i.e., r = a). As explained before (Eq. 5.17), the temperature and the value of the function of R(a) at the surface of the cylinder are zero, thus R(a) = c2 J0 (βa) = 0

(5.25)

The value of c2 cannot be zero since in that case R(r ) = 0 both values of c1 and c2 are zero; consequently, the temperature will be zero in the entire domain which is impossible. The only choice is that the value of J0 (βa) must be zero. J0 (βa) = 0

→

roots of J0 (βa) = 0 is needed

(5.26)

Infinite roots exist for β as can be seen from Fig. 5.2. Those roots of any Bessel function can be easily calculated by simple numerical methods as explained in Appendix A. In Appendix A, a table is also presented relating to the roots of J0 (x). The roots of J0 (x)) can also be calculated from this table easily. Let’s show the root of J0 (βa) = 0 as βn a where n changes from zero to ∞. Therefore, J0 (βa) can be zero when β is,

5.6 Analytical Solution

73

βn =

roots of J0 (βa) a

(5.27)

where n changes from zero to infinity. Since there are many solutions for Eq. 5.16, the function of R(r ) can be written as, R(r ) =

∞ 

cn J0 (βn r )

(5.28)

n=1

βn can be found easily by Eq. 5.27. Now, both the solutions of (t) and R(r ) are found. The multiplication of these functions yields T (r, t) which is asked. T (r, t) =

∞ 

(t).R(r ) =

n=1

∞ 

Cn e−αβn t J0 (βn r ) 2

(5.29)

n=1

The symbol appears due to multiple values of βn . The only unknown constant in Eq. 5.29 is Cn , which can be found by application of the initial condition. t =0

→

T (r, 0) = Ti

→

Ti =

∞  n=1

n0 Cn e−αβ   J0 (βn r ) 2

(5.30)

=1

The Eq. 5.30 can be written as, Ti =

∞ 

Cn J0 (βn r ) = C1 J0 (β1r ) + C2 J0 (β2 r ) + · · · + Cn J0 (βn r ) + · · · (5.31)

n=1

Similar to the previous chapter, it is possible to find the value of Cn by using the orthogonality theorem. The orthogonality theorem for the cylindrical coordinate can be written as,

a 0

⎧ ⎨0 for n = m r J0 (βn r )J0 (βm r )dr = a ⎩ r J0 (βn r )dr for n = m

(5.32)

0

If both sides of Eq. 5.31 is multiplied by J0 (βn r ) and integrated from 0 to a, the following equality is obtained.

74

5 One-Dimensional Unsteady State Heat Conduction in a Cylindrical …

a

a r T0 J0 (βn r )dr =

0

a rC1 J0 (β1r )J0 (βn r )dr +

0

rC2 J0 (β2 r )J0 (βn r )dr 0

a + ··· +

Cn rβn (r )βn (r )dr + · · · 0

Based on the orthogonality theorem, all terms of the above equation are zero except the term in which n = m, therefore,

a

a r Ti J0 (βn r )dr = c

0

r J02 (βn r )dr

(5.33)

0

Therefore, the value of Cn can be found by the following relations, a Ti 0 r J0 (βn r )dr Cn =  a 2 0 r J0 (βn r )dr

(5.34)

There are some rules on taking the integral (or taking derivatives) of Bessel functions. By using those relations, the integral of Eq. 5.34 can be calculated and the following equation for the value of Cn can be obtained, Cn =

2Ti aβn J1 (βn a)

(5.35)

Substituting Eq. 5.35 into Eq. 5.29 yields the temperature inside the cylinder. T (r, t) =

∞ 1 2Ti  2 e−αβn t J0 (βn r ) a n=1,2... J1 (βn a)βn

(5.37)

By substituting any values of r (inside the cylinder) and t in the above equation and considering a sufficient number of terms in the above series, the value of temperature at any location and any time in the cylinder can be found.

5.7 Example Consider a long cylinder with a radius of 100 mm. The material of the cylinder is plain steel (AISI 1010), as shown in Fig. 5.5. The initial temperature of the cylinder is 100 °C. Suddenly, the temperature of the outer surfaces of the cylinder decreases to 0 °C, and consequently, the temperature of the cylinder changes. Find the temperature at different locations of the cylinder at t = 10, 50, 75 and 150 s.

5.7 Example

75

Fig. 5.5 View for example of this section

Assumptions: Based on the definition of the problem, the following assumptions can be concluded, • Nothing is mentioned about heat generation in the problem; therefore, there is no heat generation. • Nothing is mentioned about the change of thermophysical properties with temperature; therefore, all thermophysical properties are constant. • The cylinder is long; therefore, most of the heat is transferred in one direction (r direction, see Fig. 5.5). Governing equation, initial and boundary conditions: Since heat transfer depends on time (i.e., t) and r direction, the heat conduction equation (Eq. 5.9) for the problem can be used. Based on the explanations in this section, the governing equation, and initial and boundary conditions for the problem can be written as, ∂2T 1 ∂T 1 ∂T = + 2 α ∂t ∂r r ∂r The boundary conditions are, ∂T =0 ∂r r =R→T =0 r =0→

and the initial condition is: t = 0 → Ti = 100 ◦ C Thermophysical properties: The thermophysical properties of plain carbon AISI 1010 can be found in tables of many heat transfer books or related internet websites as k = 60.5 W/mK, ρ = 7854

76

5 One-Dimensional Unsteady State Heat Conduction in a Cylindrical …

Fig. 5.6 Temperature distribution in the cylinder for different instants

kg/m3 , and C p = 434 J/kgK. Therefore, the value of thermal diffusivity (α = k/ρcp ) can be found as 1.775 × 10−5 m2 /s. Solution of the heat conduction equation: The solution of the one-dimensional unsteady heat conduction equation was given by Eq. 5.37. Substituting the values of the thermophysical properties, geometrical parameters, and initial condition in Eq. 5.37 yields a function in terms of r and t. ∞ 2 × 100  1 2 T (r, t) = e−αβn t J0 (βn r ) 0.1 n=1,2... J1 (0.1βn )βn

where βn is the root of J0 (0.1βn ) = 0. These roots are calculated for βn values from 1 to 10 are 314.15, 628.32, 942.48, 1256.63, 1570.79, 1884.96, 2199.11, 2513.27, 2827.43 3141.59, and 3455.75, respectively. As seen, 10 terms of the solution series are taken into account to find the solution. Demonstration of results: By substituting the values of time in the above solution, a function in terms of r can be obtained that represents the change in temperature with r for the required time. By this method, the change in temperature can be drawn for the requested instant. It should be mentioned that the above equation is valid for 0 < r < 100 mm. Figure 5.6 shows the temperature distributions in the cylinder at different requested times.

5.9 Problems

77

5.8 Review Questions

Questions (Q1) The problem studied in this chapter is two-dimensional steady state conduction heat transfer (Q2) In the problem of this chapter, the temperature is the function of t and r (Q3) The heat conduction equation solved in this chapter is the second-order partial differential equation depending only on r (Q4) The heat conduction equation of this chapter is the first-order partial differential equation depending only on time (Q5) The heat conduction equation of this chapter is solved numerically (Q6) Two boundary conditions for the r direction and two boundary conditions for t are sufficient to solve the heat conduction equation of this chapter (Q7) Two boundary conditions and one initial condition are sufficient to find the solution of the heat conduction equation of this chapter (Q8) Separation of the variable method is used to solve the heat conduction equation of this chapter (Q9) In the separation of variable method used in this chapter, the function of T (r, t) is divided into two equations as R(r ), (t) (Q10) In the method used in this chapter, the function of T (r, t) is divided into two functions as R(r ), X (x) (Q11) Ordinary differential equations that are functions of R(r ) and (t) are the second-order differential equations (Q12) The differential equation of function of the R(r ) is a second-order ODE while the differential equation of (t) is the first-order ODE (Q13) In order to find a solution for the first-order ODE in terms of (t), initial conditions should also be defined in terms of (t) (Q14) The solution of the heat conduction equation consists of only two terms with r and t variables (Q15) The solution of the heat conduction equation consists of the summation of infinite terms and each term includes r and t (Q16) An appropriate result can be found when sufficient terms from infinite terms in the solution will be taken into account

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5.9 Problems For the problems below, follow the solution style of the example explained in this chapter. All steps mentioned in the solution part of the example should be followed step by step. Problem 1: Consider a long cylinder manufactured from concrete. Initially, the concrete cylinder is at 30 °C, and suddenly the outer temperature drops to 10 °C. The diameter of the cylinder is 100 mm. Draw temperature profile for t = 1, 300, 900, and 3600 s. Problem 2: A long copper rod with a cylindrical cross-section having a diameter of 10 mm is initially at 0 °C. The surface of the rod suddenly increases to 100 °C. The

78

5 One-Dimensional Unsteady State Heat Conduction in a Cylindrical …

Fig. 5.7 Schematic view of the cylinder of Problem 3

top and bottom of the rod are thermally insulated. Plot the change in temperature at the center of the rod with respect to time. How long does it take for the entire cylinder to reach 99 °C. Problem 3: A cylinder manufactured from stainless steel with a diameter of 50 mm is shown in Fig. 5.7. As is seen, the top and bottom surfaces of the cylinder are thermally insulated. The cylinder is sufficiently long to neglect the end effects. The cylinder is at - 20 °C and suddenly the temperature of the outer surfaces increases to 20 °C. Plot temperature at 1, 100, 300 s for the cross-section of AB shown in Fig. 5.7.

Chapter 6

One-Dimensional Unsteady State Heat Conduction in a Spherical Coordinate System

6.1 Introduction Unsteady conduction heat transfer in a sphere is discussed in this chapter. Temperature changes in the radial direction and time. The angular change in temperature is ignored. The first kind of thermal boundary condition which is a constant temperature at the outer surface is applied. Further analysis of heat transfer for various boundary conditions as well as heat generation can be found in different heat transfer textbooks [7, 8, 10].

6.2 Problem Definition The problem of this chapter is heat transfer in a sphere. Heat is transferred by the conduction mechanism and changes in the radial direction and time. The radius of the sphere is a, and it is initially at Ti temperature as shown in Fig. 6.1. Suddenly the surface temperature of the sphere changes to zero. An analytical expression for the determination of temperature in the sphere at any location and any instant is needed.

6.3 Assumptions For the analysis of heat transfer of this chapter, the following assumptions can be accepted: (a) The problem is one-dimensional unsteady state heat conduction and temperature changes only in the radial direction (i.e., r ) and by time (i.e., t).

© The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2023 M. Mobedi and G. Gediz Ilis, Fundamentals of Heat Transfer, https://doi.org/10.1007/978-981-99-0957-5_6

79

80

6 One-Dimensional Unsteady State Heat Conduction in a Spherical …

Fig. 6.1 A view of a sphere studied in this chapter

(b) The thermophysical properties of the sphere are constant and do not change by temperature. (c) There is no heat generation or heat dissipation in the sphere.

6.4 Implementation of Assumptions The vector form of the heat conduction equation can be written as, ∂(ρcp )T − → − → = ∇ . (k ∇ T ) + q˙ ∂t

(6.1)

− → The definition of ∇ T for the coordinate system is given by Eq. 1.35. If it is substituted into the above equation, the general form of the heat conduction equation for a spherical coordinate system can be obtained. Since all thermophysical properties of the sphere are constant, the open form of the equation will be obtained as, 1 ∂ 1 ∂T = 2 α ∂t r ∂r

      1 1 ∂ ∂T q˙ ∂ ∂T ∂T r2 + 2 2 + 2 sin θ + ∂r r sin θ ∂θ ∂θ k r sin θ ∂φ ∂φ (6.2)

As seen from the above equation, the temperature can change in the radial and angular directions are shown by r , θ , and φ. The heat conduction equation in spherical coordinates is more complex than in a Cartesian coordinate system due to the temperature change in angular directions. The first term of Eq. 6.2 on the right side shows the change in temperature in the radial direction while the second and third

6.4 Implementation of Assumptions

81

terms represent the change in temperature in the θ and φ directions. The last term refers to the generation or dissipation of heat due to different reasons such as chemical reactions. As it was mentioned in the assumption section (Sect. 6.3), the temperature only changes by time and in the r direction. Therefore, ∂T =0 ∂θ

and

∂T =0 ∂φ

(6.3)

There is no heat generation in the sphere (q˙ = 0); therefore, the following terms are in Eq. 6.2 can be ignored. 1 ∂ 1 ∂T = 2 α ∂t r ∂r

      1 ∂ ∂T ∂ ∂T 1 q˙ 2 ∂T r + 2 2 sin θ + 2 + ∂r r sin θ ∂θ ∂θ k r sin θ ∂φ ∂φ        =0

=0

=0

(6.4) The simplified heat conduction equation for the present problem can be written as: 1 ∂ 1 ∂T = 2 α ∂t r ∂r

  ∂T r2 ∂r

(6.5)

In many cases, the use of the open form of the above PDE is also expressed as, ∂2T 1 ∂T 2 ∂T = + α ∂t ∂r 2 r ∂r

(6.6)

In order to solve the above equation, the initial and boundary conditions are needed. As explained before, initially sphere with a radius of a is at Ti temperature and suddenly the surface temperatures decrease to 0. Therefore, the following initial and boundary conditions can be written for the problem, r =0

→

r =a

→

∂T =0 ∂r T =0

t =0

→

T = Ti

(6.7)

82

6 One-Dimensional Unsteady State Heat Conduction in a Spherical …

6.5 Analytical Solution Similar to the heat conduction problems of Chaps. 3, 4, and 5, the separation of variables is used to find the solution for this chapter. Fortunately, the governing equation of this problem allows for further simplification of Eq. 6.6 and finding the solution easier. A new variable T ∗ can be introduced to Eq. 6.6 as follows: T∗ = rT

(6.8)

The derivatives of the above equation can be taken with respect to t and r as follows: ∂T ∂T ∗ =r (6.9) ∂t ∂t ∂T ∗ ∂r ∂T = T +r ∂r ∂r ∂r

(6.10)

∂T ∗ ∂T = T +r ∂r ∂r

(6.11)

Taking one derivative more from Eq. 6.11 yields the following equation, ∂2T ∗ ∂T ∂2T ∂2T ∂T 1 ∂2T ∗ 2 ∂T + +r 2 → + 2 = = 2 2 ∂r ∂r ∂r ∂r r ∂r r ∂r ∂r

(6.12)

The above equation can be rearranged as follows: ∂2T 1 ∂2T ∗ 2 ∂T = − ∂r 2 r ∂r 2 r ∂r

(6.13)

Substituting Eqs. 6.9, and 6.13 into Eq. 6.6 yields: 1 1 ∂T ∗ 1 ∂2T ∗ 2 ∂T 2 ∂T = + − 2 α r ∂t r ∂r r ∂r r ∂r

(6.14)

or briefly, the new form of the heat conduction equation for this problem is, ∂2T ∗ 1 ∂T ∗ = α ∂t ∂r 2

(6.15)

This equation is the same as the heat conduction equation in a wall solved in Chap. 3. However, the initial and boundary conditions should also be written in terms of T ∗ .

6.5 Analytical Solution

r =0 r =a t =0

→ → →

83

T∗ = rT → T∗ = 0 T = 0 therefore T ∗ = r T → T ∗ = 0 T = Ti

therefore

∗

T = rT →

Ti∗

(6.16)

= r Ti

The new form of the heat conduction equation (Eq. 6.15) and the new form of the initial and boundary conditions (Eq. 6.16) are similar to the heat conduction equation and boundary conditions of the Chap. 3 problem (Eqs. 3.6–3.8), only the initial condition slightly changed. It is not uniform and the initial temperature changes in the r direction. The separation of the variable method can be used to solve the above PDE under explained initial and boundary conditions. The procedure of separation of variables is briefly explained for the present problem, but detailed information is given in Chap. 3. Again, T ∗ (r, t) can be separated into two functions such as T ∗ (r, t) = (t)R(r )

(6.17)

Therefore, two ODEs with their initial and boundary conditions can be obtained as follows (details are given in Chap. 3). The first-order ODE is, d(t) + αβ 2 (t) = 0 dt t = 0 → Ti∗ = (0)R(r )

(6.18)

while the second-order ODE is obtained as, d2 R(r ) + β 2 R(r ) = 0 dr 2 R = 0 → R(0) = 0 R = a → R(a) = 0

(6.19)

The solution of the first-order ODE represented by Eq. 6.18 can be found as: (t) = Ce−αβ

2

t

(6.20)

The solution of the second ODE (Eq. 6.18) can be obtained as, R(r ) =

∞  n=0

Cn sin(βn r )

(6.21)

84

6 One-Dimensional Unsteady State Heat Conduction in a Spherical …

where βn =

nπ a

(6.22)

Detailed information about the solution of the second-order ODE can be found in Sect. 3.5 of Chap. 3. Therefore the general solution of Eq. 6.15 can be written as: T ∗ (r, t) =

∞ 

Cn e−αβ t sin(βn r ) 2

(6.23)

n=0

The coefficient of Cn can be determined by using the initial condition. ∞ 

t = 0 → T ∗ (r, 0) =

n=0

or since Ti∗ = r Ti ; r Ti =

∞ 

0 Cn e−αβ   sin(βn r ) 2

(6.24)

=1

Cn sin(βn r )

(6.25)

n=0

Based on the orthogonality theorem explained in Eq. 3.36, the following equation can be written: a a r Ti sin(βn r )dr = Cn sin2 (βn r )dr (6.26) 0

0

The results of the integral of the left and right sides of Eq. 6.26 can be found easily,

a Cn sin2 (βn r )dr = cn

a sin(2aβn ) − 2 4βn



0



a r Ti sin(βn r )dr = Ti 0

−aβn cos(βn a) + sin(βn a) βn2



(6.27)

Substituting the results of the integral (Eq. 6.27) into Eq. 6.26 yields,  Ti

sin(βn a) − aβn cos(βn a) βn2



= cn

sin(2aβn ) a − 2 4βn

Therefore, an expression for the coefficient of Cn can be obtained:

 (6.28)

6.6 Example

85

Ti Cn = βn



sin(βn a) − aβn cos(βn a) a 2

−

sin(2aβn ) 4βn

(6.29)

The above equation is a little bit complicated and it might be useful to simplify it. Since βn = nπ/a, the following equalities are valid, sin(βn a) = 0 sin(2aβn ) = 0 cos(βn a) = (−1)

(6.30) n

Therefore Eq. 6.23 takes the following form: T ∗ (r, t) =

∞  2 2 Ti (−1)(n+1) e−αβ t sin(βn r ) β n=0 n

(6.31)

It was assumed that T ∗ (r, t) = r T (r, t); therefore the final solution is, T (r, t) =

∞ 2Ti  (−1)(n+1) −αβ 2 t e sin(βn r ) r n=0 βn

(6.32)

where βn =

nπ a

(6.33)

Equation 6.23 yields the temperature in the sphere for any values of r and any instants of the heat transfer process (i.e., t).

6.6 Example As shown in Fig. 6.2, a sphere with R = 30 mm is initially at 800 °C. It is sunk into a large water container which is at 0 °C. The material of the sphere is iron. If it is assumed that the surface of the sphere drops to zero immediately and remains at 0 °C during this thermal process, draw the temperature profile of the sphere at different time steps. Solution: The following steps are suggested in order to find the solution of this problem, Step 1: Necessary Assumptions Based on the explained problem, the following assumptions can be written.

86

6 One-Dimensional Unsteady State Heat Conduction in a Spherical …

Fig. 6.2 Figure of example

(a) There is no heat generation since nothing is mentioned about heat generation in the problem. (b) All thermophysical properties are constant since nothing is mentioned about the change of thermophysical properties with temperature. (c) Most of the heat is transferred in the radial direction, and no change of temperature exists in θ and φ directions. (d) Temperature depends on r and t. Step 2: Governing equation, initial, and boundary conditions The governing equation for the problem is the heat conduction equation in the spherical coordinate system. The problem is one-dimensional unsteady state conduction heat transfer. The temperature in the sphere depends on the location and time. Therefore the simplified form of the governing equation for this problem can be written as, ∂2T 2 ∂T 1 ∂T = + 2 α ∂t ∂r r ∂r Based on the explanation of the problem, the initial and boundary conditions for the problem can be written as r =0

→

r =a

→

Initial condition: t =0

→

∂T =0 ∂r T = 0 ◦C

T (r, 0) = 800 ◦ C

6.6 Example

87

Fig. 6.3 Temperature change in the radial direction for different instants of the example

Step 3: Thermophysical properties The thermophysical properties of iron can be found in tables of many books or websites as k = 80.2 W/mK,ρ = 7870 kg/m3 , and Cp = 447 J/kgK. Therefore the value of thermal diffusivity α = ρCk p can be obtained as 2.28 × 10−5 m2 /s.

Step 4: Solution of the heat conduction PDE The solution of the one-dimensional unsteady heat conduction equation in which temperature changes in the r and t directions for a sphere is given by Eq. 6.33. Substituting the values of the thermophysical properties, geometrical parameters, and initial condition in Eq. 6.33 yields a function in terms of r and t. The temperature in the sphere at any point and any time can be calculated by using the equation below: T (r, t) =

∞  2 n=0

r

800 × βn (−1)(n+1) e−1.775×10

−5 2

β t

sin(βn r )

(6.34)

where six values of βn are 0, 104.72, 209.44, 314.16, 418.88, and 523.60. Step 5: Demonstration of the results By substituting the values of time in the above solution, a function in terms of r [i.e., T (r )] can be obtained. Figure 6.3 shows the temperature distributions at different time steps which are 2, 5, and 15 s. The number of terms included in the calculation of the series is 8 (i.e., n = 1 to 8).

88

6 One-Dimensional Unsteady State Heat Conduction in a Spherical …

6.7 Review Questions

Questions (Q1) The problem solved in this chapter is steady state heat conduction in a sphere (Q2) In the problem of this chapter, the temperature is the function of t and r where r shows the location of any point in the sphere (Q3) The heat conduction equation is solved by the separation of variables in this chapter (Q4) The assumption of T ∗ = r T is used in this chapter to make the heat conduction equation simpler (Q5) The obtained heat conduction equation in terms of T ∗ is very similar to the problem that was solved in Chap. 3 (Q6) In order to solve PDE in terms of T ∗ , the initial and boundary conditions should also be written in terms of T ∗ (Q7) Two boundary conditions and two initial conditions are needed to find the solution for the heat conduction equation of this chapter (Q8) Two boundary conditions and one initial condition are needed to find the solution to the heat conduction equation of this chapter (Q9) In the separation of the variable method used in this chapter, the function of T (r, t) is divided into two equations as R(r ), (t) (Q10) In the method used in this chapter, the function of T (r, t) is divided into two functions as R(r ), (θ) (Q11) The differential equations involving functions R(r ) and (t) are the second-order ordinary differential equations (Q12) Differential equation with the function of R(r ) is a secondorder ODE while the differential equation of (t) is the first-order ODE (Q13) In order to find a solution for the second-order ODE, the initial condition should also be defined in terms of R(r ) (Q14) The solution of the heat conduction equation with parameters of r and t consists of only two terms (Q15) The solution of the heat conduction equation consists of a summation of infinite terms and each term includes independent variables of r and t (Q16) An appropriate result can be found when a sufficient number of terms from infinite terms in the solution will be taken into account

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6.8 Problems

89

6.8 Problems For the problems below, follow the solution style of the example explained in this chapter. All steps mentioned in the solution part of the example should be followed one by one. Problem 1: Consider a sphere with polystyrene that is at 20 °C, and suddenly the outer temperature increases to 60 °C. The diameter of the cylinder is 10 mm. Write the necessary assumptions, governing equation, and initial and boundary conditions for the problem. Write a mathematical expression in terms of r and t to yield the temperature at any time and location inside the sphere. The draw temperature profile in the sphere for t = 1, 30, 90, and 360 s. Problem 2: A sphere from the glass with a radius of 5 mm is at 10 °C. The sphere is thrown into a large vessel. The heat transfer coefficient between the fluid and sphere is large; therefore, the surface of the sphere quickly increases to 100 °C. Plot the change of temperature of points with locations of r = 0, 1, 2, 3, 4, and 5 mm with time. Problem 3: Consider a quarter of a sphere shown in Fig. 6.4. As seen, two surfaces are insulated while the temperature of one surface can change. The temperature of the shown portion is at 5 °C and suddenly its surface drops to − 20 °C. The material of the sphere portion is copper. Plot the variation in temperature in the r direction for different instant of the cooling process.

Fig. 6.4 Schematic view of sphere portion of Problem 3

Chapter 7

Basics of Single Phase Convection Heat Transfer and Governing Equations

7.1 Introduction As explained in Chap. 2, convection is one of the modes of heat transfer. Most of the time, convection heat transfer occurs when a temperature difference exists between a solid surface and a fluid touching the surface. Heat is transferred from the fluid to the surface or vice versa due to the temperature difference. In convection heat transfer, heat is carried by the fluid from one point to another point by the convection mechanism, and at the same time, it is diffused to its surroundings by conduction heat transfer (diffusion) when it is moving as shown in Fig. 7.1. Furthermore, heat is transferred from a solid surface into the fluid by conduction heat transfer since the velocity is zero at the solid surface. Therefore, both conduction and convection modes of heat transfer exist in a convection heat transfer flow field. It should always be kept in mind that convection cannot occur without fluid motion. If there is no fluid motion, the mechanism of heat transfer is converted to pure conduction heat transfer. In this chapter, classifications of convection heat transfer, assumptions of this book, governing equations, and solution methods for the governing equations are discussed.

7.2 Single and Multi-phase Convection Heat Transfer As is well known there are three material phases, solid, liquid, and gas, for many practical problems of heat transfer. If a flow field consists of one phase of fluid (such as liquid or gas only), heat transfer in that flow domain is called single phase convective heat transfer. However, in the presence of two or more phases in a flow domain (such as gas and liquid, solid and liquid), heat transfer in the flow domain is called multiphase convection heat transfer. The governing equations and mechanism of heat transfer for single and multiphase flows are different. In this book, only single phase convection heat transfer is discussed. © The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2023 M. Mobedi and G. Gediz Ilis, Fundamentals of Heat Transfer, https://doi.org/10.1007/978-981-99-0957-5_7

91

92

7 Basics of Single Phase Convection Heat Transfer and Governing …

Fig. 7.1 A schematic view for understanding the mechanism of heat and fluid flow in convection heat transfer

7.3 Single Phase Forced, Natural, and Mixed Convection Heat Transfer Single phase convective heat transfer can be classified into three groups such as forced, natural, and mixed convection. Forced convection: Forced convective heat transfer occurs due to the existence of external mechanical power such as a fan or pump. By stopping the external power, the fluid flow stops. The friction between infinitesimal layers of flow as well as fluid and solid surfaces is overcome by an external power. In forced convection heat transfer, the Reynolds number plays an important role in the analysis of flow, that’s why this dimensionless number is widely used by researchers. Natural convection: Natural convection occurs due to the existence of two parameters: (a) Existence of an external acceleration affecting flow such as gravity in the flow domain. (b) Existence of density difference in the flow domain due to temperature difference. The body forces applied to two neighbor control volumes having different temperatures in a flow domain are not the same due to different densities. Therefore, the lighter control volume moves faster in the opposite direction of the body force (for instance the gravity direction). In natural convection, there is no external power, and the existence of an acceleration affecting the flow domain (such as gravity) causes the occurrence of flow. Grashof and Rayleigh numbers are widely used in the analysis of natural convection heat transfer and their definitions are given in Chap. 2. Natural convection is discussed in detail in Chap. 10.

7.4 Governing Equations for Convection Heat Transfer

93

Fig. 7.2 Three examples of single phase convection heat transfer, a forced convection over a flat plate, b natural convection heat transfer over a vertical flat plate, c mixed convection over a vertical flat plate

Mixed convection: Forced convection is observed in some applications such as fluid flow over a vertical flat plate by using a fan. Natural convection is also observed in some applications such as fluid flow over a flat vertical plate due to density differences and gravity. However, there are cases in which the mechanisms of both natural and forced convection exist. The effects of natural convection due to density differences and gravity as well as the effect of forced convection due to an external power exist in the flow domain. If the effects of both natural convection and forced convection are significant in a flow domain, the flow is called mixed convection heat transfer. In many practical convection heat transfer problems, the effects of both natural and forced convection exist. In general, if the influence of forced convection is considerably higher than natural convection on the occurrence of flow, a forced convection analysis should be taken into account (see Fig. 7.2a). If the effect of natural convection on the occurrence of flow is considerably higher than that of forced convection, a natural convection analysis on the flow domain should be performed (see Fig. 7.2b). If any of these mechanisms cannot be ignored, mixed convection analysis should be performed to obtain velocity, pressure, and temperature values in the flow domain (see Fig. 7.2c). In mixed convection heat transfer in addition to Reynolds and Rayleigh numbers, the Richardson number is also widely used. The definition of the Richardson number is also given in Chap. 2.

7.4 Governing Equations for Convection Heat Transfer In this book, convection heat transfer is analyzed under the following assumptions. Additionally, the definition of the following concepts are explained in Chap. 2. • The fluid is Newtonian. • The flow is incompressible. • The flow is laminar.

94

7 Basics of Single Phase Convection Heat Transfer and Governing …

• • • •

Heat and fluid flow is steady state. There is no heat generation or heat dissipation in the considered flow. No viscous dissipation. The considered problems are two dimensional (2D) in Cartesian coordinate system. • Thermophysical properties of the fluid are constant. The above assumptions are important and should be taken into account for this book. The governing equations for a convection heat transfer consist of three equations as: (a) Continuity Equation: It refers to the law of mass conservation. For a steady state, the mass entering a studied domain or any control volume in the domain should be equal to the mass leaving the same volume. (b) Momentum Equation: It refers to the conservation of momentum law explained in Chap. 2. The change in momentum of fluid in a control volume is equal to the net change in forces applied to the control volume. (c) Energy Equation: It refers to the conservation of energy. For a steady convection heat transfer problem without a heat sink/generation, the total energy entering the flow domain or a control volume in the flow domain should be equal to the total energy leaving the domain or control volume. Under the above assumptions, the vector form of the governing equations for a convection heat transfer which are continuity, momentum, and energy equations can be written as, → − →− ∇ V =0 1− − →− → − → − → →− → ( V . ∇ ) V = − ∇ P + ν∇ 2 V + Fb ρ − →− → ( V . ∇ )T = α∇ 2 T

(7.1) (7.2) (7.3)

− → where Fb shows the body force. For a two-dimensional heat and fluid flow in the − → − → Cartesian coordinate system, ∇ and V are defined as, ∂ − ∂ − → → − → i + j ∇ = ∂x ∂y − → − → − → V =u i +v j

(7.4)

Substituting Eq. 7.4 into Eqs. 7.1–7.3 and performing necessary mathematical operations yield the following open form of the governing equations under the considered assumptions of this book.

7.5 Difficulties in Solving Continuity, Momentum, and Energy Equations

∂u ∂u + ∂x ∂y ∂u ∂u u +v ∂x ∂y ∂v ∂u u +v ∂x ∂y ∂T ∂T +v u ∂x ∂y

=0

95

(7.5) 



∂ 2u 1 ∂P ∂ 2u − → + Fx i + +ν 2 2 ρ ∂x ∂x ∂y   2 ∂ 2v 1 ∂P ∂ v − → + Fy j + =− +ν 2 2 ρ ∂y ∂x ∂y  2 2  ∂ T ∂ T =α + ∂x2 ∂ y2 =−

(7.6) (7.7) (7.8)

where u, v, T , and p are the velocity component in x, velocity component in y, temperature and pressure, respectively. Fx and Fy are body forces in the x and y directions. The thermophysical properties for the fluid flow domain are density, kinematic viscosity, and thermal diffusivity shown by ρ, ν, and α. The solution of the above equations under appropriate boundary conditions yields the distribution of velocity, temperature, and pressure. As seen from Eqs. 7.5 to 7.8, the dependent variables for the problem are u and v velocity components, p pressure, and T temperatures while the independent variables are x and y. Therefore, there are four partial differential equations with four unknowns, and the solution of the above set of PDEs yields the velocity, pressure, and temperature distribution in the flow domain. However, there are difficulties with the solution of these partial differential equations which are discussed in the following section.

7.5 Difficulties in Solving Continuity, Momentum, and Energy Equations Solving of continuity, momentum in the x and y directions and energy equations yields values of u, v, p, and T in the entire flow field. Unfortunately, the solution for this set of PDEs is not easy, and many serious difficulties exist. The difficulties in solving Eqs. (7.1–7.4) can be explained as: • Multiple PDE equations: There is no one partial differential equation and there are four partial differential equations that are coupled. For instance, the u velocity component or v velocity component, which are unknowns exists in all PDEs. • Nonlinearity: The momentum in the x and y directions are nonlinear. For instance, the first term of the momentum equation in the x direction is u(x, y)

∂u(x, y) ∂x

(7.9)

96

7 Basics of Single Phase Convection Heat Transfer and Governing …

which is a nonlinear term since u(x, y) and ∂u(x, y)/∂(x) are multiplied. The analytical solution of nonlinear PDE is not easy, and this makes the analytical solution of the governing equations more difficult. • No pressure equation: The values of u and v velocity components in the flow field can be obtained by solving momentum equations in the x and y directions. The values of temperature in the flow field can also be obtained by solving the energy equation. However, there is no equation to be solved for pressure and find the values of pressure in the flow field. The continuity equation is an inactive PDE equation that does not yield the distribution of the pressure. • Determination of pressure at the solid surface: It is easy to define boundary conditions for temperature and velocity at the solid surfaces, but the value of pressure at the solid surface is not known and it is not easy to write a boundary condition for the pressure at the solid surface. The aforementioned points make not only the analytical but also the numerical solution of continuity, momentum, and energy equations difficult for heat and fluid flow in a domain. Recently, developments in computational technologies have provided numerical solutions for the continuity, momentum, and energy equations under reasonable assumptions. However, the analytical solution of these equations is almost impossible and requires further assumptions. In Chaps. 8 and 9, assumptions for a flow over a flat plate and flow inside a channel are explained and Eqs. 7.5–7.8 are simplified in order to obtain an analytical solution. The aim of the analytical solution of governing equations is to find analytical expressions for the determination of u, v, p, and T .

7.6 Example There is an internal flow in a channel as shown in Fig. 7.3. Assume that the u velocity component is uniform in the y direction and also does not change in the x direction for the entire domain. The v velocity component in the domain can be omitted. It is observed that the change in temperature in the x direction is considerably smaller than the change in temperature in the y direction. There is no significant body force affecting the flow. All assumptions considered for the convection heat transfer in this book are also valid for the heat and fluid flow of this example. (a) Write all assumptions for heat and fluid flow of this problem. (b) Write the open form of the governing equations under the above assumptions. (c) Simplify the governing equations for this problem. Solution: (a) The assumptions for the heat and fluid flow shown in Fig. 7.3 are: The fluid is Newtonian, the flow is incompressible, the flow is laminar, heat and fluid flow is steady state, there is no heat generation or heat dissipation in the considered flow, no viscous dissipation, the considered problem is two dimensional (2D heat and

7.6 Example

97

Fig. 7.3 Considered internal flow

fluid flow) in the Cartesian coordinate system and all thermophysical properties of the fluid are constant, body forces are negligible, there is no v velocity components, u velocity component does not change in x and y directions, temperature change in x direction is considerably less than the change in y direction. (b) The governing equations for the problem are continuity, momentum, and energy equations. These equations under the assumptions in this book can be written as, ∂u ∂u + ∂x ∂y ∂u ∂u u +v ∂x ∂y ∂v ∂u u +v ∂x ∂y ∂T ∂T u +v ∂x ∂y

=0  2  ∂ u ∂ 2u 1 ∂P − → + 2 + Fx i =− +ν 2 ρ ∂x ∂x ∂y   2 1 ∂P ∂ v ∂ 2v − → =− +ν + 2 + Fy j 2 ρ ∂y ∂x ∂y   2 ∂2T ∂ T + =α ∂x2 ∂ y2

where u, v, p, and T are velocity component in x, velocity component in y, pressure, and temperature, respectively. ρ, ν and α are the density, kinematic viscosity, and thermal diffusivity of the fluid, respectively. (c) Based on the definition of the problem, u velocity is uniform in y direction → u = f (y) u does not change in x direction → u = f (x) no v velocity component → v = 0 body forces are negligible → Fx = Fy = 0 The u velocity component is not a function of x and y; therefore, u is constant. The continuity and momentum equations take the following forms based on the above explanations: ∂v ∂u + =0 ∂ x ∂ y   =0

=0

98

7 Basics of Single Phase Convection Heat Transfer and Governing …

⎞

⎛ u

⎜ ∂ 2u ∂ 2u ⎟ ∂u ∂u 1 ∂p ⎟ Fx +  v =− +ν⎜ ⎝ ∂ x 2 + ∂ y 2 ⎠ +  ∂x ∂y ρ ∂x   =0  =0 =0

=0

=0

⎞

⎛ u

⎜ ∂ 2v 1 ∂P ∂u ∂v ∂ 2v ⎟ ⎟ Fy =− +ν⎜ +  v + ⎝ ∂ x 2 ∂ y 2 ⎠ +  ∂ x ∂ y ρ ∂ y    =0 =0 =0

=0

=0

The continuity equation is satisfied by default and does not need to be included in the solution. From momentum equations in the x and y directions, the following terms remain: ∂p =0 ∂x ∂P =0 ∂y The above equations show that the pressure is constant in the entire domain. For the energy equation, it is declared that the change in temperature in x direction is considerably smaller than the change in velocity in the y direction; therefore, ∂T ∂T  ∂x ∂y

∂2T ∂2T  2 ∂x ∂ y2

→

The energy equation can be simplified as follows: ⎛ u

⎞

⎜ ∂2T ∂T ∂2T ⎟ ∂T ⎟ +  v = α⎜ + ⎝ ∂x2 2⎠ ∂x ∂y ∂ y  =0 negligible

Therefore, the energy equation takes the following form which is a simplified equation for this problem: ∂2T ∂T =α 2 u ∂x ∂y

7.8 Problems

99

7.7 Review Questions

Questions (Q1) This chapter explains the fundamentals of convection heat transfer (Q2) The governing equations for a single phase convection heat transfer are continuity, momentum, and energy equations (Q3) This book teaches analytical solutions for the governing equations of two-phase flows (Q4) There is no conduction heat transfer in a flow field when forced convection exists (Q5) In convection heat transfer, heat from a solid surface can be transferred to the adjacent fluid by conduction (Q6) Natural convection occurs due to the density difference of fluid in the flow field associated with the presence of a proper acceleration affecting the flow (Q7) It is easy to analytically solve the continuity, momentum, and energy equations presented in this chapter (Q8) One of the difficulties for analytical solutions of continuity, momentum, and energy equations is the existence of a nonlinear terms in the set of PDEs (Q9) One of the difficulties of the analytical solution of continuity, momentum, and energy equations is that there is no equation for the pressure (Q10) The momentum equation is a second-order PDE, that’s why the analytical solution of continuity, momentum, and energy equations is possible (Q11) The dimensionless Reynolds number is widely used in the analysis of natural convection heat transfer (Q12) The analytical solution of turbulent heat transfer including viscous dissipation is discussed in the book (Q13) In order to find a solution for the second-order ODE, an initial condition should be defined (Q14) This book discusses the analytical solution of convection heat transfer for compressible fluids (Q15) The convection heat transfer problems discussed in this book are two dimensional (Q16) Analytical methods are the only methods for solving continuity, momentum, and energy equations

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7.8 Problems For the problems below, follow the solution style of the example explained in this chapter. All steps mentioned in the solution part of the example should be followed one by one.

100

7 Basics of Single Phase Convection Heat Transfer and Governing …

Fig. 7.4 Couette flow explained in Problem 1

Problem 1: Couette flow is a flow in a channel whose one surface moves tangentially as shown in Fig. 7.4 In Couette flow, the velocity component in the transverse direction of flow is zero. The velocity component in the flow direction does not change along the flow. There is no pressure gradient and flow occurs due to the moving of the wall. Furthermore, temperature also does not change in the flow direction. (a) Write assumptions used in this book and assumption of Couette flow separately. (b) Write the continuity, momentum, and energy equations under the assumptions of this book and simplify them by using the assumptions of Couette flow. Problem 2: There is a downward inclined flow as shown in Fig. 7.5. The thickness of the flow layer does not change and flow occurs due to gravity components in the flow direction; consequently, the pressure gradient can be neglected. The considered coordinate system is shown in the figure. The velocity component in the transverse flow is zero, and the velocity component in the flow direction does not change along the flow. (a) write assumptions used in this book and assumptions of the explained downward inclined flow separately. (b) Write the continuity, momentum, and energy equations under the assumptions of this book and simplify them by using assumptions explained for this problem. Problem 3: Two main difficulties for solving continuity and momentum equations are (a) pressure term and (b) having a set of PDEs (see Sect. 7.5). It is possible to remove the pressure term in the momentum equations and obtain a single PDE by using the definition of vorticity,

Fig. 7.5 A schematic view for Problem 2

7.8 Problems

101

ξ=

∂u ∂v − ∂x ∂y

(7.10)

Consider the momentum equations given by Eqs. 7.6 and 7.7, and derive the following motion equation in terms of vorticity. Assume that nobody forces exist in the flow domain.   2 ∂ξ ∂ξ ∂ ξ ∂ 2ξ u +v =ν + ∂x ∂y ∂x2 ∂ y2

Chapter 8

External Flow: Heat and Fluid Flow Over a Flat Plate

8.1 Introduction In order to understand heat and fluid flow over a plate, it might be useful to divide this chapter into two parts: (a) fluid flow over a flat plate and (b) heat and fluid flow over a flat plate. In the first part, it is assumed that the flow is isothermal and there is no heat transfer in the entire flow, while in the second part, the change in temperature in addition to the change in the velocity in flow field on the plate is taken into account.

8.2 Fluid Flow over a Flat Plate (Isothermal Flow) The aim of this section is to find an analytical expression to give the value of the velocity at any point in the flow field. The governing equations are written, simplified, and solved by using a method called as Similarity Method.

8.2.1 Definition of the Problem Fluid flow over a flat plate is one of the classical fluid mechanics problems. Assume a fluid flows over a flat plate. The velocity of the fluid is u ∞ at the edge of the plate. A boundary layer occurs on the plate as shown in Fig. 8.1 due to the fluid’s viscosity. The boundary layer concept is a fundamental concept in heat and fluid flow problems and should be understood well. The flow field can be divided into two regions: “inside the boundary layer” and “outside boundary layer”. The velocity at the outer boundary layer is u ∞ while the velocity value changes inside the boundary layer. The value of the velocity at the plate is zero. Furthermore, the flow field in the flow direction can also be divided into three regions: laminar, transition, and turbulent regions as seen in Fig. 8.1. The differences between laminar and turbulent flow are explained in Chap. 2. © The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2023 M. Mobedi and G. Gediz Ilis, Fundamentals of Heat Transfer, https://doi.org/10.1007/978-981-99-0957-5_8

103

104

8 External Flow: Heat and Fluid Flow Over a Flat Plate

Fig. 8.1 Flow over a flat plate and occurrence of the velocity boundary layer

The thickness of the boundary layer is also important since it separates the flow field into two regions and is generally shown by δ (i.e., delta). δ is a function of x, and it increases in the flow direction. Collaterally, the boundary layer is growing in the flow direction. All assumptions used in this chapter are explained in Sect. 7.4. The flow is two dimensional, and the unknowns for the problems are (1) Velocity component in x direction (u) (2) Velocity component in y direction (v) (3) pressure ( p). In order to find the above unknowns, the equations of continuity and momentum in the x and y directions should be solved. These equations are presented in Sect. 7.4; however, it might be useful to rewrite them in this chapter. ∂v ∂u + =0 ∂x ∂y   2 ∂u ∂u 1 ∂P ∂ u ∂ 2u u +v =− +ν + ∂x ∂y ρ ∂x ∂x2 ∂ y2   2 ∂u 1 ∂P ∂ v ∂v ∂ 2v +v =− +ν u + 2 ∂x ∂y ρ ∂y ∂x2 ∂y

(8.1) (8.2) (8.3)

There are three PDEs (i.e., partial differential equations) with three unknowns as u, v, p, and the solution of these equations yields the distribution of velocity components and pressure in the entire flow field.

8.2 Fluid Flow over a Flat Plate (Isothermal Flow)

105

8.2.2 Assumptions Since the analytical solution of the governing equations presented by Eqs. 8.1–8.3 is difficult and almost impossible to solve, some simplifications must be made. The following assumptions were observed experimentally as well as they were proven theoretically and reported in many convection heat transfer textbooks. Order of magnitude analysis is a useful method to prove the following assumptions [9]. (1) Assumptions for the momentum equation in y direction It was proved that the values of the v velocity component are smaller than the u velocity component inside the boundary layer. That is why all terms of the momentum equation in the y direction which is Eq. 8.3 involving the v velocity component or its derivative can be neglected, and only the following term remains from the momentum equation in the y direction, ∂P =0 ∂y

(8.4)

(2) Assumptions for the momentum equation in the x direction Generally, the thicknesses of the velocity and thermal boundary layers are very small. The gradient of velocity in the y direction (transverse direction) is considerably higher than the velocity gradients in the x direction (longitudinal direction). That is why, the following relation is valid for the flow region inside the boundary layer, ∂ 2u ∂ 2u  2 2 ∂x ∂y

(8.5)

(3) Pressure gradient in the x direction The pressure gradients in the x and y directions are zero at the outer region of the boundary layer due to the uniform velocity distribution. The pressure gradient is also zero in the y direction inside the boundary layer based on Eq. 8.4. As it can be seen from Fig. 8.2, points A and D are inside the boundary layer while points B and C are in the outer region of the boundary layer. The pressures of A and B points are equal (due to Eq. 8.4), B and C points also have the same pressures (uniform velocity distribution in the x direction) and the pressure of C and D points should be equal, due to Eq. 8.4. Consequently, the pressures at A and D points must be equal showing that there is no pressure gradient in the x direction even inside the boundary layer: ∂P =0 ∂x

(8.6)

Based on the above assumptions, the following terms can be neglected in the continuity and momentum equations,

106

8 External Flow: Heat and Fluid Flow Over a Flat Plate

Fig. 8.2 Analysis of pressure change in the boundary layer

∂v ∂u + =0 ∂x ∂y u

(8.7) ⎞

⎛

∂u ∂u 1 ∂P ⎜ ∂ 2u ∂ 2u ⎟ +v =− ν⎜ + 2⎟ ⎝ 2 ∂x ∂y ρ ∂x ∂x ∂y ⎠    

(8.8)

 ∂ 2v ∂ 2v + 2 ∂x2 ∂y  

(8.9)

≈0

It is small

∂v ∂v 1 ∂P u +v +ν =− ∂x ∂y ρ ∂y     ≈0



≈0

Therefore, the simplified governing equations which are continuity and momentum take the following forms for fluid flow inside the boundary layer developing over a flat plate, ∂v ∂u + =0 ∂x ∂y ∂u ∂u ∂ 2u u +v =ν 2 ∂x ∂y ∂y ∂p =0 ∂y

(8.10) (8.11) (8.12)

The comparison of the set of Eqs. (8.1–8.3) and (8.10–8.12) shows the advantages of the employed assumptions and the set of Eqs. 8.10–8.12 are easier to be solved. It is clear that the fluid flows with a uniform velocity out of the boundary layer in the x direction and the simplified motion equation for flow in the outer region of the boundary layer can be written as, ∂u =0 ∂x

(8.13)

8.2 Fluid Flow over a Flat Plate (Isothermal Flow)

107

The boundary conditions for Eqs. 8.10–8.12 can be explained as follows by considering Fig. 8.1. (a) The inlet velocity into the region of flow over a flat plate is uniform and shown by u ∞ , and there is no v velocity component in the entrance velocity profile. (b) The values of u and v velocity components are zero at the flat plate due to the no-slip boundary condition. (c) At the edge of the boundary layer (i.e., y = δ) the value of the u velocity component is u ∞ . The above explained points help us to write boundary conditions for Eqs. 8.10–8.12,

y = 0 → u = 0, v = 0 y = δ → u = u∞

(8.14) (8.15)

x =0

(8.16)

→

u = u∞, v = 0

The solution of simplified governing equations (Eqs. 8.10–8.12) under the above boundary conditions (Eqs. 8.14–8.16) yields the distribution of u(x, y) and v(x, y) in the entire region inside the boundary layer. As it can be seen, there are two PDEs and two unknowns as u and v which are the velocity components in x and y directions (see Fig. 8.1) that must be found.

8.2.3 Similarity Solution Although Eqs. 8.10–8.12 are simpler than the original Eqs. 8.1–8.3, they are still nonlinear and coupled PDEs and their analytical solution is not easy. That is why these equations are transformed into an ordinary differential equation, and the solution of the ODE yields u and v velocity components inside the boundary layer. The process for transforming the PDE (Eqs. 8.10–8.12) into an ODE is described below. As explained in Chap. 1, the u and v velocity components in terms of stream function can be defined as, ∂ψ ∂y ∂ψ v=− ∂x

u=

(8.17) (8.18)

108

8 External Flow: Heat and Fluid Flow Over a Flat Plate

If the above definitions of u, v, and the velocity components are substituted into the continuity equation, it can be seen that the continuity equation is satisfied by default and that there is no need to solve the continuity equation if we use the stream function. ∂v ∂ 2ψ ∂ 2ψ ∂u + =0→ − =0 (8.19) ∂x ∂y ∂ x∂ y ∂ y∂ x Similarly, we can substitute the definition of Eqs. 8.17 and 8.18 into the momentum equation in the x direction (Eq. 8.11) u

∂u ∂ψ ∂ 2 ψ ∂v ∂ 2u ∂ψ ∂ 2 ψ ∂ 3ψ +u =ν 2 → − =ν 3 ∂x ∂y ∂y ∂ y ∂ x∂ y ∂ x ∂ y∂ x ∂y

(8.20)

Now, a new parameter as η and a new unknown function as f (η) can be introduced and the function ψ can be assumed in terms of function f (η) and the independent variable η as, ψ=

√ u ∞ νx f (η)

(8.21)

where η is assumed as,  η=y

u∞ νx

(8.22)

We considered an unknown function as f , and we assumed that this function depends only on η [i.e., f (η)]. Our aim in this section is to transform the momentum equation [i.e., Eq. 8.11] into an ordinary differential equation in which f (η) is the dependent variable while η is the independent variable. In order to transform the momentum equation in the x direction into an ODE, u and v and their partial derivatives should be written in terms of f (η), u=

√ ∂( u ∞ νx f (η)) √ ∂ψ ∂ f (η) →u= = u ∞ νx ∂y ∂y ∂y

(8.23)

The use of the chain rule and its application to the above equation yields the following relation, ∂ f (η) ∂ f (η) ∂η ∂ f (η) = → = ∂y ∂η ∂ y ∂y

   u∞ ∂ ∂ f (η) y νx ∂η

∂y

=

∂ f (η) ∂η



u∞ νx

(8.24)

Substituting the above equation (i.e., Eq. 8.24) into Eq. 8.23 yields u velocity component in terms of f (η),

8.2 Fluid Flow over a Flat Plate (Isothermal Flow)

u=

√

∂ f (η) √ u ∞ νx = u ∞ νx ∂y



u ∞ ∂ f (η) d f (η) → u = u∞ νx ∂η dη

109

(8.25)

Similarly, the v velocity component and the partial derivatives of u and v with respect to x and y can be found as, v=

1 2



  u ∞ ν d f (η) η − f (η) x dη

∂ 2u u 2∞ d3 f (η) = ∂ y2 νx dη3

(8.26)

(8.27)

The above equations (Eqs. 8.25–8.27) can be substituted in the momentum in the x direction (Eq. 8.11), and which yields the following third-order ordinary differential equation, d3 f (η) d2 f (η) + f (η) =0 (8.28) dη3 dη2 As seen, we have an ordinary differential equation with a dependent variable of f and an independent variable of η. For simplicity, it can be written as, d3 f d2 f + f 2 =0 3 dη dη

(8.29)

In order to solve the above equations, the boundary and initial conditions for Eq. 8.29 should also be known. From the first boundary condition shown by Eq. 8.14, the first boundary condition for the ODE can be found, 

u∞ →η=0 νx df df u = 0 → u = u∞ → =0 dη dη y=0→η=y

(8.30)

Therefore, the following boundary condition is valid when the value of η = 0, η=0→

df =0 dη

(8.31)

From the second boundary condition shown by Eq. 8.15, the second boundary condition for the obtained ODE can be procured,

110

8 External Flow: Heat and Fluid Flow Over a Flat Plate



u∞ =0 νx    df 1 u∞ν η −f =0 → f =0 y=0→v= 2 x dη 

y=0→η=y

(8.32)

=0

Therefore, when η = 0, the following boundary conditions can also be used: η=0→ f =0

(8.33)

and finally, from the initial condition (Eq. 8.16), the third boundary condition for the ODE can be written by considering that when x → 0, the value of η approaches infinity. (8.34) η → ∞ → u = u∞ The following condition can be found by using Eq. 8.25, u∞ = u∞

df df → =1 dη dη

(8.35)

Therefore, the third boundary condition for Eq. 8.29 is obtained as: η=∞→

df =1 dη

(8.36)

The aim of this section which was to transform the continuity and momentum equations in the x direction (Eqs. 8.10–8.12) and their initial and boundary conditions to an appropriate ODE is achieved. We could not only transform PDEs into an ODE in terms of η but also write its initial and boundary conditions in terms of f (η). Now, the new form of the governing equation and its boundary conditions can be rewritten as follows: d3 f d2 f + f 2 3 dη dη df η=0→ dη η=0→ f df η=∞→ dη

=0 =0 =0 =1

(8.37)

8.2 Fluid Flow over a Flat Plate (Isothermal Flow)

111

8.2.4 Solution of ODE By finding the solution of the function of f in terms of η, the values of the u and v velocity components can be calculated easily by using Eqs. 8.23 and 8.26. The ODE form of the governing equation (Eq. 8.29) is simpler than PDEs (Eqs. 8.10– 8.12), but it is a third-order nonlinear ODE that cannot be solved easily. Fortunately, P.R. Heinrich Blasius (a German scientist) solved the ODE (Eq. 8.29) under the given initial and boundary conditions. The solution is in the form of a series given in different heat transfer books [9]. The graphical representation of the solution of ODE is given in Fig. 8.3 while its table representation is given in Appendix B. The vertical axis of the graph in Fig. 8.3 shows f  = u/u ∞ while the x axis shows the value of η depending on the coordinates of any point inside the boundary layer. For a flow of a fluid over a flat plate and for any point inside the boundary layer with the coordinate of x, y, the value of η can be calculated easily by using Eq. 8.22 and then by using the graphical solution given in Fig. 8.3 , the values of f  can be estimated easily. Since the value of f  and u ∞ are known, the value of the u velocity component can be easily found. If the table given in Appendix B is used, the value of the v velocity component at any point in the velocity boundary layer can also be calculated. The continuity and momentum equations are solved by using Similarity Method. Kakac and Yener [9] also used the Integral Method and suggested the following equation for the determination of the u velocity component by considering thirdorder polynom. 1  y 3 u 3y − = u∞ 2δ 2 δ where δ is the boundary layer thickness explained in the next section.

Fig. 8.3 Change in the u velocity component with η [9, 11]

(8.38)

112

8 External Flow: Heat and Fluid Flow Over a Flat Plate

8.2.5 Boundary Layer Thickness The value of the u velocity component inside the boundary layer changes between 0 and u ∞ (i.e., 0 < u < u ∞ ), and at the outside of the boundary layer, it is u ∞ . In order to predict whether a point is inside or outside of the velocity boundary layer, we should determine the value of the boundary layer thickness (i.e., δ), as shown in Fig. 8.1. The boundary layer thickness is defined as the distance from the solid surface to a point where u = 0.99u ∞ . By considering Fig. 8.3, it can be seen that the condition of u = 0.99u ∞ can be achieved when η ≈ 5. Therefore, the boundary layer thickness for any point with x distance from the inlet can be found by using the following equation.  δ = 5.0

νx u∞

(8.39)

8.3 Heat and Fluid Flow over a Flat Plate In the previous section, the analysis of fluid flow over a flat plate was explained when the entire flow field is at the same temperature (i.e., isothermal flow). In order to obtain the velocity field, some assumptions were made such as steady state, two dimensional, and incompressible flow. Those assumptions are explained in Sect. 7.4. The same assumptions are also valid for this section. However, the flow is not isothermal (non-isothermal) and there is a temperature difference between the fluid and the plate. The plate temperature may be greater or less than the fluid temperature entering the flow field. Again, the similarity solution is used in this section to obtain temperature distribution in the flow field.

8.3.1 Definition of the Problem The heat and fluid flow over the plate analyzed in this section is shown in Fig. 8.4. As can be seen, the fluid enters the region above the plate with a uniform temperature of T∞ and uniform velocity of u ∞ , while the temperature of the wall is Tw . There is a velocity boundary layer on the plate, and the velocity of fluid inside the boundary layer is between 0 and u ∞ while the velocity outside the boundary layer is u ∞ . Similar to the velocity boundary layer, there is a thermal boundary layer (i.e., temperature boundary layer) on the flat plate as shown in Fig. 8.4. The temperature of the fluid in the outer of the temperature boundary layer is at T∞ while inside the thermal boundary layer, it is between T∞ and Tw . Schematic views of the velocity and temperature profiles are shown in Fig. 8.4 for the case,

8.3 Heat and Fluid Flow over a Flat Plate

113

Fig. 8.4 Velocity and temperature boundary layers over a flat plate and temperature profile of fluid when Tw > T∞

(a) the temperature of the plate is greater than that of the fluid, and (b) the temperature boundary layer is above the velocity boundary layer. Similar to the velocity boundary layer, the definition of the thermal boundary layer thickness can be explained as the distance of the plate surface to the point at which the fluid temperature is almost equal to T∞ and it is shown in δT .

8.3.2 Energy Equation and Assumptions The governing equations for heat and fluid flow over a flat plate are continuity, momentum, and energy equations given by Eqs. 8.1–8.3. The continuity and momentum equations were discussed and simplified in the previous section when the flow is isothermal. They were transformed into an ordinary differential equation (Eq. 8.37) in the previous section, and a graphical solution was presented. In this section, the solution of the energy equation will only be found since the values of u and v can be calculated by using Fig. 8.3 or Eq. 8.38. Under the assumptions described in Sect. 7.4, the energy equation for a nonisothermal flow over a flat plate shown in Fig. 8.4 can be written as, ∂T ∂T +v =α u ∂x ∂y



∂2T ∂2T + ∂x2 ∂ y2

 (8.40)

The values of u(x, y) as well as v(x, y) in the above energy equation (Eq. 8.40) are known. Again, the above energy equation, which is a second-order nonlinear partial differential equation, cannot be solved analytically and must be simplified. The simplification process is explained in the next subsection.

114

8 External Flow: Heat and Fluid Flow Over a Flat Plate

8.3.3 Simplification of the Energy Equation Based on the theoretical analysis and experimental observations for the heat flow inside the thermal boundary layer over a flat plate, it is found that the second partial derivative of temperature with respect to x is smaller than the second partial derivative of temperature with respect to y since the thermal boundary layer is very thin. ∂2T ∂2T  2 ∂x ∂ y2

(8.41)

Therefore, the simplified form of the energy equation for the heat and fluid flow inside the boundary layer can be written as, u

∂T ∂2T ∂T +v =α 2 ∂x ∂y ∂y

(8.42)

It should be mentioned that the above equation is valid for the flow inside the boundary layer, and there is no need to solve this equation for the outer region of the thermal boundary layer since the temperature is uniform and known (i.e., T∞ ) in that region. Considering Fig. 8.4, the initial and boundary conditions for Eq. 8.42 are: • the temperature is at Tw and T∞ at the plate surface and edge of the thermal boundary layer, respectively. • the inlet fluid temperature is uniform as T∞ . Mathematically, these boundary conditions can be written as, y=0

→

x =0 → y = δT (x) →

T = Tw

(8.43)

T = T∞ T = T∞

(8.44) (8.45)

where δT (x) is the thickness of the thermal boundary layer at x point.

8.3.4 Similarity Solution Although Eq. 8.42 is the simplified form of the energy equation for this problem, however, it is still a nonlinear PDE that cannot be solved analytically. That is why it is preferred to transform Eq. 8.42 into an ordinary differential equation. The method

8.3 Heat and Fluid Flow over a Flat Plate

115

for transformation of Eq. 8.42 into an ODE is described in this section. First, we define a new parameter as θ , θ=

T − Tw T∞ − Tw

(8.46)

It should be mentioned that the values of T∞ and Tw are constant and θ depends only on T , which is the temperature at any point inside the thermal boundary layer. The derivatives of θ with respect to x and y can be found as follows: 1 ∂T ∂T ∂θ ∂θ = → = (T∞ − Tw ) ∂x T∞ − Tw ∂ x ∂x ∂x ∂θ 1 ∂T ∂T ∂θ = → = (T∞ − Tw ) ∂y T∞ − Tw ∂ y ∂y ∂y ∂ 2θ ∂2T 1 ∂2T ∂ 2θ = → = (T − T ) ∞ w ∂ y2 T∞ − Tw ∂ y 2 ∂ y2 ∂ y2

(8.47) (8.48) (8.49)

The above equations can be substituted into Eq. 8.42, u(T∞ − Tw )

∂θ ∂θ ∂ 2θ + u(T∞ − Tw ) = α(T∞ − Tw ) 2 ∂x ∂y ∂y

(8.50)

Therefore, the energy equation in terms of θ can be written as, u

∂θ ∂ 2θ ∂θ +v =α 2 ∂x ∂y ∂y

(8.51)

The boundary conditions should also be written in terms of θ by considering Eqs. 8.43–8.45, T∞ − Tw → θ =1 T∞ − Tw Tw − Tw y = 0 → T = Tw → θ = → θ =0 T∞ − Tw T∞ − Tw y = δT (x) → T = T∞ → θ = → θ =1 T∞ − Tw

x=0

→

T = T∞

→

θ=

(8.52) (8.53) (8.54)

Similar to the solution of the velocity boundary layer, it is possible to define a variable such as η,  u∞ η=y (8.55) νx

116

8 External Flow: Heat and Fluid Flow Over a Flat Plate

η depends on x and y. It can be assumed that θ is function of η [i.e., θ = f (η)]. First, based on the chain rule, the derivative of θ with respect to x can be found as    ∂ y uνx∞ ∂θ ∂θ ∂η ∂θ ∂θ ∂θ ∂θ  η  − = → = → = ∂x ∂η ∂ x ∂x ∂η ∂x ∂x ∂η 2x

(8.56)

Similarly, the first and second derivatives of θ with respect to y can be found, ∂θ = ∂y



u ∞ ∂θ νx ∂η

(8.57)

∂ 2θ u∞ ∂ 2θ = ∂ y2 νx ∂η2

(8.58)

The definitions of the u and v velocity components in terms of η were derived in the previous subsection (Eqs. 8.25 and 8.26). They can be written again in this subsection as, ∂f u = u∞ ∂η    (8.59) 1 u ∞ ν d f (η) η − f (η) v= 2 x dη Substituting the new form of the derivative of θ in terms of η and the definitions of u and v in terms of η in the simplified form of the energy equation written for the thermal boundary layer in terms of θ (Eq. 8.51) yields the following second-order ODE, d2 θ 1 dθ + Pr f =0 (8.60) dη2 2 dη The definitions of θ and η are given by Eqs. 8.46 and 8.55, respectively. The solution of f is given in Fig. 8.3. Pr shows the Prandtl number and the definition of the Prandtl number is given in Chap. 2. It is defined as the ratio of kinematic viscosity and thermal diffusivity of the flowing fluid. The second-order PDE (Eq. 8.42) could be transformed into a second-order ODE (Eq. 8.60), since solving ODE is easier than solving PDE. In order to solve the obtained ODE energy equation which is in terms of η, the boundary conditions should also be written in terms of η by considering Eqs. 8.52–8.54, 

y=0 x =0

→ →

u∞ therefore, η = 0 → θ = 0 νx  u∞ η=y → ∞ therefore, η → ∞ → νx η=y

(8.61) θ =1

8.3 Heat and Fluid Flow over a Flat Plate

117

Fig. 8.5 Graphical solution of the energy equations in ODE form (Eq. 8.62) [9]

Finally, the ODE energy equation and its boundary conditions can be written as, d2 θ dθ =0 + Pr f 2 dη η η=0→θ =0 η→∞→θ =1

(8.62) (8.63) (8.64)

As it can be seen, the transformation process from PDE to ODE could be done successfully and now a second-order ODE (Eqn. 8.62) with their boundary conditions (Eqns. 8.63 and 8.64) must be solved.

8.3.5 Solution of ODE The above ODE under the written boundary conditions is solved, and the solution is given as a graph in Fig. 8.5. By knowing the Prandtl number of the fluid flowing over the flat plate and the value of η, it is possible to determine the value of θ and consequently the value of T at any point inside the boundary layer. As it can be seen from Fig. 8.5 the Prandtl number plays an important role in the solution of the energy equation, and consequently, the temperature inside the boundary layer is highly affected by the Prandtl number.

8.3.6 Thermal Boundary Layer Thickness The value of T is at T∞ on the edge of the thermal boundary layer, therefore θ = 0. Considering the definition of η given by Eq. 8.55, the thickness of the thermal boundary layer can be found easily for any point on the wall for a given Prandtl number when θ = 0,

118

8 External Flow: Heat and Fluid Flow Over a Flat Plate

  η

θ=0

 = δT (x)

u∞ νx

(8.65)

The left term of the above equation refers to the value of η when θ = 0. This value can be easily found in Fig. 8.5. Therefore, the equation for thermal boundary layer thickness at any point on the plate such as x can be found from the following equation,   δT (x) = η

 θ=0

νx u∞

(8.66)

8.4 Comparison of Velocity and Thermal Boundary Layer Thickness The velocity boundary layer may be above or below or even on the thermal boundary layer. In other words, the thickness of the velocity boundary layer may be greater or smaller, or even equal to the thermal boundary thickness. The performed analysis on the velocity and temperature boundary layers showed that the following relation exists between the velocity and temperature boundary layer thicknesses, δ 1 ≈ Pr 3 δT

(8.67)

The following remarks can be concluded from the above relationship between δ and δT : • If the Prandtl number of the fluid is smaller than 1 (i.e., Pr < 1), the temperature boundary layer thickness is greater than the velocity boundary layer thickness. For the case of Pr < 1, α > ν, consequently heat diffusion in the y direction is comparatively strong. • If the Prandtl number of the fluid is greater than 1 (i.e., Pr > 1), the thermal boundary layer thickness is smaller than the velocity boundary layer thickness. For this case, α < ν and consequently the thermal diffusion in y is comparatively weak. • If the Prandtl number is 1 (i.e., Pr = 1), the thicknesses of the temperature boundary layer and velocity boundary layers are equal since both the propagation of velocity change and thermal diffusion in the y direction are almost the same.

8.5 Example

119

Therefore it can be predicted that for a fluid such as Mercury with a Prandtl number of 0.0252 the temperature boundary layer is considerably above the velocity boundary layer while for a fluid such as Glycerine with a Prandtl number of 2450, the velocity boundary layer thickness is considerably larger than the thermal boundary layer thickness. For air with Pr = 0.7, which can be accepted near 1 (i.e., Pr ≈ 1), the thicknesses of the temperature and velocity boundary layers are close to each other.

8.5 Example Assume a non-isothermal flow over a flat plate with a length of 200 mm and a long width. The surface temperature is at 80 °C. The water enters the upper region of the flat plate with a velocity of 0.5 m/s and a temperature of 10 °C. Briefly write the governing equations and boundary conditions, find the thermo-physical properties and mention about the solution of the governing equations. Then, answer the following questions. (a) Show that the flow is laminar for this problem. (b) What is the velocity and temperature boundary layer thickness cross-sections of x = 100 and 200 mm. (c) Draw the velocity and temperature profiles at x = 100 and 200 mm from the edge of the plate. (d) Which boundary layer (velocity or temperature boundary layer) is thicker and explains the reason. Solution: The step of the solutions are written and they should be applied to all questions in this chapter. Step 1: Necessary Assumptions It is necessary to write the assumptions. For this problem, the assumptions are as follows: flow is laminar (it will be proven later), it is well known that water is a Newtonian fluid, there is no viscous dissipation in the flow field, flow can be accepted as two dimensional, is incompressible, the effect of gravity can be neglected. The flow is steady state and it does not change by time, all thermophysical properties of the fluid are constant and finally flow can be solved in the Cartesian coordinate system. Step 2: Governing equations, initial and boundary conditions Two boundary layers (thermal and velocity boundary layers) occur on the top of the plate as seen in Fig. 8.4. The governing equations are the set of ordinary differential equations for flow inside the velocity boundary layer and temperature inside the thermal boundary layer and they can be written as,

120

8 External Flow: Heat and Fluid Flow Over a Flat Plate

d3 f d2 f + f =0 dη3 dη2 d2 θ dθ + Pr f =0 dη2 η The initial and boundary conditions for the above set of ODEs are, df = 0, θ = 0 dη df η→∞→ = 1, θ = 1 dη

η = 0 → f = 0,

where f is function of η only and η is defined as,  η=y

u∞ νx

Step 3: Thermophysical properties All the thermophysical properties of the fluid are constant and calculated in film temperature which is, Tf =

Tw + T∞ = 45 ◦ C 2

The necessary properties are ν = 0.606 × 10−6 m2 /s Pr = 3.96 Step 4: Solution of ODEs The graphical solution of the above set of ODEs is given in Figs. 8.3 and 8.4. For any values of x and y inside the boundary layer, the values of u/u ∞ and θ can be found from those graphs. Step 5: Answers to the questions Answers to the questions are given in this part, (a) The critical Re number for a flow over a flat plate for a laminar region is Re = 5 × 105 . Flow with Re < 5 × 105 is accepted as laminar flow. Let’s find the Re number for the considered plate: Re =

0.5 × 0.2 u Lc = = 1.6 × 105 < 5 × 105 ν 0.606 × 10−6

which means that the entire flow is in the laminar region (Fig. 8.1).

8.5 Example

121

Fig. 8.6 Graphical illustration of the velocity and thermal boundary layer thicknesses

(b) By using Eq. 8.39, the velocity boundary layer thickness for the cross-section at 100 mm and 200 mm can be found:     νx 0.606 × 10−6 (0.1)  = 1.74 × 10−3 m = 5.0 = 5.0 δ u∞ 0.5 x=0.1 m   δ 

 νx 0.606 × 10−6 (0.2) = 2.46 × 10−3 m = 5.0 = 5.0 u∞ 0.5 

x=0.2 m

The correlation for the laminar boundary layer between velocity boundary layer thickness and thermal boundary thickness is given in Eq. 8.67, so the thermal boundary layer thickness for the cross-section at 100 and 200 mm can be found:

Fig. 8.7 Velocity boundary layer profile at locations 0.1 and 0.2 m

122

8 External Flow: Heat and Fluid Flow Over a Flat Plate

Fig. 8.8 Temperature boundary layer profiles at location 0.1 and 0.2 m

 δ  1.74 × 10−3 1 1 ≈ Pr 3 = = (3.96) 3 → δT = 1.1 × 10−3 m  δT x=0.1 m δT  δ  2.46 × 10−3 1 1 ≈ Pr 3 = = (3.96) 3 → δT = 1.55 × 10−3 m  δT x=0.1 m δT (c) Let’s draw both the velocity and thermal boundary layer profiles at cross-sections with x = 0.1 m and x = 0.2 m. They are shown in Figs. 8.7 and 8.8. The y axis shows the location, while the x axis depicts the velocity and temperature (Fig. 8.7). (d) As seen from Figs. 8.6 and 8.7, the velocity boundary layer thickness is greater than the thermal boundary thickness. Pr = 3.65 indicates that the propagation of the velocity change is more than that of the temperature change (i.e., ν > α). Therefore, it is expected that the velocity boundary layer thickness becomes above the temperature boundary layer.

8.7 Problems

123

8.6 Review Questions

Question (Q1) Isothermal flow is a flow in which the temperature of fluid changes by location or time (Q2) In an isothermal flow over a flat plate, two boundary layers as thermal and velocity boundary layers exist (Q3) The governing equations for an isothermal flow over a flat plate are continuity and momentum equations (Q4) For the problem analyzed in this chapter, it is assumed that the fluid is Newtonian, and the flow is laminar, incompressible, and steady state (Q5) The velocity in and out of the velocity boundary layer is not uniform and changes by location (Q6) Generally a steep gradient of the u velocity component exists in the y direction (perpendicular to the plate) inside the velocity boundary layer (Q7) In the velocity boundary layer, there is no v velocity component (Q8) Out of velocity boundary layer, the v velocity component exists (Q9) The continuity and momentum equations of the velocity boundary layer can be transformed into a third-order ordinary differential equation (Q10) The partial differential equation of the thermal boundary layer can be transformed into a second-order ordinary differential equation (Q11) The Prandtl number plays an important role in the velocity distribution inside the boundary layer (Q12) The thickness of the thermal boundary layer not only depends on x but also the value of the Pr number must be known (Q13) For high values of the Prandtl number, the thickness of the thermal boundary layer is greater than the velocity boundary layer thickness (Q14) For a fluid such as liquid Glycerine with a Prandtl number of 2450, the velocity boundary layer is above the thermal boundary layer

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8.7 Problems For the problems below, follow the solution style of the example explained in this chapter. All steps mentioned in the solution part of the example should be followed one by one. Problem 1: Assume that liquid mercury flows over a flat plate with a length of 200 mm and long width. The fluid velocity is 0.5 m/s. The temperature of mercury at the inlet to flow domain is 20 °C. The plate temperature is maintained at 80 °C. (a) Write the necessary assumptions for this problem. (b) Write two ODEs by which the values of u, v, and T can be found. (b) Show that flow is laminar for the entire flow domain. (d) Draw velocity and temperature profiles at x = 50 and 150 mm.

124

8 External Flow: Heat and Fluid Flow Over a Flat Plate

(c) Calculate the thermal and velocity boundary layer thickness at x = 50 and 150 mm and compare them. (d) Explain the reason for the difference between the thermal and velocity boundary layer thicknesses. Problem 2: This problem consists of two parts: (a) repeat problem 1 when the liquid is Glycerine. (b) compare the boundary layer thickness of Problems 1 and 2 and if there is a difference between them, explain the reason. Problem 3: Assume that air flows with a velocity of 1 m/s. The length of the plate is 50 mm and the depth is too long. The inlet temperature of the air is 0 °C while the plate is 50 °C. (a) Calculate the thermal and velocity boundary layer thickness for 5 points on the plate and plot both boundary layers. (b) What takes your attention when you compare the boundary layer thickness of this problem with Problems 1 and 2.

Chapter 9

Internal Flow: Heat and Fluid Flow in a Channel

9.1 Introduction Internal flow represents a flow inside a body, such as flow in a tube or channel. There are many applications of internal flows in practice such as heat exchangers, air conditioning systems and the cooling of engines. This book focuses on heat and fluid flow in a channel with long width on which constant heat flux is applied. The aim of this chapter is to obtain two analytical expressions yielding velocity and temperature at any points in the channel under assumptions that will be explained. The analytical solutions for different boundary conditions, such as constant wall temperature as well as variable temperatures or heat fluxes, have been obtained and reported in the literature [6, 9]. In this book, heat and fluid flow under constant wall heat flux is analyzed. Similar to the previous section, the problem is analyzed in two sections: (a) fluid flow in a channel (isothermal flow) and (b) heat and fluid flow in the channel (non-isothermal flow). The concepts of “entrance and fully developed regions” are explained at the beginning of this chapter. It should be mentioned that thermally and hydraulically fully developed flow is only analyzed in this book.

9.2 Isothermal Flow in a Channel As explained before, when there is no temperature difference in the flow field, the flow is called isothermal flow. Therefore, there is no need to include a heat transfer equation and the solution of continuity and momentum equations is sufficient to yield velocity and pressure at any point in the channel.

© The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2023 M. Mobedi and G. Gediz Ilis, Fundamentals of Heat Transfer, https://doi.org/10.1007/978-981-99-0957-5_9

125

126

9 Internal Flow: Heat and Fluid Flow in a Channel

Fig. 9.1 Studied channel and occurrence of fully developed flow in a channel

9.2.1 Definition of the Problem A view of the channel for which the fluid flow analysis is performed can be seen in Fig. 9.1. The height of the channel is 2H , and the origin of the Cartesian coordinate is in the mid-height of the inlet cross-section. Fluid enters the channel with a uniform velocity of u ∞ . All assumptions described in Sect. 7.4 (such as steady state, Newtonian fluid, and incompressible flow) are also valid for analyzing the fluid flow in this channel. The aim of this section is to find an analytical expression for the determination of velocity at any point inside the channel under initial and boundary conditions.

9.2.2 Entrance and Fully Developed Regions As mentioned in Chap. 8, when a fluid flows over a flat plate, a velocity boundary layer occurs on the plate and grows in the direction of flow. The channel shown in Fig. 9.1 consists of two parallel plates (bottom and top walls with long lengths); therefore, there are two velocity boundary layers on the top and bottom surfaces growing in the flow direction. They meet each other at a point (for instance, point m) that is at L e distance from the edge of the channel, as can be seen from Fig. 9.1. The region from the inlet to point m (i.e., 0 < x < L e ) is called as the “hydraulic entrance region”. In the entrance region, both u and v velocity components exist and their values change by location in the flow direction. The region after the m point ( x > L e ) is called as the “hydraulically fully developed” region. The v velocity component is zero in the fully developed region and the change in the u velocity component in the x direction is zero. This book only studies fluid flow in the fully developed region in a channel.

9.2 Isothermal Flow in a Channel

127

There are two points that must be mentioned in this subsection, (a) In order to know that a flow in a channel or tube is laminar or turbulent, the Reynolds number should be calculated. For an internal flow, the Reynolds number can be defined as, Re Dh =

u Dh ν

(9.1)

where Dh is the hydraulic diameter and can be calculated from the following expression, Dh =

A 4P

(9.2)

where A is the cross-section of the channel or tube and P is the perimeter of the wetted area. For a flow between two parallel plates with a long depth, Dh = 2H . A flow can be laminar if Re Dh < 2300. (b) The entrance length of a flow in a channel or tube can be calculated by the following equation, Le = 0.056 Re Dh Dh

(9.3)

where L e is the length of the hydraulic entrance region.

9.2.3 Governing Equations and Implementation of Assumptions The governing equations for finding velocity and pressure in the channel are continuity and momentum equations. The Cartesian coordinate system can be used, and the continuity and momentum equations for this problem can be written as follows under assumptions explained in Sect. 7.4, ∂v ∂u + =0 ∂x ∂y   2 ∂u ∂v 1 ∂p ∂ u ∂ 2u u +v =− +ν + 2 ∂x ∂y ρ ∂x ∂x2 ∂y   2 ∂v ∂ 2v ∂v 1 ∂p ∂ v u + +v =− +ν ∂x ∂y ρ ∂y ∂x2 ∂ y2

(9.4) (9.5) (9.6)

128

9 Internal Flow: Heat and Fluid Flow in a Channel

As explained in Chaps. 7 and 8, the first equation is the continuity equation and the second and third equations are called as momentum equations in x and y directions. The dependent variables (or unknowns) are u, v, and p while the independent variables are x and y. The above partial differential equations are coupled and the momentum equations are nonlinear. Furthermore, there is no equation for pressure. Thus, it is difficult to find the analytical solution to the above PDEs, and simplification is needed. The results will be found only for fully developed regions in this chapter; therefore, a simplification according to the assumptions for the hydraulically fully developed region can be performed. For a fully developed hydraulic region in a channel, the following assumptions can be written, (a) the v velocity component is zero and only the u velocity component exists. u = 0

(9.7)

v=0 (b) The change in velocity in the flow direction is zero; therefore, ∂ 2u ∂u =0→ 2 =0 ∂x ∂x

(9.8)

The above assumptions can be implemented into the continuity equation, ∂v ∂u + =0 ∂x ∂y 

(9.9)

=0

Therefore, the continuity equation takes the following form, ∂u =0 ∂x

(9.10)

This means that there is no change in the u velocity components in x, which validates the second assumption. The implementation of the assumption of the momentum equation in the x direction can be performed as, ⎛ u

⎞

⎜ ∂ 2u ∂ 2u ⎟ ∂u 1 ∂p ∂u ⎟ =− +ν⎜ +  v ⎝ ∂ x 2 + ∂ y2 ⎠ ∂x ∂y ρ ∂x   =0 =0

(9.11)

=0

Therefore, the simplified form of the momentum equation in the x direction can be written as,

9.2 Isothermal Flow in a Channel

129

∂p d2 u =μ 2 ∂x dy

(9.12)

It is well known that ν = μ/ρ. Finally, the same assumptions can be applied to the momentum equation in the y direction, ⎛

⎞

⎜∂ v ∂x ∂v 1 ∂p ∂ v⎟ +ν⎜ +v =− + 2⎟ ⎝ 2 ∂x ∂y ρ ∂y ∂x ∂x ⎠     2

u

=0

=0

=0

2

(9.13)

=0

The final form of the momentum equation for the y direction can be found as, dp =0 dy

(9.14)

The assumptions of being fully developed in a channel are useful and simplify the governing equations considerably, and the simplified forms of the governing equations are, ∂u =0 ∂x d2 u ∂p μ 2 = dy ∂x dp =0 dy

(9.15) (9.16) (9.17)

The importance of the momentum equation in the y direction is that it proves that pressure depends on x only (i.e., pressure does not change in the y direction). Therefore, the velocity distribution can be obtained by using Eq. 9.16 which is a linear second-order ODE. In order to solve Eq. 9.16 and find an expression for the u velocity component, boundary conditions must be known. In the center of the channel, a symmetry boundary condition is valid, and the velocity is zero at the surfaces of the wall. By considering Fig. 9.1, the boundary conditions for the simplified momentum equation (Eq. 9.16) can be written as, ∂u =0 ∂y y = ∓H → u = 0 y=0→

(9.18)

x = 0 → p = pin

(9.20)

(9.19)

130

9 Internal Flow: Heat and Fluid Flow in a Channel

where pin is the inlet pressure. If the ODE of Eq. 9.16 with the written boundary conditions are solved, an expression for the u velocity component can be obtained and u can be calculated at any point of the fully developed region of the channel. For the initial condition, it is assumed that the length of developing region is short.

9.2.4 Solution of ODE The dependent variable of the right side of Eq. 9.16 is pressure (i.e., p) while the dependent variable of the left side is the u velocity component. They can be equal to each other if they are equal to an arbitrary constant such as λ. μ

dp d2 u = −λ2 = dy 2 dx

(9.21)

This treatment provides two ODE equations as, dp + λ2 = 0 dy d2 u μ 2 + λ2 = 0 dy

(9.22) (9.23)

The first equation is called as the equation for pressure while the second equation is called as the equation for u velocity. • Solution for pressure equation: The initial condition for the pressure equation (Eq. 9.22) is given by Eq. 9.20. Therefore, the solution for the first-order ODE which is for pressure (Eq. 9.22) under the given initial condition can be found as, p(x) = pin − λ2 x

(9.24)

where λ2 is a constant with a positive value. The equation shows that the pressure decreases linearly in the x direction for the fully developed region. It should be kept in mind that the pressure does not change in the y direction in the fully developed region of a channel since d p/dy = 0.

9.2 Isothermal Flow in a Channel

131

• Solution for momentum in the x direction The simplified form of the momentum equation in the x direction is Eq. 9.23 and the corresponding boundary conditions are Eqs. 9.18 and 9.19. Taking two integrals from Eq. 9.16 yields the following equation for the u velocity component. y2 d2 u(y) 2 = λ → u(y) = λ + c1 y + c2 2 dy2

(9.25)

and the implantation of boundary conditions (Eqs. 9.18–9.19) provides an expression for the u velocity component, u(y) =

1 dP 2 (y − H 2 ) 2μ dx

(9.26)

It is a second-order polynomial equation indicating that the velocity profile inside the channel is a parabola. The maximum velocity is at the center while the velocities at the walls are zero. The only difficulty of Eq. 9.26 is the value of d p/dx which is not known. Since the mass flow rate and cross-section of the channel do not change and are the same for the entire channel, the average velocity (or mean velocity) is also the same, constant, and it does not change in the flow direction. The definition of the average velocity for a cross-section of the channel considered in this problem can be written as,  um =

A

u(y)dA = A

 +H −H

u(y)dy 2H

(9.27)

where A is the cross-section of the channel and can be written as A = 2H × 1 where 1 is the depth of the channel. Consequently, for a layer of the channel, it is possible to write dA = dy × 1. The graphical definition of mean velocity is shown in Fig. 9.3. The mean velocity can be obtained in terms of the pressure drop P as,  +H um =

1 dP 2 −H 2μ dx (y

− H 2 )dy

2H

(9.28)

Since a linear pressure variation exists in the flow direction (x direction), the following equality is valid, dP P = = constant L dx

(9.29)

The result of the integral of Eq. 9.28 is, um =

dp 3u m μL P 2 P H → = = 3μL L dx H2

(9.30)

132

9 Internal Flow: Heat and Fluid Flow in a Channel

Substituting Eq. 9.30 into Eq. 9.26 yields an expression for the u velocity equation in terms of u m which is known.   y 2  3 u(y) = u m 1 − 2 H

(9.31)

Therefore, the velocity at any point inside the channel when the flow is hydraulically fully developed can be calculated and plotted for a cross-section. Eq. 9.31 can also be written in dimensionless form as, u∗ =

3 (1 − y ∗ 2 ) 2

(9.32)

where u ∗ is dimensionless velocity (u ∗ = u/u m ) and y ∗ is dimensionless coordinate in y direction (y ∗ = y/H ).

9.3 Non-isothermal Flow in a Channel The continuity and momentum equations for flow in a channel are explained and solved in the previous section. In this section, the energy equation when there is a temperature difference in the channel will be solved.

9.3.1 Definition of the Problem If there is a temperature difference between the surface of the channel and the inlet fluid temperature, there will be a heat transfer between the surface and flowing fluid in the channel. A view of the non-isothermal channel considered in this study is shown in Fig. 9.3. As seen, fluid flows into the channel with uniform velocity and temperature of u in and Tin . A constant heat flux shown by q  is applied to the top and bottom surfaces of the channel. Therefore, the temperature is not constant on the surfaces of the channel and it changes by location. The aim of this section is to find a mathematical expression to find the temperature value at any point inside the channel.

9.3.2 Thermally Entrance and Fully Developed Regions Similar to the heat and fluid flow over a flat plate explained in Chap. 8, in addition to the velocity boundary layer, there is a thermal boundary layer over the flat plate. For

9.3 Non-isothermal Flow in a Channel

133

Fig. 9.2 Studied channel and occurrence of thermally fully developed flow in a channel

the channel shown in Fig. 9.2, there are two thermal boundary layers, one develops on the bottom surface while the other develops on the top surface. The thermal boundary layers grow in the flow direction and meet each other at a point (such as the n point) which is at L t distance from the edge of the plate. After n point, the “dimensionless temperature” (let’s show it by θ ) does not change in the flow direction (i.e., x direction). The region between the edge of the channel and the n point (meeting point of two thermal boundary layers) is called as the “thermal entrance region” as shown in Fig. 9.2. The region after the n point in which the dimensionless temperature does not change in the flow direction is called as the “thermally fully developed region”. Therefore, for the thermally fully developed region the following expression is valid. ∂θ =0 ∂x

(9.33)

The dimensionless temperature used in the internal flow is defined as, θ=

T − Tw Tm − Tw

(9.34)

where θ is the dimensionless temperature, Tw is the wall temperature (or plate temperature) and Tm is the mean temperature of the cross-section perpendicular to the flow. It should be mentioned that the temperature changes in the x and y directions (i.e., T = f (x, y)) while θ , Tm and Tw are functions of x only (i.e., θ = f (x), Tm = f (x), Tw = f (x)). The methematical definition of Tm for an incompressible and steady flow with constant thermophysical properties can be written as,  +H Tm (x) =

−H

ρu(y)T (x, y)dy m˙

(9.35)

134

9 Internal Flow: Heat and Fluid Flow in a Channel

Fig. 9.3 Graphical definition of mean velocity and mean temperature

where m˙ is the mass flow rate. The graphical presentation of mean velocity and mean temperature are shown in Fig. 9.3. Here, it might be useful to mention three important points: (a) T shows the temperature at any point in the channel and it is a function of x and y. (b) Tw is the wall temperature changing only in x direction (i.e., Tw = f (x)). (c) Tm is the average temperature of a cross-section perpendicular to flow, and it changes only in the x direction (i.e., Tm = f (x)). When an analysis is done on non-isotherms flow in a channel, two boundary layers, as can be seen from Fig. 9.4, exist. Both thermal and hydraulic boundary layers develop in the flow direction. The thermal boundary layer can be below or above or even on the hydraulic boundary layer. Figure 9.4 shows the development of the velocity and temperature boundary layers in the case that the hydraulic boundary layer is above the thermal boundary layer (Pr < 1). For the flow shown in Fig. 9.4, there are three regions in the channel as: Region (1) Hydraulically and thermally developing (entrance) boundary layer region: In this region both hydraulic and thermal boundary layers develop in the flow direction and both velocity and dimensionless temperature change not only in the y direction but also in the flow direction (i.e., x direction). Region (2) Hydraulically fully developed and thermally developing region: Since the hydraulic boundary layer is above the thermal boundary layer, the top and bottom velocity boundary layers meet each other earlier than the thermal boundary layer. Therefore in Region 2, a hydraulically fully developed with developing thermal boundary layer flow exists, Region (3) Hydraulically and thermally fully developed region: Similar to the hydraulic boundary layer, two thermal boundary layers developing on the bottom and top sides also meet each other after a distance and a thermally fully developed region appears; therefore, in Region 3, a hydraulically and thermally fully developed flow exists.

9.3 Non-isothermal Flow in a Channel

135

Fig. 9.4 Studied channel and occurrence of thermally fully developed flow in a channel when (Pr > 1)

Although it is possible to find temperature and velocity in all three regions analytically, the temperature and velocity are found only for hydraulically and thermally fully developed regions in this book (i.e., Region (3)). Finally, it might be useful to mention that the length of the thermal entrance region can be calculated by using the following equation, Lt = 0.05 Re Dh Pr Dh

(9.36)

where L t is the length of the thermal entrance region and Re Dh shows Reynolds number based on the hydraulic diameter. The definition of Dh is given by Eq. 9.2, and it is 2H for the channel.

9.3.3 Governing Equation and Implementation of Assumptions The continuity and momentum equations for flow in a channel are given and solved in the previous chapter. The assumptions explained in Sect. 7.4 for a fluid flow analyzed in this book (such as no heat generation, Newtonian fluid, and incompressible flow) are also valid for the studied non-isothermal flow considered in this section. Therefore, the energy equation for heat and fluid flow can be written as,

136

9 Internal Flow: Heat and Fluid Flow in a Channel

u

∂T ∂T +v =α ∂x ∂y



∂2T ∂2T + 2 ∂x ∂ y2

 (9.37)

Based on the performed experimental works as well as theoretical studies, it was proved that the gradient of temperature in the x direction is considerably smaller than the temperature gradient in the y direction. Therefore the second derivative of temperature in the y direction is sufficiently smaller than the second derivative of temperature in the x direction, ∂2T ∂2T 359 mm); thus, the velocity distribution at x = 600 mm will be the same at x = 400 mm as shown in Fig. 9.7. Calculating of y∗ = y/H for the asked point as y∗ = 1/3 and substituting this value into velocity equation in step 4 yields the velocity at y = 1mm which is 0.0133 m/s. (e) Based on Eq. 9.30, the pressure change along the x direction which is constant can be written as, −3μu m P = L H2 All parameters on the right side of the above equations are constant showing that the pressure changes linearly in the x direction. The slope of the lines is, tan(α) =

−3 × 0.0547 × 0.01 P = = −45.58 L 0.0062

Therefore, the line for the pressure drop can be found as, p − p|x=0.4 = −45.58 × (x − 0.4) → p = −45.58x + 18.23 + p|x=0.4

148

9 Internal Flow: Heat and Fluid Flow in a Channel

Fig. 9.8 Pressure change along the x direction

The pressure change along the x direction can be schematically drawn as shown in Fig. 9.8, (f) The heat balance for the volume between the inlet (x = 0) and the mentioned point in the question (x = 500 mm) can be written as, ˙ p (Tm − Tin ) q  × L × 1 = mC where 1 is the unit depth. From the above equation, it is possible to find the value of Tm . The mass flow rate can be calculated easily as m˙ = ρ Au in = 988 × 0.06 × 1 × 0.01 = 0.059 kg/s. The heat balance between the inlet (x = 0) and the required point which is x = 500 mm yields the mean temperature at x = 500 mm ˙ p Tin q  × L × 1 + mC 4000 × 0.5 × 1 + 0.059 × 4180 × 20 → → Tm = mC ˙ p 0.059 × 4180 Tm = 28.1 ◦ C Tm =

(g) By using the definition of the Nusselt number given Eq. 9.74, the heat transfer coefficient of a fully developed region can be found, Nu =

h(0.006) h(2H ) → h = 1256.36 W/m2 K → 4.1176 = kf 0.641

(h) The wall temperature at x = 500 mm can be found easily by using Newtons cooling law, q  = h(Tw − Tm ) → Tw =

q  4000 + Tm → Tw = + 28.1 = 31.28 ◦ C h 1256.4

9.4 Example

149

Fig. 9.9 Temperature distribution in the y direction for the point of x = 0.5 mm

(i) An equation for the determination of dimensionless temperature is given in step 4. By substituting the value of θ in this equation, the value of θ at x = 500 mm can be found,    y 2 Nu ∗ 4 4.1176  y 4 ∗ ∗2 (y − 6y + 5) = −6 +5 θ (y ) = 16 16 0.003 0.003 where y∗ = y/H . The definition of θ is given by Eq. 9.34, and therefore a function for temperature can be found as, θ=

Nu T − Tw → T = (Tm − Tw ) (y ∗ 4 − 6y ∗ 2 + 5) + Tw Tm − Tw 16

Therefore, T (x, y) can be calculated easily,    y 2 4.1176  y 4 −6 + 5 + 31.28 T (x, y) = (28.1 − 31.28) 16 0.003 0.003 The equation for the temperature profile at x = 500 mm becomes as T (x, y) = −10103370370 × y 4 + 545582 × y 2 + 27.2 By substituting different values of y in the above equation, the temperature distribution in the y direction can be plotted as shown in Fig. 9.9. As it can be seen, the temperature at the wall of the channel at x = 500 mm is Tw = 31.28.

150

9 Internal Flow: Heat and Fluid Flow in a Channel

9.5 Review Questions

Question (Q1) For an isothermal flow in a channel, there are two regions in the channel as entrance and fully developed regions. (Q2) In an isothermal flow in a channel, two boundary layers as thermal and velocity boundary layers appear. (Q3) For an isothermal flow in the entrance region of a channel, the u and v velocity components do not change in the flow direction. (Q4) For an isothermal flow in the fully developed region of a channel, the v velocity component is zero while the u velocity component exists. (Q5) For an isothermal flow in the fully developed region of a channel, the u velocity component is the function of y only. (Q6) For a flow in a channel, there are two thermal boundary layers on the top and bottom walls and they meet each other at a point in the channel if the channel is sufficiently long. (Q7) The length of the hydraulic entrance region is always greater than the length of the thermal entrance region. (Q8) For a non-isothermal flow in a channel, the channel can be divided into two regions such as thermally entrance and thermally fully developed regions. (Q9) For the thermally fully developed region, temperature changes in the flow direction. (Q10) For the thermally fully developed region, dimensionless temperature changes in the flow direction. (Q11) In the entrance region, the Prandtl number plays an important role in the thickness of the velocity and temperature boundary layer. (Q12) For fully developed flow in a channel, the heat transfer coefficient is defined based on the temperature difference between the wall and the average of temperatures at the cross-section. (Q13) Nusselt number has a constant value for the thermally fully developed region. (Q14) It is possible to find the temperature profile in a channel in a hydraulically and thermally fully developed region analytically.

Correct Incorrect [] [] []

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[]

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9.6 Problems For the problems below, follow the solution style of the example explained in this chapter. All steps mentioned in the solution part of the example should be followed one by one. Problem 1: Mercury flows in a channel with an inlet temperature of Tin = 20◦ C and velocity of 0.5 m/s. The height and length of the channel are 10 and 1000 mm, respectively. A constant heat flux of q  = 5 × 106 W/m2 is applied on both sides of the channel.

9.6 Problems

(a) (b) (c) (d) (e) (f) (g) (h) (i)

151

calculate the Reynolds number for the flow in the channel, calculate the length of the entrance region for the velocity, calculate the length of the entrance region for the temperature, determine which boundary layer meets each other earlier (velocity or temperature boundary layer), and explain the reason, what is the velocity value at the point with x = 500 mm and y = 5 mm, plot the velocity profile for the hydraulically fully developed region, plot the dimensionless temperature for the hydraulically and thermally fully developed region, plot the dimensional temperature profile for a cross-section with x = 500 mm, calculate the heat transfer coefficient at a point on the wall with x = 500 mm.

Problem 2: Repeat problem 1 for water. All conditions, channel geometries, and questions are the same. Problem 3: Machine oil flows in a channel with a height of 40 mm and length of 1500 mm. The inlet velocity and temperature are 1 m/s and 80 ◦ C, respectively. The heat flux applied onto the walls is q  = 2 × 105 W/m2 . (a) What is the length of the region in which both velocity and temperature are not fully developed? (b) What is the length of the region in which velocity is fully developed but the temperature boundary layer is developing? (c) Find the first location of a point (i.e., X ) for which both velocity and temperatures are fully developed. (d) Plot dimensionless velocity and temperature for a cross-section of the thermally and hydraulically fully developed region. (e) Plot the wall and mean temperature (Tw and Tm ) as well as temperature at the points with y = 10 mm in the flow direction for the region between 1000 mm and 1500 mm (i.e., 1000 mm < x < 1500 mm).

Chapter 10

Natural Convection Over a Vertical Flat Plate

10.1 Introduction As mentioned in Chap. 2, the types of convection heat transfer are: forced convection, natural convection, boiling, condensation, or a combination of these. In Chaps. 8 and 9, forced convection heat transfer over a flat plate and in a channel was described. In this chapter, natural convection over a vertical isothermal plate is studied. After the explanation of the considered problem and assumptions, a brief discussion on the occurrence of natural convection, the boundary layer, and the involvement of gravity in the momentum equation will be presented. These subsections will be followed by transforming the governing equations, which are partial differential equations, into ordinary differential equations and presenting solution charts for the determination of velocity and temperature at any point inside the boundary layer.

10.2 Considered Problem A vertical flat plate at Tw temperature is in contact with a fluid at a temperature of T∞ as shown in Fig. 10.1. The coordinate axis of the problem is shown in Fig. 10.1, which is slightly different than traditional directions (x shows the vertical direction). Gravity exists and acts in the −x direction. Due to natural convection, a boundary layer occurs on the surface of the flat plate. The velocity and temperature at any point inside the boundary layer are considered in this chapter.

© The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2023 M. Mobedi and G. Gediz Ilis, Fundamentals of Heat Transfer, https://doi.org/10.1007/978-981-99-0957-5_10

153

154

10 Natural Convection Over a Vertical Flat Plate

Fig. 10.1 Schematic view of natural convection over a vertical flat plate

10.3 Assumptions Two points make the problem of this chapter different from than convection problems discussed in Chaps. 8 and 9, (a) All thermophysical properties of the fluid are constant except density changing with temperature. (b) There is a body force in the momentum equation in −x direction (vertical direction) due to gravity. It plays an important role in the occurrence of flow. Except for the above-stated conditions, all assumptions described in Chap. 7 are also valid for this chapter. The problem is considered in the Cartesian coordinate system, and the flow is a laminar steady state with no viscous dissipation.

10.4 Occurrence of Natural Convection As can be understood from the name “Natural Convection”, it occurs without including any external force or power (i.e., naturally). The use of a fan or pump is unavoidable in forced convection heat transfer; however, in natural convection, the motion of fluid occurs naturally. Two conditions must exist for the formation of a natural convection flow, (a) a density difference in the fluid flow region (b) gravity (or any external acceleration acting on the flow field). It is well known that there is no natural convection in a spacecraft traveling in space since there is no gravity, although a density change in the air inside the spacecraft may exist. It should be kept in mind that when there is no gravity (or acceleration)

10.4 Occurrence of Natural Convection

155

Fig. 10.2 Formation of natural convection over a vertical flat plate

natural convection does not occur even if a density difference exists. Therefore, for natural convection both aforementioned conditions must exist. Figure 10.2 shows a surface whose right region is occupied by a fluid. The surface is at a temperature of Tw while the fluid is at a lower temperature as T∞ . It is well known that the density of materials is a function of temperature, and it decreases by increasing temperature. A material expands with increasing temperature causing a decrease in density however with decreasing temperature it shrinks, and consequently its density increases. In the fluid region of Fig. 10.2, there is a temperature gradient in the fluid domain. If two neighboring control volumes of CV1 and CV2 are considered, their densities are different due to temperature differences. For example when ρ1 < ρ2 , the weight of CV2 (i.e., ρ2 g) is greater than the weight of CV1 (i.e., ρ1 g). Therefore, the fluid in CV1 starts to rise due to Archimedes’ law. In this way, fluid motion occurs in the fluid domain due to two issues: temperature difference causing density difference and the effect of gravity. In forced convection heat transfer, the Reynolds number is widely used to discover the mechanism of heat transfer. The Reynolds number represents the ratio of inertia forces to viscous forces. In natural convection, dimensionless numbers of Grashof and Rayleigh are widely used. The definition of both dimensionless parameters is explained in Chap. 2. The Grashof number for a vertical flat plate shown in Fig. 10.1 is defined as, Gr =

gβT L 3 ν2

(10.1)

156

10 Natural Convection Over a Vertical Flat Plate

and the Rayleigh number is, Ra = Pr Gr =

gβT L 3 να

(10.2)

where L is the height of the flat plate and T is the reference temperature difference. Both of these dimensionless numbers are widely used in this section.

10.5 Governing Equations for Natural Convection Heat Transfer The governing equations for natural convection heat transfer are similar to what is explained for forced convection (Chaps. 8 and 9). The governing equations for two-dimensional natural convection can be written as, ∂ρv ∂ρu + ∂x ∂y ∂ρuu ∂ρuv + ∂x ∂y ∂ρuv ∂ρvv + ∂x ∂y ∂ρuT ∂ρvT + ∂x ∂y

=0

(10.3) 



∂p ∂ 2u ∂ 2u − ρg +μ + ∂x ∂x2 ∂ y2   2 ∂P ∂ v ∂ 2v =− +μ + ∂y ∂x2 ∂ y2  2 2  ∂ T ∂ T =k + 2 ∂x ∂ y2 =−

(10.4) (10.5) (10.6)

The comparison of the above governing equations and Eqs. (7.5–7.8) shows that the only difference between them is the density which is a function of temperature and that is why they are inside the partial derivatives. Therefore, in order to solve the natural convection governing equations we need a “state equation for density” providing a relation between the density and temperature (i.e., ρ = f (T )). There are two methods to involve the change in density in the governing equations of natural convection (Eqs. (10.3–10.6) as, (a) the use of a state equation (b) Boussinesq approximation. These methods are described below.

10.5 Governing Equations for Natural Convection Heat Transfer

157

10.5.1 State Equation for Density The state equation is generally used in computational studies rather than analytical works. There are two different methods for the expression of a state equation for density, (a) Ideal gas equation It is used for gases whose thermodynamics behavior is close to an ideal gas such as air. For a domain with a small pressure change, based on the ideal gas relation, it is known that P = ρ(T )RT

(10.7)

where R is the ideal gas constant. Therefore, if the pressure value and temperature at any point are known, it is possible to find the value of density at that point. In natural convection flow field, since the temperature changes from point to point, the density also changes by location and a motion of fluid occurs due to the effect of gravity. (b) Mathematical expressions In many computational heat transfer software, a function is used to define the change in density in terms of temperature. For instance, for water, the following equations can be used [13], ρ(T ) = ρo + a1 × T − a2 × T 2 + a3 × T 3 − a4 × T 4 + a5 × T 5 ρo =

999.842594 kg/m3

a1 = 0.06793952 kg/(m3 C) a2 = 0.009095290 kg/(m3 C2 ) a3 = 1.001685 × 10−4 kg/(m3 C3 ) a4 = 1.120083 × 10−6 kg/m3 C4 ) a5 = 6.536332 × 10−9 kg/(m3 C5 ) (10.8) As mentioned before, the implementation of the state equation for density is mostly used in the computational analysis of natural convection heat transfer.

10.5.2 Boussinesq Approximation The Boussinesq approximation is widely used both in analytical and computational of natural convection. The Boussinesq approximation states that “In the equations of natural convection heat transfer, density can be accepted as constant except in source term of momentum equation” in the flow direction.

158

10 Natural Convection Over a Vertical Flat Plate

This assumption provides accurate results for natural convection heat transfer, and it is widely used by researchers. Based on this approximation, the governing equations for natural convection can be written as,  ∂v ∂u + = ρ ∂x ∂y   ∂u ∂u ρ u +v = ∂x ∂y   ∂u ∂v +v ρ u = ∂x ∂y   ∂T ∂T +v = ρ u ∂x ∂y 

0

(10.9) 



∂p ∂ 2u ∂ 2u − ρ(T )g +μ +    ∂x ∂x2 ∂ y2 source term  2  ∂ v ∂ 2v ∂p +μ + − ∂y ∂x2 ∂ y2  2  ∂2T k ∂ T + cp ∂ x2 ∂ y2 −

(10.10)

(10.11) (10.12)

As can be seen, density in all terms is constant except in the source term, which is a function of temperature. There are no difficulties with continuity, the momentum equation in the y direction as well as energy equation, while there is difficulty with the momentum equation in the x direction since density depends on temperature (ρ = f (T )). A relationship should be established to describe density in terms of temperature. There are two pressures in a flow field: static and dynamic pressure. p = pd + pst

(10.13)

Pd and Pst are the dynamic and static pressures, respectively. The static pressure can be written as: pst = ρ∞ gx

(10.14)

where ρ∞ is the density at a temperature of T∞ (see Fig. 10.1). The derivative of p in Eq. 10.4 with respect to x can be found as, ∂ pd ∂p = − ρ∞ g ∂x ∂x

(10.15)

Therefore, the momentum equation can be written as:       2 ∂u ∂u ∂ Pd ∂ u ∂ 2u − ρ(T )g (10.16) ρ u +v =− − ρ∞ g + μ + ∂x ∂y ∂x ∂x2 ∂ y2 It should be mentioned that static pressure ( pst ) is a function of x and therefore static pressure does not change in the momentum equation in the y direction. Considering Eq. 10.16, the momentum equation in the x direction becomes as,

10.6 Boundary Layer Concept and Boundary Layer Equations

159

     2  ∂u ∂ Pd ∂ u ∂u ∂ 2u +v =− +μ ρ u + 2 + g[ρ∞ − ρ(T )](10.17) ∂x ∂y ∂x ∂x2 ∂y It is possible to use the thermal expansion coefficient of fluid to find a relationship for ρ(T ). 

1 ρ(T ) − ρ∞ 1 ∂ρ  →β=− β=− ρ ∂ T P ρ∞ T − T∞

(10.18)

or the following equation for ρ(T ) can be found, ρ(T ) = ρ∞ [1 − β(T − T∞ )]

(10.19)

Substituting Eq. 10.19 into Eq. 10.16 yields the following equation for momentum in the x direction:    2  ∂u ∂p ∂ u ∂u ∂ 2u + ρ∞ gβ(T − T∞ ) (10.20) +v =− +μ ρ u + ∂x ∂y ∂x ∂x2 ∂ y2 By assuming that ρ∞ = ρ, the following equation for the momentum equation in the x direction can be obtained, ∂u ∂u +v = u ∂x ∂y

1 ∂p − +ν ρ ∂x



∂ 2u ∂ 2u + ∂x2 ∂ y2

 + gβ(T − T∞ )

(10.21)

It should be mentioned that the pressure that is found from Eq. 10.21 is the dynamic pressure.

10.6 Boundary Layer Concept and Boundary Layer Equations Similar to forced convection, in natural convection, a thermal and velocity boundary layer occurs over the solid surface. Figure 10.3 shows a thermal and velocity boundary layer over a vertical flat plate. The wall temperature is shown by Tw , and the fluid temperature in the stagnant region (far from the plate) is T∞ . It is assumed that Tw > T∞ . As can be seen from Fig. 10.3 both the temperature and velocity boundary layers overlap each other since the concept of “no temperature difference no velocity” is valid in natural convection. The temperature of the fluid gradually decreases from the surface which is at Tw to T∞ . Similar to the boundary layer in forced convection heat transfer, the temperature of the fluid is at T∞ in the region out of the boundary layer, while it changes between Tw and T∞ inside the boundary layer. The interesting point of Fig. 10.3 is the velocity profile, which is zero at the solid surface while it

160

10 Natural Convection Over a Vertical Flat Plate

Fig. 10.3 A natural convection boundary layer over a vertical flat plate

increases and after attaining a maximum value, it decreases and its value becomes zero at the edge of the boundary layer. The velocity profile for the natural convection boundary layer is completely different than the velocity profile of forced convection over a flat plate. The u velocity component in the region outside of the boundary layer is zero while inside the boundary layer it changes between 0 and u max . The set of governing equations for natural convection heat transfer under the Boussinesq assumption can be rewritten as, ∂u ∂v + ∂x ∂y ∂u ∂u +v u ∂x ∂y ∂u ∂v +v u ∂x ∂y ∂T ∂T +v u ∂x ∂y

=0

(10.22) 



1 ∂p ∂ 2u ∂ 2u + gβ(T − T∞ ) +ν + ρ ∂x ∂x2 ∂ y2   2 1 ∂p ∂ v ∂ 2v =− +ν + ρ ∂y ∂x2 ∂ y2  2 2  ∂ T ∂ T =α + 2 ∂x ∂ y2 =−

(10.23) (10.24) (10.25)

where ρ is constant in the above set of governing equations. α is the thermal diffusivity of the fluid. It is almost impossible to find the analytical solution to the above equations, that’s why some assumptions are required to simplify them and solve them analytically.

10.7 Boundary Layer Assumptions and Governing Equations

161

10.7 Boundary Layer Assumptions and Governing Equations Theoretical analysis such as “Order of Magnitude Analysis” and the results of experimental studies show that the following assumptions are valid for heat and fluid flow inside the boundary layer of natural convection for a vertical plate, (a) The v velocity component is much smaller than the u velocity component (i.e., u  v), and therefore, the momentum equation in the y direction can be omitted. The solution of continuity and momentum equation in the x direction is sufficient to find the velocity components of u and v. (b) The magnitude of the second gradient of the u velocity component in the y direction is much greater than the second gradient of the same velocity component in the x direction, ∂ 2u ∂ 2u  (10.26) ∂x2 ∂ y2 Therefore, the momentum equation in the x direction can be written as, u

∂u 1 ∂P ∂ 2u ∂u +v =− + ν 2 + gβ(T − T∞ ) ∂x ∂y ρ ∂x ∂y

(10.27)

(c) The dynamic pressure inside the boundary layer is negligible compared to the static pressure. As mentioned before p in the above equation (Eq. 10.27) represents the dynamic pressure. The static pressure is dominant; therefore, the term of ∂ p/∂ x can be neglected, and the momentum equation in the x direction takes the following form, ∂p ∂u ∂u =0→u +v =ν ∂x ∂x ∂y



∂ 2u ∂ y2

 + gβ(T − T∞ )

(10.28)

(d) The temperature gradient in the y direction is smaller than the gradient in the x direction. Therefore, the following relationship is valid inside the boundary layer, ∂2T ∂2T  (10.29) 2 ∂x ∂ y2 The energy equation for natural convection heat transfer inside the boundary can be written as, u

∂T ∂T ∂2T +v =α 2 ∂x ∂y ∂x

(10.30)

162

10 Natural Convection Over a Vertical Flat Plate

Based on the above assumptions, the continuity, momentum, and energy equations inside a boundary layer occurring over a vertical flat plate become, ∂u ∂v + =0 ∂x ∂y ∂u ∂ 2u ∂u +v = ν 2 + gβ(T − T∞ ) u ∂x ∂y ∂y ∂T ∂2T ∂T +v =α 2 u ∂x ∂y ∂x

(10.31) (10.32) (10.33)

As is seen, there are three equations with three unknowns: u, v, and T . The solution of the above set of PDEs yields velocity components in x, y, and also T inside the boundary layer. However, the solution of the above equations requires initial and boundary conditions. The initial and boundary conditions for the above equations can be written as, y = 0 → u = v = 0; T = Tw

(10.34)

y = δ → u = 0; T = T∞ x = 0 → u = v = 0; T = T∞

(10.35) (10.36)

where δ is the boundary layer thickness. The velocity and temperature boundary layers overlap each other in natural convection. δ is the thickness of both velocity and temperature boundary layers.

10.8 Similarity Solution The set of Eqs. 10.31–10.33 are partial differential equations, and it is not easy to solve them analytically. The PDEs are coupled and nonlinear. That is why it might be better to transform them into a set of ODEs and solve them. The difficulty of natural convection with respect to forced convection is the existence of the temperature source term in the momentum equation (Eq. 10.32). The equations are more coupled compared to the set of forced convection partial differential equations. The process of transforming momentum and energy equations (PDEs) into two ODEs should be performed together since they are coupled. This transformation is represented in this section. For transforming the momentum and energy equations (Eqs. 10.32–10.33) with boundary conditions (Eqs. 10.34–10.36), the concept of the stream function and a dimensionless temperature are used. The definition of velocity components in terms of stream function can be written as,

10.8 Similarity Solution

163

∂ψ ∂y ∂ψ v=− ∂x

u=

(10.37) (10.38)

The stream function can be defined in terms of a function such as f (η), ψ(x, y) = 4νC x 3/4 f (η)

(10.39)

where C is constant and can be defined as: C=

gβ(Tw − T∞ ) 4ν 2

1/4 (10.40)

and η is the independent parameter of the f (η) function and is defined as, η = C yx −1/4

(10.41)

The definition of the u velocity component in terms of f (η) can be found easily, u=

∂ ∂ψ ∂f →u= (4νC x 3/4 f (η)) → u = 4νC x 3/4 ∂y ∂y ∂y

(10.42)

The terms d f /dy must be converted to d f /dη, which can be done by using the chain rule: ∂ f ∂η ∂f ∂f ∂f = → = C x −1/4 ∂y ∂η ∂ y ∂y ∂η

(10.43)

By substituting Eq. 10.43 into Eq. 10.42, the definition of u in terms of η can only be found, u = 4νC 2 x 1/2

∂f ∂η

(10.44)

Similarly, the definition of the v velocity component can be obtained, v = −3νC x −1/4 f + νC 2 x −1/2 y

df dx

(10.45)

The derivatives of the u velocity component can also be found, and the results are,

164

10 Natural Convection Over a Vertical Flat Plate

∂u ∂f ∂2 f = 2νC 2 x −1/4 + 4νC 3 x 1/2 2 ∂x ∂η ∂η 2 ∂ f ∂u = 4νC 3 x 1/4 2 ∂y ∂η 2 3 ∂ u ∂ f = 4νC 4 3 ∂ y2 ∂η

(10.46) (10.47) (10.48)

Substituting the above terms into the momentum equations (Eq. 10.32) yields the following equation,  2 ∂3 f ∂2 f ∂f gβT +3f 2 −2 + 2 4 =0 ∂η3 ∂η ∂η 4ν C

(10.49)

The definition of C is given by Eq. 10.40. A dimensionless temperature can be defined as, θ=

T − T∞ Tw − T∞

(10.50)

Using the definition of dimensionless temperature, Eq. 10.49 takes the following form,  2 ∂2 f ∂f ∂3 f + 3 f − 2 +θ =0 ∂η3 ∂η2 ∂η

(10.51)

As seen, a second-order PDE (Eq. 10.32) is transformed into a second-order ODE. Solving Eq. 10.51 is easier than Eq. 10.32. In order to solve the above ODE, boundary conditions should also be written in terms of η and f (η). y = 0 → η = C 0 x −1/4 → η = 0

(10.52)

For η = 0 which is on the hot plate, the following equations can be written, ∂f ∂f =0 → =0 ∂η ∂η df − 3νC x −1/4 f + νC 2 x −1/2 0 =0 → f =0 dx u = 0 → 4νC x 3/4

v=0→

(10.53)

Similarly, the following qualities are valid considering the u velocity component is zero at the outer boundary, y → ∞ η = C yx −1/4 → for η → ∞

η→∞ ∂f ∂f =0→ =0 u = 0 → 4νC x 3/4 ∂η ∂η

(10.54)

10.8 Similarity Solution

165

or briefly, the boundary conditions can be written as  ∂ f  = f (0) = 0 ∂η η=0  ∂ f  =0 ∂η 

(10.55) (10.56)

η→∞

The energy equation can also be transformed into a dimensionless form. The dimensionless temperature defined by Eq. 10.50. If it is substituted into the energy equation (Eq. 10.33), the energy equation will take the following form, u

∂θ ∂ 2θ ∂θ +v =α 2 ∂x ∂y ∂x

(10.57)

The definition of ∂θ/∂ x in terms of f (η) can be found as follows: ∂θ ∂θ ∂η = ∂x ∂η ∂ x

→

∂θ ∂θ = ∂x ∂η



−C y −5/4 x 4

 (10.58)

Similarly, the definition of ∂θ/∂ y and ∂ 2 θ/∂ y 2 in terms of f (η) can be obtained as, ∂θ = ∂θ (C x −1/4 ) ∂η ∂y ∂ 2θ 2 = ∂∂ηθ2 (C x −1/2 ) ∂ y2

(10.59) (10.60)

The definitions of the u and v velocity components in terms of η and f (η) were found and shown in Eqs. (10.44 and 10.45). Therefore, by substituting Eq. 10.40 (u velocity), Eq. 10.45 (v velocity), Eq. 10.58 (for ∂θ/∂ x), Eq. 10.59 (for ∂θ/∂ y) and Eq. 10.60 (for ∂ 2 θ/∂ y 2 ) into the energy equation (Eq. 10.57) yields a third-order ordinary differential equation in terms of f (η), dθ d2 θ + 3Pr f =0 dη2 dη

(10.61)

The boundary conditions can also be written in dimensionless form, therefore, y=0→η=0  T − T∞  = 1 → θ (0) = 1 T = Tw → θ = Tw − T∞ η=0

(10.62) (10.63)

166

10 Natural Convection Over a Vertical Flat Plate

Similarly, y→∞ ; η→∞ T = Tw

→

 T − T∞  θ= =1 Tw − T∞ η→∞

→

(10.64)

θ =0

Therefore, the new forms of the continuity, momentum, and energy equations in terms of f (η) are,  2 ∂3 f ∂2 f ∂f +3f 2 −2 +θ =0 ∂η3 ∂η ∂η ∂ 2θ ∂θ =0 + 3Pr f ∂η2 ∂η

(10.65) (10.66)

with the following initial and boundary conditions, η = 0 → f (0) = f  (0) = 0; θ = 1 

η → ∞ → f (∞) = 0; θ = 0

(10.67) (10.68)

As can be seen, the above equations are ODEs and their solution will be easier than that of complicated PDEs (Eqs. 10.31–10.33).

10.9 Solution of ODE Momentum and Energy Equations The solution of Eqs. 10.65 and 10.66 and the initial and boundary conditions Eqs. 10.67 and 10.68 can be easily obtained by using numerical methods. The solution to these equations is not discussed in this book; however, detailed knowledge can be found in Ref. [9]. Two important charts were obtained and suggested by researchers in this field. Figure 10.4 shows the change in f  (η) with η. The value of η can be found easily since the coordinates of x and y inside the boundary layer are known. The value of the Gr x number can be easily calculated for the considered point. The value of Pr depends on the fluid and can be found in heat transfer tables. For a point inside the boundary layer, when the values of η and Pr are known, the value of f  (η) can be predicted by using Fig. 10.4. The value of the u velocity for the considered point in the boundary layer can be easily calculated from the value of f  (η). Similarly, the temperature at any point in the boundary layer can be predicted by using Fig. 10.5. Again, if the values of x and y and Gr x number are known for a point inside the boundary layer; therefore, the value of η can be calculated. By using the values of η and Pr, the value of θ (η) and consequently the temperature at any points in the boundary layer can be found from Fig. 10.5.

10.9 Solution of ODE Momentum and Energy Equations

167

Fig. 10.4 Graphical solution for the ODE of the momentum equation [9, 12]

Fig. 10.5 Graphical solution for the ODE of the energy equation [9, 12]

As can be seen, the values of u and T at any point inside the boundary layer can be calculated easily by using Figs. 10.4 and 10.5. There are many correlations in the literature for the determination of the local heat transfer coefficient for an isothermal vertical plate to the ambient environment by natural convection, Eqn. 10.69 represents one of them [6]. Nux =

0.75P f 1/2 hx L = k (0.609 + 1.221Pr1/2 + 1.238Pr)1/4

(10.69)

where Nux , and h x are the local Nusselt number and local heat transfer coefficient, respectively, and k is the thermal conductivity of the fluid. The average heat transfer coefficient for the plate with a length of L can also be found from the following equation, Nu =

4 hL = Nu L k 3

(10.70)

where Nu and h are the average Nusselt number and average heat transfer coefficient of the considered plate, respectively, while N u L is the local Nusselt number at the point with x = L.

168

10 Natural Convection Over a Vertical Flat Plate

10.10 Example A vertical plate at a temperature of 80 ◦ C and length of 50 mm is in contact with air at a temperature of 20 ◦ C. (a) determine whether the flow is laminar or turbulent, (b) plot the u velocity and temperature profiles when x = 10 and 40 mm, (c) what is the boundary layer thickness at x= 10 and 40 mm. Solution The solution of the governing equations given for this exercise will be employed for the all problems given at the end of this chapter. Step 1: Assumptions: The fluid is air and all assumptions explained in Chapter 7 for heat and fluid flow are valid also for this problem. For instance, with no viscous dissipation in the flow field, the flow field can be accepted as a two-dimensional domain, the flow is at a steady state, all thermophysical properties of the fluid are constant, etc. It should be mentioned that for the problem of this chapter, a body force in the x direction exists in the flow field and gravity affects the flow (see Fig. 10.3). Step 2: Governing equations and boundary conditions: The governing equations are continuity, momentum, and energy equations, but it is difficult to solve their PDE form. Therefore, they are transformed into two ODEs as,  2 d2 f df d3 f +3f 2 −2 +θ =0 dη3 dη dη d2 θ dθ =0 + 3Pr f 2 dη dη with the following initial and boundary conditions, η = 0 → f (0) = f  (0) = 0; θ = 1 η → ∞ → f  (∞) = 0; θ = 0 Step 3: Thermophysical properties: All thermophysical properties of the fluid are constant and can be found at the film temperature, which is the average temperature of the ambient environment and the plate (Tw + T∞ )/2. The film temperature for this problem is 50 °C, and therefore the values of the necessary thermophysical parameters are,

10.10 Example

169

Fig. 10.6 Plot of a u velocity profile, b temperature profile, at x = 10 and 40 mm.

ν = 1.7947 × 10−5

m2/s

ρ = 1.0925 kg/m3 C p = 1.0074 kJ/kgK Pr = 0.71053 α = 25.91.10−6 m2/s β = 1/T = 1/(50 + 273.15) = 0.003095 1/K−1 Step 4: Solution of governing equations: The graphical solution of the governing equations is given in Figs. 10.4 and 10.5. By knowing the values of x, y, Grx number, and Pr number, it is possible to predict the value of the u velocity component and T at any points inside the boundary layer. Step 5: Answers of the questions: (a) In order to check if the flow in the channel is laminar or in a turbulent region, let’s find the Ra L number. Therefore, Ra =

(9.81)(0.0030951)(80 − 20)(0.05)3 gβT L 3 = = 489717.56 να (1.7947 × 10−5 )(25.91 × 10−6 )

Since Ra < 109 the flow can be considered as laminar flow. (b) In order to find the velocity and temperature profile, as given in Figs. 10.4 and 10.5, η should be found by using Gr at x = 10 and 40 mm. Grx=10 mm =

Grx=40

mm

=

ρgβT x 3 (1.0925)(9.81)(0.0030951)(80 − 20)(0.01)3 = = 6179.2 ν2 (1.7947 × 10−5 )2

ρgβT x 3 (1.0925)(9.81)(0.0030951)(80 − 20)(0.04)3 = = 395468.8 2 ν (1.7947 × 10−5 )2

170

10 Natural Convection Over a Vertical Flat Plate

By using the Gr number for each local x value (10 and 40 mm), where y is changing, a value is found and by using Figs. 10.4 and 10.5, f  (η), and θ can be found. By using Eqns. 10.44 and 10.50, u and θ can be found, respectively. The requested Fig. 10.6 can be plotted. (c) The boundary layer thickness can also be found by using Fig. 10.6 where the velocity and temperature values reach zero and 20 °C, respectively. As seen, u velocity and temperature reach zero and 20°C where y is approximately 6 and 9 mm at x = 10 and 40 mm, respectively.

10.11 Review Questions

Question (Q1) Natural convection occurs due to external powers such as fan and pump (Q2) In order to have natural convection in a fluid domain, a temperature difference and gravity must exist (Q3) Natural convection governing equations are more coupled compared to the governing equations of forced convection (Q4) For the problem analyzed in this chapter, it is assumed that fluid flow is turbulent and incompressible and all thermophysical properties are constant (Q5) In natural convection, the temperature boundary layer overlaps the velocity boundary layer (Q6) In the natural convection boundary layer for a line perpendicular to the wall, the velocity is zero at the solid surface, then it increases by y and reaches a maximum value and finally it decreases to zero value (Q7) It is possible to numerically solve the governing equations of natural convection if a density equation in terms of temperature exists (Q8) The Boussinesq approximation is a practical way to include the effect of density change into the governing equations (Q9) In the Boussinesq approximation it is assumed that density is constant except the density in the source term of the momentum equation (Q10) The partial differential equation of the thermal boundary layer could be successfully transformed into a second-order ordinary differential equation in the x direction (Q11) The momentum and energy equations of natural convection can be transformed into 2 ODEs and solved easily by numerical methods (Q12) In this chapter, the solutions of two ODE governing equations are presented graphically to determine the u velocity component and T at any point inside the boundary layer (Q13) The values of the u velocity component and T∞ are zero and outside the boundary layer (Q14) Both the velocity and temperature values in the boundary layer depend on location (x and y), local Gr x , and Pr numbers (Q15) If the wall temperature is less than the fluid temperature natural convection occurs and fluid flows upward

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10.12 Problems

171

10.12 Problems For the problems below, follow the solution style of the example explained in this chapter. All steps mentioned in the solution part of the example should be followed one by one. Problem 1: A vertical flat plate is at 90 °C and fluid is at 10 °C. The length of the plate is 1000 mm. Gravity (−9.8 m/s2 ) acts on the flow field. (a) calculate the length of the plate portion on which the flow is laminar, (b) plot the dimensional u velocity component and temperature profiles at x = 10, 20 and 30 mm, (c) plot the boundary layer, roughly, (d) find the value of u max for x = 10 and 30 mm, approximately, does it increase or decrease in the x direction? Explain the reason for increasing or decreasing. Problem 2: Solve problem 1 for working fluid as water. (a) compare the dimensional u velocity component and temperatures of water and air for a line perpendicular to x = 10, 20, and 30 mm, (b) if there is a difference between the u and T profiles for water and air, explain the reason, (c) compare also the length of laminar flow for water and air, if there is a difference between the length of laminar flow, explain the reason, Problem 3: Repeat Problem 1, if natural convection occurs on Mars with a gravity of 3.7 m/s2 . (a) compare the u and T profiles for x = 10, 20, and 30 mm of lines perpendicular to the wall. (b) from which plate more heat is transferred to the ambient and explain the reason.

Chapter 11

Radiation Heat Transfer

11.1 Introduction As mentioned in Chap. 2, heat can be transferred by conduction, convection, and radiation. The conduction and convection heat transfer are explained in Chaps. 3–10. In this chapter, essential knowledge on radiation heat transfer is explained. Although a summary of radiation heat transfer is given in Chap. 2, it might be useful to mention about two points before the detailed explanations: (a) Radiation heat transfer is a surface phenomenon; heat is emitted from a surface and received by a surface. Therefore, the surface properties play an important role on the emitted or received radiation heat transfer rate. (b) Heat can be transferred by radiation between two or more surfaces without the presence of matter. This characteristic of radiation heat transfer allows sun rays to arrive on the earth. In this chapter, after the explanation of fundamental concepts of radiation heat transfer, radiation heat transfer between two black surfaces as well as real gray surfaces is presented. Information on radiation heat transfer can be found in various heat transfer books such as Refs. [4–6].

11.2 Emission Radiation heat transfer can be considered as the propagation of electromagnetic waves. As is well known, in general, two important parameters of a wave are frequency and wavelength. In radiation heat transfer, the following equality provides a relation between frequency and wavelength,

© The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2023 M. Mobedi and G. Gediz Ilis, Fundamentals of Heat Transfer, https://doi.org/10.1007/978-981-99-0957-5_11

173

174

11 Radiation Heat Transfer

Fig. 11.1 Emission of a spectral intensity from a surface d A1 and passing through d A2

λ=

c ν

(11.1)

where c is the speed of light in a vacuum which is 2.998 × 108 and ν is the frequency, 1/s. All surfaces whose temperature is greater than 0 K emit radiation; however the emission power depends on many parameters such as surface temperature, surface radiative properties, and ray direction. Consider a surface that is at T temperature shown in Fig. 11.1, it emits radiation in all directions; however the emission power may not be the same in all directions. The intensity of radiation emitted from the d A1 surface at the λ wavelength and passing through an infinitesimal area of d A2 of the sphere (shown in Fig. 11.1) is called the spectral intensity and shown by I (λ, θ, φ). By changing the wavelength and position of the infinitesimal surface on the hemisphere, the spectral intensity may change. In this book, considering Fig. 11.1, we focus on the total emission power which can be expressed as, ∞ 2π π/2 E(T ) =

I (λ.θ, φ)dθ dφdλ 0

0

(11.2)

0

As seen from the above equation, total emissive power is the total radiation of spectral intensity at a temperature such as T integrated over all wavelengths (from 0 to ∞) and also overall directions of the hemisphere surface (for θ from 0 to π/2 and for φ from 0 to 2π ). The unit of total emissive power is W/m2 (i.e., W ). It shows the total emitted heat transfer from a surface with unit area in all directions and an all wavelengths.

11.4 Real Surface Emission

175

Fig. 11.2 Difference between the emission of a real surface and blackbody

11.3 Blackbody Emission A blackbody (or it may also be called a black surface) is a body that can emit maximum radiation; in other words, a black surface always has maximum spectral and total emissive power. The radiated energy from a black surface is uniform in all directions. Therefore, it is independent of the direction of the hemisphere as shown in Fig. 11.2. The total emissive power of a black surface with a surface of A and at a temperature of T can be calculated by the Stefan–Boltzmann law which is, E blk = σ T 4

(11.3)

where σ is the Stefan–Boltzmann constant and its value is 5.67 × 10−8 (W/m2 K4 ). It should be kept in mind that T , in Eq. 11.3, represents the surface temperature and must be in Kelvin. Equation 11.3 shows that the total emission power of the black surface strongly depends on its temperature since the temperature is powered by 4. A small increase in the temperature may increase the total emission power considerably for a surface.

11.4 Real Surface Emission Practically many surfaces are not black surfaces and their emitted radiation is smaller than the blackbody. Figure 11.2 schematically compares emissions between a real surface, and a black surface. For a black surface spectral emission is identical for all directions and independent of wavelength (same energy in all directions); however, for a real surface, the emitted radiation may change by direction and the emitted energy may not be identical in all directions. Therefore, in order to determine the emitted radiation of a real surface, an emissivity coefficient whose value changes between 1 and 0 is suggested. The value of emissivity can be a function of direction (directional emissivity) and wavelength (spectral emissivity), but in this book, it is assumed that the emissivity is identical for all directions and independent of wavelength which is called total emissivity. Total emissivity is defined as the fraction of

176

11 Radiation Heat Transfer

total radiation emitted by a real surface and total radiation emitted by a black surface. Mathematically it can be expressed as, ε=

E real E blk

(11.4)

It should be mentioned that E real and E blk are the total hemispherical emission with a unit of W/m2 . The maximum value of ε can be 1 indicating that the considered real surface is a black surface and ε = 0 refers to no emission from the real surface. As is seen from Eq. 11.4, emissivity is a dimensionless quantity. Knowing the value of ε of a real surface is sufficient to calculate the total emission power of the surface if the temperature and radiation heat transfer areas are known. The surface emissivity of materials can be found in different heat transfer handbooks. For instance, the emissivity of a black-painted surface is around 0.98 while this value for an aluminum block can be around 0.4.

11.5 Irradiation Irradiation refers to radiation directed to a surface. Radiation emerges from other surfaces and is directed to a surface as shown in Fig. 11.3. The radiation coming to the surface may be a function of wavelength or even direction. In this book, it is assumed that the incoming radiation into a surface is independent of direction and is integrated over all wavelengths. Therefore, it can be called as total irradiation and shown by G irr with a unit of W/m2 .

11.6 Surface Radiative Properties For radiation heat transfer between surfaces, surface radiative properties such as emissivity, absorptivity, reflectivity, and transmissivity are important. Emissivity is discussed in Sect. 11.4. In this section, a brief discussion on absorptivity, reflectivity, and transmissivity is given. It should be kept in mind that all these properties are independent of direction and also wavelength in this book. As it was mentioned before, all surfaces that are at a temperature greater than 0 K emit radiation. At the same time, those surfaces also receive radiation from the surroundings which is called as irradiation. Irradiation to a surface can be managed by the surface in three ways as shown in Fig. 11.3. The coefficients of absorptivity, reflectivity, and transmissivity concern the behavior of the surface when it receives irradiation.

11.6 Surface Radiative Properties

177

Fig. 11.3 Behavior of a surface when it receives irradiation

Absorptivity Some fraction of irradiation to a real surface is absorbed, and the adsorbed heat is transferred to the body of the surface. The total absorptivity coefficient is defined as the ratio of the total adsorbed irradiation by the surface to the total irradiation incident to the surface (see Fig. 11.3). The total absorptivity coefficient is independent of wavelength and direction. Mathematically, it can be described as, α=

G abs G total

(11.5)

where G total is the total irradiation incident to the surface and G abs is the total adsorbed irradiation by the surface. The value of absorptivity changes between 0 and 1. For a black surface, the value of α is 1. The absorptivity coefficient is a radiative property of a surface and depends on many functions such as color and roughness. The total absorptivity values of surfaces can be found in many heat transfer books. For instance, the absorptivity of polished aluminum surface is 0.09 while it is 0.92 for black nickel oxide. Reflectivity Some fraction of irradiation to a surface may be reflected in the surroundings. Therefore, the total reflectivity is defined as the fraction of the total reflected irradiation by the surface and the total irradiation that is incident to the surface, as seen in Fig. 11.3. The total reflectivity coefficient is independent of direction and wavelength. Mathematically it can be written as, ρ=

G ref G total

(11.6)

where G ref is the total radiation reflected by the surface. The value of reflectivity also changes between 0 and 1. For a black surface, ρ = 0 since all irradiation is incident to the surface is absorbed. The value of the reflectivity of surfaces can also be seen in many heat transfer books. For instance, the value of the total reflectivity of polished aluminum is 0.91 while it is 0.08 for black nickel oxide.

178

11 Radiation Heat Transfer

Transmissivity Some parts of the irradiation that are incident to a surface may be transmitted through the body as seen in Fig. 11.3. The total transmissivity is defined as the fraction of the total part of irradiation transmitted through the body and the total irradiation that is incident to the surface. The total transmissivity is independent of direction and wavelength. Mathematically, it can be written as, τ=

G trn G total

(11.7)

where G trn is the total irradiation transmitted through the body whose surface is under the irritation of G total . The value of transmissivity is also between 0 and 1. The values of the transmissivity of surfaces can be found in heat transfer books. For instance, the transmissivity of glass with a thickness of 2–3 mm is 0.87.

11.7 Opaque Surface, Gray Surface and Radiosity The concepts of opaque surfaces, gray surfaces, and radiosity are widely used in radiation heat transfer between surfaces. Therefore, it might be useful to mention these concepts briefly in this section. Opaque surface Practically, always there is irradiation to a surface as shown in Fig. 11.3. The irradiation to the surface may be absorbed, and/or reflected, and/or transmitted which can be calculated by absorptivity, reflectivity, and transmissivity coefficients. If the transmissivity factor is zero for a surface, the surface is called as an opaque surface. Therefore for an opaque surface. α+ρ =1

(11.8)

Gray surface The law of Kirchhoff and the definition of gray surface are given in many heat transfer books [4–6]. Briefly, the gray surface is an opaque surface for which absorptivity and emissivity are independent of wavelength. By considering Kirchhoff’s law, the following relation can be written for a gray surface, α=ε

(11.9)

where α and ε are the total absorptivity and total emissivity, respectively. The assumption of the gray surface is implemented in many practical radiation heat transfer problems, and reasonable results can be obtained.

11.8 View Factor

179

Radiosity Some portion of irradiation to a gray surface such as the surface shown in Fig. 11.3 is reflected, which can be expressed as ρG total . At the same time, the surface emits radiation to its surroundings which is shown by εE blk . The total radiation that leaves the surface is the summation of ρG total and εE blk and it is called as “radiosity”. Radiosity is an important concept in radiation heat transfer and helps to easily analyze radiation heat transfer between surfaces. For an opaque surface, the radiation heat transfer leaves the surface and is directed to the surroundings can be written as, J = εE blk + ρG total

(11.10)

11.8 View Factor Consider the two surfaces shown in Fig. 11.4 which are at different temperatures. Assume that both surfaces emit hemispherical radiation that is independent of direction and integrated over wavelength. In order to calculate the rate of radiation leaves surface A1 and receives by surface A2 (or vice versa), a parameter called the view factor is suggested. In many heat transfer books, the view factor is shown by Fi j and it is used to calculate the radiation that leaves surface i and receives by surface j. In other words, the view factor of Fi j is the fraction of radiosity from surface Ai and receives by surface A j with the total radiosity of surface Ai to its surroundings. It might be useful to mention that Fi j may be equal to or different than F ji . The best example for understanding the view factor is heat transfer between two surfaces of a long cylinder as shown in Fig. 11.5. The inner surface is shown by A1 while A2 refers to the outer surface. All emitted radiation by surface A1 is received by surface A2 . Therefore the view factor of F12 must be 1. However a portion of the emitted radiation by surface A2 is intercepted by the surface of A1 and the remaining part is received by itself (i.e., A2 ). Therefore, F12 may not be equal to F21 due to the different areas and positions. In many heat transfer books, the value of the view factor between two surfaces is given as an equation or as a graph. For instance, the following equation is found for the view factor of two identical parallel plates with a length of L, a width of W , and a distance of H in between by Ref. [6]:   L L 21 W12 + 2L W1 arctan − arctan L ln 2 Fi j = W1 L 1 + W12 − 1 π LW   W +2W L 1 arctan − arctan W (11.11) L1 1



180

11 Radiation Heat Transfer

Fig. 11.4 View factor between two surfaces

Fig. 11.5 View factor between two long cylinders

 2 where L and W are defined as L = L/H and W = W/H , and L 1 = 1 + L , W1 =  2 1 + W , respectively. L and W are the length and width of the plate while H is the distance between two parallel plates. The same view factor can also be calculated from the graph given in Fig. 11.6. For instance, when the values of L = W = H = 10 mm the value of F12 is calculated as 0.0316, and the same value can be obtained from Fig. 11.6. Although it is possible to use suggested correlations and graphs to find the value of the view factor between two surfaces, it is also possible to calculate the view factor between two surfaces by using some rules for simple cases. There are some rules which can be used if a view factor of two simple surfaces is asked. Here, two common rules are described. • Reciprocity rule If two surfaces such as Ai and A j see each other, the following equation relating to their view factors can be written, Ai Fi j = A j F ji

(11.12)

where Fi j is the view factor between the i and j surfaces while F ji is the view factor between the j and i surfaces. If the surfaces of Ai and A j are equal to each other, Fi j and F ji will be identical but for different surface areas of Ai j and A ji , the view factors of Fi j and F ji are not equal.

11.8 View Factor

181

Fig. 11.6 View factor between two parallel plates

• Summation rule For a closed domain with n surfaces, the following relation is valid for the view factor between the surface of i and other surfaces of the domain, n 

Fi j = 1

(11.13)

j=1

The value of Fi j should be less than 1 and their summation for an enclosure should not exceed 1. For instance, there are totally two surfaces in the closed space of Fig. 11.5, and the following relations based on the summation rules can be written, F11 + F12 = 1

(11.14)

F21 + F22 = 1

(11.15)

182

11 Radiation Heat Transfer

An example relating to the use of the reciprocity and summation rules and determination of view factor is given in the example part at the end of this chapter.

11.9 Radiation Heat Transfer Between Two Black Surfaces Consider two black surfaces named A1 and A2 as shown in Fig. 11.7. Both surfaces are black with areas of the A1 and A2 . It is assumed that the temperature of A1 surface is T1 and is greater than T2 , which is the temperature of surface A2 . Based on the Stefan–Boltzmann law, the heat leaving the surface A1 and reaching the surface A2 can be written as, q1→2 = A1 F12 (σ T1 4 )

(11.16)

where σ T1 4 is the radiation emitted by the A1 black surface to the surroundings and F12 is the view factor the between surfaces A1 and A2 . Similarly, the heat leaves the surface A2 and reaches to surface A1 is, q2→1 = A2 F21 (σ T2 4 )

(11.17)

Again, σ T2 4 is the radiation emitted by the A1 surface to the surroundings and F21 is the view factor between the surfaces of A2 and A1 . Therefore, the net heat exchange between the surfaces of A1 and A2 can be easily obtained as, qnet (12) = q1→2 − q2→1 = A1 F12 (σ T1 4 ) − A2 F21 (σ T2 4 )

(11.18)

The reciprocity rule was explained in the previous section; therefore, the following relationship exists between the view factors of two surfaces,

Fig. 11.7 Radiation heat transfer between two black surfaces

11.9 Radiation Heat Transfer Between Two Black Surfaces

Ai Fi j = A j F ji

183

(11.19)

Substituting Eqs. 11.19 into 11.18 yields the following equation for the net radiation heat transfer between two surfaces, qnet(12) = A1 F12 (σ T1 4 − σ T2 4 )

(11.20)

Since the heat transfer process of this problem is steady state, the heat that is transferred from the body of the surface A1 must be equal to the net heat transfer between the surfaces of A1 and A2 . Similarly, the heat transfer from surface A2 to its body is also equal to the net heat transfer between the surfaces of A1 and A2 , by other words, (11.21) qnet(1) = qnet(12) = −qnet(2) It is also possible to write Eq. 11.20 in terms of radiation thermal resistance as follows: σ T1 4 − σ T2 4 σ T1 4 − σ T2 4 = (11.22) qnet(12) = 1/(A1 F12 ) Rth In some heat transfer books, Eq. 11.22 is written as qnet(12) =

E blk1 − E blk2 E blk1 − E blk2 = 1/(A1 F12 ) Rth

(11.23)

where E blk1 and E blk2 are the emission heat transfer of the black surfaces of A1 and A2 per unit area. Rth = 1/(A1 F12 ) is the radiation thermal resistance between surfaces A1 and A2 . An example for thermal electric circuit between two parallel plates close to each other is shown in Fig. 11.8.

Fig. 11.8 Thermal circuit for radiation heat transfer between two black surfaces close to each other

184

11 Radiation Heat Transfer

11.10 Radiation Heat Transfer Between Two Gray Surfaces Again, consider two gray surfaces of A1 and A2 shown in Fig. 11.9. Similar to the previous section, it is possible to write the radiation heat transfer leaves the surface of A1 and the radiation is captured by the surface of A2 as, q1→2 = A1 F12 (E 1 + ρG 1 ) = A1 F12 J1

(11.24)

where J1 is the radiosity of the A1 surface. G 1 represents the total irradiation of surface A1 . It should be mentioned that for a black body reflectivity is zero (ρ = 0) therefore, the radiation leaves the surface is E b . For a gray surface, the radiation that leaves the surface is the summation of emitted and reflected irradiation and consequently radiosity is used in Eq. 11.24. Similarly, the radiation leaves surface A2 and the thermal radiation captured by A1 can be written as, q2→1 = A2 F21 (E 2 + ρG 2 ) = A2 F21 J2

(11.25)

where G 2 is the total irradiation of surface A2 . Therefore, the net heat transfer between the surfaces of A1 and A2 can be written as, qnet(12) = A1 F12 J1 − A2 F21 J2

(11.26)

Again, based on the reciprocity rule explained in the previous section, Ai Fi j = A j F ji , Eq. 11.26 can take the following form, qnet(12) = A1 F12 (J1 − J2 )

Fig. 11.9 Thermal circuit for radiation heat transfer between two diffusive gray surfaces

(11.27)

11.10 Radiation Heat Transfer Between Two Gray Surfaces

185

The difficulty with the above equation is that the values of radiosity (J1 and J2 ) are not known and should be found. Considering Fig. 11.9, the following relation can be written for qnet(1) , qnet(1) = A1 (J1 − G 1 ) (11.28) The definition of radiosity was given before as, J1 = ε1 E blk1 + ρG 1

(11.29)

Since the surface is opaque, ρ = 1 − α, therefore, the above equation takes the following form, (11.30) J1 = ε1 E b1 + (1 − α)G 1 For a diffuse gray surface, it is known that α = ε, therefore, J1 = ε1 E b1 + (1 − ε1 )G 1 The G 1 can be written as G1 =

(11.31)

J1 − ε1 E b1 1 − ε1

(11.32)

Finally, if the above equation is substituted into Eq. 11.28, the following relation can be obtained, E b1 − J1 qnet(1) = 1−ε1 (11.33) ε1 A1

Equation 11.33 is a very useful equation and shows that a thermal resistance can be considered between E b1 and J1 . Considering Fig. 11.9, a similar equation can also be derived for surface A2 , E b2 − J2 (11.34) qnet(2) = 1−ε2 ε2 A2

Since the heat transfer is steady state, the following condition is valid, qnet(1) = qnet(12) = −qnet(2)

(11.35)

Substituting Eqs. 11.27, 11.33, 11.34 into Eq. 11.35 yields the following relation, E b1 − J1 1−ε1 ε1 A1

=

J1 − J2 E b2 − J2 = − 1−ε2 A1 F12 ε A 2

(11.36)

2

The summation of the numerator and denominator of Eq. 11.36 gives the following equations.

186

11 Radiation Heat Transfer

Fig. 11.10 Thermal circuit for radiation heat transfer between two diffusive gray surfaces

qnet(1) = qnet(12) = −qnet(2) =

1−ε1 ε1 A1

E b1 − E b2 + A11F12 +

1−ε2 ε2 A2

(11.37)

E b1 and E b2 are the black surface emission of the A1 and A2 surfaces which can be easily calculated from the Stefan–Boltzmann equation. Furthermore, it is also possible to draw the thermal circuit for Eq. 11.37 for a better understanding of the mechanism of radiation heat transfer between two surfaces. The thermal circuit for heat transfer between the diffusive gray surfaces of A1 and A2 is shown in Fig. 11.10. Considering Fig. 11.10, the first and third thermal resistances are the resistances causing the reduction of radiation of heat transfer of A1 and A2 surfaces since they are not black surfaces. If the surfaces were black surfaces, there would be no thermal resistance at the surfaces. The second thermal resistance in Fig. 11.10 comes from the position of two surfaces with respect to each other as well as their areas. This thermal resistance is the same for radiation heat transfer between black surfaces as well as gray surfaces. The radiative heat transfer between two diffusive gray systems can be easily determined considering the thermal circuit shown in Fig. 11.10. The same method can be used when the number of surfaces is more than two surfaces in a cavity. Detailed information for radiative heat transfer between multiple surfaces in an enclosure can be found in different heat transfer books such as Ref. [4–6].

11.11 Example Consider the inner surface of a hemisphere and a surface on which the hemisphere is located, as shown in Fig. 11.11. The radius of the hemisphere is 0.1 m. The inner surface of the hemisphere is at 1050 °C while the bottom surface is 10 °C . Both surfaces are diffusive gray. The emissivity of the sphere’s inner surface is 0. 8 while the value of the flat bottom surface is 0.3. (a) Draw the thermal circuit between the sphere’s inner surface and the bottom horizontal surface when the surfaces are gray. (b) Calculate the thermal radiation rate emitted from the bottom surface and captured by the sphere’s inner surface, if both surfaces are assumed to be black surfaces.

11.11 Example

187

Fig. 11.11 Figure of example

Fig. 11.12 Thermal circuit between the diffusive gray hemisphere’s inner and bottom surfaces

(c) Similarly, calculate the rate of radiation emitted by the sphere surface and captured by the bottom surface, if both surfaces are assumed as black surface. (d) What is the net heat transfer rate between two surfaces, if both surfaces are assumed as black surface. (e) What is the net radiation heat transfer between two surfaces for the real case. Solution The radiation heat transfer between surfaces is a surface phenomenon (without considering absorption of radiation by fluid). Therefore, there is no governing equation and all calculations will be performed by the methods explained in this chapter. Let assume the bottom surface as “surface 1” and the inner surface of the hemisphere as “surface 2”. (a) The thermal circuit between the diffusive gray hemisphere’s inner and bottom surfaces can be illustrated as shown in Fig. 11.12: “Surface 1” shows the bottom surface and “surface 2” is the inner surface of the hemisphere. (b) The thermal radiation rate that leaves the bottom surface and is captured by the inner surface can be found by using the following equation q1→2 = A1 F12 (ε1 σ T1 4 )

188

11 Radiation Heat Transfer

The value of F12 = 1 since all thermal radiation emitted by Surface 1 is captured by Surface 2 and it can be calculated as, q1→2 = π × 0.12 × 1 × 0.8 × 5.67 × 10−8 × 283.154 = 9.15 W (c) The radiation rate that leaves the sphere surface and is captured by the bottom surface can be found by a similar method, q2→1 = A2 F21 (ε2 σ T2 4 ) The value of F21 can be found by using the reciprocity rule given by Eq. 11.12. A1 F12 = A2 F21

→ F21 =

π × 0.12 × 1 = 0.5 2 × π × 0.12

Therefore, the value of q2→1 can be found as, q2→1 = A2 F21 J2 = 2 × π × 0.12 × 0.5 × 5.67 × 10−8 × 1323.154 = 5456.93 W

Since the temperature of the sphere is considerably higher than that of Surface 1, the heat emitted by Surface 2 and captured by Surface 1 is considerably high. (d) The net heat transfer between two surfaces is, qnet(12) = q2→1 − q1→2 = 5456.93 − 9.15 = 5447.78 W (e) As it as mentioned above, the net heat transfer between surfaces 1 and 2 can be calculated by different methods, and there is no doubt that the final values of all methods will be identical. It is also possible to find also qnet(12) by using Eq. 11.37 considering the thermal resistance circuit of this example, qnet(12) =

1−ε1 ε1 A1

E b1 − E b2 + A11F12 +

1−ε2 ε2 A2

The areas of surfaces 1 and 2 are, A1 = π × r 2 = π × 0.12 = 0.0314 m2 A2 = 2 × π × 0.12 = 2 × π × 0.12 = 0.0628 m2 and F12 = 1, therefore qnet(12) =

5.67 × 10−8 × 1323.154 − 5.67 × 10−8 × 283.154 = 2253.31 W 1−0.8 1 1−0.3 + 0.0314×1 + 0.3×0.0628 0.8×0.0314

11.13 Problems

189

As it can be seen, the net heat transfer of the surfaces if they are black is greater than heat transfer between the gray surfaces.

11.12 Review Questions

Question (Q1) Radiation can be transferred between two or more surfaces without presence of a matter (Q2) Any surface whose temperature is greater than 0 K emits radiation (Q3) Radiation is a phenomenon that concerns heat transfer inside a body (Q4) No surface can emit more thermal radiation than a black surface (Q5) The emitted radiation by a black surface is independent of direction (Q6) The emissivity coefficient has been developed to calculate emitted radiation from any real surface (Q7) Total irradiation is the thermal radiation integrated over all directions as well as all wavelengths (Q8) The summation of the absorptivity, reflectivity, and transmissivity coefficients of a real surface is 1 (Q9) An opaque surface is a surface that can transmit radiation (Q10) The absorptivity, reflectivity, and transmissivity are material thermal properties (Q11) The absorptivity, reflectivity, and transmissivity are the radiation surface properties of a body (Q12) View factor depends on the roughness of a surface-emitting radiation (Q13) In an enclosure, the summation of a surface’s view factor with all surfaces is 1 (Q14) A diffusive gray surface is a surface for which the emitted radiation is independent of direction and α = ε

Correct Incorrect [] [] [] []

[] []

[] [] []

[] [] []

[]

[]

[]

[]

[] []

[] []

[]

[]

[]

[]

[]

[]

[]

[]

11.13 Problems Problem 1: An aluminum block with a polished surface is at 0 °C with an area of 0.5 m2 .The total irradiation to the surface is 1000 W/m2 . (a) Calculate the rate of total emitted thermal radiation from the surface. (b) Calculate the rate of absorbed, reflected, and transmitted thermal radiation at the surface. (c) What is the rate of thermal radiosity. (d) What will be the rate of absorbed and reflected thermal radiation rate if the considered surface is covered with black Parsons paint.

190

11 Radiation Heat Transfer

Fig. 11.13 Figure of Problem 2

Problem 2: Consider a sphere with a radius of 100 mm located in a cube as shown in Fig. 11.13. The outer surface of a sphere is at 100 °C while the inner surface of a cube is at 200 °C. Both the outer surface of the sphere and the inner surface of the cube can be accepted as ideal black surfaces. (a) Calculate the view factor between two surfaces (both F12 and F21 ). (b) What is the rate of thermal radiation leaving the sphere’s surface and reaches to the cube’s inner surface? (c) What is the rate of thermal radiation that leaves the cube surface and is captured by itself? (d) Calculate the net radiation heat transfer between the sphere surface and the cube’s inner surface. (e) Draw the thermal circuit between two surfaces. Problem 3: There are two solid blocks with the same size of 0.1 × 0.1 × 0.1m3 as shown in Fig. 11.14. The material of the bottom block is concrete while the upper block is stainless steel with a polished surface. The distance between two surfaces is 0.2 mm. The bodies and surfaces of concrete and stainless steel blocks are 10

Fig. 11.14 Figure of Problem 3

11.13 Problems

191

and 90 °C, respectively. The surfaces of both blocks can be accepted as diffuse gray surfaces. (a) Plot the thermal circuit for radiation heat transfer between the surface of concrete and stainless steel. (b) Calculate the radiation heat transfer that leaves the concrete surface and is captured by the concrete surface. (c) Calculate the rate of radiation emitted from surface of stainless steel and is captured by concrete surface. (d) What is the net thermal radiation heat transfer between two surfaces. (e) What would be the net heat transfer rate between two surfaces if both surfaces were ideal black surface.

Appendix A

Finding Roots of Bessel Functions

As it was explained in Chap. 5, in order to find the value of βn , the roots of the related Bessel function must be found. In the most of cases, more than 4 or 5 roots must be known to find temperature at any point of the wall. Although there are tables providing roots of different orders of Bessel function, one way is to find the roots by bisection method. It can be done by using excel software. Information about bisection method can be found in many numerical methods book [14]. In this appendix, bisection method is explained briefly. These are the steps that must be followed to find a root of a function: step (1) Plot the function and consider two values on the left and right sides of the roots of the function as x1 and x2 , let assume that x1 is on the right and x2 is on the felt side of the root as shown in Fig. A.1. step (2) Since the points are on the left and right sides of the roots, f (x1 ). f (x2 ) < 0. step (3) Take the arithmetic average of x1 and x2 and find x3 and also function of f (x3 ). step (4) If x3 is on right side of the root, (therefore f (x1 ). f (x3 ) < 0) take the average of x3 and x1 ). It can be named as x4 and then process is repeated by the same logic to find x5 , x6 etc.

Fig. A.1 Schematic view of finding a root by using bisection method

© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2023 M. Mobedi and G. Gediz Ilis, Fundamentals of Heat Transfer, https://doi.org/10.1007/978-981-99-0957-5

193

194 Table A.1 First 10 roots of J1 (x) x x1 x2 x3 x4 x5 x6 x7 x8 x9

Appendix A: Finding roots of Bessel functions

Root value 2.4048 5.5201 11.7915 14.9309 18.0711 21.2116 24.3525 27.4935 30.6346

step (5) If x3 is on left side of the root, (therefore f (x2 ). f (x3 ) < 0) take the average of x3 and x2 . It can be named as x4 and then process is repeated by the same logic to find x5 , x6 etc. step (6) Processes in step 4 or step 5 are repeated till |xn − xn−1 | < . The value of  will be defined by user. Table A.1 also provides the first 10 roots of first kind first order of Bessel function (J1 (x) = 0).

Appendix B

Dimensionless Velocity Values in a Boundary Layer

Based on the similarity solution, u velocity component can be found from Fig. 8.3; however, Fig. 8.3 may not provide accurate values. The numerical values of Fig. 8.3 are given in Table B.1 [9].

Table B.1 Flat plate laminar boundary layer functions  η = y uνx∞ f = 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 3.8

u U∞

0 0.06641 0.13277 0.19894 0.26471 0.32979 0.39378 0.45627 0.51676 0.57477 0.62977 0.68132 0.72899 0.77246 0.81152 0.84605 0.87609 0.90177 0.92333 0.94112 (continued)

© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2023 M. Mobedi and G. Gediz Ilis, Fundamentals of Heat Transfer, https://doi.org/10.1007/978-981-99-0957-5

195

196 Table B.1 (continued)  η = y uνx∞ 4.0 4.2 4.4 4.6 4.8 5.0 5.2 5.4 5.6 5.8 6.0

Appendix B: Dimensionless Velocity Values in a Boundary Layer

f =

u U∞

0.95552 0.96696 0.97587 0.98269 0.98779 0.99155 0.99425 0.99616 0.99748 0.99838 0.99898

Appendix C

Compatibility Relation

In this appendix, compatibility relation for heat transfer in a channel is derived. The mean temperature is defined in Chap. 9 by Eq. 9.35. Mean temperature for a channel with unit depth can be written as,  +H Tm (x) =

0

ρC p u(y)T (x, y)dy  +H ρC p u(y)dy 0

(C.1)

For an incompressible flow with constant thermo-physical properties (ρ and C p are constant). The above equation takes the following form,  +H Tm (x) =

u(y)T (x, y)dy  +H u(y)dy 0

0

(C.2)

u(y) and y can be made dimensionless by multiplying both sides with 1/(u m H ), 

1 H um





+H Tm (x)

u(y)dy =

1 H um

0

or

+H Tm (x)

 +H u(y)T (x, y)dy 0

u(y) y d = um H

0

+H

u(y) y T (x, y)d um H

0

Using dimensionless velocity and dimensionless y direction (u ∗ and y ∗ ) given in Sect. 9.2 yields following equation 1 Tm (x)

∗

∗

1

u dy = 0

u ∗ T (x, y)dy ∗

(C.3)

0

© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2023 M. Mobedi and G. Gediz Ilis, Fundamentals of Heat Transfer, https://doi.org/10.1007/978-981-99-0957-5

197

198

Appendix C: Compatibility Relation

The dimensionless temperature is defined by Eq. 9.34; therefore, following equalities can be written, T − Tw θ= → T = θ (Tm − Tw ) + Tw (C.4) Tm − Tw Substituting Eq. C.4 into Eq. C.3 yields the following equation, 1

∗

1

∗

u dy =

Tm 0

∗

1

∗

u θ (Tm − Tw )dy + 0

u ∗ Tw dy ∗

0

The above equation can be rearranged, 1

∗

1

∗

u dy − Tw

Tm 0

∗

1

∗

u dy = (Tm − Tw ) 0

or   (T m − Tw ) 

1

0

  u dy =  (T m − Tw ) ∗

u ∗ θ dy ∗

∗

0

1

u ∗ θ dy ∗

0

Therefore, the above equation can be simplified into the following form, 1

∗

1

∗

u dy = 0

u ∗ θ dy ∗

0

An equation for u ∗ is found and given by Eq. 9.32. 1

∗

∗

1

u θ dy = 0

3 (1 − y ∗ 2 )dy ∗ 2

0

The result of the above integral yields the compatibility relation, 1 0

u ∗ θ dy ∗ = 1

(C.5)

References

1. Borthwick, D.: Introduction to Partial Differential Equations. Springer (2016) 2. Yakubov, S., Yakubov Y.: Differential-Operator Equations, Ordinary and Partial Differential Equations. Chapman&Hall/CRC (2000) 3. Pritchard, P.J., Mitchell, J.W.: Fox and McDonald’s Introduction to Fluid Mechanics, 9th edn. Wiley (2015) 4. Cengel, Y.: Heat Transfer: A Practical Approach. McGraw Hill (2013) 5. Cengel, Y., Ghajar A.: Heat and Mass Transfer, Fundamentals and Applications, 5th edn. McGraw Hill (2014) 6. Bergman, T.L., Lavine, A.S., Incropera, F.P., Dewitt, D.P.: Introduction to Heat Transfer, 6th edn. Wiley (2011) 7. Ozisik, M.N.: Heat Conduction, 2nd edn. Wiley (1993) 8. Ozisik, M.N.: Finite Difference Methods in Heat Transfer, 1st edn. CRC Press (1994) 9. Kakac, S., Yener, Y., Pramuanjaroenkij, A.: Convective Heat Transfer, 3rd edn. CRC Press (2014) 10. Yener, Y., Kakac, S.: Heat Conduction, 4th edn. CRC Press (2008) 11. Howarth, L.: On the solution of the laminar boundary layer equations. Proc. R. Soc. 164, 547–579 (1938) 12. Ostrach, S.: An Analysis of Laminar Free-Convection Flow and Heat Transfer About a Flat Plate Parallel to the Direction of the Generating Body Force, NACA Report 111 (1953) 13. Gill, A.E.: Atmosphere-Ocean Dynamics. Academic Press, New York, NY (1982) 14. Rao, G.S.: Mathematical Methods. I.K. International Pvt, Ltd (2007)

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